Image Production

What does the term PACS stand for in digital imaging? A Picture archiving and communication system B Pixel archiving and contrast system C Patterns archiving and communication system D Pair archiving and communication system

A

When an exposure is made using AEC but positioning is suboptimal, such that a portion of an active cell is struck by the primary beam, which of the following will occur? A The exposure will terminate prematurely and the displayed image will have mottle in the regions where the anatomy is thickest. B The exposure duration will be longer and the patient will receive excessive dose. C The exposure will terminate as expected and the displayed image will show no adverse effects. D The exposure duration will be longer and the displayed image will have mottle in the regions where the anatomy is thickest.

A AECs function as calibrated exposure kill switches. When they receive a predetermined amount of exposure, enough ionization is created to activate a circuit that diverts power from the high voltage generator. If primary (unattenuated) beam is allowed to strike an active AEC detector, that detector will reach the threshold ionization very rapidly and the exposure will terminate prematurely. In sufficient mAs will be sent through the patient to the image receptor. Areas of the image receptor behind thick portions of the anatomy will not have adequate signal and will appear mottled, as too few photons reached those areas. Patient dose will increase only as a result of having to repeat the projection due to poor image quality.

In digital radiography, patient information is displayed in the corners of the monitor for an exam. The 3 steps a technologist must take to ensure that the information is accurate are: A Verify the patient's Identity by confirming their name and date of birth after the exam is complete. B Verify the patient's Identity by confirming their name and date of birth before beginning the exam. C Match the exam images to the Technologist Image number. D Match the exam images to the exam accession number, either via query of the RIS worklist or manual entry. E Verify the exam details with the patient. Confirm type of exam, body part to be examined, and side of the body are correct.

All workflows that rely on information exchange between multiple people are prone to error. In healthcare, these errors, if left unchecked, may result in undesirable outcomes and jeopardize patient safety. To guard against these errors, the practicing technologist is ethically obligated to do the following for each exam: confirm patient identity using two patient identifiers before ANY procedure begins, confirm that the exam details match the patient's understanding regarding their care, and to match the acquired images to the correct accession number supplied by the RIS (radiography information system). When these three steps are performed consistently, the efficiency of the imaging department improves, and the institution will see a reduction in safety incidents due to medical error. The patient population served by such a facility can have confidence in the treatment they receive. (Long, Rollins, and Smith, 14th edition, Merrill's Atlas of Radiographic Positioning & Procedures, Vol. 1, pages 10 and 33)

Exposure factors of 100 kVp and 6 mAs are used with a 6:1 grid for a particular exposure. What should be the new milliampere-seconds value if a 12:1 grid is substituted? A 7.5 mAs B 10 mAs C 13 mAs D 18 mAs

B

Exposure factors of 80 kVp and 8 mAs are used for a particular nongrid exposure. What should be the new milliampere-seconds value if an 8:1 grid is added? A 16 mAs B 24 mAs C 32 mAs D 40 mAs

C

Compared to a high-ratio grid, what benefit(s) would be gained from using a low-ratio grid? 1. Increased amount of scatter radiation produced within the patient 2. Decreased patient dose 3. Reduced scatter fog seen on the radiographic image A 1 only B 2 only C 1 and 3 only D 2 and 3 only

Compared to a high-ratio grid, what benefit(s) would be gained from using a low-ratio grid? 1. Increased amount of scatter radiation produced within the patient 2. Decreased patient dose 3. Reduced scatter fog seen on the radiographic image A 1 only B 2 only C 1 and 3 only D 2 and 3 only

Which of the following is (are) tested as part of a QC program?Beam alignmentReproducibilityLinearity A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

D

When an imaging department installs new radiographic equipment with higher DQE (Detective Quantum Efficiency) than previous equipment, the most important/valuable advantage is: A Image quality improves. B Repeat rates reduce. C Exam throughput increases. D Patient dose is reduced.

DQE is a measure of the quality of the Image Receptor in respect to capturing photons for a particular radiographic system. DQE is a function of the thickness of the active layer, its atomic composition, and the spatial frequency of the image-forming beam. All of these factors affect the probability of photoelectric absorption by the active layer of the IR. When an imaging system has a higher DQE, which means it captures a greater percentage of the remnant beam, and therefore, lower mAs techniques can be applied with these machines while still yielding a diagnostic image. Improvements in image quality, reduction in repeat rate, and workflow efficiencies are all beneficial, however, they are better influenced by high quality technologists, not high quality equipment. Patient populations served by the new machines will receive less dose as these low mAs techniques are employed. (Bushong, 11th edition, p 318)

Which three of the following six pathologies require a reduction in technique during a radiographic examination: A Osteoarthritis B Pneumonia C Osteopetrosis D Emphysema E Osteomalacia F Atelectasis

Degenerative diseases are those in which the native tissue thins out or deteriorates as the disease progresses. Therefore, less technique is required as the beam can more easily penetrate the thinner tissue. Osteoarthritis and Osteomalacia both remove the mineral Calcium from the bones. In doing so, the average atomic number of the calcium depleted bony tissue is lower, and it takes less energy to penetrate than healthy bone tissue. Emphysema manifests as an over inflation of the alveoli of the lungs. When they expand in size, there are less of them per area of lung tissue, and the tissue becomes thinner. Osteopetrosis, pneumonia, and atelectasis are all additive pathologies, meaning that tissue become thicker and therefore more difficult to penetrate. Students should remember that pneumothorax and atelectasis are not the same; pneumothorax is free air in the pleura, and is sometimes, but not always, the cause of collapsed lung, or atelectasis. Pleural effusion or hematoma may also cause a lung to collapse. (Long, Rollins, and Smith, 13th edition, Volume 1, page 240 and page 486)

A decrease from 90 to 77 kVp will result in an increase inwavelengthgray scalescattered radiation A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: A As kilovoltage is decreased, fewer electrons are driven to the anode at a slower speed and with less energy. This results in production of fewer and lower energy, longer wavelength x-ray photons. Thus, kV affects both quantity and quality of the x-ray beam. However, although kilovoltage and receptor exposure are directly related, they are not directly proportional; that is, twice the receptor exposure does not result from doubling the kilovoltage. With respect to the effect of kilovoltage on ireceptor exposure, there is a convenient rule (15% rule) that can be followed. If it is desired to double the receptor exposure yet impossible to adjust the mAs, a similar effect can be achieved by increasing the kV by 15%. Conversely, the receptor exposure may be cut in half by decreasing the kV by 15%. (Shephard, pp. 194-197)

An increase in kilovoltage with appropriate compensation of milliampere-seconds will result inincreased part penetration.higher contrast.increased receptor exposure. A 1 only B 1 and 2 only C 2 and 3 only D 1 and 3 only

The Correct Answer is: A As the kilovoltage is increased, photon energy increases and more part penetration will occur. As the milliampere-seconds value is decreased to compensate for the increased kilovoltage, receptor exposure should remain the same. (Shephard, p. 204)

A 15% increase in kVp accompanied by a 50% decrease in mAs will result in A decreased patient dose B increase in contrast. C increased brightness. D spatial resolution.

The Correct Answer is: A Since mAs is directly proportional to receptor exposure and patient dose, the receptor exposure is cut in half and patient dose is cut in half. Spatial resolution is unaffected by changes in kVp. Brightness is a function of the computer software.

Which of the following devices is used to overcome severe variation in patient anatomy or tissue density, providing more uniform radiographic density? A Compensating filter B Grid C Collimator D Protective filter

The Correct Answer is: AA compensating filter is used when the part to be radiographed is of uneven thickness or tissue density (in the chest, mediastinum vs. lungs). The filter (made of aluminum or lead acrylic) is constructed in such a way that it will absorb much of the x-ray beam directed toward the low tissue-density area while not affecting the x-ray photons to directed toward the high tissue-density area. A collimator is used to decrease the production of scattered radiation by limiting the volume of tissue irradiated. The grid functions to trap scattered radiation before it reaches the IR, thus reducing scattered radiation fog. Protective filtration absorbs low energy x-ray photons that contribute only to patient (skin) dose and would never reach the image receptor. (Selman, 9th ed., pp. 254-255)

Which of the following devices converts electrical energy to mechanical energy? A Motor B Generator C Stator D Rotor

The Correct Answer is: AA motor is the device used to convert electrical energy to mechanical energy. The stator and rotor are the two principal parts of an induction motor. A generator converts mechanical energy into electrical energy. (Selman, p 78)

A photostimulable phosphor plate is used with A CR B Direct DR C fluoroscopic intensifying screens D image-intensified fluoroscopy

The Correct Answer is: AA photostimulable (light-stimulated) phosphor plate, or simply image plate (IP), is used in CR. The CR image plate (IP) contains a photostimulable phosphor that is the image receptor. On exposure, the PSP stores information. The IP is placed into a special scanner/processor where the PSP is scanned with a laser light and the stored image is displayed on the computer monitor. (Shephard, pp. 321-329)

A star pattern is used to measure1.focal spot resolution.2.scatter resolution.3.SID resolution. A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: AA quality control program requires the use of a number of devices to test the efficiency of various components of the imaging system. A star pattern is a resolution testing device that is used to test the effect of focal spot size. (Selman, p 210)

A satisfactory radiograph was made without a grid, using a 72-inch SID and 8 mAs. If the distance is changed to 40 inches and an 8:1 ratio grid is added, what should be the new mAs? A 10 mAs B 18 mAs C 20 mAs D 32 mAs

The Correct Answer is: AAccording to the inverse square law of radiation, as the distance between the radiation source and the IR decreases, the exposure rate increases. Therefore, a decrease in technical factors is first indicated to compensate for the distance change. The following formula (exposure maintenance formula) is used to determine new mAs values, when changing distance: Substituting known values, To then compensate for adding an 8:1 grid, you must multiply the 2.4 mAs by a factor of 4. Thus, 9.6 mAs is required to produce a receptor exposure similar to the original radiograph. The following are the factors used for mAs conversion from nongrid to grid: (Bushong, 8th ed., pp. 69, 252) No grid= 1 × original mAs5:1 grid = 2 × original mAs6:1 grid = 3 × original mAs8:1 grid = 4 × original mAs12:1 grid = 5 × original mAs16:1 grid = 6 × original mAs

Which of the following is/are components of the secondary, or high voltage, side of the x-ray circuit?Rectification systemAutotransformerkV meter A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: AAll circuit devices located before the primary coil of the high-voltage transformer are said to be on the primary or low-voltage side of the x-ray circuit. The timer, autotransformer, and (prereading) kilovoltage meter are all located in the low-voltage circuit. The secondary/high-voltage side of the circuit begins with the secondary coil of the high-voltage transformer. The mA meter is connected at the midpoint of the secondary coil of the high-voltage transformer. Following the secondary coil is the rectification system, and the x-ray tube. (Selman, 9th ed., pp. 150-151) Transformers are used to change the value of alternating current (AC). They operate on the principle of mutual induction. The secondary coil of the step-up transformer is located in the high-voltage (secondary) side of the x-ray circuit. The step-down transformer, or filament transformer, is located in the filament circuit and serves to regulate the voltage and current provided to heat the x-ray tube filament. The rectification system is also located on the high-voltage, or secondary, side of the x-ray circuit. (Selman, 9th ed., pp. 155-156)

An increase in kilovoltage in analog imaging is most likely to A produce a longer scale of contrast B produce a shorter scale of contrast C decrease the receptor exposure D decrease the production of scattered radiation

The Correct Answer is: AAn increase in kilovoltage increases the overall average energy of the x-ray photons produced at the target, thus giving them greater penetrability. (This can increase the incidence of Compton interaction and, therefore, the production of scattered radiation.) Greater penetration of all tissues serves to lengthen the scale of contrast. However, excessive scattered radiation reaching the IR will cause a fog and carries no useful information. (Selman, 9th ed., pp. 127-128)

Which of the following is most likely to produce a radiograph with a long scale of contrast? A Increased photon energy B Increased OID C Increased mAs D Increased SID

The Correct Answer is: AAn increase in photon energy accompanies an increase in kilovoltage. Kilovoltage regulates the penetrability of x-ray photons; it regulates their wavelength—the amount of energy with which they are associated. The higher the related energy of an x-ray beam, the greater its penetrability (kilovoltage and photon energy are directly related; kilovoltage and wavelength are inversely related). Adjustments in kilovoltage can have a big impact on radiographic contrast in analog imaging: As kilovoltage (photon energy) is increased, the number of grays increases, thereby producing a longer scale of contrast. An increase in OID would, if anything (air-gap), result in an increase in contrast. An increase in mAs is frequently accompanied by an appropriate decrease in kilovoltage, which would also shorten the contrast scale. SID and image contrast are unrelated. (Shephard, p 204)

Spatial resolution is directly related to1.source-image distance (SID).2.tube current.3.focal spot size. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: AAs SID increases, spatial resolution increases (a direct relationship) because magnification decreases. So SID is directly related to spatial resolution. As focal spot size increases, spatial resolution decreases because more blur/penumbra is produced. Focal spot size is thus inversely related to spatial resolution. Tube current affects receptor exposure and is unrelated to spatial resolution. (Fauber, 2nd ed., pp. 79, 81)

SID affects spatial resolution in which of the following way A Spatial resolution is directly related to SID. B Spatial resolution is inversely related to SID. C As SID increases, spatial resolution decreases. D SID is not a spatial resolution factor.

The Correct Answer is: AAs the distance from focal spot to IR (SID) increases, so does spatial resolution. Because the part is being exposed by more perpendicular (less divergent) rays, less magnification and blur are produced. Although the best spatial resolution is obtained using a long SID, the necessary increase in exposure factors and resulting increased patient exposure become a problem. An optimal 40-in. SID is used for most radiography, with the major exception being chest examinations. (Selman, 9th ed., pp. 206-207; Shephard, pp. 221-222)

A part whose width is 6 inches will be imaged at 44 inches SID. The part to be imaged lies 9 inches from the IR. What will be the projected image width of the part? A 8 inches B 10 inches C 12 inches D 20 inches

The Correct Answer is: AAs the object-to-image receptor distance (OID) increases, magnification of that object increases. Depending upon the information provided, we can determine the magnification factor, the percentage magnification, and image width. In the stated scenario, we are looking for image width. The formula used to determine image width is: Substituting known factors the equation becomes: 35x = 264 x = 7.5 inches projected image width (Shephard, p 230)

The functions of automatic beam limitation devices includereducing the production of scattered radiationincreasing the absorption of scattered radiationchanging the quality of the x-ray beam A 1 only B 2 only C 1 and 2 only D 1, 2, and 3

The Correct Answer is: ABeam restrictors function to limit the size of the irradiated field. In so doing, they limit the volume of tissue irradiated (thereby decreasing the percentage of scattered radiation generated in the part) and help to reduce patient dose. Beam restrictors do not affect the quality (energy) of the x-ray beam—that is, the function of kilovoltage and filtration. Beam restrictors do not absorb scattered radiation—that is a function of grids. (Shephard, p. 27)

As grid ratio is decreased, A the scale of contrast becomes longer B the scale of contrast becomes shorter C receptor exposure decreases D radiographic distortion decreases

The Correct Answer is: ABecause lead content decreases when grid ratio decreases, a smaller amount of scattered radiation is trapped before reaching the IR. More grays, therefore, are recorded, and a longer scale of contrast results. Receptor exposure would increase with a decrease in grid ratio. Grid ratio is unrelated to distortion. (Carlton and Adler, 4th ed., p. 432)

What are the effects of scattered radiation on a radiographic image?It produces fog.It increases contrast.It increases grid cutoff. A 1 only B 2 only C 1 and 2 only D 1, 2, and 3

The Correct Answer is: AScattered radiation is produced as x-ray photons travel through matter, interact with atoms, and are scattered (change direction). If these scattered rays are energetic enough to exit the body, they will strike the IR from all different angles. They, therefore, do not carry useful information and merely produce a flat, gray (low-contrast) fog over the image. Grid cutoff increases contrast and is caused by an improper relationship between the x-ray tube and the grid, resulting in absorption of some of the useful/primary beam. (Bushong, 8th ed., p. 248

Any images obtained using dual x-ray absorptiometry (DXA) bone densitometryare used to evaluate accuracy of the region of interest (ROI)are used as evaluation for various bone/joint disordersreflect the similar attenuation properties of soft tissue and bone A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: ADXA imaging is used to evaluate BMD. It is the most widely used method of bone densitometry—it is low-dose, precise, and uncomplicated to use/perform. DXA uses two photon energies—one for soft tissue and one for bone. Since bone is denser and attenuates x-ray photons more readily, their attenuation is calculated to represent the degree of bone density. Soft tissue attenuation information is not used to measure bone density. Any images obtained in DXA/bone densitometry are strictly to evaluate the accuracy of the region of interest (ROI); they are not used for further diagnostic purposes—additional diagnostic examinations are done for any required further evaluation. Bone densitometry/DXA can be used to evaluate bone mineral content of the body or part of it, to diagnose osteoporosis, or to evaluate the effectiveness of treatments for osteoporosis. (Frank, Long, and Smith, 11th ed., vol. 3, pp. 455-456)

To compensate for variations in gain across a digital receptor, which of the following maintenance steps should be taken? A Conduct a calibration correction for image nonuniformity B Increase or decrease the exposure factors to compensate C Install a variable resistor to adjust the electrical supply to the unit D Keep a log for at least 30 days to confirm consistent variations before making any adjustments

The Correct Answer is: ADigital systems require that a uniformity correction (A) be applied to compensate for variations in gain across the receptor. This calibration for nonuniformity (also called shading correction) must be repeated on a periodic basis; the frequency depends on the digital device and ranges from daily to semi-annually. The exposure factors should not be adjusted (B) as a result of gain variations. This would be an unacceptable practice, especially if the exposure is increased, as this will cause unnecessary patient radiation dosage. Technologists should never alter the electrical supply (C) to the digital unit. Gain adjustments can be made to the equipment by simply adjusting the gain setting. Keeping a log for 30 days (D) to track the variations in gain would not facilitate timely correction to ensure that optimal diagnostic images are being produced. (Seeram, p. 227).

The ability of an x-ray unit to produce constant radiation output at a given mAs, using various combinations of mA and time is called A linearity. B reproducibility. C densitometry. D sensitometry.

The Correct Answer is: AEach of the four factors are used as part of a complete quality assurance (QA) program. Linearity means that a given mAs, using different mA stations with appropriate exposure time adjustments, will provide consistent intensity. Reproducibility means that repeated exposures at a given technique must provide consistent intensity. Sensitometry and densitometry are used in evaluation of the film processor, part of a complete QA program. Linearity must not deviate more than 10% from one mA station to another. ((Bushong,11th ed, p 553))

All of the following are potential digital pre-processing problems, except: A Edge enhancement B Defective pixel C Image lag D Line noise

The Correct Answer is: AEdge enhancement (A) is a type of post-processing image manipulation, which can be effective for enhancing fractures and small, high-contrast tissues. Answers B, C and D are problems that may be encountered in pre-processed digital images. (Bushong, 10 th ed., p. 326).

The relationship between the height of a grid's lead strips and the distance between them is referred to as grid A ratio B radius C frequency D focusing distance

The Correct Answer is: AGrids are used in radiography to trap scattered radiation that otherwise would cause fog on the radiograph. Grid ratio is defined as the ratio of the height of the lead strips to the distance between them. Grid frequency refers to the number of lead strips per inch. Focusing distance and grid radius are terms denoting the distance range with which a focused grid may be used. (Selman, 9th ed., p. 236)

Which of the following groups of exposure factors would be most appropriate for a sthenic adult IVU? A 300 mA, 0.02 s, 72 kVp B 300 mA, 0.01 s, 82 kVp C 150 mA, 0.01 s, 94 kVp D 100 mA, 0.03 s, 82 kVp

The Correct Answer is: AIVU requires the use of iodinated contrast media. Low kilovoltage (about 70 kVp) is usually used to enhance the photoelectric effect and, in turn, to better visualize the renal collecting system. High kilovoltage will produce excessive scattered radiation and obviate the effect of the contrast agent. A higher milliamperage with a short exposure time generally is preferable. (Fauber, 2nd ed., p. 264)

If a lateral projection of the chest is being performed on an asthenic patient and the outer photocells are selected, what is likely to be the outcome? A Decreased receptor exposure B Increased receptor exposure C Scattered radiation fog D Motion blur

The Correct Answer is: AIf a lateral projection of the chest is being performed on an asthenic patient and the outer photocells are selected incorrectly, the outcome is likely to be an underexposed image. The patient is thin, and the lateral photocells have no tissue superimposed on them. Therefore, as soon as the lateral photocells detect radiation (which will be immediately), the exposure will be terminated, resulting in insufficient exposure. (Shephard, pp. 280-281)

To produce a just perceptible increase in receptor exposure in analog imaging, the radiographer should increase the A mAs by 30% B mAs by 15% C kV by 15% D kV by 30%

The Correct Answer is: AIf an x-ray image lacks sufficient receptor exposure, an increase in milliampere-seconds is required. The milliampere-seconds value regulates the number of x-ray photons produced at the target. An increase or decrease in milliampere-seconds of at least 30% is necessary to produce a perceptible effect. Increasing the kilovoltage by 15% will have about the same effect as doubling the milliampere-seconds. (Shephard, p. 173)

A radiograph exposed using a 12:1 ratio focused grid may exhibit a loss of receptor exposure at its lateral edges because the A SID was too great. B grid failed to move during the exposure. C x-ray tube was angled in the direction of the lead strips. D CR was off-center.

The Correct Answer is: AIf the SID is above or below the recommended focusing distance, the primary beam at the lateral edges will not coincide with the angled lead strips. Consequently, there will be absorption of the useful beam, termed grid cutoff. If the grid failed to move during the exposure, there would be grid lines throughout. CR angulation in the direction of the lead strips is appropriate and will not cause grid cutoff. If the CR were off-center, there would be uniform loss of receptor exposure. (Selman, 9th ed., p. 240)

If a radiograph exposed using a 12:1 ratio grid exhibits a loss of receptor exposure at its lateral edges, it is probably because the A SID was too great B grid failed to move during the exposure C x-ray tube was angled in the direction of the lead strips D central ray was off-center

The Correct Answer is: AIf the SID is above or below the recommended focusing distance, the primary beam will not coincide with the angled lead strips at the lateral edges. Consequently, there will be absorption of the useful beam, termed grid cutoff. If the grid failed to move during the exposure, there would be grid lines throughout. Central ray angulation in the direction of the lead strips is appropriate and will not cause grid cutoff. If the central ray were off-center, there would be uniform loss of receptor exposure. (Carlton and Adler, 4th ed., p. 260)

What pixel size has a 2,048 × 2,048 matrix with a 60-cm FOV? A 0.3 mm B 0.5 mm C 0.15 mm D 0.03 mm

The Correct Answer is: AIn digital imaging, pixel size is determined by dividing the FOV by the matrix. In this case, the FOV is 60 cm; since the answer is expressed in millimeters, first change 60 cm to 600 mm. Then 600 divided by 2,048 equals 0.29 mm: The FOV and matrix size are independent of one another; that is, either can be changed, and the other will remain unaffected. However, pixel size is affected by changes in either the FOV or matrix size. For example, if the matrix size is increased, pixel size decreases. If FOV increases, pixel size increases. Pixel size is inversely related to resolution. As pixel size increases, resolution decreases. (Fosbinder and Kelsey, p. 285)

Which of the following fluoroscopic modes delivers the smallest patient dose? A 30 cm field B 25 cm field C 12 cm field D 9 cm field

The Correct Answer is: AIn magnification fluoroscopic imaging, the charge on the electrostatic focusing lenses is increased in order to confine electrons to a smaller portion of the input phosphor. This magnifies the image, but at the expense of less brightness. In order to increase brightness to a diagnostic level, the mA is increased. Smaller input phosphor field sizes (D) produce magnified images of the anatomical areas being evaluated, but with an increase in patient dose. Larger input phosphor field sizes (A) produce little or no magnification of the anatomical areas being evaluated, and with decreased patient dose.(Seeram, p. 135).

Radiography using a collimated thin fan X-ray beam would be found in: A Scanned projection radiography (SPR) of the chest B Long bone measurement radiography C Radiography of foreign objects D Fluoroscopic evaluation of the ureters, as they are thin structures

The Correct Answer is: AIn scanned projection radiography (SPR) of the chest (A), the X-ray beam is collimated to a thin fan by pre-patient collimators. Post-patient image-forming X-rays likewise are collimated to a thin fan that corresponds to a detector array consisting of a scintillation phosphor, usually NaI or CsI, which is married to a linear array of CCDs through a fiberoptic path. Long bone measurement radiography (B) uses a special ruler (called a Bell-Thompson ruler) that is placed beneath and between the patient's legs. It contains centimeter markers that are displayed on specific collimated portions of the anatomy on a large radiographic film. Typical collimated exposures, taken one at a time, are focused on the hip joints, knee joints, and ankle joints. By taking any two centimeter markings corresponding to any two anatomical areas, the smaller number can be subtracted from the larger number to determine the length between the two anatomical areas. Any bilateral discrepancies would indicate either uneven growth or otherwise disproportionate lengths of the lower extremities. Radiography of foreign objects (C) requires either a static full-field exposure or collimated exposure on a radiographic cassette containing a radiographic film. This radiographic investigation to discover foreign objects requires a single exposure per projection. The resultant processed radiographs (minimum of two at 90 degree projections) will demonstrate the location of a foreign object, particularly if the atomic number of the foreign object differs from the surrounding anatomic tissues and organs. Fluoroscopic evaluation of the ureters (D) first, involves fluoroscopy, which involves a constant X-ray exposure to demonstrate real-time imaging of the anatomical structures. The ureters are typically examined during an intravenous urogram (IVU) after an iodinated contrast medium is injected in to the patient's venous system, usually via the antecubital vein route. Once the contrast medium is excreted by the kidneys, the ureters will begin to fill and, upon a static X-ray exposure on a 14" x 17" film, will be demonstrated as fine, white (because of the high atomic number of iodine), and linear structures running longitudinally to the urinary bladder. (Bushong, 10 th ed., p. 297).

While indirect digital detectors use a scintillator (phosphor) to convert X-ray energy, direct detectors use a: A Photoconductor B Scintillator C Charged coupled device D Histogram detector screen

The Correct Answer is: AIndirect digital systems use scintillators/phosphors (B) to convert X-ray energy, whereas direct digital systems use a photoconductor (A) to covert this energy. In the two types of indirect systems this energy is subsequently converted by either a charged coupled device (CCD) array or photodiode array (coupled with a thin film transistor array) , or by a TFT array in a direct system. Finally, both indirect and direct digital system conversions result in an analog signal that is converted to a digital signal by the analog-to-digital convertor (ADC). A scintillator (phosphor) receives X-ray energy and converts it to light in an indirect digital detector system. In direct conversion digital detectors, an X-ray photoconductor (B) is used to convert this energy. Some indirect digital detectors use charged coupled devices (CCD), but a scintillator (phosphor) converts the X-ray energy and, through light optics, transfers this energy to the CCD (C). The histogram (D) is a computerized graphic display of the X-ray intensities received by the detectors in direct or indirect digital detector systems (or by the image plate in CR systems). There is no histogram detector screen (D). (Seeram, 1 st ed., p. 106)

Which of the following examinations might require the use of 70 kV?AP abdomenChest radiographBarium-filled stomach A 1 only B 2 only C 1 and 2 only D 2 and 3 only

The Correct Answer is: AIt is appropriate to perform an AP abdomen radiograph with lower kilovoltage because it has such low subject contrast. Abdominal tissue densities are so similar that it takes high- or short-scale contrast (using low kilovoltage) to emphasize the little difference there is between tissues. However, high-kilovoltage factors are used frequently to even out densities in anatomic parts having high tissue contrast (e.g., the chest). However, since high kilovoltage produces added scattered radiation, it generally must be used with a grid. Barium-filled structures frequently are radiographed using 120 kV or more to penetrate the barium—to see through to posterior structures (Carlton and Adler, 4th ed., pp. 423-424)

Which of the following is (are) directly related to photon energy?KilovoltageMilliamperesWavelength A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: AKilovoltage is the qualitative regulating factor; it has a direct effect on photon energy. That is, as kilovoltage is increased, photon energy increases. Photon energy is inversely related to wavelength. That is, as photon energy increases, wavelength decreases. Photon energy is unrelated to milliamperage. (Shephard, pp. 173, 178)

Decreasing field size from 14 × 17 in. to 8 × 10 in., with no other changes, will A decrease receptor exposure and decrease the amount of scattered radiation generated within the part B decrease receptor exposure and increase the amount of scattered radiation generated within the part C increase receptor exposure and increase the amount of scattered radiation generated within the part D increase receptor exposure and decrease the amount of scattered radiation generated within the part

The Correct Answer is: ALimiting the size of the radiographic field (irradiated area) serves to limit the amount of scattered radiation produced within the anatomic part. As the amount of scattered radiation production decreases, so does the resultant receptor exposure. Therefore, as field size decreases, scattered radiation production decreases, and overall receptor exposure decreases. Limiting the size of the radiographic field is a very effective means of reducing the quantity of non-information-carrying scattered radiation (fog) produced. Limiting the size of the radiographic field is also the most effective means of patient radiation protection. (Shephard, p. 203)

The purpose of magnification fluoroscopy is to: A Enhance the image in order to facilitate diagnostic interpretation B Decrease patient dosage C Decrease fluoroscopy time D Increase efficiency of X-ray production

The Correct Answer is: AMagnification of the image is an important feature of the image intensifier. The purpose of magnification fluoroscopy is to enhance the image in order to assist the radiologist in diagnostic interpretation (A). Magnification mode in fluoroscopy actually increases patient dosage (B), as more radiation is necessary to produce the brightness levels needed to view the images. The magnification mode should therefore be used only when necessary to enhance diagnostic interpretation of small specific anatomical areas in question. Fluoroscopy time should be limited in order to ensure the practice of ALARA. However, the time needed to evaluate the anatomical areas in question is not limited to a certain time (C). Magnification fluoroscopy neither increases or decreases fluoroscopic evaluation time. X-ray production efficiency is a function of the generator and X-ray tube providing the necessary X-ray energy to produce the fluoroscopic image. Magnification fluoroscopy, therefore, does not alter the efficiency of X-ray production (D). (Seeram, p. 134).

Fractional-focus tubes, with a 0.3-mm focal spot or smaller, have special application in A magnification radiography B fluoroscopy C tomography D image intensification

The Correct Answer is: AMagnification radiography may be used to demonstrate small, delicate structures that are difficult to image with conventional radiography. Because OID is an integral part of magnification radiography, the problem of magnification unsharpness arises. The use of a fractional focal spot (0.3 mm or smaller) is essential to the maintenance of image sharpness in magnification films. Radiographic rating charts should be consulted because the heat load to the anode may be critical in magnification radiography. The long exposures typical of image-intensified fluoroscopy and tomography make the use of a fractional focal spot generally impractical and hazardous to the anode. (Selman, 9th ed., p. 226)

In which type of equipment does kilovoltage decrease during the actual length of the exposure?Condenser-discharge mobile equipmentBattery-operated mobile equipmentFixed x-ray equipment A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: AMobile x-ray machines are compact and cordless and are either the battery-operated type or the condenser-discharge type. Condenser-discharge mobile x-ray units do not use batteries; this type of mobile unit requires that it be charged before each exposure. A condenser (or capacitor) is a device that stores electrical energy. The stored energy is used to operate the x-ray tube only. Because this machine does not carry many batteries, it is much lighter and does not need a motor to drive or brake it. The major disadvantage of the capacitor/condenser-discharge unit is that as the capacitor discharges its electrical charge, the kilovoltage gradually decreases throughout the length of the exposure—therefore limiting tube output and requiring recharging between exposures. (Frank, Long, and Smith, 11th ed., vol. 3, p. 235)

Capacitor-discharge mobile x-ray units use capacitors to power thex-ray tubemachine locomotionbraking mechanism A 1 only B 2 only C 1 and 2 only D 1, 2, and 3

The Correct Answer is: AMobile x-ray machines are smaller and more compact than their fixed counterparts in the radiology department. It is important that they be relatively easy to move, that their size allows entry into patient rooms, and that their locks enable securing of the x-ray tube into the required positions. Mobile x-ray machines are cordless and are either the battery-operated type or the condenser-discharge type. Condenser-discharge mobile x-ray units do not use batteries; this type of mobile unit requires that it be charged before each exposure. A condenser (or capacitor) is a device that stores electrical energy. The stored energy is used to operate the x-ray tube only. Because this machine does not carry many batteries, it is much lighter and does not need a motor to drive or brake it. The major disadvantage of the capacitor/condenser-discharge unit is that as the capacitor discharges its electrical charge, the kilovoltage gradually decreases throughout the length of the exposure—hence, the need for recharging between exposures. (Frank, Long, and Smith, 11th ed., vol. 3, p. 235)

How would the introduction of a 6-in. OID affect analog image contrast? A Contrast would be increased. B Contrast would be decreased. C Contrast would not change. D The scale of contrast would not change.

The Correct Answer is: AOID can affect contrast of analog image when it is used as an air gap. If a 6-in. air gap (OID) is introduced between the part and IR, much of the scattered radiation emitted from the body will not reach the IR, as shown in Figure 7-20. The OID thus is acting as a low-ratio grid and increasing analog image contrast. (Shephard, p. 205)

Which of the following groups of exposure factors will produce the shortest scale of contrast? A 200 mA, 0.25 s, 70 kVp, 12:1 grid B 500 mA, 0.10 s, 90 kVp, 8:1 grid C 400 mA, 0.125 s, 80 kVp, 12:1 grid D 300 mA, 0.16 s, 70 kVp, 8:1 grid

The Correct Answer is: AOf the given factors, kilovoltage and grid ratio will have a significant effect on the scale of radiographic contrast. The milliampere-seconds values are almost identical. Because decreased kilovoltage and high-ratio grid combination would allow the least amount of scattered radiation to reach the IR, thereby producing fewer gray tones, (A) is the best answer. Group (D) also uses low kilovoltage, but the grid ratio is lower, thereby allowing more scatter to reach the IR and producing more gray tones. (Shephard, p. 308)

In radiography of a large abdomen, which of the following is (are) effective way(s) to minimize the amount of scattered radiation reaching the image receptor (IR)?Use of optimal collimationUse of low mAsUse of a low-ratio rather than high-ratio grid A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: AOne way to minimize scattered radiation reaching the IR is to use optimal kilovoltage; excessive kilovoltage increases the production of scattered radiation. Maximum collimation is exceedingly important because the smaller the volume of irradiated material, the less scattered radiation will be produced. The mAs selection has no impact on scattered radiation production or cleanup. Low-ratio grids allow a greater percentage of scattered radiation to reach the IR. Use of a high-ratio grid will clean up a greater amount of scattered radiation before it reaches the IR. Use of a compression band, or the prone position, in a large abdomen has the effect of making the abdomen "thinner"; it will, therefore, generate less scattered radiation. (Shephard, p. 203)

Which of the following systems function(s) to compensate for changing patient/part thicknesses during fluoroscopic procedures? A Automatic brightness control B Minification gain C Automatic resolution control D Flux gain

The Correct Answer is: AParts being examined during fluoroscopic procedures change in thickness and density as the patient is required to change positions and as the fluoroscope is moved to examine different regions of the body that have varying thickness and tissue densities. The automatic brightness control functions to vary the required milliampere-seconds and/or kilovoltage as necessary. With this method, patient dose varies, and image quality is maintained. Minification and flux gain contribute to total brightness gain. (Bushong, 8th ed., p. 360)

Which of the following focal-spot sizes should be employed for magnification radiography? A 0.2 mm B 0.6 mm C 1.2 mm D 2.0 mm

The Correct Answer is: AProper use of focal spot size is of paramount importance in magnification radiography. A magnified image that is diagnostic can be obtained only by using a fractional focal spot of 0.3 mm or smaller. The amount of blur or geometric unsharpness produced by focal spots that are larger in size render the radiograph undiagnostic. (Selman, 9th ed., p. 226)

Which of the following combinations would pose the most hazard to a particular anode? A 0.6 mm focal spot, 75 kVp, 30 mAs B 0.6 mm focal spot, 85 kVp, 15 mAs C 1.2 mm focal spot, 75 kVp, 30 mAs D 1.2 mm focal spot, 85 kVp, 15 mAs

The Correct Answer is: ARadiographic rating charts enable the operator to determine the maximum safe mA, exposure time, and kVp for a particular exposure using a particular x-ray tube. An exposure that can be made safely with the large focal spot may not be safe for use with the small focal spot of the same x-ray tube. The total number of HU that an exposure generates also influences the amount of stress (in the form of heat) imparted to the anode. The product of mAs and kVp determines HU. Groups A and C produce 2250 HU; groups B and D produce 1275 HU. Groups B and D deliver less heat load, but group D delivers it to a larger area (actual focal spot) making this the least hazardous group of technical factors. The most hazardous group of technical factors is group A because it delivers the greatest heat (2,250 HU) with the small focal spot.

Which of the following combinations would pose the least hazard to a particular anode? A 1.2-mm focal spot, 92 kVp, 1.5 mAs B 0.6-mm focal spot, 80 kVp, 3 mAs C 1.2-mm focal spot, 70 kVp, 6 mAs D 0.6-mm focal spot, 60 kVp, 12 mAs

The Correct Answer is: ARadiographic rating charts enable the operator to determine the maximum safe milliamperage, exposure time, and kilovoltage for a particular exposure using a particular x-ray tube. An exposure that can be made safely with the large focal spot may not be safe for use with the small focal spot of the same x-ray tube. The total number of heat units that an exposure generates also influences the amount of stress (in the form of heat) imparted to the anode. The product of milliampere-seconds and kilovoltage determines heat units. Group (A) produces 138 HU, group (B) produces 240 HU, group (C) produces 420 HU, and group (D) produces 720 HU. The least hazardous group of technical factors is, therefore, group (A). Group (A) is also delivering its heat to the large focal spot, thereby decreasing the heat load to the anode. (Selman, 9th ed., pp. 144-145)

A device used to ensure reproducible radiographs, regardless of tissue-density variations, is the A AEC B penetrometer C moving grid D compensating filter

The Correct Answer is: ARadiographic reproducibility is an important concept in producing high-quality diagnostic images. Radiographic results should be consistent and predictable not only in terms of positioning accuracy but also with respect to technical factors. AEC devices (ionization chambers) automatically terminate the x-ray exposure once a predetermined quantity of x-rays has penetrated the part, thus ensuring consistent results. (Shephard, p. 274)

A positive contrast agentabsorbs x-ray photonsresults in a dark area on the radiographis composed of elements having low atomic number A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: ARadiopaque contrast agents appear white on the finished image because many x-ray photons are absorbed. These are referred to positive contrast agents—composed of dense (i.e., high atomic number) material through which x-rays will not pass easily. Radiolucent contrast agents appear black on the finished image because x-ray photons pass through easily. An example of a radiolucent contrast agent is air. (Shephard, pp. 200-202)

Which of the following terms/units is used to express the resolution of a diagnostic image? A Line pairs per millimeter (lp/mm) B Speed C Latitude D Kiloelectronvolts (keV)

The Correct Answer is: AResolution describes how closely fine details may be associated and still be recognized as separate details before seeming to blend into each other and appear "as one." The degree of resolution transferred to the IR is a function of the resolving power of each of the system components and can be expressed in line pairs per millimeter (lp/mm). It can be measured using a resolution test pattern; a variety of resolution test tools are available. The star pattern generally is used for focal-spot-size evaluation, whereas the parallel-line type is used for evaluating image receptors. Resolution can also be expressed in terms of line-spread function (LSP) or modulation transfer function (MTF). LSP is measured using a 10-×m x-ray beam; MTF measures the amount of information lost between the object and the IR. (Carlton and Adler, 4th ed., p. 334)

Exposure values arising from excessive kV, insufficient collimation, or thick anatomic structures are termed A fog. B matrix. C artifact. D resolution.

The Correct Answer is: AScattered radiation produces fog, which can add unwanted exposure values to the x-ray image and impair its diagnostic value. Scattered radiation production is encouraged at high kV, insufficient beam restriction, and thick anatomic parts. Scattered radiation can be removed from the remnant beam with the use of grids.

Misalignment of the tube-part-IR relationship results in A shape distortion B size distortion C magnification D blur

The Correct Answer is: AShape distortion (e.g., foreshortening or elongation) is caused by improper alignment of the tube, part, and IR. Size distortion, or magnification, is caused by too great an OID or too short an SID. Focal-spot blur is caused by the use of a large focal spot. (Fauber, p. 93)

Better resolution is obtained with A high SNR. B low SNR. C windowing. D smaller matrix.

