Lange #3

Lakukan tugas rumah & ujian kamu dengan baik sekarang menggunakan Quizwiz!

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

(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.

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

(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

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

(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.

195. Exposure-type artifacts include 1. double exposure 2. motion 3. image fading (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(B) Artifacts can be a result of exposure, handling, and storage, or processing. Exposure artifacts include motion, double exposure, and patient clothing/jewelry—the effects of these are seen as a result of the exposure. Handling and storage artifacts include fogged PSP, image fading, upside down IP, damaged PSP—all these occur as a result of improper use or storage. Processing artifacts occur while the PSP is in the scanner/reader and include skipped scan lines, laser jitter, etc.

196. 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%

(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 radiographic density. 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 radiographic density in half (reduce kilovoltage by 15%).

219. IRs frequently have a rear lead-foil layer that functions to (A) improve penetration (B) absorb backscatter (C) preserve resolution (D) increase the screen speed

(B) Many IR have a thin lead-foil layer behind their rear section to absorb backscattered radiation that is energetic enough to exit the rear, strike the metal back, and bounce back to fog the image. The lead foil absorbs the backscatter before it can fog the PSP.

27. 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

(B) The formula for mAs is . Substituting known values:

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

(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).

An increase in added filtration will result in 1. an increase in maximum energy of the x-ray beam 2. a decrease in x-ray intensity 3. an increase in effective energy of the x-ray beam (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(C) Added aluminum filtration removes more low-energy photons; therefore there is a decrease in the number of photons in the x-ray beam, that is, beam intensity. Because low-energy photons are removed, the overall average energy of the x-ray beam is increased. This process can also be referred to as beam hardening because its average energy is increased. The maximum energy of the beam is unchanged as long as the kV remains unchanged.

18. 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

(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.

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

(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.

198. An exposure was made of a part using 300 mA and 0.06 second with a 200-speed film-screen combination. An additional radiograph is requested using a 400-speed system to reduce motion unsharpness. Using 400 mA, all other factors remaining constant, what should be the new exposure time? (A) 5 ms (B) 11 ms (C) 22 ms (D) 44 ms

(C) High-speed imaging systems are valuable for reducing patient exposure and patient motion. However, some detail will be sacrificed, and quantum mottle can cause further image impairment. In general, doubling the film-screen speed doubles the radiographic density, thereby requiring that the milliampere-seconds value be halved to maintain the original radiographic density. Changing from 200 to 400 screens requires halving the milliampere-seconds value to 9 mAs. The new exposure time, using 400 mA, is . Thus, -s exposure using 400 mA and 400-speed screens .

15. 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

(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).

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

(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. (

58. 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

(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: To adjust exposure factors, you simply compare the old with the new:

61. Which of the following technical changes would best serve to remedy the effect of very dissimilar tissue densities? (A) Use of a small focal spot (B) Use of a high-ratio grid (C) High-kilovoltage exposure factors (D) High milliampere-seconds exposure factors

(C) When tissue densities within a part are very dissimilar (e.g., chest x-ray), the radiographic result can be unacceptably high contrast. To "even out" these densities and produce a more appropriate scale of grays, exposure factors using high kilovoltage should be employed. Focal spot size is unrelated to image contrast. The higher the grid ratio, the higher is the contrast. Exposure factors using high milliampere-seconds generally result in unnecessary patient exposure.

238. Compression of the breast during mammo-graphic 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

(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 detail. As the breast tissue is compressed and essentially becomes thinner, less scattered radiation is produced. Compression serves as excellent immobilization as well.

121. What is the purpose of the thin layer of lead that is often located in the rear portion of an IP? (A) To prevent crossover (B) To increase speed (C) To diffuse light photons (D) To prevent scattered radiation fog

(D) The purpose of the thin layer of lead that is often located in the rear portion of an IP is to absorb x-ray photons that strike the rear of the IP and bounce back toward the PSP, resulting in scattered radiation fog. The thin layer of lead absorbs these x-ray photons and thus improves the radiographic image.

74. Pathologic or abnormal conditions that would require a decrease in exposure factors include all of the following except (A) osteoporosis (B) osteomalacia (C) emphysema (D) pneumonia

189. 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

(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.

104. 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

(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.

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

(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 (21) pixel will demonstrate 2 shades of gray, whereas a 6-bit (26) pixel can display 64 shades and a 7-bit (27) 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.

212. 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

(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.

133. The x-ray image seen on the computer display monitor is a (an) (A) analog image (B) digital image (C) phosphor image (D) emulsion image

(A) Remnant x-rays emerging from the patient/part are converted to electrical signals. This is an analog image. These electrical signals are sent to the ADC (analog-to-digital converter) to be converted to digital data (in binary digits or bits). Then, the digital image data is transferred to a DAC (digital-to-analog converter) to be converted to a perceptible analog image on the display monitor. The monitor image can be manipulated, postprocessed, stored, or transmitted.

85. Practice(s) that enable the radiographer to reduce the exposure time required for a particular image include 1. use of a higher milliamperage 2. use of a higher kilovoltage 3. use of a higher ratio grid (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(B) If it is desired to reduce the exposure time for a particular radiograph, as it might be when radiographing patients who are unable to cooperate fully, the milliamperage must be increased sufficiently to maintain the original milliampere-seconds value and thus image density/brightness. A higher kilovoltage could be useful because it would allow further reduction of the milliampere-seconds (exposure time) according to the 15% rule. Changing grid ratio is unrelated to desired changes in exposure time. Use of a higher-ratio grid would only necessitate an increase in mAs and not likely a decrease in exposure time.

2. What pixel size has a 512 × 512 matrix with a 20-cm field of view (FOV)? (A) 0.07 mm/pixel (B) 0.40 mm/pixel (C) 0.04 mm/pixel (D) 4.0 mm/pixel

(B) In digital imaging, pixel size is determined by dividing the field of view (FOV) by the matrix. In this case, the FOV is 20 cm; since the answer is expressed in millimeters, first change 20 cm to 200 mm. Then 200 divided by 512 equals 0.39 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 is increased, pixel size increases. Pixel size is inversely related to resolution. As pixel size increases, resolution decreases.

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

(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.

45. 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.

(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 Substituting known values: Thus, . image width.

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

(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-kilo-voltage 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.

43. Image contrast is a result of 1. differential tissue absorption 2. atomic number of tissue being traversed 3. proper regulation of milliampere-seconds (A) 1 only (B) 1 and 2 only (C) 1 and 3 only (D) 1, 2, and 3

(B) Radiographic contrast is defined as the degree of difference between adjacent densities. These density differences represent sometimes very subtle differences in the absorbing properties of adjacent body tissues. The radiographic subject, the patient, is composed of many different tissue types that have varying 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 finished radiograph. Normal tissue density may be altered significantly in the presence of pathologic processes. The technical factor used to regulate contrast is kilovoltage. Radiographic contrast is unrelated to milliampere-seconds.

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

(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.

99. A film/screen image exhibiting insufficient density might be attributed to 1. inadequate kilovoltage 2. inadequate SID 3. grid cutoff (A) 1 only (B) 1 and 2 only (C) 1 and 3 only (D) 1, 2, and 3

(C) As kilovoltage is reduced, the number of high-energy photons produced at the target is reduced; therefore, a decrease in radiographic density 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 radiographic density. If the SID is inadequate (too short), an increase in radiographic density will occur.

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

(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.

156. Characteristics of high-ratio focused grids, compared with lower-ratio grids, include which of the following? 1. They allow more positioning latitude. 2. They are more efficient in collecting SR. 3. 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

(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.

143. An increase in kilovoltage will serve to (A) produce a longer scale of contrast (B) produce a shorter scale of contrast (C) decrease the radiographic density (D) decrease the production of scattered radiation

(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.

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

(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.

6. SID affects recorded detail in which of the following ways? (A) Recorded detail is directly related to SID. (B) Recorded detail is inversely related to SID. (C) As SID increases, recorded detail decreases. (D) SID is not a detail factor.

(A) As the distance from focal spot to IR (SID) increases, so does recorded detail. Because the part is being exposed by more perpendicular (less divergent) rays, less magnification and blur are produced. Although the best recorded detail 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.

210. As grid ratio is decreased, (A) the scale of contrast becomes longer (B) the scale of contrast becomes shorter (C) radiographic density decreases (D) radiographic distortion decreases

(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. Radiographic density would increase with a decrease in grid ratio. Grid ratio is unrelated to distortion.

77. A 15% decrease in kilovoltage accompanied by a 50% increase in milliampere-seconds will result in a(n) (A) shorter scale of contrast (B) increase in exposure latitude (C) increase in radiographic density (D) decrease in recorded detail

(A) A 15% decrease in kilovoltage with a 50% increase in milliampere-seconds produces an image similar to the original but with some obvious differences. The overall blackness (radiographic density) is doubled because of the increase in milliampere-seconds. This increase in blackness is compensated for by the loss of gray shades (i.e., shorter scale contrast) from the decreased kilovoltage. The decrease in kilovoltage also decreases exposure latitude; there is less margin for error in lower-kilovoltage ranges. Recorded detail is unaffected by changes in kilovoltage.

101. 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) Added filtration

(A) A compensating filter is used when the part to be radiographed is of uneven thickness or 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 primary radiation that would expose the low-tissue-density area while allowing the primary radiation to pass unaffected to 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. Added filtration addresses patient protection, decreasing patient dose.

23. 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

(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.

164. A decrease in kilovoltage will result in (A) a decrease in optical density (B) a decrease in contrast (C) a decrease in recorded detail (D) a decrease in image resolution

(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 radiographic density are directly related, they are not directly proportional; that is, twice the radiographic density does not result from doubling the kilovoltage. With respect to the effect of kilovoltage on image density, if it is desired to double the radiographic density yet impossible to adjust the milliampere-seconds, a similar effect can be achieved by increasing the kilovoltage by 15%. Conversely, the density 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 density. Additionally, a decrease in kilovoltage will produce fewer shades of gray, that is, a shorter-scale, or higher/increased, contrast. Kilovoltage is unrelated to recorded detail and resolution.

144. The functions of automatic beam limitation devices include 1. reducing the production of scattered radiation 2. increasing the absorption of scattered radiation 3. changing the quality of the x-ray beam (A) 1 only (B) 2 only (C) 1 and 2 only (D) 1, 2, and 3

(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.

