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NRRPT Prep Applied Radiation Protection Questions With Complete Solutions 2025 Graded A+, Exams of Nursing

NRRPT Prep Applied Radiation Protection Questions With Complete Solutions 2025 Graded A+ NRRPT Prep Applied Radiation Protection Questions With Complete Solutions 2025 Graded A+

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2024/2025

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NRRPT Prep Applied Radiation Protection Questions With Complete Solutions
2025 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 - ANSWERS-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?
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NRRPT Prep Applied Radiation Protection Questions With Complete Solutions 2025 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 - ANSWERS-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 - ANSWERS-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 - ANSWERS-The correct answer is: D dps = (2000 pCi)(1 x 10-12Ci/pi )(3.7x1010 dps/Ci) dps = 74

A) 2.0 mCi B) 1.0 mCi C) 1.5 mCi D) 3.5 mCi E) 4.0 mCi - ANSWERS-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 - ANSWERS-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- decayed. How many half lives did the carbon go through? A) 1 B) 2 C) 3 D) 4 E) 5 - ANSWERS-The correct answer is: D 1 half life = 50% remaining 2 half lives = 25% remaining 3 half lives = 12.5% remaining

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 - ANSWERS-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. - ANSWERS-The 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 C) 2.5 x 107 dps D) 4.8 x 1011 dps E) 7.4 x 1010 dps - ANSWERS-The correct answer is: B By definition, 1 Ci = 3.7 x 10 10 disintegrations per second Therefore, 8 curies would have: 8 Ci * 3.7 x 10 10 dps/Ci = 2.96 x 1011 dps Question Number: 9 Fundamentals of Radiation Protection Eventually, charged particles give up their energy to the surrounding medium. In the case of the alpha particle, it becomes (a): A) Proton

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. E) braking radiation. - ANSWERS-The correct answer is: 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. Question Number: 12 Fundamentals of Radiation Protection 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

E) 2200 mr/hr - ANSWERS-The correct answer is: B Apply the inverse square law: (r2/r1) = (d1/d2)2 r2 = (r1)(d1/d2) r2 = (100 mr/hr)(10 ft/2 ft)2 r2 = 2500 mr/hr Where: r1 = exposure rate 1 r2 = exposure rate 2 d1= distance 1 d2 = distance 2 Question Number: 13 Fundamentals of Radiation Protection What is the usual unit of measurement for laser radiation? A) J/cm2 or W/cm B) H/cm2 or E/cm C) g/cm2 or m/cm D) J/min or W/min E) cm2/g or mg/cm2 - ANSWERS-----------The correct answer is: 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. Question Number: 14

D) Uranium E) Krypton-85 - ANSWERS-The correct answer is: C Iodine has a half life of eight days and collects in the thyroid gland. Uranium accumulates in the lung and kidney; Strontium accumulates in the bones. Question Number: 16 Fundamentals of Radiation Protection One RAD is equal to : A) 100 ergs/gram. B) 0.1 curies. C) 3.14 rems. D) 0.87 Roentgen. E) 0.98 rem. - ANSWERS-The correct answer is: A The RAD is a special unit for absorbed dose and is equal to 100 ergs/gram. Question Number: 17 Fundamentals of Radiation Protection Plutonium, taken into the circulatory system, will deposit in what areas of the body?

A) Bone B) Liver C) Kidney D) Thyroid E) Nasopharynx - ANSWERS-The correct answer is: A Soluble plutonium deposits mainly in the bones. The insoluble forms remain in the lungs. Question Number: 18 Fundamentals of Radiation Protection Sealed sources should be tested for leakage at least: A) weekly. B) monthly. C) quarterly. D) semiannually. E) annually. - ANSWERS-The correct answer is: D All sealed sources should be checked for leaks semiannually under NRC RAM licenses.

C) quantity of a radioisotope. D) activity of a radioisotope. E) dose equivalent from radiation. - ANSWERS-The correct answer is: D The becquerel is a unit in the S.I. system used to express the activity of a radioisotope. Question Number: 21 Fundamentals of Radiation Protection The current OSHA standard for far field exposure to microwave radiation averaged over any 0.1 hour period is: A) 0.01 W/cm B) 0.1 W/cm C) 1.0 W/cm D) 10 W/cm E) 100 W/cm2 - ANSWERS-The correct answer is: A The value is normally listed as 10 mw/cm2. Question Number: 22 Fundamentals of Radiation Protection

The electromagnetic radiation produced by the rapid deceleration of charged particles is called which of the following? A) Compton effect protons B) Bremsstrahlung C) Light D) Angular radiometric release E) Pair production - ANSWERS-The correct answer is: B The classic definition of Bremsstrahlung is the electromagnetic radiation produced by the rapid deceleration of charged particles. Question Number: 23 Fundamentals of Radiation Protection The fraction of atoms which undergo decay per unit time is which of the following? A) Half life B) Activity C) Decay constant D) Effective half life E) Effective removal constant - ANSWERS-The correct answer is: C

D) Blood cells E) Bone - ANSWERS-The correct answer is: B Damage can occur to the skin but it is of secondary importance to the eye. Question Number: 26 Fundamentals of Radiation Protection The radon daughters inhaled during uranium mining are normally deposited in what part of the human body? A) Bone B) Kidneys C) Liver D) Lungs E) Thyroid - ANSWERS-The correct answer is: D The most important radon daughters are Bi, Po, Pb. They are deposited in the lung when inhaled, where they may lead to lung cancer. Question Number: 27 Fundamentals of Radiation Protection The gray is a quantity used to express the:

A) energy deposited per unit weight. B) ionization in air. C) biological effectiveness. D) ionization in air due to gamma radiation. E) specific ionization. - ANSWERS-The correct answer is: A The gray is the SI quantity of absorbed dose, and is expressed in units of J/kg. One gray = 100 rad. Question Number: 28 Fundamentals of Radiation Protection The Roentgen is a unit used to express the: A) energy deposited in a unit weight of material. B) ionization in air due to all radiation types. C) biological effectiveness of gamma radiation. D) ionization in air due to gamma radiation. E) specific ionization of a material. - ANSWERS-The correct answer is: D The measure of ionization produced in air by X-ray or gamma radiation is called the Roentgen (R).