Rad Review: Image Production

A 15% decrease in kilovoltage accompanied by a 50% increase in milliampere-seconds will result in a(n) A increase in patient dose B increase in exposure latitude C decrease in receptor exposure D decrease in spatial resolution

The Correct Answer is: A A 15% decrease in kilovoltage with a 50% increase in milliampere-seconds produces an image similar to the original but with significant differences. The receptor exposure and patient dose are doubled because of the increase in milliampere-seconds. Exposure latitude is wide in CR and DR, controlled by computer software. Spatial resolution is unaffected by changes in kilovoltage

Which of the following groups of technical factors would be most appropriate for the radiographic examination shown in Figure 4-30? A. 400 mA, 1/30 s, 72 kV B. 300 mA, 1/50 s, 82 kV C. 300 mA, 1/120 s, 94 kV D. 50 mA, ¼ s, 72 kV

The Correct Answer is: A A 15-minute oblique image of an IVU is pictured. IVU requires the use of iodinated contrast medium. Low kilovoltage (about 70 kV) usually is employed to enhance the photoelectric effect and, in turn, 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 shorter exposure time is preferred to decrease the possibility of motion

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: A A 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.

The line-focus principle refers to the fact that A the actual focal spot is larger than the effective focal spot B the effective focal spot is larger than the actual focal spot C x-rays travel in straight lines D x-rays cannot be focused

The Correct Answer is: A A distinction is made between the actual focal spot and the effective, or projected, focal spot. The actual focal spot is the finite area on the tungsten target that is actually bombarded by electrons from the filament. The effective focal spot is the foreshortened size of the focus as it is projected down toward the image receptor. This is called line focusing or the line-focus principle. The quoted focal-spot size is the effective focal-spot size.

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

The Correct Answer is: A A motor is a 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.

A parallel-plate ionization chamber receives a particular charge as x-ray photons travel through it. This is the operating principle of which of the following devices? A AEC B Image intensifier C Cine camera D Spot camera

The Correct Answer is: A A parallel-plate ionization chamber is the most commonly used AEC. A radiolucent chamber is beneath the patient (between the patient and the IR). As photons emerge from the patient, they enter the chamber and ionize the air within it. Once a predetermined charge has been reached, the exposure is terminated automatically.

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

The Correct Answer is: A A 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.

A star pattern is used to measure 1. 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: A A 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.

In Figure 5-8, what is the maximum safe milliamperage that may be used with a 0.10-s exposure and 120 kVp, using the 3-phase, 1.0-mm focal-spot x-ray tube? A 300 mA B 400 mA C 500 mA D 600 mA

The Correct Answer is: A A radiographic rating chart enables the radiographer to determine the maximum safe milliamperage, exposure time, and kilovoltage for a given exposure using a particular x-ray tube. Because the heat load that an anode will safely accept varies with the size of the focal spot, type of rectification, and anode rotation, these variables must also be identified. Each x-ray tube has its own characteristics and its own rating chart. Find the correct chart for the three-phase, 1.0-mm focal-spot x-ray tube. Locate 0.1 s on the horizontal (seconds) axis and follow it up to where it intersects with the 120-kVp line on the vertical (kVp) axis. They intersect just below the 300-mA curve, at approximately 310 mA. Thus, 300 mA is the maximum safe milliamperage for this particular group of exposure factors and x-ray tube

Which of the illustrations in the figure below depicts the ionization-chamber type of automatic exposure control? A. Figure a. B. Figure b. C. Both are ionization-chamber-type AEC. D. Neither is ionization-chamber-type AEC.

The Correct Answer is: A AEC devices are used in today's equipment and serve to produce consistent and comparable radiographic results. In one type of AEC, there is an ionization chamber just beneath the tabletop above the IR (A). The part to be examined is centered to the sensor and radiographed. When a predetermined quantity of ionization has occurred (equal to the correct receptor exposure), the exposure automatically terminates. In the other type of AEC, the phototimer, a small fluorescent screen is positioned beneath the IR (B). When remnant radiation emerging from the patient exits the IR, the fluorescent screen emits light. Once a predetermined amount of fluorescent light is "seen" by the photocell sensor, the exposure is terminated. A special IR, one without lead foil backing, is often required with this type of AEC. In either case, the manual timer should be used as a backup timer. In case of AEC malfunction, this would terminate the exposure, thus avoiding patient overexposure and tube overload.

Foreshortening can be caused by A the radiographic object being placed at an angle to the IR B excessive distance between the object and the IR C insufficient distance between the focus and the IR D excessive distance between the focus and the IR

The Correct Answer is: A Aligning the x-ray tube, anatomic part, and IR so that they are parallel reduces shape distortion. Angulation of the long axis of the part with respect to the IR results in foreshortening of the object. Tube angulation causes elongation of the part. Size distortion (magnification) is inversely proportional to SID and directly proportional to OID. Decreasing the SID and increasing the OID serve to increase size distortion

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

The Correct Answer is: A All 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.

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: A An 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.

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: A An 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

The CR should be directed to the center of the part of greatest interest to avoid A rotation distortion B magnification C foreshortening D elongation

The Correct Answer is: A Anatomic details placed away from the path of the CR will be exposed by more divergent rays, resulting in rotation distortion. This is why the CR must be directed to the midpoint of the part of greatest interest. For example, if bilateral hands are requested, they should be examined individually; if imaged simultaneously, the CR will be directed to no anatomic part (between the two hands) and rotation distortion will occur. Magnification occurs when an OID is introduced, or with a decrease in SID. Foreshortening and elongation are the two types of shape distortion—caused by nonalignment of the x-ray tube, part/subject, and IR.

Spatial resolution is directly related to 1. 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: A As SID increases, so does spatial resolution, because magnification is decreased. Therefore, 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.

A decrease in kilovoltage will result in A a decrease in receptor exposure B a decrease in image contrast C a decrease in spatial resolution D an increase in spatial resolution

The Correct Answer is: A As kilovoltage is increased, more electrons are driven to the anode with greater speed and energy. More high-energy electrons will result in production of more high-energy x-rays. Thus, kilovoltage affects both quantity and quality (energy) of the x-ray beam. However, although kilovoltage and receptor exposure are directly related, they are not directly proportional; that is, twice the radiographic receptor exposure is not achieved by doubling the kilovoltage. If it is desired to double the receptor exposure yet impossible to adjust the mAs, a similar effect can be achieved by increasing the kilovoltage by 15%. Conversely, the receptor exposure may be cut in half by decreasing the kilovoltage by 15%. Therefore, a decrease in kilovoltage will produce fewer x-ray photons, resulting in decreased receptor exposure. A decrease in kilovoltage will produce fewer shades of gray in analog imaging, that is, a shorter-scale, or higher/increased, contrast. Kilovoltage is unrelated to spatial resolution

X-ray photon energy is inversely related to photon wavelength applied milliamperes (mA) applied kilovoltage (kV) A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: A As kilovoltage is increased, more high-energy photons are produced, and the overall energy of the primary beam is increased. Photon energy is inversely related to wavelength; that is, as photon energy increases, wavelength decreases. An increase in milliamperage serves to increase the number of photons produced at the target but is unrelated to their energy.

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: A As 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.

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: A As 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: Image size/Object size = SID/SOD (SOD=SID-OID) Substituting known factors the equation becomes: x/6 = 44/35 (44-9=35) 35x = 254 x = 7.5 inches projected image width (rounded up to 8 inches)

The functions of automatic beam limitation devices include reducing the production of scattered radiation increasing the absorption of scattered radiation changing 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: A Beam 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

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: A Because 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.

If a radiograph, made using AEC, is overexposed because an exposure shorter than the minimum response time was required, the radiographer generally should A decrease the milliamperage B increase the milliamperage C increase the kilovoltage D decrease the kilovoltage

The Correct Answer is: A Decreasing or increasing the kilovoltage will produce a change in radiographic contrast. The image was overexposed (from excessive exposure time) because the AEC wasn't capable of producing the required extremely short exposure time. To bring the required exposure to an exposure time the AEC is capable of, the mA should be decreased (thus requiring a longer exposure time, within the capability of the AEC).

The interaction between x-ray photons and matter illustrated in Figure 4-22 is most likely to be associated with 1. high kilovoltage 2. high contrast 3. high-atomic-number absorber A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The Correct Answer is: A Diagnostic x-ray photons interact with tissue in a number of ways, but mostly they are involved in the production of Compton scatter or the photoelectric effect. Compton scatter is pictured; it occurs when a relatively high-energy (kV) photon uses some of its energy to eject an outer-shell electron. In so doing, the photon is deviated in direction and becomes a scattered photon. Compton scatter causes objectionable scattered radiation fog in large structures such as the abdomen and poses a radiation hazard to personnel during procedures such as fluoroscopy. In the photoelectric effect, a relatively low-energy x-ray photon uses all its energy to eject an inner-shell electron, leaving a hole in the K shell. An L-shell electron then drops down to fill the K vacancy and in so doing emits a characteristic ray whose energy is equal to the difference between the binding energies of the K and L shells. The photoelectric effect occurs with high-atomic-number absorbers such as bone and positive contrast media and is responsible for the production of radiographic contrast. It is helpful for the production of the radiographic image, but it contributes to the dose received by the patient (because it involves complete absorption of the incident photon).

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: A Digital 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.

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: A Each 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

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: A Edge 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

Exposure rate will decrease with an increase in SID kilovoltage 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 Exposure rate decreases with an increase in SID according to the inverse-square law of radiation. The quantity of x-ray photons produced at the focal spot is the function of milliampere-seconds. The quality (i.e., wavelength, penetration, and energy) of x-ray photons produced at the target is the function of kilovoltage. The kilovoltage also has an effect on exposure rate because an increase in kilovoltage will increase the number of high-energy x-ray photons produced at the anode

Geometric unsharpness is directly influenced by OID SOD SID A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: A Geometric unsharpness is affected by all three factors listed. As OID increases, so does magnification—therefore, OID is directly related to magnification. As SOD and SID decrease, magnification increases—therefore, SOD and SID are inversely related to magnification

Grid cutoff due to off-centering would result in A overall loss of receptor exposure B both sides of the image being underexposed C overexposure under the anode end D underexposure under the anode end

The Correct Answer is: A Grids are composed of alternate strips of lead and interspace material and are used to trap scattered radiation after it emerges from the patient and before it reaches the IR. Accurate centering of the x-ray tube is required. If the x-ray tube is off-center but within the recommended focusing distance, there usually will be an overall loss of receptor exposure. Over- or under-exposure under the anode is usually the result of exceeding the focusing distance limits in addition to being off-center.

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: A Grids 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

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: A IVU 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

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: A If 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

Foreshortening of an anatomic structure means that A. it is projected on the IR smaller than its actual size B. its image is more lengthened than its actual size C. it is accompanied by geometric blur D. it is significantly magnified

The Correct Answer is: A If a structure of a given length is not positioned parallel to the recording medium (PSP or film), it will be projected smaller than its actual size (foreshortened). An example of this can be a lateral projection of the third digit. If the finger is positioned so as to be parallel to the IR, no distortion will occur. If, however, the finger is positioned so that its distal portion rests on the cassette while its proximal portion remains a distance from the IR, foreshortening will occur.

To produce a just perceptible increase in receptor exposure, 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: A If 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

A radiograph exposed using a 12:1 ratio 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: A If 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

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: A If 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.

If the center photocell were selected for a lateral projection of the lumbar spine that was positioned with the spinous processes instead of the vertebral bodies centered to the grid, how would the resulting radiograph look? A The image would be underexposed. B The image would be overexposed. C The image would be correctly exposed. D An exposure could not be made.

The Correct Answer is: A If the photocell were centered more posteriorly to a thinner and less dense structure, then the exposure received would be correct for that less-dense structure. The spinous processes would be well visualized, but the denser vertebral bodies and surrounding structures (pedicles and lamina) would be underexposed. Accurate selection of photocells and precise positioning are critical with the use of automatic exposure devices

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: A In 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: **1 cm = 10 mm 60 cm (x10) = 600 mm 600/2048 = 0.29 mm (rounded up 0.3 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.

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: A In 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

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: A Indirect digital systems use scintillators/phosphors (B) to convert X-ray energy, whereas direct digital systems use a photoconductor (A) to covert this energy. This energy is subsequently converted by either a charged coupled device (CCD) array or photodiode array (coupled with a thin film transistor array) in the two types of indirect systems, 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)

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

The Correct Answer is: A It 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

The primary function of filtration is to reduce A patient skin dose. B operator dose. C image noise. D scattered radiation.

The Correct Answer is: A It is our ethical responsibility to minimize radiation dose to patients. X-rays produced at the target make up a heterogeneous primary beam. There are many "soft" (low-energy) photons that, if not removed, would contribute only to greater patient dose. They are too weak to penetrate the patient and expose the IR. These soft x-rays penetrate only a small thickness of tissue before being absorbed.

The primary function of filtration is to reduce A patient skin dose B operator dose C image noise D scattered radiation

The Correct Answer is: A It is our ethical responsibility to minimize the radiation dose to our patients. X-rays produced at the tungsten target make up a heterogeneous primary beam. There are many "soft" (low-energy) photons that, if not removed by filters, would only contribute to greater patient skin dose. They are too weak to penetrate the patient and contribute to the image-forming radiation; they penetrate a small thickness of tissue and are absorbed. (

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

The Correct Answer is: A Kilovoltage 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.

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 exposureand 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: A Limiting 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. (

Which of the following is (are) characteristic(s) of a 5:1 grid? It allows some positioning latitude. It is used with high-kilovoltage exposures. It absorbs a high percentage of scattered radiation. A 1 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: A Low-ratio grids, such as 5:1, 6:1, and 8:1, are used with moderate-kilovolt techniques and are not recommended for use beyond 85 kV. They are not able to clean up the amount of scatter produced at high kilovoltages, but their low ratio permits more positioning latitude than high-ratio grids. High-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.

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: A Magnification 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

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: A Magnification 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.

In which type of equipment does kilovoltage decrease during the actual length of the exposure? Condenser-discharge mobile equipment Battery-operated mobile equipment Fixed 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: A Mobile 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.

Capacitor-discharge mobile x-ray units use capacitors to power the x-ray tube machine locomotion braking mechanism A 1 only B 2 only C 1 and 2 only D 1, 2, and 3

The Correct Answer is: A Mobile 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.

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: A Of 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

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 close collimation Use of low mAs Use 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: A One way to minimize scattered radiation reaching the IR is to use optimal kilovoltage; excessive kilovoltage increases the production of scattered radiation. Close 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

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: A Parts 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. (

In electronic imaging, as digital image matrix size increases pixel size decreases resolution decreases pixel depth decreases A 1 only B 2 only C 1 and 2 only D 2 and 3 only

The Correct Answer is: A Pixel depth is directly related to shades of gray—called dynamic range—and is measured in bits. The greater the number of bits, the more shades of gray. For example, a 1-bit (2 1 ) pixel will demonstrate 2 shades of gray, whereas a 6-bit (2 6 ) pixel can display 64 shades and a 7-bit (2 7 ) pixel 128 shades. However, pixel depth is unrelated to resolution. A 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 (e.g., 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 millimeter (lp/mm). As matrix size is increased (e.g., from 512 × 512 to 1,024 × 1,024) there are more and smaller pixels in the matrix and, therefore, improved resolution. Fewer and larger pixels result in poor resolution, a "pixelly" image, that is, one in which you can actually see the individual pixel boxes.

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: A Radiographic 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.

A positive contrast agent absorbs x-ray photons results in a dark area on the radiograph is 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: A Radiopaque 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

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: A Resolution 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.

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: A Scattered 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

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: A Scattered 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: A Shape 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. (

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

The Correct Answer is: A Spatial 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."

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: A The 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.

Which of the x-ray circuit devices shown in Figure 7-16 operates on the principle of self-induction? 1. Number 1 2. Number 2 3. Number 3 A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The Correct Answer is: A The autotransformer (number 1) controls/selects the amount of voltage sent to the primary winding of the high-voltage transformer and operates on the principle of self-induction. The step-up (high-voltage) transformer (primary coil is number 2; secondary coil is number 3) operates on the principle of mutual induction. The step-up transformer functions to change low voltage to the high voltage necessary to produce x-ray photons. The x-ray tube is identified as number 7.