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

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

206. Which interaction is responsible for producing the most x-ray photons at the x-ray tube target? (A) Bremsstrahlung (B) Characteristic (C) Photoelectric (D) Compton

(A) Diagnostic x-rays are produced within the x-ray tube when high-speed electrons are rapidly decelerated upon encountering the tungsten atoms of the anode/target. The source of electrons is the heated cathode filament; they are driven across to the anode focal spot when thousands of volts (kV) are applied. When the high-speed electrons are suddenly stopped at the focal spot, their kinetic energy is converted to x-ray photon energy. This happens in two ways: Bremsstrahlung ("Brems") or "braking" radiation: A high-speed electron, passing near or through a tungsten atom, is attracted and "braked" (i.e., slowed down) by the positively charged nucleus and deflected from its course with a loss of energy. This energy loss is given up in the form of an x-ray photon. The electron might not give up all its kinetic energy in one interaction; it can go on to have several more interactions deeper in the anode, each time producing an x-ray photon having less and less energy. This is one reason the x-ray beam is polyenergetic, that is, has a spectrum of energies. Brems radiation comprises 70-90% of the x-ray beam. Characteristic radiation: In this case, a high-speed electron encounters a tungsten atom within the anode and ejects a K shell electron, leaving a vacancy in that shell. An electron from the adjacent L shell moves to the K shell to fill its vacancy and in doing so emits a K characteristic ray. The energy of the characteristic ray is equal to the difference in energy between the K and L shell energy levels. Characteristic radiation comprises 10-30% of the x-ray beam. Photoelectric effect and Compton scatter are interactions that occur between x-ray photons and matter.

96. Exposure rate will decrease with an increase in 1. SID 2. kilovoltage 3. focal spot size (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(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 highenergy x-ray photons produced at the anode.

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

(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 density. Over- or under-exposure under the anode is usually the result of exceeding the focusing distance limits in addition to being off-center.

154. 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

(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.

100. 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 density (B) Increased density (C) Scattered radiation fog (D) Motion blur

(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 radiograph. The patient is thin, and the lateral cells have no tissue superimposed on them. Therefore, as soon as the lateral photocells detect radiation (which will be immediately), the exposure will be terminated, giving the lateral chest insufficient exposure.

190. To produce a just perceptible increase in radiographic density, the radiographer must increase the (A) mAs by 30% (B) mAs by 15% (C) kV by 15% (D) kV by 30%

(A) If a radiograph lacks sufficient blackening, 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.

84. If a radiograph exposed using a 12:1 ratio grid exhibits increased brightness/loss of density 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

(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 primary 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 density.

49. Causes of grid cutoff, when using focused reciprocating grids, include the following? 1. Inadequate SID 2. X-ray tube off-center with the long axis of the lead strips 3. Angling the beam in the direction of the lead strips (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(A) If the SID is above or below the recommended focusing distance, the primary beam will not coincide with the angled lead strips at their lateral edges. Consequently, there will be absorption of the primary beam termed grid cutoff. If the central ray is off-center longitudinally, there will be no ill effects. If the central ray is off-center side to side, the lead strips are no longer parallel with the divergent x-ray beam, and there will be loss of density owing to grid cutoff. Central ray angulation in the direction of the lead strips is appropriate and will not cause grid cutoff. Central ray angulation against the direction of the lead strips will cause grid cutoff.

124. 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

(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 0.35 mm: The FOV and matrix size are independent of one another; that is, either can be changed, and the other will remain unaffected. However, pixel size is affected by changes in either the FOV or matrix size. For example, if the matrix size is increased, pixel size decreases. If FOV increases, pixel size increases. Pixel size is inversely related to resolution. As pixel size increases, resolution decreases.

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

(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.

13. Decreasing field size from 14 × 17 in. to 8 × 10 in., with no other changes, will (A) decrease the amount of scattered radiation generated within the part (B) increase the amount of scattered radiation generated within the part (C) increase x-ray penetration of the part (D) decrease x-ray penetration of the part

(A) Limiting the size of the radiographic field (irradiated area) serves to limit the amount of scattered radiation produced within the anatomic part. Therefore, as field size decreases, scattered radiation production decreases, and image quality increases. Limiting the size of the radiographic field is a very effective means of reducing the quantity of non-information-carrying scattered radiation (fog) produced, resulting in improved detail visibility. Limiting the size of the radiographic field is also the most effective means of patient radiation protection.

179. Which of the following is (are) characteristic(s) of a 5:1 grid? 1. It allows some positioning latitude. 2. It is used with high-kilovoltage exposures. 3. 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

(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.

59. The main difference between direct capture and indirect capture DR is that (A) direct capture/conversion has no scintillator (B) direct capture/conversion uses a photo-stimulable phosphor (C) in direct capture/conversion, light is detected by CCDs (D) in direct capture/conversion, light is detected by TFTs

(A) One type of indirect-capture flat-panel detector uses cesium iodide or gadolinium oxysulfide as the scintillator, that is, which captures x-ray photons and emits light. That light is then transferred via a photodetector coupling agent—a CCD or TFT. In direct-capture flat-panel detector systems, x-ray energy is converted to an electrical signal in a single layer of material such as the semiconductor aSe. Electric charges are applied to both surfaces of the a-Se, electron-hole pairs are created, and charges are read by TFT arrays located on the surfaces. The electrical signal is transferred directly to the ADC. The number of TFTs is equal to the number of image pixels. Thus, the direct-capture system eliminates the scintillator step required in indirect DR. Since selenium has a relatively low Z number (compared with gadolinium [Z 64] or cesium [Z 55]), a-Se detectors are made thicker to improve detection, thus compensating for the low x-ray absorption of selenium. There is no diffusion of electrons, so spatial resolution is not affected in this manner.

38. 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)? 1. Use of close collimation 2. Use of low mAs 3. Use of a low-ratio grid (A) 1 only (B) 1 and 2 only (C) 1 and 3 only (D) 1, 2, and 3

(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.

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

(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.

70. All the following are related to recorded detail except (A) milliamperage (B) focal-spot size (C) SID (D) OID

(A) The focal-spot size selected will determine the amount of focal-spot, or geometric, blur produced in the image. OID and SID are responsible for image magnification and hence recorded detail. The milliamperage is unrelated to recorded detail; it affects only the quantity of x-ray photons produced and thus the radiographic density.

215. The line-focus principle expresses the relationship between (A) the actual and the effective focal spot (B) exposure given the IR and resulting density (C) SID used and resulting density (D) grid ratio and lines per inch

(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 recorded detail by reducing blur. The relationship between the exposure given the IR and the resulting density is expressed in the reciprocity law; the relationship between the SID and resulting density is expressed by the inverse-square law. Grid ratio and lines per inch are unrelated to the line-focus principle.

235. If 40 mAs and a 200-speed screen-film system were used for a particular exposure, what new milliampere-seconds value would be required to produce the same density if the screen-film system were changed to 800 speed? (A) 10 (B) 20 (C) 80 (D) 160

(A) The screen-film system and radiographic density are directly proportional; that is, if the system speed is doubled, the radiographic density is doubled. In this case, we started at 40 mAs with a 200-speed system. If the system speed is doubled to 400, we should decrease the milliampere-seconds to 20 mAs. If the speed is again doubled to 800, we use half the 20 mAs, or 10 mAs. Or milliampere-seconds conversion factors and the following formula may be used: Thus, mAs with a 800-speed screen-film system.

110. Which of the following would be useful for an examination of a patient suffering from Parkinson's disease? 1. Short exposure time 2. Decreased SID 3. Compensating filtration (A) 1 only (B) 1 and 2 only (C) 1 and 3 only (D) 1, 2, and 3

(A) The shortest possible exposure should be used as a matter of routine. Parkinson disease is characterized by uncontrollable tremors, and the resulting unsharpness can destroy image resolution/detail. A short exposure time is essential. That can be achieved by increasing the mA or the kV—AEC will react with a shorted exposure time. SID and compensating filtration are unrelated to the problem and are not indicated here.

146. Of the following groups of exposure factors, which will produce the shortest scale of radiographic 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

(A) The single most important factor regulating radiographic contrast 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 density, 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.

25. Using fixed milliampere-seconds and variable kilovoltage technical factors, each centimeter increase in patient thickness requires what adjustment in kilovoltage? (A) Increase 2 kV (B) Decrease 2 kV (C) Increase 4 kV (D) Decrease 4 kV

(A) When the variable-kilovoltage method is used, a particular milliampere-seconds value is assigned to each body part. As part thickness increases, the kilovoltage (i.e., penetration) is also increased. The body part being radiographed must be measured carefully, and for each centimeter of increase in thickness, 2 kV is added to the exposure.

55. The differences between CR and DR include 1. CR uses IPs. 2. CR has higher DQE and lower patient dose. 3. CR images are displayed immediately. (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(A) While CR utilizes traditional x-ray tables and IPs to enclose and protect the flexible PSP screen, DR requires the use of significantly different equipment. DR does not use IPs or a traditional x-ray table—it is a direct-capture/conversion, or indirect capture/conversion, system of x-ray imaging. Besides eliminating IPs and their handling, DR affords the advantage of immediate display of the image (compared with CR's slightly delayed image display), and DR exposures can be lower because of the detector's higher DQE (i.e., ability to perceive and interact with x-ray photons). DR, like CR, also offers the advantage of image preview and postprocessing.

203. 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

(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.

134. A focal-spot size of 0.3 mm or smaller is essential for (A) small-bone radiography (B) magnification radiography (C) tomography (D) fluoroscopy

(B) A fractional focal spot of 0.3 mm or smaller is essential for reproducing fine detail 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.

178. An exposure was made at a 36-in. SID using 300 mA, a 30-ms exposure, and 80 kV and an 8:1 grid. It is desired to repeat the radiograph using a 40-in. SID and 70 kV. With all other factors remaining constant, what new exposure time will be required? (A) 0.03 second (B) 0.07 second (C) 0.14 second (D) 0.36 second

(B) A review of the problem reveals that three changes are being made: an increase in SID, a decrease in kV, and a change in exposure time (to be considered last). The original mAs value was 9. The decrease in kV requires us to double the mAs in order to maintain sufficient exposure. Now, we must deal with the distance change. Using the density-maintenance formula (and remembering that 18 is now the old milliampere-seconds value), we find that the required new milliampere-seconds value at 42 in. is 22. Thus, mAs at 40-in. SID. Because milliamperage is unchanged, we must determine the exposure time that, when used with 300 mA, will yield 22 mAs.