The Correct Answer is: ASpatial resolution increases as SNR (signal-to-noise ratio) increases. A high SNR (e.g., 1000:1) indicates that there is far more signal than noise. A lower SNR (e.g., 200:1) indicates a "noisy" image. Windowing is unrelated to resolution; it permits post-processing image manipulation. Image matrix has a great deal to do with resolution. A larger image matrix (1800 × 1800) offers better resolution than a smaller image matrix (700 × 700). Smaller image matrices look "pixelly." (Seeram, p 112)

The movement of the IP through the transport system of a CR reader is referred to as the: A Slow-scan direction B Charge-coupled direction C Nyquist direction D Fast-scan direction

The Correct Answer is: AThe IP moves slowly through the transport system of a CR reader and this movement is considered the slow-scan direction (A). The laser light in the reader is rapidly reflected by an oscillating polygonal mirror that redirects the beam through a special lens called the f-theta lens, which focuses the light on a cylindrical mirror that reflects the light toward the PSP (photostimulable phosphor). This light moves back and forth very rapidly to scan the PSP transversely, in a raster pattern, and this movement of the laser beam across the PSP is therefore called the fast-scan direction (D). Charge-coupled direction (B) is not a term used to describe laser scanning in the CR reader. Charge-coupled devices are used in digital image receptors. Nyquist direction (C) is not a term used to describe laser scanning in the CR reader. With digital systems, the spatial resolution is related to pixel pitch. The maximum spatial resolution is equal to the Nyquist frequency, or 1/2X the pixel pitch (mm). (Seeram p. 54; Shephard p. 327; Carter and Veale p. 70)

Incomplete erasure of CR plates can contribute to a A Ghost artifact B Moiré artifact C Static artifact D Grid cutoff artifact

The Correct Answer is: AThe appearance of ghost artifacts (A) can be seen when CR image plates are incompletely erased. If an image plate has not been used for 24 hours, its phosphor storage plate should be erased again before using it for a diagnostic radiographic exposure. If a radiographic grid has a frequency that approximates the CR scan frequency and the grid strips are oriented in the same direction as the scan, the Moiré artifact may be observed (B); to decrease the possibility of this effect, high frequency grids are recommended for digital imaging. Grid cutoff artifacts (D) are seen as a decrease in receptor exposure on one side of the image when the CR is off-centered or not perpendicular to the grid - or, on both sides when the central ray is properly centered but the focal range is exceeded when using a focused grid. Static artifacts (C) may be seen on radiographic film imaging. (Bushong, 9 th ed., p. 487)

When radiographing a cross-table lateral hip or axial shoulder using CR, one method of creating a collimation margin at the bottom of the radiograph is to: A Use a narrow lead strip at the bottom edge of the IP, but out of the anatomy B Only one collimation margin is necessary, so this would not be necessary C Make two exposures with suspended respiration; one for the uppermost anatomy, then a second for the dependent anatomy D Expose the anatomy as is and use the post-processing cropping feature

The Correct Answer is: AThe difference between cross-table hips or axial shoulders is that most often only one collimated edge is visible (because soft tissue extends to edge of table/IP). If a second collimated border is not detected, the exposure field is not accurately located, processing/rescaling errors will likely occur. One may create a second collimation margin by using a narrow (approx. 1 in.) lead strip at the bottom of the IP to generate a "margin" between the exposure field and the edge of the cassette (A). If only one collimation margin is included on the receptor (B), the radiographer has improperly centered the anatomical part. This may result in misidentification of the exposure field and therefore, cause a processing error. Two exposures at different central ray locations (C) would result in two images where a misaligned image of the anatomy for both exposures would result. The cropping feature (D) is a post-processing function that will not affect the system's ability to recognize the exposure field. (Seeram, 1 st ed., p. 92)

A fill factor of 80% in direct or indirect digital radiography means that: A 20% of the pixel area is occupied by the detector electronics with 80% representing the sensing area B 80% of the pixel area is occupied by the detector electronics with 20% representing the sensing area C The saturation level will be unacceptable D Only 20% of the image will be captured

The Correct Answer is: AThe fill factor is expressed as a percentage. In this case (A), 80% means that 20% of the pixel area is occupied by the detector electronics with 80% representing the sensing area which, in turn, represents the image. Larger fill factors indicate large sensing areas; larger fill factors (and sensing areas) indicate better spatial and contrast resolution. In (B), 20% means that 80% of the pixel area is occupied by the detector electronics with 20% representing the sensing area which, in turn, represents the image. Saturation (C) means that beyond a certain exposure level, a large number of the pixels will be at the maximum digital value (black) so that there is no signal difference in the very high exposure areas, resulting in a loss of anatomical structures in that region. This is an undesirable effect. Collimation defines the exposure field, so 20% of the image would only occur if 20% of the anatomical area were to be exposed and captured (D). (Seeram, 1 st ed., p. 112)

All the following are related to spatial resolution except A milliamperage B focal-spot size C source-to-object distance D OID

The Correct Answer is: AThe focal-spot size selected will determine the amount of focal-spot, or geometric, blur produced in the image. OID is responsible for image magnification and hence spatial resolution. Source-to-object distance can vary with changes in SID and/or OID, and therefore impact magnification and resolution. The milliamperage is unrelated to spatial resolution; it affects the quantity of x-ray photons produced and thus receptor exposure and patient dose. (Selman, 9th ed., pp. 206-210)

The image intensifier's input phosphor differs from the output phosphor in that the input phosphor A is much larger than the output phosphor B emits electrons, whereas the output phosphor emits light photons C absorbs electrons, whereas the output phosphor absorbs light photons D is a fixed size, and the size of the output phosphor can vary

The Correct Answer is: AThe image intensifier's input phosphor is 6 to 9 times larger than the output phosphor. It receives the remnant radiation emerging from the patient and converts it into a fluorescent light image. Very close to the input phosphor, separated only by a thin, transparent layer, is the photocathode. The photocathode is made of a photoemissive alloy, usually a cesium and antimony compound. The fluorescent light image strikes the photocathode and is converted to an electron image, which is focused by the electrostatic lenses to the small output phosphor. (Bushong, 8th ed., pp. 360-363)

The image intensifier's input phosphor generally is composed of A cesium iodide B zinc cadmium sulfide C gadolinium oxysulfide D calcium tungstate

The Correct Answer is: AThe image intensifier's input phosphor receives the remnant beam from the patient and converts it to a fluorescent light image. To maintain resolution, the input phosphor is made of cesium iodide crystals. Cesium iodide is much more efficient in this conversion process than was the phosphor used previously, zinc cadmium sulfide. Calcium tungstate was used in intensifying screens in film screen imaging for many years prior to the development of rare earth phosphors such as gadolinium oxysulfide. (Bushong, 8th ed., p. 360)

In the CR reader, some of the laser light is redirected to a reference detector by way of a(n): A Beam splitter B Analog-to-digital converter C Photomultiplier tube D f-theta lens

The Correct Answer is: AThe laser beam in a CR reader is directed to a reference detector by way of a beam splitter (A). Optical components called beam splitters are used to divide input light into two separate parts. Beam splitters are found in many laser or illumination systems, and light can be split according to overall intensity or by wavelength. A reference detector enables the CR reader to monitor the laser beam intensity and make adjustments for any fluctuations that may occur, thereby ensuring constant laser beam intensity and uniform release of stored phosphor energy. The PMT, or photomultiplier tube (C), receives the light emitted from a CR phosphor plate as it is scanned by the laser beam, which, in turn, sends an electronic signal to the ADC. The ADC, or analog-to-digital convertor (B), receives an electrical signal from a photomultiplier tube that receives the light emitted from a CR image plate as it is scanned by the laser beam. The ADC changes this electrical (analog) signal to a binary (digital) signal to be used by the processing computer. The f-theta lens in a CR reader focuses the laser light onto a cylindrical mirror, which, in turn, reflects this light toward the image plate as it traverses the scanning section of the CR reader (D). (Shephard p. 327)

According to the line-focus principle, an anode with a small angle providesimproved spatial resolution.improved heat capacity.less heel effect. A 1 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: AThe line-focus principle states that as the target/focal track angle gets larger, the effective focal spot is larger. When the effective focal spot is larger, focal spot blur increases, and spatial resolution is decreased. With focal spot/track size (area of electron bombardment) unchanged, heat capacity remains unchanged. The smaller the anode angle, the more apparent the anode heel effect.

The following are disadvantages of a capacitor-discharge mobile unit, except: A The mAs increases during the exposure, called "mAs creep" B The capacitor may continue to discharge after the exposure C The actual kilovoltage achieved during an exposure is significantly lower than the set kVp D At lower kVp settings, the capacitors discharge more slowly and, therefore, a considerable residual kV may exist after the desired exposure time

The Correct Answer is: AThe mAs does not increase during an exposure (A) using a capacitor discharge mobile unit, but rather, the kVp decreases during the exposure. A disadvantage of a capacitor discharge unit is that the capacitor may continue to discharge after the usable exposure is made (B). Exposure begins at peak kV and then decreases during the exposure. The end of the exposure is called wavetail cutoff. The actual kilovoltage achieved during an exposure is significantly lower than the set kVp (C), approximately one kVp per mAs lower than the set kVp. At lower kVp settings, the capacitors discharge more slowly and, therefore, a considerable residual kV may exist after the desired exposure time (D). This can create a leakage of radiation, although there are several devices that are designed to avoid this problem. For instance, grid-biased X-ray tubes can be used to terminate the X-ray photon emission at a set time by reversing the charge polarity of a wire grid positioned in front of the cathode filament. Additionally, some tube collimators are designed to automatically close its lead shutters after the desired exposure is made, thus stopping radiation leakage. (Carlton and Adler, 5 th ed., p. 102).

Of the following groups of analog exposure factors, which is likely to produce the shortest scale of image contrast? A 500 mA, 0.040 second, 70 kV B 100 mA, 0.100 second, 80 kV C 200 mA, 0.025 second, 92 kV D 700 mA, 0.014 second, 80 kV

The Correct Answer is: AThe most important factor regulating radiographic contrast in analog imaging is kilovoltage. The lower the kilovoltage, the shorter is the scale of contrast. All the milliampere-seconds values in this problem have been adjusted for kilovoltage changes to maintain receptor exposure, but just a glance at each of the kilovoltages is often a good indicator of which will produce the longest scale or the shortest scale contrast. In this case, the lowest kV will produce the shortest scale (highest) contrast.(Shephard, pp. 306, 308)

Pixel size and spacing determine the spatial resolution of the digital image. This is known as: A Pixel pitch B Focal resolution C Nyquist resolution D Frequency modulation

The Correct Answer is: AThe pixel size and spacing (i.e., pixel pitch, which is the distance from the midpoint of one pixel to the midpoint of the adjacent pixel) determine the spatial resolution of the image (A). The number of pixels can be obtained by multiplying the horizontal number of pixels by the vertical number of pixels in the image matrix (A). Focal resolution (B) is not a term used to describe spatial resolution in a digital radiographic image. However, the focal "spot" size does have an influence on image resolution. The smaller focal spot sizes should be used for smaller anatomical parts, whenever involuntary motion is absent. Nyquist "resolution" (C) is not a term used to describe spatial resolution in a digital radiographic image. However, the Nyquist "frequency," which is 1/2X the pixel pitch (mm) is equivalent to the spatial resolution. Frequency modulation (D) is not a term used to describe spatial resolution in a digital radiographic image. However, modulation transfer function (MTF) measures the ability of a detector to transfer its spatial resolution characteristics to the image. (Seeram, 1 st ed., p. 110)

Which of the following absorbers has the highest attenuation coefficient? A Bone B Muscle C Fat D Air

The Correct Answer is: AThe radiographic subject, the patient, is composed of many different tissue types that have varying tissue densities, resulting in varying degrees of photon attenuation and absorption. The atomic number (Z) of the tissues under investigation is directly related to its attenuation coefficient. This differential absorption contributes to the various shades of gray (scale of radiographic contrast) on the finished x-ray image. Air has an effective Z number of 7.78, fat is about 6.46, water is 7.51, muscle is 7.64, and bone is 12.31. (Carlton and Adler, 3rd ed., p. 249)

The safe approach to avoid an exposure field recognition error when using CR is to: A Expose one image on the smallest IP available with collimation margins aligned parallel to the edges of the IP B Expose multiple images on one IP, but make sure all collimation margins overlap C Expose one image on the IP, but do not collimate D Expose multiple images on one IP, but make sure all collimation margins are parallel to each other and do not overlap

The Correct Answer is: AThe safe approach to avoid an exposure field recognition error when using CR is to acquire one image on the smallest IP available. Collimation margins should also be parallel to the edges of the cassette (A). Exposing multiple images on one image plate (B) with overlapping collimation borders can result in an exposure field recognition error. The ALARA principle should be applied for every radiographic exposure. Collimation is critical to minimize patient exposure and dose (C). It is best to expose one image on the smallest image plate that will include all pertinent anatomy. Making multiple exposures on one image plate, regardless of attention to proper collimation can result in an exposure field recognition error (D). (Seeram, 1 st ed., p. 92)

The filtering effect of the x-ray tube's glass envelope and its oil coolant are referred to collectively as A inherent filtration B added filtration C compensating filtration D port filtration

The Correct Answer is: AThe x-ray photons emitted from the anode focus are heterogeneous in nature. The low-energy photons must be removed because they are not penetrating enough to contribute to the image and because they do contribute to the patient's skin dose. The glass envelope and oil coolant provide approximately 0.5- to 1.0-mm Al equivalent filtration, which is referred to as inherent because it is a built-in, permanent part of the tube head. (Selman, 9th ed., p. 132)

All the following statements regarding three-phase current are true except A three-phase current is constant-potential direct current. B three-phase equipment produces more x-rays per milliampere-second. C three-phase equipment produces higher-average-energy x-rays than single-phase equipment. D the three-phase waveform has less ripple than the single-phase waveform.

The Correct Answer is: AThree-phase current is obtained from three individual alternating currents superimposed on, but out of step with, one another by 120 degrees. The result is an almost constant potential current, with only a very small voltage ripple (4%-13%), producing more x-rays per milliampere-second. (Bushong, 10th ed., p. 243)

Deposition of vaporized tungsten on the inner surface of the x-ray tube glass windowacts as additional filtrationresults in increased tube outputresults in anode pitting A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: AThrough the action of thermionic emission, as the tungsten filament continually gives up electrons, it gradually becomes thinner with age. This evaporated tungsten frequently is deposited on the inner surface of the glass envelope at the tube window. When this happens, it acts as an additional filter of the x-ray beam, thereby reducing tube output. Also, the tungsten deposit actually may attract electrons from the filament, creating a tube current and causing puncture of the glass envelope. (Selman, 9th ed., pp. 137-138)

The exposure factors of 300 mA, 0.017 second, and 72 kVp produce an mAs value of A 5. B 50. C 500. D 5000.

The Correct Answer is: ATo calculate mAs, multiply milliamperage times exposure time. In this case, 300 mA × 0.017 s = 5.10 mAs. Careful attention to proper decimal placement will help avoid basic math errors. (Shephard, p 170)

In digital imaging, the maximum spatial resolution is equal to: A The Nyquist frequency, which is 1/2X the pixel pitch (mm) B The wavelength of the detector system's analog-to-digital convertor's electrical signal C The distance between the silver halide crystals in the image receptor D The detective quantum efficiency of the imaging system; this should be at least 2X the frequency of the analog-to-digital convertor electrical signal

The Correct Answer is: AWith digital systems, the spatial resolution is related to pixel pitch. The maximum spatial resolution is equal to the Nyquist frequency, 1/2X the pixel pitch (mm) (A). The wavelength of the electrical signal in an analog-to-digital convertor (ADC) is constant, and not affected by the pixel pitch of the matrix (B). Digital imaging does not use receptors with silver halide crystals. These crystals are used in radiographic film (C). Spatial resolution depends on the pixel sizes and pitch in the image matrix. Detective quantum efficiency (DQE) is a measure of the efficiency of a digital system to detect the X-ray photons and convert them into an image signal, regardless of the size and pitch of the image matrix pixels (D). (Seeram, 1 st ed., p. 97)

Which of the following voltage ripples is (are) produced by single-phase equipment1.100% voltage ripple2.13% voltage ripple3.3.5% voltage ripple A 1 only B 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: AWith single-phase, full-wave-rectified equipment, the voltage drops to zero every 180° (of the AC waveform); that is, there is 100% voltage ripple. With three-phase equipment, the voltage ripple is significantly smaller. Three-phase, 6-pulse equipment has a 13% voltage ripple, and three-phase, 12-pulse equipment has only a 3.5% ripple. Three-phase, 12-pulse equipment comes closest to constant potential, as the voltage never falls below 96.5% of maximum value. (Selman, p 96)

If 85 kVp, 400 mA, and ⅛ s were used for a particular exposure using single-phase equipment, which of the following milliamperage or time values would be required, all other factors being constant, to produce a similar receptor exposure using high frequency multiphase equipment? A 200 mA B 800 mA C 0.125 s D 0.25 s

The Correct Answer is: AWith three-phase equipment, the voltage never drops to zero, and x-ray intensity is significantly greater. When changing from single-phase to high frequency multi phase equipment, a conversion factor of 0.5 is used. In this case, we were using 50 mAs. Therefore, 50 mAs x 0.5 = 25 mAs. Choice A provides us with the correct mA - 200 mA x 1/8 sec = 25 mAs. When changing from high frequency multiphase equipment to single phase equipment, the conversion factor of 2.0 is used. (Carlton and Adler, 6th ed., p. 359)

If a 4-inch collimated field is changed to a 14-inch collimated field, with no other changes, the image receptor will experience A decreased receptor exposure. B increased receptor exposure. C more spatial resolution. D less spatial resolution.

The Correct Answer is: B More scattered radiation will be generated within a part if the *size of the field is increased, if the *kilovoltage is increased, or if the *thickness and density of tissue increases. As the quantity of scattered radiation increases from any of these sources, receptor exposure increases. Beam restriction does not impact spatial resolution. (Carlton & Adler, 5th ed p 396)

Spatial resolution is inversely related toSIDOIDgrid ratio A 1 only B 2 only C 1 and 2 only D 1, 2, and 3

The Correct Answer is: B SID is directly related to spatial resolution because as SID increases, so does resolution (because magnification is decreased). OID is inversely related to spatial resolution because as OID increases, spatial resolution decreases. Grid ratio is not associated with spatial resolution. Therefore, of the given choices, OID is inversely related to spatial resolution. SID is directly related to spatial resolution. (Shephard, pp. 221-224)

Which among the following components is (are) part of the gantry of a CT imaging system?X-ray tubeDetector arrayControl panel A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BA computed tomographic (CT) imaging system has three component parts—a gantry, a computer, and an operator console. The gantry component includes an x-ray tube, a detector array, a high-voltage generator, a collimator assembly, and a patient couch with its motorized mechanism. The computer is exceedingly sophisticated, performing thousands of calculations simultaneously per second. It is responsible for image reconstruction and postprocessing functions. At the operator console, somewhat similar to a control panel used in projection radiography, has the controls for equipment operation and image manipulation. Technical factors are selected and monitored here, adjustments can be made, and the patient couch is operated. (Bushong, 8th ed., pp. 429-430; Bontrager and Lampignano, 6th ed., p. 731)

Which of the following is a device that can be used in lieu of an image intensifier/charge-coupled device combination in digital fluoroscopy? A Charge-coupled device B Flat panel image receptor C photometer D photomultiplier tube

The Correct Answer is: BA flat panel image receptor (FPIR) (B) composed of cesium iodide and amorphous silicon pixel detectors can be used in place of an image intensifier in digital fluoroscopy. There are several advantages of FPIR imaging over image intensifier/CCD imaging, including distortion free images, constant image quality and contrast resolution over the entire image, high detective quantum efficiency (DQE) at all dose levels, a rectangular image area coupled to a similar shaped image monitor, and its immunity to external magnetic fields. A charge-coupled device (CCD) (A) is mounted on the output phosphor of the image intensifier tube and is coupled via fiber optics or a lens system. The sensitive layer of crystalline silicon within the CCD responds to the light from the output phosphor, creating and electrical charge. The charges are sampled, pixel by pixel, and then manipulated to produce a digital image. A photometer (C) is used to measure the luminance response and uniformity of monitors used in digital imaging. A photomultiplier tube receives light energy from the scanned IP plate in a CR reader and converts it into an electrical (analog) signal that can then be converted to a binary signal in the analog-to-digital convertor (ADC). This binary signal is then processed by a computer to develop a diagnostic image. Newer CR readers may use a charged-coupled device (CDC) (D) to convert the light energy into an electrical signal. (Bushong, 9 th ed., pp. 441-442).

A focal-spot size of 0.3 mm or smaller is essential for which of the following procedures? A Bone radiography B Magnification radiography C Tomography D Fluoroscopy

The Correct Answer is: BA fractional focal spot of 0.3 mm or smaller is essential for rendering fine detail without focal-spot blurring in magnification radiography. As the object image is magnified, so will be the associated blur unless the fractional focal spot is used. Fluoroscopic procedures probably would cause great wear on a fractional focal spot. Use of the fractional focal spot is not essential in bone radiography, although magnification of bony structures often is helpful in locating hairline fractures. (Selman, 9th ed., pp. 226-228)

A focal-spot size of 0.3 mm or smaller is essential for A small-bone radiography B magnification radiography C long SID techniques D fluoroscopy

The Correct Answer is: BA fractional focal spot of 0.3 mm or smaller is essential for reproducing fine spatial resolution without focal-spot blurring in magnification radiography. As the object image is magnified, so will be any associated blur unless a fractional focal spot is used. Use of a fractional focal spot on a routine basis is unnecessary; it is not advised because it causes unnecessary wear on the x-ray tube and offers little radiographic advantage. (Shephard, p. 217)

A device contained within many CR readers that functions to convert light energy released by the PSP into electrical energy, is called a: A Transilluminator B Photomultiplier tube C Light gate D Penetrometer

The Correct Answer is: BA photomultiplier tube (B) receives light energy from the scanned PSP plate in a CR reader and converts it into an electrical (analog) signal that can then be converted to a binary signal in the analog-to-digital convertor (ADC). This binary signal is then processed by a computer to develop a diagnostic image. Newer CR readers may use a charged-coupled device (CCD) to convert the light energy into an electrical signal. The light gate (C), (or channeling guide,) in a CR reader channels the light energy released by the image plate as it is scanned by the laser beam to the photomultiplier tube. A penetrometer (D) (or aluminum step wedge) is a device used for quality control testing in film radiography. After an exposure of this device is made while it rests on top of a film cassette, the film within the cassette is chemically processed. The resultant image demonstrates multiple steps of densities. The densities can be measured by a densitometer to determine the film contrast index and other processing-related factors. A sensitometer, which is an electrical device, can be used in lieu of the penetrometer and projects a preset light exposure on the film in the darkroom. After the film is processed, multiple steps of densities, similar to those achieved using the penetrometer, are demonstrated and can then be measured by a densitometer in the same fashion. A transilluminator is a device used for imaging of fluorescent DNA and proteins in a molecular biology lab (A). (Seeram p. 54; Shephard p. 328-329; Carter and Veale p. 72)

The smallest digital detectors (approximately 100 microns) provide the best spatial resolution and, therefore, are best-suited for use in: A Fluoroscopy procedures B Mammography C Pediatric radiography D Long bone measurement to ensure measurement accuracy

The Correct Answer is: BAlthough spatial resolution is important in all radiographic or fluoroscopic applications, the systems affording the maximum spatial resolution are applied to those examinations such as mammography (B) where microscopic lesions must be detected. Lesions typically detected in fluoroscopic images (A) are at the macroscopic level. Maximum spatial resolution in cassetteless digital systems is limited by the size of the digital detectors. In the case of mammography, the best possible spatial resolution is required to ensure the detection and display of micro-calcifications, which may be suggestive of malignant lesions (B). Spatial resolution is important in pediatric imaging (C) and those systems used for this application provide sufficient resolution to display diagnostically acceptable images. The spatial resolution is not as important for long-bone measurement (D) as it is in mammography. These radiographic procedures, regardless of the spatial resolution, are intended to provide measurements from one joint to another, which does not require optimal spatial resolution. (Seeram, 1 st ed., p. 98)

Which of the following will improve the spatial resolution of image-intensified images?A very thin coating of cesium iodide on the input phosphorA smaller-diameter input screen/phosphorIncreased total brightness gain A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: BAn image's spatial resolution refers to the sharpness of its image details. As the input screen/phosphor layer is made thinner, spatial resolution increases. Also, the smaller the input phosphor diameter, the greater is the spatial resolution. A brighter image is easier to see but does not affect resolution. (Bushong, 8th ed., pp. 360-363)

An increase in kilovoltage will have which of the following effects?More scattered radiation will be produced.The exposure rate will increase.Radiographic contrast will increase. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BAn increase in kilovoltage (photon energy) will result in a greater number (i.e., exposure rate) of scattered photons (Compton interaction). These scattered photons carry no useful information and contribute to radiation fog, thus potentially decreasing radiographic contrast, especially in analog imaging. (Bushong, 11th ed., p. 155)

In an AP abdomen radiograph taken at 105-cm SID during an IVU series, one renal shadow measures 9 cm in width. If the OID is 18 cm, what is the actual width of the kidney? A 5 cm B 7.5 cm C 11 cm D 18 cm

The Correct Answer is: BAs OID increases, magnification increases. Viscera and structures within the body will be varying distances from the IR depending on their location within the body and the position used for the exposure. The size of a particular structure or image can be calculated using the following formula: Substituting known quantities: Thus, x = 7.45 cm (approximate actual size). The relationship between SID, SOD, and OID is illustrated in Figure 7-23. (Bushong, 10th ed., p. 174)

In an AP abdomen taken at 105-cm SID during an IV urography series, one renal shadow measures 9 cm in width. If the OID is 18 cm, what is the actual width of the kidney? A 5 cm B 7.5 cm C 11 cm D 18 cm

The Correct Answer is: BAs OID increases, magnification increases. Viscera and structures within the body will be varying distances from the image receptor, depending on their location within the body and the position used for the exposure. The size of a particular structure or image can be calculated using the following formula: Substituting known quantities, The relationship between SID, SOD, and OID and the equation for determining image or object size is illustrated in the figure below. (Bushong, 8th ed., p. 284)

Which one of the following is (are) used to control the production of scattered radiation?CollimatorsOptimal kVUse of grids A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BAs kilovoltage is increased, x-ray photons begin to interact with atoms of tissue via the Compton scattered interaction. Scattered x-ray photons result, which serve only to add unwanted, undiagnostic densities (scattered radiation fog) to the radiologic image. (While Compton scatter reduces patient dose compared with photoelectric interactions, it can pose a significant radiation hazard to personnel during fluoroscopic procedures.) Therefore, the use of optimal kilovoltage is recommended to reduce the production of scattered radiation. Scattered radiation is also a function of the size and content of the irradiated field. The greater the volume and atomic number of the tissue, the greater is the production of scattered radiation. Although there is little that can be done about the atomic number of the structure to be radiographed, every effort can be made to keep the field size restricted to the essential area of interest in an effort to decrease production of scattered radiation. Grids have no effect on the production of scattered radiation, but they are very effective in removing scattered radiation from the beam before it strikes the IR. (Fauber, 2nd ed., pp. 71, 103)

As the CR laser scanner/reader recognizes the phosphostimulated luminescence (PSL) released by the PSP storage plate, it constructs a graphic representation of pixel value distribution called a A processing algorithm B histogram C lookup table D exposure index

The Correct Answer is: BAs the CR laser scanner/reader recognizes the phosphostimulated luminescence (PSL) released by the PSP storage plate, it constructs a graphic representation of pixel value distribution called a histogram. The photostimulable storage phosphor (PSP) within the IP is the image receptor (IR). The PSP is a europium-doped barium fluorohalide coated storage plate. When the PSP is exposed by x-ray photons, the x-ray energy interacts with the crystals and a small amount of visible light is emitted, but most of the x-ray energy is stored (hence, the term storage plate). This stored energy represents the latent image. The IP is placed in the CR scanner/reader where a helium-neon, or solid-state, laser beam scans the PSP and its stored energy is released as blue-violet light (phosphostimulated luminescence [PSL]). This light signal represents varying tissue densities and the latent image that is then transferred to an analog-to-digital converter (ADC)—converting the signal to a digital (electrical) one. The PSL values will result in numerous image brightness values that represent various tissue densities (i.e., x-ray attenuation properties), for example, bone, muscle, blood-filled organs, air/gas, pathologic processes, and so on. The CR scanner/reader recognizes all these values and constructs a representative gray-scale histogram of them corresponding to the anatomical characteristics of the imaged part. Thus, all PA chest histograms will be similar, all lateral chest histograms will be similar, all pelvis histograms will be similar, and so on. A histogram is a graphic representation of pixel-value distribution. The histogram analyzes all the densities from the PSP and represents them graphically—demonstrating the quantity of exposure, the number of pixels, and their value. Histograms are generated that are unique to each body part that can be imaged. After a part is exposed/imaged, its PSP is read/scanned and its own histogram is developed and analyzed. The resulting analysis, and histogram of the actual imaged part, is compared to the programmed representative histogram for that part. Over time, if required diagnostic image characteristics change, a histogram can be updated to reflect the latest required characteristics. (Carlton and Adler, 4th ed., pp. 361-362)

An increase in the kilovoltage applied to the x-ray tube increases thepercentage of high-energy photons produced.beam intensity.patient absorption. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BAs the kilovoltage is increased, a greater number of electrons are driven across to the anode with greater force. Therefore, as energy conversion takes place at the anode, more high-energy photons are produced. However, because they are higher-energy photons, there will be less patient absorption. (Fauber, 2nd ed., p. 58)

All the following statements regarding beam restriction are true except A beam restriction improves contrast resolution B beam restriction improves spatial resolution C field size should never exceed IR dimensions D beam restriction reduces patient dose

The Correct Answer is: BBeam restriction is used to determine the size of the x-ray field. This size never should be larger than the IR size. Because the size of the irradiated area can be made smaller, patient dose is reduced. Beam restriction reduces the production of scattered radiation that leads to fog and, therefore, improves contrast resolution. Spatial resolution is related to factors affecting recorded detail, not contrast resolution. (Bushong, 8th ed., pp. 243, 244)

Which of the following imaging procedures do not require the use of ionizing radiation to produce an image?1.Ultrasound2.Computed axial tomography3.MRI A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BBoth ultrasound and magnetic resonance imaging do not require the use of ionizing radiation to produce an image. Computed axial tomography does require ionizing radiation to produce an image. Ultrasound requires the use of high-frequency sound waves to produce images of soft tissue structures and certain blood vessels within the body. Magnetic resonance imaging relies on the use of a very powerful magnet and specially designed coils that are sensitive to radio-wave signals to produce the image. (Torres et al, p 362)

Electronic imaging terms used to indicate the intensity of radiation reaching the IR includeexposure indexsensitivity (S) numberfield of view (FOV) A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BComputed radiography (CR) offers wide latitude and automatic optimization of the radiologic image. When AEC is not used, CR can compensate for about 80% underexposure and 500% overexposure. This can be an important advantage in trauma and mobile radiography. The radiographer still must be vigilant in patient dose considerations—overexposure, though correctable, results in increased patient dose; underexposure results in decreased image quality owing to increased image noise. CR systems provide an exposure indicator: an S (sensitivity) number, exposure index EI, or other relative exposure index depending on the manufacturer used. The manufacturer usually provides a chart identifying the acceptable range the exposure indicator numbers should be within for various examination types. For example, a high S number often is related to underexposure, whereas a high EI number is related to overexposure. Field of view (FOV) refers to the anatomic area being visualized. (Bontrager and Lampignano, 6th ed., p. 52)

An exposed image plate will retain its original image quality for about A 2 hours B 8 hours C 24 hours D 48 hours

The Correct Answer is: BComputed radiography image plates (IP) have a protective function (for the PSP within) and can be used in the Bucky tray or directly under the anatomic part; they need not be light-tight because the PSP is not light sensitive. The IP has a thin lead-foil backing to absorb backscatter. Inside the IP is the photostimulable phosphor (PSP) storage plate. This PSP within the IP has a layer of europium-activated barium fluorohalide that serves as the IR as it is exposed in the traditional manner and receives the latent image. The PSP can store the latent image for several hours; after about 8 hours, noticeable image fading will occur. (Carlton and Adler, 4th ed., p. 358)

Which of the following mobile radiography applications enables the radiographer to view the radiographic image before leaving the patient? A Fixed digital units of any type B Tethered or wireless flat-panel digital mobile units C Portable units using conventional radiographic film D Battery operated conventional radiography mobile units

The Correct Answer is: BDetectors in mobile digital units may use either tethered or wireless flat-panels (B), which allows the radiographer to view the radiographic image at the patient's bedside. An acceptable image may then be sent to a PACS system for physician review. Fixed digital units (A) are found in the radiology department and cannot be used for mobile applications. Portable units using conventional radiographic film (C) requires the radiographer to chemically process the film in a darkroom located in the radiology department. Battery operated conventional radiography mobile units (D) are used with conventional radiographic film and, therefore, the film must be chemically processed in the darkroom located in the radiology department. (Carlton and Adler, 5 th ed., p. 528).

The effect that differential absorption has on radiographic contrast of a high-subject-contrast part can be minimized byusing a compensating filter.using high-kilovoltage exposure factors.increased collimation. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BDifferential absorption refers to the different attenuation, or absorption, properties of adjacent body tissues. Two parts with widely differing absorption characteristics will produce a high radiographic contrast. Frequently, exposure factors that would properly expose one part will severely overexpose or underexpose the neighboring part (as with lungs vs. the thoracic spine). This effect can be minimized by the use of a compensating filter or by the use of high kilovoltage (for more uniform penetration). Increased collimation is important in the control of patient dose and scattered radiation, not differential absorption. (Bushong, 8th ed., pp. 169, 170)

Radiographic contrast is the result of A transmitted electrons B differential absorption C absorbed photons D milliampere-seconds selection

The Correct Answer is: BDifferential absorption refers to the x-ray absorption characteristics of neighboring anatomic structures—determined by the atomic number of the tissue being examined. The radiographic representation of these various tissue density structures is referred to as radiographic contrast; it may be enhanced with high-contrast technical factors, especially using low kilovoltage levels in analog imaging. At low kilovoltage levels, the photoelectric effect predominates. If photons are absorbed, there will be no contrast. The technical factor milliampere-seconds is used to regulate receptor exposure. (Bushong, pp. 181-184)

One reason why only one image is preferred per image plate is: A To ensure that the anatomical part is properly centered to avoid undercutting of the image B To allow the radiologist to split the PACS monitor and display the current image and a prior image side-by-side for comparison C To ensure optimal radiation safety, since only one image is exposed on one image plate D To reduce the chances of grid cutoff artifacts

The Correct Answer is: BExposing only one image on one image plate does not ensure proper centering (A). This is a technical skill required of the radiographer to ensure the anatomical part is properly centered. One reason to collect only one image per IP (B) is the ability of the radiologist to then split the PACS monitor and display the current image and the prior image side by side for comparison. Radiation safety is not optimized by exposing one image on one imaging plate. The required number of projections (exposures) is required, regardless (C). Grid cutoff artifacts can occur with faulty tube/image receptor alignment or improper SID for any radiographic exposure, regardless of the number of projections taken on an image plate (D). (Seeram, 1 st ed., p. 93)

What type of x-ray imaging uses an area beam and a photostimulable phosphor as the IR? A Film radiography B Computed radiography C Digital radiography D Cineradiography

The Correct Answer is: BFilm radiography used an area x-ray beam, but the IR was film emulsion sandwiched between intensifying screens in a cassette. Computed radiography (CR) also uses an area x-ray beam, but the IR is a photostimulable phosphor such as europium-activated barium fluorohalide coated on an image plate. Digital radiography (DR) can use an area x-ray beam detected by a direct-capture solid-state device. DR can also use a fan-shaped x-ray beam. The fan-shaped beam is "read" by a linear array of detectors. (Bushong, 8th ed., pp. 401, 403)

Due to the high sensitivity of digital detectors to low intensity radiation (background, scatter and/or off-focus radiation), there is likely to be scatter and off-focus radiation contributing to the image outside the collimation margins. Since many radiologists find this distracting, the most appropriate radiographer action would be to: A Use film-screen imaging only B Apply a black border to the image before it is printed or sent to PACS C Expose the anatomical parts as is; there is nothing that can be done to improve the presentation of the image(s) due to the inherent sensitivity of the system D Reduce exposure factors by one-half to ensure minimal scatter and off-focus radiation

The Correct Answer is: BFilm-screen radiography has been abandoned in most hospitals and imaging centers. Most of these institutions no longer maintain a darkroom or resources to produce film-screen images (A). Many radiologists find scatter and off-focus radiation distracting when viewing images. The appropriate response to scatter and off-focus exposure outside the collimation margin is to apply a black border to the image before it is printed or sent to PACS (B). Close collimation should be used to minimize scatter radiation (C). The exposure factors must be appropriate for the anatomical part being imaged. Halving the appropriate mAs or kVp (D) will result in image mottle or inadequate penetration of the part, respectively. (Seeram, 1 st ed., p. 94)

The exposure factors of 400 mA, 70 ms, and 78 kV were used to produce a particular receptor exposure. A similar radiograph can be produced using 500 mA, 90 kV, and A 14 ms B 28 ms C 56 ms D 70 ms

The Correct Answer is: BFirst, evaluate the change(s): The kilovoltage was increased by 15% (78 + 15% = 90). A 15% increase in kilovoltage will double the receptor exposure; therefore, it is necessary to use half the original milliampere-seconds value to maintain the original receptor exposure. The original milliampere-seconds value was 28 mAs (400 mA × 0.07 second [70 ms] 28 mAs), so we now need 14 mAs, using 500 mA. Because mA × s mAs: (Fauber, pp. 55, 59-60)

In amorphous selenium flat-panel detectors, the term amorphous refers to a A crystalline material having typical crystalline structure. B crystalline material lacking typical crystalline structure. C toxic crystalline material. D homogeneous crystalline material.

The Correct Answer is: BFlat-panel detectors used in DR are often made of an amorphous selenium (a-Se)-coated thin-film transistor (TFT) array. They function to convert the x-ray energy (emerging from the radiographed part) into an electrical signal. The TFT capacitors send the electrical signal to the analog-to-digital converter (ADC) to be changed to a digital signal. Amorphous selenium refers to a crystalline material (selenium) that lacks its crystalline structure. Amorphous selenium or silicon is used to produce the direct-conversion flat-panel detectors used in DR. (Bushong, 8th ed., p. 404)

The continued emission of light by a phosphor after the activating source has ceased is termed A fluorescence B phosphorescence C image intensification D quantum mottle

The Correct Answer is: BFluorescence occurs when an intensifying screen absorbs x-ray photon energy, emits light, and then ceases to emit light as soon as the energizing source ceases. Phosphorescence occurs when an intensifying screen absorbs x-ray photon energy, emits light, and continues to emit light for a short time after the energizing source ceases. Quantum mottle is the freckle-like appearance on some radiographs made using a very fast imaging system. The brightness of a fluoroscopic image is amplified through image intensification. (Bushong, 8th ed., p. 221)

The factors that impact spatial resolution include1.Focal spot size2.Type of rectification3.SID A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BFocal spot size affects spatial resolution by its effect on focal spot blur: The larger the focal spot size, the greater the blur produced. Spatial resolution is significantly affected by distance changes because of their effect on magnification. As SID increases, magnification decreases and spatial resolution increases. The method of rectification has no controlling effect on spatial resolution. Single-phase rectified units produce intermittent radiation at fluctuating voltage, whereas three-phase units produce almost constant potential. Single phase equipment exposures could require longer exposures, possibly resulting in motion unsharpness, though that equipment is seldom used today. (Bushong 10th ed p251)

Which of the following information is necessary to determine the maximum safe kilovoltage using the appropriate x-ray tube rating chart?Milliamperage and exposure timeFocal-spot sizeImaging-system speed A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BGiven the milliamperage and exposure time, a radiographic rating chart enables the radiographer to determine the maximum safe kilovoltage for a particular exposure. Because the heat load an anode will safely accept varies with the size of the focal spot and the type of rectification, these variables must be identified. Each x-ray tube has its own radiographic rating chart. The speed of the imaging system has no impact on the use of a radiographic rating chart. (Selman, 9th ed., p. 145)

Characteristics of low ratio focused grids include the following:1.they have a greater focal range2.they are less efficient in collecting SR3.they can be used inverted A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BGrid ratio compares the height of the lead strip to the distance between the lead strips. Focused grids have their lead strips angled so as to parallel the divergent x-ray beam. The higher the grid ratio, the greater the grid's efficiency in absorbing scattered radiation before it reaches the image receptor—but the more critical the centering and distance specifications. Although higher ratio focused grids absorb more SR they have a narrower focal range (focusing distance) and grid/tube centering becomes much more critical. Focused grids must not be accidentally inverted—to do so would cause the lead strips to be placed exactly in the path of the lead strips (grid cutoff), everywhere but in the center of the grid. (Ballinger & Frank, vol 3, p 235)

Compared to a low ratio grid, a high ratio grid will1.absorb more of the useful beam.2.absorb more scattered radiation.3.allow more centering latitude. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BGrid ratio is defined as the height of the lead strips to the width of the interspace material (see the figure below). The higher the lead strips (or the smaller the distance between the strips), the greater the grid ratio and the greater the percentage of scattered radiation absorbed. However, a grid does absorb some of the useful x-ray beam as well. The higher the lead strips, the more critical the need for accurate centering, as the lead strips will more readily trap photons whose direction do not parallel them. (Shephard, pp. 245, 255)

Of the following groups of technical factors, which will produce the greatest receptor exposure? A 10 mAs, 74 kV, 44-in. SID B 10 mAs, 74 kV, 36-in. SID C 5 mAs, 85 kV, 48-in. SID D 5 mAs, 85 kV, 40-in. SID

The Correct Answer is: BIf (A) and (B) are reduced to 5 mAs for consistency, the kilovoltage will increase to 85 kV in both cases, thereby balancing receptor exposures. Thus, the greatest receptor exposure is determined by the shortest SID (greatest exposure rate). (Shephard, pp. 306-307)

With milliamperage adjusted to produce equal exposures, all the following statements are true except A a single-phase examination done at 10 mAs can be duplicated with three-phase, 12-pulse at 5 mAs. B There is greater patient dose with three-phase equipment than with single-phase equipment. C Three-phase equipment can produce comparable radiographs with less heat unit (HU) buildup. D Three-phase equipment produces lower-contrast radiographs than single-phase equipment.

The Correct Answer is: BIf the same kilovoltage is used with single-phase and three-phase equipment, the three-phase unit will require about 50% fewer milliampere-seconds to produce similar radiographs. Because three-phase equipment has much higher effective voltage than single-phase equipment, the three-phase radiograph will have lower contrast. A lower milliampere-seconds value can be used with three-phase equipment, so heat units are not built up as quickly. When technical factors are adjusted to obtain the same receptor exposure and contrast, there is no difference in patient dose. (Selman, 9th ed., pp. 162-164)

TV camera tubes used in image intensification, such as the Plumbicon and Vidicon, function to A increase the brightness of the input-phosphor image. B transfer the output-phosphor image to the TV monitor. C focus and accelerate electrons toward the output phosphor. D record the output-phosphor image on the IR.

The Correct Answer is: BImage intensification is a process that converts the dim fluoroscopic image into a much brighter image, much like normal daylight. As x-ray photons emerge from the patient and enter the image intensifier, they first encounter the input phosphor, which is generally composed of cesium iodide phosphors. At the input phosphor, x-ray photons are converted to light photons, which, in turn, strike the photocathode. The photocathode is a photoemissive metal (usually antimony and cesium compounds); when struck by light, it emits electrons in proportion to the intensity of the light striking it. The electrons then are directed to the output phosphor via the electrostatic focusing lenses, speeded up in the neck of the tube by the accelerating anode and directed to the output phosphor for further amplification. Most image intensifiers offer brightness gains of 5,000-20,000. From the output phosphor, the image is taken by the TV camera, most often a Plumbicon or Vidicon tube, and transferred to the TV monitor. (Thompson et al., p. 370)

Which of the following is used in digital fluoroscopy, replacing the image intensifier's television camera tube? A Solid-state diode B Charge-coupled device C Photostimulable phosphor D Vidicon

The Correct Answer is: BIn digital fluoroscopy (DF), the image-intensifier output screen image is coupled via a charge-coupled device (CCD) for viewing on a display monitor. A CCD converts visible light to an electrical charge that is then sent to the analog-to-digital converter (ADC) for processing. When output screen light strikes the CCD cathode, a proportional number of electrons are released by the cathode and stored as digital values by the CCD. The CCD's rapid discharge time virtually eliminates image lag and is particularly useful in high-speed imaging procedures such as cardiac catheterizations. CCD cameras have replaced analog cameras (such as the Vidicon and Plumbicon) in new fluoroscopic equipment. CCDs are more sensitive to the light emitted by the output phosphor (than the analog cameras) and are associated with less "noise." DF eliminates the need for cassette-loaded spot films and/or 100-mm spot films. DF photo-spot images, which are simply still-frame images, need no chemical processing, require less patient dose, and offer postprocessing capability. DF also offers "road-mapping" capability. "Road-mapping" is a technique useful in procedures involving guidewire/catheter placement. During the fluoroscopic examination, the most recent fluoroscopic image is stored on the monitor, thereby reducing the need for continuous x-ray exposure. This technique can offer significant reductions in patient and personnel radiation exposure. (Bushong, 8th ed., p. 410)

What pixel size has a 1024 × 1024 matrix with a 35-cm FOV? A 30 mm B 0.35 mm C 0.15 mm D 0.03 mm

The Correct Answer is: BIn digital imaging, pixel size is determined by dividing the FOV by the matrix. In this case, the FOV is 35 cm; since the answer is expressed in millimeters, first change 35 cm to 350 mm. Then 350 divided by 1024 equals 0.35 mm. The FOV and matrix size are independent of one another, that is, either can be changed and the other will remain unaffected. However, pixel size is affected by changes in either the FOV or matrix size. For example, if the matrix size is increased, pixel size decreases. If FOV increases, pixel size increases. Pixel size is inversely related to resolution. As pixel size increases, resolution decreases. (Fosbinder & Kelsey, p 285)

Which of the following x-ray circuit devices operate(s) on the principle of mutual induction?1.High-voltage transformer2.Filament transformer3.Autotransformer A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: BIn mutual induction, two coils are in close proximity, and a current is supplied to one of the coils. As the magnetic field associated with every electric current expands and "grows up" around the first coil, it interacts with and "cuts" the turns of the second coil. This interaction, motion between magnetic field and coil (conductor), induces an emf in the second coil. This is mutual induction, the production of a current in a neighboring circuit. Transformers such as the high-voltage transformer and the filament (step-down) transformer operate on the principle of mutual induction. The autotransformer operates on the principle of self-induction. Both the transformer and the autotransformer require the use of alternating current. (Bushong, 8th ed., p. 99)

Continuous rotation of the CT x-ray tube and detector array, with simultaneous movement of the CT couch, has been accomplished through implementation of A additional cables. B slip rings. C multiple rows of detectors. D electron beam CT.