Referring to the simplified x-ray circuit shown in Figure 6-5, what is indicated by the number 3? A. Step-up transformer B. Autotransformer C. Filament circuit D. Rectification system

The Correct Answer is: A The autotransformer is labeled 1, the primary coil of the high-voltage transformer is labeled 2, the grounded milliampere meter is labeled 4, and the filament circuit is labeled 6. The rectification system, which is used to change alternating current to unidirectional current, is indicated by number 5. The rectification system is located between the secondary coil of the high-voltage (step-up) transformer (number 3) and the x-ray tube (number 7).

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: A The 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

The number 2 in Figure 5-2 indicates the A. nickel focusing cup B. actual focal spot C. effective focal spot D. anode stem

The Correct Answer is: A The figure illustrates the component parts of a rotating-anode x-ray tube enclosed within a glass envelope (number 3) to preserve the vacuum necessary for x-ray production. Number 4 is the rotating anode with its beveled focal track at the periphery (number 8) and its stem (at number 5). Numbers 6 and 7 are the stator and rotor, respectively—the two components of an induction motor—whose function it is to rotate the anode. Number 1 is the filament of the cathode assembly, which is made of thoriated tungsten and functions to liberate electrons (thermionic emission) when heated to white hot (incandescence). Number 2 is the molybdenum focusing cup, which functions to direct the liberated filament electrons to the focal spot

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: A The 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).

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: A The 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.

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: A The 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.

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: A The 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

What is the correct critique of the CR image shown in Figure 4-3? A. double exposure B. inverted IP C. incomplete erasure D. image fading

The Correct Answer is: A The image shown is a double exposure. Note the ilia and lower pelvic structures. Two pelves are clearly identifiable. Particularly noteworthy is how CR will "correct" the exposure values. The image does not appear overexposed, but the superimposed abdominal images are unmistakably evident. An inverted IP would have imaged the rear panel of the IP—a large grid-like appearance. An incomplete erasure or image fading would show only a portion of the image—here we have the entire superimposed abdomen.

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: A The 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

Which type of error results in grid cutoff at the periphery of the radiographic image? A Off-focus B Off-center C Off-level D Off-angle

The Correct Answer is: A The lead strips in a focused grid are made to parallel the x-ray beam. Therefore, scattered radiation, which radiates in directions other than that of the primary beam, will be absorbed by the grid. When the x-ray beam does not parallel the lead strips, some type of grid cutoff occurs. If the x-ray beam is not centered to the grid, or if the x-ray tube and grid surface are not parallel (level), there will be a fairly uniform decrease in receptor exposure across the entire image. However, if the grid is not used within its recommended SID (focus) range (i.e., if the SID is too great or too little), there will be a decrease in receptor exposure at the periphery of the image

The line-focus principle expresses the relationship between A the actual and the effective focal spot B exposure given the IR and resulting spatial resolution C SID used and resulting receptor exposure D grid ratio and lines per inch

The Correct Answer is: A The line-focus principle is a geometric principle illustrating that the actual focal spot is larger than the effective (projected) focal spot. The actual focal spot (target) is larger, to accommodate heat over a larger area, and is angled so as to project a smaller focal spot, thus maintaining spatial resolution by reducing blur. The relationship between the SID and resulting receptor exposure is expressed by the inverse-square law. Grid ratio and lines per inch are unrelated to the line-focus principle

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: A The 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 shortest scale contrast.

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

The Correct Answer is: A The 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

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: A The 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).

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: A The 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

Deposition of vaporized tungsten on the inner surface of the x-ray tube glass window 1. acts as additional filtration 2. results in increased tube output 3. results 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: A Through 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

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: A To 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

When using the smaller field in a dual-field image intensifier, the image is magnified the image is brighter a larger anatomic area is viewed A 1 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: A When 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

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: A With 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)

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 three-phase, 12-pulse equipment? A 200 mA B 600 mA C 0.125 s D 0.25 s

The Correct Answer is: A With 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 milliampere-seconds are required to produce a radiograph with similar receptor exposure. (When going from three-phase, six-pulse to single-phase, add one-third more milliampere-seconds.) When changing from single-phase to three-phase, 12-pulse equipment, only one-half of the original milliampere-seconds is required. (Going from three-phase, 12-pulse to single-phase requires twice the milliampere-seconds.) In this instance, we are changing from single-phase to three-phase, 12-pulse equipment; therefore, the new milliampere-seconds value should be half the original 50 mAs, or 25 mAs. The only selection that will provide 25 mAs is (A), 200 mA. (B) will produce 75 mAs (600 mA × ⅛ s = 75 mAs); (C) will produce 50 mAs (400 mA × 0.125 s = 50 mAs); (D) will produce 100 mAs (400 × 0.25 = 100 mAs). Comparison of technical factors required Single phase Three phase Three phase 6-pulse 12-pulse x mAs ⅔ x mAs ½ x mAs

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: A 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 first indicated to compensate for the distance change. The following formula (exposure maintenance formula) is used to determine new mAs values, when changing distance: mAs1/mAs2 = D1 squared/D2 squared 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:

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 magnification factor? A 1.25 B 1.86 C 4.9 D 7.3

The Correct Answer is: A As 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 magnification factor is: MF = SID/SOD (SOD = SID - OID) Substituting known factors the equation becomes: MF = 44/35 MF = 1.257 The "1" in the answer represents the actual object, while the ".257" represents the degree of magnification. The percent magnification can be determined by moving the decimal two places to the right. Thus, the percent magnification is 25.7%.

The pre-contrast image used to subtract pixel values from the post-contrast image showing contrast-filled blood vessels in digital subtraction angiography is called the : A Mask image B Ghost (phantom) image C Moiré image D Latent image

The Correct Answer is: A In digital image subtraction, the pixel values from post-contrast images are electronically subtracted from pixel values from the first pre-contrast (mask) image (A) to show contrast-filled blood vessels with the other structures (e.g., bone) removed in order to enhance the diagnostic impressions of the radiologist. A ghost (or phantom) image (B) is an image artifact. The appearance of ghost images 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 (C); to decrease the possibility of this effect, high frequency grids are recommended for digital imaging. The latent image (D) is the image that exists in a radiographic film prior to chemical processing. It represents the collection of silver atoms around the sensitivity specks within the film emulsion. Upon chemical processing in a darkroom, the film will reveal an anatomical image with densities representing the varying levels of radiation exposure to the sensitivity specks contained within the film's emulsion

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: A In 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.

What lies immediately under the phosphor layer of a PSP storage plate? A Reflective layer B Base C Antistatic layer D Lead foil

The Correct Answer is: A The PSP storage plate within the IP has a layer of europium-activated barium fluorohalide (BaFX: Eu 2 +; X = halogen) mixed with a binder substance. This layer serves as the image receptor when exposed. The barium fluorohalide is usually granular or turbid phosphors. Other examples of turbid phosphors are gadolinium oxysulfide and rubidium chloride. "Needle"-shaped, or columnar phosphors (usually cesium iodide), have the advantage of better x-ray absorption and less light diffusion. Just under the barium fluorohalide layer is a reflective layer that helps direct emitted light up toward the CR reader. Below the reflective layer is the base, behind that is an antistatic layer, and then the lead foil to absorb backscatter. Over the top of the barium fluorohalide is a protective layer.

Figure 4-19 is representative of A. the anode heel effect B. the line-focus principle C. the inverse-square law D. the reciprocity law

The Correct Answer is: A The figure represents the anode heel effect. Because the anode's focal track is beveled, x-ray photons can freely diverge toward the cathode end of the x-ray tube. However, the "heel" of the focal track prevents x-ray photons from diverging toward the anode end of the tube. This results in varying intensity with fewer photons at the anode end (A) and more photons at the cathode end (B). The line-focus principle relates to the anode's focal spot, and x-ray tube targets are constructed according to the line-focus principle—the focal spot is angled to the vertical. As the actual focal spot is projected downward, it is foreshortened; thus, the effective focal spot is always smaller than the actual focal spot. The inverse-square law deals with the changing x-ray intensity with changes in distance. This principle is important in image formation and in radiation protection.

A decrease from 90 to 77 kVp will result in an increase in wavelength gray scale scattered 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%.

An increase in kilovoltage with appropriate compensation of milliampere-seconds will result in 1. increased part penetration. 2. higher contrast. 3. 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

How would the introduction of a 6-in. OID affect 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: A OID can affect contrast 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 image contrast.

If a radiograph were made of an average-size knee using automatic exposure control (AEC) and all three photocells were selected, the resulting radiograph would demonstrate A underexposed image. B overexposed image. C poor spatial resolution. D adequate exposure.

The Correct Answer is: A Proper functioning of the AEC 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(s) to achieve the appropriate exposure. If a photocell is left uncovered, scattered radiation from the part being examined will cause premature termination of exposure and an underexposed radiograph.

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 converts electrical energy to mechanical energy? A. Motor B. Generator C. Stator D. Rotor

The Correct Answer is: A A 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.

Quality control testing of digital display monitors should be conducted A. Daily B. Weekly C. Monthly D. Annually

The Correct Answer is: A Although a comprehensive quality control program is strongly recommended, daily evaluation of digital display monitors is very important (A). This is best accomplished using the TG 18-QC test pattern. Many system performance changes can be readily spotted by the observant technologist, called to the attention of the physicist for further testing. A medical physicist must review the QC program periodically. Weekly, monthly, or annual quality control evaluations of digital display monitors is not timely enough to ensure consistencies in the diagnostic image display.

The automatic exposure device that is located immediately under the x-ray table is the A. ionization chamber B. scintillation camera C. photomultiplier D. photocathode

The Correct Answer is: A Automatic exposure control (AEC) devices are used to produce consistent and comparable radiographic results. In one type of AEC, there is an ionization chamber just beneath the tabletop above the IR. The part to be examined is centered to it (the sensor) and radiographed. When a predetermined quantity of ionization has occurred (equal to the correct receptor exposure), the exposure terminates automatically. In the other type of AEC, the phototimer/photomultiplier, a small fluorescent screen is positioned beneath the IR. When remnant radiation emerging from the patient exposes and exits the IR, the fluorescent screen emits light. Once a predetermined amount of fluorescent light is "seen" by the photocell sensor, the exposure is terminated. A scintillation camera is used in nuclear medicine. A photocathode is an integral part of the image intensification system.

Methods that help to reduce the production of scattered radiation include using 1. compression 2. beam restriction 3. a grid A. 1 and 2 only B. 1 and 3 only C. 2 and 3 only D. 1, 2, and 3

The Correct Answer is: A Limiting the size of the irradiated field is a most effective method of decreasing the production of scattered radiation. The smaller the volume of tissue irradiated, the smaller is the amount of scattered radiation generated; this can be accomplished using compression (prone position instead of supine or a compression band). Use of a grid does not affect the production of scattered radiation but rather removes it once it has been produced.

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: A Radiographic 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: A Radiographic 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.

The collimator light and actual irradiated area must be accurate to within what percentage of the SID? A. 2% B. 5% C. 10% D. 15%

The Correct Answer is: A Restriction of field size is one important method of patient protection. However, the accuracy of the light field must be evaluated periodically as part of a QA program. Guidelines set forth for patient protection state that the collimator light and actual irradiated area must be accurate to within 2% of the SID.

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: A The 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).

Backscatter on a digital image can cause an artifact called a A. Phantom image artifact B. Moiré artifact C. Static artifact D. Grid cutoff artifact

The Correct Answer is: A The appearance of phantom image artifact (A) can be seen when excessive backscatter exposes the image receptor. The back side of the image receptor should be shielded with lead to reduce exposure to backscatter radiation. 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). Grid cutoff artifacts are seen as a decrease in receptor exposure on one side of the image when the central ray is off-centered or not perpendicular to either a non-focused or focused grid, or on both sides when the central ray is properly centered but the focal range is exceeded (caused by improper SID) when using a focused grid (D). Static artifacts may be seen on radiographic film imaging

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: A The 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.

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: A The 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.

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: A Three-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.

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

The Correct Answer is: A With 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.

Double-focus x-ray tubes have two focal spots. filaments. anodes. A 1 only B 1 and 2 only C 1 and 3 only D 2 and 3 only

The Correct Answer is: B A double-focus tube has two focal-spot sizes available. These focal spots actually are two available paths on the focal track. There are also two filaments. When the small focal spot is selected, the small filament is heated, and electrons are driven across to the smaller portion of the focal track. When the large focal spot is selected, the large filament is heated, and electrons are driven across to the larger portion of the focal track

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: B A 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.

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: B A 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. (

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: B A 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.

An exposure was made using 8 mAs and 60 kV. If the kilovoltage was changed to 70 to obtain longer-scale contrast, what new milliampere-seconds value is required to maintain receptor exposure? A 2 B 4 C 16 D 32

The Correct Answer is: B According to the 15% rule, if the kilovoltage is increased by 15%, receptor exposure will be doubled. Therefore, to compensate for this change and to maintain receptor exposure, the milliampere-seconds value should be reduced to 4 mAs.

A satisfactory radiograph was made without a grid using a 72-in. SID and 8 mAs. If the distance is changed to 40 in. and an 12:1 ratio grid is added, what should be the new milliampere-seconds value? A 9.5 mAs B 12 mAs C 21 mAs D 26 mAs

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 first indicated to compensate for the distance change. The following formula (exposure-maintenance formula) is used to determine new milliampere-seconds values when changing distance: mAs1/mAs2 = D1 squared/D2 squared Substituting known values: Thus, x = 2.47 mAs at 40-in. SID. To then compensate for adding a 12:1 grid, you must multiply the 2.47 mAs by a factor of 5. Thus, 12 mAs is required to produce a receptor exposure similar to the original image. The following are the factors used for milliampere-seconds conversion from nongrid to grid:

Which of the following artifacts is occasionally associated with the use of grids in digital imaging? A Incomplete erasure B Aliasing C Image fading D Vignetting

The Correct Answer is: B An artifact associated with digital imaging and grids is "aliasing" or the "moiré effect." If the direction of the lead strips and the grid lines per inch (i.e., grid frequency) matches the scan frequency of the scanner/reader, this artifact can occur. Aliasing (or Moiré effect) appears as superimposed images slightly out of alignment, an image "wrapping" effect. This most commonly occurs in mobile radiography with stationary grids and can be a problem with DR flat panel detectors.

Which of the following will improve the spatial resolution of image-intensified images? A very thin coating of cesium iodide on the input phosphor A smaller-diameter input screen Increased 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: B An image's spatial resolution refers to ithe sharpness of its image details. As its input screen's 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.

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: B An 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 decreasing radiographic contrast. (

The radiograph seen below illustrates incorrect use of A. collimator B. grid C. AEC D. focal spot

The Correct Answer is: B An upside-down focused grid presents its lead strips in the opposite direction to that of the x-ray beam. This results in severe grid cutoff everywhere except in the central portion of the radiographic image. Severe grid cutoff of chest anatomy can be seen outside the central exposed area. A misaligned collimator would not show such symmetrical loss of receptor exposure, nor would an incorrectly selected AEC photocell. Focal spot is unrelated to receptor exposure.

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: B As 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: image size/object size = SID/SOD (= SID -OID) Substituting known quantities: 9/x = 105/87 (105-18=87) 105x = 783 x = 7.45 Thus, x = 7.45 cm (approximate actual size). The relationship between SID, SOD, and OID

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

The Correct Answer is: B As 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

OID is related to spatial resolution in which of the following ways? A Spatial resolution is directly related to OID. B Spatial resolution is inversely related to OID. C As OID increases, so does spatial resolution. D OID is unrelated to spatial resolution.

The Correct Answer is: B As the distance from the object to the IR (OID) increases, so does magnification distortion, thereby decreasing spatial resolution. Some magnification is inevitable in radiography because it is not possible to place anatomic structures directly on the IR. However, our understanding of how to minimize magnification distortion is an important part of our everyday work.

An increase in the kilovoltage applied to the x-ray tube increases the x-ray wavelength exposure rate patient absorption A 1 only B 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: B As 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 (short-wavelength) photons are produced. However, because they are higher-energy photons, there will be less patient absorption.

An increase in the kilovoltage applied to the x-ray tube increases the percentage 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: B As 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.

