Image Acquisition and Technical Evaluation - Rad Review

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?

16:1 -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.

In amorphous selenium flat-panel detectors, the term amorphous refers to a?

crystalline material lacking typical crystalline structure

What lies immediately under the phosphor layer of a PSP storage plate?

reflective layer

The CR should be directed to the center of the part of greatest interest to avoid?

rotation distortion -Anatomic details placed away from the path of the CR will be exposed by more divergent rays, resulting in rotation distortion. This is why the CR must be directed to the midpoint of the part of greatest interest. For example, if bilateral hands are requested, they should be examined individually; if imaged simultaneously, the CR will be directed to no anatomic part (between the two hands) and rotation distortion will occur.

What can result from excessive kV, insufficient collimation, and/or thick anatomic structures?

scattered radiation fog

Misalignment of the tube-part-IR relationship results in?

shape distortion -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.

Cassette-front material can be made of which of the following? 1. Carbon fiber 2. Magnesium 3. Lead

1 and 2 only

Compared to a low ratio grid, a high ratio grid will 1. absorb more primary radiation. 2. absorb more scattered radiation. 3. allow more centering latitude.

1 and 2 only

As digital image matrix size increases, for the same FOV, 1. pixel size decreases. 2. resolution increases. 3. pixel size increases.

1 and 2 only -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 (FOV) is used to describe how much of the patient (eg, 150-mm diameter) is included in the matrix. The matrix and the FOV can be changed independently, without one affecting the other, but changes in either will change pixel size. As in traditional radiography, spatial resolution is measured in line pairs per mm (lp/mm). As matrix size is increased with the same FOV, there are more and smaller pixels in the matrix, and therefore improved resolution. Fewer and larger pixels result in poor resolution, "pixelly" image, that is, one in which you can actually see the individual pixel boxes.

Which of the following will improve the spatial resolution of image-intensified images? 1. A very thin coating of cesium iodide on the input phosphor 2. A smaller-diameter input screen 3. Increased total brightness gain

1 and 2 only -An image's spatial resolution refers to ithe sharpness of its image details. As its input screen's phosphor layer is made thinner, spatial resolution increases. Also, the smaller the input phosphor diameter, the greater is the spatial resolution. A brighter image is easier to see but does not affect resolution.

Disadvantages of using low-kilovoltage technical factors include 1. insufficient penetration 2. increased patient dose 3. diminished resolution

1 and 2 only -As the kilovoltage is decreased, x-ray-beam energy (i.e., penetration) is also decreased. Consequently, a shorter scale of contrast is obtained. As kilovoltage is reduced, the milliampere-seconds value must be increased accordingly to maintain adequate receptor exposure. This increase in milliampere-seconds results in greater patient dose. Resolution is not related to kV.

The effect that differential absorption has on radiographic contrast of a high-subject-contrast part can be minimized by 1. using a compensating filter. 2. using high-kilovoltage exposure factors. 3. increased collimation.

1 and 2 only -Differential absorption refers to the different attenuation, or absorption, properties of adjacent body tissues. Two parts with widely differing absorption characteristics will produce a high radiographic contrast. Frequently, exposure factors that would properly expose one part will severely overexpose or underexpose the neighboring part (as with lungs vs. the thoracic spine). This effect can be minimized by the use of a compensating filter or by the use of high kilovoltage (for more uniform penetration). Increased collimation is important in the control of patient dose and scattered radiation, not differential absorption.

Which one of the following is (are) used to control the production of scattered radiation? 1. Collimators 2. Optimal kV 3. Use of grids

1 and 2 only -Grids have no effect on the production of scattered radiation

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

1 and 2 only -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.

Methods that help to reduce the production of scattered radiation include using 1. compression 2. beam restriction 3. a grid

1 and 2 only -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.

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

1 and 2 only -One generally thinks in terms of moving grids being totally superior to stationary grids because moving grids function to blur the images of the lead strips on the radiographic image. Moving grids do, however, have several disadvantages. First, their complex mechanism is expensive and subject to malfunction. Second, today's sophisticated x-ray equipment makes possible the use of extremely short exposures, a valuable feature whenever motion may be a problem (as in pediatric radiography). However, grid mechanisms frequently are not able to oscillate rapidly enough for the short exposure times, and as a result, the grid motion is "stopped," and the lead strips are imaged. Third, patient dose is increased with moving grids. Since the central ray is not always centered to the grid because it is in motion, lateral decentering occurs (resulting in diminished density), and consequently, an increase in exposure is needed to compensate (either manually or via AEC).

Low-kilovoltage exposure factors usually are indicated for radiographic examinations using 1. water-soluble, iodinated media 2. a negative contrast agent 3. barium sulfate

1 and 2 only -Positive contrast medium is radiopaque; negative contrast material is radioparent. Barium sulfate (radiopaque, positive contrast material) is used most frequently for examinations of the intestinal tract, and high-kilovoltage exposure factors are used to penetrate (to see through and behind) the barium. Water-based iodinated contrast media (Conray, Amipaque) are also positive contrast agents. However, the K-edge binding energy of iodine prohibits the use of much greater than 70 kV with these materials. Higher kilovoltage values will obviate the effect of the contrast agent. Air is an example of a negative contrast agent, and high-kilovoltage factors are clearly not indicated.

