NRRPT prep Radiation Fundamentals

Question Number: 314 Fundamentals of Radiation Protection An absorbed dose of 3 rad results in an energy deposition of: A) 3 x 10 -2 J/Kg B) 1 x 10 -3 erg/gm C) 3.3 x 10-3 J/kg D) 3 x 10 -2 J/gm E) 1 x 10 -2 erg/Kg

The correct answer is: A (3 rad)(1 gray/100 rad)(1 J/Kg/gray) = 3 x 10-2 J/Kg

Question Number: 185 Fundamentals of Radiation Protection The most abundant radionuclide in the earth's crust on the average is which of the following? A) Thorium B) Uranium C) Plutonium D) Carbon E) Radium

The correct answer is: A In the U.S., on the average, one square mile dug to one foot would yield: 6 tons Th-232 3 tons U-nat 1 ton K-40

Question Number: 84 Fundamentals of Radiation Protection Name the initial and final nuclide in the thorium series. A) Th-232; Pb-208 B) U-238; Pb-206 C) U-235; Pb-207 D) Np-237; Bi-209 E) Th-230; Pb-214

The correct answer is: A The four series in A through D have occurred in nature. However, the neptunium series has essentially decayed away, and no longer occurs in nature. The thorium series contains isotopes of radium and radon.

Question Number: 152 Fundamentals of Radiation Protection The SI unit for radiation absorbed dose is which of the following? A) Gray B) Rad C) Sievert D) Rem E) Becquerel

The correct answer is: A The gray (Gy) is defined as 1 Joule per kilogram. It is the SI counterpart of the rad, which is 100 ergs per gram.

Question Number: 92 Fundamentals of Radiation Protection The periodic table lists: A) all elements. B) all nuclides. C) all radionuclides. D) all naturally occurring elements. E) all naturally occurring radionuclides.

The correct answer is: A The periodic table is a list of elements (atomic number) and electron structure. It does not list nuclides or radionuclides. It also lists man-made elements such as technecium and the transuranics.

Question Number: 143 Fundamentals of Radiation Protection The Department of Transportation (DOT) label White I allows what radiation dose rate at 1 meter? A) None above background B) Less than 5 µSv per hr C) Between 5 µSv per hr and 500 µSv per hr D) Between 100 µSv per hr and 500 µSv per hr E) Greater than 500 µSv per hr

The correct answer is: A 49 CFR Part 173 describes the labels. Radioactive White I is all white with one red vertical stripe. The surface radiation rate allowed must be less than 5 µSv per hr with no measurable radiation at 1 m.

Question Number: 62 Fundamentals of Radiation Protection If an individual incurred a dose equivalent of 5000 rem, how many Sieverts would it be? A) 5 B) 50 C) 500 D) 5000 E) 50,000

The correct answer is: B One Sievert is equivalent to 100 rem; therefore 5000 rem is equivalent to 50 Sv.

Question Number: 384 Fundamentals of Radiation Protection How often are sealed sources required to be leak tested? A) Monthly B) Every six months C) Annually D) Every two years E) Every three years

The correct answer is: B See10 CFR Part 34 and 39 . Longer intervals may be allowed for certain sources, depending on the manufacturer and type of source.

Question Number: 223 Fundamentals of Radiation Protection Which of the following are considered risks of chronic low-level radiation exposure? 1. Sterility 2. Leukemia 3. Cataracts 4. Skin erythema A) 1 B) 2 C) 3 D) 1,3 E) 2,3,4

The correct answer is: B The risk of chronic, low-level exposure, such as occupational exposures, is for stochastic effects only. The only stochastic effect shown is leukemia.

Question Number: 96 Fundamentals of Radiation Protection The target organ for consideration of biological damage to humans from laser exposure is the: A) whole body. B) eye. C) skin. D) immune system. E) bone.

The correct answer is: B The target organ is the eye with a secondary organ of the skin.

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.

The correct answer is: B The wavelengths responsible for conjunctivitis are 270-280 nm in the ultraviolet area of the electromagnetic spectrum.

Question Number: 166 Fundamentals of Radiation Protection What is a disadvantage of a long sample counting time for optimization of counting statistics? A) Resolution poorer B) Instrument saturates C) Background may change D) Sample may decay E) There is no disadvantage.

The correct answer is: C Background can vary in certain instances, for example, if radioactive materials are being transported nearby while an instrument is counting or if atmospheric conditions cause background radiation fluctuation.

Question Number: 332 Fundamentals of Radiation Protection The speed of a thermal neutron (at 20° C) is: A) 2200 cm/s B) 0.025 m/s C) 2200 m/s D) 2.2 m/s E) 3 x 108 m/s

The correct answer is: C kT = 0.5 mv2 (1.38 x 10-23 J/K) (293° K) = (0.5)(1.67492 x 10 -27 kg) (v2) 4.84 x 106 = v2 2200 m/s = v

Question Number: 391 Fundamentals of Radiation Protection The most radiosensitive cells in the human body are: A) Hair B) Liver C) Platelets D) Leukocytes E) Muscle

The correct answer is: D Leukocytes are the white blood cells.

Question Number: 128 Fundamentals of Radiation Protection The term "acute radiation syndrome" describes which of the following radiation exposure scenarios? A) Long term, moderate dose B) Long term, high dose C) Short term, moderate dose D) Short term, high dose E) Short term, low dose

The correct answer is: D Radiation exposure to humans is broken into two groups: chronic and acute radiation exposure. Chronic exposure describes low dose, long term exposure, for example, background radiation exposure. duration and higher intensity. Acute exposure is of short

Question Number: 160 Fundamentals of Radiation Protection What is the mathematical relationship between the decay constant and the physical half-life? A) Inverse square B) Linear C) Direct D) Inverse E) Equal

The correct answer is: D Since the physical half-life is equal to 0.693/decay constant, the physical half life is inversely proportional to the decay constant. Therefore, materials with a short half life have large decay constants and vice versa.

Question Number: 251 Fundamentals of Radiation Protection Charged particles interact with matter by which of the following? A) Photoelectric effect and Compton scattering B) Inelastic collision C) Elastic scattering D) Excitation and ionization E) Bremsstrhlung and pair production

The correct answer is: D Charged particles lose energy by collision and radiative processes. The two collision processes are excitation and ionization. Radiative loss occurs for beta particles and secondary electrons by Bremsstrahlung.

Question Number: 394 Fundamentals of Radiation Protection The threshold dose to cause erythema due to radiation is approximately: A) 0.1 mGy B) 1 mGy C) 10 mGy D) 25 mGy E) 2 Gy

The correct answer is: E A dose of 2 Gy is 200 rads.

Question Number: 292 Fundamentals of Radiation Protection The SI unit of energy is which of the following? A) Sievert B) Roentgen C) Watt D) BTU E) Joule

The correct answer is: E The corresponding unit in the English system is the "erg". Moe, H.J. (1992) Operational Health Physics Training .

Question Number: 189 Fundamentals of Radiation Protection If an isotope has a DAC of 1 x 10 -8 µCi /ml and it will be released through a stack with a flow rate of 1 x 10 how many µCi of the isotope can be released at a constant rate in one day without exceeding the DAC? A) 240 B) 120 C) 24 D) 1.0 E) 0.4 6 liter/hr,

The correct answer is: A (1 x 10 -8 µCi /ml)(1 x 106 L/hr )(24 hr )(1 x 10 3 ml/L) = 240 µCi

Question Number: 174 Fundamentals of Radiation Protection What does an "alpha" emission consist of? A) An electron B) An electromagnetic wave C) 2 neutrons and 2 protons D) 1 neutron and 1 proton E) 1 triton and 1 neutron

The correct answer is: C An "alpha" emission consists of 2 neutrons and 2 protons (a helium nucleus). This type of ionizing radiation is easily shielded as an external source, but it is considered an internal radiation hazard as it can cause heavy damage in tissue. External shielding can consist of paper.

Question Number: 372 Fundamentals of Radiation Protection The noble gas Ar-41 is produced at a reactor as a result of which of the following? A) Fission B) Corrosion product activation C) Activation of Argon in air D) Activation of Oxygen in air E) Fission product decay

The correct answer is: C Naturally occurring argon in the air is activated by absorbing a neutron.

Question Number: 252 Fundamentals of Radiation Protection The range of an alpha particle in tissue is approximately 50 to 100: A) centimeters. B) millimeters. C) microns. D) angstroms. E) barns.

The correct answer is: C Since the alpha particle energy is not given, we assume an energy range of 4-8 MeV, which is typical for alpha decay in natural and man-made radionuclides. Alphas in this energy range will travel 50-100 µm in a material of unit density such as tissue.

Question Number: 297 Fundamentals of Radiation Protection Specific activity units are which of the following? A) Joules/kg B) Gram/curie C) Curie/gram D) Ergs/gram E) Ergs/cm

The correct answer is: C Specific activity is equal to decay constant times the number of atoms in a gram.

Question Number: 24 Fundamentals of Radiation Protection The most common exposure to ultraviolet radiation is from: A) Welding B) Germicidal lamps C) Direct sunlight D) Black lights E) X-ray diffraction units

The correct answer is: C Sunlight is the major source of exposure to ultraviolet light.

Question Number: 183 Fundamentals of Radiation Protection Which of the following government agencies regulates the classification and control measures incorporated into laser products? A) OSHA B) EPA C) NRC D) FDA E) ACGIH

The correct answer is: D The Food and Drug Administration regulates the classification and control measures incorporated into laser products, high mercury lamps, and sunlamps.

Question Number: 304 Fundamentals of Radiation Protection The stopping power for electrons as the atomic number of the absorber . A) decreases, decreases B) increases, decreases C) increases, increases D) is constant, increases E) decreases, increases

The correct answer is: E This occurs because substances of high Z have fewer electrons per gram and these are more tightly bound. Consequently, the range tends to increase as Z increases. Moe, H.J. (1992) Operational Health Physics Training pg. 3 -15.

Question Number: 340 Fundamentals of Radiation Protection The most important type of UV for causing cancer from sunlight is: A) Active UV B) UVL C) UltraUV D) UVC E) UVB

The correct answer is: E UVA and UVB are both transmitted through the atmosphere.

Question Number: 179 Fundamentals of Radiation Protection Sr-90 decays by beta to Y-90 with an atomic number of: A) 39 B) 38 C) 42 D) 37 E) 36

The correct answer is: A Beta decay causes an increase of +1 in the atomic number as a negative mass is leaving the atom. A +1 increase in the atomic number increases the atom's ranking on the periodic table to the next higher element.

Question Number: 291 Fundamentals of Radiation Protection CS-134, Cs-137 and Cs-135 are: A) isotopes. B) isotones. C) isomers. D) isobars. E) isometrics.

The correct answer is: A Isotopes have the same number of protons or atomic number. Moe, H.J. (1992) Operational Health Physics Training .

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

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: 169 Fundamentals of Radiation Protection What would be a good approximation of the quality factor (QF) of the neutrino? A) 0 B) 1 C) 5 D) 10 E) 20

The correct answer is: A Since neutrinos have essentially zero probability of interaction, they cannot cause ionization along their path through the tissue. Hence, the quality factor will be zero.

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.

The correct answer is: A The RAD is a special unit for absorbed dose and is equal to 100 ergs/gram.

Question Number: 370 Fundamentals of Radiation Protection A quantity of 3.7 * 10 10 Bq of radium 226 is approximately equal to which of the following? A) 1 Bq/g density B) 1 g mass C) 1 R/h at 1 cm D) 0.1 Ci of Sr-90 (mass only) E) 0.2 Ci of Co-60 (mass only)

The correct answer is: B A curie was originally defined as 1 g of Ra-226.

Question Number: 33 Fundamentals of Radiation Protection What is one of the special units used to express activity of radioactive materials? A) Microrad B) Curie C) Gamma D) Dose E) Becquerel

The correct answer is: B The curie is a unit of activity. One curie equals 3.7 x 10^10 disintegrations per second. The SI unit is the becquerel. One becquerel is equal to one disintegration per second.

Question Number: 365 Fundamentals of Radiation Protection The dose rate required to double the mutation rate in man has been extrapolated from mouse models. This dose was estimated by the BEIR V report to be: A) > 1 Gy. B) > 1 Gy low dose rate, low LET radiation. C) > 1 Gy high dose rate, low LET radiation. D) > 1 Gy low dose rate, high LET radiation. E) > 1 Gy high dose rate, high LET radiation.

The correct answer is: B This is called the "doubling dose". (1990). BEIR V. Washington, DC., pg. 4.

Question Number: 368 Fundamentals of Radiation Protection Non-stochastic effects are the same as whcih of the following? A) Somatic effects B) Deterministic effects C) Stochastic effects D) Extrapolation effects E) ICRP recommendations for stochastic limits

The correct answer is: B Nonstochastic effect means health effects, the severity of which varies with the dose and for which a threshold is believed to exist. Radiation-induced cataract formation is an example of a nonstochastic effect (also called a deterministic effect). 10 CFR Part 20.1003

Question Number: 164 Fundamentals of Radiation Protection The best data to use to estimate the hazard of radiation to genetic damage is: A) rat data. B) rat and mouse data. C) human data. D) mathematical modeling. E) beagle data.

The correct answer is: C Human data is the best evidence for any study of effects on humans, radiation or otherwise.

Question Number: 50 Fundamentals of Radiation Protection The SI replacement unit for the rem is the: A) Gray. B) Becquerel. C) Sievert. D) Rad. E) Coulomb per Kilogram.

The correct answer is: C The sievert is the SI unit that replaces the rem. One sievert is equal to 100 rem. 0.01 Sv is equivalent to 1.0 rem.

Question Number: 177 Fundamentals of Radiation Protection What is the chemical symbol for Radon? A) Ra B) Rd C) Ro D) Rn E) R

The correct answer is: D Radon is represented on the Periodic Table by the symbol "Rn". "Ra" denotes Radium.

Question Number: 283 Fundamentals of Radiation Protection The electron volt is a unit of energy and is defined as which of the following? A) 33.7 ergs B) 100 ergs C) 2.25 x 10-19 kw-hr D) 1.6 x 10-19 joule E) 1.0 x 10-12 coulombs

The correct answer is: D The joule and the erg are both measures of energy, or "work." Cember, H. Introduction to Health Physics .

Question Number: 341 Fundamentals of Radiation Protection The outermost layer of the skin that provides protection to the underlying tissues is the: A) dermis. B) basal layer. C) epidermis. D) stratum granulosum. E) stratum corneum.

The correct answer is: E The layers listed are from deepest to outermost. Cember, H.,Introduction to Health Physics, Chapter 7.

Question Number: 127 Fundamentals of Radiation Protection If 1 gram of radium equals 1 Ci and 1 mole of radon weighs 222 grams, what is the mass of 100 pCi of radon? -8 A) 2.22 x 10 g B) 2.22 x 10-10 g C) 2.22 x 10-12 g D) 2.22 x 10-14 g E) 2.22 x 10-16 g

The correct answer is: A 100 pCi = 1 x 10 -10 Ci (1 x 10 -10 Ci)(222g/Ci) = 222 x 10 -8 g

Question Number: 142 Fundamentals of Radiation Protection The Department of Transportation (DOT) label White I allows a dose equivalent on the surface of the package: A) less than 5 µSv per hr. B) between 5 µSv per hr and 100 µSv per hr. C) between 100 µSv per hr and 500 µSv per hr. D) greater than 500 µSv per hr. E) no greater than background.

The correct answer is: A 49 CFR Part 173 describes the labels. Radioactive White I is full white with one red vertical stripe. There can be no measurable radiation at 1 meter, and must be less than 5 µSv (0.5 mrem) per hour on the package surface.