The attenuation of x-ray photons is not influenced by 1. pathology 2. effective atomic number 3. photon quantity A. 1 only B. 3 only C. 2 and 3 only D. 1, 2, and 3

The Correct Answer is: B Attenuation (decreased intensity through scattering or absorption) of the x-ray beam is a result of its original energy and its interactions with different types and thicknesses of tissue. The greater the original energy/quality (the higher the kilovoltage) of the incident beam, the less is the attenuation. The greater the effective atomic number of the tissues (tissue type and pathology determine absorbing properties), the greater is the beam attenuation. The greater the volume of tissue (subject density and thickness), the greater is the beam attenuation

Which of the following can impact the visibility of the anode heel effect? SID IR size Screen speed A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: B Because the focal spot (track) of an x-ray tube is along the anode's beveled edge, photons produced at the target are able to diverge considerably toward the cathode end of the x-ray tube but are absorbed by the heel of the anode at the opposite end of the tube. This results in a greater number of x-ray photons distributed toward the cathode end, which is known as the anode heel effect. The effect of this restricting heel is most pronounced when the x-ray photons are required to diverge more, as would be the case with short SID, large-size IRs and steeper (smaller) target angles

Electronic imaging terms used to indicate the intensity of radiation reaching the IR include exposure index sensitivity (S) number field 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: B Computed 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

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: B Computed 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

The effect that differential absorption has on radiographic contrast of a high-subject-contrast part can be minimized by using 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: B Differential 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

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

The Correct Answer is: B Differential 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

Bone densitometry is often performed to 1. measure degree of bone (de)mineralization 2. evaluate results of osteoporosis treatment/therapy 3. evaluate condition of soft tissue adjacent to bone A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The Correct Answer is: B Dual x-ray absorptiometry (DXA) imaging is used to evaluate bone mineral density (BMD). 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. 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, the attenuation is calculated to represent the degree of bone density. Soft tissue attenuation information is not used to measure bone density

Referring to the anode cooling chart in Figure 5-9, if the anode is saturated with 300,000 heat units (HU), how long will the anode need to cool before another 160,000 HU can be safely applied? A. 3 minutes B. 4 minutes C. 5 minutes D. 7 minutes

The Correct Answer is: B Each x-ray exposure made by the radiographer produces hundreds or thousands of heat units at the target. If the examination requires several consecutive exposures, the potential for extreme heat load is increased. Just as each x-ray tube has its own radiographic rating chart, each tube also has its own anode cooling curve to describe its unique heating and cooling characteristics. An x-ray tube generally cools most rapidly during the first 2 minutes of nonuse. First, note that the tube is saturated with heat at 300,000 HU. In order for another 160,000 HU to be safely applied, the x-ray tube must first release 160,000 HU, which means that it has to cool down at least to 140,000 HU. Find the 140,000 HU point on the vertical axis and follow across to where it intersects with the cooling curve. It intersects at about the 4-minute point.

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: B Exposing 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).

For the same FOV, spatial resolution will be improved using A a smaller matrix B a larger matrix C fewer pixels D shorter SID

The Correct Answer is: B Field of view (FOV) refers to the area being viewed. The FOV can be increased or decreased. As the FOV is increased, the part being examined is magnified; as the FOV is decreased, the part returns closer to actual size. Pixel size is affected by changes in either the FOV or matrix size. For example, if the matrix size is increased, for example, from 256 × 256 to 512 × 512, pixel size must decrease. If FOV increases, pixel size must increase. Pixel size is inversely related to resolution. As pixel size decreases, resolution increases. Decreasing SID would decrease spatial resolution.

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: B Film 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.

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: B First, 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:

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: B Flat-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

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: B Fluorescence 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

Focal-spot blur is greatest A toward the anode end of the x-ray beam B toward the cathode end of the x-ray beam C directly along the course of the CR D as the SID is increased

The Correct Answer is: B Focal-spot blur, or geometric blur, is caused by photons emerging from a large focal spot. Because the projected focal spot is greatest at the cathode end of the x-ray tube, geometric blur is also greatest at the corresponding part (cathode end) of the radiograph. The projected focal-spot size becomes progressively smaller toward the anode end of the x-ray tube

Focal-spot blur is greatest A directly along the course of the central ray B toward the cathode end of the x-ray beam C toward the anode end of the x-ray beam D as the SID is increased

The Correct Answer is: B Focal-spot blur, or geometric blur, is caused by photons emerging from a large focal spot. The actual focal spot is always larger than the effective (or projected) focal spot, as illustrated by the line-focus principle. In addition, the effective focal-spot size varies along the longitudinal tube axis, being greatest in size at the cathode end of the beam and smallest at the anode end of the beam. Because the projected focal spot is greatest at the cathode end of the x-ray tube, geometric blur is also greatest at the corresponding part (cathode end) of the radiograph

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

The Correct Answer is: B Given 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

Characteristics of low ratio focused grids include the following: 1. they have a greater focal range 2. they are less efficient in collecting SR 3. 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: B Grid 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.

Compared to a low ratio grid, a high ratio grid will 1. absorb more primary radiation. 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: B Grid 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 primary radiation 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

Which of the following is (are) characteristic(s) of a 16:1 grid? It absorbs more useful radiation than an 8:1 grid. It has more centering latitude than an 8:1 grid. It is used with higher-kilovoltage exposures than an 8:1 grid. A 1 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: B High-kilovoltage exposures produce large amounts of scattered radiation, and high-ratio grids are used often with high-kilovoltage techniques in an effort to absorb more of this scattered radiation. However, as more scattered radiation is absorbed, more primary radiation is absorbed as well. This accounts for the increase in milliampere-seconds required when changing from an 8:1 to a 16:1 grid. In addition, precise centering and positioning become more critical; a small degree of inaccuracy is more likely to cause grid cutoff in a high-ratio grid.

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: B If (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).

When involuntary motion must be considered, the exposure time may be cut in half if the kilovoltage is A doubled B increased by 15% C increased by 25% D increased by 35%

The Correct Answer is: B If the exposure time is cut in half, one normally would double the milliamperage to maintain the same milliampere-seconds value and, consequently, the same receptor exposure. However, increasing the kilovoltage by 15% has a similar effect. For example, if the original kilovoltage were 85 kV, 15% of this is 13, and therefore, the new kilovoltage would be 98 kV. The same percentage value would be used to cut the receptor exposure in half (reduce kilovoltage by 15%)

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: B If 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.

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 16- or 35-mm film.

The Correct Answer is: B Image 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. A cine camera is required to record images on 16- or 35-mm film.

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: B In 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. 35cm = 350mm 350/1024 = 0.35mm 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

The term windowing describes the practice of A varying the automatic brightness control B changing the image contrast and/or brightness C varying the FOV D increasing resolution

The Correct Answer is: B In electronic imaging (CR/DR), the radiographer can manipulate the digital image displayed on the CRT through postprocessing. One way to alter image contrast and/or brightness is through windowing. This refers to some change made to window width and/or window level. Change in window width changes the number of gray shades, that is, image contrast. Change in window level changes the image brightness. Windowing and other postprocessing mechanisms permit the radiographer to produce "special effects" such as edge enhancement, image stitching, and image inversion, rotation, and reversal. 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. Automatic brightness control is associated with image intensification.

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

The Correct Answer is: B In 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 electromotive force (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

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

The Correct Answer is: B In 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.

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: B In 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

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

The Correct Answer is: B In 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

An analog x-ray exposure of a particular part is made and restricted to a 14 × 17 in. field size. The same exposure is repeated, but the x-ray beam is restricted to a 4 × 4 in. field. Compared with the first image, the second image will demonstrate less receptor exposure more contrast more receptor exposure A 1 only B 1 and 2 only C 3 only D 2 and 3 only

The Correct Answer is: B Less scattered radiation is generated within a part as the kilovoltage is decreased, as the size of the field is decreased, and as the thickness and density of tissue decrease. As the quantity of scattered radiation decreases from any of these sources, the less is the total receptor exposure

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 5.7 in. C 6.1 in. D 6.7 in.

The Correct Answer is: B Magnification 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 Image size/Object size = SID/SOD (SID - OID) Substituting known values: x/5 = 44/38 (44-6=38) 38x = 220 x = 5.78 in image width

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: B Magnification 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: SID/SOD (SID - OID) = image size/object size Substituting known values: 36/32 (36-4 =32) = x/3 32x = 108 x = 3.37 x = 3.37 inches image width

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

The Correct Answer is: B Magnification 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.

The advantages of capacitor-discharge mobile x-ray equipment include 1. compact size 2. light weight 3. high kilovoltage capability A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The Correct Answer is: B Mobile 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.

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: B Off-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).

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: B Patients 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

Low-kilovoltage exposure factors usually are indicated for radiographic examinations using water-soluble, iodinated media a negative contrast agent barium sulfate A 1 only B 1 and 2 only C 3 only D 1 and 3 only

The Correct Answer is: B Positive contrast medium is radiopaque; negative contrast material is radioparent. Barium sulfate (radiopaque, positive contrast material) is used most frequently for examinations of the intestinal tract, and high-kilovoltage exposure factors are used to penetrate (to see through and behind) the barium. Water-based iodinated contrast media (Conray, Amipaque) are also positive contrast agents. However, the K-edge binding energy of iodine prohibits the use of much greater than 70 kV with these materials. Higher kilovoltage values will obviate the effect of the contrast agent. Air is an example of a negative contrast agent, and high-kilovoltage factors are clearly not indicated

Which of the following can have an effect on radiographic contrast? 1. Beam restriction 2. Grids 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: B Radiographic 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.

The radiographic accessory used to measure the thickness of body parts in order to determine optimal selection of exposure factors is the A. fulcrum B. caliper C. densitometer D. ruler

The Correct Answer is: B Radiographic technique charts are highly recommended for use with every x-ray unit. A technique chart identifies the standardized factors that should be used with that particular x-ray unit for various examinations/positions of anatomic parts of different sizes. To be used effectively, these technique charts require that the anatomic part in question be measured correctly with a caliper. A fulcrum is of importance in tomography; a densitometer is used in sensitometry and QA

Phosphors classified as rare earth include lanthanum 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: B Rare 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

Which of the following is (are) classified as rare earth phosphors? Lanthanum 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: B Rare 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

Which of the following units is (are) used to express resolution? Line-spread function Line pairs per millimeter Line-focus principle A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: B Resolution 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), line-spread function (LSP), or modulation transfer function (MTF). Lp/mm 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. The effective focal spot is the foreshortened size of the actual focal spot as it is projected down toward the IR, that is, as it would be seen looking up into the emerging x-ray beam. This is called the line-focus principle and is not a unit used to express resolution.

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: B Resolution 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.

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: B SNR 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.

Which of the following is most likely to occur as a result of using a 30-in. SID with a 14 × 17 in. IR to radiograph a fairly homogeneous structure? A. Production of quantum mottle B. Receptor exposure variation between opposite ends of the IR C. Production of scatter radiation fog D. Excessively short-scale contrast

The Correct Answer is: B Since x-ray photons are produced at the tungsten target, they more readily diverge toward the cathode end of the x-ray tube. As they try to diverge toward the anode, they interact with and are absorbed by the anode "heel." Consequently, there is a greater intensity (quantity) of x-ray photons at the cathode end of the x-ray beam. This phenomenon is known as the anode heel effect. Because shorter SIDs and larger IR sizes require greater divergence of the x-ray beam to provide coverage, the anode heel effect will be accentuated

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

The Correct Answer is: B Some 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

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: B The 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

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

The Correct Answer is: B The 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

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: B The 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.

The process of "leveling and windowing" of digital images determines the image A spatial resolution B contrast C pixel size D matrix size

The Correct Answer is: B The 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.

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: B The exposure factor that regulates receptor exposure is milliampere-seconds (mAs). The equation used to determine mAs is mA × s = mAs. Substituting known factors: 300 mA x s = 18 mAs (divide by 300 mA) s = 0.06 (multiply by 1000 ms) ms = 60 ms

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: B The formula for mAs is mA × s = mAs. Substituting known values: mA x 0.05s = 30 mAs (30/0.05) mA = 600

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: B The 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, MF = image size/object size 1.3 = 15.2/x 1.3x = 15.2 x = 11.69 x = 11.69 cm (actual anatomic size)

If 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: 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 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, Vs/Vp = Ns/Np To determine secondary I: Nx/Np = Ip/Is Substituting known values, x/220 = 10,000/200 200x = 22,000,000 x = 110,000 V (110,000/1,000) x = 110 kV

Figure 5-7 illustrates the A. inverse-square law B. line-focus principle C. reciprocity law D. anode heel effect

The Correct Answer is: B The illustration demonstrates the actual focal spot (#1) and the effective focal spot (#2). The actual focal spot is the area on the focal track that is bombarded by the electron stream coming from the heated filament. The upper dotted line corresponds to/illustrates the angle of the actual focal spot. The illustration shows how the effective, or projected, focal spot is always smaller than the actual focal spot. This is called the line focus principle. The inverse square law deals with the relationship between distance and x-ray beam intensity. The reciprocity law relates to the relationship between and mA, time, and mAs. The anode heel effect refers to the variation in x-ray beam intensity between the anode and cathode.

An advantage of coupling the image intensifier to the TV camera or CCD via a fiber-optic coupling device is its compact size durability ability 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: B The 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

What is the device that directs the light emitted from the image intensifier to various viewing and imaging apparatus? A. Output phosphor B. Beam splitter C. Spot-film changer D. Automatic brightness control

The Correct Answer is: B The light image emitted from the output phosphor of the image intensifier is directed to the TV monitor for viewing and sometimes to recording devices such as a spot-film camera or cine film. The light is directed to these places by a beam splitter or objective lens located between the output phosphor and the TV camera tube (or CCD). The majority of the light will go to the recording device, whereas a small portion goes to the TV so that the procedure may continue to be monitored during filming.

A radiograph made using 300 mA, 0.1 second, and 75 kV exhibits motion unsharpness but otherwise satisfactory technical quality. The radiograph will be repeated using a shorter exposure time. Using 86 kV and 400 mA, what should be the new exposure time? A 25 ms B 37 ms C 50 ms D 75 ms

The Correct Answer is: B The milliampere-seconds (mAs) formula is milliamperage × time = mAs. With two of the factors known, the third can be determined. To find the milliampere-seconds value that was used originally, substitute the known values: 300 mA x 0.1 s = 30 mAs We have increased the kilovoltage to 86 kV, an increase of 15%, which has an effect similar to that of doubling the milliampere-seconds. Therefore, only 15 mAs is now required as a result of the kilovoltage increase: mA x s = mAs 400 mA x s = 15 mAs (divide both sides by 400) s = 0.0375 Thus, x = 0.0375-s exposure = 37.5 ms.

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: B The 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 rreceptor exposure, one simply doubles the milliampere-seconds; therefore, milliampere-seconds and receptor exposure are directly proportional.

A lateral radiograph of the lumbar spine was made using 200 mA, 1-second exposure, and 90 kV. If the exposure factors were changed to 200 mA, 0.5 second, and 104 kV, there would be an obvious change in which of the following? 1. Receptor exposure 2. Scale of grays/contrast 3. Distortion A. 1 only B. 2 only C. 2 and 3 only D. 1, 2, and 3

The Correct Answer is: B The original milliampere-seconds value (regulating receptor exposure) was 200. The original kilovoltage (impacting contrast) was 90. The milliampere-seconds value was cut in half, to 100, causing a decrease in receptor exposure. The kilovoltage was increased (by 15%) to compensate for the receptor exposure loss and thereby increase the scale of grays

Which of the following circuit devices operate(s) on the principle of self-induction? Autotransformer Choke coil High-voltage transformer A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: B The principle of self-induction is an example of the second law of electromagnetics (Lenz's law), which states that an induced current within a conductive coil will oppose the direction of the current that induced it. It is important to note that self-induction is a characteristic of AC only. The fact that AC is constantly changing direction accounts for the opposing current set up in the coil. Two x-ray circuit devices operate on the principle of self-induction. The autotransformer operates on the principle of self-induction and enables the radiographer to vary the kilovoltage. The choke coil also operates on the principle of self-induction; it is a type of variable resistor that may be used to regulate filament current. The high-voltage transformer operates on the principle of mutual induction

Which of the following errors is illustrated in the figure below? A. Patient not centered to IR B. X-ray tube not centered to grid C. Inaccurate collimation D. Unilateral grid cutoff

The Correct Answer is: B The radiograph shown demonstrates a 1.5-in. unexposed strip along the length of the film. This occurred because, although the patient was centered correctly to the collimator light and x-ray field, the x-ray tube was not centered to the grid. If the patient was off-center, the entire image would be exposed, and the patient's spine would be off-center. Grid cutoff would not appear as such a sharply delineated line but rather as a gradually decreasing receptor exposure.