Which of the following can have an effect on radiographic contrast? 1. Beam restriction 2. Grids 3. Focal spot size

1 and 2 only -Radiographic contrast the result of x-ray absorption difference between tissue densities resulting in scale of grays. Since the function of grids is to collect scattered radiation, they serve to shorten the scale of contrast. Beam restrictors function to limit the x-ray field size, thereby reducing the production of scattered radiation and shortening the scale of contrast. Focal spot size is one of the geometric factors affecting spatial resolution; it has no effect on the scale of contrast or receptor exposure. It is the function of radiographic contrast to make details visible.

Which of the following is (are) classified as rare earth phosphors? 1. Lanthanum oxybromide 2. Gadolinium oxysulfide 3. Cesium iodide

1 and 2 only -Rare earth phosphors have a greater conversion efficiency than do other phosphors. Lanthanum oxybromide is a blue-emitting rare earth phosphor, and gadolinium oxysulfide is a green-emitting rare earth phosphor. Cesium iodide is the phosphor used on the input screen of image intensifiers; it is not a rare earth phosphor.

Which of the following adult radiographic examinations usually require(s) use of a grid? 1. Ribs 2. Vertebrae 3. Shoulder

1, 2, and 3 -Generally speaking, anatomic parts measuring in excess of 10 cm require a grid. (The major exception to this rule can be the chest). The larger the part, the more scattered radiation is generated. To avoid degradation of the image as a result of scattered radiation fog, a grid is used to absorb scatter. Parts generally requiring the use of a grid include the skull, spine, ribs, pelvis, shoulder, and femur.

Which of the following units is (are) used to express resolution? 1. Line-spread function 2. Line pairs per millimeter 3. Line-focus principle

1 and 2 only -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.

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

1 and 2 only -Significant scattered radiation is generated within the part when imaging large or dense body parts and when using high kilovoltage. A radiographic grid is made of alternating lead strips and interspace material; it is placed between the patient and the IR to absorb energetic scatter emerging from the patient. Although a grid prevents much of the scattered radiation from reaching the radiograph, its use does necessitate a significant increase in patient exposure.

A grid is usually employed 1. when radiographing a large or dense body part. 2. when using high kilovoltage. 3. when less patient dose is required.

1 and 2 only -Significant scattered radiation is produced when radiographing large or dense body parts and when using high kilovoltage. A radiographic grid is made of alternating lead strips and interspace material; it is placed between the patient and the IR to absorb energetic scatter emerging from the patient. Although a grid prevents much scattered radiation fog from reaching the radiograph, its use does necessitate a significant increase in patient exposure.

In comparison with 60 kV, 80 kV will 1. permit greater exposure latitude 2. produce more scattered radiation 3. produce shorter-scale contrast

1 and 2 only -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

According to the line-focus principle, an anode with a small angle provides 1. improved spatial resolution. 2. improved heat capacity. 3. less heel effect.

1 and 2 only -The line-focus principle illustrates that as the target angle decreases, the effective focal spot decreases (providing improved spatial resolution), but the actual area of electron interaction remains much larger (allowing for greater heat capacity). It must be remembered, however, that a steep (small) target angle increases the heel effect, and part coverage may be compromised.

Exposure rate increases with an increase in 1. mA. 2. kVp. 3. SID.

1 and 2 only -The quantity of x-ray photons produced at the target is the function of mAs. The quality (wavelength, penetration, energy) of x-ray photons produced at the target is the function of kVp. The kVp also has an effect on exposure rate, because an increase in kVp will increase the number of high-energy x-ray photons produced at the target. Exposure rate decreases with an increase in SID.

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

1 and 2 only -When differences in absorption characteristics are decreased, body tissues absorb radiation more uniformly, and as a result, more grays are seen on the radiographic image. A longer scale of contrast is produced. High-kilovoltage and low-milliamperage factors achieve this. Compensating filtration is also used to "even out" densities in uneven anatomic parts, such as the thoracic spine. The photoelectric effect is the interaction between x-ray photons and matter that occurs at low-peak kilovoltage levels—levels that tend to produce short-scale contrast

If the x-ray image exhibits insufficient receptor exposure, this might be attributed to 1. Insufficient kilovoltage 2. Insufficient SID 3. grid cutoff

1 and 3 only

If a radiograph exhibits insufficient receptor exposure, this might be attributed to 1. inadequate kVp. 2. inadequate SID. 3. grid cutoff.

1 and 3 only -As kVp is reduced, the number of high-energy photons produced at the target is reduced; therefore, a decrease in receptor exposure occurs. If a grid has been used improperly (off-centered or out of focal range), the lead strips will absorb excessive amounts of the useful beam, resulting in grid cutoff and loss of rreceptor exposure. If the SID is inadequate (too short), an increased receptor exposure will result.

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

1 and 3 only -Focal spot size affects spatial resolution by its effect on focal spot blur: The larger the focal spot size, the greater the blur produced. Spatial resolution is significantly affected by distance changes because of their effect on magnification. As SID increases, magnification decreases and spatial resolution increases. The method of rectification has no controlling effect on spatial resolution. Single-phase rectified units produce intermittent radiation at fluctuating voltage, whereas three-phase units produce almost constant potential. Single phase equipment exposures could require longer exposures, possibly resulting in motion unsharpness, though that equipment is seldom used today

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

1, 2, and 3 -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.

