Chapter 6 - 1

Question 1

1. The product of milliamperage and exposure time is ________________ to the quantity of x-rays produced

a. directly proportional
b. inversely proportional
c. not related

Answer 1

a. directly proportional

The product of milliamperage and exposure time is directly proportional to the quantity of x-rays produced.
REF: p.148

Question 2

2. As mAs increases, the quantity of radiation reaching the IR ________________.

a. increases
b. decreases
c. is not affected

Answer 2

a. increases

As mAs increases, the quantity of radiation reaching the IR increases.
REF: p.149

Question 3

3. If you were to select 250 mA and 0.5 s on the control panel, how much mAs would be created?

a. 62.5 mAs
b. 250 mAs
c. 125 mAs
d. 500 mAs

Answer 3

c. 125 mAs

mAs is calculated by multiplying mA by time in seconds. In this case, 250 × 0.5 = 125 mAs.
REF: p.149

Question 4

4. What effect does the level of mAs have on image brightness when using digital image receptors?

a. Increases image brightness.
b. Decreases image brightness.
c. Does not directly affect image brightness.

Answer 4

c. Does not directly affect image brightness.

The level of mAs does not directly affect image brightness when using digital IRs. During computer processing, image brightness is maintained when the mAs is too low or too high.
REF: p.150

Question 5

5. Increased quantum noise is seen in a digital image with:

a. severely lower-than-needed mAs.
b. a little lower-than-needed mAs.
c. a little higher-than-needed mAs.
d. severely higher-than-needed mAs.

Answer 5

a. severely lower-than-needed mAs.

Severely lower-than-needed mAs produces an image with increased quantum noise, and higher-than-needed mAs exposes a patient to unnecessary radiation.
REF: p.152

Question 6

6. The numerical value that is displayed on the processed image to indicate the level of x-ray exposure received on the digital image receptor is the ____________.

a. image brightness
b. exposure indicator
c. accession number
d. optical density

Answer 6

b. exposure indicator

The numerical value that is displayed on the processed image to indicate the level of x-ray exposure received on the digital image receptor is the exposure indicator.
REF: p.150

Question 7

7. With film-screen imaging, which of the following technique factors directly controls the density produced on the image?

a. mAs
b. kVp
c. SID
d. Central ray angle

Answer 7

a. mAs

With film-screen imaging, mAs directly controls the density produced on the image.
REF: p.151

Question 8

8. With film-screen imaging, when mAs is increased, density is ____________; when mAs is decreased, density is ____________.

a. increased; increased
b. increased; decreased
c. decreased; decreased
d. decreased; increased

Answer 8

b. increased; decreased

With film-screen imaging, when mAs is increased, density is increased; when mAs is decreased, density is decreased.
REF: p.151

Question 9

9. When a film image is too light, it has ____________ density, and would require a(n) ____________ in mAs by 2 on repeat to obtain a diagnostic image.

a. insufficient; increase
b. insufficient; decrease
c. excessive; decrease
d. excessive; increase

Answer 9

a. insufficient; increase

When a film image is too light, it has insufficient density, and would require an increase in mAs by 2 on repeat to obtain a diagnostic image.
REF: p.151

Question 10

10. With film-screen imaging, what is the minimum change in mAs that would result in a visible change in radiographic density?

a. 10%
b. 15%
c. 30%
d. 100%

Answer 10

c. 30%

A 25% to 30% adjustment in mAs is the minimum needed to demonstrate a visible change in density.
REF: p.151

Question 11

11. If 20 mAs was used for all exposures, which kVp would produce the film-screen image with the greatest density?

a. 70 kVp
b. 80 kVp
c. 90 kVp
d. 100 kVp

Answer 11

d. 100 kVp

The higher the kVp, the greater amount of radiation reaching the image receptor and therefore, with film-screen imaging, the greater density.
REF: p.152

Question 12

12. Which of the following technical factors affects the exposure to the IR by altering the amount and penetrating ability of the x-ray beam?

a. kVp
b. mA
c. Seconds
d. SID

Answer 12

a. kVp

kVp affects the exposure to the IR by altering the amount and penetrating ability of the x-ray beam.
REF: p.152