Question Number: 180 Fundamentals of Radiation Protection What is a Roentgen? A) The amount of X or gamma radiation that produces ionization resulting in 1 electrostatic unit of charge in 1 cc of dry air at STP B) Absorption of 100 ergs of energy from any radiation in 1 gm of any material C) The "biological" dose to living tissue D) 98 ergs per gram of any absorber E) 3.34 x 10-10 Coulombs/kilogram

The correct answer is: A A Roentgen is a measurement of exposure whose definition is given in Answer choice A. The measurement of absorbed dose (rad) definition is given in Answer B. A rem is the biological dose. One Roentgen will deposit 87 ergs per gram of air, 98 ergs per gram of tissue, and will create 3.34 x 10 -10 Amperes per cc of dry air at STP.

Question Number: 397 Fundamentals of Radiation Protection The linear no-threshold model: A) implies that every additional dose carries additional risk. B) implies that risk is proportional to the dose rate for alpha exposure only. C) provides a framework for calculating dose to those exposed to line sources only. D) cannot be used for safety purposes, but is always used in epidemiology experiments. E) is no longer used.

The correct answer is: A A linear no-threshold (LNT) dose-response relationship is used to describe the relationship between radiation dose and the occurrence of cancer. This dose-response model suggests that any increase in dose, no matter how small, results in an incremental increase in risk. The U.S. Nuclear Regulatory Commission (NRC) accepts the LNT hypothesis as a conservative model for estimating radiation risk.

Question Number: 115 Fundamentals of Radiation Protection Ionization chambers have gas amplification factors: A) equal to 1. B) greater than 1. C) less than 1. D) dependent on the voltage. E) dependent on the volume.

The correct answer is: A A radiation detector used in the ionizing region has a gas amplification factor equal to one. Other instruments have gas amplification factors greater than one but not less than one.

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

The correct answer is: A Absorption of gamma ray photons occurs primarily by the photoelectric effect, Compton effect and pair production.

Question Number: 329 Fundamentals of Radiation Protection Many fission products are in secular equilibrium with their progeny. If secular equilibrium has been achieved with a 10000 GBq Sr-90 (29 year half life) source, how much Y-90 (64.1 hr half life) is present? A) 5 x 10 -4 g B) 1.4 x 10-7 g C) 5 x 10 -7 g D) 5 x 10 -8 g E) 2.33 x 10-11 g

The correct answer is: A Activity = decay constant * number of atoms {10,000 GBq / (0.693/64.1 h)} * {3600 s/1 h} * {1 mole/ 6.02 x 10 23 atoms} * {90 g/ mole} = 5 x 10 -4 g

Question Number: 390 Fundamentals of Radiation Protection If a photon undergoes Compton scattering: A) the scattered photon will be of lower energy. B) the scattered photon will be moving at a slightly lower speed than the incoming photon. C) the scattered positron will decay into two 0.511 MeV photons. D) the scatter angle will result in a neutron emission. E) Beta emission will no longer be possible.

The correct answer is: A All photons travel at the speed of light.

Question Number: 108 Fundamentals of Radiation Protection The energy distribution of alpha emission: A) is monoenergetic. B) ranges from zero to a maximum. C) ranges from 1/3 maximum to a maximum. D) depends on radionuclide activity. E) depends on bound elements to the nuclide.

The correct answer is: A Alpha emission is monoenergetic and characteristic of the radionuclide; beta emission ranges from zero to a maximum characteristic of the radionuclide. The beta particle (or positive beta) shares the energy with the antineutrino (or neutrino).

Question Number: 34 Fundamentals of Radiation Protection What type of radiation presents the greatest internal hazard? A) Alpha B) Beta C) Gamma D) Positron E) Neutron

The correct answer is: A Alpha particles can cause extensive ionization in tissues, due to their large mass and high LET. They are shielded by the skin as an external source.

Question Number: 32 Fundamentals of Radiation Protection What is the most common ionizing radiation hazard associated with antistatic devices? A) Alpha B) Beta C) Gamma D) Neutron E) All of the above

The correct answer is: A Antistatic devices often contain Polonium-210, an alpha emitter. It is a hazard only if ingested or inhaled.

Question Number: 350 Fundamentals of Radiation Protection At higher frequencies (95 GHz) the tissue(s) most STRONGLY impacted by radio-frequency radiation is/are which of the following? A) Skin B) Ankles C) Hands D) Joints E) Ears

The correct answer is: A At 95 GHz, the majority of the energy is absorbed in the surface of the skin.

Question Number: 162 Fundamentals of Radiation Protection The "lapse rate" is the: A) temperature gradient of the atmosphere. B) rate of decay. C) initial activity minus decayed activity. D) inhalation rate of noble gases. E) period between initial decay and equilibrium.

The correct answer is: A Atmospheric conditions are an important consideration for the evaluation of downwind discharges from stacks. There are many models for estimation of dispersion of radionuclides, which include the lapse rate as a variable.

Question Number: 227 Fundamentals of Radiation Protection If a linear relationship between somatic effects of radiation and absorbed dose is assumed, the number of additional cancer deaths to be expected per million persons per rad is: A) greater than 100 B) 20 C) 50 D) 10 E) 2

The correct answer is: A BEIR V (1990) estimates an additional 800 cancer mortalities in a population of 100,000 exposed to 10 rad. This can be extrapolated to 800 per 1 million per 1 rad, if effect to dose is assumed to be linear.

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

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: 111 Fundamentals of Radiation Protection An ionization chamber has voltage great enough to cause: A) ions before recombination; no secondary electrons. B) secondary electrons with a short range avalanche. C) secondary electrons with avalanche to whole anode. D) continuous discharge. E) recombination before current is produced.

The correct answer is: A Each answer above represents gas-filled detector responses in which the voltage begins at zero and is then increased. Recombination is first, then the ionization region; then the proportional region, the Geiger region, and then the continuous discharge region. Each gas-filled detector operates in one of these regions.

Question Number: 65 Fundamentals of Radiation Protection What is the formula for the relationship between energy of a particle and its mass? A) E = mc2 B) F = ma C) V = IR D) c = wavelength * frequency E) E = 1/2MV2

The correct answer is: A Einstein's formula relating mass and energy describes the fundamental relationship between the energy of a particle and its mass.

Question Number: 107 Fundamentals of Radiation Protection Full width half-maximum (FWHM) of the photopeak is a measurement of: A) resolution. B) energy of the photon. C) energy of the charged particle. D) channel width. E) channel energy.

The correct answer is: A FWHM is a standard method for the measurement of resolution.

Question Number: 119 Fundamentals of Radiation Protection The function of an "annular kinetic impactor head" is to trap: A) airborne particles. B) all surface contamination. C) surface contamination that would be permanent. D) liquid particles that are filtrable. E) particles with an AMAD less than 1.

The correct answer is: A For collecting alpha, beta, and gamma emitting contaminants, such as plutonium and fission products, it collects large airborne particles. What is particular to this collecting device is that it does not collect radon and thoron. The approximate efficiency is 95%.

Question Number: 171 Fundamentals of Radiation Protection Notices to radiation workers and instructions to workers fall under which part of 10 CFR? A) Part 19 B) Part 20 C) Part 30 D) Part 35 E) Part 50

The correct answer is: A Four parts to the Code that should be familiar are: Part 19--Notices, Instructions, Reports to Workers Part 20--Standards for Protection Against Radiation Part 30--Licensing of By-Product Materials Part 35--Medical Use of Byproduct Material

Question Number: 137 Fundamentals of Radiation Protection Non-bacterial radiopharmaceutical waste disposal: A) requires separate marked waste containers. B) needs no separation from regular waste. C) requires a separate ventilated room. D) is not regulated by the NRC. E) has no special requirement.

The correct answer is: A Generally speaking, the medical pharmaceuticals have low toxicity and short half-lives. They are disposed into a marked receptacle during the day. They are removed to a locked storage area for decay to background and then disposed of with the regular trash. Typically, if the half-life is less than 100 days, they are held for ten half-lives prior to disposal as "clean" waste.

Question Number: 201 Fundamentals of Radiation Protection Genetic effects produced by radiation in males are dependent on: 1. State of germ cell development 2. LET of the radiation 3. Interval between the exposure and conception A) 1,2,3 B) 1,2 C) 1,3 D) 1 E) None of the answers

The correct answer is: A Germ cells in the male are spermatozoa. Bergonie and Tribondeau show that cell maturity is indeed a factor in radiosensitivity. LET is certainly a factor since a higher LET causes a higher probability of interaction in a fixed volume of spermatozoa. The interval between exposure and conception is important since time could be allowed for repair of marginally damaged cells or death of very damaged cells.

Question Number: 31 Fundamentals of Radiation Protection What is the charge produced in a 250 cc free air ionization chamber by an exposure of 100 mr? A) 25 esu B) 30 esu C) 100 esu D) 250 esu E) 300 esu

The correct answer is: A If 1 R = 1 esu/cc, then how many esu's = 0.1R? X = (0.1 R)(1 esu/cc/R)(250 cc) = 25 esu Where R = Roentgen; esu = Electrostatic unit

Question Number: 215 Fundamentals of Radiation Protection Emission of a negatively charged beta particle results from the transformation of: A) a neutron into a proton. B) a proton into a neutron. C) a neutrino into a neutron. D) a positron into a proton. E) an electron into a proton.

The correct answer is: A If one thinks of the neutron as a particle having one positive charge and one negative charge, it is easy to visualize a proton being created (positively charged) when the negatively charged beta particle is emitted. Such a process occurs during beta minus decay and when a free neutron decays.

Question Number: 163 Fundamentals of Radiation Protection An atmospheric inversion is one in which the lapse rate is which of the following? A) Positive B) Negative C) Continuous D) Zero E) Lognormal

The correct answer is: A If the lapse rate is positive, then the air temperature increases with increasing height. This condition is a very stable condition and undesired since the stack effluents tend to sink. This is a consideration in stack studies involving release of radionuclides.

Question Number: 354 Fundamentals of Radiation Protection Many fission products are in secular equilibrium with their progeny. To achieve secular equilibrium, which of the following conditions must be met? A) Parent half-life >>> daughter half life B) Parent half-life > daughter half life C) Parent half-life < daughter half life D) Parent half-life <<< daughter half life E) Parent half-life same as daughter half life

The correct answer is: A If the parent half-life is only slightly greater than the daughter half-life, then "transient" equilibrium occurs. If the parent half-life is less than the daughter half-life, then there is no equilibrium. Cember, H.,Introduction to Health Physics, Chapter 4.

Question Number: 335 Fundamentals of Radiation Protection The "unattached fraction " discussed when evaluating radon concentrations is defined by which of the following statements? A) The "unattached fraction" is used by the EPA to estimate the amount of actual radon progeny present and is sometimes called the "equilibrium factor" with a typical value of 0.5. The "unattached fraction" represents the fraction of radon progeny (including radon) used to calculate the working B) level, and is based upon one liter of atoms with the highest diffusion coefficient that are not attached to particles typically found in the lungs. C) The "unattached fraction" represents the radon atoms unattached to condensation nuclei in one liter of air (at 20° C) and existing as ions, which have a much higher diffusion coefficient than the attached species. The "unattached fraction" represents the fraction of any given radon progeny (including radon) unattached to D) condensation nuclei in one liter of air and existing as free atoms or ions, which have a much lower diffusion coefficient than the attached species. The "unattached fraction" represents the fraction of any given radon progeny unattached to condensation nuclei in E) the air and existing as free atoms or ions, which have a much higher diffusion coefficient than the attached species.

The correct answer is: A In NRC regulations ( 10 CFR Part 20 Appendix B), the unattached fraction is not included in the values for radon. Turner, J.,Atoms, Radiation and Radiation Protection.

Question Number: 100 Fundamentals of Radiation Protection In a radioactive decay process in which a longer lived radionuclide decays into much shorter half-life radionuclide, the process is called which of the following? A) Secular equilibrium B) Transient equilibrium C) No equilibrium D) Transmutation E) Depends on radionuclide

The correct answer is: A In secular equilibrium, the parent radionuclide has a much longer half-life than the daughter; in transient equilibrium the half-life of the daughter is of the same order of magnitude but shorter than that of the parent. If the parent half-life is shorter than that of the daughter, then there is no equilibrium.

Question Number: 191 Fundamentals of Radiation Protection 95-Am-241 has protons and neutrons. A) 95; 146 B) 95; 241 C) 95; 336 D) 146; 95 E) 241; 95

The correct answer is: A In the term Z-X-A: Z is the number of protons X is the chemical symbol for the element A is the number of protons and neutrons (nucleons) A - Z = number of neutrons

Question Number: 300 Fundamentals of Radiation Protection As an alpha particle transverses a substance, the ionization of the particle produces: A) increases as the particle loses energy. B) decreases linearly as the particle loses energy. C) remains constant as the particle loses energy. D) is independent of the energy of the particle. E) decreases exponentially as the particle loses energy.

The correct answer is: A Ionization gradually increases as the particle loses energy, until the ionization reaches a peak value and drops to zero as all the alpha energy is lost. Moe, H.J. (1992) Operational Health Physics Training .

Question Number: 52 Fundamentals of Radiation Protection When ionizing radiation impacts a biological molecule a(n) is ejected. A) Orbital electron B) Nuclear electron C) Neutron D) Proton E) Beta particle

The correct answer is: A Ionization is the ejection of an orbital electron. This ejection results in formation of an ion pair. (i.e. a positive ion that the electron left and a negative ion that the electron became.)

Question Number: 388 Fundamentals of Radiation Protection Ionizing radiation is different from non-ionizing radiation because: A) it can remove an electron from an atom. B) itcauses things to glow. C) it releases x-rays. D) it is radioactive. E) it can remove neutrons from atoms.

The correct answer is: A Ionization is the process by which an atom or a molecule acquires a negative or positive charge by gaining or losing electrons.

Question Number: 333 Fundamentals of Radiation Protection A person is undergoing an angioplasty procedure and is exposed to an extended amount of fluoroscopic exposure. The patient presents with some bruises after a few weeks, then the affected area appears to be slightly white in appearance. This is most likely due to: A) increased vascular permeability followed by capillary occlusion due to the formation of scar tissue. B) decreased vascular permeability followed by capillary occlusion due to the formation of scar tissue. C) increased vascular permeability followed by capillary dilation due to the formation of scar tissue. D) changes in the cellular structure leading to cellular impervious dielectric changes. E) removal of the lymphocytes within the area of interest due to the high radiation dose, followed by the death of red blood cells.

The correct answer is: A Irradiation causes increased vascular permeability, especially in capillaries. This leads to the bruised appearance. Next, the capillaries are occluded by late fibrotic action (scar tissue formation in the capillaries) that slowly cuts off blood flow in the area, so that it appears "white". Hall, E., Radiobiology for the Radiologist.

Question Number: 385 Fundamentals of Radiation Protection Isotones are atoms that have the same number of: A) Neutrons B) Protons C) Electrons D) Beta emissions E) Gamma emissions

The correct answer is: A Isotones are atoms that have the same number of neutrons. This can easily be remembered as isotoNes, has an "N" in it. Similarly, isotoPes has a "P", indicating that isotopes have the same number of protons. IsoMers have the same mass number. See Gollnick, Chapter 2.

Question Number: 238 Fundamentals of Radiation Protection The symbols Pa, Ga, Sn, Ni represent which respective elements? A) Protactinium, gallium, tin, nickel B) Protactinium, gadolinium, tin, nickel C) Palladium, gallium, tin, nickel D) Palladium, gadolinium, tin, nickel E) Praseodymium, gallium, tin, nickel

The correct answer is: A It is helpful to memorize chemical symbols of the elements either from the Periodic Table or the Chart of the Nuclides.