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

1 and 3 only -High-kilovoltage exposures produce large amounts of scattered radiation, and high-ratio grids are used often with high-kilovoltage techniques in an effort to absorb more of this scattered radiation. However, as more scattered radiation is absorbed, more primary radiation is absorbed as well. This accounts for the increase in milliampere-seconds required when changing from an 8:1 to a 16:1 grid. In addition, precise centering and positioning become more critical; a small degree of inaccuracy is more likely to cause grid cutoff in a high-ratio grid.

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

1 and 3 only -Increased photon energy is caused by an increase in kVp, resulting in more penetration of the part and a longer scale of contrast. Increasing the grid ratio will result in a larger percentage of scattered radiation being absorbed and hence a shorter scale of contrast.

Exposure rate will decrease with an increase in 1. SID 2. kilovoltage 3. focal-spot size

1 only

In electronic imaging, as digital image matrix size increases 1. pixel size decreases 2. resolution decreases 3. pixel depth decreases

1 only

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

1 only

What are the effects of scattered radiation on a radiographic image? 1. It produces fog. 2. It increases contrast. 3. It increases grid cutoff.

1 only

Which of the following examinations might require the use of 70 kV? 1. AP abdomen 2. Chest radiograph 3. Barium-filled stomach

1 only

Spatial resolution is directly related to 1. SID. 2. tube current. 3. focal-spot size.

1 only -As SID increases, so does spatial resolution because magnification is decreased - a direct relationship. Therefore, SID is directly related to spatial resolution. As focal spot size increases, spatial resolution decreases because more penumbral blur is produced. Focal spot size is thus inversely related to spatial resolution - as FSS increases,resolution decreases. Tube current affects receptor exposure and is unrelated to spatial resolution

Spatial resolution is directly related to 1. source-image distance (SID). 2. tube current. 3. focal spot size.

1 only -As SID increases, so does spatial resolution, because magnification is decreased. Therefore, SID is directly related to spatial resolution. As focal spot size increases, spatial resolution decreases because more blur/penumbra is produced. Focal spot size is thus inversely related to spatial resolution. Tube current affects receptor exposure and is unrelated to spatial resolution.

X-ray photon energy is inversely related to 1. photon wavelength 2. applied milliamperes (mA) 3. applied kilovoltage (kV)

1 only -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

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

1 only -As the kilovoltage is increased, photon energy increases and more part penetration will occur. As the milliampere-seconds value is decreased to compensate for the increased kilovoltage, receptor exposure should remain the same

Terms that refer to size distortion include 1. magnification 2. attenuation 3. elongation

1 only -Distortion is misrepresentation of the actual size or shape of the object being imaged. Size distortion is magnification.

Geometric unsharpness is directly influenced by 1. OID 2. SOD 3. SID

1 only -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.

Which of the following is (are) directly related to photon energy? 1. Kilovoltage 2. Milliamperes 3. Wavelength

1 only -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.

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.

1 only -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

Which of the following factors is/are related to grid efficiency? 1. Grid ratio 2. Number of lead strips per inch 3. Amount of scatter transmitted through the grid

1, 2, and 3 -Grid ratio is defined as the ratio of the height of the lead strips to the width of the interspace material; the higher the lead strips, the more scattered radiation they will trap and the greater is the grid's efficiency. The greater the number of lead strips per inch, the thinner and less visible they will be on the finished radiograph. The function of a grid is to absorb scattered radiation in order to improve radiographic contrast. The selectivity of a grid is determined by the amount of primary radiation transmitted through the grid divided by the amount of scattered radiation transmitted through the grid.

In 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 rather than high-ratio grid

1 only -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

Geometric unsharpness will be least obvious 1. at long SIDs. 2. with small focal spots. 3. at the anode end of the image.

1, 2 and 3 -The x-ray tube anode is designed according to the line-focus principle, that is, with the focal track beveled (Figure 6-24). This allows a larger actual focal spot to project a smaller effective focal spot, resulting in improved spatial resolution with less blur. However, because of the target angle, penumbral blur varies along the longitudinal tube axis, being greater at the cathode end of the image and less at the anode end of the image. Therefore, better spatial resolution will be appreciated using small focal spots at the anode end of the x-ray beam and at longer SIDs.

A technique chart should be prepared for each AEC x-ray unit and should contain which of the following information for each type of examination? 1. Photocell(s) used 2. Optimum kilovoltage 3. Backup time

1, 2, and 3

Factor(s) that impact receptor exposure include 1. milliamperage. 2. exposure time. 3. kilovoltage.