The Correct Answer is: BIn the 1990s, the implementation of slip ring technology allowed continuous rotation of the x-ray tube (through elimination of cables) and simultaneous couch movement. Sixth-generation CT scanning is termed helical (or spiral) CT—permitting acquisition of volume multislice scanning. Today's helical multislice scanners, employing thousands of detectors (up to 60+ rows), can obtain uninterrupted data acquisition of 128 "slices" per tube rotation and can perform 3D multiplanar reformation (MPR). Fifth-generation CT is electron beam; ultra high-speed CT is used specifically for cardiac imaging. (Bushong, 9th ed., p. 375; Romans, pp. 50-51)

Greater latitude is available to the radiographer in which of the following circumstances?Using high-kV technical factorsUsing a low-ratio gridUsing low-kV technical factors A 1 only B 1 and 2 only C 2 and 3 only D 3 only

The Correct Answer is: BIn the low-kilovoltage ranges, a difference of just a few kilovolts makes a very noticeable radiographic difference, therefore offering little margin for error/latitude. High-kilovolt technical factors offer much greater margin for error; in the high-kV ranges, an error of a few kV makes little/no difference in the resulting image. Lower-ratio grids offer more tube-centering latitude than high-ratio grids. (Saia, 4th ed., p. 360)

Magnification fluoroscopy is only possible with: A Decreased patient dosage B Multifield image intensifiers C Decreased fluoroscopic time D Increased efficiency of X-ray production

The Correct Answer is: BMagnification fluoroscopy requires that a multifield image intensifier (B) be used to allow reduction of the X-ray field size to the input phosphor area. Smaller input phosphor field sizes produce magnified images of the anatomical areas being evaluated at the output phosphor. Magnification mode in fluoroscopy actually increases patient dosage (A), as more radiation is necessary to produce the brightness levels needed to view the images. The magnification mode should therefore be used only when necessary to enhance diagnostic interpretation of small anatomical areas in question (e.g., the gallbladder or duodenal bulb). Fluoroscopy time should be limited to that which is absolutely necessary in order to ensure proper practice of ALARA. However, the time needed to evaluate the anatomical areas in question is not limited to a certain time. Magnification fluoroscopy neither increases or decreases fluoroscopic evaluation time (C). X-ray production efficiency is a function of the generator and X-ray tube (D) providing the necessary X-ray energy to produce the fluoroscopic image. Magnification fluoroscopy, therefore, does not alter the efficiency of X-ray production. (Seeram, p. 134).

A 5-in. object to be radiographed at a 44-in. SID lies 6 in. from the IR. What will be the image width? A 5.1 in. B 5.8 in. C 6.1 in. D 6.7 in.

The Correct Answer is: BMagnification is part of every radiographic image. Anatomic parts within the body are at various distances from the IR and, therefore, have various degrees of magnification. The formula used to determine the amount of image magnification is Substituting known values: Thus, x = 5.78-in. image width. (Bushong, 11th ed., p. 174)

A 3-inch object to be radiographed at a 36-inch SID lies 4 inches from the image recorder. What will be the image width? A 2.6 inches B 3.3 inches C 26 inches D 33 inches

The Correct Answer is: BMagnification is part of every radiographic image. Anatomic parts within the body are at various distances from the image recorder and therefore have various degrees of magnification. The formula used to determine the amount of image magnification is: Substituting known values: x = 3.37 inches image width (Bushong, p 284)All the following statements regarding mobile radiographic equipment are true except A the exposure cord must permit the operator to stand at least 6 ft from the patient, x-ray tube, and useful beam B exposure switches must be the two-stage type C a lead apron should be carried with the unit and worn by the radiographer during exposure D the radiographer must alert individuals in the area before making the exposure

The use of which of the following is (are) essential in magnification radiography?High-ratio gridFractional focal spotDirect exposure technique A 1 only B 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: BMagnification radiography is used to enlarge details to a more perceptible degree. Hairline fractures and minute blood vessels are candidates for magnification radiography. The problem of magnification unsharpness is overcome by using a fractional focal spot; larger focal-spot sizes will produce excessive blurring unsharpness. Grids are usually unnecessary in magnification radiography because of the air-gap effect produced by the OID. Direct-exposure technique probably would not be used because of the excessive exposure required. (Selman, 9th ed., pp. 226-228)

IRs/cassettes frequently have a lead-foil layer behind the rear screen that functions to A improve penetration B absorb backscatter C preserve resolution D increase the screen speed

The Correct Answer is: BMany cassettes/IRs have a thin lead-foil layer behind the rear screen to absorb backscattered radiation that is energetic enough to exit the rear screen, strike the metal back, and bounce back to fog the image. When this happens, the IR's metal hinges or straps may be imaged in high-kilovoltage radiography. The lead foil absorbs the backscatter before it can fog the film. (Shephard, pp. 41-42)

Which of the following is (are) classified as rare earth phosphors?Lanthanum oxybromideGadolinium oxysulfideCesium iodide A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BRare earth phosphors have a greater conversion efficiency than do other phosphors. Lanthanum oxybromide is a blue-emitting rare earth phosphor, and gadolinium oxysulfide is a green-emitting rare earth phosphor. Cesium iodide is the phosphor used on the input screen of image intensifiers; it is not a rare earth phosphor. (Shephard, p. 68)

The advantages of capacitor-discharge mobile x-ray equipment includecompact sizelight weighthigh kilovoltage capability A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BMobile x-ray machines are compact and cordless and are either the battery-operated type or the condenser-discharge type. Condenser-discharge mobile x-ray units do not use batteries; this type of mobile unit requires that it be charged before each exposure. A condenser (or capacitor) is a device that stores electrical energy. The stored energy is used to operate the x-ray tube only. Because this machine does not carry many batteries, it is much lighter and does not need a motor to drive or brake it. The major disadvantage of the capacitor/condenser-discharge unit is that as the capacitor discharges its electrical charge the kilovoltage gradually decreases throughout the length of the exposure—therefore limiting tube output and requiring recharging between exposures. (Frank, Long, and Smith, 11th ed., vol. 3, p. 235)

All the following statements regarding mobile radiographic equipment are true except A the exposure cord must permit the operator to stand at least 6 ft from the patient, x-ray tube, and useful beam B exposure switches must be the two-stage type C a lead apron should be carried with the unit and worn by the radiographer during exposure D the radiographer must alert individuals in the area before making the exposure

The Correct Answer is: BNCRP Report No. 102 states that the exposure switch on mobile radiographic units shall be so arranged that the operator can stand at least 2 m (6 ft) from the patient, the x-ray tube, and the useful beam. An appropriately long exposure cord accomplishes this requirement. The fluoroscopic and/or radiographic exposure switch or switches must be of the "dead man" type; that is, the exposure will terminate should the switch be released. A lead apron should be carried with every mobile x-ray unit for the operator to wear during the exposure. Lastly, the radiographer must be certain to alert individuals in the area, enabling unnecessary occupants to move away, before making the exposure. (NCRP Report No. 102, p. 25)

Which of the following groups of analog exposure factors is most likely to produce the longest scale of contrast? A 200 mA, 0.25 second, 70 kVp, 12:1 grid B 500 mA, 0.10 second, 90 kVp, 8:1 grid C 400 mA, 0.125 second, 80 kVp, 12:1 grid D 300 mA, 0.16 second, 70 kVp, 8:1 grid

The Correct Answer is: BOf the given factors, kilovoltage and grid ratio will have a significant effect on the scale of radiographic contrast. The mAs values are almost identical. Because an increased kilovoltage and low-ratio grid combination would allow the greatest amount of scattered radiation to reach the IR, thereby producing more gray tones, B is the best answer. Group D also uses a low-ratio grid, but the kilovoltage is too low to produce as many gray tones as B. (Shepard, p. 308)

Off-focus and scatter radiation outside of the exposure field when using CR can cause: A Narrowing of the histogram B Widening of the histogram C Improper alignment of the exposure field D High contrast

The Correct Answer is: BOff-focus and scatter radiation outside of the exposure field would be detected as additional information and, therefore, would widen the histogram (B), resulting in a processing error. Histogram analysis errors can result in rescaling errors and exposure indicator determination errors. Alignment of the exposure field (C) is set by the radiographer prior to the exposure. Any off-focus and scatter radiation exposure outside of the exposure field will not change this alignment. Scatter radiation decreases image contrast (D). (Seeram, 1 st ed., p. 91).

Disadvantages of moving grids over stationary grids include which of the following?They can prohibit the use of very short exposure times.They increase patient radiation dose.They can cause phantom images when anatomic parts parallel their motion. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BOne generally thinks in terms of moving grids being totally superior to stationary grids because moving grids function to blur the images of the lead strips on the radiographic image. Moving grids do, however, have several disadvantages. First, their complex mechanism is expensive and subject to malfunction. Second, today's sophisticated x-ray equipment makes possible the use of extremely short exposures, a valuable feature whenever motion may be a problem (as in pediatric radiography). However, grid mechanisms frequently are not able to oscillate rapidly enough for the short exposure times, and as a result, the grid motion is "stopped," and the lead strips are imaged. Third, patient dose is increased with moving grids. Since the central ray is not always centered to the grid because it is in motion, lateral decentering occurs (resulting in diminished density), and consequently, an increase in exposure is needed to compensate (either manually or via AEC). (Shephard, p. 249)

The best way to control voluntary motion is A immobilization of the part. B careful explanation of the procedure. C short exposure time. D physical restraint.

The Correct Answer is: BPatients who are able to cooperate are usually able to control voluntary motion if they are provided with an adequate explanation of the procedure. Once patients understand what is needed, most will cooperate to the best of their ability (by suspending respiration and holding still for the exposure). Certain body functions and responses, such as heart action, peristalsis, pain, and muscle spasm, cause involuntary motion that is uncontrollable by the patient. The best and only way to control involuntary motion is by always selecting the shortest possible exposure time. Involuntary motion may also be minimized by careful explanation, immobilization, and (as a last resort and only in certain cases) restraint. (Ballinger & Frank, vol 1, pp 12-13)

Which of the following can have an effect on radiographic contrast?1.Beam restriction2.Grids3.Focal spot size A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BRadiographic contrast the result of x-ray absorption difference between tissue densities resulting in scale of grays. Since the function of grids is to collect scattered radiation, they serve to shorten the scale of contrast. Beam restrictors function to limit the x-ray field size, thereby reducing the production of scattered radiation and shortening the scale of contrast. Focal spot size is one of the geometric factors affecting spatial resolution; it has no effect on the scale of contrast or receptor exposure. It is the function of radiographic contrast to make details visible.

Phosphors classified as rare earth includelanthanum oxybromide.gadolinium oxysulfide.cesium iodide. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BRare earth phosphors have a greater conversion efficiency than do other phosphors. Lanthanum oxybromide is a blue-emitting phosphor, and gadolinium oxysulfide is a green-emitting phosphor. Cesium iodide is the phosphor used on the input screen of image intensifiers; it is not a rare earth phosphor. (Shephard, p. 66)

Which of the following terms is used to express spatial resolution? A Kiloelectronvolts (keV) B Modulation transfer function (MTF) C Relative speed D Latitude

The Correct Answer is: BResolution describes how closely fine details may be associated and still be recognized as separate details before seeming to blend into each other and appear as one. The degree of resolution transferred to the image receptor is a function of the resolving power of each of the system components and can be expressed in line pairs per millimeter (lp/mm), line-spread function (LSP), or modulation transfer function (MTF). Line pairs per millimeter can be measured using a resolution test pattern; a number of resolution test tools are available. LSP is measured using a 10-μm x-ray beam; MTF measures the amount of information lost between the object and the IR. (Carlton and Adler, 4th ed., p. 334)

Which of the following is most likely to produce a high-quality image? A Small image matrix B High signal-to-noise ratio (SNR) C Large pixel size D Low resolution

The Correct Answer is: BSNR can refer to home television images, magnetic resonance images (MRIs), ultrasound images, x-ray images, and so on. Noise interferes with visualization of anatomic image details, for example, scattered radiation fog, graininess from quantum mottle, and so on. The actual signal can be from x-rays, sound waves, and so on. The signal is desirable, the noise is not, therefore, a higher SNR produces a higher-quality image. Low SNR severely impairs contrast resolution. (Bushong, 8th ed., p. 412)

The device used to change alternating current to unidirectional current is a A capacitor B solid-state diode C transformer D generator

The Correct Answer is: BSome x-ray circuit devices, such as the transformer and autotransformer, will operate only on AC. The efficient operation of the x-ray tube, however, requires the use of unidirectional current, so current must be rectified before it gets to the x-ray tube. The process of full-wave rectification changes the negative half-cycle to a useful positive half-cycle. An x-ray circuit rectification system is located between the secondary coil of the high-voltage transformer and the x-ray tube. Rectifiers are solid-state diodes made of semiconductive materials such as silicon, selenium, or germanium that conduct electricity in only one direction. Thus, a series of rectifiers placed between the transformer and x-ray tube function to change AC to a more useful unidirectional current. (Bushong, 8th ed., p. 119)

To ensure proper operation of the digital image display monitor, all of the following are important in order to develop a quality control (QC) program, except: A Routine quality control tests by the QC technologist B Disassembly and cleaning of the internal monitor control devices by the QC technologist C Periodic review of the QC program by a qualified medical physicist D Annual and post-repair medical physics performance evaluations

The Correct Answer is: BThe QC technologist should never disassemble the monitor to expose its control devices for cleaning (B). Any malfunctions and undesirable results should be reported to the medical physicist and/or manufacturer. Routine quality control tests by the QC technologist (A) ensure that the display monitor accurately reveals diagnostic images. Periodic review of the integrity of the QC program should be evaluated by a qualified medical physicist (C) who may either make recommendations for revisions or approve the existing program. Annual and post-repair medical physics performance evaluations (D) should be performed by a qualified medical physicist. (Bushong, 9 th ed., p. 485).

A test pattern, such as the TG 18-CT test pattern, is used to qualitatively evaluate A Radiographic film-screen contact B The luminance response of a digital display monitor C The X-ray exposure field alignment D The exposure rate in an X-ray beam

The Correct Answer is: BThe TG 18-CT test pattern is used to qualitatively evaluate the luminance response of a digital display monitor (B). Luminance response refers to the comparison of input to the display device and the actual displayed luminance value. The displayed luminance value varies between L min and L max and is impacted by ambient light as well (L amb ). The pattern in TG 18-CT testing device includes 16 low-contrast targets that should be visible on the display. The test pattern should be viewed from a distance of approximately 30 cm. One frequent observation is inability to visualize one or more shades in the darker regions. Radiographic film-screen (A) contact is evaluated by exposing a wire mesh screen on top of a conventional radiographic cassette holding an unexposed film. Any blurred areas of the wire mesh would indicate that there is poor film-screen contact in that particular area. The X-ray exposure field alignment (C) can be tested by using a square or rectangular leaded test pattern. An exposure is made with this test pattern device positioned on top of the receptor with collimator light field adjusted to match the size of the test pattern. The resultant image is then inspected to determine if the X-ray exposure field is congruent with the borders of the test pattern. The exposure rate (D) in an X-ray beam is measured with a calibrated radiation dosimeter that contains an ionization chamber or photodiode. (Bushong, 9 th ed., p. 482)

Which of the following is a type of television camera tube that converts a visible image on the output phosphor of the image intensifier into an electronic signal? A Ionization chamber B Vidicon C Charge-coupled device D Cathode ray tube

The Correct Answer is: BThe Vidicon (B) is a television camera tube used in television fluoroscopy. It is a cylindrical glass vacuum tube that contains a cathode and anode. The cathode (also called the electron gun) is responsible for thermionic emission of electrons, which are accelerated through an electrostatic field that focuses them on a target assembly (anode). The target assembly consists of three layers: the window (thin part of the glass envelope), a metal or graphite signal plate, and a photoconductive layer called the target. When visible light from the output phosphor of the image intensifier tube strikes the anode target assembly, the photoconductive layer of the target conducts electrons. Therefore, in the presence of light, the electrons emitted from the cathode are able to pass through the target to the signal plate and, from there, out of the tube as the video signal. Ionization chambers (A) are found in an automatic exposure control (AEC) system. Air in these chambers is ionized in proportion to the number of X-rays interacting with the air and an electrical signal is generated. This signal, once it reaches a specific magnitude, initiates an exposure timer in the X-ray circuit, which terminates the exposure according to the radiographer's preselected density setting. A major change from conventional television fluoroscopy to digital fluoroscopy is the use of a charge-coupled device (CCD) (C) in lieu of a television camera tube. The CCD is mounted directly to the output phosphor of the image intensifier tube and is coupled through fiber optics or a lens system to receive the light from the output phosphor. The cathode ray tube (CRT) (D) is a television monitor tube that is viewed by the operator during fluoroscopic evaluation of the anatomy of interest. (Bushong, 9 th ed., pp. 353-354).

For which of the following examinations might the use of a grid not be necessary in an adult patient? A Hip B Knee C Abdomen D Lumbar spine

The Correct Answer is: BThe abdomen is a thick structure that contains many structures of similar tissue density, and thus it requires increased exposure and a grid to absorb scattered radiation. The lumbar spine and hip are also dense structures requiring increased exposure and use of a grid. The knee, however, is frequently small enough to be imaged without a grid. The general rule is that structures measuring more than 10 cm should be imaged with a grid. (Bontrager and Lampignano, 6th ed., p. 45)

The exposure factors used for a particular nongrid x-ray image were 300 mA, 4 ms, and 90 kV. Another image, using an 8:1 grid, is requested. Which of the following groups of factors is most appropriate? A 400 mA, 3 ms, 110 kV B 400 mA, 12 ms, 90 kV C 300 mA, 8 ms, 100 kV D 200 mA, 240 ms, 90 kV

The Correct Answer is: BThe addition of a grid will help to clean up the scattered radiation produced by higher kilovoltage, but the grid requires an adjustment of milliampere-seconds. According to the grid conversion factors listed here, the addition of an 8:1 grid requires that the original milliampere-seconds be multiplied by a factor of 4: The original milliampere-seconds value is 1.2. The ideal adjustment, therefore, requires a 4.8 mAs at 90 kV. Although 2.4 mAs with 100 kV (choice C), or 1.2 mAs with 110 kV (choice A), also might seem workable, an increase in kilovoltage would further compromise contrast, nullifying the effect of the grid. Additionally, kilovoltage exceeding 100 should not be used with an 8:1 grid. (Shephard, pp. 247-248)

Cassette-front material can be made of which of the following?Carbon fiberMagnesiumLead A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: BThe cassette-IR front material must not attenuate the remnant beam yet must be sturdy enough to withstand daily use. Bakelite has long been used as the material for tabletops and IR fronts, but now it has been replaced largely by magnesium and carbon fiber. Lead would not be a suitable material because it would absorb the remnant beam, and no image would be formed. (Shephard, p. 41)

If the distance from the focal spot to the center of the collimator's mirror is 6 in., what distance should the illuminator's light bulb be from the center of the mirror? A 3 in. B 6 in. C 9 in. D 12 in.

The Correct Answer is: BThe collimator assembly includes a series of lead shutters, a mirror, and a light bulb (Figure 5-14). The mirror and light bulb function to project the size, location, and center of the irradiated field. The bulb's emitted beam of light is deflected by a mirror placed at an angle of 45 degrees in the path of the light beam. In order for the projected light beam to be the same size as the x-ray beam, the focal spot and the light bulb must be exactly the same distance from the center of the mirror. (Carlton and Adler, 4th ed., p. 232)

The processes of "leveling and/or windowing" of digital images can participate in adjusting image A spatial resolution B contrast C pixel size D matrix size

The Correct Answer is: BThe digital images' scale of contrast, or contrast resolution, can be changed electronically through leveling and windowing of the image. It is often stated simply that window level controls density and window width controls contrast. However, the level control specifically determines the central or middensity of the scale of contrast, whereas the window control determines the total number of grays (to the right and left of the central/middensity). Matrix and pixel sizes are related to (spatial) resolution of digital images. (Fosbinder and Kelsey, p. 289; Bushong 9th ed p. 328)

If 300 mA has been selected for a particular exposure, what exposure time should be selected to produce 18 mAs? A 40 ms B 60 ms C 400 ms D 600 ms

The Correct Answer is: BThe exposure factor that regulates receptor exposure is milliampere-seconds (mAs). The equation used to determine mAs is mA × s = mAs. Substituting known factors: (Selman, 9th ed., p. 214)

An important feature of the pixel in a flat-panel TFT digital detector active matrix array is the: A Nyquist frequency B Fill factor C Image lag D Modulation transfer function

The Correct Answer is: BThe fill factor (B) is defined as the ratio of the sensing area of the pixel to the area of the pixel itself. The sensing area of the pixel receives the data from the layer above it, which captures X-rays that are subsequently converted to light (indirect flat-panel detectors) or electrical charges (direct flat-panel detectors). The Nyquist frequency (A) is 1/2X the pixel pitch (mm) and is equivalent to the spatial resolution in digital systems. A pixel contains generally three components: the TFT, the capacitor, and the sensing area. Image lag (C) is an undesirable phenomenon that refers to the persistence of the image, that is, a charge is still being produced in a digital detector after the radiation beam from the X-ray tube has been turned off. The modulation transfer function (D) is a mathematical function that measures the ability of the digital detector to transfer its spatial resolution characteristics to the image. (Seeram, 1 st ed., p. 111)

If a duration of 0.05 second was selected for a particular exposure, what milliamperage would be necessary to produce 30 mAs? A 900 B 600 C 500 D 300

The Correct Answer is: BThe formula for mAs is mA × s = mAs. Substituting known values: (Selman, 9th ed., p. 214)

In a PA projection of the chest being used for cardiac evaluation, the heart measures 15.2 cm between its widest points. If the magnification factor is known to be 1.3, what is the actual diameter of the heart? A 9.7 cm B 11.7 cm C 19.7 cm D 20.3 cm

The Correct Answer is: BThe formula for magnification factor is MF = image size/object size. In the stated problem, the anatomic measurement is 15.2 cm, and the magnification factor is known to be 1.3. Substituting the known factors in the appropriate equation, x = 11.69 cm (actual anatomic size) (Fauber, pp 90-92)

For which of the following examinations can the anode heel effect be an important consideration?Lateral thoracic spineAP femurRight anterior oblique (RAO) sternum A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: BThe heel effect is characterized by a variation in beam intensity that increases gradually from anode to cathode. This can be effectively put to use when performing radiographic examinations of large body parts with uneven tissue density. For example, the AP thoracic spine is thicker caudally than cranially, so the thicker portion is best placed under the cathode. However, in the lateral projection of the thoracic spine, the upper portion is thicker because of superimposed shoulders, and therefore, that portion is best placed under the cathode end of the beam. The femur is also uneven in tissue density, particularly in the AP position, and can benefit from use of the heel effect. However, the sternum and its surrounding anatomy are fairly uniform in thickness and would not benefit from use of the anode heel effect. The anode heel effect is most pronounced when using large IRs at short SIDs and with an anode having a steep (small) target angle.

f the primary coil of the high-voltage transformer is supplied by 220 V and has 200 turns, and the secondary coil has 100,000 turns, what is the voltage induced in the secondary coil? A 40 kV B 110 kV C 40 V D 110 V

The Correct Answer is: BThe high-voltage, or step-up, transformer functions to increase voltage to the necessary kilovoltage. It decreases the amperage to milliamperage. The amount of increase or decrease depends on the transformer ratio, that is, the ratio of the number of turns in the primary coil to the number of turns in the secondary coil. The transformer law is as follows: To determine secondary V, To determine secondary I: Substituting known values, (Selman, pp 84-85)

An advantage of coupling the image intensifier to the TV camera or CCD via a fiber-optic coupling device is itscompact sizedurabilityability to accommodate auxilary imaging devices A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: BThe image intensifier can be coupled to the TV camera via a fiber-optic bundle or via a lens coupling device. The fiber-optic connection offers less fragility, more compactness, and ease of maneuverability. The objective lens can use the, now infrequently used, auxiliary imaging devices such as a cine camera or spot-film camera. (Bushong, 8th ed., p. 366)

How are mAs and receptor exposure related in the process of image formation? A mAs and receptor exposure are inversely proportional B mAs and receptor exposure are directly proportional C mAs and receptor exposure are related to image unsharpness D mAs and receptor exposure are unrelated

The Correct Answer is: BThe milliampere-seconds value regulates the number of x-ray photons produced at the target and thus regulates receptor exposure. If it is desired to double the receptor exposure, one simply doubles the milliampere-seconds; therefore, milliampere-seconds and receptor exposure are directly proportional. (Selman, 9th ed., p. 214)

A lateral (analog) radiograph of the lumbar spine was made using 200 mA, 1/2 second exposure, and 90 kV. If the exposure factors were changed to 200 mA, 0.25 second, and 104 kV, there would be an obvious change in which of the following?Receptor exposureScale of grays/contrastDistortion A 1 only B 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BThe original milliampere-seconds value (regulating receptor exposure) was 100. The original kilovoltage (impacting contrast) was 90. The milliampere-seconds value was cut in half, to 50. The kilovoltage was increased (by 15%) to compensate for the receptor exposure loss and thereby increase the scale of grays. (Carlton and Adler 6th ed p 358)

The output phosphor can be coupled with the Vidicon TV camera or charge-coupled device (CCD) viafiber optics.an image distributor or lens.closed-circuit TV. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BThe output phosphor of the image intensifier displays the brighter, minified, and inverted image. From the output phosphor, the light image is conveyed to its destination by some kind of image distributor—either a series of lenses and a mirror or via fiber optics. Fiber optics is often the method of choice where equipment size is of concern (e.g., mobile equipment). The image distributor, that is, the lens or fiber optics, then sends the majority of light to the TV monitor for direct viewing and the remaining light (about 10%) to the IR (e.g., photospot camera). (Bushong, 8th ed., p. 366)

Exposure rate increases with an increase in1.mA.2.kVp.3.SID. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BThe quantity of x-ray photons produced at the target is the function of mAs. The quality (wavelength, penetration, energy) of x-ray photons produced at the target is the function of kVp. The kVp also has an effect on exposure rate, because an increase in kVp will increase the number of high-energy x-ray photons produced at the target. Exposure rate decreases with an increase in SID. (Selman, p 117)

Which of the following combinations can potentially offer the greatest heat-loading capability? A 17-degree target angle, 0.9-mm actual focal spot B 12-degree target angle, 1.2-mm actual focal spot C 17-degree target angle, 0.6-mm actual focal spot D 10-degree target angle, 0.6-mm actual focal spot

The Correct Answer is: BThe smaller the focal spot, the more limited the anode is with respect to the quantity of heat it can safely accept. Therefore, group (B) offers the greatest heat-loading potential, with a 1.2 mm focal spot, even though the anode bevel is less. It must be remembered, however, that a steep target angle increases the heel effect, and IR coverage may be compromised. As target angle decreases, the actual focal spot could be increased while still maintaining a small effective focal spot. (Selman, 9th ed., pp. 145-146)

An exposure was made at 40-in. SID using 5 mAs and 105 kVp with an 8:1 grid. In an effort to improve image contrast, the image is repeated using a 12:1 grid and 90 kVp. Which of the following exposure times will be most appropriate, using 400 mA, to maintain the original receptor exposure? A 0.01 s B 0.03 s C 0.1 s D 0.3 s

The Correct Answer is: BThe use of high kilovoltage with a fairly low-ratio grid will be ineffective in ridding the remnant beam of scattered radiation. To improve contrast in this example, it has been decided to decrease the kilovoltage by 15%, thus making it necessary to increase the milliampere-seconds from 5 mAs to 10 mAs. Because an increase in the grid ratio to 12:1 is also desired, another change in milliampere-seconds will be required (remember, 10 mAs is now the old mAs): Thus, x = 12.5 mAs at 90 kVp. Now determine the exposure time required with 400 mA to produce 12.5 mAs: (Selman, 9th ed., p. 214)

The direction of electron travel in the x-ray tube is A filament to cathode B cathode to anode C anode to focus D anode to cathode

The Correct Answer is: BThe x-ray tube is a diode tube; that is, it has two electrodes—a negative and a positive. The cathode assembly is the negative terminal of the x-ray tube, and the anode is the positive terminal. Electrons are released by the cathode filament (thermionic emission) as it is heated to incandescence. When kilovoltage is applied, the electrons are driven across to the anode's focal spot. Upon sudden deceleration of electrons at the anode surface, x-rays are produced. Hence, electrons travel from cathode to anode within the x-ray tube. (Bushong, 9th ed., pp. 122-125)

A satisfactory radiograph of the abdomen was made at a 38-in. SID using 400 mA, 60-ms exposure, and 80 kV. If the distance is changed to 42 in., what new exposure time would be required? A 25 ms B 50 ms C 73 ms D 93 ms

The Correct Answer is: CAccording to the exposure maintenance formula, as the distance between the radiation source and the IR decreases, the exposure rate increases. Therefore, a decrease in technical factors is indicated, and the exposure-maintenance formula is used to determine new milliampere-seconds values when changing distance: Thus, x = 29.31 mAs at 42-in. SID. Then, to determine the new exposure time (mA × s = mAs), Thus, x = 0.073 second (73 ms) at 400 mA. (Selman, 9th ed., p. 214)

Fluorescent light is collected from the image intensifier output phosphor and converted to an electronic video signal by the1.TV camera tube.2.CCD.3.coaxial cable. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BThere are two devices that can take the fluorescent image from the image intensifier output phosphor and convert it to an electronic video signal: a TV camera tube and a CCD. A TV camera tube is found on older fluoroscopic equipment. Today's newer fluoroscopic equipment uses a CCD (charge-coupled device) to accomplish this task. The CCD is a solid-state device that offers much better spatial resolution and less image noise. A coaxial cable follows the TV camera or CCD in the fluoroscopic chain. It is used to connect the TV camera or CCD to the TV monitor. (Seeram, p 98)

The exposure factors of 400 mA, 17 ms, and 82 kV produce a milliampere-seconds value of A 2.35 B 6.8 C 23.5 D 68

The Correct Answer is: BTo calculate milliampere-seconds, multiply milliamperage times exposure time. In this case, 400 mA × 0.017 second (17 ms) = 6.8 mAs. Careful attention to proper decimal placement will help to avoid basic math errors. (Shephard, p. 170)

Using a short (25-30 in.) SID with a large (14 × 17 in.) IR is likely to A increase the scale of contrast B increase the anode heel effect C cause malfunction of the AEC D cause premature termination of the exposure

The Correct Answer is: BUse of a short SID with a large-size IR (and also with anode angles of 10 degrees or less) causes the anode heel effect to be much more apparent. The x-ray beam needs to diverge more to cover a large-size IR, and it needs to diverge even more for coverage as the SID decreases. The x-ray beam has no problem diverging toward the cathode end of the beam, but as it tries to diverge toward the anode end of the beam, it is eventually stopped by the anode (x-ray photons are absorbed by the anode). This causes a decrease in beam intensity at the anode end of the beam and is characteristic of the anode heel effect. (Carlton and Adler, 4th ed., p. 407)

A backup timer for the AEC serves toprotect the patient from overexposureprotect the x-ray tube from excessive heatadjust image contrast A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BWhen an AEC is installed in an x-ray circuit, it is calibrated to deliver the most appropriate receptor as required by the radiologist. Once the part being radiographed has been exposed to produce the correct receptor exposure, the AEC automatically terminates the exposure. The manual timer should be used as a backup timer; in case the AEC fails to terminate the exposure, the backup timer would protect the patient from overexposure and the x-ray tube from excessive heat load. Image contras in CR/DR is determined by computer software.

Which of the following methods can be used effectively to decrease differential absorption, providing a longer scale of contrast in the diagnostic range?Using high peak kilovoltage and low milliampere-seconds factorsUsing compensating filtrationUsing factors that increase the photoelectric effect A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BWhen differences in absorption characteristics are decreased, body tissues absorb radiation more uniformly, and as a result, more grays are seen on the radiographic image. A longer scale of contrast is produced. High-kilovoltage and low-milliamperage factors achieve this. Compensating filtration is also used to "even out" densities in uneven anatomic parts, such as the thoracic spine. The photoelectric effect is the interaction between x-ray photons and matter that occurs at low-peak kilovoltage levels—levels that tend to produce short-scale contrast. (Shephard, pp. 193, 197, 199)

What information must be included on an x-ray image for it to be considered as legitimate legal evidence?Name of facility where exam performedExamination dateDate of birth A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: BX-ray images are often subpoenaed as court evidence in cases of medical litigation. In order to be considered as legitimate legal evidence, each x-ray image must contain certain essential and specific patient information. Essential information that must be included on each image is patient identification, the identity of the facility where the x-ray study was performed, the date that the study was performed, and a right- or left-side marker. Other useful information that may be included, but that is not considered essential, is additional patient demographics such as their date of birth, the identity of the referring physician, the time of day that the study was performed, and the identity/initials of the radiographer performing the examination.

Conditions that contribute to x-ray tube damage includelengthy anode rotationexposures to a cold anodelow-milliampere-seconds/high- kilovoltage exposure factors A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: BX-ray tube life may be extended by using exposure factors that produce a minimum of heat, that is, a lower milliampere-seconds and higher kilovoltage combination, whenever possible. When the rotor is activated, the filament current is increased to produce the required electron source (thermionic emission). Prolonged rotor time, then, can lead to shortened filament life as a result of early vaporization. Large exposures to a cold anode will heat the anode surface, and the big temperature difference can cause cracking of the anode. This can be avoided by proper warming of the anode prior to use, thereby allowing sufficient dispersion of heat through the anode. (Selman, 9th ed., pp. 143-145)

The type of x-ray tube designed to turn on and off rapidly, providing multiple short, precise exposures, is A high speed B grid-controlled C diode D electrode

The Correct Answer is: BX-ray tubes are diode tubes; that is, they have two electrodes—a positive electrode called the anode and a negative electrode called the cathode. The cathode filament is heated to incandescence and releases electrons—a process called thermionic emission. During the exposure, these electrons are driven by thousands of volts toward the anode, where they are suddenly decelerated. That deceleration is what produces x-rays. Some x-ray tubes, such as those used in fluoroscopy and in capacitor-discharge mobile units, are required to make short, precise—sometimes multiple—exposures. This need is met by using a grid-controlled tube. A grid-controlled tube uses the nickel focusing cup as the switch, permitting very precise control of the tube current (flow of electrons between cathode and anode). (Bushong, 8th ed., p. 132)

If 85 kV and 20 mAs were used for a particular abdominal exposure with single-phase equipment, what mAs would be required to produce a similar radiograph with 3-phase, 12-pulse equipment. A 40 B 25 C 20 D 10

The Correct Answer is: DSingle-phase radiographic equipment is much less efficient than three-phase equipment because it has a 100% voltage ripple. With three-phase equipment, voltage never drops to zero, and x-ray intensity is significantly greater. To produce similar receptor exposure with three phase equipment, the original mAs would be cut in half; thus, 10 mAs should be used. (Bushong 11th ed p 120)

Which of the following is most likely to result from the introduction of a grid with appropriate technical factor change? A Increased patient dose and increased scattered radiation fog B Decreased patient dose and decreased scattered radiation fog C Increased patient dose and decreased scattered radiation fog D Decreased patient dose and increased scattered radiation fog

The Correct Answer is: C A grid is a device interposed between the patient and image receptor that absorbs a large percentage of scattered radiation before it reaches the image receptor. It is constructed of alternating strips of lead foil and radiolucent filler material. X-ray photons traveling in the same direction as the primary beam pass between the lead strips. X-ray photons, having undergone interactions within the body and deviated in various directions, are absorbed by the lead strips; this is referred to as "clean-up" of scattered radiation. When a grid is introduced, there is a very significant decrease in receptor exposure. To maintain a diagnostic image, the addition of a grid must be accompanied by an appropriately substantial increase in mAs, hence, increased patient dose.

Cassetteless digital systems have a fixed spatial resolution determined by: A The image plate laser divergence B The focal spot size used C The thin film transistor (TFT) detector element (DEL) size D The proximity of the phosphor screen crystals

The Correct Answer is: CA cassetteless system refers to direct or indirect digital systems where no cassette/IP is used. CR uses an IP containing a PSP (photostimulable phosphor plate). Laser divergence is a negative factor that occurs in computed radiography (CR) readers (A). However, with cassetteless digital systems, the spatial resolution of the detector elements (DEL) determines the maximum image resolution that can be obtained and is important in both CR and direct or indirect digital imaging systems. The thin film transistor (TFT) or detector element (DEL) size is fixed and, therefore, the maximum spatial resolution is defined by the physical size of the individual elements and their proximity to each other (C). The focal spot size (B) appropriate for the anatomical part being imaged is important to render optimal image resolution. Proximity of the phosphor screen crystals refers to conventional film-screen radiography (D). (Seeram, 1 st ed., p. 97)

With all other factors constant, as digital image matrix size increases,1.pixel size decreases.2.resolution increases.3.pixel size increases. A 1 only B 2 only C 1 and 2 only D 2 and 3 only

The Correct Answer is: CA digital image is formed by a matrix of pixels (picture elements) in rows and columns. A matrix that has 512 pixels in each row and column is a 512 × 512 matrix. The term field of view is used to describe how much of the patient (eg, 150-mm diameter) is included in the matrix. The matrix and the field of view can be changed independently, without one affecting the other, but changes in either will change pixel size. As in traditional radiography, spatial resolution is measured in line pairs per mm (lp/mm). As matrix size is increased, there are more and smaller pixels in the matrix, and therefore improved resolution. Fewer and larger pixels result in a poor resolution, "pixelly" image, that is, one in which you can actually see the individual pixel boxes. (Fosbinder & Kelsey, p 286; Shephard, p 336)

Which of the following materials may be used as grid interspace material?LeadPlasticAluminum A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: CA grid is composed of alternate strips of lead and interspace material. The lead strips serve to trap scattered radiation before it fogs the IR. The interspace material must be radiolucent; plastic or sturdier aluminum usually is used. Cardboard was used in the past as interspace material, but it had the disadvantage of being affected by humidity (moisture). (Selman, 9th ed., p. 234)

A graphic diagram of signal values representing various absorption properties within the part being imaged is called a A processing algorithm B DICOM C histogram D window

The Correct Answer is: CA histogram is a graph usually having several peaks and valleys representing the pixel values/absorbing properties of the various tissues, and so on that make up the imaged part. These various attenuators include such things as bone, muscle, air, contrast agents, foreign bodies, and pathology. The various pixel values, then, represent image contrast. If the histogram has a rather flat "tail," this represents underexposed areas at the periphery of the image, which can skew the overall histogram analysis. The radiographer selects the particular processing algorithm on the computer/control panel that corresponds to the anatomic part and projection being performed. DICOM (Digital Imaging and Communications in Medicine) refers to the standard for communication between PACS and HIS/RIS systems. Windowing refers to the radiographer's postprocessing adjustment of contrast and brightness (at the workstation). (Shephard, pp. 341, 345)

Combinations of milliamperage and exposure time that produce a particular milliampere-seconds value will produce identical receptor exposure. This statement is an expression of the A inverse-square law B line-focus principle C reciprocity law D D log E curve

The Correct Answer is: CA number of milliamperage and exposure time settings can produce the same milliampere-seconds value. Each of the following milliamperage and time combinations produces 10 mAs: 100 mA and 0.1 s, 200 mA and 0.05 s, 300 mA and, and 400 mA and 0.025 s. These milliamperage and exposure-time combinations should produce identical receptor exposures. This is known as the reciprocity law. The radiographer can make good use of the reciprocity law when manipulating exposure factors to decrease exposure time and decrease motion unsharpness. (Selman, 9th ed., p. 214)

All of the following are components of a television picture tube (cathode ray tube), except: A Electron gun B Glass envelope C Signal plate D Focusing coil

The Correct Answer is: CA signal plate (C) is a component of a television camera tube, such as the Vidicon. An electron gun (A) is used in the cathode section of a television picture tube (cathode ray tube (CRT)) to generate electrons that are accelerated onto the output phosphor and converted to visible light. An outer glass envelope (B) is necessary in a CRT to contain a vacuum, thereby eliminating air molecules that would otherwise impede the electrons traveling from the electron gun to the output phosphor. The focusing coil (D) is a component of a CRT. Its function is to keep the electron beam produced by the electron gun confined to a narrow stream. (Bushong, 9 th ed., p. 353).

The reduction in x-ray photon intensity as the photon passes through material is termed A absorption B scattering C attenuation D divergence

The Correct Answer is: CAbsorption occurs when an x-ray photon interacts with matter and disappears, as in the photoelectric effect. Scattering occurs when there is partial transfer of energy to matter, as in the Compton effect. The reduction in the intensity of an x-ray beam buy absorption and scattering as it passes through matter is called attenuation. (Bushong, 11th ed., p. 139)

A satisfactory radiograph was made using a 40-inch SID, 10 mAs, and a 12:1 grid. If the examination will be repeated at a distance of 48 inches and using an 8:1 grid, what should be the new mAs to maintain the original receptor exposure? A 5.6 B 8.8 C 11.5 D 14.4

The Correct Answer is: CAccording to the exposure maintenance formula, if the SID is changed to 48 inches, 14.4 mAs is required to maintain the original receptor exposure. Then, to compensate for changing from a 12:1 grid to an 8:1 grid, the mAs becomes 11.5: Thus, 11.5 mAs is required to produce a receptor exposure similar to that of the original image. The following are the factors used for mAs conversion from nongrid to grid: No grid = 1 × the original mAs 5:1 grid = 2 × the original mAs 6:1 grid = 3 × the original mAs 8:1 grid = 4 × the original mAs 12:1 grid = 5 × the original mAs 16:1 grid = 6 × the original mAs

All the following x-ray circuit devices are located between the incoming power supply and the primary coil of the high-voltage transformer except A the timer B the kilovoltage meter C the milliamperage meter D the autotransformer

The Correct Answer is: CAll circuit devices located before the primary coil of the high-voltage transformer are said to be on the primary or low-voltage side of the x-ray circuit. The timer, autotransformer, and (prereading) kilovoltage meter are all located in the low-voltage circuit. The milliampere meter, however, is connected at the midpoint of the secondary coil of the high-voltage transformer. When studying a diagram of the x-ray circuit, it will be noted that the milliampere meter is grounded at the midpoint of the secondary coil (where it is at zero potential). Therefore, it may be placed in the control panel safely. (Selman, 9th ed., pp. 150-151)

All the following x-ray circuit devices are located between the incoming power supply and the primary coil of the high-voltage transformer except A the circuit breaker. B the kilovoltage selector. C the rectifiers. D the autotransformer.

The Correct Answer is: CAll circuit devices located before the primary coil of the high-voltage transformer are said to be on the primary, or low-voltage, side of the x-ray circuit. The timer, circuit breaker, autotransformer, kilovoltage selector switch, and (prereading) kilovoltage meter are all located in the low-voltage circuit. The rectifiers, however, are placed after the secondary coil of the high-voltage transformer and before the x-ray tube. (Selman, pp. 150-151)

All of the following are advantages of digital fluoroscopic imaging systems over conventional fluoroscopic imaging systems, except: A Post-processing capability to enhance image contrast B Increased image acquisition speed C No need for pulsed or continuous radiation exposure D Higher milliamperage settings can be used

The Correct Answer is: CAll fluoroscopic imaging (conventional and digital) requires either pulsed or continuous X-ray exposure (C) to provide a dynamic image of the anatomical area of interest. In digital fluoroscopic units, the X-ray tube actually operates in the radiographic mode. However, multiple exposures are made in succession to produce the dynamic image. In these systems, the X-ray generator must be capable of switching on (also called interrogation time) and off (also called extinction time) rapidly in less than 1 ms. The digitized image in a digital fluoroscopy system can be post-processed to enhance image contrast (A), similar to the post-processing that can be done with computed and direct capture static radiographic images. One of the advantages of a digital fluoroscopic system over a conventional fluoroscopic system is the elimination of the television camera tube from the imaging chain, thereby increasing image acquisition speed (B). Either a charge-coupled device or a flat panel image receptor is used to generate electrical signals that can be digitized in a much faster and efficient way, when compared to conventional fluoroscopy. During digital fluoroscopy, the X-ray tube actually operates in the radiographic mode using higher milliamperage settings (D). Tube current is measured in hundreds of milliamperes (mA) rather than less than 5 mA, as in image intensified fluoroscopy. This is not a problem, as the exposures are made in rapid succession and in a pulsed manner (also called pulsed progression fluoroscopy). (Bushong, 9 th ed., pp. 437-438).