236. 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 density? (A) 2 (B) 4 (C) 16 (D) 32

(B) According to the 15% rule, if the kilovoltage is increased by 15%, radiographic density will be doubled. Therefore, to compensate for this change and to maintain radiographic density, the milliampere-seconds value should be reduced to 4 mAs.

71. 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 density? (A) 5.6 (B) 6.5 (C) 9.7 (D) 13

(B) According to the density-maintenance formula, if the SID is changed to 42 in., 16.33 mAs is required to maintain the original radiographic density: Thus, 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 an image density similar to that of the original radiograph. The following are the factors used for milliampere-seconds conversion from non-grid to grid:

93. 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 shorter SID should be used. (D) A lower-ratio grid should be used.

(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 and grid ratio are unrelated to magnification.

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

(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.

109. The intensity of ionizing radiation decreases as (A) distance from the source of radiation decreases (B) distance from the source of radiation increases (C) frequency increases (D) wavelength decreases

(B) As distance from a light source increases, the light diverges and covers a larger area; the quantity of light available per unit area becomes less and less as distance increases. The intensity (quantity) of light decreases according to the inverse square law, that is, the intensity of light at a particular distance from its source is inversely proportional to the square of the distance. For example, if you decreased the distance between a book you were reading and your illuminating lamp from 6 to 3 feet, you would have four times as much light available. Similarly, SID has a significant impact on x-ray beam intensity. As the distance between the x-ray tube and IR increases, exposure rate decreases according to the inverse square law.

140. OID is related to recorded detail in which of the following ways? (A) Radiographic detail is directly related to OID. (B) Radiographic detail is inversely related to OID. (C) As OID increases, so does radiographic detail. (D) OID is unrelated to radiographic detail.

(B) As the distance from the object to the IR (OID) increases, so does magnification distortion, thereby decreasing recorded detail. 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.

224. Exposed silver halide crystals are changed to black metallic silver by the (A) preservative (B) reducers (C) activators (D) hardener

(B) As the film emulsion is exposed to light or x-rays, latent image formation takes place. The exposed silver halide crystals are reduced to black metallic silver by the developer/reducing agents in the automatic processor's developer solution. The preservative helps to prevent oxidation of the developer solution. The activator provides the necessary alkalinity for the developer solution, and hardener is added to the developer in automatic processing to keep emulsion swelling to a minimum.

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

(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.

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

(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.

89. Which of the following can impact the visibility of the anode heel effect? 1. SID 2. IR size 3. Focal spot size (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(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.

26. Unopened boxes of radiographic film should be stored away from radiation and (A) in the horizontal position (B) in the vertical position (C) stacked with the oldest on top (D) stacked with the newest on top

(B) Boxes of x-ray film, especially the larger sizes, should be stored in the vertical (upright) position. If film boxes are stacked on one another, the sensitive emulsion can be affected by pressure from the boxes above. Pressure marks are produced and result in loss of contrast in that area of the radiographic image. When retrieving x-ray film from storage, the oldest should be used first.

21. Which of the following is/are true when comparing film-screen imaging to CR imaging? 1. CR DQE is better than film-screen DQE. 2. CR has a wider exposure range than film-screen. 3. CR has better spatial resolution than film-screen. (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(B) CR systems convert x-ray photons into useful information much more efficiently than film-screen systems, hence a far better DQE. CR also converts that information over a far wider exposure range (about 104 times wider) than screen film. The single negative aspect of CR is its limited spatial resolution (image detail). While film-screen systems resolve about 10-15 lp/mm, CR resolution is about 3-5 lp/mm.

117. Bone densitometry is often performed to 1. measure degree of bone (de) mineralization 2. evaluate the results of osteoporosis treatment/therapy 3. evaluate the 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

(B) DXA imaging is used to evaluate bone mineral density (BMD). Bone densitometry (i.e., 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, their attenuation is calculated to represent the degree of bone density. Soft tissue attenuation information is not used to measure bone density.

98. Diagnostic x-rays are generally associated with (A) high frequency and long wavelength (B) high frequency and short wavelength (C) low frequency and long wavelength (D) low frequency and short wavelength

(B) Electromagnetic radiation can be described as wave-like fluctuations of electric and magnetic fields. There are many kinds of electromagnetic radiation; visible light, microwaves, and radio waves, as well as x-ray and gamma rays, are all part of the electromagnetic spectrum. All the electromagnetic radiations have the same velocity, that is, 3 × 108 m/s (1,86,000 miles per second); however, they differ greatly in wavelength and frequency. Wavelength refers to the distance between two consecutive wave crests. Frequency refers to the number of cycles per second; its unit of measurement is hertz (Hz), which is equal to 1 cycle per second. Frequency and wavelength are closely associated with the relative energy of electromagnetic radiations. More energetic radiations have shorter wavelength and higher frequency. The relationship among frequency, wavelength, and energy is graphically illustrated in the electromagnetic spectrum. Some radiations are energetic enough to rearrange atoms in materials through which they pass, and they can therefore be hazardous to living tissue. These radiations are called ionizing radiation because they have the energetic potential to break apart electrically neutral atoms, resulting in the production of negative and/or positive ions.

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

(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.

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

(B) First, evaluate the change(s): The kilovoltage was increased by 15% . A 15% increase in kilovoltage will double the radiographic density; therefore, it is necessary to use half the original milliampere-seconds value to maintain the original density. The original milliampere-seconds value was 28 mAs (300 mA × 0.07 second [70 ms] 28 mAs), so we now need 14 mAs, using 500 mA. Because :

220. 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

(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.

80. 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

(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.

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

(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.

150. Of the following groups of technical factors, which will produce the greatest radiographic density? (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

(B) If (A) and (B) are reduced to 5 mAs for consistency, the kilovoltage will increase to 85 kV in both cases, thereby balancing radiographic densities. Thus, the greatest density is determined by the shortest SID (greatest exposure rate). (Shephard, pp. 306-307) In electronic imaging (CR, DR), the term brightness is used instead of density. A film image with little density would be described as a monitor digital image as having greater brightness.

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

(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, contrast scale/contrast resolution. 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.

181. An 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 1. more contrast 2. higher contrast 3. more density (A) 1 only (B) 1 and 2 only (C) 3 only (D) 2 and 3 only

(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 higher/greater will be the contrast and the less is the overall density (greater brightness) of the resulting image.

115. HVL is affected by the amount of 1. kVp 2. beam filtration 3. tissue density (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(B) Manufacturers of x-ray equipment must follow guidelines that state maximum x-ray output at specific distances, total quantities of filtration, positive beam limitation, and other guidelines. Radiographers must practice safe principles of operation; preventive maintenance and quality control (QC) checks must be performed at specific intervals to ensure continued safe equipment performance. Radiologic QC involves monitoring and regulating the variables associated with image production and patient care. HVL testing provides beam quality information that is different from that obtained from kV testing. HVL is defined as the thickness of any absorber that will reduce x-ray beam intensity to one-half its original value. It is determined by measuring the beam intensity without an absorber and then recording the intensity as successive millimeters of aluminum are added to the radiation field. It is influenced by the type of rectification, total filtration, and kV. An x-ray tube HVL should remain almost constant. If HVL decreases, it is an indication of a decrease in the actual kV. If the HVL increases, it indicates the deposition of vaporized tungsten on the inner surface of the glass envelope (as a result of tube aging) or an increase in the actual kV.

200. HVL is affected by the amount of 1. kVp 2. beam filtration 3. tissue density (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

202. If, upon QC testing, the HVL of the x-ray beam produced by a particular x-ray tube increases, it is an indication of 1. vaporized tungsten deposited on the inner surface of the glass envelope 2. an increase in the kilovoltage being produced by the tube 3. a decrease in the kilovoltage being produced by the tube (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

73. Which of the following groups of exposure factors will produce the shortest scale of contrast? (A) 200 mA, 0.08 second, 95 kV, 12:1 grid (B) 500 mA, 0.03 second, 70 kV, 8:1 grid (C) 300 mA, 0.05 second, 95 kV, 8:1 grid (D) 600 mA, 1/40 seconds, 70 kV, 6:1 grid

(B) Of the given factors, kilovoltage and grid ratio will have a significant effect on radiographic contrast. Remember that the milliampere-seconds value has no effect on contrast scale. Because a combination of lower kilovoltage and a higher-ratio grid will allow the least amount of scattered radiation to reach the IR, thereby producing fewer gray tones, (B) is the best answer. (D) also uses low kilovoltage, but the grid ratio is lower—allowing more scatter to reach the IR (therefore, longer-scale contrast).

97. Which of the following pathologic conditions probably will require a decrease in exposure factors? (A) Osteomyelitis (B) Osteoporosis (C) Osteosclerosis (D) Osteochondritis

(B) Osteoporosis is a condition, often seen in the elderly, marked by increased porosity and softening of bone. The bones are much less dense, and thus a decrease in exposure is required. Osteomyelitis and osteochondritis are inflammatory conditions that usually have no effect on bone density. Osteosclerosis is abnormal hardening of the bone, and an increase in exposure factors would be required.