1, 2, and 3

Factors that determine AEC exposure determination include 1. the thickness and density of the object 2. positioning of the object with respect to the photocell 3. beam restriction

1, 2, and 3

In which of the following ways might higher image contrast be obtained in abdominal radiography? 1. By using lower kilovoltage 2. By using a contrast medium 3. By limiting the field size

1, 2, and 3

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

1, 2, and 3

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

1, 2, and 3

Typical examples of digital imaging include 1. MRI 2. CT 3. CR

1, 2, and 3

Which of the following can affect histogram appearance? 1. Centering accuracy 2. Positioning accuracy 3. Processing algorithm accuracy

1, 2, and 3

Which of the following is (are) associated with subject contrast? 1. Patient thickness 2. Tissue density 3. Kilovoltage

1, 2, and 3

Which of the following may be used to reduce the effect of scattered radiation on the radiographic image? 1. Grids 2. Collimators 3. Compression bands

1, 2, and 3

Which of the following pathologic conditions are considered additive conditions with respect to selection of exposure factors? 1. Osteoma 2. Bronchiectasis 3. Pneumonia

1, 2, and 3 -All these conditions are considered technically additive because they all involve an increase in tissue density. Osteoma, or exostosis, is a (usually benign) bony tumor that can develop on bone. Bronchiectasis is a chronic dilatation of the bronchi with accumulation of fluid. Pneumonia is inflammation of the lung(s) with accumulation of fluid. Additional bony tissues and the pathologic presence of fluid are additive pathologic conditions and require an increase in exposure factors. Destructive conditions such as osteoporosis require a decrease in exposure factors.

Which of the following pathologic conditions require(s) a decrease in exposure factors? 1. Pneumothorax 2. Emphysema 3. Multiple myeloma

1, 2, and 3 -All three pathologic conditions involve processes that render tissues more easily penetrated by the x-ray beam. Pneumothorax is a collection of air or gas in the pleural cavity. Emphysema is a chronic pulmonary disease characterized by an increase in the size of the air-containing terminal bronchioles. These two conditions add air to the tissues, making them more easily penetrated. Multiple myeloma is a condition characterized by infiltration and destruction of bone and marrow. Each of these conditions requires that factors be decreased from the normal to avoid overexposure.

Which of the following factors influence(s) the production of scattered radiation? 1. Kilovoltage level 2. Tissue density 3. Size of field

1, 2, and 3 -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.

Which of the following factors impact(s) spatial resolution? 1. Focal spot size 2. Subject motion 3. SOD

1, 2, and 3 -Focal-spot size affects spatial resolution by its effect on focal-spot blur: The larger the focal-spot size, the greater is the blur produced. Spatial resolution is affected significantly by distance changes because of their effect on magnification. As SID increases and as OID decreases, magnification decreases and spatial resolution increases. SOD is determined by subtracting OID from SID.

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

1, 2, and 3 -The chest is composed of tissues with widely differing densities (bone and air). In an effort to "even out" these tissue densities and better visualize pulmonary vascular markings, high kilovoltage generally is used. This produces more uniform penetration and results in a longer scale of contrast with visualization of the pulmonary vascular markings as well as bone (which is better penetrated) and air densities. The increased kilovoltage also affords the advantage of greater exposure latitude (an error of a few kilovolts will make little, if any, difference). The fact that the kilovoltage is increased means that the milliampere-seconds value is reduced accordingly, and thus patient dose is reduced as well. A grid usually is used whenever high kilovoltage is required.

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

1, 2, and 3 -The radiographic subject, the patient, is composed of many different tissue types that have varying tissue densities, resulting in varying degrees of photon attenuation and absorption. The atomic number of the tissues under investigation is directly related to their attenuation coefficient. This differential absorption contributes to the various shades of gray (scale of radiographic contrast) on the radiographic image. Normal tissue density may be altered significantly in the presence of pathologic processes. For example, destructive bone disease can cause a dramatic decrease in tissue density (and subsequent increase in receptor exposure). Abnormal accumulation of fluid (as in ascites) will cause a significant increase in tissue density. Muscle atrophy or highly developed muscles similarly will decrease or increase tissue density.

Geometric unsharpness is influenced by which of the following? 1. Distance from object to image 2. Distance from source to object 3. Distance from source to image

1, 2, and 3 -Geometric unsharpness is affected by all three factors listed. As OID increases, so does magnification. OID is directly related to magnification; i.e. as OID increases, so does magnification. Focal-object distance and SID are inversely related to magnification. As focal-object distance and SID decrease, magnification increases.

If 300 mA has been selected for a particular exposure, what exposure time would be required to produce 60 mAs?

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

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

10 mAs, 74 kV, 36-in. SID -If (A) and (B) are reduced to 5 mAs for consistency, the kilovoltage will increase to 85 kV in both cases, thereby balancing receptor exposures. Thus, the greatest receptor exposure is determined by the shortest SID (greatest exposure rate).

An exposure was made using 600 mA, 0.04 sec exposure, and 85 kVp. Each of the following changes will serve to reduce the receptor exposure by one-half except change to?

18 mAs -Receptor exposure is directly proportional to milliampere-seconds. 600 mA with 0.04 sec = 24 mAs. It is desired to reduce receptor exposure to 12 mAs, or its near equivalent. If exposure time is halved from 0.04 sec to 0.02 (1/50) sec, receptor exposure will be cut in half. Changing to 300 mA also will halve the milliampere-seconds, effectively halving the receptor exposure. If the kilovoltage is decreased by 15%, from 85 to 72 kVp, receptor exposure will be halved according to the 15% rule. To cut the receptor exposure in half, the milliampere-seconds must be reduced to 12 mAs (not 18 mAs).

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

2 and 3 only -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.

