NRRPT PREP COMPREHENSIVE
ASSESSMENT EXAM VERIFIED
QUESTIONS AND ANSWERS 100%
CORRECT – LATEST 2025 /2026- GRADE A+
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 - 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.
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 - answer..The correct answer is: B
After 10 half lives, the remaining activity is approximately
1000th of the original amount. Therefore, if there was 1µCi 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 - answer..The correct answer is: D
dps = (2000 pCi)(1 x 10-12Ci/pi )(3.7x1010 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: 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 - 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.
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 - 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.
, 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 - answer..The correct answer is: D
dps = (2000 pCi)(1 x 10-12Ci/pi )(3.7x1010 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
ASSESSMENT EXAM VERIFIED
QUESTIONS AND ANSWERS 100%
CORRECT – LATEST 2025 /2026- GRADE A+
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 - 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.
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 - answer..The correct answer is: B
After 10 half lives, the remaining activity is approximately
1000th of the original amount. Therefore, if there was 1µCi 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 - answer..The correct answer is: D
dps = (2000 pCi)(1 x 10-12Ci/pi )(3.7x1010 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: 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 - 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.
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 - 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.
, 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 - answer..The correct answer is: D
dps = (2000 pCi)(1 x 10-12Ci/pi )(3.7x1010 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