Question Number: 327 Fundamentals of Radiation Protection Kerma is a commonly misunderstood radiation dosimetry term. Kerma is actually an acronym for which of the following? A) Kinetic energy of radiation produced per unit mass in matter B) Kinetic energy radiated in materials C) Kinetic energy deposited in mass D) Kinetic energy that is mass absorbed E) Kinetic energy released in mass and absorbed

The correct answer is: A KERMA includes all energy released, whereas absorbed dose only includes energy released by particle collision. Cember, H. Introduction to Health Physics . Chapter 6.

Question Number: 323 Fundamentals of Radiation Protection Which of the following is true for a charged particle traveling through a material? A) The linear rate of energy loss is directly proportional to the electrical charge and inversely proportional to the velocity. B) The linear rate of energy loss is directly proportional to the velocity and inversely proportional to the electrical charge. C) The linear rate of energy loss is directly proportional to the velocity and electrical charge. D) The linear rate of energy loss is inversely proportional to the velocity and electrical charge. E) Answering this question correctly is directly proportional to my understanding of the question and indirectly proportional to my knowledge.

The correct answer is: A Multiple-charged particles will lose energy more rapidly than singly-charged particles. As charged particles slow down, they give up energy more readily.

Question Number: 205 Fundamentals of Radiation Protection The dose equivalent to the U.S. population from all man-made sources of radiation is given by the NCRP as approximately: A) 60 mrem/year B) 300 rem/year C) 1.5 rem/year D) 5 rem/year E) 12 rem/year

The correct answer is: A NCRP Report No. 160 shows the average annual radiation exposure to the population of the United States to be 620 millirem.

Question Number: 268 Fundamentals of Radiation Protection Cancer induced by radon daughters in underground mines is found primarily in the: A) upper trachea and bronchial tree. B) upper pulmonary parenchyma. C) lower trachea and bronchial tree. D) nasal passage. E) lower pulmonary parenchyma.

The correct answer is: A NCRP Report No. 78: "Attributed to inhalation of the airborne, short-lived daughters Po-218, Pb-214, Bi-214 and Po-214."

Question Number: 270 Fundamentals of Radiation Protection Unlike many man made materials, where radon release is very low, most soils have the ability to release more than of the radon formed. A) 10% B) 20% C) 30% D) 40% E) 60%

The correct answer is: A NCRP Report No. 78: "The fraction of radon released from a solid material depends on its porosity and whether radiation is on or near the surface of the material. It is currently believed that the source of most of the radon is the soil beneath a structure."

Question Number: 66 Fundamentals of Radiation Protection What is a "neutrino"? A) Massless particle B) Photon C) Neutron undergoing decay D) Beta antiparticle E) Positron antiparticle

The correct answer is: A Neutrinos are products of beta decay which share energy with the beta particle. They must be defined to suit the physics conservation laws and are termed "massless particles that travel at the speed of light." Neutrinos are of no consequence in Health Physics because their probability of interaction is zero and thus give no radiation dose.

Question Number: 87 Fundamentals of Radiation Protection Activation analysis occurs when: A) radiation occurs after absorption of a neutron. B) nucleus is bombarded by a linear accelerator. C) betatron scans nucleus. D) leptons are produced. E) quarks are produced.

The correct answer is: A Neutron activation analysis occurs after a nucleus has absorbed a neutron and then radiates. The radiation is characteristic of that nuclide.

Question Number: 13 Fundamentals of Radiation Protection 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 E) cm2/g or mg/cm2

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: 63 Fundamentals of Radiation Protection An activity of 3.7 x 10 10 becquerel is equivalent to how many Ci? A) 1.0 B) 0.1 C) 0.01 D) 0.001 E) 0.0001

The correct answer is: A One becquerel is equivalent to one disintegration per second. Since one Curie (Ci) has an activity of 3.7 x 10^10 disintegrations per second; any material that had this activity would equal one curie.

Question Number: 74 Fundamentals of Radiation Protection Which effect predominates in the 10 keV - 100 keV range for photons? A) Photoelectric (PE) B) Compton C) Pair production (PP) D) Elastic scattering E) Inelastic scattering

The correct answer is: A PE predominates in the above range. There can be some Compton effect, but due to the 1.02 MeV threshold in PP, there is no PP in this region. Elastic scattering can occur in any region, but does not predominate in this region.

Question Number: 51 Fundamentals of Radiation Protection Smokers have increased exposure to several radionuclides relative to non-smokers. This increased dose to the lungs is due to the presence of in tobacco. A) Polonium-210 and Lead-210 B) Thorium-232 C) Radon-222 D) Radium-226 E) Potassium-40

The correct answer is: A Polonium 210 and lead 210 are thought to be responsible for a three fold dose equivalent rate for the lungs of smokers versus those of non-smokers.

Question Number: 121 Fundamentals of Radiation Protection What is the approximate efficiency for the detection of Xenon-133 on activated charcoal? A) 1% B) 10% C) 50% D) 95% E) 99%

The correct answer is: A Since it is a noble gas, the efficiency is usually less than 1%. Xenon concentration would have to be measured with a Marinelli-type grab sample or by direct gamma measurement in the atmosphere. This radioisotope of xenon is a by-product in the fission process in a nuclear reactor and is used in Nuclear Medicine as a lung scanning agent.

Question Number: 44 Fundamentals of Radiation Protection Which of the following control methods would not be effective in controlling exposure to radon? A) Mechanical air filters B) Building pressurization C) Increase dilution ventilation in building D) Increase natural ventilation in building E) Sealing foundation cracks

The correct answer is: A Since radon is a gas, mechanical filtering devices would not be a good control method. Pressurizing the building, increasing dilution ventilation and natural ventilation would be methods for controlling radon gas exposure.

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

The correct answer is: A Soluble plutonium deposits mainly in the bones. The insoluble forms remain in the lungs.

Question Number: 356 Fundamentals of Radiation Protection More than 90% of the energy absorbed in tissue from neutrons less than 20 MeV is from: A) protons recoiling from inelastic collisions. B) elastic collisions during which the neutrons are slowed down to thermal energies. C) spallation. D) H-1 (n, gamma) H-2. E) N-14 (n, proton) C-14.

The correct answer is: A Spallation is typically only important above 20 MeV, H-1 and C-14 reactions deliver the majority of thermal neutron dose. (1990). BEIR V. Washington, DC., pg.16.

Question Number: 308 Fundamentals of Radiation Protection The term used to describe a process in which a number of fragments are emitted from an excited nucleus is: A) spallation. B) kerma. C) fission. D) fragmentation. E) scattering.

The correct answer is: A Spallation literally means "to break apart". Spallation occurs when a nucleus is bombarded with a particle that exceeds its binding energy, such as in an accelerator target. Moe, H.J. (1992) Operational Health Physics Training .

Question Number: 296 Fundamentals of Radiation Protection The formula for calculating specific activity is: A) λ * N B) (λ * N) / g C) (λ * g) / N D) N * g / λ E) g * λ

The correct answer is: A Specific activity is equal to decay constant ( λ) times the number of atoms (N) in a gram. Moe, H.J. (1992) Operational Health Physics Training .

Question Number: 275 Fundamentals of Radiation Protection During criticality accidents, exposure to neutrons will induce radioactivity such as activation of a normal body element to the gamma emitting isotope: A) Na-24 B) P-32 C) Cs-137 D) Co-60 E) I-131

The correct answer is: A Stable Na-23 absorbs a neutron, producing the gamma-emitting Na-24. NCRP 65, pg 29.

Question Number: 373 Fundamentals of Radiation Protection Stochastic effects in radiation generally refer to which of the following? A) Cancer B) Genetic defects C) Blood changes D) Events where increasing dose will increase the injury E) Physical injuries, such as radiation burns

The correct answer is: A Stochastic effects means health effects that occur randomly and for which the probability of the effect occurring, rather than its severity, is assumed to be a linear function of dose without threshold. Hereditary effects and cancer incidence are examples of stochastic effects. 10 CFR Part 20.1003

Question Number: 248 Fundamentals of Radiation Protection The antineutrino is associated with what process? A) Beta minus decay B) Beta plus decay C) Electron capture decay D) Alpha decay E) Neutron capture

The correct answer is: A The "Q" value in a beta minus decay is shared by the beta particle and the antineutrino, and unless the nuclide is a "pure beta emitter", also a gamma photon. In positron decay (beta plus) or electron capture decay, the neutrino shares this energy.

Question Number: 155 Fundamentals of Radiation Protection The underlying principle for measurement of radiation absorbed dose is the principle. A) Bragg-Gray B) Bergonie and Tribondeau C) Becquerel and Curie D) Einstein E) Coulomb

The correct answer is: A The Bragg-Gray principle relates absorbed dose in a medium (usually the detector wall material) to a measurement of a known amount of ionization in a small volume of a gas-(usually air) filled cavity.

Question Number: 148 Fundamentals of Radiation Protection KERMA is a unit of radiation measurement expressed in which of the following? A) Joule per kilogram B) Joule per coulomb C) Coulomb per kilogram D) Disintegrations per kilogram E) Disintegrations per liter

The correct answer is: A The KERMA (kinetic energy released in material) and absorbed dose differ in magnitude only if the energy transferred by the charged ionizing particle to the medium is not totally absorbed by the material. The KERMA is less than the absorbed dose, for instance, when radiative loss of energy (bremsstrahlung) is substantial. In this case, energy escapes via photons, with low probability of subsequent interaction in the material. KERMA is equal to absorbed dose when all energy loss is by collision of the charged particle in the material.

Question Number: 225 Fundamentals of Radiation Protection "Below a certain radiation dose, no effects whatsoever occur in the human body." This statement supports which dose/effect theory? A) Threshold B) Linear Non-Threshold C) Non-Linear Non-Threshold D) Linear Quadratic E) The "Hormesis" Effect

The correct answer is: A The Threshold Theory is not commonly subscribed to for radiation-induced stochastic effects.

Question Number: 240 Fundamentals of Radiation Protection If the maximum beta particle energy for a radionuclide is 7.0 MeV, the average beta particle energy is approximately: A) 2.33 MeV B) 0.223 MeV C) 1.43 MeV D) 0.256 MeV E) 700 keV

The correct answer is: A The average beta particle energy is calculated as 1/3 Emax, where Emax is the maximum beta particle (or "endpoint") energy.

Question Number: 105 Fundamentals of Radiation Protection Semiconductor detectors are used primarily for their: A) high energy resolution. B) low cost. C) portability. D) atmospheric resilience. E) All of the above

The correct answer is: A The best current health physics detectors are made from semiconductor material. They operate with low voltage but may require cryostatic temperatures.

Question Number: 250 Fundamentals of Radiation Protection The three types of blood cells in the human body are: A) erythrocytes, leukocytes, platelets. B) erythrocytes, red blood cells, platelets. C) white blood cells, leukocytes, platelets. D) leukocytes, granulocytes, platelets. E) red blood cells, white blood cells, granulocytes.

The correct answer is: A The blood is composed of red blood cells, white blood cells, platelets, and plasma (water). Red blood cells are clinically named erythrocytes and white blood cells are leukocytes. Granulocytes are a subgroup of white blood cells.

Question Number: 103 Fundamentals of Radiation Protection The build-up factor in radiation shielding is caused by: A) broad beam conditions. B) narrow beam conditions. C) particle emission. D) increments in shielding thickness. E) poor shielding materials.

The correct answer is: A The build-up factor, B, is equal to or greater than one. It is caused by incident photons scattered back to the detector by bad geometry, i.e., broad beam conditions. The B equals one under narrow beam conditions, i.e., good geometry.

Question Number: 124 Fundamentals of Radiation Protection The function of a cascade impactor is to separate: A) particles according to mass. B) radiation emissions (alpha, beta, gamma). C) solid material from liquid. D) liquid material from gas. E) particulate radiation emission (alpha, beta).

The correct answer is: A The cascade impactor separates particles according to their mass (diameter is also a determining factor). This is useful in characterizing the composition of airborne contaminants.

Question Number: 317 Fundamentals of Radiation Protection The approximate diameter of an atomic nucleus is: A) 1 x 10 -12 cm B) 1 x 10 -8 cm C) 1 x 10 -12 m D) 1 x 10 -10 m E) 1 x 10 -8 m

The correct answer is: A The diameter of the nucleus is roughly 1 x 10 -12 cm, while the diameter of the entire atom is roughly 1 x 10 -8 cm. Cember, Introduction to Health Physics.

Question Number: 64 Fundamentals of Radiation Protection A "pig" is a: A) container used to ship or store radioactive items. B) nuclear reactor. C) small ionization chamber to measure radiation. D) Survey instrument to integrate dose. E) metal alloy used in detector walls.

The correct answer is: A The etymology of this word is Celtic and not related to an animal. It is normally a lead container used for shielding.

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.

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: 37 Fundamentals of Radiation Protection Which of the following is a Beta emitter? A) Strontium-90 B) Americium-241 C) Thorium-232 D) Uranium-238 E) Neptunium-237

The correct answer is: A The heavier elements Am, Th, U, and Np are all alpha emitters. Strontium is a beta emitter.

Question Number: 136 Fundamentals of Radiation Protection The pancake GM probe is designed for: A) checking of surfaces. B) high radiation dose measurement. C) high radiation dose rate measurement. D) pulsed radiation fields. E) fixed area radiation monitoring.

The correct answer is: A The large surface area of a pancake detector is designed to make the GM sensitive enough to measure bench tops and personnel. This high sensitivity precludes its use in high radiation dose rate areas.

Question Number: 224 Fundamentals of Radiation Protection Which of the following radionuclides is a bone seeker? A) Ca-45 B) I-125 C) I-131 D) Kr-85 E) Co-60

The correct answer is: A The metabolism of radionuclides in the body is dependent upon chemical form. Since stable calcium goes to the bone, so does radioactive calcium.

Question Number: 133 Fundamentals of Radiation Protection Which radiation instrument below operates on the principle of discharging of a capacitor to estimate radiation dose? A) Pocket ionization chamber B) X-ray sensitive film C) Thermoluminescent dosimeter D) Proportional counter E) Bubble detector

The correct answer is: A The pocket chamber is a charged capacitor. As ionization occurs due to ionizing radiation exposure, the capacitor discharges and indicates the amount of exposure. Physical shock can likewise discharge the capacitor, yielding a false high dose.

Question Number: 94 Fundamentals of Radiation Protection A radionuclide that is constantly being released by cosmogenic action is: A) C-14 B) C-12 C) K-40 D) V-50 E) He-4

The correct answer is: A The two isotopes of carbon are present naturally, carbon-12 (stable) and carbon-14 (radioactive). C-14 is produced by the (n,p) reaction on N-14. Another common cosmogenic nuclide is H-3, which is a spallation product from primary cosmic interaction with O and N. The neutrons which initiate C-14 production are a portion of secondary cosmic radiation. Potassium-40 is a very long lived naturally occurring radionuclide and is present in soil and food.

Question Number: 181 Fundamentals of Radiation Protection The target organ for radium is which of the following? A) Bone B) GI tract C) Lung D) Thyroid E) Spleen

The correct answer is: A The uptake of radium is incorporated into the bone, possibly leading to bone cancers. Other target organs for radionuclides include: ISOTOPE TARGET ORGAN Tritium Total body Strontium Bone Iodine Thyroid Plutonium Bone

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/cm2 B) 0.1 W/cm2 C) 1.0 W/cm2 D) 10 W/cm2 E) 100 W/cm2

The correct answer is: A The value is normally listed as 10 mw/cm2.