37. Pathologic or abnormal conditions that would require an increase in exposure factors include all of the following except (A) atelectasis (B) pneumoperitoneum (C) Paget disease (D) congestive heart failure

(B) Pathologic processes and abnormal conditions that alter tissue composition or thickness can have a significant effect on image density. The radiographer must be aware of these variants and processes to make an appropriate and accurate adjustment of technical factors. Examples of additive pathologic conditions: • Ascites • Rheumatoid arthritis • Paget disease • Pneumonia • Atelectasis • Congestive heart failure • Edematous tissue Examples of destructive pathologic conditions: • Osteoporosis • Osteomalacia • Pneumoperitoneum • Emphysema • Degenerative arthritis • Atrophic and necrotic conditions

166. What feature is used to display RIS information on current patients? (A) HIS (B) Modality work list (C) PACS (D) DICOM

(B) Patient demographic and examination information originates from the hospital/facility HIS, where it is obtained when the patient is initially registered. That information is available or retrievable when the patient is scheduled, or arrives, for imaging services. Typical patient information includes name, DOB or age, sex, ID number, accession number, examination being performed, date and time of examination. Additional information may be available on the examination requisition; more information is usually entered by the technologist at the time of the examination. A feature that is useful in sorting examinations and decreasing (but not eliminating) errors is the Modality Work List (MWL). The MWL "brings up" existing RIS information, that is, the examinations scheduled for each imaging area—for example, x-ray, CT, MR, mammography, ultrasound, etc. The technologist selects the correct patient, which includes that patient's particular demographics, from the particular modality work list. It is essential that the technologist is attentive to detail and accuracy when entering patient information; errors in patient demographics entry, and entry duplication, must be avoided. PACS is used by health care facilities to economically store, archive, exchange, transmit digital images from multiple imaging modalities; it replaces the need to manually file, retrieve, and transport film and film jackets. RIS and HIS can be integrated with PACS for electronic health information storage. The purpose of HIS is to manage health care information and documents electronically, and to ensure data security and availability. RIS is a system for tracking radiological and imaging procedures. RIS is used for patient registration and scheduling, radiology workflow management, reporting and printout, manipulation and distribution and tracking of patient data, and billing. RIS complements HIS and is critical to competent workflow to radiologic facilities.

132. 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

(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.

208. Which of the following is (are) classified as rare earth phosphors? 1. Lanthanum oxybromide 2. Gadolinium oxysulfide 3. Cesium iodide (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(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.

177. Beam attenuation characteristics, or density values, in CT are expressed as 1. Hounsfield units 2. CT numbers 3. heat units (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(B) Sir Godfrey Newbold Hounsfield created the first CT unit, describing the reconstruction of data taken from multiple projection angles. Alan MacLeod Cormack worked with the complex mathematical algorithms required for image reconstruction. Their first commercial CT head scanner was available in 1971. In 1979, Hounsfield and Cormack shared the Nobel Prize in Medicine for their historic work with this new imaging science. To express the beam attenuation characteristics of various tissues, the Hounsfield unit (HU) is used. HUs can also be referred to as CT numbers or density values. Godfrey Hounsfield assigned a value of 0 to distilled water, a value of +1,000 to dense osseous tissue, and a value of -1,000 to air. There is a direct relationship between the HU and tissue attenuation coefficient. The greater the attenuation coefficient of the particular tissue, the higher the HU value. One HU represents a 0.1% difference between the particular tissue attenuation characteristics and that of distilled water. HU value accuracy can be affected by equipment calibration, volume averaging, and image artifacts.

242. What is the relationship between tissue attenuation coefficient in CT and its related Hounsfield unit (HU)? (A) The greater the tissue attenuation coefficient, the lower the HU value. (B) The greater the tissue attenuation coefficient, the higher the HU value. (C) Tissue attenuation coefficient and HU value are unrelated. (D) Tissue attenuation coefficient and HU value are identical.

173. 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

(B) The abdomen is a thick structure that contains many structures of similar 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 radiographed without a grid. The general rule is that structures measuring more than 10 cm should be radiographed with a grid.

8. 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

(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.

120. The process of "windowing" of digital images determines the image (A) spatial resolution (B) contrast (C) pixel size (D) matrix size

(B) The digital images' scale of contrast, or contrast resolution, can be changed electronically through leveling and windowing of the image. The level control determines the central or middensity of the scale of contrast, whereas the window control determines the total number of densities/grays (to the right and left of the central/middensity). Matrix and pixel sizes are related to (spatial) resolution of digital images.

116. In a posteroanterior (PA) projection of the chest being used for cardiac evaluation, the heart measures 14.7 cm between its widest points. If the magnification factor is known to be 1.2, what is the actual diameter of the heart? (A) 10.4 cm (B) 12.25 cm (C) 13.5 cm (D) 17.64 cm

(B) The formula for magnification factor (MF) = image size/object size. In the stated problem, the anatomic measurement is 14.7 cm, and the magnification factor is known to be 1.2. Substituting the known factors in the appropriate equation: (actual anatomic size).

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

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

82. Image plate front material can be made of which of the following? 1. Carbon fiber 2. Magnesium 3. Lead (A) 1 only (B) 1 and 2 only (C) 1 and 3 only (D) 1, 2, and 3

(B) The image plate 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.

148. Magnification fluoroscopy is accomplished by 1. moving the image intensifier focal point further from the output phosphor 2. selecting a smaller portion of the input phosphor 3. decreasing the voltage to the electrostatic lenses (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(B) The input phosphor of image intensifiers is usually made of cesium iodide. For each x-ray photon absorbed by cesium iodide, approximately 5,000 light photons are emitted. As the light photons strike a photoemissive photocathode, a number of electrons are released from the photocathode and focused toward the output side of the image tube by voltage applied to the negatively charged electrostatic focusing lenses. The electrons are then accelerated through the neck of the tube where they strike the small (0.5-1 in.) output phosphor that is mounted on a flat glass support. The entire assembly is enclosed within a 2-4-mm thick vacuum glass envelope. Remember that the image on the output phosphor is minified, brighter, and inverted (electron focusing causes image inversion). Input screen diameters of 5-12 in. are available. Although smaller diameter input screens improve resolution, they do not permit a large FOV, that is, viewing of large patient areas. Dual- and triple-field image intensifiers are available that permit magnified viewing of fluoroscopic images. To achieve magnification, the voltage to the focusing lenses is increased and a smaller portion of the input phosphor is used, thereby resulting in a smaller FOV. Because minification gain is now decreased, the image is not as bright. The mA is automatically increased to compensate for the loss in brightness when the image intensifier is switched to magnification mode. Entrance skin exposure (ESE) can increase dramatically as the FOV decreases (i.e., as magnification increases). As FOV decreases, magnification of the output screen image increases, there is less noise because increased mA provides a greater number of x-ray photons, and contrast and resolution improve. The focal point in the magnification mode is further away from the output phosphor (as a result of increased voltage applied to the focusing lenses) and therefore the output image is magnified.

162. Which of the following is/are associated with magnification fluoroscopy? 1. Increased mA 2. Smaller portion of the input phosphor is used. 3. Image intensifier focal point moves closer to the output phosphor. (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

191. 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

(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: 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: Thus, -s exposure = 37.5 ms.

81. How are mAs and patient dose related? (A) mAs and patient dose are inversely proportional. (B) mAs and patient dose are directly proportional. (C) mAs and patient dose are unrelated. (D) mAs and patient dose are inversely related.

(B) The milliampere-seconds value regulates the number of x-ray photons produced at the target and thus regulates patient dose. If the milliampere-seconds is doubled, dose is doubled; therefore, mAs and patient dose are directly proportional.

163. A lateral projection 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. Radiographic density 2. Scale of radiographic contrast 3. Distortion (A) 1 only (B) 2 only (C) 2 and 3 only (D) 1, 2, and 3

(B) The original milliampere-seconds value (regulating radiographic density) was 200. The original kilovoltage (regulating radiographic contrast) was 90. The milliampere-seconds value was cut in half, to 100, causing a decrease in density. The kilovoltage was increased (by 15%) to compensate for the density loss and thereby increase the scale of contrast.

113. An anteroposterior (AP) projection of the femur was made using 300 mA, 0.03 second, 76 kV, 40-in. SID, 1.2-mm focal spot, and a 400-speed film-screen system. With all other factors remaining constant, which of the following exposure times would be required to maintain radiographic density/brightness at a 44-in. SID using 500 mA? (A) 12 ms (B) 22 ms (C) 30 ms (D) 36 ms

(B) The original milliampere-seconds value was 9 (300 mA × 0.03 second). Using the density-maintenance formula, the new milliampere-seconds value must be determined for the distance change from 40 to 44 in. of the SID: Thus, (11) mAs at 44-in. SID. Then, if 500 is the new milliamperage, we must determine what exposure time is required to achieve 8.1 mAs: Thus, second (22 ms) at 500 mA and 44-in. SID.

153. The processing algorithm represents the (A) pixel value distribution (B) anatomical part and projection (C) image grayscale (D) screen speed

(B) The radiographer selects a processing algorithm by selecting the anatomic part and particular projection on the computer/control panel. The CR unit then matches that information with a particular lookup table (LUT)—a characteristic curve that best matches the anatomic part being imaged. The observer is able to review the image and, if desired, change its appearance (through "windowing"); doing so changes the LUT. Hence, histogram analysis and use of the appropriate LUT together function to produce predictable image quality in CR.

60. 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

(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.

130. 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

(B) To calculate milliampere-seconds, multiply milliamperage times exposure time. In this case, 400 mA × 0.017 second . Careful attention to proper decimal placement will help to avoid basic math errors.

51. 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

(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.

118. The differences between CR and DR include 1. DR images are displayed immediately. 2. DR has higher DQE and lower patient dose. 3. DR uses IPs. (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(B) While CR utilizes traditional x-ray tables and IPs to enclose and protect the flexible PSP screen, DR requires the use of significantly different equipment. DR does not use IPs or a traditional x-ray table—it is a direct-capture/conversion, or indirect-capture/conversion, system of x-ray imaging. Besides eliminating IPs and their handling, DR affords the advantage of immediate display of the image (compared with CR's slightly delayed image display), and DR exposures can be lower because of the detector's higher DQE (i.e., ability to perceive and interact with x-ray photons). DR, like CR, also offers the advantage of image preview and postprocessing.

211. X-ray film emulsion is most sensitive to safelight fog (A) before exposure and development (B) after exposure (C) during development (D) at low humidity

(B) X-ray film emulsion becomes more sensitive to safelight fog following exposure to fluorescent light from intensifying screens. Care must be taken not to leave exposed film on the darkroom workbench for any length of time because its sensitivity to safelight fog is now greatly heightened.

204. Boxes of film stored in too warm an area may be subject to (A) static marks (B) film fog (C) high contrast (D) loss of density

(B) X-ray film emulsion is sensitive and requires proper handling and storage. It should be stored in a cool (40-60°F), dry (40-60% humidity) place. Exposure to excessive temperatures or humidity can lead to film fog and loss of contrast. Static marks are a result of low humidity.