Question 13

13. With film-screen imaging, what is the primary controller of image contrast?

a. kVp
b. mA
c. Seconds
d. SID

Answer 13

a. kVp

kVp, by controlling the penetrating power of the x-ray beam, is considered the primary controller of film-screen radiographic contrast.
REF: p.152

Question 14

14. Increasing the kVp will __________ the IR exposure and density produced on the film image.

a. increase
b. decrease
c. not affect

Answer 14

a. increase

Increasing the kVp increases the IR exposure and density produced on a film image, and decreasing the kVp decreases the IR exposure and density produced on a film image.
REF: p.153

Question 15

15. With film-screen imaging, if the kVp were decreased by 15% and no changes were made to the mAs:

a. density would be unchanged, and contrast would decrease.
b. density would increase, and contrast would decrease.
c. density would decrease, and contrast would increase.
d. there would be no change in either contrast or density.

Answer 15

c. density would decrease, and contrast would increase.

Decreasing kVp results in photons with lower penetrating power. Fewer photons are transmitted, resulting in decreased density, and even fewer penetrate the more dense parts, resulting in a film-screen image with increased contrast.
REF: p.154

Question 16

16. In order to reduce patient exposure, ______ kVp and ______ mAs should be used when possible.

a. lower, higher
b. higher, higher
c. higher, lower
d. It makes no difference.

Answer 16

c. higher, lower

Using a higher kVp and lower mAs is best, because the higher kVp provides more penetration, requiring less patient exposure.
REF: p.155

Question 17

17. With film-screen imaging, what would be the appropriate change in mAs if the kVp were decreased by 15% and the density needed to be maintained?

a. Double the mAs.
b. Halve the mAs.
c. Use one fourth of the mAs.
d. No change would be necessary.

Answer 17

a. Double the mAs.

Decreasing the kVp by 15% would require two times the mAs in order to maintain film-screen image density.
REF: p.154

Question 18

18. With film-screen imaging, which of the following technique factors will produce an image with the greatest density?

a. 300 mA, 0.2 s
b. 400 mA, 0.2 s
c. 100 mA, 1 s
d. 100 mA, 0.001 s

Answer 18

c. 100 mA, 1 s

mA and exposure time, also expressed as their product (mAs), control radiographic density. 100 mA × 1 s (100 mAs) is the highest amount listed and will produce the greatest density.
REF: p.149

Question 19

19. With film-screen imaging, which of the following technique factors will produce an image with the greatest density?

a. 300 mA, 0.5 s
b. 400 mA, 0.5 s
c. 100 mA, 0.5 s
d. 800 mA, 0.01 s

Answer 19

b. 400 mA, 0.5 s

mA and exposure time, also expressed as their product (mAs), control radiographic density. 400 mA × 0.5 s (200 mAs) is the highest amount listed and will produce the greatest density.
REF: p.149

Question 20

20. When a film-screen image needs to be repeated because it is too dark, the minimum change in mAs needed is:

a. reduce the mAs by 30%.
b. reduce the mAs by 50%.
c. increase the mAs by 30%.
d. increase the mAs by a factor of two.

Answer 20

b. reduce the mAs by 50%.

When the image density is so far off that the image must be repeated, the minimum amount of change needed is to either double or halve the mAs.
REF: p.151

Question 21

21. With film-screen imaging, an abdomen image is done using 60 kVp and 30 mAs. Image density is appropriate, but the image has too high (short scale) contrast. Which of these exposure factors would be the best change to make?

a. 51 kVp and 60 mAs
b. 51 kVp and 30 mAs
c. 69 kVp and 30 mAs
d. 69 kVp and 15 mAs

Answer 21

d. 69 kVp and 15 mAs

In order to lower the contrast, a higher kVp is needed. However, because the density was appropriate, the mAs has to be reduced to maintain image density.
REF: p.157

Question 22

22. A high kVp results in:

a. less absorption and more transmission in anatomic tissues.
b. less variation in the x-ray intensities exiting the patient.
c. increase interactions from Compton scattering.
d. low-contrast image.
e. all of the above.

Answer 22

e. all of the above.

