NRRPT prep Radiation Fundamentals:Exam Complete Questions With Verified Answers || Explained Answers || Study Guide || Frequently Tested Qs || Graded A+ || 100% Pass

NRRPT prep Radiation Fundamentals:Exam Complete Questions With Verified Answers || Explained Answers || Study Guide || Frequently Tested Qs || Graded A+ || 100% Pass .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 )(3.7x10^10 dps/Ci) dps = 74 dpm = (74 dps) (60 sec/min) dpm = 4440 Remember to convert to disintegrations per minute not DISINTEGRATIONS PER SECOND. .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 .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 - Answer-The 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 - Answer-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. - Answer-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 10^4 dps B) 2.96 x 10^11 dps C) 2.5 x 10^7 dps D) 4.8 x 10^11 dps E) 7.4 x 10^10 dps - Answer-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 10^11 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 B) Neutron C) Tritium D) Helium E) Deuteron - Answer-The correct answer is: 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. .Question Number: 10 Fundamentals of Radiation Protection Gamma radiation produces ionization by which of the following? A) Photoelectric effect, Compton effect, pair production B) Photoelectric effect, Compton effect, bremsstrahlung C) Bremsstrahlung, photoelectric effect D) Excitation, photoelectric effect, pair production E) Excitation and bremsstrahlung - Answer-The correct answer is: A Absorption of gamma ray photons occurs primarily by the photoelectric effect, Compton effect and pair production. .Question Number: 11 Fundamentals of Radiation Protection 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. - Answer-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.