Which of the following combinations will offer the greatest heat-loading capability? A 17-degree target angle, 1.2-mm actual focal spot B 10-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: B The smaller the focal spot, the more limited the anode is with respect to the quantity of heat it can safely accept. As the target angle decreases, the actual focal spot can be increased while still maintaining a small effective focal spot. Therefore, group (B) offers the greatest heat-loading potential, with a steep target angle and a large actual focal spot. It must be remembered, however, that a steep target angle increases the heel effect, and IR coverage may be compromised

Star and wye configurations are related to A autotransformers B three-phase transformers C rectification systems D AECs

The Correct Answer is: B The terms star and wye (or delta) refer to the configuration of transformer windings in three-phase equipment. Instead of having a single primary coil and a single secondary coil, the high-voltage transformer has three primary and three secondary windings—one winding for each phase (Figure 5-13). Autotransformers operate on the principle of self-induction and have only one winding. Three-phase x-ray equipment often has three autotransformers

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: B The 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): 10 (old mAs)/x (new mAs) = 4 (8:1)/5 (12:1) 4x = 50 x = 12.5 mAs Thus, x = 12.5 mAs at 90 kVp. Now determine the exposure time required with 400 mA to produce 12.5 mAs: 400x = 12.5 x = 0.03 seconds

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: B The 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

Fluorescent light is collected from the image intensifier output phosphor and converted to an electronic video signal by the 1. 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: B There 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.

What is the correct critique of the CR image seen below? A. double exposure B. grid centering error C. incorrect AEC photocell D. inverted focused grid

The Correct Answer is: B This is an example of both off-focus and lateral decentering errors. Note the asymmetric cutoff from right to left. The individual grid errors, as well as the result of both errors together, is summarized below. Off-focus errors: Grid cutoff will occur if the SID is below the lower limits, or above the upper limits, of the specified focal range. This type of error is also referred to as focus-grid distance decentering. Off-focus errors are usually characterized by loss of receptor exposure at the periphery of the image. Off-center errors: If the x-ray beam is not centered to the grid (i.e., if it is shifted laterally) grid cutoff will occur. This type of error is referred to as lateral decentering and characterized by a uniform receptor exposure loss across the radiographic image. If the x-ray beam is both off-center and off-focus below the focusing distance, the portion of the image below the focus will show increased receptor exposure; if the x-ray beam is off-center and off-focus above the focusing distance, the image below the focus will show decreased receptor exposure.

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: B To 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.

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: B Use 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.

The radiograph illustrated in the figure below was made using a single-phase, full-wave-rectified unit with a timer and rectifiers that are known to be accurate and functioning correctly. What exposure time was used to produce this image? A. 1/10 second B. 0.05 second C. 1/12 second D. 0.025 second

The Correct Answer is: B When a spinning top is used to test the timer efficiency of full-wave-rectified single-phase equipment, the result is a series of dots or dashes, with each dot representing a pulse of radiation. With full-wave-rectified current and a possible 120 dots (pulses) available per second, one should visualize 12 dots at 1/10 second, 6 dots at 0.05 second, 10 dots at 1/12 second, and 3 dots at 0.025 second. Because three-phase equipment is at almost constant potential, a synchronous spinning top must be used for timer testing, and the result is a solid arc (rather than dots). The number of degrees covered by the arc is measured and equated to a particular exposure time.

A backup timer for the AEC serves to protect the patient from overexposure protect the x-ray tube from excessive heat adjust image contrast A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: B When 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 factors Using compensating filtration Using 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: B When 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.

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

The Correct Answer is: B X-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 include lengthy anode rotation exposures to a cold anode low-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: B X-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

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: B X-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 molybdenum focusing cup as the switch, permitting very precise control of the tube current (flow of electrons between cathode and anode)

A satisfactory radiograph was made using a 36-in. SID, 12 mAs, and a 12:1 grid. If the examination will be repeated at a distance of 42 in. and using a 5:1 grid, what should be the new milliampere-seconds value to maintain the original receptor exposure? A 5.6 B 6.5 C 9.7 D 13

The Correct Answer is: B According to the exposure-maintenance formula, if the SID is changed to 48 in., 16.33 mAs is required to maintain the original radiographic receptor exposure: Thus, x = 16.33 mAs at 42 in. SID. Then, to compensate for changing from a 12:1 grid to a 5:1 grid, the milliampere-seconds value becomes 6.53 mAs: Thus, x 6.53 mAs with 5:1 grid at 42 in. SID. Hence, 6.53 mAs is required to produce a receptor exposure similar to that of the original radiograph. The following are the factors used for milliampere-seconds conversion from nongrid to grid:

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: B As 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: image size/object size= SID/SOD (SOD = SID -OID) 9/x = 105/87 (105-18=87) 105x = 783 x = 7.45 (~7.5) 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.

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: B 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 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.

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: B In 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.

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: B Magnification 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

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: B The 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.

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: B The 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

Disadvantages of using low-kilovoltage technical factors include insufficient penetration increased patient dose diminished resolution A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: B As the kilovoltage is decreased, x-ray-beam energy (i.e., penetration) is also decreased. Consequently, a shorter scale of contrast is obtained. As kilovoltage is reduced, the milliampere-seconds value must be increased accordingly to maintain adequate receptor exposure. This increase in milliampere-seconds results in greater patient dose. Resolution is not related to kV.

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 is generated within a part as the kilovoltage is increased, as the size of the field is increased, and as 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.

Spatial resolution is inversely related to SID OID grid 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.

Saturation" of an image in CR means that: A. The CR reader has difficulty converting insufficient exposure signals to produce a diagnostic image; all electronic enhancement mechanisms are maximized B. Beyond a certain exposure level, a large number of pixels will be at maximum digital value (black), resulting in loss of visibility of anatomical structures in that region. C. There is a large amount of scatter radiation that is contributing to loss of anatomical image detail due to loss of contrast, thus decreasing diagnostic quality of the image D. The CR image reader has adequately, and completely, read all the exposure intensities from the pixels resulting from the exposure

The Correct Answer is: B "Saturation" of the CR image plate refers to overexposure of the image plate, not underexposure (A). "Saturation" means that, beyond a certain exposure level, a large number of the pixels will be at the maximum digital value, i.e., black (B). Therefore, there is no signal difference in the very high exposure areas, resulting in a loss of anatomical structures visualized in that region. It is therefore important for radiographers to use the appropriate exposure level for any particular anatomical part and not rely on the CR reader to adjust for overexposure. This is also critical to ensure proper adherence to the ALARA principle and to honor professional ethics. Although scatter radiation contributes to the exposure of the image plate (C), it is mainly the excessive exposure setting that "saturates" the pixels and prevents the CR reader from displaying densities darker than the blackest areas. The "saturation" exposure prevents the CR reader from reading exposure intensities beyond the blackest densities in the pixels (D).

The part of a CT imaging system made of thousands of solid-state photodiodes is the A. gantry. B. detector array. C. collimator assembly. D. x-ray tube.

The Correct Answer is: B A 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 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 1000 mA. The scintillation detector array is made of thousands of solid-state photodiodes. These scintillation crystal (cadmium tungstate or rare earth oxide ceramic crystals) photodiode assemblies convert the transmitted x-ray energy into light. That light is then 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 percent. 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 500-25,000 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 that 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. It's motorized movement should be smooth and accurate. Inaccurate indexing can result in missed anatomy and/or double-exposed anatomy.

A compensating filter is used to A. absorb the harmful photons that contribute only to patient dose B. even out widely differing tissue densities C. eliminate much of the scattered radiation D. improve fluoroscopy

The Correct Answer is: B A compensating filter is used to make up for widely differing tissue densities. For example, it is difficult to obtain a satisfactory image of the mediastinum and lungs simultaneously without the use of a compensating filter to "even out" the densities. With this device, the chest is radiographed using mediastinal factors, and a trough-shaped filter (thicker laterally) is used to absorb excess photons that would overexpose the lungs. The middle portion of the filter lets the photons pass to the mediastinum almost unimpeded. Filters that absorb the photons contributing to skin dose are inherent and added filters. Compensating filtration is unrelated to elimination of scattered radiation or fluoroscopy.

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: B A 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 (CDC) 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).

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: B Although 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.

If the radiographer is unable to achieve a short OID because of the structure of the body part or patient condition, which of the following adjustments can be made to minimize magnification distortion? A. A smaller focal-spot size should be used. B. A longer SID should be used. C. A smaller FOV should be used. D. A lower-ratio grid should be used.

The Correct Answer is: B An increase in SID will help to decrease the effect of excessive OID. For example, in the lateral projection of the cervical spine, there is normally a significant OID that would result in obvious magnification at a 40-in. SID. This effect is decreased by the use of a 72-in. SID. However, especially with larger body parts, increased SID usually requires a significant increase in exposure factors. Focal-spot size, FOV size, and grid ratio are unrelated to magnification.

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: B Beam 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.

Which of the following cells is the least radiosensitive? A. Myelocytes B. Myocytes C. Megakaryocytes D. Erythroblasts

The Correct Answer is: B Bergonié and Tribondeau theorized in 1906 that all precursor cells are particularly radiosensitive (e.g., stem cells found in bone marrow). There are several types of stem cells in bone marrow, and the different types differ in degree of radiosensitivity. Of these, red blood cell precursors, or erythroblasts, are the most radiosensitive. White blood cell precursors, or myelocytes, follow. Platelet precursor cells, or megakaryocytes, are even less radiosensitive. Myocytes are mature muscle cells and are fairly radioresistant.

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: B Detectors 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.

The factors that impact spatial resolution include 1. Focal spot size 2. Type of rectification 3. 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: B Focal 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.

Which of the following requires two exposures to evaluate focal-spot accuracy? A. Pinhole camera B. Slit camera C. Star pattern D. Bar pattern

The Correct Answer is: B Focal-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.

Types of moving grid mechanisms include 1. oscillating. 2. reciprocating. 3. synchronous. A. 1 only B. 1 and 2 only C. 1 and 3 only D. 2 and 3 only

The Correct Answer is: B Grids are devices constructed of alternating strips of lead foil and radiolucent interspacing material. They are placed between the patient and the IR, and they function to remove scattered radiation from the remnant beam before it forms the latent image. Stationary grids will efficiently remove scattered radiation from the remnant beam; however, their lead strips will be imaged on the radiograph. If the grid is made to move (usually in a direction perpendicular to the lead strips) during the exposure, the lead strips will be effectively blurred. The motion of a moving grid, or Potter-Bucky diaphragm, may be reciprocating (equal strokes back and forth), oscillating (almost circular direction), or catapult (rapid forward motion and slow return). Synchronous refers to a type of x-ray timer.

Before a flat-panel detector can be used for a radiographic exposure, it must be prepared. This preparation is referred to as: A. Propagation B. Initialization C. Augmentation D. Instrumentation

The Correct Answer is: B In order to prepare a flat-panel detector for an X-ray exposure, it must be initialized, where all switching elements are held in an "off" state by the appropriate control voltage (e.g., -5 volts) (B). Once the x-ray exposure is made, the pixel's sensing area contains the image information. That information is obtained, line by line, by changing the control voltage (e.g., +10 volts). The resulting signal is digitized and stored. Propagation (A) refers to energy travelling through a medium, such as an anatomical part. In medical imaging, the term "augmentation" (C) refers to either forced accelerated venous blood return to the heart by manually compressing a patient's leg during a venous ultrasound Doppler procedure or, in mammography, when imaging augmented breasts. Instrumentation (D) is a general term that refers to devices used in medical procedures or the development, and safe and effective use of medical technology.

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: B Many 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.

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: B NCRP 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.

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: B Of 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.

Off-focus, or extrafocal, radiation is minimized by A. avoiding the use of very high kilovoltages B. restricting the x-ray beam as close to its source as possible C. using compression devices to reduce tissue thickness D. avoiding extreme collimation

The Correct Answer is: B Off-focus, or extrafocal, radiation is produced as electrons strike metal surfaces other than the focal track and produce x-rays that emerge with the primary beam at a variety of angles. This radiation is responsible for indistinct images outside the collimated field. Mounting a pair of shutters as close to the source as possible minimizes off-focus radiation.

Disadvantages of moving grids over stationary grids include which of the following? 1. They can prohibit the use of very short exposure times. 2. They increase patient radiation dose. 3. 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: B One 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).

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: B The 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.

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: B The 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.

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: B The 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.

To maintain image clarity in an image-intensifier system, the path of electron flow from the photocathode to the output phosphor is controlled by A. the accelerating anode B. electrostatic lenses C. the vacuum glass envelope D. the input phosphor

The Correct Answer is: B The 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 carefully focused to maintain image resolution by the electrostatic focusing lenses through the accelerating anode and to the output phosphor for conversion back to light.

To maintain image clarity, the path of electron flow from photocathode to output phosphor is controlled by A. the accelerating anode B. electrostatic lenses C. the vacuum glass envelope D. the input phosphor

The Correct Answer is: B The 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 carefully focused, to maintain image resolution, by the electrostatic focusing lenses, through the accelerating anode and to the output phosphor for conversion back to light.

The output phosphor can be coupled with the Vidicon TV camera or charge-coupled device (CCD) via 1. fiber optics. 2. an image distributor or lens. 3. 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: B The 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).

Exposure rate increases with an increase in 1. 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: B The 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.

The electron cloud within the x-ray tube is the product of a process called A. electrolysis B. thermionic emission C. rectification D. induction

The Correct Answer is: B The thoriated tungsten filament of the cathode is heated by its own filament circuit. The x-ray tube filament is made of thoriated tungsten and is part of the cathode assembly. Its circuit provides current and voltage to heat it to incandescence, at which time it undergoes thermionic emission—the liberation of valence electrons from the filament atoms. Electrolysis describes the chemical ionization effects of an electric current. Rectification is the process of changing alternating current to unidirectional current.

Advantages of battery-powered mobile x-ray units include their 1. ability to store a large quantity of energy 2. ability to store energy for extended periods of time 3. lightness and ease of maneuverability A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The Correct Answer is: B There are two main types of mobile x-ray equipment—capacitor-discharge and battery-powered. Although capacitor-discharge units are light, and therefore fairly easy to maneuver, the battery-powered mobile unit is very heavy (largely because it carries its heavy-duty power source). It is, however, capable of storing a large milliampere-seconds capacity for extended periods of time. These units frequently have a capacity of 10,000 mAs, with 12 hours required for a full charge.

A three-phase timer can be tested for accuracy using a synchronous spinning top. The resulting image looks like a A. series of dots or dashes, each representative of a radiation pulse B. solid arc, with the angle (in degrees) representative of the exposure time C. series of gray tones, from white to black D. multitude of small, mesh-like squares of uniform sharpness

The Correct Answer is: B When a spinning top is used to test the efficiency of a single-phase timer, the result is a series of dots or dashes, with each representing a pulse of radiation. With full-wave-rectified current and a possible 120 dots (pulses) available per second, one should visualize 12 dots at 1/10 s, 24 dots at 1/5 s, 6 dots at 1/20 s, and so on. However, because three-phase equipment is at almost constant potential, a synchronous spinning top must be used, and the result is a solid arc (rather than dots). The number of degrees formed by the arc is measured and equated to a particular exposure time. A multitude of small, mesh-like squares describes a screen contact test. An aluminum step wedge (penetrometer) may be used to demonstrate the effect of kilovoltage on contrast (demonstrating a series of gray tones from white to black), with a greater number of grays demonstrated at higher kilovoltage levels.

Histogram data that is skewed relative to the values of interest (VOI) of the histogram analysis used for a particular exam may be caused by all of the following, except: A. Anatomical structures not centered to the IP B. Excessive windowing C. The X-ray beam is not correctly aligned to the edges of the IP D. The beam edge is irregular because of overlap of a radiopaque shadow, such as a gonadal shield

The Correct Answer is: B Windowing (B) is a post-processing method of adjusting the brightness and contrast in the digital image. Histogram analysis errors occur prior to post-processing of the 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 darker. When decreased, the image becomes brighter. 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). Answers A, C and D can cause histogram data to be skewed relative to the values of interest (VOI) of the histogram analysis used for a particular exam.

The fact that x-ray intensity across the primary beam can vary as much as 45% describes the A line-focus principle. B transformer law. C anode heel effect. D inverse-square law.