A high kVp results in less absorption and more transmission in anatomic tissues, which results in less variation in the x-ray intensities exiting the patient (lower subject contrast), producing a low-contrast image. High kVp results in a greater proportion of interactions from Compton scattering than x-ray absorption (photoelectric effect).
REF: p.157

Question 23

23. During selection of the focal spot size, the radiographer is really determining the:

a. angle of the anode used.
b. actual size of the filament used.
c. energy of electrons available for tube current.
d. distance the electrons travel from cathode to anode.

Answer 23

b. actual size of the filament used.

When selecting large or small focal spot at the console, what’s really being selected is the large or small cathode filament.
REF: p.158

Question 24

24. The distance between the radiation source and the image receptor is the:

a. magnification factor.
b. SID.
c. OID.
d. SOD.

Answer 24

b. SID.

The distance between the radiation source and the image receptor is the SID, or source to image receptor distance.
REF: p.158

Question 25

25. The relationship between distance and x-ray beam intensity, specifically that the intensity of the x-ray beam is inversely proportional to the square of the distance from the source, is the: a. direct square law. b. Gurney-Mott theory. c. inverse square law. d. magnification law.

Answer 25

c. inverse square law. The relationship between distance and x-ray beam intensity, specifically that the intensity of the x-ray beam is inversely proportional to the square of the distance from the source, is the inverse square law. REF: p.158

Question 26

26. If a person stands 3 feet from the source of exposure, receives an exposure of 160 mR, and then moves to 6 feet from the source of exposure, what would be the new exposure according to the inverse square law? a. 10 mR b. 40 mR c. 640 mR d. 2560 mR

Answer 26

b. 40 mR According to the inverse square law, increasing the distance from the x-ray source from 3 feet to 6 feet will result in a reduction of exposure from 160 to 40 mR. REF: p.158

Question 27

27. If a person stands 12 feet from the source of exposure, receives an exposure of 60 mR, and then moves to 3 feet from the source of exposure, what would be the new exposure according to the inverse square law? a. 15 mR b. 45 mR c. 240 mR d. 960 mR

Answer 27

d. 960 mR According to the inverse square law, decreasing the distance from the x-ray source from 12 to 3 feet (factor of four) will result in an increase of exposure from 60 to 960 mR. REF: p.158

Question 28

28. The distance between the object being radiographed and the IR is the: a. magnification factor. b. SID. c. OID. d. SOD.

Answer 28

c. OID. The distance between the object being radiographed and the IR is the OID or object to image distance. REF: p.162

Question 29

29. When the SID is divided by the SOD, what is the result called? a. Magnification factor b. Degree of unsharpness c. OID d. Image width

Answer 29

a. Magnification factor The magnification factor formula is SID divided by SOD. REF: p.163

Question 30

30. If the first radiograph of a chest is done using 72 in and 12 mAs, and a second radiograph is done using 40 in, how much mAs should be used to maintain exposure to the IR? a. 2 mAs b. 4 mAs c. 7 mAs d. 9 mAs

Answer 30

b. 4 mAs According to the mAs/distance compensation formula, decreasing the distance from 72 to 40 will require a decrease of mAs from 12 to 4 in order to maintain exposure to the IR. REF: p.160

Question 31

31. A quality radiograph is done using 10 mAs, 70 kVp, and a 12:1 ratio grid. How much mAs is needed to produce an image with the same exposure to the IR when the grid is removed? a. 2 mAs b. 5 mAs c. 15 mAs d. 50 mAs

Answer 31

a. 2 mAs According to the grid conversion chart, when a 12:1 grid is removed, only one fifth of the mAs is needed. REF: p.169

Question 32

32. The misrepresentation of the size of an object is: a. shape distortion. b. magnification. c. foreshortening. d. elongation.

Answer 32

b. magnification. One aspect of distortion is magnification, or size distortion. REF: p.161

Question 33

33. In order to image a structure that is located anteriorly in the body, it is best radiographed to minimize magnification by doing a(n) __________ projection. a. posterior-anterior b. anterior-posterior c. lateral d. oblique

Answer 33

a. posterior-anterior The posterior-anterior (PA) projection will put the anterior surface of the body closest to the image receptor, reducing OID and magnification. REF: p.164

Question 34

34. Magnification is affected by: a. focal spot size. b. OID. c. SID. d. B and C.

Answer 34

d. B and C. Both the OID and SID affect magnification, although OID has the greatest effect. REF: p.165