Comprehensive Exam Study Guide on
NRRPT Prep Complete Study Guide
Exam Latest Updated 2026/2027
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 - ansThe 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 - ansThe 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 - ansThe correct answer is: D
,Comprehensive Exam Study Guide on
NRRPT Prep Complete Study Guide
Exam Latest Updated 2026/2027
Verified Questions and Solutions A+
Pass Guaranteed
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 - ansThe 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 - ansThe 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
,Comprehensive Exam Study Guide on
NRRPT Prep Complete Study Guide
Exam Latest Updated 2026/2027
Verified Questions and Solutions A+
Pass Guaranteed
disintegrations per minute.
A) 370 dpm
B) 900 dpm
C) 3770 dpm
D) 4440 dpm
E) 5320 dpm - ansThe 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
E) 5 - ansThe 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?
7
A) 3.7 x 10 dps
B) 2.5 x 103 dps
C) 1.7 x 108 dps
, Comprehensive Exam Study Guide on
NRRPT Prep Complete Study Guide
Exam Latest Updated 2026/2027
Verified Questions and Solutions A+
Pass Guaranteed
D) 2.2 x 106 dps
E) 3.7 x 1010 dps - ansThe 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.
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 - ansThe 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. - ansThe correct answer is: B
The wavelengths responsible for conjunctivitis are 270-280 nm in the ultraviolet area of the
electromagnetic spectrum.
Question Number: 8
Fundamentals of Radiation Protection
Eight curies of tritium has a disintegration rate of: A) 12.5 x 104 dps
B) 2.96 x 1011 dps
NRRPT Prep Complete Study Guide
Exam Latest Updated 2026/2027
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 - ansThe 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 - ansThe 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 - ansThe correct answer is: D
,Comprehensive Exam Study Guide on
NRRPT Prep Complete Study Guide
Exam Latest Updated 2026/2027
Verified Questions and Solutions A+
Pass Guaranteed
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 - ansThe 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 - ansThe 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
,Comprehensive Exam Study Guide on
NRRPT Prep Complete Study Guide
Exam Latest Updated 2026/2027
Verified Questions and Solutions A+
Pass Guaranteed
disintegrations per minute.
A) 370 dpm
B) 900 dpm
C) 3770 dpm
D) 4440 dpm
E) 5320 dpm - ansThe 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
E) 5 - ansThe 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?
7
A) 3.7 x 10 dps
B) 2.5 x 103 dps
C) 1.7 x 108 dps
, Comprehensive Exam Study Guide on
NRRPT Prep Complete Study Guide
Exam Latest Updated 2026/2027
Verified Questions and Solutions A+
Pass Guaranteed
D) 2.2 x 106 dps
E) 3.7 x 1010 dps - ansThe 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.
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 - ansThe 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. - ansThe correct answer is: B
The wavelengths responsible for conjunctivitis are 270-280 nm in the ultraviolet area of the
electromagnetic spectrum.
Question Number: 8
Fundamentals of Radiation Protection
Eight curies of tritium has a disintegration rate of: A) 12.5 x 104 dps
B) 2.96 x 1011 dps