105. The function(s) of the developer in film processing is (are) to 1. remove the unexposed silver bromide crystals 2. change the exposed silver bromide crystals to black metallic silver 3. harden the emulsion (A) 1 only (B) 2 only (C) 1 and 3 only (D) 2 and 3 only

(B) X-ray film processing consists of four parts: development, fixing, wash, and dry. Developing agents change the exposed silver bromide crystals (latent image) to black metallic silver, thus producing a manifest image. The fixer solution removes the unexposed silver bromide crystals from the emulsion and hardens the gelatin emulsion, thus ensuring permanence of the radiograph.

114. An AP radiograph of the femur was made using 300 mA, 30 ms, 76 kV, 40-in. SID, and 1.2-mm focal spot. With all other factors remaining constant, which of the following exposure times would be required to maintain brightness/density using 87 kV and the addition of a 12:1 grid? (A) 38 ms (B) 60 ms (C) 75 ms (D) 150 ms

(C) A 15% increase in kilovoltage was made, increasing the kilovoltage to 87 kV. Because the kilovoltage change effectively doubles the density, the milliampere-seconds value now must be cut in half (from 9 to 4.5 mAs) to compensate. Grids are used to absorb scattered radiation from the remnant beam before it can contribute to the x-ray image. Because the grid removes scattered radiation (and some useful photons as well) from the beam, an increase in exposure factors is required. The amount of increase depends on the grid ratio; the higher the grid ratio, the higher is the correction factor. The correction factor for a 12:1 grid is 5; therefore, the milliampere-seconds value (4.5) is multiplied by 5 to arrive at the new required milliampere-seconds value (22.5). Using the milliampere-seconds equation , it is determined that 0.15 second will be required at 300 mA:

33. Subject/object unsharpness can result from all of the following except when (A) object shape does not coincide with the shape of x-ray beam (B) object plane is not parallel with x-ray tube and/or IR (C) anatomic object(s) of interest is/are in the path of the CR (D) anatomic object(s) of interest is/are a distance from the IR

(C) A certain amount of object unsharpness is an inherent part of every radiographic image because of the position and shape of anatomic structures within the body. Structures within the three-dimensional human body lie in different planes. Additionally, the three-dimensional shape of solid anatomic structures rarely coincides with the shape of the divergent beam. Consequently, some structures are imaged with more inherent distortion than others, and shapes of anatomic structures can be entirely misrepresented. Structures farther from the IR will be distorted (i.e., magnified) more than those closer to the IR; structures closer to the x-ray source will be distorted (i.e., magnified) more than those farther from the x-ray source. For the shape of anatomic structures to be accurately recorded, the structures must be parallel to the x-ray tube and the IR, and aligned with the central ray (CR). The shape of anatomic structures lying at an angle within the body or placed away from the CR will be misrepresented on the IR. There are two types of shape distortion. If a linear structure is angled within the body, that is, not parallel with the long axis of the part/body and not parallel to the IR, that anatomic structure will appear smaller—it will be foreshortened. On the other hand, elongation occurs when the x-ray tube is angled. Image 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 part of greatest interest. Unless the edges of a three-dimensional object conform to the shape of the x-ray beam, blur or unsharpness will occur at the partially attenuating edge of the object. This can be accompanied by changes in radiographic/image density, according to the thickness of areas traversed by the x-ray beam.

247. Which of the following is (are) methods used for x-ray film silver reclamation? 1. Photoelectric method 2. Metallic replacement method 3. Electrolytic method (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(C) About half the silver in a film emulsion remains to form the image. The other half is removed from the film during the fixing process. Therefore, fixer solution has a high silver content. Silver is a toxic metal and cannot simply be disposed of into the public sewer system. Since silver is also a precious metal, it becomes financially wise to recycle the silver removed from x-ray film. The three most commonly used silver recovery systems are the electrolytic, metallic replacement, and chemical precipitation methods. In electrolytic units, an electric current is passed through the fixer solution. Silver ions are attracted to and become plated onto the negative electrode of the unit. The plated silver is periodically scraped from the cathode and accurately measured so that the hospital can be reimbursed appropriately. The electrolytic method is a practical recovery system for moderate-and high-use processors. The metallic replacement (or displacement) method of silver recovery uses a steel mesh/steel wool type of cartridge that traps silver as fixer is run through it. This system is useful for low-volume processors and is often also used as a backup to the electrolytic unit. Chemical precipitation adds chemicals that release electrons into the fixer solution. This causes the metallic silver to precipitate out, fall to the bottom of the tank, and form a recoverable sludge. This method is used principally by commercial silver dealers.

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

(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.

170. 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

(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 density-maintenance formula is used to determine new milliampere-seconds values when changing distance: Thus, mAs at 42-in. SID. Then, to determine the new exposure time (), Thus, second (73 ms) at 400 mA.

155. 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

(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.

103. 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

(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.

112. Potential digital image postprocessing tasks include 1. PACS 2. annotation 3. inversion/reversal (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(C) Digital image postprocessing provides the opportunity for image optimization. Image annotation permits placement of labels, arrow indicators, etc. Windowing allows adjustment of image contrast d/or brightness to diagnostic requirements. Contrast scale enhancement is the most valuable tool in digital imaging. Image minification, with larger matrix sizes, enables us to see tiny anatomic details and improve spatial resolution. Image inversion, or reversal, provides a different perspective by changing white to black and black to white. Image flip also provides another perspective by enabling us to rotate the image. Edge enhancement is useful for small and high-contrast tissues. Other postprocessing functions include highlighting, zoom, pan, and scroll. The pixel shift feature is important in DSA. Image subtraction is used to enhance contrast. If the part moves during acquisition of serial images, misregistration occurs, making the required exact superimposition impossible. Pixel shift is a function that can correct misregistration. Another emerging postprocessing task used in diagnostic functions is determining numeric pixel value for particular ROI (region of interest). This feature has proved useful in bone densitometry, renal calculus recognition, and calcified lung nodule identification.

92. Which of the following is the correct order of radiographic film processing? (A) Developer, wash, fixer, dry (B) Fixer, wash, developer, dry (C) Developer, fixer, wash, dry (D) Fixer, developer, wash, dry

(C) During automatic processing (Figure 4-33), radiographic film is first immersed in the developer solution, which functions to reduce the exposed silver bromide crystals in the film emulsion to black metallic silver (which constitutes the image). Next, the film goes directly into the fixer, which functions to remove the unexposed silver bromide crystals from the emulsion. The film then is transported to the wash tank, where chemicals are removed from the film, and then into the dryer section, where it is dried before leaving the processor.

83. Which of the following pathologic conditions would require a decrease in exposure factors? (A) Congestive heart failure (B) Pneumonia (C) Emphysema (D) Pleural effusion

(C) Emphysema is abnormal distension of the pulmonary alveoli (or tissue spaces) with air. The presence of abnormal amounts of air makes a decrease from normal exposure factors necessary to avoid excessive density. Congestive heart failure, pneumonia, and pleural effusion all involve abnormal amounts of fluid in the chest and, therefore, would require an increase in exposure factors.

234. What information, located on each box of film, is important to note and has a direct relationship to image quality? (A) Number of films in the box (B) Manufacturer's name (C) Expiration date (D) Emulsion lot

(C) Every box of film comes with the expiration date noted. Film used after its expiration date usually will suffer a loss of speed and contrast and will exhibit fog. Film should be ordered in quantities that will ensure that it is used before it becomes outdated, and it should be rotated in storage so that the oldest is used first.

131. All the following statements regarding CR IPs are true except (A) IPs do not contain radiographic film. (B) IPs use no intensifying screens. (C) IPs must exclude all white light. (D) IPs function to protect the PSP.

(C) Externally, IPs (Image Plates) appear very much like traditional film-screen cassettes. However, the main function of an IP is to support and protect the PSP (SPS) that lies within the IP. IPs do not contain intensifying screens or film and, therefore, do not need to be lighttight. The photostimulable PSP is not affected by light.

142. In digital imaging, as DEL size decreases (A) brightness increases (B) brightness decreases (C) spatial resolution increases (D) spatial resolution decreases

(C) Factors influencing recorded detail in digital imaging are very much the same as those factors affecting recorded detail in analog imaging, that is, motion, geometric factors (focal spot size, OID, and SID), and spatial resolution. In analog imaging, spatial resolution is related to the speed of the imaging system (intensifying screen speed) and is significantly better than digital imaging resolution. The spatial resolution of direct digital systems, however, is fixed and is related to the detector element (DEL) size of the thin film transistor (TFT). The smaller the TFT DEL size, the better the spatial resolution. DEL size of 100 microns provides a spatial resolution of about 5 lp/mm (available only in some digital mammography systems). DEL size of 200 microns provides a spatial resolution of about 2.5 lp/mm (general radiography)—lower than that achieved with 400 speed intensifying screen system. A 100 speed intensifying screen system offers a spatial resolution of about 10 lp/mm—significantly greater than, and currently unachievable in, digital imaging. Spatial resolution in digital imaging is fixed, but it is very important that radiographers are alert to the opportunity they have to utilize and control the remaining recorded detail factors (motion and geometric factors).

182. In digital imaging, TFT DEL size is related to (A) contrast (B) brightness (C) spatial resolution (D) plate size

(C) Factors influencing recorded detail in digital imaging are very much the same as those factors affecting recorded detail in analog imaging, that is, motion, geometric factors (focal spot size, OID, and SID), and spatial resolution. The spatial resolution of direct digital systems is fixed and is related to the detector element (DEL) size of the thin film transistor (TFT). The smaller the TFT DEL size, the better the spatial resolution. DEL size of 100 microns provides a spatial resolution of about 5 lp/mm (available only in some digital mammography systems). DEL size of 200 microns provides a spatial resolution of about 2.5 lp/mm (general radiography)—lower than that achieved with 400 speed intensifying screen system. A 100 speed intensifying screen system offers a spatial resolution of about 10 lp/mm—significantly greater than, and currently unachievable in, digital imaging. Spatial resolution in digital imaging is fixed, but it is very important that radiographers are alert to the opportunity they have to utilize and control the remaining recorded detail factors (motion and geometric factors).

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

(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.

32. 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

(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.

7. Grid interspace material can be made of 1. carbon fiber 2. aluminum 3. plastic fiber (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(C) Grids are composed of alternating strips of lead and radiolucent interspace material. The interspace material is either aluminum or plastic fiber. Aluminum resists moisture, is sturdier, provides a "smoother" appearance with less visible grid lines, but requires a higher mAs and therefore increases patient dose. Plastic fiber interspace material can be affected by moisture, resulting in warping. Carbon fiber is often used as image plate front material because of its durability and homogeneity.

