Page 1 of 187
“NRRPT PREP RADIATION FUNDAMENTALS 2026
”LATEST EXAM 2026 – 2027 SOLVED QUESTIONS &
ANSWERS VERIFIED 100% GRADED A+ (LATEST
VERSION) WELL REVISED 100% GUARANTEE PASS
NRRPT prep Radiation Fundamentals
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
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
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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
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
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.
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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
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
The correct answer is: A
By definition, 1 mCi = 3.7 x 10^7 dps. Dps stands for disintegrations per second.
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Therefore, if one mCi of tritium is ingested, the number of disintegrations per second
must be 3.7 x 10^7.
Question Number: 6
Fundamentals of Radiation Protection
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
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
Where D = absorbed dose rate
Question Number: 7
Fundamentals of Radiation Protection
Conjunctivitis may result from a welding arc due to:
A) intense visible light radiation.
B) UV radiation.
C) IR radiation.
D) soft x-ray radiation.
E) spark.
The correct answer is: B
“NRRPT PREP RADIATION FUNDAMENTALS 2026
”LATEST EXAM 2026 – 2027 SOLVED QUESTIONS &
ANSWERS VERIFIED 100% GRADED A+ (LATEST
VERSION) WELL REVISED 100% GUARANTEE PASS
NRRPT prep Radiation Fundamentals
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
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
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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
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
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.
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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
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
The correct answer is: A
By definition, 1 mCi = 3.7 x 10^7 dps. Dps stands for disintegrations per second.
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Therefore, if one mCi of tritium is ingested, the number of disintegrations per second
must be 3.7 x 10^7.
Question Number: 6
Fundamentals of Radiation Protection
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
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
Where D = absorbed dose rate
Question Number: 7
Fundamentals of Radiation Protection
Conjunctivitis may result from a welding arc due to:
A) intense visible light radiation.
B) UV radiation.
C) IR radiation.
D) soft x-ray radiation.
E) spark.
The correct answer is: B
