NRRPT Prep Radiation
Fundamentals Study Guide Exam
Latest 2026/2027 Updated
Verified Questions and Solutions
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Question Number: 1
Fundamentals of Radiation Protection
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
- ANS - 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.
Question Number: 2
,Fundamentals of Radiation Protection
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
D) 3.5 mCi
E) 4.0 mCi
- ANS - 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.
Question Number: 3
Fundamentals of Radiation Protection
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
- ANS - 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.
Question Number: 4
Fundamentals of Radiation Protection
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
- ANS - 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
Question Number: 5
Fundamentals of Radiation Protection
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
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
- ANS - 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.
Fundamentals Study Guide Exam
Latest 2026/2027 Updated
Verified Questions and Solutions
A+ Pass Guaranteed
Question Number: 1
Fundamentals of Radiation Protection
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
- ANS - 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.
Question Number: 2
,Fundamentals of Radiation Protection
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
D) 3.5 mCi
E) 4.0 mCi
- ANS - 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.
Question Number: 3
Fundamentals of Radiation Protection
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
- ANS - 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.
Question Number: 4
Fundamentals of Radiation Protection
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
- ANS - 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
Question Number: 5
Fundamentals of Radiation Protection
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
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
- ANS - 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.