63. 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

(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.

240. As window level increases (A) contrast scale increases (B) contrast scale decreases (C) brightness increases (D) brightness decreases

(C) In electronic/digital imaging, changes in window width affect changes in contrast scale, while changes in window level affect changes in brightness (in analog terms, density). As window width increases, the scale of contrast increases (i.e., contrast decreases). As window level increases, brightness increases. It should be noted that, to describe an increase in brightness, our old analog terminology would say that density decreases. This can be easily illustrated as you postprocess/window your own digital photographs or scanned documents. As you slide the brightness scale to increase brightness, the photo/image gets darker, that is, increased density.

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

(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.

14. Which of the following groups of exposure factors will produce the most radiographic density? (A) 100 mA, 50 ms (B) 200 mA, 40 ms (C) 400 mA, 70 ms (D) 600 mA, 30 ms

(C) Milliampere-seconds (mAs) is the exposure factor that is used to regulate radiographic density. Using the equation milliamperage × time = mAs, determine each mAs: , , , . Group C will produce the most radiographic density.

205. Which of the following will influence recorded detail? 1. Dynamic range 2. Part motion 3. Focal spot (A) 1 and 2 only (B) 1 and 3 only (C) 2 and 3 only (D) 1, 2, and 3

(C) Motion is the greatest enemy of resolution. While everything else may be perfect, if motion is introduced, resolution is lost. Focal spot blur is related to focal spot size; smaller focal spots produce less blur and thus better recorded detail/resolution. 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. Dynamic range is not related to resolution.

237. Recorded detail can be improved by decreasing 1. the SID 2. the OID 3. patient/part motion (A) 1 only (B) 3 only (C) 2 and 3 only (D) 1, 2, and 3

(C) Motion, voluntary or involuntary, is most detrimental to good recorded detail. Even if all other factors are adjusted to maximize detail, if motion occurs during exposure, detail 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 recorded detail.

222. How often are radiographic equipment collimators required to be evaluated? (A) Annually (B) Biannually (C) Semiannually (D) Quarterly

(C) Quality Control refers to our equipment and its safe and accurate operation. Various components must be tested at specified intervals, and test results must be within specified parameters. Any deviation from those parameters must be corrected. Examples of equipment components that are tested annually are the focal spot size, linearity, reproducibility, filtration, kV, and exposure time. Congruence is a term used to describe the relationship between the collimator light field and the actual x-ray field—they must be congruent (i.e., match) to within 2% of the SID. Radiographic equipment collimators should be inspected and verified as accurate semiannually, that is, twice a year. Kilovoltage settings can most effectively be tested using an electronic kV meter; to meet required standards, the kV should be accurate to within +/- 4 kV. Reproducibility testing should specify that radiation output be consistent to within +/- 5%.

123. 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

(C) Quantum noise, or mottle, is a grainy appearance; it has a spotted or freckled appearance. It looks very similar to a low-resolution photograph/image that has been enlarged. Low mAs and high kV factors are most likely to be the cause of quantum noise/mottle. SID is unrelated to quantum noise.

175. 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

(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.

245. An exposure was made using 300 mA, 40 ms exposure, and 85 kV. Each of the following changes will decrease the radiographic density by one half except a change to (A) 1/50-s exposure (B) 72 kV (C) 10 mAs (D) 150 mA

(C) Radiographic density is directly proportional to milliampere-seconds. If exposure time is halved from 40 ms (0.04 or 1/25 s) to 0.02 (1/50) s, radiographic density will be cut in half. Changing to 150 mA also will halve the milliampere-seconds, effectively halving the radiographic density. If the kilovoltage is decreased by 15%, from 85 to 72 kV, radiographic density will be halved according to the 15% rule. To cut the density in half, the milliampere-seconds value must be reduced to 6 mAs (rather than 10 mAs).

216. Which of the following has the greatest effect on radiographic density/brightness? (A) Aluminum filtration (B) Kilovoltage (C) SID (D) Scattered radiation

(C) Radiographic density 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 radiographic density decreases. Exposure rate is inversely proportional to the square of the SID. Aluminum filtration, kilovoltage, and scattered radiation all have a significant effect on density, but they are not the primary controlling factors.

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

(C) Recorded detail is evaluated by how sharply tiny anatomic details are imaged on the radiograph. The area of blurriness that may be associated with small image details is termed geometric blur. The blurriness can be produced by using a large focal spot or by diffused fluorescent light from intensifying screens. The image proper (i.e., without blur) is termed the umbra. Mottle is a grainy appearance caused by fast imaging systems.

145. Recorded detail/resolution is inversely related to 1. SID 2. OID 3. part motion (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(C) SID is directly related to recorded detail/resolution because as SID increases, so does recorded detail/resolution (because magnification is decreased). OID is inversely related to recorded detail because as OID increases, recorded detail/resolution decreases. Motion is also inversely related to recorded detail/resolution because as motion increases, recorded detail/resolution decreases—as a result of motion blur, the greatest enemy of resolution. Therefore, of the given choices, OID and motion are inversely related to recorded detail. SID is directly related to recorded detail.

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

(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. 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 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. Imaging system speed is unrelated to scattered radiation.

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

(C) Significant scattered radiation is generated within the part when imaging large or dense body parts and when using high kilo-voltage. 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.

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

(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.

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

(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/SPS phosphors. Smaller phosphor size improves resolution in ways similar to that of intensifying screens—anything that causes an increase in light diffusion will result in a decrease in resolution. Smaller phosphors in the PSP (SPS) 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.

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

(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 image density and is the factor of choice for regulating radiographic/optical density). 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 radiographic density. If the milliampere-seconds value is doubled, twice the exposure rate and twice the density occur. If the milliampere-seconds value is cut in half, the exposure rate and resulting density are cut in half. Kilovoltage is the qualitative exposure factor—it determines beam quality by regulating photon energy (i.e., wavelength).

122. The purpose of the electroconductive layer of a CR PSP plate is to (A) provide support to the PSP layer (B) provide mechanical strength (C) facilitate transportation through the scanner/reader (D) provide better resolution

(C) The PSP plate within the CR cassette/IP has several layers. Its uppermost layer is a protective coat for the phosphor layer below. This layer affords durability and must be translucent to allow passage of photostimulable luminescent light. The phosphor layer is the "active" layer that responds to the x-ray photons that reach it. Under the phosphor layer is the electroconductive layer that serves to facilitate transportation through the scanner/reader and prevent image artifacts resulting from static electricity. Below the electroconductive layer is the plate support layer. Below the support layer is a light-shield layer that serves to prevent light from erasing image plate data or from approaching through the rear protective layer. Behind the light-shield layer is the rear protective layer of the PSP plate.

187. The purpose of the electroconductive layer of a CR PSP plate is to (A) provide support to the PSP layer (B) provide mechanical strength (C) facilitate transportation through the scanner/reader (D) provide better resolution

69. 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

(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.

19. The luminescent light emitted by the PSP is transformed into the image seen on the CRT by the (A) PSP (B) scanner-reader (C) ADC (D) helium-neon laser

(C) The exposed IP 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 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 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.

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

(C) The formula is used to determine each milliampere-second setting (remember to first change milliseconds to seconds). The greatest radiographic density will be produced by the combination of highest milliampere-seconds value and shortest SID. The groups in choices (B) and (D) should produce identical radiographic density, 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 density. 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 radiographic density.

249. 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

(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.

139. A particular milliampere-seconds value, regardless of the combination of milliamperes and time, will reproduce the same radiographic density. This is a statement of the (A) line-focus principle (B) inverse-square law (C) reciprocity law (D) law of conservation of energy

(C) The reciprocity law states that a particular milliampere-seconds value, regardless of the milliamperage and exposure time used, will provide identical radiographic density. This holds true with direct exposure techniques, but it does fail somewhat with the use of intensifying screens. However, the fault is so slight as to be unimportant in most radiographic procedures.

17. 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 (B) 800 (C) 1,000 (D) 1,200

(C) When exposure rate decreases (as a result of increased SID), an appropriate increase in milliampere-seconds is required to maintain the original radiographic density. Unless exposure is increased, the resulting radiograph will be underexposed. The formula used to determine the new milliampere-seconds value (density-maintenance formula) is substituting known values: Substituting known values: Thus, mAs at 72 in. SID. To determine the required milliamperes ,

106. 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

(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.

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

(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.

193. X-ray tubes used in CT differ from those used in x-ray, in that CT x-ray tubes must 1. have a very high short-exposure rating 2. be capable of tolerating several million heat units 3. have a small focal spot for optimal resolution (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(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. While the CT x-ray tube is similar to direct-projection x-ray tubes, it has several special requirements. The CT x-ray tube must have a very high short-exposure rating and must be capable of tolerating several million heat units while still having a small focal spot for optimal resolution. To help tolerate the very high production of heat units, the anode must be capable of high-speed rotation. The x-ray tube produces a pulsed x-ray beam (1-5 ms) using up to about 1,000 mA.

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

(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.

180. How is SID related to exposure rate and image density? (A) As SID increases, exposure rate increases and image density increases. (B) As SID increases, exposure rate increases and image density decreases. (C) As SID increases, exposure rate decreases and image density increases. (D) As SID increases, exposure rate decreases and image density decreases.

(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 image density are directly proportional, if the exposure rate of a beam directed to an IR is decreased, the resulting image density would be decreased proportionately.

16. 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

(D) All the factor changes affect recorded detail, 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.

213. If a 6-in. OID is introduced during a particular radiographic examination, what change in SID will be necessary to overcome objectionable magnification? (A) The SID must be increased by 6 in.. (B) The SID must be increased by 18 in.. (C) The SID must be decreased by 6 in.. (D) The SID must be increased by 42 in..

(D) As OID is increased, recorded detail is diminished as a result of magnification distortion. If the OID cannot be minimized, an increase in SID is required to reduce the effect of magnification distortion. However, the relationship between OID and SID is not an equal relationship. In fact, to compensate for every 1 in. of OID, an increase of 7 in. of SID is required. Therefore, an OID of 6 in. requires an SID increase of 42 in.. This is why a chest radiograph with a 6-in. air gap usually is performed at a 10-ft SID

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

(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.

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

(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 recorded detail.

