NRRPT PREP COMPLETE STUDYGUIDE EXAM WITH CORRECT
ACTUAL QUESTIONS AND CORRECTLY WELL DEFINED ANSWERS
LATEST ALREADY GRADED A+
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
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
ACTUAL QUESTIONS AND CORRECTLY WELL DEFINED ANSWERS
LATEST ALREADY GRADED A+
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
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