The Correct Answer is: C A beveled focal track extends around the periphery of the anode disk; when a small angle is used, the beveled edge allows for a smaller effective focal spot and better detail. The disadvantage, however, is that photons are 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 beam. This is known as the anode heel effect and can cause a primary beam variation of up to 45%. The anode heel effect becomes more pronounced as the SID decreases, as IR size increases, and as target angle decreases.

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: C A 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).

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: C A 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.

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

The Correct Answer is: C A 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)

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: C A 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

Delivery of large exposures to a cold anode or the use of exposures exceeding tube limitation can result in increased tube output cracking of the anode rotor-bearing damage A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: C A large quantity of heat applied to a cold anode can cause enough surface heat to crack the anode. Excessive heat to the target can cause pitting or localized melting of the focal track. Localized melts can result in vaporized tungsten deposits on the glass envelope, which can cause a filtering effect, decreasing tube output. Excessive heat also can be conducted to the rotor bearings, causing increased friction and tube failure

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: C A 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.

In the radiographic rating charts shown below, what is the maximum safe kVp that may be used with the 1-mm focal spot, single-phase x-ray tube, using 300 mA and 1/50-second exposure? A. 80 B. 95 C. 105 D. 112

The Correct Answer is: C A radiographic rating chart enables the radiographer to determine the maximum safe mA, exposure time, and kVp for a given exposure using a particular x-ray tube. Because the heat load that an anode will safely accept varies with the size of the focal spot, type of rectification, and anode rotation, these variables must also be identified. Each x-ray tube has its own characteristics and its own rating chart. First, find the chart with the identifying single-phase sine wave in the upper right corner of the chart and the correct focal spot size in the upper left corner of the chart (chart C). Once the correct chart has been identified, locate 1/50 (0.02) second on the horizontal axis and follow its line up to where it intersects with the 300-mA curve. Then draw a line to where this point meets the vertical (kVp) axis; it meets at between 100 and 110 kVp, or approximately 107 kVp. This is the maximum permissible kVp exposure at the given mAs for this x-ray tube. The radiographer should always use somewhat less than the maximum exposure. This same procedure is followed to answer the next two questions.

In the radiographic rating charts shown in Figure 5-8, what is the maximum safe kilovoltage that may be used with the 1.0-mm focal-spot, single-phase x-ray tube using 400 mA and a 0.02-s exposure? A 70 kVp B 75 kVp C 80 kVp D 85 kVp

The Correct Answer is: C A radiographic rating chart enables the radiographer to determine the maximum safe milliamperage, exposure time, and kilovoltage for a given exposure using a particular x-ray tube. Because the heat load that an anode will safely accept varies with the size of the focal spot, type of rectification, and anode rotation, these variables must also be identified. Each x-ray tube has its own characteristics and its own rating chart. First, find the chart with the identifying single-phase sine wave in the upper right corner and the correct focal-spot size in the upper left corner (chart C). Once the correct chart has been identified, locate 0.02 s on the horizontal axis, and follow its line up to where it intersects with the 400-mA curve. Then draw a line to where this point meets the vertical (kVp) axis; it intersects at exactly 80 kVp. This is the maximum permissible kilovoltage exposure at the given milliampere-seconds for this x-ray tube. The radiographer should always use somewhat less than the maximum exposure.

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: C A 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

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: C Absorption 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 as it passes through matter is called attenuation.

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: C 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 milliampere-seconds values when changing distance: mAs1/mAs2 = D1 squared/D2 squared 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.

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: C All 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.

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: C All 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.

Which of the following will serve to increase the effective energy of the x-ray beam? Increase in added filtration Increase in kilovoltage Increase in milliamperage A 1 only B 2 only C 1 and 2 only D 1, 2, and 3

The Correct Answer is: C As filtration is added to the x-ray beam, the lower-energy photons are removed, and the overall energy or wavelength of the beam is greater. As kilovoltage is increased, more high-energy photons are produced, and again, the overall, or average, energy of the beam is greater. An increase in milliamperage serves to increase the number of photons produced at the target but is unrelated to their energy.

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: C As x-ray photons travel through a part, they either pass all the way through to expose the image receptor, or they undergo interaction(s) that may result in their being absorbed by the part or deviated in direction. It is those that change direction (scattered radiation) that undermine the image. With respect to the radiographic image, it is responsible for the scattered radiation that reaches the image receptor. Scattered radiation adds unwanted, degrading exposure to the radiographic image. The single most important way to reduce the production of scattered radiation is to collimate. Although collimation, use of lower kVp (with appropriately higher mAs), and compression can be used, a large amount of scattered radiation can still be generated within the part being radiographed. Because scattered radiation adds unwanted noninformation-carrying photons, it can have a degrading effect on image quality...thus the need for grids.

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

The Correct Answer is: C Because mammographic techniques operate at very low kilovoltage levels, the cassette 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 cassette front material

Because of the anode heel effect, the intensity of the x-ray beam is greatest along the A path of the central ray B anode end of the beam C cathode end of the beam D transverse axis of the IR

The Correct Answer is: C Because the anode's focal track is beveled (angled, facing the cathode), x-ray photons can freely diverge toward the cathode end of the x-ray tube. However, the "heel" of the focal track prevents x-ray photons from diverging toward the anode end of the tube. This results in varying intensity from anode to cathode, with fewer photons at the anode end and more photons at the cathode end. The anode heel effect is most noticeable when using large IRs, short SIDs, and steep target angles

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: C By 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).

Advantages of direct digital radiography over computed radiography (CR) include direct digital is less expensive. direct digital has immediate readout. IPs 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: C Computed 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 IP is erased and reused. DR offers the advantage of immediate visualization of the x-ray image; in CR there is a short delay

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: C Control 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.

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: C Distortion 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

The most common cause of x-ray tube failure is A a cracked anode. B a pitted anode. C vaporized tungsten on glass envelope. D insufficient heat production.

The Correct Answer is: C Excessive heat production is a major problem in x-ray production. Of the energy required to produce x-rays, 0.2% is transformed to x-rays, and 99.8% is transformed to heat. The copper anode stem and the oil surrounding the x-ray tube help to move heat away from the face of the anode. Excessive heat can cause pitting of the anode (resulting in decreased output) or actual cracking of the anode or damage to the rotor bearings (resulting in tube failure). As the cathode filament is heated for exposure after exposure, some of its tungsten is vaporized and deposited on the inner surface of the glass envelope near the tube window. After a time, this can cause electric arcing and tube failure. This is the most common cause of tube failure because it can occur even with normal use

In the radiographic rating charts shown below, what is the maximum safe mA that may be used with 0.1-second exposure and 120 kVp, using the three-phase, 2-mm focal spot x-ray tube? A 400 B 500 C 600 D 700

The Correct Answer is: C Find the correct chart for the three-phase, 2-mm focal spot x-ray tube. Locate 0.1 second on the horizontal (seconds) axis and follow it up to where it intersects with the 120-kVp line on the vertical (kVp) axis. They intersect midway between the 600- and 700-mA curves, at approximately 650 mA. Thus, 600 mA is the maximum safe milliamperage for this particular group of exposure factors and x-ray tube.

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: C First, 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

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: C Focal-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

What grid ratio is represented in Figure 4-8? A. 3:1 B. 5:1 C. 10:1 D. 16:1

The Correct Answer is: C Grid ratio is defined as the height of the lead strips compared with (divided by) the width of the interspace material. The width of the lead strips has no bearing on the grid ratio. The height of the lead strips is 5 mm; the width of the interspace material (same as the distance between the lead strips) is 0.5 mm. Therefore, the grid ratio is 5/0.5, or a 10:1 grid ratio

Compared with a low-ratio grid, a high-ratio grid will 1. allow more centering latitude 2. absorb more scattered radiation 3. absorb more primary radiation A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The Correct Answer is: C Grid ratio is defined as the height of the lead strips to the width of the interspace material (Figure 4-32). The higher the lead strips (or the smaller the distance between the strips), the higher the grid ratio, and the greater the percentage of scattered radiation absorbed. However, a grid does absorb some primary/useful radiation as well. The higher the lead strips, the more critical is the need for accurate centering because the lead strips will more readily trap photons whose direction does not parallel them.

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: C Grids 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.

Which of the following are methods of limiting the production of scattered radiation? 1. Using moderate ratio grids 2. Using the prone position for abdominal examinations 3. Restricting the field size to the smallest practical size A. 1 and 2 only B. 1 and 3 only C. 2 and 3 only D. 1, 2, and 3

The Correct Answer is: C If a fairly large patient is turned prone, the abdominal measurement will be significantly different from the AP measurement as a result of the effect of compression. Thus, the part is essentially "thinner," and less scattered radiation will be produced. If the patient remains supine and a compression band is applied, a similar effect will be produced. Beam restriction is probably the single most effective means of reducing the production of scattered radiation. Grid ratio affects the cleanup of scattered radiation; it has no effect on the production of scattered radiation.

Central ray angulation may be required for magnification of anatomic structures foreshortening or self-superimposition superimposition of overlying structures A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: C If structures are overlying or underlying the area to be demonstrated (e.g., the medial femoral condyle obscuring the joint space in the lateral knee projection), CR angulation is used (e.g., 5-degree cephalad angulation to see the joint space in the lateral knee). If structures are likely to be foreshortened or self-superimposed (e.g., the scaphoid in a PA wrist), CR angulation may be employed to place the structure more closely parallel with the IR. Another example is the oblique cervical spine, where cephalad or caudad angulation is required to "open" the intervertebral foramina. Magnification is controlled by object-to- image-receptor distance (OID) and SID; it is unrelated to CR angulation

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: C If 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

The radiograph shown in the image below exhibits a loss of receptor exposure as a result of A. x-ray tube angulation across grid lines. B. exceeding the focusing distance. C. incorrect grid placement. D. insufficient SID.

The Correct Answer is: C If the x-ray tube is angled significantly across the lead strips of a focused grid, there is uniform loss of receptor exposure (grid cutoff). Insufficient or excessive distance with focused grids causes loss of receptor exposure (grid cutoff) along the periphery of the image. Figure 6-10 demonstrates grid cutoff everywhere except along a central vertical strip of the image. This receptor exposure loss is due to the focused grid's being placed upside down. Thus, the middle vertical lead strips allow x-rays to pass, but because the lead strips cant laterally, they are directly opposite to the direction of the x-ray photons (rather than parallel to them), and severe grid cutoff results

Which of the following statements is (are) true regarding the images below? 1. Image A was made using a higher kilovoltage than image B. 2. Image A was made with a higher-ratio grid than image B. 3. Image A demonstrates shorter-scale contrast than image B. A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The Correct Answer is: C Image A was made using 80 kV at 75 mAs; image B was made using 100 kV at 18 mAs; all other exposure factors remained the same. As kilovoltage is increased, the percentage of scattered radiation relative to primary radiation increases—hence, the grayer appearance of image B. Use of optimal kilovoltage for each anatomic part is helpful in keeping scatter to a minimum. The production of scattered radiation also will be limited if the field size is as small as possible. A grid is the most effective way to remove scattered photons from those exiting the patient. Grids are designed to selectively absorb scattered radiation while absorbing as little of the useful beam as possible. Images produced with higher-ratio grids are likely to evidence the effect of less scattered radiation than those made with lower-ratio grids.

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: C Image 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. When the window level is increased, the image becomes darker. When decreased, the image becomes brighter. 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.

Both radiographic images seen in the figure below were made of the same subject using identical exposure factors. Which of the following statements correctly describes these images? 1. Image A illustrates less receptor exposure because a shorter SID was used. 2. Image A illustrates more receptor exposure because the subject was turned PA. 3. Image B illustrates more receptor exposure because a shorter SID was used. A 1 only B 2 only C 3 only D 1 and 2 only

The Correct Answer is: C In the figure shown, image B is darker because the receptor exposure was greater. Receptor exposure is significantly affected by mAs, SID, and exposure rate. In this case, there is a difference in SID between the two images. As SID decreases, exposure rate increases and receptor exposure increases. Image B is darker (received a greater exposure) than image A because image B was exposed at a shorter SID (and therefore a higher exposure rate).

Both radiographic images shown in the figure below were made of the same subject using identical exposure factors. Which of the following statements correctly describe(s) these images? 1. Image A demonstrates less receptor exposure because a shorter SID was used. 2. Image A demonstrates more receptor exposure because the subject was turned PA. 3. Image B demonstrates more receptor exposure because a shorter SID was used. A. 1 only B. 2 only C. 3 only D. 1 and 2 only

The Correct Answer is: C In the figure, image B is darker and, therefore, has grater receptor exposure. Receptor exposure is largely determined by milliampere-seconds, SID, and exposure rate. In this case, there is a difference in SID between the two images. As SID decreases, exposure rate increases and receptor exposure increases. Image B is darker (demonstrates greater receptor exposure) than image A because image B was exposed at a shorter SID (and, therefore, a higher exposure rate)

Referring to the simplified x-ray circuit shown in Figure 7-16, what is indicated by the number 4? A. Step-up transformer B. Kilovoltage meter C. Grounded milliamperage meter D. Rectification system

The Correct Answer is: C In the simplified x-ray circuit shown, the autotransformer is labeled number 1, the primary coil of the high-voltage transformer is number 2, and the secondary coil is labeled number 3. The autotransformer selects the voltage that will be sent to the high-voltage transformer to be stepped up to the thousands of volts required for x-ray production. At the midpoint of the secondary coil is the grounded milliamperage meter (number 4). Since the milliamperage meter is in the control panel and is associated with high voltage, it must be grounded. The rectification system, which is used to change alternating current to unidirectional current, is indicated by the number 5. The rectification system is located between the secondary coil of the high-voltage transformer (number 3) and the x-ray tube (number 7).

Which of the following will contribute to the production of longer-scale radiographic contrast? 1. An increase in kV 2. An increase in grid ratio 3. 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: C Increased 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.

The component of a CR image plate (IP) that records the radiologic image is the A emulsion B helium-neon laser C photostimulable phosphor D scanner-reader

The Correct Answer is: C Inside the IP is the photostimulable phosphor (PSP). This PSP (or SPS—Storage Phosphor Screen), with its layer of europium-activated barium fluorohalide, serves as the IR because 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. Once the IP is placed into the CR processor (scanner or reader), the PSP plate is removed automatically. The latent image on the PSP is changed to a manifest image as it is scanned by a narrow, high-intensity helium-neon laser to obtain the pixel data. As the PSP is scanned in the reader, it releases a violet light—a process referred to as photostimulated luminescence (PSL). (

Which of the following affect(s) both the quantity and the quality of the primary beam? Half-value layer (HVL) Kilovoltage (kV) Milliamperage (mA) A 1 only B 2 only C 1 and 2 only D 1, 2, and 3

The Correct Answer is: C Kilovoltage and the HVL affect both the quantity and the quality of the primary beam. The principal qualitative factor for the primary beam is kilovoltage, but an increase in kilovoltage will also create an increase in the number of photons produced at the target. HVL is defined as the amount of material necessary to decrease the intensity of the beam to one-half its original value, thereby effecting a change in both beam quality and quantity. The milliampere-seconds value is adjusted to regulate the number of x-ray photons produced at the target. X-ray-beam quality is unaffected by changes in milliampere-seconds.

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: C Medical 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.

Which of the following groups of exposure factors will produce the greatest receptor exposure? A 100 mA, 50 ms B 200 mA, 40 ms C 400 mA, 70 ms D 600 mA, 30 ms

The Correct Answer is: C Milliampere-seconds (mAs) is the exposure factor that determines receptor exposure. Using the equation milliamperage × time = mAs, determine each mAs: (A) = 5 mAs, (B) = 8 mAs, (C) = 28 mAs, (D) = 18 mAs. Group C will produce the greatest receptor exposure

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: C Most 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.

Spatial resolution can be improved by decreasing the SID the OID patient/part motion A 1 only B 3 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: C Motion, 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.

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: C Patient 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.