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

(D) As the kilovoltage is decreased, x-ray-beam energy (i.e., penetration) is also decreased. Consequently, a shorter scale of contrast is obtained, and at lower kilovoltage levels, there is less exposure latitude (less margin for error in exposure). As kilovoltage is reduced, the milliampere-seconds value must be increased accordingly to maintain adequate density. This increase in milliampere-seconds results in greater patient dose.

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

(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.

232. Which of the following may be used to reduce the effect of scattered radiation on a finished radiograph? 1. Grids 2. Collimators 3. Compression bands (A) 1 only (B) 1 and 3 only (C) 2 and 3 only (D) 1, 2, and 3

(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.

169. The effect described as differential absorption is 1. responsible for radiographic contrast 2. a result of attenuating characteristics of tissue 3. 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

(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.

47. The term pixel is associated with all of the following except (A) two dimensional (B) picture element (C) measured in xy direction (D) how much of the part is included in the matrix

(D) Digital image storage is located in a pixel, which is a two-dimensional "picture element," measured in the "XY" direction. The third dimension, "Z" direction, in the matrix of pixels is the depth that is referred to as the voxel (volume element). The depth of the block is the number of bits required to describe the gray level that each pixel can take on—known as the bit depth. Bit depth in CT is approximately 212 with a dynamic range of almost 5,000 gray shades, approximately 214 in CR/DR with a dynamic range of more than 16,000 gray shades, and approximately 216 in digital mammography with a dynamic range of more than 65,500 gray shades. The matrix is the number of pixels in the XY direction. As matrix size increases, for a fixed FOV, pixel size is smaller and better spatial resolution results. An electronic/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 (a typical CT image). The term FOV is used to describe how much of the patient is included in the matrix. Either the matrix or the FOV can be changed without one affecting the other, but changes in either will change pixel size. As FOV increases, for a fixed matrix size, the size of each pixel increases and spatial resolution decreases. Fewer and larger pixels result in a poor-resolution "pixelly" or "mosaicked" image, that is, one in which you can actually see the individual pixel boxes.

56. The term voxel is associated with all of the following except (A) bit depth (B) volume element (C) measured in Z direction (D) field of view

68. For the same FOV, as the matrix size increases 1. spatial resolution increases 2. image quality increases 3. pixel size decreases (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(D) Digital image storage is located in a pixel, which is a two-dimensional "picture element," measured in the "XY" direction. The third dimension, "Z" direction, in the matrix of pixels is the depth that is referred to as the voxel (volume element). The depth of the block is the number of bits required to describe the gray level that each pixel can take on—known as the bit depth. The matrix is the number of pixels in the XY direction. As matrix size increases, for a fixed FOV, pixel size is smaller and better spatial resolution results. An electronic/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 (a typical CT image). The term FOV is used to describe how much of the patient is included in the matrix. Either the matrix or the FOV can be changed without one affecting the other, but changes in either will change pixel size. As FOV increases, for a fixed matrix size, the size of each pixel increases and spatial resolution decreases. Fewer and larger pixels result in a poor-resolution "pixelly" or "mosaicked" image, that is, one in which you can actually see the individual pixel boxes.

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

(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.

52. Which of the following groups of factors would produce the least radiographic density? (A) 400 mA, 0.010 second, 94 kV, 100-speed screens (B) 500 mA, 0.008 second, 94 kV, 200-speed screens (C) 200 mA, 0.040 second, 94 kV, 50-speed screens (D) 100 mA, 0.020 second, 80 kV, 200-speed screens

(D) Each milliampere-second setting is determined [; ; ; ] and numbered in order of greatest to least density [; (A) and ; ]. Then, the kilovoltages are reviewed and also numbered in order of greatest to least density [(A), (B), and ; ]. Next, screen speeds are numbered from greatest density-producing to least density-producing [(D) and ; ; ]. Finally, the numbers assigned to the milliampere-seconds, kilovoltage, and screen speed are added up for each of the four groups [; (A) and ; ]; the lowest total (B) indicates the group of factors that will produce the greatest radiographic density; the highest total (D) indicates the group of factors that will produce the least radiographic density. This process is illustrated as follows:

79. 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

(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.

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

(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.

128. Factors that determine recorded detail in digital imaging include 1. part motion 2. geometric factors 3. spatial resolution (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(D) Factors influencing recorded detail in digital imaging are very much the same as those factors affecting recorded detail in analog imaging, that is, motion, geometric factors (focal spot size, OID, and SID), and spatial resolution. In analog imaging, spatial resolution is related to the speed of the imaging system (intensifying screen speed) and is significantly better than digital imaging resolution. The spatial resolution of direct digital systems, however, is fixed and is related to the detector element (DEL) size of the thin film transistor (TFT). The smaller the TFT DEL size, the better the spatial resolution. DEL size of 100 microns provides a spatial resolution of about 5 lp/mm (available only in some digital mammography systems). DEL size of 200 microns provides a spatial resolution of about 2.5 lp/mm (general radiography)—lower than that achieved with 400 speed intensifying screen system. A 100 speed intensifying screen system offers a spatial resolution of about 10 lp/mm—significantly greater than, and currently unachievable in, digital imaging. Spatial resolution in digital imaging is fixed, but it is very important that radiographers are alert to the opportunity they have to utilize and control the remaining recorded detail factors (motion and geometric factors).

184. Factors that determine recorded detail in digital imaging include 1. focal spot size 2. SID 3. DEL size (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

28. Factors that contribute to film fog include 1. the age of the film 2. excessive exposure to safelight 3. processor chemistry (A) 1 only (B) 1 and 2 only (C) 1 and 3 only (D) 1, 2, and 3

(D) Film age is an important consideration when determining the causes of film fog. Outdated film will exhibit loss of contrast in the form of fog and loss of speed. A safelight is "safe" only for practical periods of time required for the necessary handling of film. Films that are left out on the darkroom counter can be fogged by excessive exposure to the safelight. Film emulsion is much more sensitive to safelight fog after exposure. The high temperatures required for automatic processors' rapid processing are a source of film fog. Daily QA ensures that fog levels do not exceed the upper limit of 0.2 density.

78. Which of the following factors impact(s) recorded detail? 1. Focal-spot size 2. Subject motion 3. SOD (A) 1 and 2 only (B) 1 and 3 only (C) 2 and 3 only (D) 1, 2, and 3

(D) Focal-spot size affects recorded detail by its effect on focal-spot blur: The larger the focal-spot size, the greater is the blur produced. Recorded detail is affected significantly by distance changes because of their effect on magnification. As SID increases and as OID decreases, magnification decreases and recorded detail increases. SOD is determined by subtracting OID from SID.

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

(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.

214. 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

(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.

39. Which of the following factors contribute(s) to the efficient performance of a grid? 1. Grid ratio 2. Number of lead strips per inch 3. Amount of scatter transmitted through the grid (A) 1 only (B) 2 only (C) 1 and 2 only (D) 1, 2, and 3

(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.

129. Examples of health care informatics include 1. HIS 2. RIS 3. PACS (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(D) Health care information technology, or health informatics, has ever-increasing application and use in the imaging sciences. PACS (Picture Archiving and Communication Systems) is used by health care facilities to economically store, archive, exchange, and transmit digital images from multiple imaging modalities. PACS replaces the need to manually file, retrieve, and transport film and film jackets. RIS (Radiology Information Systems) and HIS (Hospital Information Systems) can be integrated with PACS for electronic health information storage. The purpose of HIS is to manage health care information and documents electronically, and to ensure data security and availability. RIS is a system for tracking radiological and imaging procedures. RIS is used for patient registration and scheduling, radiology workflow management, reporting and printout, manipulation and distribution and tracking of patient data, and billing. RIS complements HIS and is critical to competent workflow to radiologic facilities.

42. Factors that determine the production of scattered radiation include 1. field size 2. beam restriction 3. kilovoltage (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(D) High kV may be desirable in terms of patient dose, tube life, and making more details visible, but use of excessively high kV will result in production of excessive amounts of scattered radiation and fog, resulting in diminished visibility of image details. Much of the scattered radiation produced is highly energetic and exits the patient along with useful image-forming radiation. Scattered radiation carries no useful information but adds noise in the form of fog, thereby impairing detail visibility. Because scattered radiation can have such a devastating effect on image contrast, it is essential that radiographers are knowledgeable about methods of reducing its production. The three factors that have a significant effect on the production of scattered radiation are beam restriction (i.e., size of irradiated field), kV, and thickness/volume and density of tissues. Perhaps the most important way to limit the production of scattered radiation and improve contrast is by limiting the size of the irradiated field, that is, through beam restriction. As the size of the x-ray field is reduced, there is less area and tissue volume for scattered radiation to be generated. As the volume and/or density of the irradiated tissues increase(s), so does scattered radiation. Thicker and denser anatomic structures will generate more scattered radiation. Compression of certain parts can occasionally be used to minimize the effect of scatter, but close collimation can always be used effectively.

53. Chemical fog may be attributed to 1. excessive developer temperature 2. oxidized developer 3. excessive replenishment (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(D) If developer temperature is too high, some of the less exposed or unexposed silver halide crystals may be reduced, thus creating chemical fog. If the developer solution has become oxidized from exposure to air, chemical fog also results. If developer replenishment is excessive, and too much new solution is replacing the deteriorated developer, chemical fog is again the result.

91. Which of the following statements about histograms is/are true? 1. A histogram illustrates pixel value distribution. 2. There is a default histogram for each/different body parts. 3. A histogram is representative of the image grayscale. (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(D) In digital imaging, as in film/screen radiography, there are numerous density/brightness values that represent various tissue densities (i.e., x-ray attenuation properties), for example, bone, muscle, fat, blood-filled organs, air/gas, metal, contrast media, and pathologic processes. In CR, the CR scanner/reader recognizes all these values and constructs a representative grayscale histogram of them, corresponding to the anatomic characteristics of the imaged part. Thus, all PA chest histograms are similar, all lateral chest histograms are similar, all pelvis histograms are similar, etc. A histogram is a graphic representation of pixel value distribution. The histogram is an analysis and graphic representation of all the densities from the PSP screen, demonstrating the quantity of exposure, the number of pixels, and their value. Histograms are unique to each body part imaged. Histogram appearance and patient dose can be affected by the radiographer's knowledge and skill using digital imaging, in addition to their 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 anatomic 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. Poor collimation can affect exposure level and exposure latitude; these changes are reflected in the images' informational numbers ("S number," "exposure index," etc.). 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), changes in scatter, source-to-image-receptor distance (SID), object-to-image-receptor distance (OID), and collimation—in short, anything that affects scatter and/or dose.

157. Factors that can affect histogram appearance include 1. beam restriction 2. centering errors 3. incorrect SID (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(D) In digital imaging, as in film/screen radiography, there are numerous density/brightness values that represent various tissue densities (i.e., x-ray attenuation properties), for example, bone, muscle, fat, blood-filled organs, air/gas, metal, contrast media, and pathologic processes. In CR, the CR scanner/reader recognizes these values and constructs a representative grayscale histogram of them, corresponding to the anatomic characteristics of the imaged part. Thus, all PA chest histograms are similar, all lateral chest histograms are similar, all pelvis histograms are similar, etc. A histogram is a graphic representation of pixel value distribution. The histogram is an analysis and graphic representation of all the densities from the PSP screen, demonstrating the quantity of exposure, the number of pixels, and their value. Histograms are unique to each body part imaged. Histogram appearance and patient dose can be affected by the radiographer's knowledge and skill using digital imaging, in addition to their 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 anatomic 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. Poor collimation can affect exposure level and exposure latitude; these changes are reflected in the images' informational numbers ("S number," "exposure index," etc.). 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), changes in scatter, source-to-image-receptor distance (SID), object-to-image-receptor distance (OID), and collimation—in short, anything that affects scatter and/or dose.

197. 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

(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.

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

(D) Insufficient milliamperage and/or exposure time will result in lack of radiographic density. Insufficient kilovoltage will result in underpenetration and excessive contrast. Insufficient SID, however, will result in increased exposure rate and radiographic overexposure.

57. 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

(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.

159. Changes in milliampere-seconds can affect all the following except (A) quantity of x-ray photons produced (B) exposure rate (C) optical density (D) recorded detail

(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 (i.e., a particular quantity of x-ray photons) will produce identical optical density. This is known as the reciprocity law. Density is a quantitative factor because it describes the amount of image blackening. The milliampere-seconds value is also 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 image density and is the factor of choice for regulating radiographic/optical density). 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 radiographic density. If the milliampere-seconds value is doubled, twice the exposure rate and twice the density occur. If the milliampere-seconds value is cut in half, the exposure rate and resulting density are cut in half. The milliampere-seconds value has no effect on recorded detail.

141. 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

(D) Milliampere-seconds (mAs) is the exposure factor that regulates radiographic density. The equation used to determine mAs is . Substituting the known factors:

125. Which of the following possesses the widest dynamic range? (A) Film/screen imaging (B) Beam restriction (C) AEC (D) CR

(D) One of the biggest advantages of CR is the dynamic range, or latitude, it offers. The characteristic curve of x-ray film emulsion has a certain "range of correct exposure," limited by the toe and shoulder of the curve. In CR, there is a linear relationship between the exposure, given the PSP (SPS) 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. The same exposure factors as screen-film systems, or less, generally are recommended for CR. Intensifying screens used in screen-film x-ray imaging tend to produce high contrast. The faster the screens, the higher is the contrast; higher contrast often is associated with decreased latitude. AEC refers to automatic exposure control and is unrelated to dynamic range or latitude.

172. Typical patient demographic and examination information include(s) 1. type of examination 2. accession number 3. date and time of examination (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(D) Patient demographic and examination information originates from the hospital/facility HIS, where it is obtained when the patient is initially registered. That information is available or retrievable when the patient is scheduled, or arrives, for imaging services. Typical patient information includes name, DOB or age, sex, ID number, accession number, examination being performed, date and time of examination. Additional information may be available on the examination requisition; more information is usually entered by the technologist at the time of the examination. A feature that is useful in sorting examinations and decreasing (but not eliminating) errors is the Modality Work List (MWL). The MWL "brings up" existing RIS information, that is, the examinations scheduled for each imaging area—for example, x-ray, CT, MR, mammography, ultrasound, etc. The technologist selects the correct patient, which includes that patient's particular demographics, from the particular modality work list. It is essential that the technologist is attentive to detail and accuracy when entering patient information; errors in patient demographics entry, and entry duplication, must be avoided.

35. To be suitable for use in an image intensifier's input screen, a phosphor should have which of the following characteristics? 1. High conversion efficiency 2. High x-ray absorption 3. High atomic number (A) 1 only (B) 3 only (C) 1 and 2 only (D) 1, 2, and 3

(D) Phosphors that have a high atomic number are more likely to absorb a high percentage of the incident x-ray photons and convert x-ray photon energy to fluorescent light energy. How efficiently the phosphors detect and interact with the x-ray photons is termed quantum detection efficiency. How effectively the phosphors make this energy conversion is termed conversion efficiency.

171. 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

(D) Radiographic contrast is the sum of film emulsion contrast and subject contrast. Subject contrast has by far the greatest influence on radiographic contrast. 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.

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

(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.

185. The photostimulable phosphor (PSP) plates used in CR are constructed in layers that include 1. light shield layer 2. support layer 3. electroconductive layer (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(D) The PSP plate within the CR image plate has several layers. Its uppermost layer is a protective coat for the phosphor layer below. This layer affords durability and must be translucent to allow passage of photostimulable luminescent light. The phosphor layer is the "active" layer that responds to the x-ray photons that reach it. Under the phosphor layer is the electroconductive layer that serves to facilitate transportation through the scanner/reader and prevent image artifacts resulting from static electricity. Below the electroconductive layer is the plate support layer. Below the support layer is a light-shield layer that serves to prevent light from erasing image plate data or from approaching through the rear protective layer. Behind the light-shield layer is the rear protective layer of the PSP plate.

226. 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

(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.

231. Which of the following is/are associated with magnification fluoroscopy? 1. Less noise 2. Improved contrast resolution 3. Improved spatial resolution (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(D) The input phosphor of image intensifiers is usually made of cesium iodide. For each x-ray photon absorbed by cesium iodide, approximately 5,000 light photons are emitted. As the light photons strike a photoemissive photocathode, a number of electrons are released from the photocathode and focused toward the output side of the image tube by voltage applied to the negatively charged electrostatic focusing lenses. The electrons are then accelerated through the neck of the tube where they strike the small (0.5-1 in.) output phosphor that is mounted on a flat glass support. The entire assembly is enclosed within a 2-4-mm thick vacuum glass envelope. Remember that the image on the output phosphor is minified, brighter, and inverted (electron focusing causes image inversion). Input screen diameters of 5-12 in. are available. Although smaller diameter input screens improve resolution, they do not permit a large FOV, that is, viewing of large patient areas. Dual- and triple-field image intensifiers are available that permit magnified viewing of fluoroscopic images. To achieve magnification, the voltage to the focusing lenses is increased and a smaller portion of the input phosphor is used, thereby resulting in a smaller FOV. Because minification gain is now decreased, the image is not as bright. The mA is automatically increased to compensate for the loss in brightness when the image intensifier is switched to magnification mode. Entrance skin exposure (ESE) can increase dramatically as the FOV decreases (i.e., as magnification increases). As FOV decreases, magnification of the output screen image increases, there is less noise because increased mA provides a greater number of x-ray photons, and contrast and spatial resolution improve. The focal point in the magnification mode is further away from the output phosphor (as a result of increased voltage applied to the focusing lenses) and therefore the output image is magnified.

137. A radiograph made with a parallel grid demonstrates decreased density 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

(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 density at the lateral edges of the radiograph. 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 density loss on one side or the other.

126. 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

(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.

165. Brightness and contrast resolution in digital imaging can be influenced by 1. window level (WL) 2. window width (WW) 3. look-up table (LUT) (A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1, 2, and 3

(D) The radiographer selects a processing algorithm by selecting the anatomic part and particular projection on the computer/control panel. The CR unit then matches that information with a particular lookup table (LUT)—a characteristic curve that best matches the anatomic part being imaged. The observer is able to review the image and, if desired, change its appearance (through "windowing"); doing so changes the LUT. Histogram analysis and use of the appropriate LUT together function to produce predictable image quality in CR. Additionally, the radiographer can manipulate, that is, change and enhance, the digital image displayed on the display monitor through postprocessing. One way to alter image contrast and/or brightness is through windowing. The term windowing refers to some change made to window width and/or window level, that is, a change in the LUT. Change in window width affects change in the number of gray shades, that is, image contrast. Change in window level affects change in the image brightness, that is, density. Therefore, windowing and other postprocessing mechanisms permit the radiographer to affect changes in the image and produce "special effects," such as contrast enhancement, edge enhancement, image stitching.

75. Which of the following is likely to contribute to the radiographic contrast present on the finished radiograph? 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

(D) The radiographic subject, the patient, is composed of many different tissue types that have varying 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 finished radiograph. 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 radiographic 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.

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

(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 for decreasing involuntary motion. Immobilization is also very useful in eliminating motion unsharpness.

244. 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

(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.

31. 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

(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.

86. If 400 mA, 10 ms, and 90 kV were used for a particular exposure using three-phase, 12-pulse equipment, which of the following exposure changes would be most appropriate for use on single-phase equipment to produce a similar image? (A) Use 200 mA (B) Use 20 mAs (C) Use 70 kV (D) Use 0.02 second

(D) 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 value is required to produce a radiograph with similar density. When changing from single-phase to three-phase, 12-pulse equipment, only half the original milliampere-seconds value is required. In this problem, we are changing from three-phase, 12-pulse to single-phase equipment; therefore, the milliampere-seconds value should be doubled (from 4-8 mAs).

Lange #3

Set pelajaran terkait

Sociology Midterm

Mod 12: Disorders of the Liver, Biliary Tract, and Pancreas

Chapter 1

Strategy Chapter 9

Midterm review

Leadership Module 1 Practice Questions

Chapter 3 Accounting

Module 5

SOC337 Test 3

CITI program IRB Social and Behavioral Responsible Conduct of Research

Simulate Your Exam - Missed Questions

Module 8

11,12,14,15

Psych and Criminal Justice

life insurance chapter 2

unit 5: breed identification

PU Unit III

Series 7 Options and Quicksheet, Series 7 - Master Exam

GRE Math

Ahip 2024 module 1