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NRRPT PREP RADIATION PRACTICE
QUESTIONS AND QUESTIONS NEWLY
MODIFIED 2025/2026 NEW UPDATE
A sample of radioactive material is reported to contain 2000 picocuries of activity. Express
this value as disintegrations per minute.
A) 370 dpm
B) 900 dpm
C) 3770 dpm
D) 4440 dpm
E) 5320 dpm -- ANSWER--The correct answer is: D
dps = (2000 pCi)(1 x 10^-12Ci/pi )(3.7x10^10 dps/Ci) dps = 74
dpm = (74 dps) (60 sec/min) dpm = 4440
Remember to convert to disintegrations per minute not DISINTEGRATIONS PER SECOND.
A worker accidentally ingested one mCi of tritium. Tritium has a half life of 12 years. The
number of disintegrations per second in the worker's body is which of the following?
A) 3.7 x 10^7dps
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B) 2.5 x 10^3 dps
C) 1.7 x 10^8 dps
D) 2.2 x 10^6 dps
E) 3.7 x 10^10 dps -- ANSWER--The correct answer is: A
By definition, 1 mCi = 3.7 x 10^7 dps. Dps stands for disintegrations per second. Therefore,
if one mCi of tritium is ingested, the number of disintegrations per second must be 3.7 x
10^7.
Calculate the absorbed dose rate produced in bone (f = 0.922) by a 1MeV gamma radiation
source which produced an exposure rate of 0.5mr/hr.
A) 0.37 mrads/hr
B) 0.4 mrads/hr
C) 0.32 mrads/hr
D) 0.004 mrads/hr
E) 0.002 mrads/hr -- ANSWER--The correct answer is: B
D = 0.87 * f * X (in rads)
= 0.869 * 0.922 * 5 x 10^-3rads/hr
= 0.4 x 10^-3rads/hr
= 0.4 mrads/hr
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Where D = absorbed dose rate
A low energy alpha detector is usually effective if the detector is distant from the source.
A) 1/4 inch
B) 1/2 inch
C) 1 inch
D) 1 1/2 inches
E) 2 inches -- ANSWER--The correct answer is: A
Low energy alpha particles can only travel less than 1/2 inch in air. Therefore, one must be
closer than this to detect them.
A sample of I-131 (half life = 8 days) is kept for 80 days, at which time the activity is 1 µCi .
What was the original activity?
A) 2.0 mCi
B) 1.0 mCi
C) 1.5 mCi
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D) 3.5 mCi
E) 4.0 mCi -- ANSWER--The correct answer is: B
After 10 half lives, the remaining activity is approximately 1000th of the original amount.
Therefore, if there was 1 µCi left after 10 half lives, then there must have been 1000 times
more to start with. Hence, 1 µCi * 1000 = 1 mCi.
A sample of wood from an ancient forest showed 93.75% of the Carbon-14 decayed. How
many half lives did the carbon go through?
A) 1
B) 2
C) 3
D) 4
E) 5 -- ANSWER--The correct answer is: D
1 half life = 50% remaining
2 half lives = 25% remaining
3 half lives = 12.5% remaining 4 half lives = 6.25% remaining
Conjunctivitis may result from a welding arc due to: A)
intense visible light radiation.
B) UV radiation.
4
NRRPT PREP RADIATION PRACTICE
QUESTIONS AND QUESTIONS NEWLY
MODIFIED 2025/2026 NEW UPDATE
A sample of radioactive material is reported to contain 2000 picocuries of activity. Express
this value as disintegrations per minute.
A) 370 dpm
B) 900 dpm
C) 3770 dpm
D) 4440 dpm
E) 5320 dpm -- ANSWER--The correct answer is: D
dps = (2000 pCi)(1 x 10^-12Ci/pi )(3.7x10^10 dps/Ci) dps = 74
dpm = (74 dps) (60 sec/min) dpm = 4440
Remember to convert to disintegrations per minute not DISINTEGRATIONS PER SECOND.
A worker accidentally ingested one mCi of tritium. Tritium has a half life of 12 years. The
number of disintegrations per second in the worker's body is which of the following?
A) 3.7 x 10^7dps
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B) 2.5 x 10^3 dps
C) 1.7 x 10^8 dps
D) 2.2 x 10^6 dps
E) 3.7 x 10^10 dps -- ANSWER--The correct answer is: A
By definition, 1 mCi = 3.7 x 10^7 dps. Dps stands for disintegrations per second. Therefore,
if one mCi of tritium is ingested, the number of disintegrations per second must be 3.7 x
10^7.
Calculate the absorbed dose rate produced in bone (f = 0.922) by a 1MeV gamma radiation
source which produced an exposure rate of 0.5mr/hr.
A) 0.37 mrads/hr
B) 0.4 mrads/hr
C) 0.32 mrads/hr
D) 0.004 mrads/hr
E) 0.002 mrads/hr -- ANSWER--The correct answer is: B
D = 0.87 * f * X (in rads)
= 0.869 * 0.922 * 5 x 10^-3rads/hr
= 0.4 x 10^-3rads/hr
= 0.4 mrads/hr
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Where D = absorbed dose rate
A low energy alpha detector is usually effective if the detector is distant from the source.
A) 1/4 inch
B) 1/2 inch
C) 1 inch
D) 1 1/2 inches
E) 2 inches -- ANSWER--The correct answer is: A
Low energy alpha particles can only travel less than 1/2 inch in air. Therefore, one must be
closer than this to detect them.
A sample of I-131 (half life = 8 days) is kept for 80 days, at which time the activity is 1 µCi .
What was the original activity?
A) 2.0 mCi
B) 1.0 mCi
C) 1.5 mCi
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D) 3.5 mCi
E) 4.0 mCi -- ANSWER--The correct answer is: B
After 10 half lives, the remaining activity is approximately 1000th of the original amount.
Therefore, if there was 1 µCi left after 10 half lives, then there must have been 1000 times
more to start with. Hence, 1 µCi * 1000 = 1 mCi.
A sample of wood from an ancient forest showed 93.75% of the Carbon-14 decayed. How
many half lives did the carbon go through?
A) 1
B) 2
C) 3
D) 4
E) 5 -- ANSWER--The correct answer is: D
1 half life = 50% remaining
2 half lives = 25% remaining
3 half lives = 12.5% remaining 4 half lives = 6.25% remaining
Conjunctivitis may result from a welding arc due to: A)
intense visible light radiation.
B) UV radiation.
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