Question Number: 265 Fundamentals of Radiation Protection Which of the following radionuclides is not included in the definition of Working Level? A) Rn-222 B) Po-218 C) Pb-214 D) Bi-214 E) Po-214

The correct answer is: A The working level is defined as that amount of radon daughters which produces 1.3 x 10^5 MeV alpha energy per liter of air. It does not include the contribution of radon itself. All other distractors are Rn-222 daughters.

Question Number: 70 Fundamentals of Radiation Protection What is the "photoelectric effect" in regard to ionizing radiation? A) Total absorption of incident photon energy B) Partial absorption of incident photon energy C) Conversion of incident photon to mass D) Change in direction with no change in energy E) Change in direction with 2% energy absorption

The correct answer is: A There are three processes to transfer energy from ionizing photons to matter: photoelectric, Compton, and pair production. In photoelectric interaction the incident photon energy is totally absorbed; in Compton the energy is partially absorbed with a scattered photon of less energy; in equivalents. pair production the energy is converted into mass

Question Number: 280 Fundamentals of Radiation Protection The relationship between the mass of radioactive material and the activity of a material is which of the following? A) Specific activity B) Decay constant C) Source activity D) Transformation constant E) Plancks constant

The correct answer is: A This quantity is typically expressed in curies per gram. Cember, Introduction to Health Physics.

Question Number: 263 Fundamentals of Radiation Protection The ICRP 26 risk factor used for Stochastic effects is: A) 0.01 per Sievert. B) 0.001 per Sievert. C) 0.0001 per Sievert. D) 0.002 per Sievert. E) 0.05 per Sievert.

The correct answer is: A This risk factor is based upon the annual mortality rate in safe industries, which is 1 in 10,000 or 1 x 10^-4. A risk of 1x 10^-4 per rem would be the same as a risk of 1 x 10^-2 per 100 rem. Since 100 rem equals 1 Sievert, the risk per Sievert is 1 x 10^-2, or 0.01.

Question Number: 245 Fundamentals of Radiation Protection A researcher orders 10 mCi of I-131, which has an 8-day half-life. If it takes 16 days for the shipment to reach its destination, then the minimum quantity that must be shipped is: A) 40 mCi B) 20 mCi C) 60 mCi D) 80 mCi E) 100 mCi

The correct answer is: A Using A = Ao e - ( t), solve for A o Ao = A/[e - ( t)] = 10 mCi/[e-((0.693/8 d) * 16 d) ] = 40 mCi

Question Number: 378 Fundamentals of Radiation Protection A beam of neutrons on a sample of pure water appears to be causing the emission of gamma rays. The most likely reason is: A) the prompt capture gammas from absorption of neutrons in the water. B) contamination of the water. C) imperfections in the neutron beam that cause the production of gamma rays with the neutrons. D) problems with the gamma ray detector causing it to respond to the neutrons. E) neutron disintegration.

The correct answer is: A When a nucleus absorbs a neutron, the binding energy of that neutron is released as a gamma ray. See Cember, Chapter 6.

Question Number: 242 Fundamentals of Radiation Protection The major source of radiation exposure averaged over the United States population is from: A) natural background radiation. B) diagnostic medical applications. C) therapeutic medical applications. D) fallout from nuclear weapons. E) releases from the fuel cycle necessary to support the civilian nuclear power program.

The correct answer is: A According to NCRP Report No.160, of the 620 mrem annual average exposure to the population of the U.S., 300 mrem is due to natural radiation. The components of natural radiation include cosmic and terrestrial sources. The largest contributor is radon/thoron daughters, at about 200 mrem.

Question Number: 200 Fundamentals of Radiation Protection Effects of acute radiation exposure on the central nervous system can be observed at doses of: A) 1000 rads. B) 500 rads. C) 250 rads. D) 100 rads. E) 25 rads.

The correct answer is: A CNS syndrome, where an electrolytic imbalance from excessive ionization causes nervous system dysfunction, does not occur until about 1000 rads of total body irradiation.

Question Number: 129 Fundamentals of Radiation Protection Chronic radiation exposure is the result of: A) low dose over a long time. B) moderate dose over a long time. C) low dose in a short time. D) high dose in a long time. E) moderate dose in a short time.

The correct answer is: A Chronic radiation exposure usually involves low level exposures over a long period of time. Acute radiation syndrome is caused by exposure to high levels of radiation over a short period of time.

Question Number: 342 Fundamentals of Radiation Protection C-band transmission optical networking systems (fiber optic) use lasers that operate from 1520 to 1570 nm. If a technician repairing this fiber optic has been exposed to this system, the tissue most likely injured would be the: A) cornea. B) lens. C) retina. D) fovea. E) macula.

The correct answer is: A Light at wavelengths greater than 1400 nm is absorbed in the cornea.

Question Number: 122 Fundamentals of Radiation Protection For what reason is a mica window used on a portable radiation detector? A) Low Z material to allow detection of beta radiation B) Maintains a partial vacuum well C) Light weight and inexpensive D) Machinable to correct geometry E) Deep dose tissue equivalence

The correct answer is: A Mica is fragile and subject to fracture and requires a protective cap when the instrument is not in use. It is low Z that will allow low energy beta radiation through. It can be made shallow dose equivalent if used at a density thickness of 7 mg/cm2.

Question Number: 286 Fundamentals of Radiation Protection An active person who eats a high fiber diet and has adequate fluids will have a gastrointestinal tract transit time of: A) 24 to 36 hours. B) 4 to 6 hours. C) 5 days or longer. D) 10 to 12 hours. E) 2 to 3 days.

The correct answer is: A The mean emptying time for the: Stomach: 1 hour Small Intestine: 4.0 hours Upper large intestine: 13 -20 hours Lower large intestine: 24 hours NCRP Report No. 65

Question Number: 198 Fundamentals of Radiation Protection A 7 mg/cm2 mylar window is used on a radiation detector. What is the minimum beta particle energy required to reach the detector fill gas? A) 7 keV B) 70 keV C) 700 keV D) 900 keV E) 1 MeV

The correct answer is: B 7 mg/cm2 is significant because it is the density thickness used for shallow dose equivalent. Thus, it requires a 70 keV beta particle to penetrate the dead layer of skin.

Question Number: 46 Fundamentals of Radiation Protection A classic type interaction occurs when a positron and an electron interact to produce 2 photons each with an energy of 511 KeV. The reaction is called: A) Compton effect. B) Annihilation. C) Fusion. D) Electron fission. E) Pair production.

The correct answer is: B A positron is a particle with the same mass as an electron but carries a positive charge. When a positron collides with an electron, both particles are annihilated and converted to two photons of energy, each with 511 KeV.

Question Number: 141 Fundamentals of Radiation Protection The "working level" (WL) used in the USA and Canada as a unit of radon decay is: 5 A) 1 liter water with emission of 1.3 x 10 MeV alpha. B) 1 liter air with emission of 1.3 x 10 5 MeV alpha. C) amount received averaged over 40 hours. D) volume of air averaged over 40 hours. E) 1 x 10 -7 pCi/L.

The correct answer is: B A working level is any combination of radon daughters in one liter of air that will result in the ultimate emission of 1.3 x 105 MeV of alpha energy. 1 working level is equivalent to 100 pCi/L air of Rn-222 in equilibrium with its daughters. 1 working level is also equivalent to 3 times the DAC of Rn-222 from 10 CFR Part 20 Appendix B.

Question Number: 42 Fundamentals of Radiation Protection How long will it take for 20 Ci of Polonium 210 to decay to less than 1 mCi? The halflife of Polonium 210 is 138 days. Polonium 210 is common in static removal devices. A) 3.0 years B) 5.4 years C) 9.2 years D) 10.1 years E) 12.3 years

The correct answer is: B Activity at t = Activity at to e^-((ln2/t1/2) * time) At = Ao e^-((0.693/138) * time) 1.0 mCi = 20 Ci e ^-((0.693/138) * time) Solve for Days: Days = - (ln 1 E-3/20) (138/0.693) Days = 9.9 * 199 = 1970 days = 5.4 years

Question Number: 78 Fundamentals of Radiation Protection What is the SI unit of activity? A) Ci B) Bq C) R D) Gy E) Sv

The correct answer is: B Activity is the amount of disintegrations per unit time or a reciprocal time unit. There are two such units for radiation: curie and becquerel. The SI unit is the becquerel. 1 Bq = 1 dps.

Question Number: 41 Fundamentals of Radiation Protection What will the activity of 10 Ci of I -131 be in 60 days? (Given that the halflife of I -131 is 8 days). A) 5.5 mCi B) 55 mCi C) 550 mCi D) 37.5 mCi E) 375 mCi

The correct answer is: B Activity: = Ci * e^- ((ln2/halflife) * time) = 10 Ci * e^- ((ln2/8) * 60) = 10 * e^- ((0.693/8) * 60) = 10 * e^- (5.20) = 10 (0.0055) = 55 mCi

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

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: 75 Fundamentals of Radiation Protection Which effect predominates in the 100 keV - 10 MeV region for photons? A) Photoelectric effect (PE) B) Compton effect C) Pair production (PP) D) Elastic scattering E) Inelastic scattering

The correct answer is: B All effects occur in this region. PE and elastic scattering are minimal due to the high momentum of the photon. PP has a threshold at 1.02 MeV, but the effect does not dominate until very high energies. Compton predominates, and since it is also Z independent, is the region of choice for radiation therapy and why lead aprons don't improve shielding.

Question Number: 110 Fundamentals of Radiation Protection Radium is considered to be a seeker when internal to the body. A) Lung B) Bone C) Total body water D) Thyroid E) Liver

The correct answer is: B All group II elements are considered bone seekers. They have a valence of 2+. Radium is especially important as the data from the radium dial workers has shown.

Question Number: 371 Fundamentals of Radiation Protection Self shielding is important when: A) counting beta emitters. B) counting alpha emitters. C) counting gamma emitters. D) entering a high radiation area. E) using x-ray machines.

The correct answer is: B Alpha emitters can be shielded by the medium in which they are collected. For example, collecting an air sample of alpha emitters can result in the particulate being embedded into the fibers of the air filter, and cause the alphas to be absorbed in the filter and not detected.

Question Number: 260 Fundamentals of Radiation Protection Based on the laws of Bergoinie and Tribondeau, rate the following cells from most to least sensitive. 1. Intestinal crypt cells 2. Nerve cells 3. Mature spermatocytes 4. Lymphocytes 5. Erythrocytes A) 4,5,1,3,2 B) 4,1,3,5,2 C) 1,4,3,2,5 D) 1,4,5,3,2 E) 3,4,1,5,2

The correct answer is: B Also use the heirarchy of effects from the Acute Radiation Syndrome to rank cells in their order of radiosensitivity. Knowing that white blood cells are the most radiosensitive immediately narrows the choice to only Answers A or B. Answer B is the better choice, since erythrocytes (no nucleus, specialized) are less radiosensitive than intestinal crypt cells (unspecialized, rapidly dividing).

Question Number: 318 Fundamentals of Radiation Protection The electron volt is: A) a unit of power equal to 1.6 x 10 -19 joules per second. B) a unit of energy equal to 1.6 x 10 -19 joules. C) a unit of electrostatic charge equal to 1.6 x 10 -19 coulombs. D) a unit of electrical capacitance equal to 1.6 x 10 -19 coulomb per volt. E) a unit of energy equal to 1.6 x 10 -19 ergs.

The correct answer is: B An electron volt is a unit of energy equal to 1.6 x 10 -19 joules.

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

The correct answer is: 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 Where: r1 = exposure rate 1 r2 = exposure rate 2 d1= distance 1 d2 = distance 2

Question Number: 178 Fundamentals of Radiation Protection Sr-90 decays by beta to Y-90 with a mass number of: A) 91 B) 90 C) 88 D) 87 E) 86

The correct answer is: B As beta decay causes no change in the mass number it remains at 90 for Yttrium.

Question Number: 287 Fundamentals of Radiation Protection In accidents that release fission products, radioactive iodine uptake by the thyroid can be checked by holding a beta-gamma survey probe over the thyroid. Peak thyroid uptake values will not be reached until after exposure. A) 2 hours B) 12 hours C) 2 days D) 30 minutes E) 36 hours

The correct answer is: B At 12 hours, equilibrium of iodine will be reached in the thyroid. NCRP 65.

Question Number: 258 Fundamentals of Radiation Protection Based on the 1990 BEIR V Report, the weighted average risk of death from cancer following an acute dose equivalent of 0.1 Sv of low-LET radiation to all body organs is estimated to be: A) 1 x 10^-4 B) 8 x 10^-3 C) 2 x 10^-3 D) 8 x 10^-4 E) 1 x 10^-2

The correct answer is: B BEIR V estimates 800 additional cancer deaths in a population of 100,000 exposed to 10 rad. Since the radiation weighting factor is 1 for low-LET radiation, we can assume that 10 rad is equal to 10 rem. 0.1 Sv equals 10 rem. Therefore, the risk is 800/100,000 or 8 x 10^-3.

Question Number: 182 Fundamentals of Radiation Protection Why are the low atomic number materials such as aluminum and organic plastics used to shield against Beta? A) Cheaper to use B) Eliminate Bremsstrahlung C) Higher atomic number materials are not effective D) Lighter than lead E) They can be transparent

The correct answer is: B Beta particles can generate Bremsstrahlung (braking radiation) x-rays when they are slowed too quickly by hitting materials of high atomic numbers. Therefore, aluminum and plastics with atomic numbers < 14 are utilized for shielding.

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

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: 19 Fundamentals of Radiation Protection The average person contains about 0.1 µCi of K-40 (half life = 1.27 x 10^6 years). How many dps is this? A) 3.7 x 10^4dps B) 3.7 x 10^3dps C) 4.6 x 10^6dps D) 2.6 x 10^3dps E) 1.3 x 10^6dps

The correct answer is: B By definition: 1 µCi = 3.7 x 10^4dps Therefore: 0.1 µCi = 3.7 x 10^3dps.

Question Number: 90 Fundamentals of Radiation Protection What is a use of a chelating agent? A) Change decay products to stable ones B) Chemically bind materials C) Change the activation energy of a radionuclide D) Measure radiation absorbed dose chemically E) To induce vomiting

The correct answer is: B Chelating agents are used to bind radioactive materials. The chelating agent is used in medicine to give the patient a correct chemical form of radionuclide for diagnostics. It is also used to remove excess uptakes of nuclides from the body. An example of chelating agent for Pu and Am is DTPA. Sometimes chelating agents are used to clean surfaces by binding the contamination in a manner that makes cleanup easier or more effective. They may also be used in the decontamination of contaminated systems.

Question Number: 360 Fundamentals of Radiation Protection Photon energy is deposited in tissue in many different ways. Which photon interaction dominates for photons between 1.02 MeV and 10 MeV incident upon the human body? A) Photoelectric Effect B) Compton Scattering C) Pair Production D) Photoneutron effect E) A combination of pair production and photoneutron

The correct answer is: B Compton scattering dominated from 0.1 MeV to almost 20 MeV in human tissues. (1990). BEIR V. Washington, DC., pg.10.

Question Number: 279 Fundamentals of Radiation Protection The minimum mass of material which sustains a nuclear chain reaction for a given set of conditions is called: A) mass defect. B) critical mass. C) enriched mass. D) keff mass. E) fissile material.

The correct answer is: B Critical mass depends on the fissile isotope, the isotopes enrichment, its geometry and the presence and types of moderator and reflector material. Cember, Introduction to Health Physics

Question Number: 288 Fundamentals of Radiation Protection Cesium-137 decays by emitting: A) two gammas: 1.17 Mev and 1.13 Mev. B) two betas: 0.51 Mev and 1.17 Mev. C) one gamma: 0.662 Mev. D) one positron: 0.662 Mev. E) alpha and gamma at 0.662 Mev.