The x-ray tube in a CT imaging system is most likely to be associated with A low-energy photons B an unrestricted x-ray beam C a pulsed x-ray beam D a large focal spot

The Correct Answer is: CAlthough the CT x-ray tube is similar to direct-projection x-ray tubes, it has several special requirements. The CT x-ray tube must have a very high short-exposure rating and must be capable of tolerating several million heat units while still having a small focal spot for optimal resolution. To help tolerate the very high production of heat units, the anode must be capable of high-speed rotation. The x-ray tube produces a pulsed x-ray beam (1-5 ms) using up to about 1,000 mA. The collimator assembly has two parts: The prepatient, or predetector, collimator is at the x-ray tube and consists of multiple beam restrictions so that the x-ray beam diverges little. This reduces patient dose and reduces the production of scattered radiation, thereby improving the CT image. The postpatient collimator, or predetector collimator, confines the exit photons before they reach the detector array and determines slice thickness. (Bushong, 8th ed., pp. 429-430)

To determine how quickly an x-ray tube will disperse its accumulated heat, the radiographer uses a(n) A technique chart. B radiographic rating chart. C anode cooling curve. D spinning top test.

The Correct Answer is: CAn anode cooling curve identifies how many HU the anode can accommodate and the length of time required for adequate cooling between exposures. A radiographic rating chart is used to determine if the selected mA, exposure time, and kVp are within safe tube limits. A technique chart is used to determine the correct exposure factors for a particular part of the body of a given thickness. A spinning top test is used to test for timer inaccuracy or rectifier failure. (Selman, p 147)

If the x-ray image exhibits insufficient receptor exposure, this might be attributed toInsufficient kilovoltageInsufficient SIDgrid cutoff A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: CAs kilovoltage is reduced, the number of high-energy photons produced at the target is reduced; therefore, a decrease in receptor exposure occurs. If a grid has been used improperly (off-centered or out of focal range), the lead strips will absorb excessive amounts of primary radiation, resulting in grid cutoff and loss of receptor exposure. If the SID is inadequate (too short), an increase in receptor exposure will occur. (Selman, 9th ed., pp. 214, 240-242)

In which of the following examinations would a IR front with very low absorption properties be especially desirable? A Extremity radiography B Abdominal radiography C Mammography D Angiography

The Correct Answer is: CBecause mammographic techniques operate at very low kilovoltage levels, the IR front material becomes especially important. The use of soft, low-energy x-ray photons is the underlying principle of mammography; any attenuation of the beam would be most undesirable. Special plastics that resist impact and heat softening, such as polystyrene and polycarbonate, are used frequently as IR front material. (Shephard, p. 49)

Recently, dual-sided reading technology has become available in more modern CR readers, in which two sets of photodetectors are used to capture light released from the front and back sides of the phosphor storage plate, or PSP (photostimulable phosphor). This technology enables improved: A Slow-scan direction speed B Modulation transfer function C Signal-to-noise ratio D Fast-scan direction speed

The Correct Answer is: CBy incorporating two sets of light guides and photodetectors on either side of the IP as it travels through the CR reader, a single laser beam can effectively stimulate release of stored energy from both sides of the phosphor plate. This increases the amount of energy that may be released and used in the form of light to be converted by the photodetectors to an electrical (analog) signal. Therefore, the higher signal intensity increases the SNR, i.e. signal-to-noise ratio (C). Slow scan direction speed refers to the linear travel speed of the phosphor plate through the CR reader (A). The laser light in the CR reader is rapidly reflected by an oscillating polygonal mirror that redirects the beam through a special lens called the f-theta lens, which focuses the light on a cylindrical mirror that reflects the light toward the IP. This light moves back and forth very rapidly to scan the plate transversely, in a raster pattern, and this movement of the laser beam across the IP is therefore called the fast-scan direction (D). The modulation transfer function is a mathematical function that measures the ability of the digital detector to transfer its spatial resolution characteristics to the image (B). (Seeram p. 56)

Advantages of direct digital radiography over computed radiography (CR) includedirect digital is less expensive.direct digital has immediate readout.PSPs are not needed for direct digital . A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: CComputed radiography (CR) is less expensive primarily because it is compatible with existing equipment. Direct digital radiography requires existing equipment to be modified or new equipment purchased. The image plate (IP) can also be used for mobile studies, though direct digital is currently available for mobile imaging as well. After image processing, the PSP is erased and reused. DR offers the advantage of immediate visualization of the x-ray image; in CR there is a short delay. (Shephard, p. 335)

Which of the following groups of exposure factors would be most appropriate to control involuntary motion? A 400 mA, 0.03 second B 200 mA, 0.06 second C 600 mA, 0.02 second D 100 mA, 0.12 second

The Correct Answer is: CControl of motion, both voluntary and involuntary, is an important part of radiography. Patients are unable to control certain types of motion, such as heart action, peristalsis, and muscle spasm. In these circumstances, it is essential to use the shortest possible exposure time in order to have a "stop action" effect. The shortest exposure time (0.02) will accomplish this. (Carlton, Adler, and Balac 6th ed., p. 396)

Which of the following will produce the greatest distortion? A AP projection of the skull B PA projection of the skull C 37° AP axial of the skull D 20° PA axial of the skull

The Correct Answer is: CDistortion is the result of misalignment of the x-ray tube, the anatomic part, and the IR. If these three parts are not parallel with one another, shape distortion occurs. The greater the misalignment, the greater the distortion. In the example cited, the image made with the greatest tube angle will produce the greatest distortion. Distortion is often introduced intentionally to visualize some structure to better advantage. The 37° (caudad) AP axial projection of the skull, for example, projects the facial bones inferiorly so that the occipital bone can be visualized to better advantage. (Shephard, pp 231-234)

The exposure factors of 300 mA, 0.07 second, and 95 kVp were used to deliver a particular analog receptor exposure and contrast. A similar analog x-ray image can be produced using 500 mA, 80 kVp, and A 0.01 second. B 0.04 second. C 0.08 second. D 0.16 second.

The Correct Answer is: CFirst, evaluate the change(s): The kVp was decreased by about 15% [95-15% = 80.7]. A 15% decrease in kVp will cut the receptor exposure in half; therefore, it is necessary to use twice the original mAs to maintain the original receptor exposure. The original mAs was 21, and so we now need 42 mAs, using the 500-mA station. Because mA × s = mAs, 500x = 42 x = 0.084 second (Fauber, pp 55, 59-60)

Geometric blur can be evaluated using all the following devices except A star pattern B slit camera C penetrometer D pinhole camera

The Correct Answer is: CFocal-spot size accuracy is related to the degree of geometric blur, that is, edge gradient or penumbra. Manufacturer tolerance for new focal spots is 50%; that is, a 0.3-mm focal spot actually may be 0.45 mm. Additionally, the focal spot can increase in size as the x-ray tube ages—hence the importance of testing newly arrived focal spots and periodic testing to monitor focal-spot changes. Focal-spot size can be measured with a pinhole camera, slit camera, or star-pattern-type resolution device. The pinhole camera is rather difficult to use accurately and requires the use of excessive tube (heat) loading. With a slit camera, two exposures are made; one measures the length of the focal spot, and the other measures the width. The star pattern, or similar resolution device such as the bar pattern, can measure focal-spot size as a function of geometric blur and is readily adaptable in a QA program to monitor focal-spot changes over a period of time. It is recommended that focal-spot size be checked on installation of a new x-ray tube and annually thereafter. A penetrometer, or aluminum step wedge, is used to demonstrate the effect of kV ion contrast. (Bushong, 11th ed., p. 347, 181)

The absorption of useful radiation by a grid is called A grid selectivity. B grid cleanup. C grid cutoff. D latitude.

The Correct Answer is: CGrids are used in radiography to absorb scattered radiation before it reaches the IR (grid "cleanup"), thus improving radiographic contrast. Contrast obtained with a grid compared with contrast without a grid is termed contrast-improvement factor. The greater the percentage of scattered radiation absorbed compared with absorbed primary radiation, the greater is the "selectivity" of the grid. If a grid absorbs an abnormally large amount of useful radiation as a result of improper centering, tube angle, or tube distance, grid cutoff occurs. (Selman, 9th ed., p. 370)

All of the following are factors that cause low contrast in CR images, except: A High kVp B Inadequate grid efficiency or no grid C Excessive beam limiting (collimation) D Incomplete erasure of the image plate

The Correct Answer is: CHigh kVp (beyond that which is optimal for the anatomical part being imaged) provides scattered X-ray photons enough energy to exit the anatomical part in various directions to strike the image receptor (A). This scatter radiation contributes nothing to the "true" anatomical image, but causes decreased contrast in the image. Inadequate grid efficiency, or not using a grid when needed (B), allows scatter radiation to strike the image receptor, causing decreased contrast. Many of the factors that cause low contrast in film-screen systems also cause low contrast in CR images: high kVp, inadequate grid efficiency or no grid, insufficient beam limiting, and incomplete erasure of the image plate (C). Incomplete erasure of an image plate from a previous exposure or background radiation will result in extraneous exposure data that reduces image contrast in the successive image (D). (Seeram, 1 st ed., p. 96)

Which of the following combinations will result in the most scattered radiation reaching the image receptor? A Using more mAs and compressing the part B Using more mAs and a higher ratio grid C Using less mAs and more kVp D Using less mAs and compressing the part

The Correct Answer is: CHigher kV increases x-ray photon energy and the production of scattered radiation. The number of x-ray photons produced (beam intensity/quantity) is controlled by mAs - which has no effect on scattered radiation. Grids are used to decrease the amount of scattered radiation reaching the IR. Compression can decrease the amount of scattered radiation reaching the IR, but adjusting the mAs at the same time will not affect a decrease in SR reaching the IR. (Bushong, 8th ed, p 236)

Which of the following may occur if the X-ray exposure field is not properly collimated, positioned, and sized? A Modulation transfer function failure B Moiré artifact C Exposure field recognition errors may occur D Ghost artifact

The Correct Answer is: CIf the X-ray exposure field is improperly collimated, positioned, and sized, exposure field recognition errors can occur (C). These can lead to histogram analysis errors due to signals generated from outside of the exposure field. This may result in dark, light, or noisy images. The MTF, or modulation transfer function (A) is a mathematical function that measures the ability of a digital detector to transfer its spatial resolution characteristics of the image. If a radiographic grid has a frequency that approximates the CR scan frequency and the grid strips are oriented in the same direction as the scan, the Moiré artifact may be observed (B). The appearance of ghost artifacts can be seen when CR image plates are incompletely erased. If an image plate has not been used for 24 hours, it should be erased again before using it for a diagnostic radiographic exposure (A). (Bushong, 9 th ed., p. 493)

Which of the following digital post-processing methods remove high-frequency noise from the image? A Edge enhancement B Windowing C Smoothing D Aliasing

The Correct Answer is: CImage smoothing (C) is a type of spatial frequency filtering performed during digital image post-processing. Also known as low-pass filtering, smoothing can be achieved by averaging each pixel's frequency with surrounding pixel values to remove high-frequency noise. The result is reduction in noise and contrast. Smoothing (low-pass filtering) is useful for viewing small structures such as fine bone tissues. Edge enhancement (A) is a type of post-processing image manipulation, which can be effective for enhancing fractures and small, high-contrast tissues. In digital imaging, after the signal is obtained for each pixel, the signals are averaged to shorten processing time and decrease storage needs. The larger the number of pixels involved in the averaging, the smoother the image appears. The signal strength of one pixel is averaged with the strength of its neighboring pixels. Edge enhancement is achieved when fewer neighboring pixels are included in the signal average. Therefore, the smaller the number of neighboring pixels, the greater the edge enhancement. Windowing (B) is a post-processing method of adjusting the brightness and contrast in the digital image. There are two types of windowing: level and width. Window level adjusts the overall image brightness. Window width adjusts the ratio of white to black, thereby changing image contrast. Narrow window width provides higher contrast (short-scale contrast), whereas wide window width will produce an image with less contrast (long-scale contrast). Aliasing (D) is an image artifact that occurs when the spatial frequency is greater than the Nyquist frequency and the sampling occurs less than twice per cycle. This causes loss of information and a fluctuating signal and wrap-around image is produced, which appears as two superimposed images that are slightly out of alignment, resulting in a moiré effect. The Nyquist theorem states that when sampling a signal (such as the conversion from the analog to digital image), the sampling frequency must be greater than twice the bandwidth of the input signal so that reconstruction of the original image properly displays the anatomy of interest. (Carter and Veale, p. 118).

All of the following are equipment options that may be used to record the anatomical image in mobile radiography, except: A Tethered flat panel B Remote (wireless) digital flat panel array C Scanned projection radiography (SPR) D Conventional radiographic film

The Correct Answer is: CIn scanned projection radiography (SPR) (A), typically of the chest, the X-ray beam is collimated to a thin fan by pre-patient collimators. Post-patient image-forming X-rays likewise are collimated to a thin fan that corresponds to a detector array consisting of a scintillation phosphor, usually NaI or CsI, which is married to a linear array of CCDs through a fiberoptic path. This type of unit is a fixed unit located in the radiology department. Answers A, B and D can be used with mobile radiographic units to record the anatomical image. (Carlton and Adler, 5 th ed., p. 528).

Which of the following will contribute to the production of longer-scale radiographic contrast?1.An increase in kV2.An increase in grid ratio3.An increase in photon energy A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: CIncreased photon energy is caused by an increase in kVp, resulting in more penetration of the part and a longer scale of contrast. Increasing the grid ratio will result in a larger percentage of scattered radiation being absorbed and hence a shorter scale of contrast. (Shephard, pp 203-204)

The x-ray imaging system that uses a flat panel detector is A film emulsion system. B computed radiography. C direct digital radiography. D fluoroscopy.

The Correct Answer is: CMedical imaging is experiencing rapid technological growth, and x-ray images can be obtained in a number of ways. Film/screen systems are rarely used today. Imaging systems used today include computed radiography (CR) and direct digital radiography (DR). CR uses an Image Plate (IP) that encloses the photostimulable phosphor (PSP). When the PSP is exposed, it stores the image; a scanner-reader then converts the PSP image to a digital image; the image is then displayed on a computer monitor. Direct digital radiography (DR) eliminates the IP and PSP. The x-ray image is captured by a flat panel detector in the x-ray table and converts it to a digital image; the x-ray image is displayed immediately on a computer monitor. Fluoroscopy is a "live action" or "real-time" examination where the dynamics (motion) of parts can be evaluated; "still" images can be made during the fluoroscopic exam. (Ballinger and Frank, 10th ed, vol 1, p 3)

Which of the following modes of a trifield image intensifier will result in the highest patient dose? A Its 25-cm. mode B Its 17-cm. mode C Its 12-cm. mode D Diameter does not affect patient dose

The Correct Answer is: CMost image-intensifier tubes are either dual-field or trifield, indicating the diameter of the input phosphor. When a change to a smaller-diameter mode is made, the voltage on the electrostatic focusing lenses is increased, and the result is a magnified but dimmer image. The milliamperage will be increased automatically to compensate for the loss in brightness with a magnified image, resulting in higher patient dose in the smaller-diameter modes. (Bushong, 10th ed., p. 407)

Spatial resolution can be improved by decreasingthe SIDthe OIDpatient/part motion A 1 only B 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: CMotion, voluntary or involuntary, is most detrimental to good spatial resolution. Even if all other factors are adjusted to maximize detail, if motion occurs during exposure, resolution is lost. The most important ways to reduce the possibility of motion are using the shortest possible exposure time, careful patient instruction (for suspended respiration), and adequate immobilization when necessary. Minimizing magnification through the use of increased SID and decreased OID functions to improve spatial resolution. (Carlton and Adler, 4th ed., p. 451)

The functions of a picture archiving and communication system (PACS) includestorage of analog imagesretrieval of digital imagesstorage of digital images A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: CPACS refers to a picture archiving and communication system. Analog images (conventional images) can be digitized with a digitizer. PACS systems receive digital images and display them on monitors for interpretation. These systems also store images and allow their retrieval at a later time. (Shephard, pp. 365-367)

All of the following are steps that should be used to accomplish quality control (QC) in digital radiography, except: A Acceptance testing B Establishment of baseline performance C Monitoring patient size to evaluate variations in equipment performance D Diagnosis of changes in performance

The Correct Answer is: CPatient size variations are expected in a radiology department. Equipment operation is expected to respond accordingly to variations in the size of the patient, although image quality may vary, as expected (C). Acceptance testing (A) is the initial opportunity to determine whether the imaging equipment meets the requirements of state and regulatory agencies, as well as special requirements that may be included in the purchasing contract. It is important to determine acceptable performance before the imaging device is used for patients. It is also important to conduct the acceptance testing with the vendor service engineer present, so that deficiencies may be corrected immediately. Establishment of baseline performance (B) is an important QC step. New equipment is expected to perform well, but it is important to monitor indicators of change and establish control limits with subsequent use. Control limits determine the maximum deviation from normal that is considered allowable before initiating corrective action. Diagnosing changes in equipment performance (D) is an important component of a QC program. When a decrease in performance expectations is observed and corrective action is taken, it is important to verify that performance has returned to normal levels. This may require more comprehensive tests than the usual performance indicators and possibly a repeat of the complete acceptance testing procedures. (Seeram, pp. 219-220).

Which of the following combinations is most likely to be associated with quantum mottle? A Decreased milliampere-seconds, decreased SID B Increased milliampere-seconds, decreased kilovoltage C Decreased milliampere-seconds, increased kilovoltage D Increased milliampere-seconds, increased SID

The Correct Answer is: CQuantum mottle is a grainy appearance on a finished image that is seen especially in fast-imaging systems. It is similar to the "pixelated" appearance of an enlarged digital image; it has a spotted or freckled appearance. Fast imaging systems using low-milliampere-seconds and high-kilovoltage factors are most likely to be the cause of quantum mottle. (Bushong, 8th ed., p. 273)

Congruence of the x-ray beam with the light field is tested using A a pinhole camera B a star pattern C radiopaque objects D a slit camera

The Correct Answer is: CRadiographic results should be consistent and predictable with respect to positioning accuracy, exposure factors, and equipment operation. X-ray equipment should be tested and calibrated periodically as part of an ongoing quality assurance (QA) program. The focal spot should be tested periodically to evaluate its size and its impact on spatial resolution; this is accomplished using a slit camera, a pinhole camera, or a star pattern. To test the congruence of the light and x-ray fields, a radiopaque object such as a paper clip or a penny is placed at each corner of the light field before the test exposure is made. Upon viewing, the corners of the x-ray field should be exactly delineated by the radioopaque objects. (Carlton and Adler, 4th ed., p. 484)

An exposure was made using 300 mA, 40 ms exposure, and 85 kV. Each of the following changes will cut the receptor exposure in half except a change to A 1/50 sec exposure B 72 kV C 10 mAs D 150 mA

The Correct Answer is: CReceptor exposure is directly proportional to milliampere-seconds. If exposure time is halved from 40 ms (0.04 or 1 /25) sec to 0.02 ( 1 / 50 ) sec, receptor exposure will be cut in half. Changing to 150 mA also will halve the milliampere-seconds, effectively halving the receptor exposure. If the kilovoltage is decreased by 15%, from 85 to 72 kV, rreceptor exposure will be halved according to the 15% rule. To cut the receptor exposure in half, the mAs value must be reduced to 6 mAs (rather than 10 mAs). (Johnston and Fauber, 3rd ed., p. 134)

Which of the following has the greatest effect on receptor exposure? A Aluminum filtration B Kilovoltage C SID D Scattered radiation

The Correct Answer is: CReceptor exposure is greatly affected by changes in the SID, as expressed by the inverse-square law of radiation. As distance from the radiation source increases, exposure rate decreases, and receptor exposure decreases. Exposure rate is inversely proportional to the square of the SID. Aluminum filtration, kilovoltage, and scattered radiation all have a significant effect on receptor exposure, but they are not the primary controlling factors. (Selman, 9th ed., p. 214)

If 92 kV and 12 mAs were used for a particular abdominal exposure with single-phase equipment, what mAs would be required to produce a similar radiograph with three-phase equipment? A 36 B 24 C 8 D 6

The Correct Answer is: DSingle-phase radiographic equipment is much less efficient than three-phase equipment because it has a 100% voltage ripple. With three-phase equipment, voltage never drops to zero, and x-ray intensity is significantly greater. To produce similar receptor exposure, only one half of the original mAs would be used for (both 6 pulse and 12 pulse) three-phase equipment . (Bushong 11th ed p 100)

A grid usually is employed in which of the following circumstances?When radiographing a large or dense body partWhen using high kilovoltageWhen a lower patient dose is required A 1 only B 3 only C 1 and 2 only D 1, 2, and 3

The Correct Answer is: CSignificant scattered radiation is generated within the part when imaging large or dense body parts and when using high kilovoltage. A radiographic grid is made of alternating lead strips and interspace material; it is placed between the patient and the IR to absorb energetic scatter emerging from the patient. Although a grid prevents much of the scattered radiation from reaching the radiograph, its use does necessitate a significant increase in patient exposure. (Bushong, 8th ed., p. 248)

A grid is usually employed1.when radiographing a large or dense body part.2.when using high kilovoltage.3.when less patient dose is required. A 1 only B 3 only C 1 and 2 only D 1, 2, and 3

The Correct Answer is: CSignificant scattered radiation is produced when radiographing large or dense body parts and when using high kilovoltage. A radiographic grid is made of alternating lead strips and interspace material; it is placed between the patient and the IR to absorb energetic scatter emerging from the patient. Although a grid prevents much scattered radiation fog from reaching the radiograph, its use does necessitate a significant increase in patient exposure. (Shephard, pp 244-245)

An exposure was made using 600 mA and 18 ms. If the mA is changed to 400, which of the following exposure times would most closely approximate the original receptor exposure? A 16 ms B 0.16 second C 27 ms D 0.27 second

The Correct Answer is: CSince 18 ms is equal to 0.018 s, and since mA × time = mAs, the original mAs was 10.8. Now it is only necessary to determine what exposure time must be used with 400 mA to provide the same 10.8 mAs (and thus the same receptor exposure) because mA × time = mAs, 400x = 10.8 x = 0.027 second (27 milliseconds) (Selman, p 214)

Which of the following terms is used to describe unsharp edges of tiny radiographic details? A Diffusion B Mottle C Blur D Umbra

The Correct Answer is: CSpatial resolution is evaluated by how sharply tiny anatomic details are imaged on the x-ray image. The area of blurriness that may be associated with tiny image details is termed geometric blur. The blurriness can be produced by using a large focal spot, increased OID, or decreased SID. The image proper (i.e., without blur) is often termed the umbra. Mottle is a grainy appearance associated with insufficient receptor exposure.

If obtaining multiple images on one image plate, it is important to: A Allow for X-ray tube cooling between successive exposures B Avoid shielding of the image plate at all times to avoid field recognition errors C Properly shield each exposed and unexposed area during the imaging of each individual image D Expose the AP or PA projection in the right lower portion of the image plate

The Correct Answer is: CSuccessive static exposures taken on one or more image plates rarely would cause overheating of the X-ray tube (A). Shielding of the image plate for multiple exposures is important to avoid intrafield scatter radiation exposure and a possible field recognition error (B). The keys to multiple fields on one IP are symmetry and uniform distribution. One should only use 3-on-1 distribution for fingers and toes where the amount of intrafield scatter is low. If larger body structures are done 3-on-1, the intrafield scatter will reduce the contrast unless the unexposed areas are shielded between exposures (C). The specific location of any projection on an image plate does not discount the importance of including one projection on one image plate (D). (Seeram, 1 st ed., p. 93)

Which of the following will occur as a result of a decrease in the anode target angle?1.Less pronounced anode heel effect2.Decreased effective focal spot size3.Greater photon intensity toward the cathode side of the x-ray tube A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: CTarget angle has a pronounced geometric effect on the effective, or projected, focal spot size. As the target angle decreases, the effective (projected) focal spot becomes smaller. This is advantageous because it will improve spatial resolution without creating a heat-loading crisis at the anode (as would occur if the actual focal spot size were reduced to produce a similar resolution improvement). There are disadvantages, however. With a smaller target angle, the anode heel effect increases; photons are more noticeably absorbed by the "heel" of the anode, resulting in a smaller percentage of x-ray photons at the anode end of the x-ray beam and a concentration of x-ray photons at the cathode end of the radiograph. (Shephard, p 221)

A decrease from 80 mAs to 40 mAs will result in a decrease in which of the following?WavelengthExposure rateBeam intensity A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: CTechnical factors can be expressed in terms of milliampere-seconds rather than milliamperes and time. The milliampere-seconds value is a quantitative factor because it regulates x-ray-beam intensity, exposure rate, quantity, or number of x-ray photons produced (the milliampere-seconds value is the single most important technical factor associated with receptor exposure and patient dose). The milliampere-seconds value is directly proportional to the intensity (i.e., exposure rate, number, and quantity) of x-ray photons produced and the resulting receptor exposure. If the milliampere-seconds value is doubled, twice the exposure rate and twice the receptor exposure occurs. If the milliampere-seconds value is cut in half, the receptor exposure and patient dose are cut in half. Kilovoltage is the qualitative exposure factor—it determines beam quality by regulating photon energy (i.e., wavelength). (Carlton and Adler, 5th ed., p. 183)

Which part of an induction motor is located outside the x-ray tube glass envelope? A Filament B Focusing cup C Stator D Rotor

The Correct Answer is: CThe anode is made to rotate through the use of an induction motor. An induction motor has two main parts, a stator and a rotor. The stator is the part located outside the glass envelope and consists of a series of electromagnets occupying positions around the stem of the anode. The stator's electromagnets are supplied with current and the associated magnetic fields function to exert a drag or pull on the rotor within the glass envelope. The anode is a 2- to 5-in. diameter molybdenum or graphite disc with a beveled edge. The beveled surface has a focal track of tungsten and rhenium alloy. The anode rotates at about 3,600 rpm (high-speed anode rotation is about 10,000 rpm), so that heat generated during x-ray production is evenly distributed over the entire track. Rotating anodes can withstand delivery of a greater amount of heat for a longer period of time than stationary anodes. (Carlton and Adler, 4th ed., p. 73)

If 0.05 second was selected for a particular exposure, what mA would be necessary to produce 15 mAs? A 900 B 600 C 500 D 300

The Correct Answer is: DThe formula for mAs is mA × s = mAs. Substituting known values, 0.05x = 15 x = 300 mA (Selman, p 214)

During CR imaging, the latent image present on the PSP is changed to a digital signal by the A PSP B Scanner-reader C ADC D helium-neon laser

The Correct Answer is: CThe exposed CR cassette is placed into the CR scanner/reader, where the PSP (SPS) is removed automatically. The latent image appears as the PSP is scanned by a narrow, high-intensity helium-neon laser to obtain the pixel data. As the plate is scanned in the CR reader, it releases a violet light—a process referred to as photostimulated luminescence (PSL). The luminescent light is converted to electrical energy representing the analog image. The electrical energy is sent to an analog-to-digital converter (ADC), where it is digitized and becomes the digital image that is displayed eventually (after a short delay) on a high-resolution monitor and/or printed out by a laser printer. The digitized images can also be manipulated in postprocessing, transmitted electronically, and stored/archived. (Carlton and Adler, 4th ed., p. 358)

Of the following groups of exposure factors, which will produce the greatest receptor exposure? A 400 mA, 30 ms, 72-in. SID B 200 mA, 30 ms, 36-in. SID C 200 mA, 60 ms, 36-in. SID D 400 mA, 60 ms , 72-in. SID

The Correct Answer is: CThe formula mA × s = mAs is used to determine each milliampere-second setting (remember to first change milliseconds to seconds). The greatest receptor exposure will be produced by the combination of highest milliampere-seconds value and shortest SID. The groups in choices (B) and (D) should produce identical receptor exposure, according to the inverse-square law, because group (D) includes twice the distance and 4 times the milliampere-seconds value of group (B). The group in (A) has twice the distance of the group in (B) but only twice the milliampere-seconds; therefore, it has the least receptor exposure. The group in (C) has the same distance as the group in (B) and twice the milliampere-seconds, making group in (C) the group of technical factors that will produce the greatest receptor exposure. (Selman, 9th ed., p. 214)

In comparison with 60 kV, 80 kV willpermit greater exposure latitudeproduce more scattered radiationproduce shorter-scale contrast A 1 only B 2 only C 1 and 2 only D 2 and 3 only

The Correct Answer is: CThe higher the kilovoltage range, the greater is the exposure latitude (margin of error in exposure). Higher kilovoltage produces more energetic photons, is more penetrating, and produces more grays on the radiographic image, lengthening the scale of contrast. As kilovoltage increases, the percentage of scattered radiation also increases. (Bushong, 9th ed., pp. 256-257)

All the following statements regarding CR IPs are true except A IPs have a thin lead foil backing. B IPs can be placed in the Bucky tray. C IPs must exclude all white light. D IPs function to protect the PSP.

The Correct Answer is: CThe image plate has a protective function for the flexible photostimulable storage phosphor within; it can be conveniently placed in a Bucky tray or under the anatomic part, and comes in a variety of sizes. The PSP within the IP is the image receptor/detector. IPs do not contain a lightsensitive material and, therefore, do not need to be light-tight. The photostimulable PSP is not affected by light. (Shephard, p. 51)

All the following are components of the image intensifier except A the photocathode B the focusing lenses C the TV monitor D the accelerating anode

The Correct Answer is: CThe input phosphor of an image intensifier receives remnant radiation emerging from the patient and converts it to a fluorescent light image. Directly adjacent to the input phosphor is the photocathode, which is made of a photoemissive alloy (usually a cesium and antimony compound). The fluorescent light image strikes the photocathode and is converted to an electron image. The electrons are focused carefully, to maintain image resolution, by the electrostatic focusing lenses, through the accelerating anode and to the output phosphor for conversion back to light. The TV monitor is not part of the image intensifier but serves to display the image that is transmitted to it from the output phosphor. (Bushong, 8th ed., pp. 360-363)

In which of the following examinations should 70 kV not be exceeded? A Upper GI (UGI) B Barium enema (BE) C Intravenous urogram (IVU) D Chest

The Correct Answer is: CThe iodine-based contrast material used in IVU gives optimal opacification at 60 to 70 kV. Use of higher kilovoltage will negate the effect of the contrast medium; a lower contrast will be produced, and poor visualization of the renal collecting system will result. GI and BE examinations employ high-kilovoltage exposure factors (about 120 kV) to penetrate through the barium. In chest radiography, high-kilovoltage technical factors are preferred for maximum visualization of pulmonary vascular markings made visible with long-scale contrast. (Saia, 4th ed., p. 347)

Major components of a CR reader include all of the following, except: A Laser source B Image plate transport mechanism C Thin-film transistor D Analog-to-digital convertor

The Correct Answer is: CThe laser source (A) is a major component of a CR reader because it is this light energy that, when distributed on the image plate's PSP (photostimulable phosphor), releases the stored energy from the X-ray exposure to the PSP, which can then be used to produce the diagnostic anatomical image. The major components of a computed radiography (CR) reader include the laser source, image plate (IP) transport mechanism (B), light channeling guide, photodetector (photomultiplier tube), and the analog-to-digital convertor (ADC). The TFT, i.e. thin-film transistor (C), is a component found in flat-panel detector type digital systems. The analog-to-digital convertor (D) is a device that receives the analog signal from the CR reader and converts this signal into binary code to be used by the computer for read-out and post-processing. (Seeram, p. 54)

A radiograph made using 300 mA, 0.1 second, and 75 kVp exhibits motion unsharpness, but otherwise satisfactory technical quality. The radiograph will be repeated using a shorter exposure time. Using 86 kV and 500 mA, what should be the new exposure time? A 0.12 second B 0.06 second C 0.03 second D 0.01 second

The Correct Answer is: CThe mAs formula is milliamperage × time = mAs. With two of the factors known, the third can be determined. To find the mAs that was originally used, substitute the known values: 300 × 0.1 = 30 We have increased the kilovoltage to 86, an increase of 15%, which has an effect similar to that of doubling the mAs. Therefore, only 15 mAs is now required as a result of the kV increase: mA × s = mAs 500 x = 15 x = 0.03-second exposure (Selman, p 214)

If 300 mA has been selected for a particular exposure, what exposure time would be required to produce 60 mAs? A 1/60 second B 1/30 second C 1/10 second D 1/5 second

The Correct Answer is: DThe mAs is the technical factor that regulates receptor exposure. The equation used to determine mAs is mA × s = mAs. Substituting the known factors, 300x = 60 x = 0.2 (1/5) second (Fauber, p 55)

A test radiograph like the one pictured in Figure A would be made by the radiation safety officer (RSO) or equipment service person and is used to evaluate A focal spot size. B linearity. C collimator alignment. D spatial resolution.

The Correct Answer is: CThe radiograph illustrates testing done to evaluate the x-ray beam and light beam alignment. Light-localized collimators must be tested periodically and must be accurate to within 2% of the SID. Linearity means that a given mA, using different mA stations with appropriate exposure time adjustments, will provide consistent intensity. A star pattern would be used to evaluate focal spot resolution, and a parallel line-type resolution pattern could also be used to evaluate spatial resolution. (Bushong, p 568)

The term differential absorption is closely related tobeam intensitysubject contrastpathology A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: CThe radiographic subject, the patient, is composed of many different tissue types of varying densities (i.e., subject contrast), resulting in varying degrees of photon attenuation and absorption. This differential absorption contributes to the various shades of gray in the x-ray image, as does the energy/penetration characteristic of the beam as regulated by the kV. Beam intensity/quantity/mAs is not related to differential absorption. Normal tissue density may be significantly altered in the presence of pathology. For example, destructive bone disease can cause a dramatic decrease in tissue density. Abnormal accumulation of fluid (as in ascites) will cause a significant increase in tissue density. Muscle atrophy or highly developed muscles similarly will decrease or increase tissue density. (Carlton and Adler, 4th ed., p. 245)

The primary parts of the cathode include the 1. focal track. 2. filament. 3. focusing cup. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: CThe typical diode x-ray tube consists of a positive electrode (the anode) and a negative electrode (the cathode). Electrons are released from the cathode's filament, directed toward the anode by the cathode's focusing cup, and delivered at very high speed to the anode's focal track. (Bushong, 8th ed, p 131)

Which of the following function(s) to reduce the amount of scattered radiation reaching the IR?1.Grid devices2.Restricted focal spot size3.Beam restrictors A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: CThere are several ways to reduce the amount of scattered radiation reaching the IR. First, the use of optimum kVp is essential; excessive kVp will increase the production of scattered radiation. Second, conscientious use of the beam restrictor (collimator) will reduce scattered radiation; the smaller the volume of irradiated tissue, the less scattered radiation is produced. The use of grids helps clean up scattered radiation before it reaches the IR. The size of the tube focus has an impact on image geometry and spatial resolution, but it has no effect on scattered radiation. (Shephard, p 203)

Capacitor-discharge mobile x-ray unitsuse a grid-controlled x-ray tubeare typically charged before the day's workprovide a direct-current output A 1 only B 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: CThere are two main types of mobile x-ray units—capacitor-discharge and battery-powered. The capacitor-discharge units consist of a capacitor, or condenser, which is given a charge and then stores energy until the x-ray tube uses it to produce x-rays. The charge may not be stored for extended periods, however, because it tends to "leak" away; the capacitor must be charged just before the exposure is made. Its x-ray tube is grid-controlled, permitting very fast (short) exposure times. Capacitors discharge a direct current (as opposed to single- or three-phase pulsating current) in which the kilovoltage decreases by a value of approximately 1 kV/mAs. Thus, although the value at the onset of the exposure may be 20 mAs and 80 kVp, at the end of the exposure, the kilovoltage value will be approximately 60 kVp. In addition, capacitor-discharge units permit only limited milliampere-seconds values, usually 30 to 50 mAs per charge. (Bushong, 8th ed., pp. 123-124)

What is the best way to reduce magnification distortion? A Use a small focal spot. B Increase the SID. C Decrease the OID. D Avoid tube angle techniques.

The Correct Answer is: CThere are two types of distortion: size and shape. Shape distortion relates to the alignment of the x-ray tube, the part to be radiographed, and the image recorder. There are two kinds of shape distortion: elongation and foreshortening. Size distortion is magnification, and it is related to the OID and the SID. Magnification can be reduced by either increasing the SID or decreasing the OID. However, an increase in SID must be accompanied by an increase in mAs to maintain receptor exposure. It is therefore preferable, in the interest of exposure, to reduce OID whenever possible. (Fauber, p 90)

A particular radiograph was produced using 6 mAs and 110 kVp with an 8:1 ratio grid. The radiograph is to be repeated using a 16:1 ratio grid. What should be the new mAs? A 3 B 6 C 9 D 12

The Correct Answer is: CTo change nongrid exposures to grid exposures, or to adjust exposure when changing from one grid ratio to another, you must remember the factor for each grid ratio: No grid = 1 × the original mAs 5:1 grid = 2 × the original mAs 6:1 grid = 3 × the original mAs 8:1 grid = 4 × the original mAs 12:1 grid = 5 × the original mAs 16:1 grid = 6 × the original mAs To adjust exposure factors, you simply compare the old with the new: x = 9 mAs using 16:1 grid.

Exposure factors of 90 kVp and 4 mAs are used for a particular nongrid exposure. What should be the new mAs if an 8:1 grid is added? A 8 B 12 C 16 D 20

The Correct Answer is: CTo change nongrid to grid exposure or to adjust exposure when changing from one grid ratio to another, it is necessary to recall the factor for each grid ratio: No grid = 1 × the original mAs 5:1 grid = 2 × the original mAs 6:1 grid = 3 × the original mAs 8:1 grid = 4 × the original mAs 12:1 (or 10:1) grid = 5 × the original mAs 16:1 grid = 6 × the original mAs Therefore, to change from nongrid to an 8:1 grid, multiply the original mAs by a factor of 4. A new mAs of 16 is required. (Saia, p 328)

An x-ray image of the ankle was made at 40-SID, 200 mA, 50 ms, 70 kV, 0.6 mm focal spot, and minimal OID. Which of the following modifications would result in the greatest increase in magnification? A 1.2 mm focal spot B 36-in. SID C 44-in. SID D 4-in. OID

The Correct Answer is: DAll the factor changes affect spatial resolution, but focal spot size does not affect magnification. An increase in SID would decrease magnification. Although a decrease in SID will increase magnification, it does not have as significant an effect as an increase in OID. In general, it requires an increase of 7 in. SID to compensate for every inch of OID. (Carlton and Adler, 4th ed., pp. 458-460)

Characteristics of high-ratio focused grids, compared with lower-ratio grids, include which of the following?They allow more positioning latitude.They are more efficient in collecting SR.They absorb more of the useful beam. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: CTwo of a grid's physical characteristics that determine its degree of efficiency in the removal of scattered radiation are grid ratio (the height of the lead strips compared with the distance between them) and the number of lead strips per inch. As the lead strips are made taller or the distance between them decreases, scattered radiation is more likely to be trapped before reaching the IR. A 12:1 ratio grid will absorb more scattered radiation than an 8:1 ratio grid. An undesirable but unavoidable characteristic of grids is that they do absorb some primary/useful photons as well as scattered photons. The higher the ratio grid, the more scatter radiation the grid will clean up, but more useful photons will be absorbed as well. The higher the primary to scattered photon transmission ratio, the more desirable is the grid. Higher-ratio grids restrict positioning latitude more severely—grid centering must be more accurate (than with lower-ratio grids) to avoid grid cutoff. (Shephard, pp. 245-246)

The voltage ripple associated with a three-phase, six-pulse rectified generator is about A 100% B 32% C 13% D 3%

The Correct Answer is: CVoltage ripple refers to the percentage drop from maximum voltage each pulse of current experiences. In single-phase rectified equipment, the entire pulse (half-cycle) is used; therefore, there is first an increase to the maximum (peak) voltage value and then a decrease to zero potential (90-degree past-peak potential). The entire waveform is used; if 100 kV were selected, the actual average kilovoltage output would be approximately 70 kV. Three-phase rectification produces almost constant potential, with just small ripples (drops) in maximum potential between pulses. Approximately a 13% voltage ripple (drop from maximum value) characterizes the operation of three-phase, six-pulse generators. Three-phase, 12-pulse generators have about a 3.5% voltage ripple. (Selman, 9th ed., p. 162)

A lateral radiograph of the cervical spine was made at 40 in. using 300 mA and 0.03 second exposure. If it is desired to increase the distance to 72 in., what should be the new milliampere (mA) setting, all other factors remaining constant? A 400 mA B 800 mA C 1000 mA D 1200 mA

The Correct Answer is: CWhen exposure rate decreases (as a result of increased SID), an appropriate increase in milliampere-seconds is required to maintain the original receptor exposure. The formula used to determine the new milliampere-seconds value (exposure-maintenance formula) is substituting known values: Substituting known values: Thus, x = 29.16 mAs at 72 in. SID. To determine the required milliamperes (mA × s = mAs), 0.03 x = 29.16

If 82 kVp, 300 mA, and 0.05 second were used for a particular exposure using 3-phase, 12-pulse equipment, what mAs would be required, using single-phase equipment, to produce a similar radiograph? A 7.5 B 20 C 30 D 50

The Correct Answer is: CWith three-phase equipment, the voltage never drops to zero and x-ray intensity is significantly greater. When changing from single-phase to three-phase, six-pulse equipment, two-thirds of the original mAs is required to produce a radiograph with similar receptor exposure. When changing from single-phase to three-phase, 12-pulse equipment, only one-half of the original mAs is required. In this problem, we are changing from three-phase, 12-pulse to single-phase equipment; therefore, the mAs should be doubled (from 15 to 30 mAs). (Carlton & Adler, p 98)

Although the stated focal-spot size is measured directly under the actual focal spot, focal-spot size actually varies along the length of the x-ray beam. At which portion of the x-ray beam is the effective focal spot the largest? A At its outer edge B Along the path of the central ray C At the cathode end D At the anode end

The Correct Answer is: CX-ray tube targets are constructed according to the line-focus principle—the focal spot is angled (usually 12-17 degrees) to the vertical (Figure 4-34). As the actual focal spot is projected downward, it is foreshortened; thus, the effective focal spot is always smaller than the actual focal spot. As it is projected toward the cathode end of the x-ray beam, the effective focal spot becomes larger and approaches the actual size. As it is projected toward the anode end, it gets smaller because of the anode heel effect. (Selman, 9th ed., p. 139)

Objectionable widening of the histogram in CR can be caused by all of the following, except: A Off-focus and scatter radiation outside of the exposure field B Windowing C Improper pre-exposure anatomical selection D Subtraction

The Correct Answer is: D Off-focus and scatter radiation outside of the exposure field would be detected as additional information and, therefore, would widen the histogram (A), resulting in a processing error. Histogram analysis errors can result in rescaling errors and exposure indicator determination errors. Windowing (B) is a post-processing method of adjusting the brightness and contrast in the digital image. There are two types of windowing: level and width. Window level adjusts the overall image brightness. When the window level is increased, the image becomes less bright. When decreased, the image brightness increases. Window width adjusts the ratio of white to black, thereby changing image contrast. Narrow window width provides higher contrast (short-scale contrast), whereas wide window width will produce an image with less contrast (long-scale contrast). Improper pre-exposure anatomical selection (C) (e.g., selecting chest versus the intended foot selection) can interfere with proper histogram assignment (and display) for the anatomical part of interest. In digital image subtraction (D), the pixel values from post-contrast images are electronically subtracted from pixel values from the first pre-contrast (mask) image to show contrast-filled blood vessels with the other structures (e.g., bone) removed in order to enhance the diagnostic impressions of the radiologist, and is unrelated to histogram changes. (Seeram, 1 st ed., p. 37; Carroll, 3rd ed p 469; Carlton and Adler, 6th ed, p279).