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

2 and 3 only -If a fairly large patient is turned prone, the abdominal measurement will be significantly different from the AP measurement as a result of the effect of compression. Thus, the part is essentially "thinner," and less scattered radiation will be produced. If the patient remains supine and a compression band is applied, a similar effect will be produced. Beam restriction is probably the single most effective means of reducing the production of scattered radiation. Grid ratio affects the cleanup of scattered radiation; it has no effect on the production of scattered radiation

Central ray angulation may be required for 1. magnification of anatomic structures 2. foreshortening or self-superimposition 3. superimposition of overlying structures

2 and 3 only -Magnification is controlled by object-to- image-receptor distance (OID) and SID; it is unrelated to CR angulation.

Types of shape distortion include 1. magnification 2. elongation 3. foreshortening

2 and 3 only -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.

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

2 and 3 only -Technical factors can be expressed in terms of milliampere-seconds rather than milliamperes and time. The milliampere-seconds value is a quantitative factor because it regulates x-ray-beam intensity, exposure rate, quantity, or number of x-ray photons produced (the milliampere-seconds value is the single most important technical factor associated with receptor exposure and patient dose). The milliampere-seconds value is directly proportional to the intensity (i.e., exposure rate, number, and quantity) of x-ray photons produced and the resulting receptor exposure. If the milliampere-seconds value is doubled, twice the exposure rate and twice the receptor exposure occurs. If the milliampere-seconds value is cut in half, the receptor exposure and patient dose are cut in half. Kilovoltage is the qualitative exposure factor—it determines beam quality by regulating photon energy (i.e., wavelength).

In digital imaging, as the size of the image matrix increases, 1. FOV increases 2. pixel size decreases 3. spatial resolution increases

2 and 3 only -The FOV and matrix size are independent of one another; that is, either can be changed, and the other will remain unaffected. However, pixel size is affected by changes in either the FOV or matrix size. For example, if the matrix size is increased, pixel size decreases. If FOV increases, pixel size increases. Pixel size is inversely related to resolution. As pixel size decreases, resolution increases. FOV and matrix size are related to pixel size according to the equation Pixel size = FOV / Matrix

An incorrect relationship between the primary beam and the center of a focused grid results in 1. an increase in scattered radiation production 2. grid cutoff 3. insufficient receptor exposure

2 and 3 only -The lead strips of a focused grid are angled to correspond to the configuration of the divergent x-ray beam. Thus, any radiation that is changing direction, as is typical of scattered radiation, will be trapped by the lead foil strips. However, if the central ray and the grid center do not correspond, the lead strips will absorb useful radiation. The absorption of useful radiation is termed cutoff and results in diminished receptor exposure.

The term differential absorption is related to 1. beam intensity 2. subject contrast 3. pathology

2 and 3 only -The radiographic subject, the patient, is composed of many different tissue types of varying densities (i.e., subject contrast), resulting in varying degrees of photon attenuation and absorption. This differential absorption contributes to the various shades of gray in the x-ray image. Normal tissue density may be significantly altered in the presence of pathology. For example, destructive bone disease can cause a dramatic decrease in tissue density. Abnormal accumulation of fluid (as in ascites) will cause a significant increase in tissue density. Muscle atrophy or highly developed muscles similarly will decrease or increase tissue density

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

2 only

An increase in the kilovoltage applied to the x-ray tube increases the 1. x-ray wavelength 2. exposure rate 3. patient absorption

2 only

The use of which of the following is (are) essential in magnification radiography? 1. High-ratio grid 2. Fractional focal spot 3. Direct exposure technique

2 only -Magnification radiography is used to enlarge details to a more perceptible degree. Hairline fractures and minute blood vessels are candidates for magnification radiography. The problem of magnification unsharpness is overcome by using a fractional focal spot; larger focal-spot sizes will produce excessive blurring unsharpness. Grids are usually unnecessary in magnification radiography because of the air-gap effect produced by the OID. Direct-exposure technique probably would not be used because of the excessive exposure required.

Spatial resolution is inversely related to 1. SID 2. OID 3. grid ratio

2 only -SID is directly related to spatial resolution because as SID increases, so does resolution (because magnification is decreased). OID is inversely related to spatial resolution because as OID increases, spatial resolution decreases. Grid ratio is not associated with spatial resolution. Therefore, of the given choices, OID is inversely related to spatial resolution. SID is directly related to spatial resolution.

Using a 48-in. SID, how much OID must be introduced to magnify an object two times?

24-in. OID -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.

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

25 mA, 7/10 s, 70 kV -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.

An exposure was made using 600 mA and 18 ms. If the mA is changed to 400, which of the following exposure times would most closely approximate the original receptor exposure?

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

The attenuation of x-ray photons is not influenced by 1. pathology 2. effective atomic number 3. photon quantity

3 only -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.

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

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

If 0.05 second was selected for a particular exposure, what mA would be necessary to produce 15 mAs?

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

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

300 mA, 0.02 s, 72 kVp -IVU requires the use of iodinated contrast media. Low kilovoltage (about 70 kVp) is usually used to enhance the photoelectric effect and, in turn, to better visualize the renal collecting system. High kilovoltage will produce excessive scattered radiation and obviate the effect of the contrast agent. A higher milliamperage with a short exposure time generally is preferable.

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

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

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

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

The exposure factors of 300 mA, 0.017 second, and 72 kVp produce an mAs value of?

5 -To calculate mAs, multiply milliamperage times exposure time. In this case, 300 mA × 0.017 s = 5.10 mAs. Careful attention to proper decimal placement will help avoid basic math errors.