The device shown in Figure 4-16 is used for A. tomographic QA testing B. timer and rectifier testing C. mammographic QA testing D. kilovoltage calibration testing

The Correct Answer is: C Quality control in mammography includes scrupulous testing of virtually all component parts of the mammographic imaging system. It includes processor checks, screen maintenance, accurate and consistent viewing conditions, and evaluation of phantom images, to name a few. The device shown constitutes the structures to be imaged within a mammography phantom. A mammographic phantom contains Mylar fibers, simulated masses, and specks of simulated calcifications. The American College of Radiology accreditation criteria state that a minimum of 10 objects (4 fibers, 3 specks, and 3 masses) must be visualized on test images. Changes in any part(s) of the imaging system (e.g., image receptors, x-ray equipment, filtration, or viewer) can result in unsuccessful results

The image quality seen in the figure below is most likely the result of A. an off-level grid B. pronounced anode heel effect C. low-milliampere-seconds and high-kilovoltage factors D. low-kilovoltage and high-milliampere-seconds factors

The Correct Answer is: C Quantum noise, or mottle, is a grainy appearance having a spotted or freckled appearance. Low-milliampere-seconds and high-kilovoltage factors are most likely to be the cause of quantum noise/mottle. Grid cutoff is absorption of the useful x-ray beam by the grid and usually results in loss of receptor exposure and visibility of grid lines. The anode heel effect is most pronounced using short SIDs, large IRs, small anode angles, and imaging parts having uneven tissue densities; it is represented by a noticeable receptor exposure difference between the anode and cathode ends of the image.

The radiograph of the pelvis shown in the figure below is unacceptable because of A. motion. B. inadequate penetration. C. scattered radiation fog. D. double exposure.

The Correct Answer is: C Radiographic contrast, especially in analog images, can be greatly affected by changes in kilovoltage (see figures below). As kilovoltage increases, a greater number of high-energy photons are produced at the target. These photons are more penetrating, but they also produce more scattered radiation, contributing to lower radiographic contrast as a result of scattered radiation fog. Radiograph B was made using 100 kVp and 18 mAs. Radiograph A was made of the same part using 80 kVp and 75 mAs, all other factors constant. The image details in radiograph A are far more perceptible as a result of the production of less scattered radiation.

Which of the following combinations would deliver the least amount of heat to the anode of a three-phase, 12-pulse x-ray unit? A. 400 mA, 0.12 s, 90 kVp B. 300 mA, ½ s, 70 kVp C. 500 mA, 1/30 s, 85 kVp D. 700 mA, 0.06 s, 120 kVp

The Correct Answer is: C Radiographic 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 using the large focal spot may not be safe when the small focal spot of the same x-ray tube is used. The total number of heat units an exposure generates also influences the amount of stress (in the form of heat) imparted to the anode. Single-phase heat units are determined by the product of milliampere-seconds and kilovoltage. A correction factor is required to determine the HU of three-phase equipment and high frequency equipment. Unless the equipment manufacturer specifies otherwise, three-phase and high frequency equipment heat units are determined by multiplying mA × second × kV × 1.4. In the examples given, then, group (A) produces 6,048 HU, group (B) produces 14,700 HU, group (C) produces 1,983 HU, and group (D) produces 7,056 HU. Therefore, group (C) exposure factors will deliver the least amount of heat to the anode

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: C Radiographic 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

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: C Receptor 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)

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: C Receptor 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

A grid usually is employed in which of the following circumstances? When radiographing a large or dense body part When using high kilovoltage When 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: C Significant 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.

A grid is usually employed 1. 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: C Significant 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

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: C Since 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)

Types of shape distortion include magnification elongation foreshortening A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: C Size distortion (magnification) is inversely proportional to SID and directly proportional to OID. Increasing the SID and decreasing the OID decreases size distortion. Aligning the tube, part, and IR so that they are parallel reduces shape distortion. There are two types of shape distortion—elongation and foreshortening. Angulation of the part with relation to the IR results in foreshortening of the object. Tube angulation causes elongation of the object

Resolution in CR increases as laser beam size decreases monitor matrix size decreases PSP crystal size decreases A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: C Spatial resolution in CR is impacted by the size of the PSP, the size of the scanning laser beam, and monitor matrix size. High-resolution monitors (2-4 MP, megapixels) are required for high-quality, high-resolution image display. The larger the matrix size, the better is the image resolution. Typical image matrix size (rows and columns) used in chest radiography is 2,048 × 2,048. As in traditional radiography, spatial resolution is measured in line pairs per millimeter. As matrix size is increased, there are more and smaller pixels in the matrix and, therefore, improved spatial resolution. Other factors contributing to image resolution are the size of the laser beam and the size of the PSP phosphors. Smaller phosphor size improves resolution —anything that causes an increase in light diffusion will result in a decrease in resolution. Smaller phosphors in the PSP plate allow less light diffusion. Additionally, the scanning laser light must be of the correct intensity and size. A narrow laser beam is required for optimal resolution.

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: C Spatial 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: C Successive 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)

Which of the following will occur as a result of a decrease in the anode target angle? 1. Less pronounced anode heel effect 2. Decreased effective focal spot size 3. 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: C Target 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.

A decrease from 200 to 100 mA will result in a decrease in which of the following? Wavelength Exposure rate Beam intensity A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: C Technical 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).

In digital imaging, as the size of the image matrix increases, FOV increases pixel size decreases spatial resolution increases A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: C 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 decreases, resolution increases. FOV and matrix size are related to pixel size according to the equation Pixel size = FOV / Matrix.

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

The Correct Answer is: C The 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

What is the function of the x-ray tube component numbered 2 in Figure 7-18? A. To release electrons when heated B. To release light when heated C. To direct electrons to the focal track D. To direct light to the focal track

The Correct Answer is: C The figure illustrates the x-ray tube component parts. Number 1 indicates the thoriated tungsten filament, which functions to release electrons when heated. Number 2 is the molybdenum focusing cup, which directs these electrons toward the anode's focal track. Number 4 is the rotating anode, and number 5 is the anode's focal track. The focal track is made of thoriated (for extra protection from heat) tungsten. When high-speed electrons are suddenly decelerated at the target, their kinetic energy is changed to x-ray photon energy.

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

The Correct Answer is: C The 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

In comparison with 60 kV, 80 kV will permit greater exposure latitude produce more scattered radiation produce shorter-scale contrast A 1 only B 2 only C 1 and 2 only D 2 and 3 only

The Correct Answer is: C The 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

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: C The 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.

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: C The 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

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: C The 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.

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: C The 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.

An incorrect relationship between the primary beam and the center of a focused grid results in an increase in scattered radiation production grid cutoff insufficient receptor exposure A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: C The lead strips of a focused grid are angled to correspond to the configuration of the divergent x-ray beam. Thus, any radiation that is changing direction, as is typical of scattered radiation, will be trapped by the lead foil strips. However, if the central ray and the grid center do not correspond, the lead strips will absorb useful radiation. The absorption of useful radiation is termed cutoff and results in diminished receptor exposure

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: C The 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.

Which of the following statements is (are) true regarding Figure 7-10? 1. Excessive kilovoltage was used. 2. High contrast is demonstrated. 3. Insufficient penetration is evident. A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The Correct Answer is: C The radiograph shows evidence of very few grays; this is short scale, or high, contrast. There is inadequate penetration of the denser structures—the heart and the lung bases and apices. Penetration and contrast are a function of kilovoltage. Inadequate penetration and high contrast are a result of insufficient peak kilovoltage

The chest radiograph shown in the figure below demonstrates A. motion B. focal-spot blur C. double exposure D. grid cutoff

The Correct Answer is: C The radiographic image seen in the figure demonstrates double exposure. Notice the double image of the ribs, humerus, and clavicle, especially on the left side of the chest. The anatomic parts and diaphragm are sharply defined, not blurry, as they would be in the case of motion. Focal-spot blur would also cause a slight blur/loss of resolution of anatomic details. Grid cutoff would appear loss of receptor exposure in part or all of the image

The term differential absorption is related to beam intensity subject contrast pathology A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: C The 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. 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

The device that receives the remnant beam, converts it into light, and then increases the brightness of that light is the A. charge-coupled device (CCD) B. spot camera C. image intensifier D. television monitor

The Correct Answer is: C The visual apparatus that is responsible for visual acuity and contrast perception is the cones within the retina. Cones are also used for daylight vision. Therefore, the most desirable condition for fluoroscopic viewing is to have a bright enough image to permit cone (daylight) vision for better perception of anatomic details. The image intensifier accomplishes this. The intensified image is then transferred to a TV monitor for viewing. Spot cameras record fluoroscopic events.

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

The Correct Answer is: C There 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. (

Capacitor-discharge mobile x-ray units use a grid-controlled x-ray tube are typically charged before the day's work provide a direct-current output A 1 only B 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: C There 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

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: C There 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.

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: C To 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: mAs1/mAs2 = GCF1/GCF2 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: C To 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.

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: C Two 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.

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: C Voltage 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.

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: C When 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 x = 972 mA

X-ray tube life may be extended by using high milliampere-second, low- kilovoltage exposure factors. avoiding lengthy anode rotation. avoiding exposures to a cold anode. A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: C X-ray tube life may be extended by using exposure factors that produce a minimum of heat (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 owing to early vaporization. Large exposures to a cold anode will heat the anode surface, and the temperature difference between surface and interior 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.

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: C X-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.

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: C According 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 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: C All 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).

The figure below is representative of A. the anode heel effect B. the line-focus principle C. the inverse-square law D. the reciprocity law

The Correct Answer is: C The figure illustrates that as distance from a light/x-ray source increases, the light/x-rays diverge and cover a larger area; the quantity of light/x-ray available per unit area becomes less and less as distance increases. The intensity (quantity) of light/x-ray decreases according to the inverse-square law; that is, the intensity at a particular distance from its source is inversely proportional to the square of the distance. As the distance between the x-ray tube and image receptor increases, exposure rate (and, therefore, receptor exposure) decreases according to the inverse-square law. Because the anode's focal track is beveled, x-ray photons can freely diverge toward the cathode end of the x-ray tube. However, the "heel" of the focal track prevents x-ray photons from diverging toward the anode end of the tube. This results in varying intensity with fewer photons at the anode end and more photons at the cathode end X-ray tube targets are constructed according to the line-focus principle—the focal spot is angled to the vertical. As the actual focal spot is projected downward, it is foreshortened; thus, the effective focal spot is always smaller than the actual focal spot.

A particular radiograph was produced using 12 mAs and 85 kV with a 16:1 ratio grid. The radiograph is to be repeated using an 8:1 ratio grid. What should be the new milliampere-seconds value? A 3 B 6 C 8 D 10

The Correct Answer is: C To 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 5:1: 2 6:1: 3 8:1: 4 10:1/12:1: 5 16:1: 6 To adjust exposure factors, you simply compare the old with the new: mAs1/mAs2 = GCF1/GCF2 12/x = 6/4 6x = 48 x = 8 mAs

Which of the following is most likely to result from the introduction of a grid to a particular radiographic examination? 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.

The radiograph shown in Figure 5-5 was made using a A. three-phase, 6-pulse rectified unit B. three-phase, 12-pulse rectified unit C. single-phase, full-wave rectified unit D. high-frequency rectified unit

The Correct Answer is: C A spinning top is used to test the timer efficiency of full-wave-rectified single-phase equipment. The result should be a series of dots or dashes, with each dot representing a pulse of x-radiation. With full-wave-rectified current there should be 120 dots/pulses seen per second. One should visualize 12 dots/pulses at 1 / 10 s, 6 dots at 0.05 s, 10 dots at 1 / 12 s, and 3 dots at 0.025 s. If an incorrect number of dots/pulses is obtained, it is an indication of either timer malfunction or rectifier failure. Because three-phase equipment is at almost constant potential, a synchronous spinning top must be used for timer testing, and the result is a solid arc (rather than dots). The number of degrees covered by the arc is measured and equated to a particular exposure time; one second exposure should demonstrate 360 degrees.

The mottled appearance of the radiograph shown in Figure 4-26 is most likely representative of A. Paget disease B. osteoporosis C. exposure artifacts

The Correct Answer is: C The lateral skull radiograph shown illustrates the result of great care taken by the radiographer to secure patient comfort. A skull examination was requested for this patient of advanced years. The radiographer positioned the patient, taking care to position and collimate accurately, all the while trying to ensure the patient's comfort by letting her keep her pillow. The foam stuffing of the pillow is nicely imaged. Although the artifacts somewhat resemble Paget disease or osteoporosis, note that they extend outside the anatomic part.

The AEC device operates on which of the following principles? 1. Delivery of the required exposure time 2. A parallel-plate ionization chamber charged by x-ray photons 3. Motion of magnetic fields inducing current in a conductor A. 1 only B. 2 only C. 1 and 2 only D. 1, 2, and 3

The Correct Answer is: C A parallel-plate ionization chamber is the most commonly used AEC. A radiolucent chamber is beneath the patient (between the patient and the IR). As photons emerge from the patient, they enter the chamber and ionize the air within it. Once a predetermined charge has been reached, the exposure is automatically terminated. AEC determines exposure time; the radiographer must determine optimum kV and mA. Motion of magnetic fields inducing a current in a conductor refers to the principle of mutual induction.

Grid interspace material can be made of 1. plastic 2. lead 3. aluminum A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The Correct Answer is: C A grid is a thin wafer placed between the patient and the IR to collect scattered radiation. It is made of alternating strips of lead and a radiolucent material such as plastic or aluminum. If the interspace material also were made of lead, little or no radiation would reach the IR, and no image would be formed.

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: C An 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.

If the x-ray image exhibits insufficient receptor exposure, this might be attributed to 1. a Insufficient kilovoltage 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 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.

Which of the following statements regarding dual x-ray absorptiometry is (are) true? 1. Radiation dose is low. 2. Only low-energy photons are used. 3. Photon attenuation by bone is calculated. A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The Correct Answer is: C Dual x-ray absorptiometry (DXA) imaging is used to evaluate bone mineral density (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, photon attenuation is calculated to represent the degree of bone density. 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.

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: C High 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).

Europium-activated barium fluorohalide is associated with A. fluoroscopy B. image intensifiers C. PSP storage plates D. filament material

The Correct Answer is: C Image Plates (IPs) have a protective function (for the PSP/storage plate 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 (similar to traditional cassettes) to absorb backscatter. Inside the IP is the photostimulable phosphor (PSP) storage plate. This PSP storage plate 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.

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: C In 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. 2

Use of a portion of the input phosphor during fluoroscopy, rather than the entire input phosphor, will result in 1. a larger field of view (FOV). 2. a magnified image. 3. improved spatial resolution. A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The Correct Answer is: C Multifield image intensifier tubes are usually either dual-field or trifield 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 used. As a result, the FOV decreases and magnification increases, producing better spatial resolution. At the same time, brightness is decreased, requiring an increase in milliamperage (therefore increased patient dose). This increase in milliamperage increases image quality. It can be likened to an increase in signal-to-noise ratio (SNR), with milliamperes being the signal.

The functions of a picture archiving and communication system (PACS) include 1. storage of analog images 2. retrieval of digital images 3. storage 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: C PACS 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.

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: C Quantum 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.

Why is a very short exposure time essential in chest radiography? A. To avoid excessive focal-spot blur B. To maintain short-scale contrast C. To minimize involuntary motion D. To minimize patient discomfort

The Correct Answer is: C Radiographers usually are able to stop voluntary motion using suspended respiration, careful instruction, and immobilization. However, involuntary motion also must be considered. To have a "stop action" effect on the heart when radiographing the chest, it is essential to use a short exposure time.

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, six-pulse equipment? A. 36 B. 24 C. 8 D. 6

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 significantly greater. To produce similar receptor exposure, only two thirds of the original mAs would be used for three-phase, six-pulse equipment (2/3 × 12 = 8 mAs). With 3-phase, 12-pulse equipment, the original mAs would be cut in half.

Which of the following would be appropriate IP front material(s)? 1. Tungsten 2. Magnesium 3. Bakelite A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The Correct Answer is: C The IP is used to support the PSP plate. The IP front should be made of a sturdy material with a low atomic number, because attenuation of the remnant beam is undesirable. Bakelite (the forerunner of today's plastics) and magnesium (the lightest structural metal) are commonly used for IP fronts. The high atomic number of tungsten makes it inappropriate as an IP front material.

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: C The 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.

Which of the following devices is used to control voltage by varying resistance? A. Autotransformer B. High-voltage transformer C. Rheostat D. Fuse

The Correct Answer is: C The autotransformer operates on the principle of self-induction and functions to select the correct voltage to be sent to the high-voltage transformer to be "stepped up" to kilovoltage. The high-voltage transformer increases the voltage and decreases the current. The rheostat is a type of variable resistor that is used to change voltage or current values. It is found frequently in the filament circuit. A fuse is a device used to protect the circuit elements from overload by opening the circuit in the event of a power surge.