Which of the following pathologic conditions are considered additive conditions with respect to selection of exposure factors?OsteomaBronchiectasisPneumonia A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DAll these conditions are considered technically additive because they all involve an increase in tissue density. Osteoma, or exostosis, is a (usually benign) bony tumor that can develop on bone. Bronchiectasis is a chronic dilatation of the bronchi with accumulation of fluid. Pneumonia is inflammation of the lung(s) with accumulation of fluid. Additional bony tissues and the pathologic presence of fluid are additive pathologic conditions and require an increase in exposure factors. Destructive conditions such as osteoporosis require a decrease in exposure factors. (Carlton and Adler, 4th ed., p. 249)

Component parts of a CT gantry includehigh-voltage generatormultidetector arrayx-ray tube A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DA CT imaging system has three component parts—a gantry, a computer, and an operating console. The gantry component includes an x-ray tube, a detector array, a high-voltage generator, a collimator assembly, and a patient couch with its motorized mechanism. Although the CT x-ray tube is similar to direct-projection x-ray tubes, it has several special requirements. The CT x-ray tube must have a very high short-exposure rating and must be capable of tolerating several million heat units while still having a small focal spot for optimal resolution. To help tolerate the very high production of heat units, the anode must be capable of high-speed rotation. The x-ray tube produces a pulsed x-ray beam (1-5 ms) using up to about 1,000 mA. The scintillation detector array is made of thousands of solid-state photodiodes. These scintillation crystal photodiode assemblies (cadmium tungstate or rare earth oxide ceramic crystals) convert the transmitted x-ray energy into light. The light then is converted into electrical energy and finally into an electronic/digital signal. If the scintillation crystals are packed tightly together so that there is virtually no distance between them, efficiency of x-ray absorption is increased, and patient dose is decreased. Detection efficiency is extremely high—approximately 90%. The high-voltage generator provides high-frequency power to the CT x-ray tube, enabling the high-speed anode rotation and the production of high-energy pulsed x-ray photons. Similar to the high-frequency x-ray tubes used in projection radiography, conventional 60-Hz full-wave-rectified power is converted to a higher frequency of 500 to 25,000 Hz. The high-frequency generator is small in size, in addition to producing an almost constant potential waveform. The CT high-frequency generator is often mounted in the gantry's rotating wheel. The collimator assembly has two parts: The prepatient, or predetector, collimator is at the x-ray tube and consists of multiple beam restrictions so that the x-ray beam diverges little. This reduces patient dose and reduces the production of scattered radiation, thereby improving the CT image. The postpatient collimator, or predetector collimator, confines the exit photons before they reach the detector array and determines slice thickness. The patient table, or couch, provides positioning support for the patient. Its motorized movement should be smooth and accurate. Inaccurate indexing can result in missed anatomy and/or double-exposed anatomy. (Bushong, 8th ed., pp. 429-430; Bontrager and Lampignano, 6th ed., p. 731)

X-ray tubes used in CT differ from those used in x-ray, in that CT x-ray tubes musthave a very high short-exposure ratingbe capable of tolerating several million heat unitshave a small focal spot for optimal resolution A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DA CT imaging system has three component parts—a gantry, a computer, and an operating console. The gantry component includes an x-ray tube, a detector array, a high-voltage generator, a collimator assembly, and a patient couch with its motorized mechanism. While the CT x-ray tube is similar to direct-projection x-ray tubes, it has several special requirements. The CT x-ray tube must have a very high short-exposure rating and must be capable of tolerating several million heat units while still having a small focal spot for optimal resolution. To help tolerate the very high production of heat units, the anode must be capable of high-speed rotation. The x-ray tube produces a pulsed x-ray beam (1-5 ms) using up to about 1,000 mA. (Bushong, 8th ed., pp. 429-430; Bontrager and Lampignano, 6th ed., p. 731)

A QC program serves tokeep patient dose to a minimumkeep radiographic quality consistentensure equipment efficiency A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: DA QC program includes regular overseeing of all components of the imaging system—instrumentation and equipment. With regular maintenance, testing, and repairs, equipment should operate efficiently and consistently. In turn, image quality will be consistent, and repeat exposures will be minimized, thereby reducing patient exposure. (Bushong 11th ed., p. 344)

In conventional fluoroscopy, all of the following are methods of permanently recording the images, except: A Cassette-loaded spot film B Photospot camera film C Cine film D Flat panel image receptor

The Correct Answer is: DA flat panel image receptor (FPIR) (D) composed of cesium iodide and amorphous silicon pixel detectors can be used in place of an image intensifier in digital fluoroscopy for real-time imaging. Images created from this device are digitized and therefore can be stored in a PACS, but this device is not considered a recording system in itself. It only generates the image to be recorded. A cassette-loaded spot film (A) is positioned in a lead-lined compartment between the patient and the image intensifier. When a spot-film exposure is desired, the radiologist must actuate a control that properly positions the cassette in the X-ray beam and changes the operation of the X-ray tube from low fluoroscopic milliamperes (mA) to high radiographic mA, at which time the rotating anode is energized to a higher rotation speed. Photospot camera film (B) is similar to a movie camera except only one frame is exposed when activated. This film receives its light image from the output phosphor of the image intensifier tube and therefore requires less patient exposure than that required when using the cassette-loaded spot film image recording method. Cine film is almost exclusively used in cardiac catheterization fluoroscopic procedures. Cine film (C) typically comes in 35 mm rolls of 100 and 500 feet in length and is exposed by the light from the output phosphor of the image intensifier tube, similar to that of the photospot camera film, but while rapidly moving to expose each frame of the film strip. The exposed frames can then be played back as a continuous strip of images to produce a dynamic reproduction of the fluoroscopic images, similar to how one would draw various, slightly different images, on the same spot on multiple blank pieces of paper, and then flip these pieces of paper rapidly to produce what appears to be a moving image. Because of the rapid transition to digital imaging, the use of cine film is rapidly declining. (Bushong, 9 th ed., pp. 441-442).

How is source-to-image distance (SID) related to exposure rate and receptor exposure? A As SID increases, exposure rate increases and receptor exposure increases. B As SID increases, exposure rate increases and receptor exposure decreases. C As SID increases, exposure rate decreases and receptor exposure increases. D As SID increases, exposure rate decreases and receptor exposure decreases.

The Correct Answer is: DAccording to the inverse-square law of radiation, the intensity or exposure rate of radiation from its source is inversely proportional to the square of the distance. Thus, as distance from the source of radiation increases, exposure rate decreases. Because exposure rate and receptor exposure are directly proportional, if the exposure rate of a beam directed to the IR is decreased, the resulting receptor exposure would be decreased proportionally. (Selman, 9th ed., p. 117)

How is SID related to exposure rate and receptor exposure? A As SID increases, exposure rate increases and radiographic receptor exposure increases. B As SID increases, exposure rate increases and radiographic receptor exposure decreases. C As SID increases, exposure rate decreases and radiographic receptor exposure increases. D As SID increases, exposure rate decreases and radiographic receptor exposure decreases.

The Correct Answer is: DAccording to the inverse-square law of radiation, the intensity or exposure rate of radiation is inversely proportional to the square of the distance from its source. Thus, as distance from the source of radiation increases, exposure rate decreases. Because exposure rate and receptor exposure are directly proportional, if the exposure rate of a beam directed to an IR is decreased, the resulting receptor exposure would be decreased proportionately. (Selman, 9th ed., p. 117)

Which of the following can affect radiographic contrast?1.LUT2.Pathology3.OID A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: DAll three factors can affect radiographic contrast. The look up table (LUT) can alter the contrast. Since pathology can alter the degree of attenuation of the x-ray beam, it can affect contrast. The type of pathology will determine how contrast is affected. An additive pathology such as Paget's disease will increase contrast, while a destructive disease such as osteoporosis will decrease contrast. OID can affect contrast when it is used as an air gap. If a 6-inch air gap (OID) is introduced between the part and the IR, much of the scattered radiation emitted from the body will not reach the IR; the air gap thus acts as a grid and increases image contrast. (Carlton & Adler, pp 397-398)

Which of the following pathologic conditions require(s) a decrease in exposure factors?PneumothoraxEmphysemaMultiple myeloma A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DAll three pathologic conditions involve processes that render tissues more easily penetrated by the x-ray beam. Pneumothorax is a collection of air or gas in the pleural cavity. Emphysema is a chronic pulmonary disease characterized by an increase in the size of the air-containing terminal bronchioles. These two conditions add air to the tissues, making them more easily penetrated. Multiple myeloma is a condition characterized by infiltration and destruction of bone and marrow. Each of these conditions requires that factors be decreased from the normal to avoid overexposure. (Carlton and Adler, 4th ed., pp. 251-252)

Which of the following is (are) characteristics of the x-ray tube?The target material should have a high atomic number and a high melting point.The useful beam emerges from the port window.The cathode assembly receives both low and high voltages. A 1 only B 2 only C 1 and 2 only D 1, 2, and 3

The Correct Answer is: DAnode target material with a high atomic number produces higher-energy x-rays more efficiently. Because a great deal of heat is produced at the target, the material should have a high melting point so as to avoid damage to the target surface. Most of the x-rays generated at the focal spot are directed downward and pass through the x-ray tube's port window. The cathode filament receives low-voltage current to heat it to the point of thermionic emission. Then, high voltage is applied to drive the electrons across to the focal track. (Selman, 9th ed., p. 111)

All the following have an impact on radiographic contrast except A photon energy B grid ratio C OID D focal-spot size

The Correct Answer is: DAs photon energy increases, more penetration and greater production of scattered radiation occur, producing a longer scale of contrast. As grid ratio increases, more scattered radiation is absorbed, producing a shorter scale of contrast. As OID increases, the distance between the part and the IR acts as a grid, and consequently, less scattered radiation reaches the IR, producing a shorter scale of contrast. Focal-spot size is related only to spatial resolution. (Shephard, p. 203)

Anode angle will have an effect on theseverity of the heel effectfocal-spot sizeheat-load capacity A 1 only B 2 only C 1 and 2 only D 1, 2, and 3

The Correct Answer is: DAs the anode angle is decreased (made steeper), a larger actual focal spot may be used while still maintaining the same small effective focal spot. Because the actual focal spot is larger, it can accommodate a greater heat load. However, with steeper (smaller) anode angles, the anode heel effect is accentuated and can compromise film coverage. (Selman, 9th ed., pp. 138-139)

As the image intensifier's FOV is reduced, how is the resulting image affected?1.Magnification increases2.Brightness decreases3.Quality increases A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DAs voltage is applied to the electrostatic focusing lenses, the focal point moves back—closer to the input phosphor—and, as a result, the FOV decreases and magnification increases. At the same time, brightness is decreased requiring an increase in mA (therefore increased patient dose). This increase in mA increases image quality—it can be likened to an increase in signal-to-noise ratio (SNR), with mA being the signal. (Selman, 9th ed, pp 263-264)

A type of laser used in CR scanners is A Cesium-iodide B Helium-halide C Barium-fluorohalide D Helium-neon

The Correct Answer is: DCesium iodide (A) is used in the scintillation layer of an indirect flat-panel digital detector (FPD). Helium halide (B) is not used in either computed or digital radiography. Barium fluorohalide (C) is a phosphor used in the CR PSPs which are housed within the image plate (IP). Energy is stored in a PSP plate after X-ray exposure and is then released in the CR reader when stimulated by a helium-neon laser (D) beam striking it in a raster pattern (transversely across the plate). In some newer units, solid-state laser diodes may be used to achieve the same purpose. (Seeram p. 54; Shephard p. 327; Carter and Veale p. 70)

Which of the following may be used to reduce the effect of scattered radiation on the radiographic image?GridsCollimatorsCompression bands A 1 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DCollimators restrict the size of the irradiated field, thereby limiting the volume of irradiated tissue, and hence less scattered radiation is produced that could ultimately degrade the image. Once radiation has scattered and emerged from the body, it can be trapped by the grid's lead strips. Grids effectively remove much of the scattered radiation in the remnant beam before it reaches the IR. Compression can be applied to reduce the effect of excessive fatty tissue (e.g., in the abdomen), in effect reducing the thickness of the part to be radiographed. (Selman, 9th ed., pp. 234, 247, 289)

If exposure factors of 85 kVp, 400 mA, and 12 ms yield an output exposure of 150 mR, what is the milliroentgens per milliampere-seconds (mR/mAs)? A 0.32 B 3.1 C 17.6 D 31

The Correct Answer is: DDetermining milliroentgens per milliampere-seconds output is often done to determine linearity among x-ray machines. However, all the equipment being compared must be of the same type (e.g., all single-phase or all three-phase, six-pulse). If there is linearity among these machines, then identical technique charts can be used. In the example given, 400 mA and 12 ms were used, equaling 4.8 mAs. If the output for 4.8 mAs was 150 mR, then 1 mAs is equal to 31.25 mR (150 mR ÷ 4.8 mAs = 31.25 mR/mAs). (Bushong, 8th ed., pp. 248-249)

The effect described as differential absorption isresponsible for radiographic contrasta result of attenuating characteristics of tissueminimized by the use of a high peak kilovoltage A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: DDifferential absorption refers to the x-ray absorption characteristics of neighboring anatomic structures. The radiographic representation of these structures is referred to as radiographic contrast; it may be enhanced with high-contrast technical factors, especially using low kilovoltage levels. At low-kilovoltage levels, the photoelectric effect predominates and differential absorption is emphasized. High kV actors provide more uniform penetration of all tissues and minimizes the effect of differential absorption. (Bushong, 8th ed., pp. 181-184)

Which of the following is (are) tested as part of a quality control (QC) program?Beam alignmentReproducibilityLinearity A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: DEach of the three is included in a good QC program. Beam alignment must be accurate to within 2% of the SID. Reproducibility means that repeated exposures at a given technique must provide consistent intensity. Linearity means that a given milliampere-second setting, using different milliampere stations with appropriate exposure-time adjustments, will provide consistent intensity. (Bushong, 8th ed., p. 460)

All the following affect the exposure rate of the primary beam except A milliamperage B kilovoltage C distance D field size

The Correct Answer is: DExposure rate is regulated by milliamperage. Distance significantly affects the exposure rate according to the inverse-square law of radiation. Kilovoltage also has an effect on exposure rate because an increase in kilovoltage will increase the number of high-energy photons produced at the target. The size of the x-ray field determines the volume of tissue irradiated, and hence the amount of scattered radiation generated, but is unrelated to the exposure rate. (Selman, 9th ed., p. 117)

Factor(s) that impact receptor exposure include1.milliamperage.2.exposure time.3.kilovoltage. A 1 only B 2 only C 1 and 2 only D 1, 2, and 3

The Correct Answer is: DFactors that regulate the number of x-ray photons produced at the target determine receptor exposure, namely milliamperage and exposure time (mAs). Receptor exposure is directly proportional to mAs; if the mAs is cut in half, the receptor exposure will decrease by one-half. Although kilovoltage is usually considered to regulate radiographic contrast (in analog imaging), it may also be used to impact receptor exposure in variable-kVp techniques, according to the 15% rule. (Selman, pp 213-214)

Which of the following factors impact(s) spatial resolution?Focal spot sizeSubject motionSOD A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DFocal-spot size affects spatial resolution by its effect on focal-spot blur: The larger the focal-spot size, the greater is the blur produced. Spatial resolution is affected significantly by distance changes because of their effect on magnification. As SID increases and as OID decreases, magnification decreases and spatial resolution increases. SOD is determined by subtracting OID from SID. (Shephard, p. 215)

Geometric unsharpness is influenced by which of the following?Distance from object to imageDistance from source to objectDistance from source to image A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: DGeometric unsharpness is affected by all three factors listed. As OID increases, so does magnification. OID is directly related to magnification; i.e. as OID increases, so does magnification. Focal-object distance and SID are inversely related to magnification. As focal-object distance and SID decrease, magnification increases. (Bushong, 10th ed pp. 174-175)

If a particular grid has lead strips 0.40 mm thick, 4.0 mm high, and 0.25 mm apart, what is its grid ratio? A 8:1 B 10:1 C 12:1 D 16:1

The Correct Answer is: DGrid ratio is defined as the ratio between the height of the lead strips and the width of the distance between them (i.e., their height divided by the distance between them). If the height of the lead strips is 4.0 mm and the lead strips are 0.25 mm apart, the grid ratio must be 16:1 (4.0 divided by 0.25). The thickness of the lead strip is unrelated to grid ratio. (Selman, 9th ed., p. 236)

Which of the following factors is/are related to grid efficiency?Grid ratioNumber of lead strips per inchAmount of scatter transmitted through the grid A 1 only B 2 only C 1 and 2 only D 1, 2, and 3

The Correct Answer is: DGrid ratio is defined as the ratio of the height of the lead strips to the width of the interspace material; the higher the lead strips, the more scattered radiation they will trap and the greater is the grid's efficiency. The greater the number of lead strips per inch, the thinner and less visible they will be on the finished radiograph. The function of a grid is to absorb scattered radiation in order to improve radiographic contrast. The selectivity of a grid is determined by the amount of primary radiation transmitted through the grid divided by the amount of scattered radiation transmitted through the grid. (Selman, 9th ed., pp. 236-237)

In order to erase a CR PSP storage plate, it must be exposed to high-intensity: A Heat B X-radiation C Microwaves D Light

The Correct Answer is: DHigh intensity visible light (D) produces the wavelength energy necessary to release residual stored energy from these imaging plates. Some residual energy remains stored in the IP after it has been scanned in a CR reader. In order to prevent artifacts on successive radiographic images, it is important to rid the IP of all stored energy. To do this, a high intensity light that is brighter than the stimulating laser light is exposed to the release of any residual stored energy (signal) in the IP. Heat (A) is used in thermal printers used to print hard copy digital images. X-radiation (B) would deposit energy within the image plate, rendering it useless for subsequent diagnostic radiographic exposures. Microwaves (C) are not used as an energy source to erase CR image plates. (Seeram p. 56; Shephard p. 329)

Which of the following is (are) characteristic(s) of a 16:1 grid?1.It absorbs a high percentage of scattered radiation.2.It has little positioning latitude.3.It is used with high-kVp exposures. A 1 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DHigh-kilovoltage exposures produce large amounts of scattered radiation, and therefore high-ratio grids are used in an effort to trap more of this scattered radiation. However, accurate centering and positioning become more critical to avoid grid cutoff. (Selman, p 243)

In which of the following ways might higher image contrast be obtained in abdominal radiography?1.By using lower kilovoltage2.By using a contrast medium3.By limiting the field size A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: DHigher/shorter scale contrast has few shades of gray between white and black. It is partly a result of lower energy photons (lower kVp). High contrast is also more likely when imaging parts having high tissue contrast, such as the chest. The abdomen has low tissue contrast, and abdominal radiographs can exhibit very low contrast unless technical factors are selected to increase contrast. To produce higher contrast in abdominal radiography, lower kVp can be used. To better demonstrate high contrast within a viscus, a contrast medium such as barium, iodine, or air can be used. Restricting the size of the field will also function to increase contrast because less scattered radiation will be generated. (Carlton & Adler, p 397)

In digital imaging, artifacts arise from a number of sources, including which of the following? A Imaging hardware B Image processing software C Operator error artifacts D All of these may be sources of image artifacts

The Correct Answer is: DIn digital imaging, artifacts arise from a number of sources. Imaging hardware artifacts include aged, cracked phosphor storage plates and mishandled and poorly cared for IPs. Image processing software artifacts can arise from incorrectly selected processing algorithms or from exposure field recognition issues from improper collimation, positioning, or sizing. Operator error artifacts can arise from incorrectly stored IPs, incorrect use of equipment, inaccurate selection of factors, etc. (Seeram, p. 67, 68)

What pixel size has a 2048 × 2048 matrix with an 80-cm FOV? A 0.04 mm B 0.08 mm C 0.2 mm D 0.4 mm

The Correct Answer is: DIn digital imaging, pixel size is determined by dividing the field of view (FOV) by the matrix. In this case the FOV is 80 cm; since the answer is expressed in mm, first change 80 cm to 800 mm. Then 800 divided by 2048 equals 0.4 mm. 80 cm = 800 mm The FOV and matrix size are independent of one another, that is, either can be changed and the other will remain unaffected. However, pixel size is affected by changes in either the FOV or matrix size. For example, if the matrix size is increased, pixel size decreases. If FOV increases, pixel size increases. Pixel size is inversely related to resolution. As pixel size increases, resolution decreases. (Fosbinder & Kelsey, p 285)

All of the following are daily quality control (QC) steps to ensure optimal diagnostic images on a digital display monitor, except: A Turn on the monitor and allow it to warm up B Evaluate luminance, reflection, noise and glare C Make sure that the monitor is dust-free on the viewing surface D Evaluate the QC patterns at the periphery and verify that all letters and numbers appear

The Correct Answer is: DIn order to ensure optimal diagnostic images on a digital display monitor, the QC pattern should be evaluated at the center and corners to verify that all letters and numbers can be visualized, not simply the periphery as (D) states. Turning on the monitor and allowing it to warm up (A) prior to evaluating images is an important step in daily quality control efforts. Evaluating the luminance, reflection, noise and glare (B) at the beginning of each day is important to ensure optimal image clarity for the radiologists and other physicians. It is important that the monitor viewing surface is dust-free (C) in order to provide a clear optical evaluation. (Carter and Veale, 1 st ed., pp. 199-200).

Underexposure of a radiograph can be caused by all the following except insufficient A milliamperage (mA) B exposure time C Kilovoltage D SID

The Correct Answer is: DInsufficient milliamperage and/or exposure time will result in decreased receptor exposure. Insufficient kilovoltage will result in underpenetration and decreased receptor exposure. Insufficient SID, however, will result in increased exposure rate and overexposure of the IR. (Carlton, Adler and Balac, 6th ed., p. 164)

In order to avoid background radiation artifacts when using CR, it is important to: A Erase all image plates that have not been used for 48 hours B Erase all image plates if there is any question about how long it has been since the plate has been erased C Erase an image plate if there is any doubt as to when it was last erased, especially in the case of pediatric radiography D All of these are correct actions

The Correct Answer is: DIt is important to be aware of the necessity of erasing image plates that have not been used for 24 hours. If there is any question about how long it has been since the plate has been through the read/erase cycle, one should erase the plate, especially if pediatric images are being performed. One should also be aware if images suddenly begin to exhibit low contrast, because the erasure system may have failed (D). (Bushong, 9 th ed., p. 487; Seeram, p. 96).

Magnification fluoroscopy provides: A Decreased resolution and decreased patient dose B Decreased resolution and increased patient dose C Increased resolution and decreased patient dose D Increased resolution and increased patient dose

The Correct Answer is: DMagnification fluoroscopy requires that a multifield image intensifier be used to allow reduction of the x-ray field size to the input phosphor area. Smaller input phosphor field sizes produce magnified images of the anatomical areas being evaluated at the output phosphor. Magnification fluoroscopy provides increased resolution, but at the expense of increased patient dosage (D). In fact, the increase in dosage is about 2.2 times that used in the full-field operation mode. The magnification mode should therefore be used only when necessary to enhance diagnostic interpretation of small anatomical areas in question (e.g., the gallbladder or duodenal bulb). (Seeram, p. 135).

Using a 48-in. SID, how much OID must be introduced to magnify an object two times? A 8-in. OID B 12-in. OID C 16-in. OID D 24-in. OID

The Correct Answer is: DMagnification radiography may be used to delineate a suspected hairline fracture or to enlarge tiny, contrast-filled blood vessels. It also has application in mammography. To magnify an object to twice its actual size, the part must be placed midway between the focal spot and the IR. (Selman, pp. 223-225; Shephard, pp. 229-231)

Too much edge enhancement of the image in digital systems can cause an unwanted artifact called the: A Photoelectric effect artifact B Scaling artifact C Hounsfield artifact D Halo effect artifact

The Correct Answer is: DOne of the problems seen with too much edge enhancement is an effect called the "halo" effect (D). This effect can cause loss of anatomical information and artifacts that may interfere with proper diagnoses. The photoelectric effect (A) occurs when an X-ray photon interacts with an inner shell electron of an atom and its energy is completely absorbed by the atom. This interaction is responsible for producing diagnostic information in a radiographic image and contributing to patient dosage. A scaling artifact is not an artifact seen in digital imaging (B). A Hounsfield artifact is not an artifact seen in digital imaging (C). (Seeram p. 120)

Changes in milliampere-seconds can affect all the following except A quantity of x-ray photons produced B exposure rate C receptor exposure D spatial resolution

The Correct Answer is: DMilliampere-seconds (mAs) are the product of milliamperes (mA) and exposure time (seconds). Any combinations of milliamperes and time that will produce a given milliampere-seconds value will produce identical receptor exposure. This is known as the reciprocity law. The milliampere-seconds value is a quantitative factor. The mAs is directly proportional to x-ray-beam intensity, exposure rate, quantity, or number of x-ray photons produced and patient dose. If the mAs value is doubled, twice the receptor exposure and twice the patients dose. If the milliampere-seconds value is cut in half, the receptor exposure and patient dose are cut in half. The milliampere-seconds value has no effect on spatial resolution. (Shephard, p. 170)

The batteries in battery-operated mobile x-ray units provide power tothe x-ray tubemachine locomotionthe braking mechanism A 1 only B 2 only C 1 and 2 only D 1, 2, and 3

The Correct Answer is: DMobile x-ray machines are smaller and more compact than their fixed counterparts in the radiology department. It is important that they be relatively easy to move, that their size allows entry into patient rooms, and that their locks enable securing of the x-ray tube into the required positions. Mobile x-ray machines are cordless and are either the battery-operated type or the condenser-discharge type. The battery-operated type is probably the most commonly used where consistent and high-energy output is required. Two sets of batteries are used in these mobile units: One set is used for operating the motor that drives the unit and operates the braking mechanism ("dead man" brake), and the other set is used for operating the x-ray tube. Periodic recharging of the batteries is required. (Frank, Long, and Smith, 11th ed., vol. 3, p. 234)

Using a multifield image intensifier tube, which of the following input phosphor diameters will provide the best spatial resolution? A 35 cm B 25 cm C 17 cm D 12 cm

The Correct Answer is: DMultifield image intensifier tubes are usually either dual-field or tri-field and are designed this way in order to permit magnification imaging. As voltage is applied to the electrostatic focusing lenses, the focal point moves back—closer to the input phosphor—and a smaller portion of the input phosphor is utilized. As a result, the FOV decreases and magnification increases, producing better spatial resolution. At the same time, brightness is decreased requiring an increase in mA (therefore increased patient dose). This increase in mA increases image quality. It can be likened to an increase in signal-to-noise ratio (SNR), with mA being the signal. (Seeram, p 103)

Using a multifield image intensifier tube, which of the following input phosphor diameters will provide the greatest magnification? A 35 cm B 25 cm C 17 cm D 12 cm

The Correct Answer is: DMultifield image intensifier tubes are usually either dual-field or tri-field and are designed this way in order to permit magnification imaging. As voltage is applied to the electrostatic focusing lenses, the focal point moves back—closer to the input phosphor—and a smaller portion of the input phosphor is utilized. As a result, the FOV decreases and magnification increases, producing better spatial resolution. At the same time, brightness is decreased requiring an increase in mA (therefore increased patient dose). This increase in mA increases image quality. It can be likened to an increase in signal-to-noise ratio (SNR), with mA being the signal. (Seeram, p 103)

Using a multifield image intensifier tube, which of the following input phosphor diameters will require the highest patient dose? A 35 cm B 25 cm C 17 cm D 12 cm

The Correct Answer is: DMultifield image intensifier tubes are usually either dual-field or tri-field and are designed this way in order to permit magnification imaging. As voltage is applied to the electrostatic focusing lenses, the focal point moves back—closer to the input phosphor—and a smaller portion of the input phosphor is utilized. As a result, the FOV decreases and magnification increases, producing better spatial resolution. At the same time, brightness is decreased requiring an increase in mA (therefore increased patient dose). This increase in mA increases image quality. It can be likened to an increase in signal-to-noise ratio (SNR), with mA being the signal. (Seeram, p 103)

One advantage of a battery-powered mobile radiographic unit is: A It requires less kilovoltage to penetrate the anatomical part of interest B It produces radiographic images of much better image quality C It is much lighter than other mobile units D Electrical power is available to drive itself

The Correct Answer is: DOne advantage of a battery-powered unit is that electrical power is available to drive itself (D). Some of the stored electrical power resulting from charging the unit can be used to operate the motor that propels the unit to the patient's room. Since this unit is self-driven, the radiographer must be especially careful when driving the unit down a hallway or around corners to avoid injury to others and structures. The kilovoltage (A) produced with battery-operated mobile radiography units is similar to that of other mobile units. The radiographer is responsible for selecting an adequate kilovoltage necessary to penetrate the anatomical part of interest. The battery-operated unit produces X-ray beams and image quality (B) similar to that produced with other mobile radiographic units. The quality of the radiographic image depends on the radiographer's control settings, positioning, X-ray beam alignment, and source distance, as it does when using any other mobile unit. Since the battery-operated mobile units contain several large batteries, the unit is very heavy (C). Care must be taken when driving these units to ensure the safety of others and to avoid damaging structures within the facility. (Carlton and Adler, 5 th ed., p. 528).

Which of the following possesses the widest dynamic range? A ALARA B PBL C AEC D CR

The Correct Answer is: DOne of the biggest advantages of CR is the dynamic range, or latitude, it offers. In CR, there is a linear relationship between the exposure given the PSP and its resulting luminescence as it is scanned by the laser. This affords much greater exposure latitude, and technical inaccuracies can be effectively eliminated. Overexposure of up to 500% and underexposure of up to 80% are reported as recoverable, thus eliminating most retakes. This surely affords increased efficiency; however, this does not mean that images can be exposed arbitrarily. The radiographer must keep dose reduction in mind. AEC refers to automatic exposure control and is unrelated to dynamic range or latitude. PBL refers to positive beam limitation and is unrelated to dynamic range or latitude. ALARA is a radiation protection concept referring to keeping occupational dose as low as reasonably achievable.

A patient is being positioned for a particular radiographic examination. The x-ray tube, image recorder, and grid are properly aligned, but the body part is angled. Which of the following will result? A Grid cutoff at the periphery of the image B Grid cutoff along the center of the image C Increased receptor exposure at the periphery D Image distortion

The Correct Answer is: DProper alignment of the x-ray tube, body part, and image recorder is required to avoid image distortion in the form of foreshortening or elongation. Foreshortening will usually result when the part is out of alignment. Elongation is often a result of angulation of the x-ray tube. Grid lines or grid cutoff will occur when the grid itself is off-center or not in alignment with the x-ray tube. Grid lines/grid cut off indicates absorption of the useful beam by the misaligned grid. (Shephard, pp 228-231)

Which of the following is (are) correct regarding care of protective leaded apparel?Lead aprons should be fluoroscoped yearly to check for cracks.Lead gloves should be fluoroscoped yearly to check for cracks.Lead aprons should be hung on appropriate racks when not in use. A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: DProper care of leaded protective apparel is required to ensure its continued usefulness. If lead aprons and gloves are folded, cracks will develop, and this will decrease their effectiveness. Both items should be fluoroscoped annually to check for the formation of cracks. (Bushong, 8th ed., pp. 596, 597)

Which of the following combinations would pose the least hazard to a particular anode? A 0.6-mm focal spot, 75 kVp, 30 mAs B 0.6-mm focal spot, 85 kVp, 15 mAs C 1.2-mm focal spot, 75 kVp, 30 mAs D 1.2-mm focal spot, 85 kVp, 15 mAs

The Correct Answer is: DRadiographic rating charts enable the operator to determine the maximum safe milliamperage, exposure time, and peak kilovoltage for a particular exposure using a particular x-ray tube. An exposure that can be made safely with the large focal spot may not be safe for use with the small focal spot of the same x-ray tube. The total number of heat units that an exposure generates also influences the amount of stress (in the form of heat) imparted to the anode. The product of milliampere-second and peak kilovolts determines HU. Groups (A) and (C) produce 2250 HU; groups (B) and (D) produce 1275 HU. Groups (B) and (D) deliver less heat load, but group (D) delivers it to a larger area (actual focal spot), making this the least hazardous group of technical factors. The most hazardous group of technical factors is group (A). (Selman, 9th ed., p. 145)

To be used more efficiently by the x-ray tube, alternating current is changed to unidirectional current by the A filament transformer. B autotransformer. C high-voltage transformer. D rectifiers.

The Correct Answer is: DRectifiers (solid-state or the older valve tubes) permit the flow of current in only one direction. They serve to change AC, which is needed in the low-voltage side of the x-ray circuit, to unidirectional current. Unidirectional current is necessary for the efficient operation of the x-ray tube. The rectification system is located between the secondary coil of the high-voltage transformer and the x-ray tube. The filament transformer functions to adjust the voltage and current going to heat the x-ray tube filament. The autotransformer varies the amount of voltage being sent to the primary coil of the high-voltage transformer so that the appropriate kVp can be obtained. The high-voltage transformer "steps up" the voltage to the required kilovoltage and steps down the amperage to milliamperage. (Carlton & Adler, p 78)

Which combination of exposure factors most likely will contribute to producing the shortest-scale contrast? A mAs: 10; kV: 70; Grid ratio: 5:1; Field size: 14 × 17 in. B mAs: 12; kV: 90; Grid ratio: 8:1; Field size: 14 × 17 in. C mAs: 15; kV: 90; Grid ratio: 12:1; Field size: 11 × 14 in. D mAs: 20; kV: 80; Grid ratio: 10:1; Field size: 8 × 10 in.

The Correct Answer is: DReview the groups of factors. First, because the milliampere-seconds value has no effect on the scale of contrast produced, eliminate milliampere-seconds from consideration by drawing a line through the column. Then check the two entries in each column that are likely to produce shorter-scale contrast. For example, in the kilovoltage column, because lower kilovoltage can produce shorter-scale contrast, place checkmarks next to the 70 and 80 kV. Because higher-ratio grids permit less scattered radiation to reach the IR, the 10:1 and 12:1 grids can produce a shorter scale of contrast than the lower-ratio grids; check them. As the volume of irradiated tissue decreases, so does the amount of scattered radiation produced, and consequently, the shorter is the scale of radiographic contrast; therefore, check the 11 × 14 and 8 × 10 in. field sizes. An overview shows that the factors in groups (A) and (C) have more checkmarks, than the factors in group (D), indicating that group (D) is more likely to produce the shortest-scale contrast. (Shephard, pp. 306-308)

Which of the following is (are) associated with subject contrast?Patient thicknessTissue densityKilovoltage A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: DSeveral factors influence subject contrast, each as a result of beam-attenuation differences in the irradiated tissues. As patient thickness and tissue density increase, attenuation increases, and subject contrast is increased. As kilovoltage increases, a wider range and higher-energy photons is produced, attenuation differential is reduced and subject contrast decreases. (Carlton and Adler, 6th ed., p. 216, 373)

What is the single most important factor controlling size distortion? A Tube, part, IR alignment B IR dimensions C SID D OID

The Correct Answer is: DShape distortion (foreshortening, elongation) is caused by improper alignment of the tube, part, and image receptor. Size distortion, or magnification, is caused by too great an object-image distance or too short a source-image distance. OID is the primary factor influencing magnification, followed by SID. (Bushong, 8th ed, p 284)

Shape distortion is influenced by the relationship between thex-ray tube and the part to be imaged.body part to be imaged and the IR.IR and the x-ray tube. A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DShape distortion is caused by misalignment of the x-ray tube, the body part to be radiographed, and the IR. An object can be falsely imaged (foreshortened or elongated) as a result of incorrect placement of the tube, the part, or the IR. Only one of the three need be misaligned for distortion to occur. (Selman, 9th ed., pp. 225-226)

An exposure was made using 300 mA and 50 ms. If the exposure time is changed to 22 ms, which of the following milliamperage selections would most closely approximate the original receptor exposure? A 300 mA B 400 mA C 600 mA D 700 mA

The Correct Answer is: DSince 50 ms is equal to 0.050 s, and since mA × time mAs, the original milliampere-seconds value was 15 mAs. Now, it is only necessary to determine what milliamperage must be used with 22 ms (0.022 sec) to provide the same 15 mAs (and thus the same receptor exposure). 0.022 x = 15; x = 681 mA, (Selman, 9th ed., p. 214)

If 84 kV and 8 mAs were used for a particular abdominal exposure with single-phase equipment, what milliampere-seconds value would be required to produce a similar radiograph with three-phase, 12-pulse equipment? A 24 mAs B 16 mAs C 8 mAs D 4 mAs

The Correct Answer is: DSingle-phase radiographic equipment is much less efficient than three-phase equipment because it has a 100% voltage ripple. With three-phase equipment, voltage never drops to zero, and x-ray intensity is significantly greater. To produce similar receptor exposure, only two-thirds of the original milliampere-seconds would be used for three-phase, six-pulse equipment ( 2 / 3 × 8 = 5.3 mAs). With three-phase, 12-pulse equipment, the original milliampere-seconds would be cut in half ( 1 / 2 × 8 = 4 mAs). (Bushong, 8th ed., p. 124)

The long axis of the laser beam moving transversely back and forth across the image plate in a CR reader is called the: A Scan/translation mode B Zig-zag scan mode C Slow scan direction D Fast scan direction

The Correct Answer is: DSlow scan direction (C) speed refers to the linear travel speed of the image plate through the CR reader. The IP moves slowly through the transport system of a CR reader and this movement is considered the slow-scan direction. The laser light in the reader is rapidly reflected by an oscillating polygonal mirror that redirects the beam through a special lens called the f-theta lens, which focuses the light on a cylindrical mirror that reflects the light toward the IP. This light moves back and forth very rapidly to scan the plate transversely, in a raster pattern, and this movement of the laser beam across the IP is therefore called the fast-scan direction (D). Scan/translation mode (A) and Zig-zag mode (B) are not terms used to describe the laser beam movement back and forth across the image plate while it travels through the CR reader (A). (Seeram p. 54; Shephard p. 327; Carter and Veale p. 70)

Dedicated radiographic units are available for1.chest radiography.2.dental radiography.3.mammography. A 1 only B 2 only C 1 and 2 only D 1, 2, and 3

The Correct Answer is: DSpecial units have been designed to accommodate examinations with high patient volume. Dedicated chest units are available with high frequency generator and digital flat-panel detector. Dedicated head units are available for cone beam and panoramic digital dental imaging. High-quality mammographic examinations are available with dedicated mammographic units having molybdenum or rhodium target material and other beneficial features.

A technique chart should be prepared for each AEC x-ray unit and should contain which of the following information for each type of examination?Photocell(s) usedOptimum kilovoltageBackup time A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DThe AEC automatically adjusts the exposure required for adjacent body tissues/parts that have different thicknesses and tissue densities. Proper functioning of the AEC (phototimer or ionization chamber) depends on accurate positioning by the radiographer. The correct photocell(s) must be selected, and the anatomic part of interest must completely cover the photocell to achieve the desired receptor exposure. If collimation is inadequate and a field size larger than the part is used, excessive scattered radiation from the body or tabletop can cause the AEC to terminate the exposure prematurely, resulting in an underexposed image. Backup time always should be selected on the manual timer to prevent patient overexposure and to protect the x-ray tube from excessive heat production should the AEC malfunction. Selection of the optimal kilovoltage for the part being radiographed is essential—no practical amount of milliampere-seconds can make up for inadequate penetration (kilovoltage), and excessive kilovoltage can cause the AEC to terminate the exposure prematurely. A technique chart, therefore, is strongly recommended for use with AEC; it should indicate the optimal kilovoltage for the part, the photocells that should be selected, and the backup time that should be set. (Carlton and Adler, 4th ed., p. 540)

Which of the following conditions will require an increase in x-ray photon energy/penetration? A Fibrosarcoma B Osteomalacia C Paralytic ileus D Ascites

The Correct Answer is: DThe ability of x-ray photons to penetrate a body part has a great deal to do with the composition of that part (e.g., bone vs. soft tissue vs. air) and the presence of any pathologic condition. Pathologic conditions can alter the normal nature of the anatomic part. Some conditions, such as osteomalacia, fibrosarcoma, and paralytic ileus (obstruction), result in a decrease in body tissue density. When body tissue density decreases, x-rays will penetrate the tissues more readily; that is, there is more x-ray penetrability. In conditions such as ascites, where body tissue density increases as a result of the accumulation of fluid, x-rays will not readily penetrate the body tissues; that is, there is less x-ray penetrability. (Carlton and Adler, 4th ed., p. 250)

The advantage(s) of high-kilovoltage chest radiography is (are) thatexposure latitude is increasedit produces long-scale contrastit reduces patient dose A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DThe chest is composed of tissues with widely differing densities (bone and air). In an effort to "even out" these tissue densities and better visualize pulmonary vascular markings, high kilovoltage generally is used. This produces more uniform penetration and results in a longer scale of contrast with visualization of the pulmonary vascular markings as well as bone (which is better penetrated) and air densities. The increased kilovoltage also affords the advantage of greater exposure latitude (an error of a few kilovolts will make little, if any, difference). The fact that the kilovoltage is increased means that the milliampere-seconds value is reduced accordingly, and thus patient dose is reduced as well. A grid usually is used whenever high kilovoltage is required. (Carlton and Adler, 4th ed., pp. 423-424)

Which of the following can affect histogram appearance?Centering accuracyPositioning accuracyProcessing algorithm accuracy A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DThe computed radiography (CR) laser scanner recognizes the various tissue-density values and constructs a representative grayscale histogram. A histogram is a graphic representation showing the distribution of pixel values. Histogram analysis and use of the appropriate LUT together function to produce predictable image quality in CR. Histogram appearance can be affected by a number of things. Degree of accuracy in positioning and centering can have a significant effect on histogram appearance (as well as patient dose). Change is effected in average exposure level and exposure latitude; these changes will be reflected in the images informational numbers (i.e., S number and exposure index). Other factors affecting histogram appearance, and therefore these informational numbers, include selection of the correct processing algorithm (e.g., chest vs. femur vs. cervical spine) and changes in scatter, SID, OID, and collimation. Figure 7-21 illustrates the effect of incorrect collimation on histogram appearance—in short, anything that affects scatter and/or dose. (Carlton and Adler, 4th ed., pp. 361-362)

Which of the following matrix sizes is most likely to produce the best image resolution? A 128 × 128 B 512 × 512 C 1,024 × 1,024 D 2,048 × 2,048

The Correct Answer is: DThe matrix is the number of pixels in the xy direction. The larger the matrix size, the better is the image resolution. Typical image matrix sizes used in radiography are A digital image is formed by a matrix of pixels in rows and columns. A matrix having 512 pixels in each row and column is a 512 × 512 matrix. The term field of view is used to describe how much of the patient (e.g., 150-mm diameter) is included in the matrix. The matrix or field of view can be changed without affecting the other, but changes in either will change pixel size. As in traditional radiography, spatial resolution is measured in line pairs per millimeter (lp/mm). As matrix size is increased, there are more and smaller pixels in the matrix and, therefore, improved spatial resolution. Fewer and larger pixels result in a poor-resolution "pixelly" image, that is, one in which you can actually see the individual pixel boxes. (Fosbinder and Kelsey, p. 286)

The optimal alignment of the anatomical part being imaged for all digital receptors should be: A Field centered to IP with at least two collimation margins and parallel to the IP edges B Centered anywhere on the IP, but having four distinct collimation margins, regardless of parallel orientation to the IP edges C Centered to the IP with at least one collimation margin aligned to nearest edge of IP D Field centered to IP with four collimation margins parallel to the IP edges

The Correct Answer is: DThe optimal alignment when using CR is field centered to the plate with four collimation margins parallel to the IP edges (D). Otherwise, the exposure field may not be correctly identified, resulting in a processing error. An exposure field with only two collimation margins and parallel to the IP edges (A) results in extraneous radiation exposure in the top and bottom portions of the receptor. This exposure information may cause misidentification of the exposure field, causing a processing error. Simply exposing an anatomical part anywhere on the receptor (B) has the potential to cause misidentification of the exposure field, causing a processing error. If only one collimation margin is included on the receptor (C), the radiographer has improperly centered the anatomical part. This may result in misidentification of the exposure field and therefore, cause a processing error. (Seeram, 1 st ed., p. 91).