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

500 mA, 0.10 second, 90 kVp, 8:1 grid -Of the given factors, kilovoltage and grid ratio will have a significant effect on the scale of radiographic contrast. The mAs values are almost identical. Because an increased kilovoltage and low-ratio grid combination would allow the greatest amount of scattered radiation to reach the IR, thereby producing more gray tones, B is the best answer. Group D also uses a low-ratio grid, but the kilovoltage is too low to produce as many gray tones as B.

The exposure factors of 400 mA, 17 ms, and 82 kV produce a milliampere-seconds value of?

6.8 -To calculate milliampere-seconds, multiply milliamperage times exposure time. In this case, 400 mA × 0.017 second (17 ms) = 6.8 mAs. Careful attention to proper decimal placement will help to avoid basic math errors

If 300 mA has been selected for a particular exposure, what exposure time should be selected to produce 18 mAs?

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

If a duration of 0.05 second was selected for a particular exposure, what milliamperage would be necessary to produce 30 mAs?

600 -The formula for mAs is mA × s = mAs. Substituting known values: 0.05x = 30 x = 600 mA

If 92 kV and 15 mAs were used for a particular abdominal exposure with single-phase equipment, what milliampere-seconds value would be required to produce a similar radiograph with three-phase, 12-pulse equipment?

7.5 -Single-phase radiographic equipment is less efficient than three-phase equipment because it has a 100% voltage ripple. With three-phase equipment, voltage never drops to zero, and x-ray intensity is significantly greater. To produce similar receptor exposure, only two-thirds of the original milliampere-seconds would be used for three-phase, six-pulse equipment. With three-phase, 12-pulse equipment, the original milliampere-seconds would be cut in half (one-half of 15 mAs = 7.5).

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 longer SID should be used

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

A. 30 cm field -In magnification fluoroscopic imaging, the charge on the electrostatic focusing lenses is increased in order to confine electrons to a smaller portion of the input phosphor. This magnifies the image, but at the expense of less brightness. In order to increase brightness to a diagnostic level, the mA is increased. Smaller input phosphor field sizes (D) produce magnified images of the anatomical areas being evaluated, but with an increase in patient dose. Larger input phosphor field sizes (A) produce little or no magnification of the anatomical areas being evaluated, and with decreased patient dose.

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

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

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

A. Spatial resolution is directly related to SID -As the distance from focal spot to IR (SID) increases, so does spatial resolution. Because the part is being exposed by more perpendicular (less divergent) rays, less magnification and blur are produced. Although the best spatial resolution is obtained using a long SID, the necessary increase in exposure factors and resulting increased patient exposure become a problem. An optimal 40-in. SID is used for most radiography, with the major exception being chest examinations.

The fact that x-ray intensity across the primary beam can vary as much as 45% describes the?

Anode heel effect -A beveled focal track extends around the periphery of the anode disk; when a small angle is used, the beveled edge allows for a smaller effective focal spot and better detail. The disadvantage, however, is that photons are noticeably absorbed by the "heel" of the anode, resulting in a smaller percentage of x-ray photons at the anode end of the x-ray beam and a concentration of x-ray photons at the cathode end of the beam. This is known as the anode heel effect and can cause a primary beam variation of up to 45%. The anode heel effect becomes more pronounced as the SID decreases, as IR size increases, and as target angle decreases.

How is SID related to exposure rate and receptor exposure?

As SID increases, exposure rate decreases and radiographic receptor exposure decreases. -According to the inverse-square law of radiation, the intensity or exposure rate of radiation is inversely proportional to the square of the distance from its source. Thus, as distance from the source of radiation increases, exposure rate decreases. Because exposure rate and receptor exposure are directly proportional, if the exposure rate of a beam directed to an IR is decreased, the resulting receptor exposure would be decreased proportionately.

How is source-to-image distance (SID) related to exposure rate and receptor exposure?

As SID increases, exposure rate decreases and receptor exposure decreases

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. Intravenous urogram (IVU)

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

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

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

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

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

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

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

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

Which of the following absorbers has the highest attenuation coefficient?

Bone

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

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

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. Decreased milliampere-seconds, increased kilovoltage

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

C. IPs must exclude all white light -The image plate has a protective function for the flexible photostimulable storage phosphor within; it can be conveniently placed in a Bucky tray or under the anatomic part, and comes in a variety of sizes. The PSP within the IP is the image receptor/detector. IPs do not contain a lightsensitive material and, therefore, do not need to be light-tight. The photostimulable PSP is not affected by light

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

C. SID

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

C. Using less mAs and more kVp -As x-ray photons travel through a part, they either pass all the way through to expose the image receptor, or they undergo interaction(s) that may result in their being absorbed by the part or deviated in direction. It is those that change direction (scattered radiation) that undermine the image. With respect to the radiographic image, it is responsible for the scattered radiation that reaches the image receptor. Scattered radiation adds unwanted, degrading exposure to the radiographic image.

Because of the anode heel effect, the intensity of the x-ray beam is greatest along the?

Cathode end of the beam

Which of the following devices is used to overcome severe variation in patient anatomy or tissue density, providing more uniform radiographic density?