The total brightness gain of an image intensifier is the product of 1. flux gain 2. minification gain 3. focusing gain A. 1 only B. 2 only C. 1 and 2 only D. 1 and 3 only

The Correct Answer is: C The brightness gain of image intensifiers is 5,000 to 20,000. This increase is accounted for in two ways. As the electron image is focused to the output phosphor, it is accelerated by high voltage (about 25 kV). The output phosphor is only a fraction of the size of the input phosphor, and this decrease in image size represents brightness gain, termed minification gain. The ratio of the number of x-ray photons at the input phosphor compared to the number light photons at the output phosphor is termed flux gain. Total brightness gain is equal to the product of minification gain and flux gain.

The luminescent light emitted by the PSP is converted to a digital image by the A. DAC B. scanner-reader C. ADC D. helium-neon laser

The Correct Answer is: C The exposed IP is placed into the CR scanner/reader, where the PSP 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 PSP 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 and sent to the analog-to-digital converter (ADC), where it is digitized and becomes the digital image. After a short delay the DAC (digital to analog converter) displays the recognizable analog image 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.

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: C The 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

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: C The 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.

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: C With 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).

Although the stated focal-spot size is measured directly under the actual focal spot, focal-spot size in fact 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 CR . C. At the cathode end D. At the anode end

The Correct Answer is: C X-ray tube targets are constructed according to the line-focus principle—the focal spot is angled (usually 12-17 degrees) to the vertical. As the actual focal spot is projected downward, it is foreshortened; thus, the effective focal spot is smaller than the actual focal spot. As the focal spot is projected toward the cathode end of the x-ray beam, it becomes larger and approaches its actual size. Figure 7-22 illustrates the variation of the effective focal-spot size along the longitudinal tube axis. As the effective focal spot becomes larger toward the cathode end, the images of the phalanges illustrate gradual loss of spatial resolution.

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

The Correct Answer is: C. To change nongrid to grid exposure, or to adjust exposure when changing from one grid ratio to another, remember the factor for each grid ratio: Therefore, to change from nongrid exposure to an 8:1 grid, multiply the original milliampere-seconds value by a factor of 4. Thus, a new setting of 32 mAs is required.

The x-ray tube used in CT must be capable of high-speed rotation short pulsed exposures withstanding millions of heat units A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: D A 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.

A QA program serves to keep patient dose to a minimum keep radiographic quality consistent ensure equipment efficiency A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: D A QA program includes regular overseeing of all components of the imaging system—equipment calibration, film and cassettes, processor, x-ray equipment, and so on. With regular maintenance, testing, and repairs, equipment should operate efficiently and consistently. In turn, radiographic quality will be consistent, and repeat exposures will be minimized, thereby reducing patient exposure.

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: D According 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.

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: D According 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.

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: D All 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

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

The Correct Answer is: D All 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

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

The Correct Answer is: D All 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.

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

The Correct Answer is: D All 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

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: D Anode 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.

Which of the following factors influence(s) the production of scattered radiation? Kilovoltage level Tissue density Size of field A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: D As photon energy (kV) increases, so does the production of scattered radiation. The greater the density of the irradiated tissues, the greater is the production of scattered radiation. As the size of the irradiated field increases, there is an increase in the volume of tissue irradiated, and the percentage of scatter again increases. Beam restriction is the single most important way to limit the amount of scattered radiation produced.

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: D As 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.

Anode angle will have an effect on the severity of the heel effect focal-spot size heat-load capacity A 1 only B 2 only C 1 and 2 only D 1, 2, and 3

The Correct Answer is: D As 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

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

The Correct Answer is: D As 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

Which of the following pathologic conditions would require an increase in exposure factors? A. Pneumoperitoneum B. Obstructed bowel C. Renal colic D. Ascites

The Correct Answer is: D Because pneumoperitoneum is an abnormal accumulation of air or gas in the peritoneal cavity, it would require a decrease in exposure factors. Obstructed bowel usually involves distended, air- or gas-filled bowel loops, again requiring a decrease in exposure factors. With ascites, there is an abnormal accumulation of fluid in the abdominal cavity, necessitating an increase in exposure factors. Renal colic is the pain associated with the passage of renal calculi; no change from the normal exposure factors is usually required.

Typical examples of digital imaging include MRI CT CR A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: D CT (Computed Tomography), MRI (Magnetic Resonance Imaging), and CR (Computed Radiography) are three common examples of digital imaging. Special equipment is also available for direct digital radiography (DR)—images produced by either a fan-shaped x-ray beam received by linearly arrayed radiation detectors or a traditional fan-shaped x-ray beam received by a light-stimulated phosphor plate. Digital images can also be obtained in digital subtraction angiography (DSA), nuclear medicine, and diagnostic sonography. Analog images are conventional images; they can be converted to digital images with a device called a digitizer.

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

The Correct Answer is: D Collimators restrict the size of the irradiated field, thereby limiting the volume of irradiated tissue, and hence less scattered radiation is produced. 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.

Advantages of high-frequency generators include small size decreased patient dose nearly constant potential A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: D Conventional 60-Hz full-wave rectified power is converted to a higher frequency of 500 to 25,000 Hz in the most recent generator design—the high-frequency generator. The high-frequency generator is small in size, in addition to producing an almost constant potential waveform. High-frequency generators first appeared in mobile x-ray units and were then adopted by mammography and CT equipment. Today, more and more radiographic equipment uses high-frequency generators. Their compact size makes them popular, and the fact that they produce nearly constant potential voltage helps to improve image quality and decrease patient dose (fewer low-energy photons to contribute to skin dose)

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: D Determining 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)

The effect described as differential absorption is responsible for radiographic contrast a result of attenuating characteristics of tissue minimized 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: D Differential 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

Distortion can be caused by tube angle the position of the organ or structure within the body the radiographic positioning of the part A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: D Distortion is caused by improper alignment of the tube, body part, and IR. Anatomic structures within the body are rarely parallel to the IR in a simple recumbent position. In an attempt to overcome this distortion, we position the part to be parallel with the IR or angle the central ray to "open up" the part. Examples of this technique are obliquing the pelvis to place the ilium parallel to the IR or angling the central ray cephalad to "open up" the sigmoid colon.

Which of the following groups of technical factors will produce the least receptor exposure? A 400 mA, 0.010 second, 94 kV B 500 mA, 0.008 second, 94 kV C 200 mA, 0.040 second, 94 kV D 100 mA, 0.020 second, 80 kV

The Correct Answer is: D Each milliampere-second setting is determined [(A) = 4; (B) = 4; (C) = 8; (D) = 2] and numbered in order of greatest to least receptor exposure [(C) = 1; (A) and (B) = 2; (D) = 3]. Then, the kilovoltages are reviewed and also numbered in order of greatest to least receptor exposure [(A), (B), and (C) = 1; (D) = 2]. Finally, the numbers assigned to the milliampere-seconds and kilovoltage are added for each of the four groups [ (A) and (B) = 3; (C) = 2; (D) = 5]; the lowest total (C) indicates the group of factors that will produce the greatest receptor exposure; the highest total (D) indicates the group of factors that will produce the least receptor exposure. This process is illustrated as follows: (A) 4 mAs (2) + 94 kV (1) =3 (B) 4 mAs (2) + 94 kV (1) ) = 3 (C) 8 mAs (1) + 94 kV (1) = 2 (D) 2 mAs (3) + 80 kV (2) = 5

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: D Exposure 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.

Factor(s) that impact receptor exposure include 1. 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: D Factors 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. (

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

The Correct Answer is: D Focal-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.

Focusing distance is associated with which of the following? A Computed tomography B Chest radiography C Magnification radiography D Grids

The Correct Answer is: D Focusing distance is the term used to specify the optimal SID used with a particular focused grid. It is usually expressed as focal range, indicating the minimum and maximum SID workable with that grid. Lesser or greater distances can result in grid cutoff. Although proper distance is important in computed tomography and chest and magnification radiography, focusing distance is unrelated to them

Which of the following adult radiographic examinations usually require(s) use of a grid? Ribs Vertebrae Shoulder A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: D Generally speaking, anatomic parts measuring in excess of 10 cm require a grid. (The major exception to this rule can be the chest). The larger the part, the more scattered radiation is generated. To avoid degradation of the image as a result of scattered radiation fog, a grid is used to absorb scatter. Parts generally requiring the use of a grid include the skull, spine, ribs, pelvis, shoulder, and femur.

Geometric unsharpness is influenced by which of the following? Distance from object to image Distance from source to object Distance 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: D Geometric 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

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: D Grid 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. Grid Ratio = Height of lead strip/width of the distance between the leadstrips Grid Ratio = 4/0.25 Grid Ratio = 16

Which of the following factors is/are related to grid efficiency? Grid ratio Number of lead strips per inch Amount 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: D Grid 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. (

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: D High 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

The advantages of high-frequency generators over earlier types of generators include smaller size. nearly constant potential. lower patient dose. A 1 only B 1 and 2 only C 1 and 3 only D 1, 2 and 3

The Correct Answer is: D High-frequency generators first appeared in mobile x-ray units and were then adopted by mammography and CT equipment. Today, more and more radiographic equipment uses high-frequency generators. Their compact size makes them popular, and the fact that they produce nearly constant potential voltage helps to improve image quality and decrease patient dose (fewer low-energy photons to contribute to skin dose).

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: D High-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.

In which of the following ways might higher image contrast be obtained in abdominal radiography? 1. By using lower kilovoltage 2. By using a contrast medium 3. 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: D Higher/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.

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: D In 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

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: D In 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 (D). 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

Which of the following groups of exposure factors would be most effective in eliminating prominent pulmonary vascular markings in the RAO position of the sternum? A 500 mA, 1/30 s, 70 kV B 200 mA, 0.04 second, 80 kV C 300 mA, 1/10 s, 80 kV D 25 mA, 7/10 s, 70 kV

The Correct Answer is: D In the RAO position, the sternum must be visualized through the thorax and heart. Prominent pulmonary vascular markings can hinder good visualization. A method frequently used to overcome this problem is to use a milliampere-seconds value with a long exposure time. The patient is permitted to breathe normally during the (extended) exposure and by so doing blurs out the prominent vascularities

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: D Insufficient milliamperage and/or exposure time will result in decreased receptor exposure. Insufficient kilovoltage will result in underpenetration. Insufficient SID, however, will result in increased exposure rate and overexposure of the IR.

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: D Magnification 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).

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: D Magnification 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.

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: D Milliampere-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.

If 300 mA has been selected for a particular exposure, what exposure time would be required to produce 6 mAs? A. 5 ms B. 10 ms C. 15 ms D. 20 ms

The Correct Answer is: D Milliampere-seconds (mAs) is the exposure factor that regulates receptor exposure. The equation used to determine mAs is mA × s = mAs. Substituting the known factors: 300 mA x s = 6 mAs s = 0.02 (0.02/1000 = 20 ms)

The batteries in battery-operated mobile x-ray units provide power to 1. the x-ray tube 2. machine locomotion 3. the braking mechanism A. 1 only B. 2 only C. 1 and 2 only D. 1, 2, and 3

The Correct Answer is: D Mobile 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

Acceptable method(s) of minimizing motion unsharpness is (are) suspended respiration. short exposure time. patient instruction. A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: D 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 to decrease involuntary motion. Immobilization can also be very useful in eliminating motion unsharpness

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: D Multifield 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.

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: D Multifield 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.

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: D One 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

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: D Proper 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.

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: D Proper 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.

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: D Rectifiers (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

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: D Review 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.

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: D Shape 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.

Shape distortion is influenced by the relationship between the x-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: D Shape 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

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

The Correct Answer is: D Shape distortion is caused by misalignment of the x-ray tube, the part to be radiographed, and the IR/film. An object can be falsely imaged (foreshortened or elongated) by incorrect placement of the tube, the body part, or the IR. Only one of the three need be misaligned for distortion to occur.

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: D Since 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 to provide the same 15 mAs (and thus the same receptor exposure). Because mA × time = mAs, 300 mA x 50 ms = 15,000 mAms mA x 22 ms = 15,000 mAms (divide by 22ms) mA = 681.81 answer: round up 700 mA

Which of the following waveforms has the lowest percentage voltage ripple? A Single-phase B Three-phase, six-pulse C Three-phase, 12-pulse D High-frequency

The Correct Answer is: D Single-phase current has a 100% voltage drop between peak voltages. Three-phase current decreases this voltage drop considerably. Three-phase, six-pulse current has about a 13% voltage drop between peak voltages, and three-phase, 12-pulse current has only about a 4% drop between peak voltages. However, high-frequency current is almost constant potential, having less than 1% voltage ripple

If 92 kV and 15 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 36 B 24 C 10 D 7.5

The Correct Answer is: D Single-phase radiographic equipment is 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. With three-phase, 12-pulse equipment, the original milliampere-seconds would be cut in half (one-half of 15 mAs = 7.5).

Dedicated radiographic units are available for 1. 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: D Special 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.

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: D The 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.

A quality-control (QC) program includes checks on which of the following radiographic equipment conditions? Reproducibility Linearity Positive beam limitation/automatic collimation A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: D The accuracy of all three is important to ensure adequate patient protection. Reproducibility means that repeated exposures at a given technique must provide consistent intensity. Linearity means that a given milliampere-seconds value, using different milliamperage stations with appropriate exposure time adjustments, will provide consistent intensity. PBL is automatic collimation and must be accurate to 2% of the SID. Light-localized collimators must be available and must be accurate to within 2%

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

The Correct Answer is: D The 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 disk 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 distributed evenly over the entire track. Rotating anodes can withstand delivery of a greater amount of heat for a longer period of time than stationary anodes

The advantage(s) of high-kilovoltage chest radiography is (are) that 1. exposure latitude is increased 2. it produces long-scale contrast 3. it 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: D The 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.

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

The Correct Answer is: D The 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

What x-ray tube component does the number 8 in Figure 5-11 indicate? A. Anode stem B. Rotor C. Stator D. Focal track

The Correct Answer is: D The figure illustrates the component parts of a rotating-anode x-ray tube enclosed within a glass envelope (number 3) to preserve the vacuum necessary for x-ray production. Number 4 is the rotating anode with its beveled focal track at the periphery (number 8) and its stem (at number 5). Numbers 6 and 7 are the stator and rotor, respectively—the two components of an induction motor—whose function it is to rotate the anode. Number 1 is the filament of the cathode assembly, which is made of thoriated tungsten and functions to liberate electrons (thermionic emission) when heated to white hot (incandescence). Number 2 is the molybdenum focusing cup, which functions to direct the liberated filament electrons to the focal spot.

Of what material is the x-ray tube component numbered 5 (Anode's focal track) in Figure 7-18 made? A. Cesium B. Nickel C. Molybdenum D. Tungsten

The Correct Answer is: D The figure illustrates the x-ray tube component parts. Number 1 indicates the thoriated tungsten filament, which functions to release electrons when heated. Number 2 is the nickel focusing cup, which directs these electrons toward the anode's focal track. Number 4 is the rotating anode, and number 5 is the anode's focal track. The focal track is made of a tungsten-rhenium alloy (for extra protection from heat). When high-speed electrons are suddenly decelerated at the target, their kinetic energy is changed to x-ray photon energy

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

The Correct Answer is: D The 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.

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: D The 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

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: D The 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.

The radiograph shown in Figure 4-12 demonstrates an example of A. overexposure B. motion C. laser jitter D. exposure artifact

The Correct Answer is: D The radiograph shown is that of an adult PA erect chest. The image is well positioned and exposed, but observe the braids of hair that extend past the neck and superimpose on the pulmonary apices. Braided hair should be pinned up or otherwise removed from superimposition on thoracic structures. The braided hair was imaged during the exposure of the PA chest and is, therefore, referred to as an exposure artifact. Laser jitter is an example of a processing artifacts occurring in the PSP scanner/leader.

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: D The 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

Recommended method(s) of minimizing motion unsharpness include suspended respiration short exposure time patient instruction A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: D The shortest possible exposure time should be used to minimize motion unsharpness. Motion causes unsharpness that destroys detail. Careful and accurate patient instruction is essential for minimizing voluntary motion. Suspended respiration eliminates respiratory motion. Using the shortest possible exposure time is essential to decreasing involuntary motion. Immobilization also can be useful in eliminating motion unsharpness.

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

The Correct Answer is: D The 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.