Which of the following can contribute to the image contrast?Tissue densityPathologyMuscle development A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DThe radiographic subject (the patient) is composed of many different tissue types of varying tissue densities, resulting in varying degrees of photon attenuation and absorption. This differential absorption contributes to the various shades of gray. Normal tissue density may be significantly altered in the presence of pathology. For example, destructive bone disease can cause a dramatic decrease in tissue density. Abnormal accumulation of fluid (as in ascites) will cause a significant increase in tissue density. Muscle atrophy or highly developed muscles similarly will decrease or increase tissue density. (Shephard, p. 203)

Which of the following is likely to contribute to the radiographic contrast present on an analog x-ray image?Atomic number of tissues radiographedAny pathologic processesDegree of muscle development A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DThe radiographic subject, the patient, is composed of many different tissue types that have varying tissue densities, resulting in varying degrees of photon attenuation and absorption. The atomic number of the tissues under investigation is directly related to their attenuation coefficient. This differential absorption contributes to the various shades of gray (scale of radiographic contrast) on the radiographic image. Normal tissue density may be altered significantly in the presence of pathologic processes. For example, destructive bone disease can cause a dramatic decrease in tissue density (and subsequent increase in receptor exposure). Abnormal accumulation of fluid (as in ascites) will cause a significant increase in tissue density. Muscle atrophy or highly developed muscles similarly will decrease or increase tissue density. (Bushong, 8th ed., pp. 290, 298-299)

All the following are associated with the anode except A the line-focus principle B the heel effect C the focal track D thermionic emission

The Correct Answer is: DThe rotating anode has a target (or focal spot) on its beveled edge that forms the target angle. As the anode rotates, it constantly turns a new face to the incoming electrons; this is the focal track. The portion of the focal track that is bombarded by electrons is the actual focal spot, and because of the target's angle, the effective or projected focal spot is always smaller (line-focus principle). The anode heel effect refers to decreased beam intensity at the anode end of the x-ray beam. The electrons impinging on the target have "boiled off" the cathode filament as a result of thermionic emission. (Selman, 9th ed., pp. 138-139)

Acceptable method(s) of minimizing motion unsharpness is (are)suspended respirationshort exposure timepatient instruction A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: DThe shortest possible exposure time should be used to minimize motion unsharpness. Motion causes unsharpness that destroys spatial resolution. Careful and accurate patient instruction is essential for minimizing voluntary motion. Suspended respiration eliminates respiratory motion. Using the shortest possible exposure time is essential for decreasing involuntary motion. Immobilization is also very useful in eliminating motion unsharpness. (Carlton and Adler, 4th ed., p. 451)

Image quality in digital fluoroscopy is influenced by1.pixel size.2.contrast.3.noise. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DThe smaller the pixel size and larger the matrix, the better the image's spatial resolution. For example, an image matrix of 1024 × 1024 will provide better resolution than a matrix of 700 × 700. The 1024 × 1024 matrix has a larger number of smaller pixels, therefore a less "pixelly" image. As in analog x-ray imaging, a range of diagnostic grays representing the various tissue densities is desirable. In CR and DR the image can be manipulated (i.e., "windowed") to provide the desired scale of grays and brightness. Noise is degrading to both traditional and digital images. It can result from a number of causes including insufficient mA (i.e., signal) causing graininess/mottle, and scattered radiation fog. (Selman, 9th ed, pp 311-312)

The device used to test the accuracy of the x-ray timer is the A densitometer B sensitometer C penetrometer D spinning top

The Correct Answer is: DThe spinning-top test may be used to test timer accuracy in single-phase equipment. A spinning top is a metal disk with a small hole in its outer edge that is placed on a pedestal about 6 in. high. An exposure is made (e.g., 0.1 s) while the top spins. Because a full-wave-rectified unit produces 120 x-ray photon impulses per second, in 0.1 s the film should record 12 dots (if the timer is accurate). Because three-phase equipment produces almost constant potential rather than pulsed radiation, the standard spinning top cannot be used. An oscilloscope or synchronous spinning top must be employed to test the timers of three-phase equipment. (Selman, 9th ed., p. 106)

Geometric unsharpness will be least obviousat long SIDs.with small focal spots.at the anode end of the image. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DThe x-ray tube anode is designed according to the line-focus principle, that is, with the focal track beveled (Figure 6-24). This allows a larger actual focal spot to project a smaller effective focal spot, resulting in improved spatial resolution with less blur. However, because of the target angle, penumbral blur varies along the longitudinal tube axis, being greater at the cathode end of the image and less at the anode end of the image. Therefore, better spatial resolution will be appreciated using small focal spots at the anode end of the x-ray beam and at longer SIDs. (Bushong, 8th ed., p. 287)

One advantage of digital imaging in fluoroscopy is the ability to perform "road-mapping." Road-mapping 1. keeps the most recent fluoroscopic image on the screen. 2. aids in the placement of guidewires and catheters. 3. reduces the need for continuous x-ray exposure to the patient. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DThere are several advantages of electronic/digital fluoroscopy. Electronic/digital fluoroscopic images are produced with less patient exposure and can be postprocessed (windowed to improve/enhance the image). The fluoroscopic still-frame images can be stored and/or transmitted to a TV monitor. Another advantage is the ability to perform "road-mapping." In this procedure, the most recent fluoroscopic image is retained on the screen/monitor (last image hold) is retained on the screen/monitor. Road-mapping is particularly useful in procedures that require guidewire/catheter placement. The frame-hold function eliminates the need for continuous fluoroscopy, thereby reducing patient exposure. (Hendee and Ritenour, p. 244)

Excessive anode heating can cause vaporized tungsten to be deposited on the port window. This can result in1.decreased tube output.2.tube failure.3.electrical arcing. A 1 only B 2 only C 1 and 2 only D 1, 2, and 3

The Correct Answer is: DVaporized tungsten may be deposited on the inner surface of the glass envelope at the tube (port) window. It acts as an additional filter, thereby reducing tube output. The tungsten deposit may also attract electrons from the filament, creating arcing and causing puncture of the glass envelope and subsequent tube failure. (Selman, pp 137-138)

When using the smaller field in a dual-field image intensifier,1.a smaller patient area is viewed.2.the image is magnified.3.the image is less bright. A 1 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: DWhen a dual-field image intensifier is switched to the smaller field, the electrostatic focusing lenses are given a greater charge to focus the electron image more tightly. The focal point, then, moves further from the output phosphor (the diameter of the electron image is therefore smaller as it reaches the output phosphor), and the brightness gain is somewhat diminished. Hence, the patient area viewed is somewhat smaller and is magnified. However, the minification gain has been reduced and the image is somewhat less bright. (Bushong, 8th ed., p. 363)

When the collimated field must extend past the edge of the body, allowing primary radiation to strike the tabletop, as in a lateral lumbar spine radiograph, what may be done to prevent excessive receptor exposure owing to undercutting? A Reduce the milliampere-seconds. B Reduce the kilovoltage. C Use a shorter SID. D Use lead rubber to absorb tabletop primary radiation.

The Correct Answer is: DWhen the primary beam is restricted to an area near the periphery of the body, sometimes part of the illuminated area overhangs the edge of the body. If the exposure is then made, scattered radiation from the tabletop (where there is no absorber) will undercut the part, causing excessive receptor exposure. If, however, a lead rubber mat is placed on the overhanging illuminated area, most of this scatter will be absorbed. This is frequently helpful in lateral lumbar spine and AP shoulder radiographs. (Carlton and Adler, 4th ed., pp. 233-234)

With three-phase equipment, the voltage across the x-ray tubedrops to zero every 180 degreesis 87% to 96% of the maximum valueis at nearly constant potential A 1 only B 2 only C 1 and 2 only D 2 and 3 only

The Correct Answer is: DWith single-phase, full-wave-rectified equipment, the voltage is constantly changing from 0% to 100% of its maximum value. It drops to 0 every 180 degrees (of the AC waveform); that is, there is 100% voltage ripple. With three-phase equipment, the voltage ripple is significantly smaller. Three-phase, six-pulse equipment has a 13% voltage ripple, and three-phase, 12-pulse equipment has a 3.5% ripple. Therefore, the voltage never falls below 87% to 96.5% of its maximum value with three-phase equipment, and it closely approaches constant potential [direct current (DC)]. (Carlton and Adler, 4th ed., pp. 91-93)

Regardless of pixel size, which of the following factors influence spatial resolution? (select the two that apply) A Focal spot selection B kVp selection C Adding a grid, with a compensating change in mAs D Increasing collimation E Changing SID

The answer is A and E. Spatial resolution is the ability to see shapes, outlines, and forms clearly within an image. In the digital age, spatial resolution iscontrolledby pixel size. However, geometric factors still influence the amount of blur encoded within the remnant beam captured by the image receptor. Utilizing a large focal spot and changing SID will affect the production of blur within the signal (A and E). kVp change, grid use, and collimation change will affect the amount of scatter radiation that reaches the IR (B, C, and D). As scatter increases, noise increases on the image, and contrast resolution suffers. (Bushong, 11thed., p. 177)

The benefits of utilizing a smaller target angle include which of the following? (select the three that apply) A Improved image quality for unequal thickness anatomical parts B Decreased image unsharpness C Increased image unsharpness D Smaller effective focal spot E Larger effective focal spot

The answer is A, B, and D. Because diagnostic x-ray tube anodes are angled, the filament projectile electrons traveling from the cathode during an exposure can be spread over a larger area of the anode target (the actual focal spot) and the x-rays produced can be projected toward the patient in a narrower beam, called the effective, or projected, focal spot. Therefore, asmaller effective focal spotcan be produced as a result of the anode angle (D). This describes theline-focus principle. As the anode target angle decreases, theanode heel effectis more pronounced. However, this can be used to the radiographer's benefit when performing radiographic procedures such as an AP thoracic spine or AP femur. The anode heel attenuates some of the x-rays produced, whereas those produced toward the cathode end can more easily escape the anode. Applying this knowledge, the radiographer can position the less intense portion of the beam (anode side) over the thinnest portion of the anatomy, and the most intense portion of the beam (cathode side) over the thickest part of the anatomy. Therefore,improved image quality for anatomical parts of unequal thicknesscan be achieved, as uniform exposure over the entire part can be delivered (A).Decreased image unsharpness(increased, or improved, spatial resolution) is also realized with decreased anode angles, as penumbra is also decreased (B). The other answer choices are incorrect as per the previous explanation (C and E). (Bushong, 11thed., pp. 113-115, 178-179)

The figure's X axis illustrates the following (select the three that apply):Reproduced with permission from Saia, Radiography PREP, 9th edition. New York, NY: McGraw-Hill, 2018. Figure 12-22B. A Exposure values B Pixel value distribution C Spatial resolution D Tonal distribution E Frequency/number of pixels having each particular value

The answer is A, B, and D. The histogram is an analysis and graphic representation of the information from the IR, demonstrating the number of pixels and their value. The Y axis illustrates the frequency/number of pixels that display a particular tone/value (A and D). The X axis provides information about the tonal values of the pixels and their distribution within the image—blacks are to the left, whites are to the right (B and C). In a narrow histogram, the blacks from the left and the whites from the right are squeezed to the center of the histogram, resulting in a very gray low contrast image. In a wide histogram, the mid-histogram gray pixels have moved laterally toward the blacks and whites, resulting in a higher contrast image. (Saia, PREP 9th ed., pp. 377, 378)

Significant over- or underexposure of the digital receptor is likely to result in (select the three that apply) A quantum mottle B increased brightness C information loss D saturation E magnification F structural blur

The answer is A, C, and D. Digital processing usually compensates for over- or underexposure. An exposure can exceed the processing capabilities. Significant overexposure will result in saturation (very dark image); underexposure results in quantum mottle (A and D). In both cases, there is loss of diagnostic information (C). Magnification and structural blur are spatial resolution factors and not associated with over- or underexposure of the digital receptor (E and F). (Carroll, 2nd ed., p. 546)

Histogram analysis errors can result from (select the three that apply): A selection of incorrect anatomical part/algorithm B excessively long backup time C incorrect centering D inadequate beam restriction E use of large focal spot

The answer is A, C, and D. Factors affecting histogram appearance, and reflected in the EI, include selection of the correct processing algorithm (e.g., chest vs. femur vs. cervical spine), and changes in scatter, SID, OID, and collimation—in short, anything that affects scatter and/or dose (A, C, and D). Inadequate collimation results in signals outside the anatomic area being included in the exposure data recognition (EDR)/ histogram analysis, potentially resulting in a variety of analysis errors including excessively light, dark, or noisy images. One other factor is delay in PSP processing. A delay in processing can result in fading of the image. Histogram analysis errors are less common in DR systems where only exposed pixels are analyzed, rather than entire plate analysis as in CR systems. Backup time is related to AEC (B). Focal spot size is related to spatial resolution (E). Neither of these is related to histogram analysis. (Saia, PREP 9th ed., p. 378; Carroll, 2nd ed., p. 487)

Which of the following are the principle components of the x-ray imaging system? (select the three that apply) A X-ray tube B Patient table C Operating console D High voltage generator E Printer/CD burner

The answer is A, C, and D. The primary components of the x-ray imaging system are the x-ray tube, the high voltage generator, and the operating console (A, C, and D). While most systems come with a patient table, it is not a necessary component (B). X-ray rooms that are designated for chest imaging and U-arm equipment do not require tables for use. Additionally, most x-ray imaging systems do not directly print their images; rather, images are printed from a different unit connected to the PACS system (E).(Bushong, 9th ed., pp. 86-87)

Perhaps the simplest of beam restriction devices, a sheet of lead with a single fenestration, is also known as A an aperture diaphragm. B a fenestrated drape. C a cylinder. D a variable-aperture collimator.

The answer is A. Beam restricting devices come in three basic forms: cylinders and cones, aperture diaphragms, and variable aperture collimators (A). Cylinders and cones are lead or lead lined structures, which restrict the useful beam to a required diameter (C). Variable aperture collimators are the most common device, which feature pairs of lead shutters that can be adjusted to configure the desired length and width of the beam (D). Finally, the aperture diaphragm is, just as the question says, a sheet of lead with a hole in the middle. The hole, or fenestration, is of a fixed size. A fenestrated drape is not a beam restricting device, but is a sterile drape commonly used in surgery to frame the area of the body to be surgically repaired (B). (Bushong, 11th ed., p. 193)

Which criteria from the list below are essential to the production of x-radiation? (select the three that apply) A A source of electrons B A vacuum (absence of air molecules) enclosed in a glass tube C A means to accelerate electrons D A focusing cup, to confine the electron stream E A target material capable of causing the electrons to decelerate F A rotating target anode

The answer is A, C, and E. Essential preconditions to the production of x-radiation are the following: a source of electrons (A), a means to accelerate them (C), and a target to slow them down (E). In the modern x-ray tube, a hot tungsten filament is the most commonly used source of electrons, a high voltage transformer provides the electromotive force to pull the electrons toward the anode, and a mass of tungsten metal functions as the target. Choices (B), (D), and (F) all make the modern tube more efficient and less likely to fail, but are not necessary components in the phenomena known as bremsstrahlung. A vacuum ensures that the electrons will not collide with air molecules on the way to the target, but early Crookes tubes were partially evacuated at best. A focusing cup allows for the electrons to strike a smaller area of the anode, which reduces penumbra within the x-ray image. A lower image quality beam would still be generated in a tube without a focusing cup. Finally, while a rotating anode wheel improves the heat capacity of modern tubes, many mobile x-ray units still feature stationary anode tubes. (Bushong, 11th ed., pp. 106-111)

Which of the following statements regarding direct digital imaging are correct? (select the three that apply) A The TFT array collects and stores electrical charges B Spatial resolution decreases with an increase in fill factor C Spatial resolution is variable in direct digital D The DEL is the sensing element of the TFT E Spatial resolution increases with a decrease in DEL size F Direct digital utilizes amorphous silicon

The answer is A, D, and E. Direct conversion FPDs utilize no scintillation; x-rays are directly converted to electric signals by amorphous selenium (a-Se)—most commonly used because of its excellent x-ray absorption/conversion properties and spatial resolution (F). Below the a-Se layer is a TFT array that functions to collect and store the electrical charges, which are transmitted to the ADC for digitization (A). There is no diffusion of electrons, so spatial resolution is unaffected. The spatial resolution of direct digital systems is fixed and is related to the detector element (DEL) size of the TFT and its fill factor (C). The DEL is the sensing element of the TFT, and it is desirable that its largest portion is used for just that purpose (D). If 25% of the DEL is used for other functions, the DEL is said to have a fill factor of 75%. The greater the fill factor, the better the spatial resolution and SNR (B). The smaller the TFT DEL size, and the larger the fill factor, the better the spatial resolution (E). A DEL size of 100 μm provides a spatial resolution of about 5 lp/mm (available only in some digital mammography systems). A DEL size of 200 μm provides a spatial resolution of about 2.5 lp/mm (general radiography)—lower than that achieved with a 400-speed intensifying screen system in analog systems. A 100-speed-intensifying screen system offers a spatial resolution of about 10 lp/mm—significantly greater than, and currently unachievable in, digital imaging. Spatial resolution in digital imaging is fixed, but it is very important that radiographers are alert to the opportunities they have to utilize and control the remaining recorded detail factors (motion and geometric factors). (Saia, PREP 9th ed., p. 444)

The surface of a rotating anode is beveled with a target angle θθ. As θθ increases from 8° to 15°, what 3 changes from the following list will occur (select the three that apply)? A Effective focal spot increases B Effective focal spot decreases C Spatial resolution increases D Spatial resolution decreases E Anode heat capacity is unchanged F Anode heat capacity decreases

The answer is A, D, and E. The line-focus principle states that as the target/focal track angle gets larger, the effective focal spot is larger (A and B). When the effective focal spot is larger, focal spot blur increases, and spatial resolution is decreased (C and D). With focal spot/track size (area of electron bombardment) unchanged, heat capacity remains unchanged (E). If the area of electron bombardment increases, heat capacity will increase because heat is being spread over a large area. If the area of electron bombardment decreases, heat capacity will decrease because heat is being concentrated to a smaller area. In this case, thearea of bombardment is unchanged, therefore heat capacity is unchanged. (Bushong 11thed., pp. 113, 179 ; Carroll 3rd ed p 322)

Which of the following factors are considered geometric factors that affect radiographic image quality? (select the three that apply) A Magnification B Kilovoltage C Contrast resolution D Distortion E Focal-spot blur F Exposure time

The answer is A, D, and E. There are three geometric factors that affect radiographic quality:magnification,distortion, andfocal-spot blur. Magnification is the misrepresentation of the true size of an anatomical part. Some magnification exists in all radiographic images, as the anatomical parts are three-dimensional, with some areas positioned farther away from the image receptor than others (vary in object-to-image-receptor distance, or OID). To minimize magnification, the radiographer should use the longest practical source-to-image-receptor distance (SID) and place the object as close to the image receptor as possible (short object-to-image-receptor distance (OID)).Distortionis a misrepresentation of the true shape of the anatomical part in the radiographic image. Distortion depends on the object (part) thickness, relative position with respect to the image receptor, the angle of incidence of the x-ray beam on the object and image receptor, and the object shape. Two types of shape distortion caused when the object plane and image receptor plane are not parallel are calledelongationandforeshortening, which also causes unequal magnification of opposite ends of the object in the image.Focal-spot blur(also called geometric unsharpness, or penumbra) is caused by undercutting of the edges of anatomical structures by x-rays emanating from various portions of the anode focal spot (x-rays do not emanate from a single point source in the anode). Focal-spot blur decreases spatial resolution. To minimize focal-spot blur, the radiographer should use a small focal spot size, the longest practical SID, and the shortest OID possible (A, D, and E).Kilovoltagedetermines the quality, or penetrating power, of the x-ray beam (B).Exposure timedetermines how long the x-ray beam is emitted from the x-ray tube and is measured in milliseconds or seconds. The product of exposure time and milliamperage (mA) determines the quantity, or intensity, of the x-ray beam (F). (Bushong, 11thed., pp. 175-178)

A technologist prepares to perform an exam on an x-ray machine they have never used before. Which of the following three patient factors should be considered in the selection of appropriate kVp and mAs for an optimal image of the projections involved? A Patient habitus B SID C Estimated part thickness D Part composition E Voltage waveform of the equipment F Measured part thickness G Beam filtration

The answer is A, D, and F. Exposure technique is a complex formulation for a technologist. The x-ray beam needs to be adjusted for each projection to ensure enough of the beam transmits to the image receptor. Within the consideration, certain patient factors affect the decision-making process. The overall build or habitus of the patient should be considered, as well as part composition (% bone vs % soft tissue) and measured part thickness, utilizing calipers (A, D, and F). SID, beam filtration, and voltage waveform of the equipment are worthy of consideration, but those are geometric and equipment factors, respectively, not patient factors (B, E, and G). Finally, technologists should not estimate part thicknesses in the pursuit of an optimal image, as measuring thickness with calipers yields a more accurate result (C). (Bushong, 11th ed., p. 243)

Higher kVp settings can provide which of the following benefits? 1. Improved penetration 2. Patient dose reduction 3. Increased spatial resolution A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The answer is A. Higher kVp settings increase x-ray energy, thereby providingimproved penetrationthrough tissues, particularly those that are thicker or denser. Improved penetration means less absorption (fewer photoelectric absorption interactions) andless patient dose. Additionally, higher kVp can enable lower mAs settings, also resulting in dose reduction. These settings also improve x-ray tube life as lower filament heat production may be achieved by using lower mAs settings (A). The kVp setting is considered an image visibility factor in that it determines the average x-ray photon energy and, therefore, the penetrating ability of the incident x-ray beam, to expose the image receptor. Spatial resolution is considered an image geometric factor.Increased spatial resolutionis determined by such factors as the focal spot size, geometric unsharpness (penumbra), laser beam diameter in the CR reader, and pixel size (B, C, and D). (Bushong, 11th ed., pp. 139-140, 150, 153, 183, 188, 255)

If the required technique is 10 mAs, 70 kVp, and the mA station is 500 mA, which of the following exposure times would be needed? A 20 ms B 33 ms C 50 ms D 100 ms

The answer is A. If the mAs and mA settings are known, the exposure time can be determined by dividing the mAs by the mA setting. In this question, 10 mAs divided by 500 mA would yield an exposure time of 0.02 sec or 20 ms (A). Ten mAs divided by 300 mA would yield 0.033 s or 33 ms (B). Ten mAs divided by 200 mA would yield 0.05 sec or 50 ms (C). Ten mAs divided by 100 mA would yield 0.1 sec or 100 ms (D). (Bushong, 11th. ed., pp. 238-239)

What does the "V" in VOI stand for in digital imaging? A Values B Variables C View D Visible

The answer is A. In digital imaging, the computer software has histogram models for all menu choices and is stored as histograms. The histogram models havevaluesof interest (VOI) and determine what section of the histogram data set should be included on the displayed image (A). Values of interest are established within the histogram models that determine what part of the data set should be incorporated into the displayed image.Variablespertain to kVp and mAs in the technique (B).Viewpertains to viewing the image (C).Visiblepertains to the appearance of the image (D).(Johnston and Fauber, 2nded., p. 166)

When performing an axial projection, which of the following is NOT true? A The amount of elongation distortion is affected by focal spot size. B The amount of elongation distortion is affected by the alignment of the long axis of the object to the direction of the angle. C The amount of elongation distortion is proportional to the size of the angle. D The amount of elongation distortion is proportional to the object's OID.

The answer is A. OID, tube angle, and angle direction relative to the object all affect the degree of elongation of a particular object. It is helpful to remember that an x-ray image is a shadowgraph, and therefore behaves like a shadow; a clever student might be able to conduct a quick experiment with a cell phone LED light source and a pen, projecting a shadow of the pen on a wall and observing changes (B, C, and D). Focal spot size affects edge blur; confusion on the part of the student may arise because OID is a factor in both penumbra and magnification and distortion effects (A).

Prolonged heating of the x-ray tube filament when preparing for a radiographic exposure may lead to which of the following? 1. Electrical arcing 2. Premature filament breakage 3. Anode surface melting A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The answer is A. Prolonged heating of the filament during the preparation stage when the exposure button is depressed half way (prior to the actual x-ray exposure) can lead topremature filament breakage(2), caused by excessive heating of the filament. Prolonged "boost-and-hold" not only causes filament tungsten evaporation, but the rotor bearings can be unnecessarily worn while being spun by the induction motor (stator). For these reasons, the radiographer should carefully communicate to the patient breathing instructions and the importance of remaining still when the exposure is made. Tungsten evaporated from the filament coats the inside wall of the glass tube insert. Because it is a metallic material, the filament projectile electrons may jump to the glass, a process calledelectrical arcing, during the final stage of the exposure (when the exposure button is completely depressed). This may cause a crack in the glass, and the vacuum is lost as air enters. Electrical arcing is the most common cause of tube failure (A). During the preparation stage of an x-ray exposure, the anode is not being bombarded by the filament projectile electrons, and thereforeanode surface meltingis not possible (B, C, and D). Only during the final stage of the exposure, when the exposure button is completely depressed, do the electrons bombard the anode target, creating x-rays. (Bushong, 11thed., pp. 107, 113)

When detector elements reach their maximum electrical charge and their pixels appear completely black with complete loss of information, this is termed A image saturation B dynamic range C optical density D ratio image

The answer is A. Saturationoccurs when pixels that should be gray become black, and no post processing can change the color (A). If the image is overexposed by 8‒10 times, saturation occurs.Dynamic rangeis the range of exposure intensities an image receptor can accurately detect (B). Digital imaging has a wide range when compared to film/screen.Optical densityis a measure of the degree of film darkening on an analog image (C).Ratio imagedoes not pertain to digital radiography (D).(Carroll, 2nded., pp. 549-550; Carter and Veale, 2nded., p. 25)

Dynamic range is a function of theimage receptorcomputer softwarekVbrightness level A 1 and 2 only B 1 and 3 only C 2 and 4 only D 2, 3, and 4 only

The answer is A. The ability of the computer software and the detector to distinguish and make available various pixel values for image formation is termed dynamic range (1 and 2). Contrast and brightness are controlled principally by computer software in digital imaging (3 and 4). (Carroll, 2nd ed., p. 461)

When describing the overall efficiency of CR photostimulabe phosphor plates converting x-ray signals into a useful image, which of the following terms best fit this definition? A Modulation transfer function B Detective quantum efficiency C Absorption efficiency D Conversion efficiency

The answer is B. Modulation transfer function or MTF, describes the system's ability to display signal contrast as a function of spatial resolution (A). Absorption efficiency describes the ratio of the number of x-ray photons absorbed by the phosphor crystals compared to the number of x-rays that hit the phosphor layer (C). Conversion efficiency describes the actual percentage of x-ray energy that is actually converted to light (D). Detective quantum efficiency or DQE, describes the overall efficiency of converting x-ray signals into a useful image. Ideally the CR system would reach a DQE of 1.0 or 100 percent (B).(Carroll, 2nd ed., p. 604-606)

X-ray energy is directly controlled by which of the following factors? A kVp B mAs C SID D OID

The answer is A. The controlling factor of x-ray energy (quality) is the applied kilovoltage used during a radiographic exposure (A). Increased kilovoltage in the x-ray circuit produces higher kinetic energy filament (projectile) electrons bombarding the anode of the x-ray tube. The increased kinetic energy results in the production of bremsstrahlung x-rays that are of higher energy (shorter wavelength) and can therefore penetrate the inherent and added filtration of the tube and collimator assembly, penetrate the patient, and expose the image receptor. The quantity (intensity) of x-ray photons in the x-ray beam is controlled by the mA, time, or mAs settings (B). The SID controls the quantity (intensity) of x-ray photons, based on the inverse square law, reaching the anatomical area of interest and affecting magnification distortion, geometric unsharpness (penumbra), and spatial resolution (detail) of the projected image, and is therefore considered both an exposure factor and geometric image factor (it is primarily classified as a geometric image factor, however) (C). The OID primarily affects magnification distortion of the anatomical area of interest and secondarily affects the exposure level from scatter radiation if an air-gap technique is used. SID and OID should be primarily regarded as geometric image factors (D). (Bushong, 11th ed., pp. 130-131)

To activate magnification mode in the image intensifier A more voltage is applied to the electrostatic lenses B the gap between the input phosphor and the photocathode is increased C an aperture collimator is collapsed in front of the output phosphor D the input phosphor is polarized by applying a bias voltage

The answer is A. The function of the electrostatic lenses within the image intensifier is to constrict the conical electron stream as it transitions from the photocathode to the output phosphor. The lenses are constructed as rings of conductive wire imbued with a negative charge. When the voltage of these rings is increased, there is a repulsive force that tightens the stream of electrons into a smaller cone (A). The apex of the cone is the focal point, and the cone expands again as it meets the output phosphor. In magnified mode, only electrons emitted from the center of the photocathode actually reach the output phosphor (B). Conventionally constructed image intensifiers have no gap between the input phosphor and the photocathode, as this would introduce light spread and a loss of spatial resolution. They do not include an aperture collimator at the output phosphor either (C). The notion that the input phosphor is polarized by applying a bias voltage is pseudoscience; while polarization, input phosphors, and bias voltages are real terms, a phosphor cannot be polarized by applying a voltage of any size. If there was a way to polarize the input phosphor, it would only diminish the number of light photons sent to the photocathode (D).(Bushong 11th ed., p. 41

Which two (2) of the following statements are true regarding image distortion?The least amount of image distortion occurs along the path of the CRCR alignment has little/no impact on image distortionX-ray tube angle impacts shape distortionThe least amount of image distortion occurs at the periphery of the x-ray beamSID impacts shape distortion as well as size distortion A 1 and 3 are correct B 2 and 3 are correct C 4 and 5 are correct D 2 and 5 are correct

The answer is A. Two types of image distortion are size and shape. Size distortion is magnification, which is impacted by OID and SID. Shape distortion is impacted by alignment of the x-ray tube, anatomic part, and IR. If the anatomic part is at an angle with respect to the x-ray tube and IR, foreshortening occurs. If the x-ray tube is angled with respect to the part and IR, elongation occurs. CR alignment also impacts image distortion. Anatomic parts in the path of the actual central ray have the least distortion, while parts imaged further and further away from the central ray by divergent x-ray photons experience more and more distortion. (Lampignano and Kendrick, 9th ed., pp. 46, 48)

When performing quality control checks on the x-ray beam collimator, each dimension and edge of the light field must be congruent within _____ percent of the source-image-distance (SID). A 2 B 3 C 4 D 5

The answer is A. U.S. federal guidelines specify that all x-ray beam collimators must demonstrate light field‒x-ray field congruency within 2 percent of the source-image-distance (SID) (A). Quality control test results, which identify congruency beyond 2 percent, would not be in regulation and would require service. The other answer choices are incorrect as per the previous explanation (B, C, and D). (Johnston and Fauber, 2nd ed., p. 145)

From the following list, select the effective methods of reducing scatter radiation production in the patient (select the two that apply). A Using an air-gap technique B Increasing collimation C Using a high ratio grid D Using a direct conversion digital system E Placing a lead apron over the anatomy of interest F Using anatomical part compression

The answer is B and F. Compton scatter radiation production occurs when an incident x-ray interacts with an outer shell electron in the atoms of the patient. Increasing collimation reduces the exposure field size, thereby reducing the number of scattered photons produced within the patient (B). Using anatomical part compression reduces the thickness of the part, which reduces the number of atoms with which the incident x-rays can interact (F). An air-gap technique, such as that used for a cross-table lateral c-spine radiograph, only reduces the number of scattered x-rays striking the image receptor, not the number of scattered x-rays produced within the patient (A). Using a high ratio grid would also reduce the number of scattered x-rays reaching the image receptor, but not reduce the number of scattered x-rays produced within the patient (C). For an appropriate exposure, using a direct conversion digital system does not reduce the number of scattered x-rays produced within the patient, but does improve spatial resolution when compared to an indirect conversion digital system (D). Placing a lead apron over the anatomy of interest would not be considered an effective method of reducing the number of scattered x-rays produced within the patient, as it would absorb (via photoelectric interaction) the incident x-rays that are intended to create the diagnostic image (E). (Bushong, 11th. ed., pp. 148; 190-193)

The average frequency of photons contained within the x-ray beam can be increased by 1. Increasing mAs 2. Increasing kVp 3. Increasing filtration 4. Decreasing kVp A 1 only B 1 and 2 C 2 and 3 D 1, 3, and 4

The answer is C. Average beamenergyis influenced by both kVp and filtration. As kVp increases, the KE of electrons in the tube increases, and therefore Bremsstrahlung-generated photons have a higher average energy (B and D). As filtration increases, low energy photons are removed from the beam, and so the average increases, like dropping a lowest quiz score (C). The Planck equation states that there is a direct relationship between energy and frequency for all EMR, and thus any changes that influence average energy will do likewise to average frequency. mAs only changes the number of photons produced, not their energy (A).(Bushong 11th ed., pp. 139-141)

The anode heel effect results in variation of the intensity of the beam across the field of view. Select four items from the list that will reduce the appearance of this variation on the displayed image. A Smaller anode target angle B Larger anode target angle C Longer SID D Automatic rescaling E Flat fielding F Aligning a tapering body part so that the thicker portion is on the anode side of the beam G Aligning a tapering body part so that the thicker portion is on the cathode side of the beam

The answer is B, C, E, and G. A larger anode target angle, longer SID, flat fielding, and aligning a tapering body part so that the thicker portion is on the cathode side of the beam will all reduce the appearance of the variation of beam intensity caused by the heel effect (B, C, E, and G). A smaller anode angle creates a larger heel, and more of the useful beam will be attenuated by the heel (A). A longer SID minimizes the percentage of the total field that suffers a loss of intensity. Portable radiography, often performed with shorter SIDs, exacerbates the heel effect. In structures that taper, it is the wise technologist that aligns the part so that the thinner aspect is aligned with the thinner beam, at the anode end. Automatic rescaling is an adjustment of signal strength applied to the entire matrix, typically to correct for overexposure, and the software applies this correction without regard to the cathode or anode side of the image data (D). Finally, preprocessing software allow digital systems to handicap the signals from pixels on the cathode side of the receptor before the raw data is converted to an image file via the LUT. This correction, called flat fielding, virtually eliminates the appearance of the anode heel effect (F).(Bushong 11th ed., p. 114, 327)

Receptor exposure decreases with an increase in (select the three that apply): A kV B SID C focal spot size D mAs E part thickness F grid ratio G field size

The answer is B, E, and F. As the SID increases, exposure rate and receptor exposure decrease (B). As part thickness/tissue density increases, receptor exposure decreases (E). Receptor exposure is significantly impacted with the use of grids; as grid ratio increases, receptor exposure decreases (F). Because grids remove many x-ray photons that would have contributed to receptor exposure, the addition of a grid requires a significant increase in mAs. Increased kV produces more high-energy x-ray photons and will result in an increase in receptor exposure (A). Focal spot size affects spatial resolution only, and it has no impact on receptor exposure (C). Receptor exposure and mAs are directly proportional (i.e., if mAs is doubled, receptor exposure is doubled) (D). As field size is increased and a greater volume of tissue is exposed, more scattered radiation is produced and receptor exposure increases (G). (Saia, PREP 9th ed., pp. 315, 317, 321, 330, 332)

Select the three factors from the choices below that determine the size of the actual focal spot on the target anode. A kVp selection B Filament size C SID D Voltage waveform E Focusing cup depth F Target angle G Effective focal spot size

The answer is B, E, and F. The actual focal spot describes the area of electron bombardment on the target anode. Electrons are formed by thermionic emission at the cathode filament. Specifically, when the filament heats up due to excess current, electrons are boiled off into a "cloud" surrounding the filament. Using a larger filament creates a larger cloud, as the electrons within the cloud repel each other (B). The natural tendency of this cloud would be to spread out. The focusing cup is a metal bowl that is electrified with a negative charge to repel electrons in the cloud and thus limit the size of the cloud. A deeper bowl will keep the cloud confined more effectively and steer the electrons toward a single direction as they leave the filament (E). The cloud is the source of the electron stream, which forms when the kVp is applied. The kVp, while supplying the potential for acceleration, does not cause the stream to spread leaving the focusing cup. Each electron in the stream is repelled by all the other electrons in the stream (like charges repel), and so the stream expands as it travels across the tube toward the anode. The final determination of actual focal spot size is the target angle; a larger target angle produces a larger area of bombardment (F). A flashlight and a piece of cardboard can demonstrate this effectively. If the beam intersects at a perpendicular, it is a small circle, but as the cardboard is angled, the area where the beam shines on the cardboard expands to an ellipse. kVp, voltage waveform, SID, and effective focal spot size do not cause any change in the actual focal spot(A, C, D, and G). Effective focal spot is an effect determined by target angle and electron beam width.(Bushong 11th ed., p. 113)

The x-ray tube is a DC electrical device, while the transformers that power it are AC devices. An arrangement of diodes between the two converts the AC voltage waveform into a pulsed DC output. The process of converting an electrical signal from AC into DC is called A electrification B rectification C amplification D ionization

The answer is B. Arectifieris an arrangement of diodes that converts AC to pulsating DC. In full wave rectification, the diamond shape four diode coupling reverses the polarity on the troughs of an AC wave, making them "flip" into additional crests, and effectively doubling the number of crests per cycle. Half wave rectification, which only uses two diodes, effectively eliminates the voltage for the trough portion of the cycle. Between the two, full wave rectification is a more efficient process, delivering twice as much x-ray energy for equal electrical energy input (B). Electrification is a process in static electricity in which an object acquires excessive charge, either positive or negative (A). Amplification is an increase in the voltage peak, which is the function of the step-up transformer (C). Finally, ionization is the removal of electrons from a target atom by radiation of particles: alpha, beta, and neutrons, or waves: high energy ultraviolet rays, x-rays, and gamma rays (D).(Bushong, 11th ed., pp. 61, 93-95)

An increase in the milliamperage (mA) setting increases 1. X-ray beam quantity 2. X-ray beam quality 3. Image receptor exposure A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The answer is B. As the milliamperage (mA) setting is increased on the control panel, the filament circuit increases the amperage to the filament and creates a larger thermionic emission of filament electrons. This increased electron quantity results in a larger number of projectile electrons that can bombard the anode target during an exposure, thereby increasing the x-ray beam quantity, or number of x-rays produced. This increase in x-ray quantity (exposure rate) means that more x-rays will expose the patient and subsequently increase image receptor exposure (B). X-ray beam quality is controlled by the applied kilovoltage. An increase in kilovoltage increases the kinetic energy of the filament projectile electrons, thereby increasing the average energy of the brems x-rays produced (A, C, and D). (Bushong, 11th ed., pp. 89-90, 137-138)

Which of the following pathologies would justify a decrease in technique? A Ascites B Pneumothorax C Hydrocephalus D Atelectasis

The answer is B. Ascites (A), Hydrocephalus (C), and Atelectasis (D) are all additive conditions that require an increase in technique. A patient who has a Pneumothorax results in the lung field being easier to penetrate. To compensate for this condition, the technique should be decreased so the anatomy of interest is not overexposed (B).(Johnston & Fauber, 2nd ed., p. 200)

The standard developed to establish a universal, standardized public format and protocol for communicating biomedical imaging files is A PACS B DICOM C RIS D HIS

The answer is B. DICOM is an acronym for Digital Imaging and Communications in Medicine standard (B). It originated in 1983 as a collaborative effort between the American College of Radiology (ACR) and the National Electrical Manufacturers' Association (NEMA) to establish a standardized public format and protocol for communicating biomedical imaging files. PACS is an acronym for Picture, Archiving and Communication System and denotes a networked group of computers, servers, and archives to manage digital images for storing, retrieving, and distributing medical images (A). RIS is an acronym for Radiology Information System and HIS is an acronym for Hospital Information System; both of these relate to Biomedical Informatics (BMI) platforms used in radiology departments and healthcare facilities respectively (C and D). (Carlton/Adler/Balac, 6th ed., p. 329)

Which of the following will improve the sharpness of the edges of small anatomical parts? 1. Small focal spot size 2. Increased OID 3. Increased SID A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The answer is B. Geometric unsharpness (focal spot blur, or penumbra) of an image is caused by undercutting of the edges of the anatomical structures by x-rays emanating from various portions of the anode focal spot (x-rays do not emanate from a single point source in the anode). Essentially, the edges of the anatomical structures are blurred, thereby decreasing spatial resolution. In order to minimize this blurring effect, the radiographer should use asmall focal spot size, when a lower mAs setting can be used to provide the exposure needed to the image receptor. Lower mA settings to obtain the desired mAs setting energize the smaller filament in a dual-filament x-ray tube. By using a small filament size, the actual focal spot (where the filament projectile electrons strike the anode target) is smaller, thereby producing a smaller effective (projected) focal spot size of the primary x-ray beam. Smaller effective focal spot sizes produce less penumbra and project the edges of small anatomical structures more accurately toward the image receptor. By using anincreased SID, the radiographer can also minimize the effects of penumbra, as more of the undercutting x-rays in the primary beam are absorbed by the collimator shutters when increased collimation is used to obtain the appropriate field size (B).Increased OIDincreases magnification of the anatomical part and therefore also increases, or accentuates, penumbra, resulting in increased image unsharpness (A, C, and D). (Bushong, 11thed., pp. 109-110, 113-115, 173-179)

Which of the following pathologic conditions would require a decrease in exposure factors? A Empyema B Osteoporosis C Pleural effusion D Hemothorax

The answer is B. In radiography, a pathologic condition is categorized as either destructive or additive pathology. Destructive pathology refers to an abnormal decrease in mass density in the affected area, and therefore requires the radiographer to decrease the exposure factors from what would be used for a healthy patient with an anatomical part of the same thickness. Osteoporosis is a destructive pathological condition that requires the radiographer to use a lower kVp setting than that used for healthy bone. Healthy bone is constantly being broken down and replaced. However, osteoporosis occurs when the creation of new bone fails to keep up with the removal of old bone. This results in a decrease in bone density and strength, which places the affected person at higher risk for fractures (B).Additive pathology refers to an abnormal increase in mass density, such as that caused by additional tissue (e.g., tumor mass), fluid, or other condition that effectively increases the mass density of the affected area. Therefore, the radiographer must increase the exposure factors to properly expose the IR. Empyema is a collection of pus in the pleural cavity caused by infection (A). A pleural effusion is a collection of fluid in the pleural space caused by a variety of underlying abnormal conditions including inflammation and congestive heart failure (C). Hydrothorax is a type of pleural effusion in which transudate accumulates in the pleural cavity, most commonly caused by congestive heart failure, pneumonia, malignancies, and pulmonary embolism (D). (Eisenberg and Johnson, 6th. ed., pp. 108-109)

Which of the following pathologic conditions would require an increase in exposure factors? 1. Empyema 2. Osteoporosis 3. Pleural effusion A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The answer is B. In radiography, a pathologic condition is categorized as eitherdestructiveoradditivepathology.Additivepathology refers to an abnormal increase in mass density, such as that caused by additional tissue (e.g., tumor mass), fluid, or another condition. Therefore, the radiographer must increase the exposure factors (especially kVp) to penetrate the tissues and properly expose the image receptor.Empyemais a collection of pus in the pleural cavity caused by infection. Apleural effusionis a collection of fluid in the pleural cavity caused by a variety of underlying abnormal conditions including inflammation and congestive heart failure (B).Destructivepathology refers to an abnormal decrease in mass density in the affected area, and therefore requires the radiographer to decrease the exposure factors.Osteoporosisis an example of destructive pathology, which requires the radiographer to use lower kVp settings than those used for patients with healthy bone structures. Osteoporosis occurs when the creation of new bone fails to keep up with the removal of old bone. This causes decreased bone density and strength, which places the patient at higher risk for fractures (A, C, and D). (Eisenberg and Johnson, 6th. ed., pp. 108-109)

Compton scatter production and photoelectric absorption within the patient both increase at the same rate as A beam filtration increases. B tissue thickness increases. C kVp increases. D OID increases.