Compensating filter -A compensating filter is used when the part to be radiographed is of uneven thickness or tissue density (in the chest, mediastinum vs. lungs). The filter (made of aluminum or lead acrylic) is constructed in such a way that it will absorb much of the x-ray beam directed toward the low tissue-density area while not affecting the x-ray photons to directed toward the high tissue-density area.

What type of x-ray imaging uses an area beam and a photostimulable phosphor as the IR?

Computed Radiography

How would the introduction of a 6-in. OID affect image contrast?

Contrast would be increased -OID can affect contrast when it is used as an air gap. If a 6-in. air gap (OID) is introduced between the part and IR, much of the scattered radiation emitted from the body will not reach the IR

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. 2,048 x 2,048 -The matrix is the number of pixels in the xy direction. The larger the matrix size, the better is the image resolution.

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. 4-in. OID

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

D. 700 mA -Since 50 ms is equal to 0.050 s, and since mA × time mAs, the original milliampere-seconds value was 15 mAs. Now, it is only necessary to determine what milliamperage must be used with 22 ms to provide the same 15 mAs (and thus the same receptor exposure). Because mA × time = mAs

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

D. Ascites -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.

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

D. Field size

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

D. Grids -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.

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

D. SID

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

D. Spatial resolution

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

D. Three-phase equipment produces lower-contrast radiographs than single-phase equipment

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

D. mAs: 20; kV:80; Grid ratio 10:1; Field size: 8 x 10 -Review the groups of factors. First, because the milliampere-seconds value has no effect on the scale of contrast produced, eliminate milliampere-seconds from consideration by drawing a line through the column. Then check the two entries in each column that are likely to produce shorter-scale contrast. For example, in the kilovoltage column, because lower kilovoltage can produce shorter-scale contrast, place checkmarks next to the 70 and 80 kV. Because higher-ratio grids permit less scattered radiation to reach the IR, the 10:1 and 12:1 grids can produce a shorter scale of contrast than the lower-ratio grids; check them. As the volume of irradiated tissue decreases, so does the amount of scattered radiation produced, and consequently, the shorter is the scale of radiographic contrast; therefore, check the 11 × 14 and 8 × 10 in. field sizes. An overview shows that the factors in groups (A) and (C) have more checkmarks, than the factors in group (D), indicating that group (D) is more likely to produce the shortest-scale contrast.

What is the best way to reduce magnification distortion?

Decrease the OID

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?

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

Better resolution is obtained with?

High SNR

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

Off-focus -The lead strips in a focused grid are made to parallel the x-ray beam. Therefore, scattered radiation, which radiates in directions other than that of the primary beam, will be absorbed by the grid. When the x-ray beam does not parallel the lead strips, some type of grid cutoff occurs. If the x-ray beam is not centered to the grid, or if the x-ray tube and grid surface are not parallel (level), there will be a fairly uniform decrease in receptor exposure across the entire image.

Which of the following technical changes would best serve to remedy the effect of very dissimilar tissue densities?

High-kilovoltage exposure factors -When tissue densities within a part are very dissimilar (e.g., the chest), the radiographic result (especially analog) can be unacceptably high contrast. To "even out" these exposure values and produce a more appropriate scale of grays, exposure factors using high kilovoltage should be employed. The higher the grid ratio, the higher is the resulting contrast. Use of short exposure time is always encouraged to reduce the possibility of motion unsharpness but has no impact on varying tissue densities. Exposure factors using high milliampere-seconds generally result in excessive receptor exposure, frequently obliterating much of the gray scale.

A patient is being positioned for a particular radiographic examination. The x-ray tube, image recorder, and grid are properly aligned, but the body part is angled. Which of the following will result?

Image distortion

What should be done to correct for magnification when using air-gap technique?

Increase SID -OID is used to effect an increase in contrast in the absence of a grid, usually in chest radiography. If a 6-in. air gap (OID) is introduced between the part and the IR, much of the scattered radiation emitted from the body will not reach the IR; thus, the OID acts as a low-ratio grid and increases image contrast. However, the 6-in. OID air gap will make a very noticeable increase in magnification. To correct for this, the SID must be increased. Generally speaking, the SID needs to be increased 7 in. for every 1 in. of OID. With a 6-in. OID, the SID usually is increased from 6 to 10 ft (120 in.).

Using a short (25-30 in.) SID with a large (14 × 17 in.) IR is likely to?

Increase the anode heel effect -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.

When involuntary motion must be considered, the exposure time may be cut in half if the kilovoltage is?

Increased by 15%

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

Increased patient dose and decreased scattered radiation fog

Which of the following terms/units is used to express the resolution of a diagnostic image?

Line pairs per millimeter (lp/mm)

A focal-spot size of 0.3 mm or smaller is essential for which of the following procedures?

Magnification radiography -A fractional focal spot of 0.3 mm or smaller is essential for rendering fine detail without focal-spot blurring in magnification radiography. As the object image is magnified, so will be the associated blur unless the fractional focal spot is used

In which of the following examinations would a cassette front with very low absorption properties be especially desirable?

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

What is the single most important factor controlling size distortion?

OID

A radiograph exposed using a 12:1 ratio grid may exhibit a loss of receptor exposure at its lateral edges because the?