The correct answer is: B Cs-137 decays by emitting beta particles of two different energies, 0.51 Mev (95%) and 1.17 Mev (5%) and is accompanied by a 0.662 Mev gamma from its daughter product Ba-137m.

Question Number: 204 Fundamentals of Radiation Protection Which of the following radionuclides is NOT a naturally occurring radionuclide? A) K-40 B) Cs-137 C) Ra-226 D) U-235 E) Ra-228

The correct answer is: B Cs-137 is a fission product. It may appear directly as a fission fragment, or occur from the beta-minus decay of Xe-137. K-40 is a primordial radionuclide. The remaining nuclides are members of naturally occurring decay series.

Question Number: 135 Fundamentals of Radiation Protection Both Cs-137 and I-131 are released as fission products by nuclear power plants. The biopathways for these two radionuclides can be traced via: A) grain intake (cereals). B) milk. C) leafy vegetables. D) potatoes. E) water.

The correct answer is: B Cs-137, I-131, Sr-90, and Sr-89 are picked up by cows from the grass and turn up in the milk which is consumed by humans. All of these nuclides are beta emitters. Cereals are a good indicator for Cs-137 and Sr-90, but not for I-131.

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

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: 25 Fundamentals of Radiation Protection The portion of the body most susceptible to laser damage is which of the following? A) Gonads B) Eye C) Skin D) Blood cells E) Bone

The correct answer is: B Damage can occur to the skin but it is of secondary importance to the eye.

Question Number: 374 Fundamentals of Radiation Protection When the severity of injury is directly related to increasing dose, it is referred to as which of the following? A) Stochastic effects B) Deterministic effects C) Teratogenic effects D) Normal effects E) Natural effects

The correct answer is: B Deterministic effects are also called non-stochastic effects.

Question Number: 112 Fundamentals of Radiation Protection A proportional counter has enough voltage to cause: A) ionization before recombination; no secondary electrons. B) secondary electrons with short range avalanche. C) secondary electrons with long range avalanche. D) continuous discharge. E) recombination prior to pulse.

The correct answer is: B Each answer above represents some part of the curve for gas-filled detectors. First is recombination, then the ionization region, the proportional region, the Geiger region, and the continuous discharge region.

Question Number: 56 Fundamentals of Radiation Protection One of the processes that increases the neutron/proton ratio in a nucleus involves the conversion of a proton to a neutron. This process is called: A) Beta minus decay B) Electron capture C) Fission D) Fusion E) Isomeric transition

The correct answer is: B Electron capture is sometimes called K capture because it involves the nucleus "capturing" an electron from the K or L shell. The electron converts one of the protons to a neutron and the neutron/proton ratio is increased.

Question Number: 139 Fundamentals of Radiation Protection Generally, to protect personnel against neutrons, use which of the following? A) High Z shielding (Z > 50) B) Low Z shielding (Z < 20) C) Middle Z shielding (20 < Z < 50) D) Albedo dosimetry E) Bubble detectors

The correct answer is: B For neutrons, a hydrogenated material such as paraffin or lucite is used. The hydrogen atoms are good neutron moderators. Care should be taken to shield the resulting gammas from hydrogen activation with a high Z material.

Question Number: 93 Fundamentals of Radiation Protection The process of combining two light nuclei into a third new nucleus is called which of the following? A) Fission B) Fusion C) Neutron activation D) Recombination E) Transmutation

The correct answer is: B Fusion is the process in which two low Z nuclei combine and produce a third nuclei. For example, two hydrogen atoms can combine to form helium. This is fusion.

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.

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: 187 Fundamentals of Radiation Protection Which two nuclides are produced by cosmogenic action? A) Ar-39 and Sr-90 B) H-3 and C-14 C) C-14 and Sr-90 D) H-3 and Ar-40 E) N-1 and Ar-40

The correct answer is: B H-3 is a spallation product from primary cosmic interactions of heavy nuclei with the atmosphere. C-14 is created by secondary cosmic interactions (in this case, neutrons) in the N-14 (n,p) C-14 reaction.

Question Number: 47 Fundamentals of Radiation Protection Higher atomic weight (Z) elements tend to release upon radioactive decay. A) Beta particles B) Alpha particles C) Gamma photons D) X-rays E) Positrons

The correct answer is: B High Z elements tend to release alpha particles upon radioactive decay. The lower Z elements tend to produce beta particles.

Question Number: 82 Fundamentals of Radiation Protection Which of the following is isobaric decay? A) Alpha emission B) Beta emission C) Gamma emission D) Characteristic X-ray emission E) Fission

The correct answer is: B In the beta decay processes, the Z of the decaying nucleus changes but not the mass. A beta particle or a positive beta is emitted with the corresponding neutrino. The mass of the original nucleus remains the same. This is isobaric decay.

Question Number: 197 Fundamentals of Radiation Protection In an elastic scattering interaction between a neutron and an atomic nucleus, which one of the following statements is true? A) A gamma photon will always be produced. B) A proton or a neutron will be emitted from the nucleus. C) Interaction with a heavy nucleus will result in proton recoil. D) Characteristic X radiation is always produced. E) Kinetic energy transferred to the nucleus is independent of the mass of the nucleus.

The correct answer is: B In the case of elastic scattering with a light nucleus such as hydrogen, a recoil proton is produced. In elastic scattering with a heavier nucleus, the neutron simply scatters off. Elastic scattering is often represented by the "marble and bowling ball" analogy.

Question Number: 117 Fundamentals of Radiation Protection Geiger counters have gas amplification factors: A) equal to 1. B) greater than 1. C) less than 1. D) dependent on the voltage. E) dependent on the volume.

The correct answer is: B Instruments that operate in the ionizing region have gas amplification factors equal to one; proportional counters and Geiger counter, greater than one. Proportional counters can distinguish between alpha and beta radiation, whereas Geiger counters cannot. Detectors with gas amplification factors greater than 1 operate in a pulse, rather than current mode.

Question Number: 353 Fundamentals of Radiation Protection Which of the following types of decay "competes" with isomeric transition? A) Auger electrons B) Internal conversion electrons C) Beta decay D) Electron capture E) X-ray emission

The correct answer is: B Internal conversion electrons are produced when a gamma from the nucleus liberates an electron in that atom. Turner, J.,Atoms, Radiation and Radiation Protection. Chapter 3.

Question Number: 159 Fundamentals of Radiation Protection In the fission process in a nuclear reactor, many radionuclides are produced; yet some radionuclides are considered environmental indicators of contamination, e.g. Sr-90 and I-131. Why? A) They are toxic. B) The exposure pathways are significant. C) They are volatile. D) They are short-lived. E) They are long-lived.

The correct answer is: B Iodine-131 has an 8 day half-life and Strontium-90, a 29.1 yr half-life. Their presence is easily detected in milk. This is largely due to the broad area a cow traverses to obtain her food.

Question Number: 116 Fundamentals of Radiation Protection Proportional counters have gas amplification factors that are: A) equal to 1. B) greater than 1. C) less than 1. D) dependent on the voltage. E) dependent on the volume.

The correct answer is: B Ionization detectors in the ionizing region have gas amplification factors equal to one; other instruments have factors greater than one. Proportional counters in particular distinguish between alpha and beta radiation. Gas amplification factor greater than 1 means secondary electrons are produced, and that the detector operates in the pulse mode, not the current mode. The pulse occurs at a lower voltage for alpha than for beta, or, if counted at the same voltage, the alpha produces a larger pulse than the beta.

Question Number: 247 Fundamentals of Radiation Protection Isobars are: A) different radionuclides decaying by the same mode. B) radioactive species containing the same number of nucleons. C) forms of the same chemical element containing different numbers of neutrons. D) elements with the same number of neutrons. E) gradients in radiation levels.

The correct answer is: B Isobars have the same number of nucleons (same "A" number). Isotopes have the same number of protons, but a different number of neutrons. Isotones have the same number of neutrons, but a different number of protons.

Question Number: 193 Fundamentals of Radiation Protection Isotopes are defined as: A) different radionuclides decaying by the same decay mode. B) forms of the same element containing different numbers of neutrons. C) forms of the same element containing different numbers of electrons. D) topes that have a temperature of 32°F. E) radioactive species containing the same number of neutrons PLUS electrons.

The correct answer is: B Isotopes have the same number of protons and a different number of neutrons. Isotones have the same number of neutrons and a different number of protons. Isobars have the same "A" number. Isomers have a parent nuclide in a metastable state.

Question Number: 212 Fundamentals of Radiation Protection Isotopes are: A) different radionuclides decaying by the same decay mode. B) forms of the same element containing different numbers of neutrons. C) forms of the same element containing different numbers of protons. D) radioactive species containing the same number of neutrons plus electrons. E) daughter products of radioisotopes in equilibrium.

The correct answer is: B Isotopes have the same number of protons and a different number of neutrons. Isotones have the same number of neutrons and a different number of protons. Isobars have the same number of nucleons.

Question Number: 236 Fundamentals of Radiation Protection An isotope is one of two or more atoms with: A) the same number of protons and a different number of electrons. B) the same number of protons and a different number of neutrons. C) the same number of neutrons and a different number of protons. D) the same atomic weight. E) the same mass number.

The correct answer is: B Isotopes have the same number of protons and a different number of neutrons. Isotones have the same number of neutrons and a different number of protons. Isobars have the same number of nucleons.

Question Number: 267 Fundamentals of Radiation Protection Radioactive atoms that have a neutron to proton ratio that is too large will decay by: A) beta plus decay. B) beta minus decay. C) electron capture decay. D) alpha decay. E) isometric transition.

The correct answer is: B It is helpful to visualize the Chart of the Nuclides Line of Stability to solve this problem. If the number of neutrons is too large, the nuclide lies to the right of the Line, and must rise one block to reach stability. A neutron is essentially decaying to a proton and a beta particle in this process within the nucleus.

Question Number: 367 Fundamentals of Radiation Protection Rn-222 is a part of which of the following decay chains? A) Thorium B) Uranium C) Neptunium D) Actinium E) Americium

The correct answer is: B It is part of the 4n+2 chain, U-238. Use the AUNT thumbrule.

Question Number: 326 Fundamentals of Radiation Protection It has been proposed that potassium iodide pills be stockpiled in the vicinity of nuclear power plants to protect the population from the deleterious effects of radiation. How does this pill protect a person? A) By interacting with oxygen free radicals to scavenge peroxides in the thyroid B) By causing an excess of stable iodine in a persons system, so that radio-iodine is not easily taken up by the body C) By causing a dearth of stable iodine in a persons system, so that radio-iodine is not easily taken up by the body D) By interacting with radio-iodine to "scavenge" the radio-iodine from a persons body, but mainly in the thyroid E) Stable iodine will cause the radio-iodine to not be absorbed at all in the body, and in fact, will replace any that is taken up prior to administration.

The correct answer is: B KI causes an excess of iodine such that the body "sheds" iodine and minimizes uptake of radio-iodine. It is most effective if taken prior to radio-iodine exposure.

Question Number: 91 Fundamentals of Radiation Protection What is a "nuclide"? A) Element B) Nucleus with specific Z and A numbers C) Radioactive element D) Nucleus with specific mass number E) Radioactive isotope

The correct answer is: B Nuclides specify the exact number of protons, neutrons, energy state of each. For example, Tc-99m and Tc-99 have different energy states. Also, more simply, Ac-227 and Ra-228 are different nuclides. A nuclide does NOT have to be radioactive.

Question Number: 315 Fundamentals of Radiation Protection Why is it acceptable to establish gamma dose equivalent rates (mrem/hr) using an exposure rate instrument that reads out in mR/hr? A) A conversion factor relating mR to mrem is always applied. B) One mR/hr is roughly equivalent to one mrem/hr for photons in tissue. C) Exposure rate instruments are constructed of tissue equivalent materials. D) In tissue, one roentgen of photons deposits exactly the same energy as one rad of photons. E) There is only a 10% difference between a reading of one mR/hr and one mrem/hr for photons in tissue.

The correct answer is: B One roentgen deposits 98 erg/gm in tissues roughly equivalent to the 100 erg/gm deposited by one rad. One rad = one rem for gamma photons, so one R = one rem for photons in tissue.

Question Number: 244 Fundamentals of Radiation Protection In the energy range from 0.1 to 5 MeV in either air or water, the most predominant photon interaction is: A) Pair production. B) Compton scattering. C) Photoelectric effect. D) Rayleigh scattering. E) Thomson scattering.

The correct answer is: B Photoelectric effect predominates below 200 keV, Compton scattering between 200 keV and 5 MeV, and Pair production above 5 MeV.

Question Number: 88 Fundamentals of Radiation Protection Plutonium 239: A) occurs naturally. B) is very soluble in all chemical forms. C) readily undergoes nuclear fusion. D) is easily produced in large quantities. E) can undergo thermal fission.

The correct answer is: B Plutonium is created in a nuclear reactor and is manmade. It is radioactive with a long half-life and is generally insoluble. It can be fissioned with thermal neutrons.

Question Number: 79 Fundamentals of Radiation Protection A positron has the same mass as the: A) Proton B) Beta particle C) Charged nucleon D) Alpha particle E) Any of the above

The correct answer is: B Positrons and beta particles differ in charge not mass. Positrons are positively charged electrons of nuclear origin.

Question Number: 303 Fundamentals of Radiation Protection Bremsstrahlung predominantly occurs at beat energies greater than: A) 6 Mev. B) 1 Mev. C) 70 kev. D) 100 kev. E) 2 Mev.

The correct answer is: B Probability of bremsstrahlung also increases with the density of the absorbing medium. Moe, H.J. (1992) Operational Health Physics Training .

Question Number: 59 Fundamentals of Radiation Protection Radioactive decay is said to be first order decay. It is also temperature. A) dependent on B) independent of C) dependent on enthalpy and D) dependent on entropy and E) dependent on high

The correct answer is: B Radioactive decay is first order decay and is independent of temperature.

Question Number: 269 Fundamentals of Radiation Protection The absorbed alpha dose to cells in the in the upper airways of the tracheobronchial tree is the significant dose for cancer induction in miners. A) larynx epithelium B) bronchial epithelium C) pulmonary epithelial D) trachea epithelium E) bronchial epidermis

The correct answer is: B See the NCRP Report regarding absorbed alpha radiation by the bronchial epithelium. NCRP Report No. 78

Question Number: 188 Fundamentals of Radiation Protection In an absorbing medium, one gray (100 rads) would produce: A) 1 x 102 ergs/gm B) 1 x 104 ergs/gm C) 1 x 106 ergs/gm D) 1 x 1010 ergs/gm E) 1 x 1012 ergs/gm

The correct answer is: B Since 1 rad produces 100 ergs per gram of absorber, 1 Gray (100 rads) produces 10,000 ergs per gram of absorber, or 1 x 104. 1 Gray is also equal to 1 Joule per kilogram of absorber.

Question Number: 217 Fundamentals of Radiation Protection Which of the following radionuclides requires the least mass to comprise one curie? A) U-238 B) Rb-88 C) Co-60 D) Mn-54 E) Cs-137

The correct answer is: B Since the curie is measured by the decay rate, the nuclide with the highest decay rate will require the least mass to make a curie. Since decay rate is inversely related to half-life, the shortest half-life will yield the highest decay rate. The half-lives are as follows: U-238 half-life: 4.47 x 109 years Rb-88 half-life: 17.7 minutes Co-60 half-life: 5.27 years Mn-54 half-life: 312 days Cs-137 half-life: 30.17 years

Question Number: 154 Fundamentals of Radiation Protection The SI unit of activity is the: A) Hertz. B) Becquerel. C) Curie. D) Gray. E) Sievert.