The answer is B. Increasing beam filtration effectively increases the average energy of the beam, often called beam hardening. The beam acts as if you increased kVp. Therefore, beam filtration increasing and kVp increasing are essentially the same change, and will increase Compton scatter relative to the incidence of photoelectric absorption (A and C). Increased OID limits scatter from reaching the image receptor but does nothing to affect its production in tissue (D). Only increasing tissue thickness increases both Compton and photoelectric phenomena uniformly (B). Throw a ball through a single line of trees, and it has a good chance of passing through; throw the same ball through a forest 100 m deep, and your ball is likely to bounce off a tree (Compton), disappear into a bush (absorption), or both. There are more objects in the pathway of the ball and the photon to interact with in both cases. (Bushong, 11th ed., pp. 131, 151-158)

Of the three modes that can be used in a trifield image intensifier, which would result in the least amount of image magnification? A 17-cm mode B 25-cm mode C 12-cm mode D 5-cm mode

The answer is B. Most image intensifiers used during fluoroscopic procedures are of the multifield type. Commonly, three modes are used: 25-cm, 17-cm, and 12-cm modes. In the25-cm mode, photoelectrons from the entire input phosphor are accelerated to the output phosphor and are focused by the electrostatic lenses at a point closest to the output phosphor. This results in theleastamount of image magnification when compared to that seen with the other modes (B). When the17-cm or 12-cmmodes are used, the voltage of the electrostatic lenses increases and causes the photoelectron focal point to move farther from the output phosphor, with the 12-cm mode moving the focal point farthest. The further the focal point is from the output phosphor, the more magnified the image will be (A and C). (Bushong, 11thed., pp. 407-408)

Post exposure cropping or masking of a radiographic image is performed to A take the place of collimation before exposure B reduce veiling glare C reduce the appearance of scatter within the image D change the matrix size

The answer is B. Post exposure cropping should only be utilized to reduce the effect of veiling glare (B). Veil glare is defined as a light signal broadcast by a monitor outside the collimated field incorrectly, which alters the viewer's perception of the brightness of the adjacent radiograph, often making it appear darker. Post exposure cropping will not alter the matrix size, nor will it change noise in the image from scatter (C and D). Technologists should not crop viable image data post exposure to cover for a lack of appropriate collimation, as they may conceal pathology and prevent accurate diagnosis (A). The ASRT position statement indicates that concealing exposed anatomy is the equivalent of a medical decision made by a physician and is outside of the scope of practice for a technologist. (Carter, 3rded., p. 33)

Which three (3) of the following statements are true regarding image manipulation?Window level controls brightness and darknessThe wider the window width, the lower the contrastWindow level controls contrast (i.e., white and black ratio)The higher the window level, the darker the imageWindow width controls brightness and darkness A 1, 2, and 3 are correct B 1, 2, and 4 are correct C 2, 3, and 5 are correct D 1, 4, and 5 are correct

The answer is B. Post processing image manipulation and enhancement functions can include annotating, orientation functions such as rotation and flipping, window level and/or width adjustments, magnification, and zoom. Adjustments of window level and/or width are probably most common. The window level controls the degree of brightness; the higher the window level, the darker the image (1, 3, and 4). The window width controls contrast; the wider the window width, the lower the contrast (2 and 5). (Carter and Veale, Digital Radiography and PACS, 3rd ed., p. 32)

Grids improve the radiographic image by reducing the amount of scattered radiation fog. How does the use of grids impact exposure factors? A It has no effect on exposure factors B It increases exposure factors C It decreases exposure factors D It doubles exposure factors

The answer is B. Radiographic grids are used when the thickness of the body part to be radiographed is 10 cm or greater. The use of grids increases patient dose because the amount of exit radiation reaching the IR is reduced. Grids are made of strips of lead with an interspace having an aluminum or other spacer. Adding grids always requires an increase in exposure factors (B). The higher the grid ratio the greater the increase in mAs required. The other answer choices are incorrect as per the previous explanation (A, C, and D).(Sherer, Visconti, Ritenour, and Haynes, 8thed., p. 213)

If an acceptable image is obtained utilizing 20 mAs, 75 kVp at a 40-in. SID, what adjustment must be made to mAs with a 35-in. SID to maintain image receptor exposure? A Use 25 mAs, 75 kVp B Use 15 mAs, 75 kVp C Use 18 mAs, 75 kVp D Use 5 mAs, 75 kVp

The answer is B. The SID controls the quantity (intensity) of x-ray photons, based on the inverse square law and reaching anatomical area of interest. It also affects magnification distortion, geometric unsharpness (penumbra), and spatial resolution (detail) of the projected image. Therefore, the SID is considered both an exposure factor and geometric image factor. For example, if a radiographer arrives at a patient's room to perform a femur radiograph and notices that the patient's leg is suspended by an overhead traction bar, the radiographer must place the x-ray tube under the bar at one-third less SID than desired to prevent projecting the bar into the image. If the intended technique were to be used at the new SID of 35 inches, the intensity of the x-ray beam at 20 mAs would be excessive due to the inverse square law. Based on this law, the intensity of electromagnetic radiation is inversely related to the square of the distance from the source. The formula used to determine the appropriate mAs (intensity) change needed to compensate for changes in SID (called the Square Law formula) is as follows: (new distance/original distance)2× original mAs. Therefore, a 35-inch SID divided by a 40-inch SID squared multiplied by the original mAs of 20 yields15 mAs. Since the kVp is the controlling factor of x-ray energy and, therefore, penetration, the original75 kVpsetting should remain unchanged (B). Twenty-five mAs would be the answer if the original mAs was mistakenly placed in the numerator instead of the denominator in the equation (A). Eighteen mAs would be the answer if the dividend was not squared (C). Five mAs would be the answer if the new distance was half (20-inch SID instead of 35-inch SID) of the original distance (D). (Bushong, 11th ed., p. 240)

Which of the following best makes up the graphic representation of the amount of exposure and the frequency of pixels for each exposure? A Image graph B Histogram C Image sampling D Values of images

The answer is B. The data within the collimated area produce a graphic representation of the optimal densities. Ahistogramis a graphic representation of a data set (B). This graph represents the number of digital pixel values versus the relative prevalence of those values in the image. The x-axis represents the amount of exposure and the y-axis represents the incidence of pixels for each exposure level. The computer then analyzes the histogram using processing algorithms.Image graphdoes not pertain to the field of radiology (A).Image samplingis the process used to digitize the spatial information in an image (C).Values of imagesare the numeric values that are assigned to each pixel and represent the intensity of the gray level of each location in the image (D).(Johnston and Fauber, 2nded., p. 161)

The figure's Y axis illustrates which of the following? Reproduced with permission from Saia, Radiography PREP, 9th edition. New York, NY: McGraw-Hill, 2018. Figure 12-22B. A Pixel value distribution B Spatial resolution C Tonal distribution D Frequency/number of pixels having each particular value

The answer is D. The histogram is an analysis and graphic representation of the information from the IR, demonstrating the number of pixels and their value. The Y axis illustrates the frequency/number of pixels that display a particular tone/value (A and D). The X axis provides information about the tonal values of the pixels and their distribution within the image—blacks are to the left, whites are to the right (B and C). In a narrow histogram, the blacks from the left and the whites from the right are squeezed to the center of the histogram, resulting in a very gray low-contrast image. In a wide histogram, the mid-histogram gray pixels have moved laterally toward the blacks and whites, resulting in a higher contrast image. (Saia, PREP 9th ed., pp. 377, 378)

Radiation spreads from a source in an expanding sphere, the area given by 4/3πr34/3πr34/3πr3. So long as the rate of emission is constant, the intensity of the radiation will reduce by a factor squared as an observer moves away from the source. This describes A ALARA B the inverse square law C the line focus principle D the half value layer

The answer is B. The inverse square law states the following:The intensity of radiation is inversely proportional to the square of the distance between the observer and the source(B). A critically thinking student may have seen the connection between the distance between the observer and the source andr, the radius of the expanding sphere. Ifris increasing, then the area of said sphere is increasing, and with a constant emission rate, the same quantity of radiation is now spread over a wider area, so its intensity reduces. The line focus principle establishes a relationship between actual focal spot size and effective focal spot size, and the half value layer calibrates penetration power of the beam (C and D). ALARA is the foundational principle for radiation protection—keep all exposuresas low as reasonably achievable(A).(Bushong 11th ed., pp. 10, 54, 113, 140)

Which of the following is the formula used to determine magnification factor (MF)? A MF = source-image-distance / object-image-distance B MF = source-image-distance / source-object-distance C MF = source-object-distance / source-image-distance D MF = object-image-distance / source-image-distance

The answer is B. The magnification factor (MF) indicates how much size distortion or magnification is demonstrated on a radiograph. The magnification factor (MF) is found by applying the following formula, MF = source-image-distance (SID) / source-object-distance (SOD) (B). If, for example, the SID was 40 inches and there was a source-object-distance (SOD) of 30 inches, the MF would equal 40" / 30" or 1.33. SOD is found by subtracting OID from SID (SOD = SID ‒OID). The other answer choices are incorrect as per the previous explanation (A, C, and D). (Johnston and Fauber, 2nd ed., p. 128)

What determines the numbers assigned to pixel values resulting in brightness and contrast changes during image processing? A Image values B Lookup table C Displays of image D Image noise

The answer is B. The primary factor that influences contrast with digital imaging is thelookup table(LUT)(B). The lookup tables are histograms of luminance values and are used as a reference to evaluate the input intensities and assign predetermined grayscale values.Image valuesare represented by the pixels that make up the image (A). Thedisplays of imageare where the image is actually displayed on the computer monitor (C).Image noiserefers to anything that impairs visualization of the image such as scattered radiation fog (D). Quantum noise has the appearance of an irregular granular pattern that shows up on the image.(Johnston and Fauber, 2nded., p. 166)

What is the spatial frequency of a digital system with a pixel pitch of 100 um? A 10.0 lp/mm B 5.0 lp/mm C 0.05 lp/mm D 0.01 lp/mm

The answer is B. The smallest line (singular) displayed on the image is one pixel wide. A line pair implies two lines, one white and one black, adjacent to each other. Therefore, a line pair is two pixels wide. Change 100um to 0.1 mm. If pixel pitch is 0.1 mm, a line pair is 0.2 mm wide. Therefore, there are 5 line pairs in 1 mm (5 x 0.2). (Johnston and Fauber 3rd ed p 103)

Which of the following is affected by the size of the x-ray field? 1. Amount of scattered radiation produced 2. Motion unsharpness 3. Patient dose A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The answer is B. The x-ray field size is directly related to theamount of scattered radiation producedin the patient. As the field size increases, the amount of scattered radiation produced increases, whereas a smaller field size decreases scattered radiation production.Patient doseincreases with larger field sizes, as there are more primary x-rays striking the patient's tissues. Also, some of the scattered radiation produced within the patient is absorbed in nearby tissues, thus increasing internal dose (B).Motion unsharpnessis affected by both the exposure time and patient cooperation. It is important for the radiographer to use the shortest practical exposure time and provide clear patient instructions prior to making an exposure. Immobilization methods may also be needed for patients experiencing involuntary motion (A, C, and D). (Bushong, 11th ed., pp. 189; Statkiewicz, Visconti, Ritenour & Haynes, 8th ed., pp. 49, 208)

During the processing of the PSP image, what layer of the imaging plate is responsible for sending the emitted blue light toward the processor's light collecting optics and photodetector? A Conductive layer B Reflective layer C Barium fluorohalide phosphor layer D Backing layer

The answer is B. To minimize signal loss, a reflective layer is immediately underneath the active phosphor layer in a PSP imaging plate system (B). While it is true that the blue light originates in the barium fluorohalide layer, remember than it spreads in all directions, or isotropically, as it forms (C). The light photon heading away from the reader optics will hit a mirror-like surface at the reflective layer and return the light photon from the direction it came. The net result is that most of the blue light exits the front surface of the phosphor layer, passes through the protective layer, and is captured as useful signal. The conductive layer captures stray electrons that choose to wander off after getting loose from the trap, which keeps static electricity within the plate low (A). The backing layer is typically an aluminum sheet, to offer structural rigidity and back shielding, which reduces unwanted scatter radiation from being recorded in the phosphor (D). (Carter &Veale, 3rd ed., pp. 47)

While using a digital radiography system, suppose a radiographer uses exposure factors of 10 mAs and 70 kVp with an 8:1 grid for an AP shoulder radiograph with acceptable anatomical part penetration and detector element (DEL) exposure. If the radiographer desires to increase scatter absorption using a 12:1 grid, what new exposure factors should be used to maintain the same DEL exposure? A 15 mAs, 70 kVp B 12.5 mAs, 70 kVp C 7.5 mAs, 70 kVp D 2.5 mAs, 70 kVp

The answer is B. When changing grid ratios, mAs must also be changed to maintain the same acceptable DEL exposure. When changing from an 8:1 grid to a 12:1 grid, more scatter radiation and image-forming radiation is absorbed. Therefore, increased mAs (quantity of radiation) is required to maintain the same DEL exposure. The formula for changing grid ratios is the new grid Bucky factor divided by the original grid Bucky factor, then multiplied by the original mAs used. An 8:1 ratio grid carries a "4" Bucky factor, whereas a 12:1 grid carries a "5" Bucky factor. Therefore, 5/4 x 10 mAs = 12.5 mAs (B).A change from an 8:1 grid ("4" Bucky factor) to a 16:1 grid ("6" Bucky factor) would require a change in mAs from 10 mAs to 15 mAs (A). A change from an 8:1 grid ("4" Bucky factor) to a 5:1 grid ("2" Bucky factor) would require a change in mAs from 10 mAs to 7.5 mAs (C). A change from an 8:1 grid ("4" Bucky factor) to a non-grid ("1" factor) would require a change in mAs from 10 mAs to 2.5 mAs (D). (Bushong, 11th ed., p. 198)

A new full-wave rectified x-ray machine is being installed at the outpatient imaging center downtown. Engineers and electricians from the manufacturer recommend wiring to a 220 V power supply from the building's electrical system. The operator's manual states that the maximum kilovoltage for the unit will be 150 kVp, and the rectifier will contain 4 diodes. It further states that the autotransformer uses 300 turns when it supplies unadjusted line voltage for maximum output to the step up transformer. What is the turns ratio (S:P) for the high voltage transformer for this machine? A 600:1 B 682:1 C 8,250:1 D 440:1

The answer is B. Using the Transformer law, and employing the G.U.E.S.S. method (Givens, Unknowns, Equation, Substitute, Solve), we can frame the word problem as follows: Givens: Vp = 220 V, Vs = 150,000 V Unknowns: Np/Ns (also known as the turns ratio) Equation: Vp/Vs = Np/Ns (The ratio of the voltage input to voltage output is proportional to the turns ratio: The Transformer Law) Substitute: 220/150,000 = Np/Ns Solve: 1/681.81 = Np/Ns Expressed asSecondary to Primary(S:P): 682:1 (B). The 300 turns on the autotransformer and the 4 diodes were extra information unnecessary to the solution. Utilizing it in any way may yield some of the other choices. A full-wave rectified machine is the standard schematic in the ARRT content outline. It creates high voltage by sending a measured and selected voltage from the autotransformer to the step-up, or high voltage, transformer. The high voltage transformer is constructed with a great deal more coil windings on the secondary side than the primary side; the transformer law explains that such a ratio will lead to an increase in voltage, proportional to the magnitude of the turns ratio. If we have the maximum output voltage and the maximum input voltage, we have the voltage ratio, which is said to be equal to the turns ratio. Remember to convert 150 kilovolts back to volts (multiply by 103) before you begin. The G.U.E.S.S. method is a systematic way of solving word problems in a standardized, ordered process. The student may find it useful, but like any tool, you should practice with it a couple times to get the feel for it. The other answer choices are incorrect as per the previous explanation (A, C, and D).(Bushong, 11th ed., pp. 79-80)

Which of the following formulas should be used to determine the mAs that should be used when the grid ratio is changed?1. mAs1mAs2=grid correction factor1grid correction factor21. mAs1mAs2=grid correction factor1grid correction factor2 2. Intensity1Intensity2=grid correction factor2grid correction factor12. Intensity1Intensity2=grid correction factor2grid correction factor1 3. New gridOld grid×original mAs = new mAs3. New gridOld grid×original mAs = new mAs A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The answer is B. When a radiographer switches from a non-grid exposure to a grid exposure or from one grid ratio exposure to another grid ratio exposure, the mAs must be changed to maintain the same exposure to the IR. To accurately adjust the technique, the radiographer may use a mathematical formula, including the following: mAs1mAs2=grid correction factor1grid correction factor2mAs1mAs2=grid correction factor1grid correction factor2 (1) or New grid factorOld grid factor×original mAs = new mAsNew grid factorOld grid factor×original mAs = new mAs (3) With either formula, the appropriate mAs change can be determined (B). Grid and non-grid exposures utilize an assigned "grid" factor as follow: Non-grid: (1) 5:1 grid: (2) 6:1 grid: (3) 8:1: (4) 12:1: (5) 16:1: (6) For example, if a quantity of 10 mAs is required for an appropriate exposure with no grid, and an 8:1 grid was employed, the new mAs to maintain the same exposure would be 40 mAs. In another example, if a radiographer chose to employ a 12:1 grid when an 8:1 grid exposure would require a quantity of 10 mAs, the new mAs required to maintain the same exposure would be 12.5 mAs. Therefore, when using any grid or when increasing the grid ratio, an increase in patient dose is necessary. However, the benefit of minimizing scatter radiation fog in the image is diagnostically important, especially for larger anatomical structures that inherently produce more scatter radiation. The remaining formula in this question refers to the inverse square law, which calculates the changes in x-ray beam intensity that proportionately vary inversely to the square of the distance from the x-ray source (A, C, and D). (Bushong, 11th ed., p. 198; Fauber, 5th ed., p. 168)

Suppose a radiographer would normally use 5 mAs and 65 kVp on a neonate mobile AP abdomen radiograph using a non-grid DR detector array. However, upon arriving to the NICU, it is discovered that the neonate's abdomen is significantly distended due to ascites. The radiographer decides to use an 8:1 slip-on grid to reduce the amount of scatter radiation reaching the IR. Based on the condition of the patient and the use of an 8:1 grid, which of the following would be the best selection of technical factors the radiographer should use to maintain the desired exposure to the IR? A 20 mAs, 65 kVp B 10 mAs, 75 kVp C 10 mAs, 65 kVp D 12.5 mAs, 65 kVp

The answer is B. When changing from a non-grid technique to a grid technique, the radiographer must use the grid factor to calculate the change in mAs needed to compensate for the grid to maintain the desired IR exposure. The formula for changing grid ratios is the following: new grid Bucky factor divided by the original grid Bucky factor, multiplied by the original mAs. A non-grid exposure is assigned a factor of 1, since no grid is used. An 8:1 grid is assigned a factor of 4. Therefore, 4/1 × 5 mAs = 20 mAs and the original kVp would normally remain the same. However, ascites is an additive pathology, which would require an increase in kVp to properly penetrate the abdomen. By applying the 15% rule, the radiographer can half the mAs and add 10 kVp to ensure proper penetration and the desired IR exposure (B). While the grid factor calculation would normally require technical factors of 20 mAs, 65 kVp, this technique would not consider the additive pathological condition of the patient and the need for a more penetrating x-ray beam (A). The remaining selections would be incorrect, as either the incorrect grid factor or formula arrangement is incorrect, and the pathological condition was not considered (C and D). (Bushong, 11thed., pp. 132,198)

Which of the following best allows optimal visualization of both soft tissue and bony structures in a given image? A Narrow dynamic range B Wide dynamic range C Brightness gain D Structures recorded

The answer is B. When the x-ray beam exits the patient, it contains a wide range of x-ray intensities, often more than 1000-fold. To capture these intensity extremes, a receptor with awide dynamic rangeis required (B). Since digital imaging provides a wider dynamic range than film-screen imaging and can adjust for exposure errors, a greater margin of error exists. Dynamic range is the range of exposure intensities that an image receptor can respond to and acquire image data.A narrow dynamic rangeis a much shorter range that should not be used in digital radiology (A).Brightness gainis the output luminance level or brightness of an image intensifier (C).Structures recordedrefer to the actual image that is recorded on the displayed screen (D).(Johnston and Fauber, 2nded., pp. 101-102, 252)

Increasing the overall brightness of a radiographic digital image is accomplished by A raising the window level B lowering the window level C increasing the window width D decreasing the window width

The answer is B. Windowing is a postprocessing operation that changes the contrast and brightness of the digital image on the monitor. The brightness and contrast of the digital image depends on the shades of gray, which are controlled by varying the numerical values of each pixel. Modifying the window level will change the brightness of the image; raising/increasing the window level will decrease brightness and lowering/decreasing the level will increase brightness. Changing the window width will modify the grayscale (contrast) of the image; increasing the window width lengthens the grayscale (lower contrast) and decreasing the window width will shorten the grayscale (higher contrast). (Carroll 3rd ed p469, Carter and Veale 3rd ed p32)

Filament size can have an effect on image (select the three that apply) A contrast B brightness C spatial resolution D detail E gray scale F unsharpness

The answer is C, D, and F. Most x-ray tubes actually have two or more filaments and are called double-focus tubes. The typical x-ray tube has two filaments, one small and one large, to direct electrons to either the small or large anode focal spot. Focal spot size is a factor influencing image geometry, and has significant impact on spatial resolution (C). If x-ray photons were emitted from a single-point source, structures would be recorded and resolved with great clarity. However, because x-ray photons emerge from a measurable focus, image details are represented with unsharp edges (D). Saia, Radiography PREP, 9th edition. New York, NY: McGraw-Hill, 2018. Figure 13-36 A and B. Photons emerging from various points on a measurable focal spot are responsible for producing anatomic details having blurred, unsharp edges. The extent or size of the unsharp area is directly related to focal spot size and OID and inversely related to the SID (F). (Saia, PREP 9th ed., pp. 356, 408)

Tungsten is a common choice for anode targets. It has several favorable chemical properties that are useful to the production of x-rays. From the following list, select 3 properties that tungsten exhibits that make for a better radiographic x-ray tube. A High atomic number to increase photoelectric absorption B Ferromagnetism to attract electrons C High k-shell binding energy that yields a 69.5 keV photon during characteristic emission D High melting point, to better absorb heat generated E Radioactivity, to add natural gamma emissions of 125 keV to the artificially created beam F Conductivity, to shed valence electrons and create a net positive charge during exposure

The answer is C, D, and F. Tungsten is a great choice for a target material because of its high melting point, its electrical conductivity, and its k-shell binding energy (C, D, and F). The conservation of energy for the x-ray tube states that for 100% of electrical energy input, 99% of that energy is lost as heat production, and only 1% of the energy is actually transformed into x-ray photons. Therefore, a lot of heat will be generated, primarily at the actual focal spot. A material with a high melting point will be able to sustain more electron bombardment than other materials with lower melting points. In order to accelerate electrons, they must be pulled in by a considerable voltage. Outer shell electrons of tungsten have a low binding energy, and are removed as the high voltage transformer circuit is closed (second stage of exposure switch). If tungsten wasn't conductive, no net positive charge would manifest at the anode, and electrons at the filament would not be motivated to move. Magnetic fields don't attract electrons, they make them change direction (B). Finally, the k-shell binding energy yields a photon near the middle of the diagnostic range during characteristic x-ray production, and these photons can contribute to the useful portion of the beam. Tungsten does have a high atomic number and that does cause it to reabsorb some of the beam (A). Commonly known as the anode heel effect, this would be better classified as a disadvantage. The tungsten used in the anode is not radioactive, but cobalt 60 in a single aperture metal box has been used as a gamma beam source in radiation therapy (E).(Bushong 11th ed., p. 110)

Causes of overexposure using AEC include (select the three that apply) A insufficient backup time selected B selected photocell is under tissue less dense than area of interest C plus density control incorrectly selected D required exposure time more than minimum response time E selection of incorrect Bucky F selected photocell is under a radiopaque appliance/prosthesis

The answer is C, E, and F. The +/- density (intensity) control on an AEC functions to decrease or increase the exposure that would normally be delivered to the IR. Incorrect selection of + would increase receptor exposure (C). If wall Bucky is selected for tabletop exposure overexposure will result, as exposure continues because wall photocells are not activated by tabletop exposure (E). If the activated photocell is under a surgical appliance such as a hip pin, overexposure will occur because extended exposure is required for the metallic object while tissues are extremely overexposed (F). Selection of insufficient backup time is more likely to cause underexposure (A). If the selected photocell is under tissue less dense than the areas of interest, underexposure is more likely (B). If more exposure time than the selected minimum response time is required, underexposure is again the likely outcome (D). (Carroll, 2nd ed., p. 419)

Occasionally, it is difficult for the radiographer to obtain a diagnostic image of the odontoid (dens) process using the open-mouth technique. An alternative method, called the Fuch's method, may be used where the patient extends their head to place the mental and mastoid plane perpendicular to the image receptor. However, if the patient is unable to obtain the full extension needed, a cephalic angulation of the x-ray tube may be warranted. What would be the potential disadvantage(s) of such an angulation of the tube? A The chin would be projected cephalically to prevent superimposition over the odontoid process. B The odontoid process would appear within the opening of the foramen magnum. C The tube angulation would foreshorten the odontoid process and therefore may superimpose the borders of a fracture gap in such a way that the fracture may go undetected. D The thyroid dose would significantly increase.

The answer is C. A linear bony structure such as the odontoid process, which may have a transverse fracture, should be radiographed with the CR perpendicular to the long axis of the part and IR. This ensures passage of the x-rays through such a fracture gap, skimming its walls in a tangential fashion, thereby demonstrating the fracture gap on the radiographic image. Tube angulation would foreshorten the odontoid process and therefore may superimpose the borders of a fracture gap in such a way that the fracture may go undetected (C). With a cephalic angulation, the chin would be projected cephalically to prevent superimposition over the odontoid process (A). The odontoid process would be expected to project within the opening of the foramen magnum using the Fuch's method or a modification thereof using a cephalic angle (B). The thyroid dose would be similar for a given technique if the CR was perpendicular or angled cephalically, as the thyroid gland is in the path of the primary beam in both situations (D). (Long, Rollins, Smith, 13th. ed., vol. 1, p. 383; Bushong, 11th. ed., pp. 176-177)

What will cause the electrons from the photocathode to be focused on the output phosphor? 1. Light from the input phosphor 2. Electrostatic lenses 3. Concave curvature of the cathode end of the image intensifier A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The answer is C. An image intensifier (II) consists of several components, including the input phosphor, photocathode, electrostatic lenses, and output phosphor. To increase brightness in the fluoroscopic image, thecathode end of the IIhas aconcave curvature internally, wherein thelight from the input phosphorgenerated as it is struck by remnant x-rays interacts with the photocathode, which also has aconcave curvatureinternally. The photocathode emits electrons in proportion to the light energy received by the input phosphor. Since it is concavely curved, the electrons are propelled inwardly across the II as they travel toward the positively charge anode, where the output phosphor is located. Theelectrostatic lensesassist with this focus toward the output phosphor, as they are negatively charged and repel the electrons inward. Therefore, both the curvature of the anode end of the II and the negatively charged electrostatic lenses focus the electrons onto the output phosphor and control minification gain, a contributor to overall brightness gain (C).Light from the input phosphordoes not have the ability to redirect, or focus, the electrons, but do determine the number of electrons emitted from the photocathode (A, B, and D). (Bushong, 11thed., pp. 405-407)

Filtration of the radiographic x-ray beam functions to do which of the following? Reduce patient skin doseDecrease average beam energyProvide uniform exposure of an anatomical part with different thicknesses A 1 only B 2 and 3 only C 1 and 3 only D 1, 2, and 3

The answer is C. By filtering low-energy x-rays from the primary x-ray beam using inherent and added filtration, the patient's skin dose is reduced. Inherent filtration includes the x-ray tube window, oil in the x-ray tube housing, and the housing port. Added filtration includes an aluminum plate and the collimator mirror. The plastic cover on the bottom of the collimator can also filter some low-energy x-rays. Low-energy x-rays cannot penetrate the patient to reach the image receptor and are primarily absorbed in the skin. By absorbing these low-energy x-rays before they reach the patient, skin or entrance dose is reduced. Anatomical structures such as the foot have unequal thickness. The thickness is greater proximally and less distally. Therefore, a compensating filter such as a wedge filter properly attached to the bottom of the collimator provides the benefit of decreasing the x-ray beam intensity in the thicker portion of the filter positioned over the toes (thinnest) area and increasing the beam intensity in the thinner portion of the filter positioned over the tarsal (thickest) area. Such an arrangement will ensure a more uniform exposure over the entire foot (C). Inherent and added filtration serve to filter the low-energy x-rays from the primary beam, thereby increasing the average beam energy (A, B, D). (Bushong, 11th. ed., pp. 141-143)

Breast tissue is compressed in mammographic procedures to improve contrast resolution by reducing the production of Compton scatter. Which of the following also reduces Compton scatter production? A Using a grid B Increasing kVp C Decreasing tissue density (mass/volume) D Increasing average atomic number of tissue irradiated

The answer is C. Compton scatter is a quantum mechanical phenomenon, and therefore is a randomly occurring event. The probability that Compton scatter occurs can be influenced by several factors. Tissue density and tissue thickness both have a direct relationship, and as they decrease, the probability of Compton scatter decreases (C). Atomic number is a factor that influences photoelectric interaction, but does not increase or decrease Compton scatter production (D). Finally, increasing kVp increases Compton scatterrelativeto photoelectric absorption (B). A better way to say this is kVp exponentially decreases photoelectric effects, but only mildly decreases Compton scattering. Grids do not affect the production of scatter, they clean up scatter that has already formed (A).(Bushong 11th ed., p. 148-149)

An anatomical part calls for a kilovoltage (kVp) setting necessary to penetrate the part. If a lower than acceptable kVp were selected using a fixed milliampere-seconds (mAs) setting, this would result in which of the following? Increased patient doseIncreased exposure to the image receptorDecreased penetration of the anatomical part A 1 only B 2 and 3 only C 1 and 3 only D 1, 2, and 3

The answer is C. Decreased kilovoltage settings using a fixed mAs setting increases patient dose because of the increased number of photoelectric interactions. Decreased kVp results in the production of lower energy x-rays; therefore, the beam has less penetrating ability through the anatomy of interest (C).Because lower kVp settings result in an increased number of photoelectric interactions (absorption), fewer x-ray photons penetrate the patient and expose the image receptor (A, B, and D). (Bushong, 11th. ed., pp. 131, 139)

which of the following measures the percentage of x-rays absorbed by the detector? A FPD B PPI C DQE D DXR

The answer is C. Detective quantum efficiency measures the efficiency of a system to convert x-ray input signal into a useful output image (C). Systems with high DQE can convert more signal over a wider range of exposures resulting in higher-quality images and lower doses. A flat panel detector is a system used in digital radiography (A). Pixels per inch is the measure of the resolution (B). Digital X-ray radiogrammetry is a method for measuring bone mineral density (D). (Carter and Veale, 2nd ed., p. 33)

Two x-ray machines are being installed at competing outpatient clinics. While both have four diode diamond rectifiers, the first uses a high voltage transformer that has two vertical coils. The second machine utilizes a star and wye stack transformer to create high voltage. Neither machine features an inverter. What can be said regarding the x-ray beams created by both machines? A The first machine will produce a continuous stream of x-rays as the tube is energized. B The second machine will use more electricity to deliver the same quantity of photons in the beam than the first for a given technique. C The first machine will use more electricity to deliver the same quantity of photons in the beam than the second for a given technique. D The second machine will produce a continuous stream of X-rays with less than 1% voltage ripple during exposure.

The answer is C. Fundamental to answering the question is correctly classifying the two machines according to their output voltage waveforms. The first machine is single phase, full wave rectified power. The conventional two coil transformer indicates a single waveform, while a four-diode rectifier configured in a diamond shape will produce full-wave rectification. There will be two pulses of voltage per cycle, delivering an x-ray beam pulsed at a rate of 120 pulses per second. The second machine, utilizing a star and wye configuration, will produce a three-phase AC wave. All three troughs will be flipped to crests via the same four diode rectifier, and this will produce three-phase rectified power. Voltage ripple will be either 4% or 14%, and the tube will produce a continuous stream of x-rays, as the voltage will never drop to zero. Since machine 2 has a more efficient process, it uses less electric energy, and less exposure time, to deliver an equal quantity of radiation as machine 1. Since neither machine includes an inverter, high frequency generation is not possible (D). The other answer choices are incorrect as per the previous explanation (A, B, and C).(Bushong 11th ed., p. 96)

Characteristics of an effective grid interspace material include 1. Low atomic number 2. Absorbs moisture 3. Radiolucent A 1 only B 2 only C 1 and 3 only D 1, 2, and 3

The answer is C. Grid interspace material is used to precisely separate the lead grid strips of a radiographic grid. Whereas the lead strips have a high atomic number and are therefore radiopaque, the interspace material must have alow atomic numberand must beradiolucent. The lead in the grid strips effectively absorb Compton scattered x-rays emanating from the patient, thereby preventing image fog and improving visibility of detail. The interspace material must enable the transmission of image-forming x-rays traveling through the patient without interaction toward the image receptor (C). The interspace material must also be non-hygroscopic (does not absorb moisture). If moisture were absorbed, the interspace material could become warped and cause misalignment of the adjacent lead grid strips (A, B, and D).(Bushong, 11th ed., pp. 196-197)

Within the x-ray tube, why do we need a robust current passing through the cathode filament during exposure? A To pull electrons toward the anode B To rectify the voltage waveform C To create thermionic emission D To limit the space charge effect

The answer is C. Heat generated within a filament is directly proportional to the current passing through it. Breaking the word apart, in similar fashion to medical terminology, yields "therm-," which means heat, "ion-," which means an unbound charged particle, and "-ic," which means pertaining to. Thus, the entire word means "pertaining to the formation of unbound charged particles by applying heat." Robust current implies robust heat; robust heat implies many available electrons (C). Electrons are only pulled toward the anode if there is a large positive charge, created by activating the high voltage circuit (A). Rectification is undertaken in the high voltage circuit, between the step-up transformer and the tube, not within the tube (B). Finally, the space charge effect is more likely with large mA (robust) and is the natural result of "like" charges repelling (D). Excited electrons in the space charge inhibit additional electrons from joining them, via the coulomb force. It's not something we need to create for the production of x-rays, it's more of an unwanted side effect. (Bushong, 11th ed., p. 107)

Which of the following refers to anatomical shape distortion? 1. Magnification 2. Elongation 3. Foreshortening A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The answer is C. In radiography,shape distortionis a misrepresentation of the true shape of an anatomical part in the radiographic image. Distortion depends on the object (anatomical part) thickness, and its relative position with respect to the image receptor, the angle of the x-ray tube and image receptor, and the object's shape. Two types of shape distortion seen when the object plane and image receptor plane are not properly aligned are calledelongationandforeshortening(C).Magnificationis the misrepresentation of the true size of an anatomical part. Although there is some magnification in all radiographic images due to the fact that anatomical parts are three-dimensional and some portions of the anatomy are positioned farther away from the image receptor than others, the radiographer should use the longest practical source-to-image-receptor distance (SID) and place the object as close to the image receptor as possible (short object-to-image-receptor distance (OID)) to minimize magnification of the structure of interest (A, B, and D). (Bushong, 11thed., pp. 176-177)

The radiographer receives an x-ray order for a 1-year-old child. Without adequate immobilization techniques or accessories such as a Pigg-O-Stat™, which of the following image qualities could be affected? A Receptor exposure B Contrast C Spatial resolution D Distortion

The answer is C. Receptor exposure describes the amount of exposure reaching the IR (A). Contrast describes the visible difference between two brightness levels (B). Distortion is further defined as either size or shape distortion. Size distortion or magnification results in an image size that is larger than its actual size. Shape distortion occurs when the x-ray tube, body part, and image receptor are not in proper alignment (D). Motion is the worst enemy of detail...spatial resolution will be impacted negatively if motion occurs during the exposure (C). (Johnston & Fauber, 2nd ed., p. 96-97)

Once the values of interest numbers are assigned, what is the process that then adjusts an over/underexposed image? A Values B Brightness C Rescaling D Image contrast

The answer is C. Rescalingis the adjustment of the image by the computer program to present an image of predetermined image brightness (C). An image receptor that is overexposed is rescaled lighter, and an image that is underexposed is rescaled darker. The display of the image on the computer presents a consistent and uniform image over a wide range of exposures.Valuesare based on a numeric number for the image (A).Brightnessis a measure of image luminance on the display monitor and impacts the lightness and darkness of the displayed image and not the process of rescaling (B).Image contrastis the degree of density difference between two areas on a radiograph (D). Contrast makes it easier to distinguish between the anatomy.(Johnston and Fauber, 2nded., p. 166; Fauber, 5thed., p. 99)

What does HIS stand for under hospital organizations? A Healthcare information system B Health information system C Hospital information system D Housing information system

The answer is C. The hospital information system holds the patient's full medical information, from hospital billing to the inpatient ordering system (C). Information systems used throughout the hospital include direct patient care information, billing systems, and reporting systems. The other answer choices are incorrect as per the previous explanation (A, B, and D).(Carter and Veale, 2nded., p. 147)

What portion of the x-ray tube is used by an induction motor? A Filament B Focusing cup C Rotor D Housing

The answer is C. The induction motor creates an electromagnetic field that permeates the x-ray tube, and its magnetic flux engages the iron corerotorof the x-ray tube and causes it to rotate the anode disk during both the preparation stage of an x-ray exposure and during an exposure. This type of motor ensures that the x-ray tube vacuum is maintained, as a mechanical motor connected to the rotor would otherwise compromise the vacuum seal (C). Thefilamentis part of the x-ray tube cathode and is energized (heated) by the filament circuit (A). Thefocusing cupis also part of the cathode, is stationary, and is energized by an electrical current that imparts a negative charge that repels the filament electrons into a confined space (B). The x-ray tubehousingis stationary and is not energized. It contains the insulating and cooling oil and the x-ray tube (D). (Bushong, 11thed., pp. 78-79, 113)

The secondary side of the autotransformer features two multiple channel selection switches, to choose the number of coils that will supply voltage to the primary side of the high voltage transformer. Choosing more or less coils exploits A Ohm's Law B Coulomb's Law C the Transformer Law D E = mc2

The answer is C. The kVp selector is an array of multiple connection points to the autotransformer coils on the theoretical secondary side of the device. The autotransformer operates on self-induction, having one column of coiled wire that functions as both the primary and secondary side. By changing the number of coils selected, the technologist is setting the value for Ns, the number of turns (coils, loops, etc.) of the secondary side of the transformer. Remember that the Transformer law is represented as a proportion: Vs/Vp = Ns/Np By changing Ns, the technologist is indirectly changing secondary voltage (C). Ohm's Law is exploited in a similar way at the mA selector; Coulomb's Law is exploited at the focusing cup within the tube (A and B). To exploit E = mc2, the mass to energy conversion formula, you would need a PET scanner, nuclear power plant, or nuclear weapon (D). (Bushong 11th ed., p. 88)

Order the following PSP layers correctly from top to bottomReflective layerProtective coatConductive layerSupport layerPhosphor layer A 4, 5, 3, 1, 2 B 4, 2, 5, 1, 3 C 2, 5, 1, 3, 4 D 2, 1, 5, 4, 3

The answer is C. The layers of the PSP include the outer layer, the protective coat of clear plastic(2). Next is the phosphor layer (5). The thinner the phosphor layer, the less the light spread and the better the spatial resolution/detail. A common PSP phosphor is europium-doped barium fluorohalide (BaF:Eu2 mixed with a binder). Needle-shaped phosphors also reduce diffusion of light, improving resolution. The phosphor layer is followed by the reflective layer, which directs the emitted light toward the CR reader (1). Next is the conductive layer that functions to reduce static electrical buildup, and then the color layer, which also reflects the emitted light (PSL/ photostimulable luminescence) while absorbing the stimulating laser light (3). The backing layer is a support layer that provides a reinforcement for the entire PSP (4). A narrow monochromatic high-intensity (red) helium-neon laser or a solid-state laser beam scans back and forth multiple times across the PSP while it is simultaneously being pulled under the laser to progressively read the entire PSP. The PSP movement is referred to as translation. The PSL signal represents varying tissue densities and the latent image. The PMT or photodiode (PD) detects the PSL and converts it to electrical signals, which is then transferred to an analog-to-digital converter (ADC)—converting the analog electrical signal to digital data. This digital data is then transferred to a digital-to-analog converter (DAC) to be converted to a perceptible analog image on the display monitor. (Carter and Veale, Digital Radiography and PACS 3rd ed., p. 51)

What does the term RIS stand for in a Radiology department? A Radiography image system B Radiography information system C Radiology information system D Radiology image system

The answer is C. The radiology information system, or RIS, is the name of the computer system that tracks radiological procedure ordering and scheduling, patient database maintenance, reporting and transcription, and billing (C). Radiology information systems (RISs) serve to handle textual and other information portions stored on the PACS so that information and images are integrated and can be retrieved from any number of places. The other answer choices are incorrect as per the previous explanation (A, B, and D).(Johnston and Fauber, 2nded., p. 169)

Which of the following best describes the distance from the center of one pixel to the center of the adjacent pixel? A Pixel density B Pixel size C Pixel pitch D Pixel wide

The answer is C. Thepixel pitchis the spacing or distance measured from the center of a pixel to an adjacent pixel, which affects the spatial resolution of the digital image (C). Increasing thepixel densityand decreasing the pixel pitch increases spatial resolution. Pixel density is related to the display size in inches of the image (A).Pixel sizepertains to the size of the pixels; there are typically several thousand pixels in each direction (B).Pixel wideis not a term that is used in digital imaging (D).(Johnston and Fauber, 2nded., p. 105)

Which of the following decrease attenuation of the x-ray beam? 1. Positive contrast agents 2. Air 3. Negative contrast agents A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The answer is C. To adequately visualize anatomic detail on a radiographic image, the structure of interest must vary in brightness from its adjacent structures. If the area of interest has a similar atomic number, mass density, or thickness to those of its adjacent structures, the structures blend in with each other, and visibility of detail decreases. This decreased visibility of detail is due to inherent lowsubject contrast. To artificially enhance subject contrast such as in the vascular structures or alimentary tract, apositive contrast agentmay be introduced. A positive contrast agent (or medium) is one that has a high atomic number and mass density and increases x-ray attenuation in the structure through the process of photoelectric absorption. Fewer x-rays penetrate the structure, creating a brighter area on the radiographic image compared to that of its surrounding tissues.Iodinated agentsandbarium sulfate suspensionsare types of positive contrast agents that artificially make anatomical structures moreradiopaque(A, B, and D).Airis classified as anegative contrast agentand decreases attenuation of the x-ray beam. It is commonly introduced into the alimentary tract to decrease the average atomic number and mass density of the structures being evaluated. Since fewer photoelectric interactions occur, more of the x-rays penetrate the structures and strike the image receptor, creating a darker (less bright) area in the image and thereby distinguishing the structure from its surrounding tissues. Negative contrast agents include air, oxygen, and carbon dioxide, and when introduced into the alimentary tract decrease attenuation of the x-ray beam and create a darker area in the radiographic image, enabling the physician to visually examine and distinguish the structures from their surrounding tissues. Since air is considered a negative contrast agent and such agents artificially alter the atomic number and mass density of the anatomical structures in which they are instilled (making them moreradiolucent), all negative contrast agents therefore decrease attenuation of the x-ray beam (C). (Adler and Carlton, 6th ed., p. 303)

In order to double the x-ray quantity to the image receptor (IR) using kilovoltage-peak (kVp), the radiographer would need to raise the kVp by ______ percent. A 5 B 10 C 15 D 20

The answer is C. To double x-ray quantity, the radiographer would need to raise the kVp by 15 percent (C). This is governed by the 15 percent rule, which was developed through research performed on the effect of kVp on x-ray quantity and quality. The other answer choices are incorrect as per the previous explanation (A, B, and D). (Carlton/Adler/Balac, 6th ed., p. 162)

According to the line-focus principle, to produce an effective focal spot smaller than the actual focal spot the anode target angle must be less than ______ degrees. A 15 B 35 C 45 D 60

The answer is C. When the anode target angle is less than 45 degrees the effective focal spot is smaller than the actual focal spot (C). This is accomplished by the line-focus principle. A target angle of 60 degrees would not produce an effective focal spot smaller than the actual focal. The other answer choices are incorrect as per the previous explanation (A, B, and D). (Carlton, Adler, and Balac, 6th ed., p. 84)

For an AP projection of the knee, if the patient is unable to fully extend the lower extremity, what could the radiographer do to minimize magnification distortion? Use a compression band to extend the lower extremityIncrease the SID Perform a PA projection A 1 only B 1 and 2 only C 2 and 3 only D 2 only

The answer is C.When a patient presents with an injury or pathology that prevents them from fully obtaining the desired position, the radiographer should never use force to place the part into position. This would be considered intentional negligence and battery (A). Increasing the SID would enable the radiographer to use more of the central, less divergent, x-rays to create the radiographic image. An alternative would be to place the patient in a prone position and perform a PA projection of the affected knee, which would significantly reduce the OID. Of the two modified methods described, placing the patient in a prone position using a 40-inch SID would be best to take advantage of the divergent beam, which would cast the femoral structures superiorly and the lower leg inferiorly. The central ray would be best projected through the joint space using this method (C). (Long, Rollins, & Smith, 13th ed., vol. 1, p. 298)

Which of the following are potential undesirable effect(s) of scattered radiation? 1. Image fog 2. Increased patient dose 3. Increased radiographer dose A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The answer is D. After an incoming x-ray photon interacts with an intermediate or outer shell electron in a tissue atom (Compton interaction), it imparts energy to the electron to eject it from its shell and positively ionizes the atom. Next, and with lower energy (equal to the difference between the energy of the incident x-ray and the energy of the ejected electron), the x-ray photon is scattered in a different direction, sometimes having enough energy to exit the patient and strike the image receptor. This producesimage fog, which does not provide diagnostic information, but rather produces a supplemental gray brightness over the image, thereby decreasing contrast resolution (visibility of anatomical details). If the radiographer is in the x-ray room during an exposure, those scattered photons exiting the patient can causeincreased radiographer dose. Some of the scattered x-rays, however, do not have enough residual energy to exit the patient and therefore are absorbed in nearby tissue atoms, causing increased patient dose (A, B, C, and D).