SID was too great -If the SID is above or below the recommended focusing distance, the primary beam at the lateral edges will not coincide with the angled lead strips. Consequently, there will be absorption of the useful beam, termed grid cutoff. If the grid failed to move during the exposure, there would be grid lines throughout. CR angulation in the direction of the lead strips is appropriate and will not cause grid cutoff. If the CR were off-center, there would be uniform loss of receptor exposure.

Geometric unsharpness is most likely to be greater?

at the cathode end of the image

If a 6-in. OID is introduced during a particular radiographic examination, what change in SID will be necessary to overcome objectionable magnification?

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

When the collimated field must extend past the edge of the body, allowing primary radiation to strike the tabletop, as in a lateral lumbar spine radiograph, what may be done to prevent excessive receptor exposure owing to undercutting?

Use lead rubber to absorb tabletop primary radiation

A decrease in kilovoltage will result in?

a decrease in receptor exposure

Decreasing field size from 14 × 17 in. to 8 × 10 in., with no other changes, will?

decrease receptor exposure and decrease the amount of scattered radiation generated within the part

If a radiograph, made using AEC, is overexposed because an exposure shorter than the minimum response time was required, the radiographer generally should?

decrease the milliamperage -The image was overexposed (from excessive exposure time) because the AEC wasn't capable of producing the required extremely short exposure time. To bring the required exposure to an exposure time the AEC is capable of, the mA should be decreased (thus requiring a longer exposure time, within the capability of the AEC).

A radiograph made with a parallel grid demonstrates decreased receptor exposure on its lateral edges. This is most likely due to?

decreased SID -The lead strips in a parallel grid are parallel to one another and, therefore, are not parallel to the x-ray beam. The more divergent the x-ray beam, the more likely there is to be cutoff/decreased receptor exposure at the lateral edges of the image. This problem becomes more pronounced at short SIDs. If there were a centering or tube angle problem, there would be more likely to be a noticeable receptor exposure loss on one side or the other.

A 15% increase in kVp accompanied by a 50% decrease in mAs will result in?

decreased patient dose

Radiographic contrast is the result of?

differential absorption

A compensating filter is used to?

even out widely differing tissue densities

The absorption of useful radiation by a grid is called

grid cutoff

A graphic diagram of signal values representing various absorption properties within the part being imaged is called a?

histogram -A histogram is a graph usually having several peaks and valleys representing the pixel values/absorbing properties of the various tissues, and so on that make up the imaged part. These various attenuators include such things as bone, muscle, air, contrast agents, foreign bodies, and pathology. The various pixel values, then, represent image contrast. If the histogram has a rather flat "tail," this represents underexposed areas at the periphery of the image, which can skew the overall histogram analysis.

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

increased receptor exposure. -More scattered radiation is generated within a part as the kilovoltage is increased, as the size of the field is increased, and as the thickness and density of tissue increases. As the quantity of scattered radiation increases from any of these sources, receptor exposure increases. Beam restriction does not impact spatial resolution.

Foreshortening of an anatomic structure means that?

it is projected on the IR smaller than its actual size

How are mAs and receptor exposure related in the process of image formation?

mAs and receptor exposure are directly proportional

To produce a just perceptible increase in receptor exposure, the radiographer should increase the?

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

A focal-spot size of 0.3 mm or smaller is essential for?

magnification radiography

Grid cutoff due to off-centering would result in?

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

The component of a CR image plate (IP) that records the radiologic image is the?

photostimulable phosphor

An increase in kilovoltage in analog imaging is most likely to?

produce a longer scale of contrast -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.

Combinations of milliamperage and exposure time that produce a particular milliampere-seconds value will produce identical receptor exposure. This statement is an expression of the?

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

A particular milliampere-seconds value, regardless of the combination of milliamperes and time, will reproduce the same receptor exposure. This is a statement of the?

reciprocity law -The reciprocity law states that a particular milliampere-seconds value, regardless of the milliamperage and exposure time used, will provide identical receptor exposure. Milliampere-seconds is directly proportional to beam intensity and receptor exposure

Recently, dual-sided reading technology has become available in more modern CR readers, in which two sets of photodetectors are used to capture light released from the front and back sides of the phosphor storage plate, or PSP (photostimulable phosphor). This technology enables improved:

signal-to-noise ratio

The movement of the IP through the transport system of a CR reader is referred to as the:

slow-scan direction -The IP moves slowly through the transport system of a CR reader and this movement is considered the slow-scan direction (A).

The line-focus principle expresses the relationship between?

the actual and the effective focal spot

Foreshortening can be caused by?

the radiographic object being placed at an angle to the IR

As grid ratio is decreased,

the scale of contrast becomes longer -Because lead content decreases when grid ratio decreases, a smaller amount of scattered radiation is trapped before reaching the IR. More grays, therefore, are recorded, and a longer scale of contrast results. Receptor exposure would increase with a decrease in grid ratio. Grid ratio is unrelated to distortion.

Focal-spot blur is greatest

toward the cathode end of the x-ray beam -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.

If a radiograph were made of an average-size knee using automatic exposure control (AEC) and all three photocells were selected, the resulting radiograph would demonstrate?

underexposed image -Proper functioning of the AEC depends on accurate positioning by the radiographer. The correct photocell(s) must be selected, and the anatomic part of interest must completely cover the photocell(s) to achieve the appropriate exposure. If a photocell is left uncovered, scattered radiation from the part being examined will cause premature termination of exposure and an underexposed radiograph.