The correct answer is: B The Becquerel is the SI unit for activity and is equal to one disintegration per second.

Question Number: 273 Fundamentals of Radiation Protection ICRP 26, Recommendations of the ICRP, identify radiosensitive cells in the bone as: A) haematopoietic and trabecular cells in the bone. B) endosteal and epitheal cells on the bone surface. C) epithelial and epedermial cells on the bone surface. D) epedermial and lymphocyte cells in the bone. E) platlets and lymphocyte cells in bone marrow.

The correct answer is: B The Commission recommends that where possible, dose equivalent in the bone should apply to the endosteal cells and cells on bone surfaces, and should be calculated as an average over tissue up to a distance of 10 micrometer from bone surface. ICRP 26

Question Number: 156 Fundamentals of Radiation Protection The underlying principle for radioresponse of a mammalian tissue is called the Law of . A) Bragg and Gray B) Bergonie and Tribondeau C) Becquerel and Curie D) Curie and Joliet E) Bragg-Gray

The correct answer is: B The Law of Bergonie and Tribondeau states that radioresponsiveness is directly proportional to mitotic activity and inversely proportional to degree of differentiation of the cell.

Question Number: 161 Fundamentals of Radiation Protection What is the SI unit for the decay constant? A) Second B) 1/second C) Bq D) Hz E) ln 2

The correct answer is: B The SI decay constant unit is 1/second or second -1. Bq (becquerel) is the SI unit for activity; Hz is a unit for frequency.

Question Number: 184 Fundamentals of Radiation Protection The TLV for exposure to radiation from a microwave oven is: A) 10 W/cm2 B) 10 mW/cm2 C) 100 W/cm2 D) 100 mW/cm2 E) 1000 mW/cm2

The correct answer is: B The TLV for microwave radiation from a microwave oven is 10 mW/cm2. This is based on a frequency of 2450 MHz.

Question Number: 99 Fundamentals of Radiation Protection Strontium-90 is a very important radionuclide when discovered in the environment. It is a(n) emitter. A) Alpha B) Beta C) Gamma D) Beta-gamma E) Positron

The correct answer is: B The absorption of Strontium into the bone matrix is as serious as the energy of the beta is high (E max = 546 keV), and the physical half-life is long (29.1 yr).

Question Number: 73 Fundamentals of Radiation Protection What does the "Z" represent in reference to an atom or an element or a nuclide? A) Amount of energy required to ionize the atom B) The proton number or effective proton number C) Atomic weight or atomic mass D) Probability of decay E) Neutron absorption cross section

The correct answer is: B The atomic number, the number of protons, is referred to as the "Z" number. For example, helium has a Z equal to 2. The photoelectric effect is highly Z Z dependent; the Compton effect is Z independent and pair production is slightly Z dependent.

Question Number: 104 Fundamentals of Radiation Protection A build-up factor equal to one in a radiation shielding equation occurs under: A) broad beam conditions. B) narrow beam conditions. C) particle emission. D) increments in shielding thickness. E) use of poor shielding materials.

The correct answer is: B The build-up factor, B, in a radiation shielding equation is equal to or greater than one. For good geometry, B equals one; for poor geometry B is greater than one and is caused by the scattering of incident photons back into the detector area.

Question Number: 98 Fundamentals of Radiation Protection If Kr-90 is released from the stack of nuclear plant, what radionuclide would be detected from its decay chain? A) Kr-90 B) Sr-90 C) Zr-90 D) Sr-89 E) Kr-89

The correct answer is: B The chain: Kr-90 to Rb-90 to Sr-90 to Y-90 to Zr-90. This chain is environmentally important as Strontium-90 has a 29.1 yr half-life and a plus two ionic charge. It appears in milk and grain products, is a bone seeker and can present a threat to children.

Question Number: 396 Fundamentals of Radiation Protection The children of the survivors of Hiroshima and Nagasaki: A) have seen an increase in genetic effects due to radiation. B) have not seen any increase in untoward genetic defects due to radiation. C) were born sterile due to radiation. D) were legally required not to have children. E) have mostly passed away.

The correct answer is: B The children of the Hiroshima and Nagasaki Atomic Bomb Survivors (F1 generation) and their children (F2 generation) have shown no increase in untoward genetic defects compared with the control population. See BEIR Report VII.

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

The correct answer is: B The classic definition of Bremsstrahlung is the electromagnetic radiation produced by the rapid deceleration of charged particles.

Question Number: 249 Fundamentals of Radiation Protection Background radiation contributing to external dose in a wooden house versus a concrete building is probably: A) greater because of the greater shielding provided by the concrete building. B) less because of the natural radioactivity in the concrete. C) equal to the external radiation in the concrete building. D) less because of the external radiation from radon and thoron emanations from the concrete. E) greater because of the natural Potassium-40 in the wood.

The correct answer is: B The closest distractor in this question is Answer choice D, however, radon and thoron emanations pose an internal, rather than external, hazard.

Question Number: 89 Fundamentals of Radiation Protection Technecium-99m is: A) a noble radioactive gas. B) artificially produced and does not occur naturally. C) used in medicine in Magnetic Imaging. D) a positron emitter. E) a beta minus emitter.

The correct answer is: B The element Tc does not occur in nature. The metastable state decays from Mo-99 with a half-life of 6 hours and is used extensively in the area of nuclear medicine. It has no known toxicity to man, has excellent chelating chemistry, and can be produced in a radiopharmacy in the hospital with little cost or radiation problems. As a metastable nuclide, it emits only gamma radiation of 140.5 keV. This is a desirable characteristic of a diagnostic radionuclide.

Question Number: 48 Fundamentals of Radiation Protection The SI unit replacing the rad is the: A) Coulomb per kilogram. B) Gray. C) Rem. D) Lambert. E) Sievert.

The correct answer is: B The gray is the SI unit replacing the rad. One gray is equivalent to 100 rad or 1 joule/kilogram. The gray is denoted by Gy.

Question Number: 330 Fundamentals of Radiation Protection The "source" of the majority of radon in the environment is generally from which of the following decay series? A) Actinide B) Uranium C) Neptunium D) Thorium E) Polonium

The correct answer is: B The isotope of radon in the Uranium decay series is 222Rn. Cember, H.,Introduction to Health Physics, Chapter 4.

Question Number: 336 Fundamentals of Radiation Protection The maximum range in air of 20 MeV electrons from an electron beam accelerator is approximately: A) 20m B) 100m C) 500m D) 1000m E) 5000m

The correct answer is: B The range in air of an electron from an accelerator is approximately five times the electron energy in MeV. NCRP Report No.144.pg. 41.

Question Number: 80 Fundamentals of Radiation Protection The equivalent energy for the rest mass of a proton is about: A) 0.511 MeV B) 938 MeV C) 1 amu D) Z = 1 E) 1837 keV

The correct answer is: B The ratio of the mass of the proton to the electron is 1837, something you should know from chemistry. The ratio of the energies must be the same as per E = mc2. From annihilation radiation processes, you know that the rest mass of the electron equals 0.5ll MeV. Multiply that number by 1837.

Question Number: 186 Fundamentals of Radiation Protection Which of the following series is an artificially produced decay series? A) Thorium B) Neptunium C) Actinium D) Uranium E) Radium

The correct answer is: B The three natural decay series are Thorium, Actinium, and Uranium. They each begin with Th-232, U-235, and U-238, respectively. They are also denoted as 4n, 4n+3, and 4n+2, respectively. The Neptunium decay series begins with Np-237 and is denoted as 4n+1. Once a naturally occurring series due to natural sustained fission reactions, the Np series has decayed away and no longer exists in nature.

Question Number: 67 Fundamentals of Radiation Protection What is the mathematical relationship of the wavelength of a photon to its frequency? A) Direct B) Inverse C) Independent D) Equal E) Depends on the energy

The correct answer is: B The wave equation: speed of light = frequency * wavelength. The speed of light is a constant, therefore the relationship is inverse, i.e. as the wavelength decreases the frequency will increase. Conversely, if the wavelength increases the frequency will decrease.

Question Number: 234 Fundamentals of Radiation Protection The range of a 2 MeV alpha particle in air is about: A) 0.2 cm B) 1.0 cm C) 6.5 cm D) 7.5 cm E) 9.5 cm

The correct answer is: B There are classical calculations available to estimate alpha particle range based on energy. A simple thumb rule is .5 cm/MeV below 2.5 MeV, and .75 cm/MeV above 2.5 MeV.

Question Number: 71 Fundamentals of Radiation Protection What happens to the incident photon in the Compton process? A) Total absorption of the incident energy B) Partial absorption of the photon energy C) Conversion of photon to mass D) Change in the direction with no energy change E) Change in direction with 2% absorption

The correct answer is: B There are three absorption processes of incident photons in ionizing radiation: photoelectric effect, Compton effect, and pair production. In the first, the total energy of the photon is absorbed; in the second, the photon loses some energy to the medium and continues as a lesser energetic photon; and in the third, the energy is converted into mass.

Question Number: 101 Fundamentals of Radiation Protection In a radioactive decay process in which a long lived radionuclide decays into another radionuclide with a shorter but same order of magnitude half life, the process is called: A) Secular equilibrium B) Transient equilibrium C) No equilibrium D) Transmutation E) Depends on radionuclide

The correct answer is: B There is equilibrium whenever the parent radio-nuclide has a longer half-life than that of the daughter. For a difference larger than an order of magnitude, the process is secular. If the half-life of the parent is shorter than that of the daughter, there is no equilibrium.

Question Number: 130 Fundamentals of Radiation Protection A fundamental law in radiobiology that has been in use for over a century is the Law of: A) Becquerel. B) Bergonie & Tribondeau. C) Bragg-Gray. D) Curie. E) Andersson-Braun.

The correct answer is: B This "law" states: radiosensitivity of a cell is directly proportional to metabolic activity and inversely proportional to degree of differentiation. It was Pierre Curie who approached Bergonie and Tribondeau to irradiate frog cells to do the initial radiobiological work as Madam Curie never accepted that radiation was damaging.

Question Number: 241 Fundamentals of Radiation Protection The mass defect is equal to: A) the difference in binding energies between parent and daughter nuclides. B) the difference between the mass of the nucleus as whole and the sum of the component nucleus masses. C) the rest mass of particulate radiations emitted by the daughter nuclide. D) the sum of the binding energies of the parent and daughter nuclides. E) the rest mass energy equivalence of the daughter nuclide.

The correct answer is: B This difference in mass is equivalent to the "binding energy" of the nucleus, an illustration of E = mc 2.

Question Number: 359 Fundamentals of Radiation Protection As it pertains to dosimetry and energy transfer, the absorption by a medium of KINETIC energy from a high energy charged particle (electron) through excitation and ionization is BEST described by which of the following? A) Kerma B) Dose C) Mass stopping power D) Specific Ionization E) Linear energy transfer

The correct answer is: B This is the definition of dose. (1990). BEIR V. Washington, DC., pg.12.

Question Number: 343 Fundamentals of Radiation Protection The replacement of chlorine by UV light as a means to kill micro-organisms in water has been proposed. The wavelength of light that has been identified as best for killing micro-organisms is: A) active UV. B) UVB ~270 nm. C) UltraUV. D) 300 to 400 nm. E) 400 to 700 nm.

The correct answer is: B This is the most biologically effective wavelength.

Question Number: 310 Fundamentals of Radiation Protection Rn-220 is a member of which of the following decay series? A) Uranium B) Thorium C) Actinium D) Neptunium E) Radium

The correct answer is: B Thorium is the 4n series. The mass number 220 is divisible evenly by 4. (220 / 4 = 55, no remainder)

Question Number: 195 Fundamentals of Radiation Protection If a radionuclide undergoes eight half-lives, what percentage of the original activity remains? A) 0.15% B) 0.4% C) 0.78% D) 1.5% E) 12.5%

The correct answer is: B Use A = Ao (1/2)n A = 100% (1/2)8 A = 0.39%

Question Number: 222 Fundamentals of Radiation Protection Which of the following organ systems is the MOST radiosensitive? A) Central nervous system B) Bone marrow C) Gastrointestinal system D) Muscle system E) Skeletal system

The correct answer is: B Use the Law of Bergonie and Tribondeau, as well as the heirarchy of effects from the Acute Radiation Syndrome to solve this question.

Question Number: 106 Fundamentals of Radiation Protection Radiation in the visible spectrum that results from the slowing down of charged particles in a medium is called radiation. A) Bremsstrahlung B) Cerenkov C) Bragg-Gray D) Optical E) Infrared

The correct answer is: B When a charged particle moves through a medium at a velocity greater than the phase velocity of light in that medium, the phenomenon observed is called Cerenkov radiation. Instrumentation for this effect is not used in health physics but the effect can be observed when a beta source is stored in water.

Question Number: 157 Fundamentals of Radiation Protection Which of the below quality factors (QF) apply to diagnostic x-rays? A) 0 B) 1 C) 5 D) 10 E) 20

The correct answer is: B X-rays and gamma rays have quality factors of 1.

Question Number: 57 Fundamentals of Radiation Protection A scintillation counter takes advantage of properties of materials that after being exposed to energy from radioactive sources. A) emit fast neutrons B) fluoresce C) undergo beta decay D) undergo alpha decay E) undergo fission

The correct answer is: B Zinc sulfide and Sodium Iodide have been used as detector material in scintillation counters. These materials emit bursts of light after exposure to energy from radioactive sources. The bursts of light are multiplied in intensity and counted. The count is a measure of radioactivity.

Question Number: 190 Fundamentals of Radiation Protection A one liter Marinelli air sample is counted on a system which indicates 30,000 counts per minute. If the system has a counting efficiency of 17%, the activity of the sample is: A) 1.76 x 103 Bq/ml B) 2.90 Bq/ml C) 7.95 x 101 Bq/ml D) 1.06 x 1010 Bq/ml E) 1.06 x 107 Bq/ml

The correct answer is: B [30,000 c/(1 L * 1 x 10 3 mL/L)(.17)] (1 Bq/60 dpm) = 2.9 Bq/mL

Question Number: 35 Fundamentals of Radiation Protection When a charged particle strikes an electron knocking it out of its orbital, creating an ion pair, it is called which of the following? A) Indirect ionization B) Direct ionization C) Specific ionization D) Linear energy transfer E) Spallation

The correct answer is: B Direct ionization is ion pair formation by the incident charged particle.

Question Number: 213 Fundamentals of Radiation Protection What are isotones? A) Nuclides with the same mass number B) Nuclides with the same number of neutrons C) Nuclides with the same atomic number D) Same nuclides at different excited states E) Activated isotopes

The correct answer is: B Isotones have the same number of neutrons, but a different number of protons. Isotopes have the same number of protons, but a different number of neutrons. Isobars have the same number of nucleons.

Question Number: 83 Fundamentals of Radiation Protection Name the initial and final member of the uranium decay series. A) Th-232; Pb-208 B) U-238; Pb-206 C) U-235; Pb-207 D) Np-237; Bi-209 E) U-234; Rn-222

The correct answer is: B The four series in A through D have all occurred in nature. The neptunium series has decayed to the final product due to its relatively short half-life, and no longer occurs in nature. The uranium series is very important as it contains Rn-222 and Ra-226.