Image quality in digital fluoroscopy is influenced by 1. 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: D The 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

The brightness level of the fluoroscopic image can vary with milliamperage kilovoltage patient thickness A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: D The thicker and more dense the anatomic part being studied, the less bright will be the fluoroscopic image. Both milliamperage and kilovoltage affect the fluoroscopic image in a way similar to the way in which they affect the radiographic image. For optimal contrast, especially taking patient dose into consideration, higher kilovoltage and lower milliamperage are generally preferred

Which of the following functions to increase the milliamperage? A Increasing the speed of anode rotation B Increasing the transformer turns ratio C Using three-phase rectification D Increasing the heat of the filament

The Correct Answer is: D The thoriated tungsten filament of the cathode assembly is heated by its own filament circuit. This circuit provides current and voltage to heat the filament to incandescence, at which time it undergoes thermionic emission (the liberation of valence electrons from filament atoms). The greater the number of electrons flowing between the cathode and the anode, the greater is the tube current (mA). Rectification (single- or three-phase) is the process of changing alternating current to unidirectional current. A greater number of secondary transformer turns functions to increase voltage and decrease current

Tungsten alloy is the usual choice of target material for radiographic equipment because it has a high atomic number has a high melting point can readily dissipate heat A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: D The x-ray anode may be a molybdenum disk coated with a tungsten-rhenium alloy. Tungsten, with a high atomic number (74), produces high-energy x-rays quite efficiently. Since a great deal of heat is produced at the target, its high melting point (3410°C) helps to avoid damage to the target surface. Heat produced at the target should be dissipated readily, and tungsten's conductivity is similar to that of copper. Therefore, as heat is applied to the focus, it can be conducted throughout the disk to equalize the temperature and thus avoid pitting, or localized melting, of the focal track.

Geometric unsharpness is most likely to be greater A at long SIDs. B at the anode end of the image. C with small focal spots. D at the cathode end of the image.

The Correct Answer is: D The x-ray tube anode is designed according to the line focus principle, that is, with the focal track beveled (see the figure below). 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

The x-ray beam and collimator light field must coincide to within A 10% of the OID B 2% of the OID C 10% of the SID D 2% of the SID

The Correct Answer is: D There are many radiation protection devices and laws associated with today's x-ray equipment. For example, the collimator light must accurately indicate the size and location of the x-ray beam to within 2% of the SID. Equipment that does not function properly contributes to excessive patient exposure, in the form of repeat examinations, and to poor image quality.

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: D There 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

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: D When 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

With three-phase equipment, the voltage across the x-ray tube 1. drops to zero every 180 degrees 2. is 87% to 96% of the maximum value 3. is at nearly constant potential A. 1 only B. 2 only C. 1 and 2 only D. 2 and 3 only

The Correct Answer is: D With 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)].

X-ray tube life may be extended by using low-milliampere-seconds/high- kilovoltage exposure factors avoiding lengthy anode rotation avoiding exposures to a cold anode A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The Correct Answer is: D X-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

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: D A 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.

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: D One 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

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: D Slow 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).

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) used Optimum kilovoltage Backup time A 1 only B 1 and 2 only C 2 and 3 only D 1, 2, and 3

The Correct Answer is: D The 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.

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: D The formula for mAs is mA × s = mAs. Substituting known values, 0.05x = 15 x = 300 mA

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 darker. When decreased, the image becomes brighter. 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.

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: D Cesium 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.

Compression of the breast during mammographic imaging improves the technical quality of the image because 1. geometric blurring is decreased 2. less scattered radiation is produced 3. patient motion is reduced A. 1 only B. 3 only C. 2 and 3 only D. 1, 2, and 3

The Correct Answer is: D Compression of the breast tissue during mammographic imaging improves the technical quality of the image for several reasons. Compression brings breast structures into closer contact with the IR, thus reducing geometric blur and improving resolution. As the breast tissue is compressed and essentially becomes thinner, less scattered radiation is produced. Compression serves as excellent immobilization as well.

Which of the following is (are) tested as part of a quality assurance (QA) program? 1. Beam alignment 2. Reproducibility 3. Linearity A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The Correct Answer is: D Each of the three is included in a good QA 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.

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

The Correct Answer is: D Each of the three is included in a good QC program. The QC deals with imaging equipment. (QA deals with people and management practices.) Beam alignment must be accurate to 2% of the SID. Reproducibility means that repeated exposures at a given technique must provide consistent intensity. Linearity means that a given milliampere-seconds value, using different milliamperage stations with appropriate exposure-time adjustments, will provide consistent intensity.

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: D In 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 800/2048 = 0.4mm 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.

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: D It 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).

Inadequate collimation in CR imaging can result in an image that is too 1. light 2. dark 3. noisy A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The Correct Answer is: D It is important to note that histogram appearance as well as patient dose can be affected by the radiographer's knowledge and skill using digital imaging, in addition to his or her degree of accuracy in positioning and centering. Collimation is exceedingly important to avoid histogram analysis errors. Lack of adequate collimation can result in signals outside the anatomical area being included in the exposure data recognition/histogram analysis. This can result in a variety of histogram analysis errors, including excessively light, dark, or noisy images.

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: D Multifield 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.

What should be done to correct for magnification when using air-gap technique? A. Decrease OID B. Increase OID C. Decrease SID D. Increase SID

The Correct Answer is: D OID is used to effect an increase in contrast in the absence of a grid, usually in chest radiography. If a 6-in. 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; thus, the OID acts as a low-ratio grid and increases image contrast. However, the 6-in. OID air gap will make a very noticeable increase in magnification. To correct for this, the SID must be increased. Generally speaking, the SID needs to be increased 7 in. for every 1 in. of OID. With a 6-in. OID, the SID usually is increased from 6 to 10 ft (120 in.).

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

The Correct Answer is: D One 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 being limitation and is unrelated to dynamic range or latitude. ALARA is a radiation protection concept referring to keeping one's dose as low as reasonably achievable.

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: D Radiographic 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).

Which of the following is (are) associated with subject contrast? 1. Patient thickness 2. Tissue density 3. Kilovoltage A. 1 only B. 1 and 2 only C. 1 and 3 only D. 1, 2, and 3

The Correct Answer is: D Several 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, higher-energy photons are produced, beam attenuation is decreased, and subject contrast decreases.

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: D 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 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).

Sampling frequency in computed radiography (CR) is expressed as: A. The TFT array size B. An inverse relationship between focal spot size and matrix size C. The light spread between the image plate and the light guide of the scanner D. Pixels/mm or pixel density

The Correct Answer is: D The TFT array is found in direct and indirect digital detector systems, not in computed radiography (CR) (A). The focal spot size (B) influences image resolution but has no influence on the number of pixels per millimeter that determines the sampling frequency. The light spread between the image plate and the light guide of the scanner (C) refers to computed radiography (CR) systems, not direct or indirect digital detector systems. Sampling frequency, also referred to as pixel density (pixels/mm), is expressed as pixels per millimeter. The sampling frequency determines the pixel pitch, which, in turn, determines the spatial resolution (D).

The electron beam in a television cathode ray tube (CRT) is projected onto the output phosphor in a: A. Vertical pattern B. Fixed direct beam C. Broad field D. Raster pattern

The Correct Answer is: D The electron beam in a CRT is projected horizontally in a raster pattern (D) on the output phosphor of the CRT, similar to how your eyes move back and forth to read the lines of this text. The electron beam begins in the upper left corner of the phosphor screen and moves to the upper right corner, creating a line of varying intensity of light as it moves. This is called the active trace. The electron beam is then turned off and it returns to the left side of the phosphor screen. This is called the horizontal retrace. A series of active traces, followed by horizontal retraces, until the entire output phosphor screen is scanned, thereby completing a television field. This entire process is repeated to create a second television field to interlace the two fields into one television frame. Thirty of these television frames are produced each second. The electron beam in a CRT is scanned in a horizontal raster pattern (A), from left to right, onto the output phosphor. A fixed direct beam (B) would only covert electrons to light in a specific spot on the output phosphor. A broad electron field (C) would strike a large portion of the output phosphor, preventing the individual lines of varying light intensities to build the anatomical image (C).

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: D The 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.

Which of the following can contribute to the image contrast? 1. Tissue density 2. Pathology 3. 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: D The 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.

Which of the following is likely to contribute to the radiographic contrast present on an analog x-ray image? 1. Atomic number of tissues radiographed 2. Any pathologic processes 3. Degree 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: D The 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.

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: D The 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.

Characteristics of the metallic element tungsten include 1. ready dissipation of heat 2. high melting point 3. high atomic number A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The Correct Answer is: D The x-ray anode may be a molybdenum disk coated with a tungsten-rhenium alloy. Because tungsten has a high atomic number (74), it produces high-energy x-rays more efficiently. Since a great deal of heat is produced at the target, tungsten's high melting point (3,410°C) helps to avoid damage to the target surface. Heat produced at the target should be dissipated readily, and tungsten's conductivity is similar to that of copper. Therefore, as heat is applied to the focus, it can be conducted throughout the disk to equalize the temperature and thus avoid pitting, or localized melting, of the focal track.

Exposure factors of 90 kV and 3 mAs are used for a particular nongrid exposure. What should be the new milliampere-seconds (mAs) value if a 12:1 grid is added? A. 86 B. 9 C. 12 D. 15

The Correct Answer is: D To 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: Therefore, to change from nongrid to a 12:1 grid, multiply the original milliampere-seconds value by a factor of 5. A new milliampere-seconds value of 15 is required.

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

The Correct Answer is: D Vaporized 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 sparking and causing puncture of the glass envelope and subsequent tube failure.

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: D When 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.

Characteristics of DR imaging include 1. solid-state detector receptor plates 2. a direct-capture imaging system 3. immediate image display A. 1 only B. 1 and 3 only C. 2 and 3 only D. 1, 2, and 3

The Correct Answer is: D Whereas CR uses traditional x-ray devices to enclose and protect the PSP/SPS, digital radiography (DR) requires the use of somewhat different equipment. DR does not use cassettes or a traditional x-ray table; it is a direct-capture system of x-ray imaging. DR uses solid-state detector plates as the x-ray IR (instead of a cassette in the Bucky tray) to intercept the collimated x-ray beam and form the latent image. The solid-state detector plates are made of barium fluorohalide compounds similar to that used in CR's PSP/SPSs. DR affords the advantage of immediate display of the image, compared with CR's delayed image display.

Geometric unsharpness will be least obvious at 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: D. The 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.

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, Vs/Vp = Ns/Np To determine secondary I: Ns/Np = Ip/Is 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)

Which of the following statements is (are) most likely true regarding the figure below? 1 .Image A was made using a higher kVp than image B. 2. Image A was made with a higher ratio grid than image B 3. Image A demonstrates shorter scale contrast than image B. A. 1 only B. 1 and 2 only C. 2 and 3 only D. 1, 2, and 3

The correct answer is (C). Image A was made using 80 kVp at 75 mAs; Image B was made using 100 kVp at 18 mAs; all other exposure factors remained the same. As kVp is increased, the percentage of scattered radiation relative to primary radiation increases, hence the grayer appearance of image B. Use of optimal kilovoltage for each anatomic part is helpful in keeping scatter to a minimum. The production of scattered radiation will also be limited if the field size is as small as possible. A grid is the most effective way to remove scattered photons from those exiting the patient. Grids are designed to selectively absorb scattered radiation while absorbing as little of the primary radiation as possible. Images produced with higher ratio grids will possess fewer grays than those made with lower ratio grids

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, six-pulse equipment? A 36 B 24 C 8 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, only two thirds of the original mAs would be used for three-phase, six-pulse equipment (2/3 × 12 = 8 mAs). With 3-phase, 12-pulse equipment, the original mAs would be cut in half. Single phase Three phase Three phase 6-pulse 12-pulse x mAs ⅔ x mAs ½ x mAs

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)

The voltage ripple associated with a three-phase, 12-pulse rectified generator is about A 100%. B 32%. C 13%. D 4%.

The correct answer is (D). Voltage 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° past peak potential). The entire waveform is used; if 100 kV were selected, the actual average kilovoltage output would be approximately 70. 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 voltage ripple: single phase: 100% 3-phase, 6-pulse: 13% 3-phase, 12-pulse: 4% high frequency: >1%

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.

Spatial resolution is directly related to 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, so does spatial resolution 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.

Terms that refer to size distortion include magnification attenuation elongation A 1 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3

The correct answer is: (A) Distortion is misrepresentation of the actual size or shape of the object being imaged. Size distortion is magnification. Shape distortion is a result of improper alignment of the x-ray tube, the part being radiographed, and the IR; the two types of shape distortion are foreshortening and elongation. The shapes of various structures can be misrepresented radiographically as a result of their position in the body, when the part is out of the central axis of the x-ray beam, or when the CR is angled (Figure 7-19). Parts sometimes are elongated intentionally for better visualization (e.g., sigmoid colon). Some body parts, because of their position in the body, are foreshortened, such as the carpal scaphoid. Attenuation refers to decreasing beam intensity and is unrelated to distortion

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

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.

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)

If a radiograph exhibits insufficient receptor exposure, this might be attributed to 1. inadequate kVp. 2. inadequate 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.

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. It is desired 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)

The quantity of scattered radiation reaching the IR can be reduced through the use of a fast imaging system an air gap a stationary grid A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2, and 3

The correct answer is: (C) Scattered radiation adds unwanted degrading densities to the x-ray image. The single most important way to reduce the production of scattered radiation is to collimate. Although collimation, optimal kilovoltage, and compression can be used, a large amount of scattered radiation is still generated within the part being imaged, and because it adds unwanted non-information-carrying densities, it can have a severely degrading effect on image quality. A grid is a device interposed between the patient and IR that functions to absorb a large percentage of scattered radiation before it reaches the IR. Imaging system speed is unrelated to scattered radiation. A grid 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 cleanup of scattered radiation. An air gap introduced between the object and IR can have an effect similar to that of a grid. As energetic scattered radiation emerges from the body, it continues to travel in its divergent fashion and much of the time will bypass the IR. It is usually necessary to increase the SID to reduce magnification caused by increased OID.

Of what material is the x-ray tube component numbered 2 in Figure 7-18 made? A. Cesium B. Copper C. Nickel D. Tungsten

The correct answer is: (C) The figure illustrates the x-ray tube component parts. Number 1 indicates the thoriated tungsten filament, which functions to release electrons when heated. Number 2 is the nickel focusing cup, which directs these electrons toward the anode's focal track. Number 4 is the rotating anode, and number 5 is the anode's focal track. The focal track is made of thoriated (for extra protection from heat) tungsten. When high-speed electrons are suddenly decelerated at the target, their kinetic energy is changed to x-ray photon energy.

The housing surrounding an x-ray tube functions to 1. retain heat within the glass envelope 2. protect from electric shock 3. keep leakage radiation to a minimum A. 1 and 2 only B. 1 and 3 only C. 2 and 3 only D. 1, 2, and 3

The correct answer is: (C) When high-speed electrons strike surfaces other than the tungsten target, x-rays may be produced and emitted in all directions. X-ray tubes, therefore, have a lead-lined metal protective housing to absorb much of this "leakage radiation." Leakage radiation must not exceed 100 mR/h at a distance of 1 m from the tube. Because the production of x-radiation requires the use of exceedingly high voltage, the tube housing also serves to protect from electric shock. The production of x-rays involves the production of large quantities of heat, which can be damaging to the x-ray tube. Therefore, an oil coolant surrounds the x-ray tube to further insulate it and to absorb heat from the x-ray tube structures

Periodic equipment care includes evaluation of the 1. 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.

Which of the following statements is (are) true with respect to the radiograph shown in Figure 4-29? 1. The image exhibits long-scale contrast. 2. The image exhibits a clothing artifact. 3. The image demonstrates motion blur. A. 1 and 2 only B. 1 and 3 only C. 2 and 3 only D. 1, 2, and 3

The correct answer is: (D) The abdomen radiograph shown in the figure demonstrates motion blur. This can be seen particularly in the upper abdomen and in the bowel gas patterns. Motion obliterates spatial resolution. Patients who are in pain often are unable to cooperate as fully as patients who are not in pain. Careful positioning and patient instruction are helpful, but it remains useful to use the shortest exposure time possible. The radiograph also demonstrates good long-scale contrast that enables visualization of many tissue densities. The dark horizontal line across the abdomen is a clothing artifact resulting from a taut elastic underwear waist-band.

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: A Proper 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

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: C When 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.