Which of the following best describes the number of pixels per unit area? A Pixel size B Pixel pitch C Pixel wide D Pixel density

The answer is D. An image consisting of a greater number of pixels per unit area orpixel densityprovides improved spatial resolution (D). Pixels are measured per inch; a higher pixel density per inch allows for more sharpness and clarity for the image. Pixel density is related to the display size in inches of the image.Pixel sizepertains to the size of the pixels; there are typically several thousand pixels in each direction (A).Pixel pitchdescribes the distance between pixels in an image (B). Pixel wide is not a term that is used in digital imaging (C). (Johnston and Fauber, 2nded., p. 105)

What feature of digital imaging provides correction of over- or underexposure by way of histogram shift? A Lookup table B Display of image C Histograms D Automatic rescaling

The answer is D. Automatic rescalingis the adjusting of the image by the computer program to present an image of predetermined image brightness (D). Automatic rescaling means that images are produced with uniform brightness and contrast, regardless of the amount of exposure. Thelookup tableis a program within the computer system that gives an output value for each of a range of index values (A). Thedisplay of imageis where you would display the radiographic image on a computer screen (B). Ahistogramis a graph showing the number of pixels in an image (C).(Carter and Veale, 2nded., p. 41)

Which of the following is not considered a radiographic geometric factor? A Magnification B Focal spot size C OID D Beam filtration

The answer is D. Beam filtrationexists in the form of inherent and added filtration. Inherent filtration includes the x-ray tube window, oil in the x-ray tube housing, and housing port. Added filtration includes an aluminum plate and the collimator mirror. Filtration removes the lower-energy x-rays that would otherwise be absorbed within the patient's skin or other superficial tissues and would not contribute to the diagnostic image (D).Magnification, or size distortion, is the misrepresentation of the true size of the anatomical part being radiographed. Magnification also exaggerates penumbra (geometric unsharpness or image blur). By using a long SID and short OID, the radiographer can effectively minimize magnification (A).Focal spot sizeselection affects the amount of penumbra seen in the radiographic image. Larger focal-spot sizes increase penumbra, whereas smaller focal-spot sizes reduce penumbra and increase spatial resolution (B). Minimizing the object-to-image receptor distance (OID) reduces magnification distortion and improves spatial resolution (C).(Bushong, 11th ed., pp. 141-142)

An example of an indirect DR image receptor with high photoelectric capture of the remnant beam and a low amount of light spread is: A GdOS/a-Si (gadolinium oxysulfide/amorphous silicon) B CsI/a-Se (cesium iodide/amorphous selenium) C a-Se (amorphous selenium) D CsI/a-Si (cesium iodide/amorphous silicon)

The answer is D. DR image receptor technology is subdivided into two main classifications: direct capture and indirect capture. A direct capture system is choice (C); it uses an amorphous selenium capture layer to convert the remnant beam directly to charges captured in a capacitor. Indirect systems are those that convert the remnant beam into visible light first, and then convert the light to electrons in an amorphous silicon layer. Choices (A) and (D) are both indirect capture devices. Of these, the CsI, with atomic numbers of 55 and 53, respectively, increase the probability of photoelectric absorption greatly. Additionally, CsI crystals grow in needle-like structures, and trap light photons in ways similar to fiber optic cable. For these reasons, (D) is a better answer than (A), as GdOS is a turbid phosphor unable to control the spread of the light signal. Lastly, choice (B) is not a common arrangement of capture and coupling layers would not effectively capture the image. The energy associated with visible light would not be sufficient to create ionizations in the amorphous selenium layer. (Bushong, 11th ed., p. 301)

During prolonged use, such as fluoroscopy, the x-ray tube may acquire an abundance of heat. If a three-phase rectified fluoroscopy tube is used with a technique of 90 kVp and 4 mA for a total beam on time of 3 minutes and 5 seconds, what are the total heat units (Joules) produced? A 66,600 HU B 1.946 HU C 740 HU D 93,240 HU

The answer is D. Every x-ray tube has an upper threshold of heat capacity. If this capacity is exceeded, structural damage can manifest in the anode, the rotor bearings, and possibly the anode stem. It is important that the practicing technologist be aware of heat capacity and the calculation of heat units. The formula to determine total thermal energy has the general form: HU = kVp × mA × s × VW where kVp is kilovolts peak, mA is milliamperes, s is exposure time, and VW is the voltage waveform correction factor. For single-phase equipment, the correction factor is 1; for three-phase and high frequency equipment, the correction factor is 1.4. Multiplying 90 kV × 4 mA × 185 sec × 1.4 correction factor = 93,240 HU. Minutes were converted to seconds as a preliminary step: (3 × 60) + 5 = 185. Other incorrect answer choices can be arrived at by excluding the 1.4 modifier (A), dividing by the exposure time (B), or only calculating the total mAs(C). (Bushong 11th ed., p. 119-120)

Which of the following scenarios is most likely to result in grid cutoff? A Using a parallel grid with a large SID B Using a 48" focal length grid with a 50" SID C Angling the beam in the same direction as the length of the grid strips D Using a vertical beam with a grid that is not level

The answer is D. Grid cutoff can occur when the diverging rays of the x-ray beam are not in geometric alignment with the grid strips. A parallel grid will produce this effect with short SID (A). A focused grid may exhibit cutoff for a SID that is well outside its focal range; using a 48" focal length grid with a 50" SID is not a significant enough difference to create noticeable cutoff (B). Angling the beam in the same direction as the length of the grid strips and using a vertical beam with a grid that is not level are two sides of the same coin, with either an angle applied to the tube or the grid. Angling the beam in the same direction as the length of the grid strips is the only correct way to produce an axial projection with a grid (C). The off-level error described by using a vertical beam with a grid that is not level is often a hazard during a portable examination on a soft mattress (D).(Bushong, 11thed., p. 198)

What is the purpose of the aluminum layer in a CR (photostimuable phosphor) cassette? A The aluminum layer is used to filter low energy x-rays in the primary beam, saving patient dose. B The aluminum layer reflects blue light toward the photomultiplier tube during image processing. C The aluminum layer traps the electrons, which are later released via stimulation with red laser light. D The aluminum layer absorbs backscatter radiation, preserving the signal-to-noise ratio for the image.

The answer is D. In the CR cassette, an aluminum backing layer is employed to absorb scatter radiation generated when a high energy x-ray photon transmits through room air, patient, image receptor, and only then interacts with something behind the image receptor (D). If this interaction is Compton scattering, there is a probability that a lower energy x-ray photon will travel back toward the IR. This phenomenon is called backscattering and can add noise and artifact to the image. For example, a metal bed railing, used to hold a CR cassette vertically during a cross-table projection, may send backscatter into the cassette to project an image of the railing structure over the patient's anatomy, ruining the image. An aluminum layer behind the active phosphor layer will absorb much of this lower energy backscatter. Aluminum is also used in the collimator assembly, but the question is specifically asking about the metal's use within a CR cassette (A). Choices (B) and (C) describe functions of the reflective and barium fluorohalide layers, respectively. (Carter and Veale, 3rd ed., p. 44)

An Indirect DR system will always have a lower detective quantum efficiency and a lower spatial resolution than an equivalent Direct DR system. This is because A the indirect DR detector utilizes an electron trap to capture the signal. B the direct DR system converts the remnant beam to visible light. C the indirect DR system stores the image data as an analog signal. D the indirect DR system converts the remnant beam to visible light.

The answer is D. Indirect DR exposure capture and processing is a two-step process: The remnant x-ray beam is converted to light at the scintillator, and then the light is converted to an electronic signal in the photodiodes of the DELs on the TFT array (D). While the signal is light, it spreads from its point of origin within the scintillator, via the inverse-square law. A thicker scintillation layer will cause a noticeable decrease in spatial frequency. Each time energy is converted, some of the energy is lost as heat. The direct system, with less steps in the data capture process, necessarily will transmit more of the original signal to the CPU with less heat loss; thus, it will have a higher DQE than the indirect process (B). Electron traps are utilized only in PSP systems (A). Neither system is capable of storing an analog signal in memory (C). Direct systems do not convert the remnant beam to light, but rather straight to an electronic signal via an amorphous selenium layer. (Carter and Veale, 3rd ed., pp. 23, 66-67)

Which of the following can minimize magnification seen in a radiographic image? 1. Long SID 2. Short OID 3. Compression A 1 only B 2 only C 1 and 3 only D 1, 2, and 3

The answer is D. Magnification, or size distortion, is the misrepresentation of the true size of the anatomical part being radiographed. Magnification exaggerates penumbra (geometric unsharpness or image blur), which also decreases spatial resolution. By using along SIDand ashort OID, magnification can be minimized. Anatomical partcompressioncan reduce the thickness of some parts being radiographed, bringing the parts closer to the image receptor; therefore, magnification and image blur are reduced, and spatial resolution is improved (D). The remaining answer choices do not include all correct factors and are therefore incorrect (A, B, and C).(Bushong, 11th ed., pp. 174-175, 190, 382-383)

Misrepresentation of the true size of an anatomical structure is called A elongation B foreshortening C geometric unsharpness D magnification

The answer is D. Magnification, or size distortion, is the misrepresentation of the true size of the anatomical part being radiographed. To minimize magnification, the radiographer should use the longest practical source-to-image-receptor distance (SID) and object-to-image-receptor distance (OID) (D). Misrepresentation of the true shape of the anatomical part in the radiographic image is called shapedistortion. Shape distortion depends on the object (part) thickness, its relative position with respect to the image receptor, the angle of incidence of the x-ray beam on the part and image receptor, and the object's anatomical shape. Two types of shape distortion produced when the object plane and image receptor plane are not parallel are calledelongationandforeshortening(A and B).Geometric unsharpness(focal spot blur, or penumbra) is caused by undercutting of the edges of anatomical structures by x-rays emanating from different points of the anode focal spot (C). (Bushong, 11thed., pp. 173-175)

Each manufacturer's software system determines the gray scale available for when the x-ray image is displayed. Which of the following terms best describes this definition? A Modulation transfer function B Detective quantum efficiency C Exposure latitude D Dynamic range

The answer is D. Modulation transfer function or MTF, describes the system's ability to display signal contrast as a function of spatial resolution (A). Detective quantum efficiency or DQE, describes the overall efficiency of converting x-ray signals into a useful image (B). Exposure latitude is the range of exposure values that will produce an acceptable image (C). Dynamic range determines the gray scale available for the image (D). (Carroll, 2nd ed., p. 461)

A technologist is about to perform a trauma cross-table cervical spine on a hypersthenic patient who is short of breath. As the technologist is setting up for the projection, it becomes clear that the patient will not be able to hold their breath through the exposure. How should the technologist adjust technique to preserve image quality? A Change to a grid with a lower grid ratio. B Use a shorter SID. C Increase kVp by 15% and divided mA by 2. D Increase kVp by 15% and divide exposure time by 2.

The answer is D. Motion, whether in the midst of an x-ray exposure or not, requires the passage of time to manifest. It says so within the equation for velocity: v = d/t. If time has not lapsed, no motion has occurred. A trauma patient with labored, hyperventilated breathing may take two breaths per second, only one breath in a half-second, and only inhale in a quarter of a second, etc. For each smaller increment, the breath motion reduces. Reducing exposure time will reduce signal strength, so the technologist should compensate with increasing penetration power via kVp (D). The hypersthenic habitus already makes a kVp increase necessary, and now the tech has two good reasons to do so instead of just one. A lower grid ratio will increase scatter, which will not address motion and add noise, making the image look worse (A). A shorter SID may boost signal strength at the IR, but it will not correct breath motion (B). (Bushong, 11th ed., pp. 181-182)

Which of the following pathologies would justify an increase in technique? A Multiple myeloma B Pneumothorax C Emphysema D Atelectasis

The answer is D. Multiple myeloma (A), Pneumothorax (B), and Emphysema (C) are all destructive conditions that require a decrease in technique. Atelectasis makes the area of interest more difficult to penetrate. To compensate for this condition, the technique should be increased so the anatomy of interest can be properly penetrated (D). (Johnston & Fauber, 2nd ed., p. 200)

Which of the following represents the image acquisition contrast resolution? A Pixel B Pixel size C Pixel pitch D Bit depth

The answer is D. Pixel bit depth is known as the number of bits within a pixel which presents the number of gray tones that pixel can produce (D). The number of shades of gray can be calculated by raising 2 to the power of the bit depth, or 210 or 1,024 shades of gray. A pixel is the smallest element in a digital image (A). Pixel size is related to the spatial resolution (B). Pixel pitch is the distance between the centers of two adjacent pixels (C). The smaller the pixel and pixel pitch the greater the resolution. (Carter and Veale, 2nd ed., p. 26)

Biomedical repair technicians often perform repairs of diagnostic imaging equipment without a complaint of malfunction, but rather to test and replace parts of the machine known to experience wear and tear with normal operation. These types of repairs are called A acceptance testing. B quality assurance repairs. C service call repairs. D preventive maintenance.

The answer is D. Preventive maintenance is a quality control practice where equipment that is operating normally is tested and evaluated to find minor malfunctions before they affect the diagnostic quality of the images produced (D). It is commonly employed in diagnostic imaging departments to keep the equipment operating smoothly and avoid extended maintenance downtimes. Acceptance testing refers to testing that is conducted independently of the manufacturer, as the equipment is first installed, but before it is utilized on patients (A). While manufacturers are trusted to deliver, assemble, and test new exam rooms, it is recommended that the purchaser independently verify that the equipment is functioning properly before it is utilized in a clinical setting. Quality assurance refers to people, training, communication, and workflow issues, and would not be used to describe equipment repair (B). Service call repairs begin with a complaint of malfunction; usually by the radiologic technologist noticing erratic behavior or a loss in diagnostic quality while using the equipment (C). (Carter and Veale, 3rd ed., p. 164)

If insufficient x-ray quantity is supplied to a digital image receptor, which of the following will be seen on the final radiographic image? A Decreased image brightness B Increased contrast (shorter grayscale) C Decreased spatial resolution D Quantum mottle (noise)

The answer is D. Quantum mottle (noise) results from an insufficient quantity of x-ray photons supplied to the image receptor from improperly set exposure factors (mAs and kVp) (D). This produces a grainy image. This is corrected by increasing either the mAs or kVp, depending on which is appropriate for the exam. Image brightness, contrast (grayscale), and spatial resolution are image qualities controlled by the characteristics of the digital image receptors (A, B, and C). (Carlton, Adler, and Balac, 6th ed., p. 282-283)

Quantum mottle manifests within an image because A there was not enough filtration of the primary beam. B too much scatter radiation was produced. C a grid was not employed to effectively clean up the scatter. D a poor signal was captured by the image receptor.

The answer is D. Quantum mottle is a phenomenon that occurs when the data processing algorithm cannot distinguish between the true signal of the remnant beam and the background and inherent noise captured by the IR. The signal strength should be several multiples greater than the noise, so that the software can identify it clearly (D). Inadequate technique, with examples as follows, is usually the root cause: • Low kVp • Low mAs • SID too long • Technologist has not adjusted for hypersthenic habitus • Technologist has not adjusted for inclusion of a grid, or an increase in grid ratio (C) • Body part tapers excessively; technique appropriate for thinner aspect, rather than using a wedge filter Scatter radiation, both the production of and the cleanup of, do affect the amount of noise present on the image, but that alone does not create a quantum mottle appearance—there has to be a loss of signal as well (B). Filtration of the primary beam plays no role; if the filtration wasn't used, the low energy x-rays would be absorbed by the patient, and they would never reach the image receptor (A). By the way, it's against federal law to operate x-ray equipment without filtration.

Which of the following denotes undesirable fluctuations in brightness, dependent on the number of x-ray photons reaching the IR? A Grainy image B Artifacts C Degrades D Quantum noise

The answer is D. Quantum noiserefers to the noise on an x-ray image (D). The fewer the photons that reach the image receptor to form the image, the greater the quantum noise will be on a digital image. This is an undesirable fluctuation in image brightness.Grainy imageis not a correct term that is used in radiography; it is a term that states the appearance of an image (A).Artifactsare unwanted features in an image that mask or mimic a clinical feature (B).Degradespertain to an image that loses image quality and diagnostic information is lost (C).(Johnston and Fauber, 2nded., p. 99)

For which of the following procedures must the radiographer carefully consider exposure time? 1. Patients who are unable to cooperate 2. UGI radiography 3. Lateral thoracic spine radiography A 1 only B 2 only C 1 and 3 only D 1, 2, and 3

The answer is D. Some radiographic procedures require the radiographer to be particularly careful when considering and selecting the appropriate exposure time. Using a very short exposure time is important to avoid motion unsharpness caused by voluntary motion forpatients who are unable to cooperate, such as young children, the elderly, and those who are mentally challenged (A). When performingupper gastrointestinal (UGI) radiography, the radiographer should use an exposure time of no more than 0.2 seconds for patients with normal peristaltic motility and no more than 0.1 seconds for patients with hypermotility (B). When performinglateral thoracic spine radiography, a breathing technique is commonly used to blur the vascular pulmonary structures and the ribs during the exposure (C). When the breathing technique is used, the patient should be instructed to remain still but breathe normally during the exposure. An exposure time of 2-3 seconds should be used to allow for enough breaths to adequately blur the pulmonary vasculature and ribs, which improves the visibility of detail of the thoracic vertebrae (D).(Long, Rollins, & Smith, 13th ed., vol. 1 and 2, pp. 13, 214, 461)

There are two types of automatic exposure control: photodiode and the more common parallel plate ionization chamber. How are they positioned? A Both are between the image receptor and the patient B Both are behind the image receptor C The photodiode type is situated between the patient and the image receptor, while the ionization chamber is behind the image receptor D The ionization chamber type is situated between the patient and the image receptor, while the photodiode is behind the image receptor

The answer is D. The AEC measures the quantity of radiation and is calibrated to terminate the exposure when enough radiation to yield a diagnostic image has been deposited on the image receptor. The ionization chamber type is situated between the patient and the image receptor (D). The ionization chamber is relatively radiolucent compared to the photodiode type and will not interfere with the capture of the remnant beam significantly. By contrast, the photodiode type has to be coupled to a fluorescent screen, which will absorb a large percentage of the remnant radiation, and therefore has to be placed behind the image receptor so as not to degrade the remnant beam signal. The other answer choices are incorrect as per the previous explanation (A, B, and C). (Bushong, 11thed., p. 91)

Which of the following radiographic examinations can use the anode heel effect to best advantage? A AP foot B AP knee C PA skull D AP thoracic spine

The answer is D. The anode heel effect results in a reduction (increased absorption) in the number of x-rays exiting the area of the metallic heel (inferior) portion of the anode disk and is most pronounced with a small anode target angle. Because of this, the intensity of the incident x-ray beam on the anatomical part is decreased at the anode end of the x-ray tube. Therefore, the most intense portion of the beam (the cathode end) should be directed over the thickest part of the anatomy of interest to ensure uniform exposure to the image receptor. However, the anode heel effect can only be used advantageouslywhen the field size is large, such as that used when performing anAP thoracic spineradiograph (D). Smaller anatomical parts require increased collimation (smaller field sizes); therefore, the less intense portion of the beam is absorbed by the lead shutters and the anode heel effect cannot be applied advantageously in these situations (A, B, and C). (Bushong, 11thed., pp. 115-116)

On what principle does the stator located within the x-ray tube housing operate? A Thermal conduction B Line-focus C Kinetic energy D Electromagnetic induction

The answer is D. The anode of most diagnostic x-ray tubes used in medical imaging departments is of the rotating type. Rotating anodes allow the electron beam from the cathode to interact with a much larger anode target area compared to that of a stationary anode. Therefore, the rotating anode has a higher heat capacity than does a stationary anode, which enables higher exposure factor combinations. Anelectromagnetic inductionmotor is used to rotate the anode. It is composed of two parts separated by the enclosure of the x-ray tube insert. The part located outside of the enclosure is thestator, which consists of a series of electromagnets equally spaced around the anode end of the x-ray tube (specifically around the anode rotor). The windings of the iron bars of the stator are sequentially energized such that the induced magnetic field rotates on the stator axis. This rotating magnetic field permeates the enclosure of the x-ray tube insert, dragging the ferromagnetic rotor of the anode until it reaches a spinning speed of 3,400 to 10,000 revolutions per minute (RPM). By applying theelectromagnetic inductionprinciple to rotate the anode during x-ray exposures, a mechanical motor drive can be eliminated, thus preserving the vacuum seal in the x-ray tube (D).Thermal conductionis the transfer of thermal energy (heat) from one area of an object to another (A). Theline-focusprinciple enables high anode heating with small effective focal spots. As the target angle decreases, so does the effective focal spot size, which reduces penumbra and improves image resolution (B).Kinetic energyis the energy of motion. Whereas the rotating anode exhibits kinetic energy under the influence of the rotating electromagnetic fields, the stator remains stationary (C).(Bushong, 11th ed., pp. 111-112)

Which of the following cassette-less digital image receptors uses a scintillator to produce visible light?Indirect capture with charge-coupled device (CCD)Indirect capture with amorphous silicon photodetectorDirect capture with amorphous selenium photoconductor A 1 only B 2 only C 3 only D 1 and 2 only

The answer is D. The cassette-less digital imaging systems, which use the indirect capture method, utilize scintillators or phosphor elements, which emit light when struck by x-rays. Light emitted is then read by either a charge-coupled (CCD) device or an amorphous silicon photoconductor to convert the light energy to an electrical signal (1 and 2). In the direct capture method, x-rays are converted directly to an electrical signal in the amorphous selenium photoconductor (3). This method does not utilize scintillators or phosphor elements to convert the x-ray energy to light. (Johnston and Fauber, 2nd ed., p. 158-160)

If a 30/23/15 cm trifocus image intensifier is operated in the 15-cm mode, the fluoroscopic image will be magnified by a factor of A 1.0 B 1.3 C 1.5 D 2.0

The answer is D. The degree of magnification (magnification factor [MF]) is found by dividing the full-size input diameter (30 cm) by the selected input diameter (15 cm). MF = 30 cm / 15 cm = 2.0 (D). If the image intensifier was operated in the 23-cm mode, the MF would be 30 cm / 23 cm = 1.3 (B). If the image intensifier was operated in the 30-cm mode, the MF would be 30 cm / 30 cm = 1.0 (A). (Johnston and Fauber, 2nd ed., p. 208)

Which of the following is NOT true of spatial resolution, FOV, and matrix size as they relate to each other? A Increasing matrix size with no change to FOV will improve spatial resolution. B Changing FOV has no effect on matrix size. C Decreasing FOV with no other changes will improve spatial resolution. D Changing matrix size has no effect on spatial resolution.

The answer is D. The matrix is composed of rows and columns of pixels; the larger the matrix, the more pixels. The FOV (field of view) is the physical area of the remnant beam as it is recorded on the image receptor. The FOV, regardless of its size, is chosen by the technologist at the time of image exposure and applied to the matrix. The system software determines the matrix size, and cannot be adjusted by the technologist without switching to another machine. The pixel size is directly proportional to the change in FOV, when there are no other changes. Pixel size is inversely proportional to changes in matrix size when there are no other changes. Small pixels mean better spatial resolution. Therefore, it is not true that changing matrix size has no effect on spatial resolution (D). The other choices are true (A, B, and C). A good critical thinker would see that both (A) and (D) cannot be true, as they contradict each other. This is a useful tactic to eliminate other choices if you're unsure of the right answer. (Carter & Veale, 3rd ed., p. 19)

Which digital imaging characteristic is defined as the "sharpness of the structural edges recorded in an image"? (Carroll, Quinn B., 2nd ed., p.225) A Contrast resolution B Subject contrast C Short-scale contrast D Spatial resolution

The answer is D. The spatial resolution will determine the smallest size object that can be reproduced (D). Contrast resolution is measured by modulation transfer function (MTF), and is defined as a ratio of an image's recorded contrast to the actual subject's contrast (A,B). Short-scale contrast is determined by the scale of gray tones not the sharpness of the edges recorded in an image (C). (Carroll, 2nd ed., p. 382-387)

In preparing to use Automatic Exposure Control in combination with a PSP image receptor for an AP lumbar spine projection on a patient with a hypersthenic habitus, an optimal image can best be produced with which of the following changes to technical factors? A kVp is decreased to compensate for decreased patient size. B The outer cells of the AEC are active. C The AEC calibration is adjusted via the "+" density controls on the control panel, creating a longer exposure. D kVp is increased to compensate for increased part thickness.

The answer is D. Utilizing AEC with a patient that is hypersthenic, larger than average, can present a challenge to a novice technologist. Increasing kilovoltage for the exposure ensures that the beam arrives at the patient with greater penetration power, to compensate for the greater attenuation from the increased tissue thickness (D). Choice (A) was exactly the opposite of choice (D) and may indicate the student is unfamiliar with habitus nomenclature. Hyposthenic refers to a habitus smaller than average and asthenic is thinner yet. The root word sthenic means average build. Turning on the outer cells will cause the exposure to terminate prematurely, which may cause quantum mottle to appear in the spine structures. AEC is best utilized when the area of interest directly superimposes the active cells (B). If the density controls are used with PSP, the increased attenuation due to habitus will not be effectively compensated with an increase in exposure time (C). Instead, both patient dose and the chance of motion increase, while the PSP system software will automatically rescale any intensity gains achieved by the larger mAs. (Bushong, 11th ed., pp. 256-257)

A technologist setting a manual technique on a DR imaging system is contemplating changing mA from 300 down to 150, with no other changes to other factors. Which of the following effects will happen as a result of this change? A The patient dose will double. B The image will have increased brightness. C The image will have a shorter scale of contrast. D The signal-to-noise ratio will decrease.

The answer is D. mA stands for milliamperage and is a measure of the electric current passing through the tube during exposure. Multiplying the current with the duration of the exposure determines mAs. The number of photons in the beam is directly proportional to the mAs; therefore, a mAs that is halved reduces both the quantity of photons in the beam and the quantity of photons absorbed by the patient (dose) by half as well (A and D). In the digital age, image characteristics such as brightness and contrast are software controlled (B and C). The SNR is the magnitude of the useful remnant beam compared to the magnitude of "noise": random signals from unintended sources. The most common sources of noise are Compton scatter, created within the patient and image receptor characteristics, where the internal circuitry of the IR is creating small disturbances to the electromagnetic field. (Bushong 11th ed., pp. 313, 317)

A radiographic technique of 80 kilovoltage peak (kVp), 500 milliamperes (mA), and 0.25 seconds is selected for a radiographic exposure. How many heat units (HU) would be produced from this one exposure on a high frequency x-ray unit? A 10,000 B 13,500 C 14,100 D 14,000

The answer is D.Heat units (HU) for single phase radiographic tubes are calculated by using the formula: kVp × mA × time (in seconds). Heat units for three phase and high frequency x-ray equipment are determined by using the formula kVp x mA x time (in seconds) x a correction factor of 1.4.The calculation for this exposure would be as follows: 80 kVp x 500 mA x 0.25 second x 1.4 -= 14,000 (D). Choice (A) was found by using correction factor of 1, choice (B) was found by using correction factor of 1.35, and choice (C) was found by using a correction factor of 1.41. (Bushong 11th ed p 120) (Carlton, Adler, and Balac 6th ed p 92)

If a high-voltage transformer has 100 primary turns and 35,000 secondary turns, and is supplied by 220 V and 75 A, what are the secondary voltage and current? A 200 A and 77 V B 200 mA and 77 kVp C 20 A and 77 V D 20 mA and 77 kVp

The correct answer is (B). The high-voltage, or step-up, transformer functions to increase voltage to the necessary kilovoltage. It decreases the amperage to milliamperage. The amount of increase or decrease is dependent on the transformer ratio-the ratio of the number of turns in the primary coil to the number of turns in the secondary coil. The transformer law is as follows: To determine secondary V, To determine secondary I: Substituting known factors, 35,000/100 = x/220 100x = 7,700,000 x = 77,000 V (77kVp) 35,000/100 = 75/x 35,000x = 7,500 x = 0.214 Amps (214 mA) (Selman, pp 84-85)

If 92 kV and 12 mAs were used for a particular abdominal exposure with single-phase equipment, what mAs would be required to produce a similar radiograph with three-phase equipment? A 36 B 24 C 6 D 3

The correct answer is (C). Single-phase radiographic equipment is much less efficient than three-phase equipment because it has a 100% voltage ripple. With three-phase equipment, voltage never drops to zero, and x-ray intensity is noticeably greater. To produce similar receptor exposure, the original mAs would be cut in half. (Bushong 11th ed p133)

Decreasing field size from 14 × 17 into 8 × 10 inches will A decrease receptor exposure and increase the amount of scattered radiation generated within the part. B increase receptor exposure and increase the amount of scattered radiation generated within the part. C increase receptor exposure and decrease the amount of scattered radiation generated within the part. D decrease receptor exposure and decrease the amount of scattered radiation generated within the part.

The correct answer is (D). Limiting the size of the radiographic field serves to limit the amount of scattered radiation produced within the anatomic part. As the amount of scattered radiation generated within the part decreases, so does the resultant signal or amount of radiation received by the image receptor. Hence, beam restriction is a very effective means of reducing the quantity of non-information-carrying scattered radiation (fog) produced.

A device used to measure the luminance response and uniformity of monitors used in digital imaging is called a A Penetrometer B Densitometer C Sensitometer D Photometer

The correct answer is (D). Two types of photometers (D) are commonly used to measure the luminance response and uniformity of monitors used in digital imaging: near-range and telescopic. Near-range photometers are used for measuring the monitor's luminance at close range, whereas telescopic photometers measure this from a distance of one meter. Background ambient light should be kept constant when either photometer is used. A penetrometer (or aluminum step wedge) (A) is a device used for quality control testing in film radiography. After making an exposure of this device while it rests on top of a film cassette, the film within the cassette is chemically processed. The resultant image demonstrates multiple steps of densities. The densities can be measured by a densitometer (B) to determine the film contrast index and other processing-related factors. A sensitometer (C), which is an electrical device, can be used in lieu of the penetrometer and projects a preset (visible light) exposure on the film in the darkroom. After the film is processed, multiple steps of densities, similar to those achieved using the penetrometer, are demonstrated and can then be measured by a densitometer in the same fashion (A, B, C). (Bushong, 9 th ed., p. 480).

There is direct relationship between spatial resolution and SID.tube current.focal-spot size. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The correct answer is: (A) As SID increases, spatial resolution increases because magnification is decreased - a direct relationship. Therefore, SID is directly related to spatial resolution. As focal spot size increases, spatial resolution decreases because more penumbral blur is produced. Focal spot size is thus inversely related to spatial resolution - as FSS increases,resolution decreases. Tube current affects receptor exposure and is unrelated to spatial resolution. (Fauber, 2nd ed., pp. 79, 81)

The X-ray scintillator layer used with indirect flat-panel digital detectors is usually either _____________ or ______________. A Silicon dioxide, silver halide B Cesium iodide, gadolinium oxysulfide C Yttrium oxysulfide, barium fluoride D Amorphous silicon dioxide, barium platinocyanide

The correct answer is: (B) The X-ray scintillator used in the indirect flat-panel digital detector is usually either cesium iodide (CsI) or gadolinium oxysulfide (Gd 2 O 2 S). These phosphors are not new to x-ray imaging; they have been used in x-ray image intensifiers (CsI) and in rare-earth intensifying screens (Gd 2 O 2 S) for many years (B). Silicon dioxide is a substance used in sonographic imaging, whereas silver halide was found in radiographic film emulsions (A). Yttrium oxysulfide was a phosphor material used in rare earth radiographic screens, whereas barium fluoride is a component of barium fluoride bromide crystals coated with europium in computed radiography (CR) PSPs (C). Amorphous silicon dioxide is a material used in the photodiodes used in indirect digital flat panel detectors, whereas barium platinocyanide was a phosphor material used in experiments conducted by Wilhelm Roentgen. (Seeram, 1 st ed., p. 108)

A satisfactory radiograph of the abdomen was made at a 42-inch SID using 300 mA, 0.06-second exposure, and 80 kVp. If the distance is changed to 38 inches, what new exposure time would be required? A 0.02 second B 0.05 second C 0.12 second D 0.15 second

The correct answer is: (B)According to the inverse square law of radiation, as the distance between the radiation source and the IR decreases, the exposure rate increases. Therefore, a decrease in technical factors is indicated. The exposure maintenance formula is used to determine new mAs values when changing distance: Then, to determine the new exposure time (mA × s = mAs), 300x = 14.7 x = 0.049 second at 300 mA (Selman, p 214)

If a radiograph exhibits insufficient receptor exposure, this might be attributed to1.insufficient kVp.2.insufficient SID.3.grid cutoff. A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The correct answer is: (C) As kVp is reduced, the number of high-energy photons produced at the target is reduced; therefore, a decrease in receptor exposure occurs. If a grid has been used improperly (off-centered or out of focal range), the lead strips will absorb excessive amounts of the useful beam, resulting in grid cutoff and loss of rreceptor exposure. If the SID is inadequate (too short), an increased receptor exposure will result. (Selman, pp 214, 240-242)

An exposure was made using 600 mA, 0.04 sec exposure, and 85 kVp. Each of the following changes will serve to reduce the receptor exposure by one-half except change to A 1/50 sec exposure B 72 kVp C 18 mAs D 300 mA

The correct answer is: (C) Receptor exposure is directly proportional to milliampere-seconds. 600 mA with 0.04 sec = 24 mAs. Therefore, we need to reduce receptor exposure to 12 mAs, or its near equivalent. If exposure time is halved from 0.04 sec to 0.02 (1/50) sec, receptor exposure will be cut in half. Changing to 300 mA also will halve the milliampere-seconds, effectively halving the receptor exposure. If the kilovoltage is decreased by 15%, from 85 to 72 kVp, receptor exposure will be halved according to the 15% rule. To cut the receptor exposure in half, the milliampere-seconds must be reduced to 12 mAs (not 18 mAs). (Carlton, Adler, and Balac, 6th ed., p. 358)

A particular milliampere-seconds value, regardless of the combination of milliamperes and time, will reproduce the same receptor exposure. This is a statement of the A line-focus principle B inverse-square law C reciprocity law D law of conservation of energy

The correct answer is: (C) The reciprocity law states that a particular milliampere-seconds value, regardless of the milliamperage and exposure time used, will provide identical receptor exposure. Milliampere-seconds is directly proportional to beam intensity and receptor exposure .(Shephard, p. 193)

Periodic equipment care includes evaluation of the1.kV.2.milliamperage.3.timer. A 1 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The correct answer is: (D) Radiographic results should be consistent and predictable, not only with regard to positioning accuracy, but with respect to exposure factors and image clarity as well. X-ray equipment and accessories must be calibrated periodically as part of an ongoing QA program. Image receptors should be cleaned and evaluated regularly. The quantity (mAs) and quality (kVp) of the primary beam have a big impact on the quality of the image, and their accuracy, along with that of the x-ray timer, should be assessed regularly. Kilovoltage accuracy can be evaluated with a Wisconsin test tool or digital meter and must be accurate to within 5 kV (+/` 10%). The focal spot should be tested periodically to evaluate its impact on image sharpness.

If a 6-in. OID is introduced during a particular radiographic examination, what change in SID will be necessary to overcome objectionable magnification? A The SID must be increased by 6 in.. B The SID must be increased by 18 in.. C The SID must be decreased by 6 in.. D The SID must be increased by 42 in..

`The Correct Answer is: DAs OID is increased, spatial resolution is diminished as a result of magnification distortion. If the OID cannot be minimized, an increase in SID is required to reduce the effect of magnification distortion. However, the relationship between OID and SID is not an equal relationship. In fact, to compensate for every 1 in. of OID, an increase of 7 in. of SID is required. Therefore, an OID of 6 in. requires an SID increase of 42 in.. This is why a chest radiograph with a 6-in. air gap usually is performed at a 10-ft SID. (Saia, 4th ed., p. 290)

Which of the following are methods to prolong x-ray tube life? (select the three that apply) A Perform the x-ray tube warm-up procedure when the anode is cold B Use long exposure time settings C Limit exposure prep time D Avoid performing certain projections in the department's exam protocol E Reduce repeat radiographs F Keep the kVp setting constant for all exams

`The answer is A, C, and E. The radiographer shouldperform the x-ray tube warm-up procedure when the anode is cold, according to the manufacturer's recommendations, to avoid rapid overheating of the cold anode during an exposure, which may lead to cracking of the anode. Typically, a low kVp, low mA, and long exposure time is used for multiple warm-up exposures (A). The radiographer shouldlimit exposure prep time, when possible, as "boost-and-hold" causes anode-bearing wear and evaporation of the cathode filament (C). The radiographer should alsoreduce repeat radiographs, not only to prolong x-ray tube life, but to also minimize patient dose (E).Using long exposure timedoes not prolong x-ray tube life. Shorter exposure times are generally required to prevent patient motion unsharpness. However, some exposures should be lengthened during some exams, such as a lateral thoracic spine radiographic projection, to take advantage of the blurring of ribs and lung vasculature during the exposure while the patient is breathing (B). The radiographer should neveravoid performing certain projections in the department's exam protocol.Such a decision is beyond the radiographer's scope of practice (D). If the radiographer were tokeep the kVp setting constant for all exams, proper x-ray penetration of the anatomical parts, particularly thicker or more dense ones, would not be achieved (F). (Bushong, 11th ed., pp. 12,113, 118)

Positive beam limitation devices must be accurate to within A 5% of the SID B 2% of the OID C 10% of the mAs D 2% of the SID

`The answer is D. The lead shutters of a positive beam-limiting (PBL) collimator automatically adjust the x-ray field size to match that of the image receptor being used. The x-ray field consisting of the primary beam must properly align with the light field of the collimator. Misalignment will prevent demonstration of the entire anatomical area of interest and a portion of the beam will expose unintended anatomy. The PBL function must be periodically evaluated as part of a quality control program. When the PBL function is engaged, the x-ray field must not be larger than the image receptor dimensions and any misalignment must not exceed 2 percent of the SID (D). Five percent of the SID misalignment would cause the beam to miss even more of the anatomy of interest and expose more unintended anatomy (A).The object-to-image receptor distance (OID)affects magnification of the anatomy of interest but does not affect the primary beam dimensions affected by the SID (B). The x-ray intensity (mAs) should remain constant when changing mA stations, and when appropriately adjusting the exposure time to maintain the same mAs. Similarly, when adjacent mA stations are used (e.g., changing from 50 mA to 200 mA), but the exposure time is kept the same, a doubling of x-ray intensity (expressed in units of mGya/mAs) would be expected. This is called exposure linearity and should not exceed a tolerance level of 10% variations in exposure (C).(Bushong, 11th ed., p. 346)

Exposure factors of 110 kVp and 12 mAs are used with an 8:1 grid for a particular exposure. What should be the new mAs if a 12:1 grid is substituted? A 3 B 9 C 15 D 18

c

A radiograph made with a parallel grid demonstrates decreased receptor exposure on its lateral edges. This is most likely due to A static electrical discharge B the grid being off-centered C improper tube angle D decreased SID

qThe Correct Answer is: DThe lead strips in a parallel grid are parallel to one another and, therefore, are not parallel to the x-ray beam. The more divergent the x-ray beam, the more likely there is to be cutoff/decreased receptor exposure at the lateral edges of the image. This problem becomes more pronounced at short SIDs. If there were a centering or tube angle problem, there would be more likely to be a noticeable receptor exposure loss on one side or the other. (Carlton and Adler, 4th ed., p. 260)

Which of the following technical changes would best serve to remedy the effect of very dissimilar tissue densities? A Use of short exposure time B Use of a high-ratio grid C High-kilovoltage exposure factors D High milliampere-seconds exposure factors

the The Correct Answer is: CWhen tissue densities within a part are very dissimilar (e.g., the chest), the radiographic result (especially analog) can be unacceptably high contrast. To "even out" these exposure values and produce a more appropriate scale of grays, exposure factors using high kilovoltage should be employed. The higher the grid ratio, the higher is the resulting contrast. Use of short exposure time is always encouraged to reduce the possibility of motion unsharpness but has no impact on varying tissue densities. Exposure factors using high milliampere-seconds generally result in excessive receptor exposure, frequently obliterating much of the gray scale. (Bushong, 8th ed., p. 273; Shephard, p. 200)