Question Number: 235 Fundamentals of Radiation Protection For a radionuclide with a decay constant of 0.693 per minute, the fraction of atoms that undergoes decay in three minutes is which of the following? A) 0.250 B) 0.875 C) 0.693 D) 0.069 E) 0.125

The correct answer is: B -( t) Use A = Ao e But remember, we are looking for the fraction which has decayed, NOT the fraction remaining. The fraction which has decayed will be represented by (Ao - A)/Ao. Therefore: let Ao = 1 and: A = 1 e- (0.693/min * 3 min) = 0.125 Then: (1 - 0.125)/1 = 0.875

Question Number: 145 Fundamentals of Radiation Protection The Department of Transportation (DOT) label Yellow II allows what radiation dose rate on the surface of the package? A) None above background B) Less than 5 µSv per hr C) Between 5 µSv per hr and 500 µSv per hr D) Between 100 µSv per hr and 500 µSv per hr E) Up to 2000 µSv per hr

The correct answer is: C 49 CFR Part 173 describes the labels. Radioactive Package II has two vertical red stripes. The upper half is yellow; bottom white. Surface radiation dose rate allowed is up to 500 µSv (50 mrem) per hr. Dose rate at 1 meter may be up to 10 µSv (1 mrem) per hr.

Question Number: 144 Fundamentals of Radiation Protection The Department of Transportation (DOT) label Yellow II allows what radiation dose rate at 1 meter? A) None above background B) Up to 5 µSv per hr C) Up to 10 µSv per hr D) Between 10 µSv per hr and 100 µSv per hr E) Between 100 µSv per hr and 500 µSv per hr

The correct answer is: C 49 CFR Part 173 describes the labels. Radioactive Yellow II has two vertical red stripes with upper half yellow and bottom white. Surface radiation rate allowed lies between 5 and 500 µSv per hr. Dose rate at 1 meter cannot exceed 10 µSv (1 mrem) per hr.

Question Number: 264 Fundamentals of Radiation Protection Genetic mutations are one possible result of exposure to ionizing radiation. Select the correct response regarding genetic effects. A) Within a population, a mutation is usually manifested in the first generation of offspring. B) Since many mutations are recessive, a large dose to a small population will cause more genetic damage than a small dose to a larger population. C) Hereditary defects are relatively rare, occurring < 0.1% of live-born infants. D) Radiation damage is particularly harmful because the body has no mechanism for repairing radiation induced mutation. E) For a given dose, the probability of genetic effect is assumed to be proportional to the rate at which the dose is received.

The correct answer is: C According to the ICRP, all genetic effects are distributed 50% in the first two generations and 50% in subsequent generations. Radiation-induced genetic effects have been estimated to be less than 4 x 10^-5 per rem. Since average lifetime population exposures (from all sources including occupational) which could cause genetic effects are probably on the order of magnitude of 1 x 10^1 rems or less per person, it could be argued that hereditary defects would be less than 0.1% (1 x 10^-3).

Question Number: 325 Fundamentals of Radiation Protection The minimum dose required to cause permanent male sterility, according to the BEIR V report, is an acute dose to the testes of: A) 0.1 to 0.3 Gy. B) 1 Gy. C) 3 to 5 Gy. D) 5 to 7 Gy. E) 18 mGy per day for one week.

The correct answer is: C Acute doses of 3 to 5 Gy are required to cause permanent male sterility. Acute doses below this may cause temporary sterility. Note that the LD50 is approximately 3 to 5 Gy. A local dose to the reproductive organs would be implied in this question. Radiation therapy patients would be an important population to consider with this type of dose. BEIR V Report Washington, DC. p. 365.

Question Number: 134 Fundamentals of Radiation Protection Which radiation monitoring device uses a crystal with an added impurity to trap electrons during radiation exposure? A) Pocket dosimeter B) X-ray sensitive film C) Thermoluminescent dosimeter D) Plastic scintillator E) Long counter

The correct answer is: C After the electrons are trapped, the TLD is subjected to a heating process. The trapped electrons emit optical photons (light) which is measured to determine exposure. Since heat is used and light is emitted, the name thermoluminescence is appropriate.

Question Number: 176 Fundamentals of Radiation Protection Radium-226 decays by alpha to Radon with an atomic number of: A) 88 B) 87 C) 86 D) 85 E) 84

The correct answer is: C An alpha particle contains 2 neutrons and 2 protons, thereby the release of one alpha particle decreases the element's atomic number by 2, as only protons are counted for atomic number.

Question Number: 256 Fundamentals of Radiation Protection Workers are exposed to a Cs-137 concentration of 5 x 10^-4 µCi /cc for 2 hours. Calculate their approximate committed effective dose equivalent as a result of inhalation. GIVEN: Mass of lung: 5809 grams Reference Man Breathing Rate: 137-Cs DAC: 1 x 10^7 cc/8 hr 6 x 10^-8 µCi i/cc A) 2.1 x 10^3 mrem B) 8.3 x 10^3 mrem C) 4.2 x 10^4 mrem D) 8.3 x 10^4 mrem E) 4.2 x 10^5 mrem

The correct answer is: C Assuming the DAC is based on the sALI, a simple DAC-hour calculation will be sufficient to calculate the CEDE. When the DAC is based on the sALI, one DAC-hour is equal to 2.5 millirem. When the DAC is based on the nALI, one DAC-hour is equal to 25 mrem to the target tissue. Therefore CEDE: = [(concentration/DAC) * hrs] * 2.5 mrem/DAC-hr = [(5 x 10 -4/6 x 10 -8) * 2] * 2.5 = 4.16 x 10^4 mrem

Question Number: 239 Fundamentals of Radiation Protection The primary mechanism for charged particles to lose energy in a medium are: A) nuclear interactions involving ejection of multiple particles from the target nucleus. B) Bremsstrahlung effects arising in the electronic structure of the medium. C) ionization and excitation affecting the electronic structure of the medium. D) gravitation interactions. E) pair production interactions.

The correct answer is: C Charged particles interact by ionization and excitation, and in the case of beta particles, Bremsstrahlung. Effects on electrons can alter the structure of atoms in the absorbing medium.

Question Number: 158 Fundamentals of Radiation Protection As a charged particle loses energy in a medium and begins to slow down and then stop, where along the path from entry to stop is most of the energy deposited? A) Entry (0 depth) B) Electron equilibrium C) End of path D) Equally along path E) Beginning 1/3 of path

The correct answer is: C Charged particles lose most of the energy at end of path, particularly the monoenergetic ones. The Bragg Peak is an example of the large dose deposited at the end. Betas lose most of the energy at 2/3 maximum range.

Question Number: 211 Fundamentals of Radiation Protection The radius of the nucleus of an atom as compared to the atom as a whole is: A) some 10,000 times greater. B) some 1,000 times greater. C) some 10,000 times smaller. D) some 1,000 times smaller. E) some 100 times smaller.

The correct answer is: C Contrary to many pictorial representations of the atom, the nucleus is relatively small. If the nucleus were the size of an orange, the diameter of the atom would be on the order of 1000 yards.

Question Number: 309 Fundamentals of Radiation Protection An individual living in Denver, Colorado (elevation 5280 ft above sea level) could expect to receive an average annual dose equivalent due to cosmic radiation of approximately: A) 13 mrem. B) 26 mrem. C) 50 mrem. D) 76 mrem. E) 200 mrem.

The correct answer is: C Cosmic radiation accounts for approximately 26 mrem annual average dose equivalent at sea level in the U.S. Each 2000 meter increase in altitude above sea level roughly doubles this value. NCRP Report No.93 and 160.

Question Number: 282 Fundamentals of Radiation Protection The NRC defines a hot particle as a: A) discrete radioactive fragment that is insoluble in water, is less than 1 cm and exhibits a dose rate of greater than 100 mrad/hr. B) small radioactive particle that is soluble in water and is less than 1 mm. C) discrete radioactive fragment that is insoluble in water and is less than 1 mm. D) small radioactive particle that is insoluble in water, is less the 1/4" and exhibits a dose rate of greater than 50 mrad/hr. E) discrete radioactive particle that is less than 1 mm and exhibits a dose rate of greater than 100 mrad/hr at 1 inch.

The correct answer is: C Criteria for Hot Particle Detection and Control can also be found in NCRP Report No.106 and 130. NRC Information notice IN 90-48.

Question Number: 357 Fundamentals of Radiation Protection Neutrons interact with tissue in all the following ways EXCEPT: A) Elastic collisions B) Inelastic collisions C) Recoil capture with delta ray formation internally D) Capture reactions E) Spallation

The correct answer is: C Delta rays are insignificant in producing dose in tissue. (1990). BEIR V. Washington, DC., pg. 15.

Question Number: 43 Fundamentals of Radiation Protection Radon has been estimated to account for 5k-20k lung cancers in the United States alone. Controlling radon exposures in a building means that of the building is to be avoided. A) pressurization B) congestion C) depressurization D) dilution ventilation E) open ventilation

The correct answer is: C Depressurization of the building would tend to draw radon gas in from the surrounding soils. Depressurization therefore is to be avoided.

Question Number: 172 Fundamentals of Radiation Protection The occupational derived air concentration (DAC) is calculated based on the average airborne concentration that a person would receive in what period of time? A) 40 hr week B) 168 hr week C) 2000 hr year D) 2080 hr year E) Depends on the radionuclide

The correct answer is: C Derived air concentration is the average concentration over a 2000 hr period. It is calculated by dividing the ALI by the product of the reference man breathing rate (1.2 cubic meters per hour) and 2000 hrs.

Question Number: 76 Fundamentals of Radiation Protection What effect predominates in the greater than 10 MeV ionizing region? A) Photoelectric Effect (PE) B) Compton effect C) Pair production (PP) D) Elastic scattering E) Inelastic scattering

The correct answer is: C Do not confuse 10 keV, the lower level for the ionizing region, with 10 MeV, the beginning of the energy range in which PP begins to dominate the absorption processes. The lower energy PE dominates until about 100 keV when Compton effect is about 50% of absorption process. Elastic scattering can occur but is of little importance in this range.

Question Number: 68 Fundamentals of Radiation Protection What is elastic scattering of a photon? A) Energy increase of scattered photon B) Energy decrease of scattered photon C) No energy change of scattered photon D) Total energy transfer to an electron E) Depends on incident energy

The correct answer is: C Elastic scattering, also called Thomson scattering or reflection in optics, is an example of a photon which changes direction but not energy.

Question Number: 77 Fundamentals of Radiation Protection What is the energy equivalent of the electron? A) 1.6 x 10-19 C B) Depends on sign of the charge C) 0.511 MeV D) 0.536 amu E) All of the above

The correct answer is: C Electrons, positrons, positive betas, and betas all are equivalent to 0.511 MeV of energy, independent of charge. Einstein's formula, E = mc2, is used to calculate the equivalents. A proton is equivalent to 938.2 MeV. One atomic mass unit (amu) is equivalent to 931.5 MeV.

Question Number: 302 Fundamentals of Radiation Protection The average energy of a positron emitted during positron decay is: A) 1/3. B) 1/4. C) 4/10. D) 1/2. E) 3/4.

The correct answer is: C For positrons, it is 4/10 or 0.4 of the maximum energy. Moe, H.J. (1992) Operational Health Physics Training .

Question Number: 196 Fundamentals of Radiation Protection The difference between x-rays and gamma photons of the same energy is their: A) frequency. B) wave-length. C) origins. D) properties. E) mass.

The correct answer is: C Gammas originate from the nucleus, while x-rays originate from an electron shell, or in the case of Bremsstrahlung, at some point outside of the nucleus.

Question Number: 259 Fundamentals of Radiation Protection All of the following are applicable to the lung cancer observed in uranium miners EXCEPT: A) Miners who also smoke are observed to have a much higher incidence of lung cancer than non-smokers. B) Cancer is observed primarily in the bronchial epithelium region. C) Cancer is probably associated with the radiation dose from "hot particles". D) Cancer is believed to be the result of the dose delivered by radiation emitted from radon daughter products deposited in the respiratory tract. E) Cancer is observed primarily in workers who have a number of years in the industry.

The correct answer is: C Hot particles are not encountered in the mining of natural uranium. All other distractors are valid data.

Question Number: 337 Fundamentals of Radiation Protection When electrons are accelerated above approximately 1.5 MeV, the intensity of the bremsstrahlung generally peaks in: A) a pattern that generally corresponds to a 90 degree wedge shape that can be approximated by assuming a hemi-sphere. B) a reverse pattern that is similar to that seen in a comet. C) the forward direction. D) a manner that can be predicted based on the tangent of the angle through which the electron is deflected by the interacting materials. E) a manner that results in an isotropic dispersion of photons.

The correct answer is: C In other words, the intensity of the bremsstrahlung generally peaks in the direction of the beam. NCRP Report No.144.pg. 46.

Question Number: 395 Fundamentals of Radiation Protection An increased probability of radiation induced teratogenic effects are only expected: A) at very high doses to a man's epithelial tissues. B) at very high doses to children after they have experienced puberty but before age 18. C) at very high doses to pregnant women. D) when beta doses to the eye are above legal limits. E) when older radiation workers approach lifetime doses of 50 Gy or more.

The correct answer is: C Increased probability of teratogenic effects are observed in children exposed during fetal and embryonic stages to doses above 10 rads.

Question Number: 15 Fundamentals of Radiation Protection 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 E) Krypton-85

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: 295 Fundamentals of Radiation Protection Xe-133 undergoes isomeric transition; the daughter product is: A) Cs-133. B) Te-133. C) Xe-133. D) I-134. E) I-133.

The correct answer is: C Isomeric transition involves an radionuclide existing in a metastable state. When it transitions to a stable state it emits a gamma and is the same element. Moe, H.J. (1992) Operational Health Physics Training .

Question Number: 109 Fundamentals of Radiation Protection Neutron activation is a process in which a "free" (unbound) neutron: A) is produced in fission. B) is elastically scattered. C) is captured by a nucleus. D) decays. E) accelerates.

The correct answer is: C Many nuclei can absorb a neutron for a finite amount of time and produce an isotope of that nuclide. This nuclide is usually radioactive and decays with a characteristic radiation, which serves to identify the original nucleus. Arsenic in Napoleon's hair was identified by neutron activation analysis.

Question Number: 206 Fundamentals of Radiation Protection Most of the dose received by man from man-made sources of radiation is due to: A) the operation of nuclear power reactors. B) industrial applications of radiation and radioactive materials. C) medical applications of radiation and radioactive materials. D) fallout from nuclear weapons testing. E) Ra-226

The correct answer is: C NCRP Report No.160 shows the average annual radiation exposure to the population of the United States to be 620 mrem.

Question Number: 39 Fundamentals of Radiation Protection Which one of the following types of radiation has the highest quality factor? A) Beta particles B) Gamma radiation C) Alpha particles D) High speed neutrinos E) Neutrons of unknown energy

The correct answer is: C Quality factors measure linear energy transfer. Since alpha particles are large, they would have a large quality factor. Beta particles are much smaller and hence a smaller quality factor.

Question Number: 272 Fundamentals of Radiation Protection Deleterious effects caused by radiation exposure to gonads may take the form of: A) fertility problems and stochastic effects. B) tumor reduction and somatic effects. C) fertility problems and hereditary effects. D) tumor induction and tetrogenic effects. E) hereditary and tetrogenic effects.

The correct answer is: C Radiation exposure may cause effects of three different types: tumour induction, fertility problems and hereditary effects. ICRP 26 pg. 9.

