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1. The minimum requirement for lead-equivalent content in protective aprons is

0.05 mm Pb.

0.50 mm Pb.

0.25 mm Pb.

1.0 mm Pb.
•
•
• 0.25 mm Pb.
2. How much protection is provided from a 100-kVp x-ray beam when using a 0.50-mm lead-equivalent apron?

40%

75%

88%

99%
•
•
• 75%
3. 1.0-mm lead equivalent apron will provide close to ____% protection at most kilovoltage levels, but it is rarely used because it weighs anywhere from 12 to 24 lb.
100
4. A 0.25-mm lead-equivalent apron will attenuate about ___% of a 50-kVp x-ray beam
97
5.  A 0.25-mm lead-equivalent apron will attenuate about ____% of a 75-kVp x-ray beam
66
6.  m A 0.5-mm apron will attenuate about ___% of a 50-kVp beam,
99%
7. A 0.5-mm apron will attenuate about ___% of a 75-kVp beam,
88
8. A 0.5-mm apron will attenuate about __% of a 100-kVp beam,
75
9. According to NCRP regulations, leakage radiation from the x-ray tube must not exceed

10 mR/h

100 mR/h

10 mR/min

100 mR/min
•
•
• 100 mR/h
10. If the ESE for a particular exposure is 25 mrad, what will be the intensity of the scattered beam perpendicular to and 1 m from the patient?

•
•
• In general, at 1 m from the patient, the intensity is reduced by a factor of 1,000 to about 0.1% of the original intensity. Successive scatterings can reduce the intensity to unimportant levels. Calculate that 0.1% of 25 mrad is 0.025 mrad.
11. If an exposure dose of 50 mR/h is delivered from a distance of 3 ft, what would be the dose delivered after 20 minutes at a distance of 5 ft from the source?

6 mR

18 mR

46 mR

138 mR
•
•
• 6 mR
• inverse-square law of radiation. The formula is
• Substituting known values:

• Thus, x = 18 mR/h (60 minutes) and, therefore, 6 mR in 20 minutes. Distance has a profound effect on dose received and, therefore, is one of the cardinal rules of radiation protection. As distance from the source increasesdose received decreases.
12. f the exposure rate to a body standing 7 ft from a radiation source is 140 mR/h, what will be the dose to that body at a distance of 8 ft from the source in 30 minutes?

182.8 mR

107 mR

91.4 mR

53.6 mR
•
•
• 53.6 mR
13. If the exposure rate at 2.0 m from a source of radiation is 18 mR/min, what will be the exposure rate at 5 m from the source?

2.8 mR/min

4.5 mR/min

18 mR/min

85 mR/min
•
•
• 2.8 mR/min
14. A time of 1.5 minutes is required for a particular fluoroscopic examination, whose exposure rate is 275 mR/h. What is the approximate radiation exposure for the radiologic staff present in the fluoroscopy room during the examination?

183 mR

68.7 mR

18.33 mR

6.87 mR
•
•
• 6.87 mR
• If the exposure rate for the examination is 250 mR/h (60 minutes), then a 3-minute examination would be proportionally less—as the following equation illustrates:
• Thus, x = 6.87-mR dose in 1.5 minutes.
15. If the exposure rate at 3 ft from the fluoroscopic table is 40 mR/h, what will be the exposure rate for 30 minutes at a distance of 5 ft from the table?

7 mR

12 mR

14 mR

24 mR
•
•
• 7 mR
16. If the exposure rate to a body standing 3 feet from a radiation source is 12 mR/min, what will be the exposure rate to that body at a distance of 7 feet from the source?

2.2 mR/min

5.1 mR/min

28 mR/min

36 mR/min
•
•
• 2.2 mR/min
17. What percentage of x-ray attenuation does a 0.5-mm lead-equivalent apron at 75 kVp provide?

51%

66%

75%

88%
•
•
• 88%
18. controlled area is defined as one

1. that is occupied by people trained in radiation safety

2. that is occupied by people who wear radiation monitors

3. whose occupancy factor is 1

1 and 2 only

2 only

1 and 3 only

1, 2, and 3
•
•
• 1, 2, and 3
19. In the production of characteristic radiation at the tungsten target, the incident electron

ejects an inner-shell tungsten electron

ejects an outer-shell tungsten electron

is deflected, with resulting energy loss

is deflected, with resulting energy gain
•
•
• ejects an inner-shell tungsten electron
20. The exposure rate to a body 4 ft from a source of radiation is 16 R/h. What distance from the source would be necessary to decrease the exposure to 6 R/h?

5 ft

7 ft

10 ft

14 ft
•
•
• 7 ft
• inverse-square law of radiation. The formula is
• Substituting known values:

• Thus, x = 6.5 ft (necessary to decrease the exposure to 6 R/h). Note that in order for the exposure rate to decrease, the distance from the source of radiation must increase. (Bushong,8th ed.p. 68)
21. What percentage of x-ray attenuation does a 0.5-mm lead equivalent apron at 100 kVp provide?

51%

66%

75%

94%
•
•
• 75%
22. How much protection is provided from a 75-kVp x-ray beam when using a 0.50-mm lead equivalent apron?

51%

66%

88%

99%
•
•
• 88%
23. What is the intensity of scattered radiation perpendicular to and 1 m from a patient compared with the useful beam at the patient's surface?

0.01%

0.1%

1.0%

10.0%
•
• 0.1%
24. If an individual receives an exposure of 150 mR/h at a distance of 2 feet from a radiation source, what will be their dose after 30 minutes at a distance of 5 feet from the source?

60 mR

30 mR

24 mR

12 mR
•
•
• 12 mR
25. Protective devices such as lead aprons function to protect the user from

2. the primary beam

1 only

1 and 2 only

1 and 3 only

1, 2, and 3
•
• 1 only
 Author: Anonymous ID: 302821 Card Set: Radiation Protectiom Updated: 2015-05-17 00:00:00 Tags: rad Folders: n Description: Personnel protection Show Answers: