1|Page
NRRPT EXAM AND PRACTICE EXAM NEWEST 2026
TEST BANK| NRRPT EXAM PREP WITH COMPLETE
1000+ REAL EXAM QUESTIONS AND CORRECT
DETAILED ANSWERS (VERIFIED ANSWERS)
ALREADY GRADED A+ (BRAND NEW!!)
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 - Correct Answer - D
dps = (2000 pCi)(1 x 10-12Ci/pi )(3.7x1010 dps/Ci) dps = 74
dpm = (74 dps) (60 sec/min)
dpm = 4440
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 - Correct Answer - D
1 half life = 50% remaining
2 half lives = 25% remaining
3 half lives = 12.5% remaining
pg. 1
,2|Page
4 half lives = 6.25% remaining
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 - Correct Answer - 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
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. - Correct answer - B
The wavelengths responsible for conjunctivitis are 270-280 nm in the
ultraviolet area of the electromagnetic spectrum.
Eventually, charged particles give up their energy to the surrounding
medium. In the case of the alpha particle, it becomes (a):
A) Proton
B) Neutron
pg. 2
,3|Page
C) Tritium
D) Helium - Correct answer – D
An alpha particle consists of 2 neutrons and 2 protons carrying two
positive charges. It abstracts two electrons from the surrounding atoms
and becomes a helium atom.
Gamma rays get their energy from:
A) electrons outside the nucleus.
B) nuclear disintegration.
C) cosmic rays that change as they enter the atmosphere.
D) high energy meson particles. - Correct Answer - B
Gamma rays are similar to x-rays but differ in origin and wavelength.
Gamma rays get their energy from nuclear disintegration while x-rays
are produced from dislodging inner electrons.
Given a reading of 100mr/hr, gamma, at 10 feet, what would be the
reading at 2 feet, assuming a point source geometry?
A) 6400 mr/hr
B) 2500 mr/hr
C) 5000 mr/hr
D) 3000 mr/hr - Correct Answer - B
Apply the inverse square law: (r2/r1) = (d1/d2)2 r2 = (r1)(d1/d2)2
r2 = (100 mr/hr)(10 ft/2 ft)2 r2 = 2500 mr/hr
r1 = exposure rate 1 r2 = exposure rate 2 d1= distance 1
d2 = distance 2
pg. 3
, 4|Page
What is the usual unit of measurement for laser radiation?
A) J/cm2 or W/cm2
B) H/cm2 or E/cm2
C) g/cm2 or m/cm2
D) J/min or W/min - Correct Answer - A
Normally equated to the aperture of the eye, i.e. 7 mm, the limits of the
laser are normally expressed as joules/cm 2 or watts/cm2.
If the HVL of Pb for Cobalt-60 gamma radiation is 11 mm, the thickness
of Pb that would reduce a narrow beam of Cobalt-60 gamma rays to 1/32
its original value is .
A) 352 mm
B) 0.34 mm
C) 2.91 mm
D) 55 mm - Correct Answer – D
1/32 is equal to (1/2)5. Therefore, it takes 5 HVL's to accomplish a
reduction of 1/32. Hence, 5 x 11 mm = 55 mm.
Normal background radiation exposures are partially due to
ingestion/inhalation and subsequent accumulation in the thyroid of
which of the following isotopes?
A) Strontium-90
B) Cesium-137
C) Iodine-131
D) Uranium - Correct Answer - C
pg. 4
NRRPT EXAM AND PRACTICE EXAM NEWEST 2026
TEST BANK| NRRPT EXAM PREP WITH COMPLETE
1000+ REAL EXAM QUESTIONS AND CORRECT
DETAILED ANSWERS (VERIFIED ANSWERS)
ALREADY GRADED A+ (BRAND NEW!!)
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 - Correct Answer - D
dps = (2000 pCi)(1 x 10-12Ci/pi )(3.7x1010 dps/Ci) dps = 74
dpm = (74 dps) (60 sec/min)
dpm = 4440
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 - Correct Answer - D
1 half life = 50% remaining
2 half lives = 25% remaining
3 half lives = 12.5% remaining
pg. 1
,2|Page
4 half lives = 6.25% remaining
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 - Correct Answer - 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
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. - Correct answer - B
The wavelengths responsible for conjunctivitis are 270-280 nm in the
ultraviolet area of the electromagnetic spectrum.
Eventually, charged particles give up their energy to the surrounding
medium. In the case of the alpha particle, it becomes (a):
A) Proton
B) Neutron
pg. 2
,3|Page
C) Tritium
D) Helium - Correct answer – D
An alpha particle consists of 2 neutrons and 2 protons carrying two
positive charges. It abstracts two electrons from the surrounding atoms
and becomes a helium atom.
Gamma rays get their energy from:
A) electrons outside the nucleus.
B) nuclear disintegration.
C) cosmic rays that change as they enter the atmosphere.
D) high energy meson particles. - Correct Answer - B
Gamma rays are similar to x-rays but differ in origin and wavelength.
Gamma rays get their energy from nuclear disintegration while x-rays
are produced from dislodging inner electrons.
Given a reading of 100mr/hr, gamma, at 10 feet, what would be the
reading at 2 feet, assuming a point source geometry?
A) 6400 mr/hr
B) 2500 mr/hr
C) 5000 mr/hr
D) 3000 mr/hr - Correct Answer - B
Apply the inverse square law: (r2/r1) = (d1/d2)2 r2 = (r1)(d1/d2)2
r2 = (100 mr/hr)(10 ft/2 ft)2 r2 = 2500 mr/hr
r1 = exposure rate 1 r2 = exposure rate 2 d1= distance 1
d2 = distance 2
pg. 3
, 4|Page
What is the usual unit of measurement for laser radiation?
A) J/cm2 or W/cm2
B) H/cm2 or E/cm2
C) g/cm2 or m/cm2
D) J/min or W/min - Correct Answer - A
Normally equated to the aperture of the eye, i.e. 7 mm, the limits of the
laser are normally expressed as joules/cm 2 or watts/cm2.
If the HVL of Pb for Cobalt-60 gamma radiation is 11 mm, the thickness
of Pb that would reduce a narrow beam of Cobalt-60 gamma rays to 1/32
its original value is .
A) 352 mm
B) 0.34 mm
C) 2.91 mm
D) 55 mm - Correct Answer – D
1/32 is equal to (1/2)5. Therefore, it takes 5 HVL's to accomplish a
reduction of 1/32. Hence, 5 x 11 mm = 55 mm.
Normal background radiation exposures are partially due to
ingestion/inhalation and subsequent accumulation in the thyroid of
which of the following isotopes?
A) Strontium-90
B) Cesium-137
C) Iodine-131
D) Uranium - Correct Answer - C
pg. 4