Question Number: 54 Fundamentals of Radiation Protection Cobalt 60 has a half life of 5.27 years. What amount of a 0.100 gram sample will remain after 1.0 years? A) 8.8 g B) 0.876 g C) 0.0876 g D) 0.98 g E) 0.098 g

The correct answer is: C Rate constant k = 0.693/half life Rate constant k = 0.693/5.27 years Rate constant k = 0.131 years Log (No/N) = kt/2.3 Log (No/N) = (0.131)(1.0 year)/2.3 Log (No/N) = 0.057 Log (N/No) = -0.057 (N/No) = 0.0876 (fraction remaining) (Fraction remaining)(Sample size) = Amount remaining (0.876)(0.1g) = 0.0876g

Question Number: 301 Fundamentals of Radiation Protection For an alpha particle to penetrate the dead layer of skin, it must have an energy of at least: A) 2 Mev. B) 70 kev. C) 7.5 Mev. D) 5 Mev. E) 100 kev.

The correct answer is: C Since few alphas are produced at energies exceeding 7 MeV, alphas are not generally considered to be an external radiation hazard. Moe, H.J. (1992) Operational Health Physics Training pg.3-8.

Question Number: 276 Fundamentals of Radiation Protection Sr-85 is used in the treatment of bone cancer because it is a: A) beta emitter that eliminates carcinomas in the bone marrow. B) gamma emitter that is absorbed in the bone. C) beta emitter that is taken up by the bone. D) gamma emitter that concentrates in sites of active osteogresis. E) beta emitter that remains in healthy bone for a period of years.

The correct answer is: C Sr-89 is a beta emitter which localizes in the bone. The major feature is that it is calcium analog, meaning very little uptake in the bone marrow. Marrow is spared high doses from absence of gamma emissions. Sr-89 is taken up by tumors and cancer sites and rapidly washes out of healthy bone. Early, P., Lasada, E., Health Physics Journal, 1995, Vol 69, No5.

Question Number: 202 Fundamentals of Radiation Protection What is the "threshold" of radiation dose effects? A) The dose of radiation that will kill any organism B) The dose of radiation that will kill 50% of those exposed C) The dose of radiation below which there are no effects whatsoever on the body D) The dose of radiation that will begin to alter cell structure E) The dose of radiation at which the damage effect is just balanced out by recovery

The correct answer is: C The "threshold", as used here, only applies to non-stochastic effects of radiation. The current dose effect theories (linear and linear-quadratic) assume that some biological effect occurs, no matter how small the dose.

Question Number: 376 Fundamentals of Radiation Protection If the exposure rate from a source is 2.58 × 10-4 C/kg/hr: A) the dose rate is 1 R/hr. B) the dose equivalent is 1 R/hr. C) the exposure rate is 1 R/hr. D) the exposure rate in traditional units can only be found if the quality factor for the radiation is known. E) the dose rate is 1 Sv/h.

The correct answer is: C The C/kg unit is a unit of exposure, not dose.

Question Number: 230 Fundamentals of Radiation Protection If a large group of people receive an acute deep dose of 450 rads, what fraction of the people would you expect to be alive in 2 months if they receive no medical treatment? A) Less than 25% B) 25% C) 50% D) 75% E) 100%

The correct answer is: C The LD50/60 is currently accepted to be about 450 rads. This is the amount of total body irradiation required to cause death of 50% of the exposed population in 60 days. The term is discussed in BEIR V (1990) and ICRP Publication 60 (1990).

Question Number: 58 Fundamentals of Radiation Protection In a Geiger-Muller counter, gammas and x-rays pass through a chamber containing argon. The argon atoms are in the chamber and move through an electric field. A) converted to alpha particles B) left unchanged C) ionized D) annihilated E) activated

The correct answer is: C The argon atoms are ionized to Ar+. The chamber where the ionization takes place is in an electric field of 1000-1200 volts. The positively charged argon ions cause an avalanche of ions to reach the detector anode as a pulse. This pulse is amplified and translated into a count rate, and the clicking sound heard when using the instrument.

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

The correct answer is: C The decay constant is expressed in disintegrations per second.

Question Number: 126 Fundamentals of Radiation Protection The "end window" on a Geiger-Mueller probe: A) measures output from the aperture of x-ray machines. B) filters out alpha and beta radiation. C) is the low Z material for alpha and beta detection. D) discriminates alpha from beta radiation. E) discriminates alpha from gamma radiation.

The correct answer is: C The end window is the low Z material that allows penetration of alpha, beta, and gamma. Some GM instruments are also equipped with a movable window to discriminate beta from gamma. The movable window, however, is not part of the detector wall as is the end window.

Question Number: 97 Fundamentals of Radiation Protection All the natural radioactive decay series: A) have very long half-lives. B) include Bismuth as a transmutant. C) end in a stable isotope of lead. D) include primordial radionuclides. E) include fission fuels.

The correct answer is: C The neptunium series has decayed away with its relatively short half-life in comparison to that of the earth. It's half-life is about 1 x 10^6yr and terminates with Bi-209. The other three series; thorium, uranium and actinium, terminate in a stable isotope of lead.

Question Number: 125 Fundamentals of Radiation Protection A pancake probe is used to: A) filter out beta radiation from photons. B) be wedged between two others to detect betas. C) provide a large surface area. D) be applied outside of a faulty detector. E) perform fixed area radiation monitoring.

The correct answer is: C The pancake probe was designed for rapid checking of laboratory bench top and "frisking" of personnel. The large surface area provides a large sensitive volume to measure alpha and beta that could go undetected as background. It is required by NRC for some laboratories.

Question Number: 102 Fundamentals of Radiation Protection In a radioactive decay process in which the parent radionuclide has a shorter half-life than that of the daughter, but same order of magnitude, the process is called: A) Secular equilibrium B) Transient equilibrium C) No equilibrium D) Transmutation E) Depends on radionuclide

The correct answer is: C The parent must have a longer half-life than that of the daughter for equilibrium to exist. In secular equilibrium, the half-life is much larger for that of the parent; transient equilibrium, the same order of magnitude.

Question Number: 232 Fundamentals of Radiation Protection What is the main factor determining the amount of energy that can be transferred to the electron during Compton scattering? A) The resultant characteristic x-ray B) Differences in energy between "K" and "L" shell sites C) The scattering angle D) The energy of the scattered gamma photon E) Nuclear binding energy

The correct answer is: C The scattering angle of the incident photon is inversely proportional to the energy of the scattered photon, and therefore directly proportional to the energy transferred to the electron. In other words, the greater the scattering angle, the greater the energy transferred to the electron, and the less the energy of the scattered photon.

Question Number: 153 Fundamentals of Radiation Protection The SI unit for dose equivalent in man is the: A) Gray. B) Rad. C) Sievert. D) Rem. E) Becquerel.

The correct answer is: C The sievert is defined as a quality factor times the dose in gray. The rem is the "special unit" counterpart and is equal to 0.01 Gy.

Question Number: 339 Fundamentals of Radiation Protection Ultraviolet light is a known carcinogen. The ultraviolet portion of the EM spectrum is commonly divided into three parts. Those three parts are: A) Active UV, Actinic UV, and Blue-UV. B) Vacuum UV, Occlusive UV, Transmissive UV. C) UVA, UVB, UVC. D) UVK, UVL, UVM. E) UV, UltraUV, InfraUV.

The correct answer is: C The spectrum is based upon the frequency of the electromagnetic field. Cember, H.,Introduction to Health Physics, Chapter 14.

Question Number: 85 Fundamentals of Radiation Protection Name the initial and final members of the actinium series. A) Th-232; Pb-208 B) U-238; Pb-206 C) U-235; Pb-207 D) Np-237; Bi-209 E) Ac-227; Bi-209

The correct answer is: C The three naturally occurring decay series (and the parents) are Thorium (232-Th). Uranium (238-U), and the Actinium Series (235-U). The fourth series is from nuclear reactions is the neptunium series (Np-237). Thorium series is called the 4n series; the neptunium series, 4n+1; the uranium series, 4n+2; and the actinium series, 4n+3

Question Number: 72 Fundamentals of Radiation Protection What happens to the incident photon in the pair production process? A) Total energy absorption B) Partial absorption C) Conversion of photon into mass D) Change in direction with no energy loss E) Change in direction with 2% absorption

The correct answer is: C There are three absorption processes: photoelectric, Compton, and pair production. In PE, the incident photon is totally absorbed; in Compton, the photon gives some energy to the medium and is scattered with less energy. In PP, the photon energy, which must exceed 1.02 MeV, is converted into two electrons of opposite charge, each with 0.511 MeV of energy.

Question Number: 228 Fundamentals of Radiation Protection Why is a child more radiosensitive than an adult? A) An adult has better health. B) A child's cells are weaker. C) A child's cells are multiplying more rapidly. D) A child's cells have less ability to repair damage. E) A child's cells are smaller.

The correct answer is: C This is a simple application of the Law of Bergonie and Tribondeau.

Question Number: 306 Fundamentals of Radiation Protection Kerma is the sum of the initial kinetic energy of all charged ionizing particles liberated by: A) photoelectric effect. B) pair production. C) uncharged ionizing radiation. D) elastic scattering. E) ionization.

The correct answer is: C This kinetic energy is liberated by collision (ionization and excitation) and radiative (bremsstrahlung) energy losses. Cember, Introduction to Health Physics.

Question Number: 243 Fundamentals of Radiation Protection The desired quantity of Technetium-99m (t1/2 = 6 hrs) for a medical procedure is 100 millicuries. If the shipment takes one day to reach the facility, what quantity of Technetium-99m should be shipped? A) 200 millicuries B) 800 millicuries C) 1.6 curies D) 1.8 curies E) 2.2 curies

The correct answer is: C Using A = Ao e - ( t), solve for A o Ao = A/[e - ( t)] = 100 mCi/[e - (0.693/6 hr * 24 hr) ] = 1600 mCi = 1.6 Ci

Question Number: 253 Fundamentals of Radiation Protection Which of the following is NOT an isobaric transformation? A) Beta minus decay B) Beta plus decay C) Alpha decay D) Electron capture E) Isomeric transition

The correct answer is: C We know that isobars have the same number of nucleons (same "A" number). The only type of decay shown which does not yield a daughter of the same "A" is alpha decay, where "A" decreases by 4.

Question Number: 69 Fundamentals of Radiation Protection What is inelastic scattering of a photon? A) No energy change in scattered photon B) Increase in energy of scattered photon C) Decrease in energy of scattered photon D) Total energy transfer to an electron E) Depends on incident energy

The correct answer is: C When a photon scatters, it can lose energy to the medium or it can be scattered elastically with no loss of energy. Inelastic scattering is the decrease in energy of the scattered photon.

Question Number: 81 Fundamentals of Radiation Protection Which particle is considered to be radioactive by itself? A) Alpha B) Beta C) Neutron D) Positron E) Proton

The correct answer is: C When removed from the nucleus of its atom, the neutron decays with a half-life of about 10.4 minutes into a beta particle and a proton.

Question Number: 131 Fundamentals of Radiation Protection Localized (l cm2) high radiation burns sometimes occur in which of the following? A) Glove box workers B) Patients undergoing radiation therapy C) X-ray crystallographers D) Nuclear power reactor operators E) Diagnostic radiologists

The correct answer is: C X-ray diffraction units can produce radiation burns (usually occurring when the interlocks are by-passed). Radiation monitors for these personnel are NOT worn nor required since only the index finger is the site of injury, and body monitoring cannot indicate dose. Radiation injury produced by such units can result in loss of the fingertip. X-ray diffraction units are used to examine the crystalline structure of materials. The materials are sometimes rotated by hand in the operating beam. This is an inappropriate practice in terms of radiation safety.

Question Number: 123 Fundamentals of Radiation Protection What scintillator material has a high efficiency for alpha measurement and is not especially affected by temperature, humidity, and barometric pressure? A) Naphthalene B) Mica C) ZnS D) NaI E) LiF

The correct answer is: C ZnS has been used since the beginning of this century for alpha detection. A polycarbonate filter (about 0.8 micron) is recommended to minimize contamination of the crystal if used for radon measurement. It is NaI that is used for gamma spectroscopy.

Question Number: 61 Fundamentals of Radiation Protection How long will it take for 200 Ci of Carbon 14 to decay to less than 1 Ci? (The half life is 5770 years). A) 4000 years B) 40 years C) 44,000 years D) 80,000 years E) 88,000 years

The correct answer is: C t1/2 = 5770 A1 = activity at time in question Ao = activity at time zero Time to reach = (-ln A1/Ao)(t1/2/0.693) activity A1 = -ln (1/200)(5770/0.693) = -ln (.005)(8326) = 44,114 years

Question Number: 113 Fundamentals of Radiation Protection A Geiger counter has enough voltage to cause: A) electrons before recombination; no secondary electrons. B) secondary electrons with a short range avalanche. C) secondary electrons with a long range avalanche. D) continuous discharge. E) recombination prior to pulse.

The correct answer is: C Each answer above represents part of the curve for gas-filled detectors. The first is recombination, then ionization, the proportional region, the Geiger region, and the continuous discharge region.

Question Number: 219 Fundamentals of Radiation Protection The dose to tissue from 6 MeV neutrons is delivered primarily by: A) high speed electrons. B) lower energy neutrons. C) recoil protons. D) alpha particles. E) activation products.

The correct answer is: C Fast neutrons interact by scattering. Since tissue is comprised largely of hydrogen, elastic scattering with hydrogen is producing high-LET recoil protons. Since the interaction of the protons is much more probable than for the neutrons, or resulting capture gammas, these deliver most of the dose.

Question Number: 389 Fundamentals of Radiation Protection When removing shielding that has been exposed to a neutron flux: A) always check the specific gravity of it prior to removal. B) never allow it to touch water. C) always check it for activation. D) never move it without notifying the NRC using Form 14-N. E) never move it without notifying the NRC using Form 14-S.

The correct answer is: C Neutrons can cause activation of most materials.

Question Number: 207 Fundamentals of Radiation Protection One Roentgen is equal to: A) 1 coulomb/kg air. B) 33.7 eV/kg air. C) 2.58 x 10-4 coulomb/kg air. D) 100 ergs/kg air. E) 87 ergs/kg air.

The correct answer is: C The Roentgen is the measure of ionization of air by photons less than 3 MeV. Thus, we are looking for an answer with units of electrostatic charge in a volume of air. This eliminates answers D and E. Answer B is a distractor that uses the value 33.7, which when expressed as eV is the "W" value of air. The value of 2.58 x 10 -4 coulomb/kilogram should be remembered. Other values of the Roentgen: 1R = 1 esu/cc air 1R = 87 ergs/gm air 1R = 98 ergs/gm tissue The air measurements refer to dry air at STP, which is 0° C, 760 mm Hg.

Question Number: 49 Fundamentals of Radiation Protection Under the SI system there have been a number of changes in terminology in radiation protection. The SI term for activity is the: A) Gray. B) Foot-lambert. C) Becquerel. D) Sievert. E) Coulomb per Kilogram.

The correct answer is: C The becquerel is the unit that is replacing the disintegration per second. One becquerel is one disintegration per second and is 2.703 x 10^-11 Ci. The becquerel is denoted by Bq.

Question Number: 40 Fundamentals of Radiation Protection Production of a positron is the result of the conversion of a: A) neutron to a proton. B) proton to a Pi meson. C) proton to a neutron. D) proton to an electron. E) electron to a proton.

The correct answer is: C When a proton converts to a neutron, it releases a positively charged particle called a positron.

Question Number: 285 Fundamentals of Radiation Protection An individual has received an uptake of Co-60 by inhalation. The approximate time it takes for the particle to traverse the trachea is: A) 4 hours. B) 1 day. C) 0.1 hour. D) 1 hour. E) 3 hours.

The correct answer is: C Clearance times for the respiratory tract are: Trachea: 0.1 hour Bronchi: 1.0 hour Bronchioles: 4.0 hours Alveoli: 10 to 1500+ days NCRP Report No. 65

Question Number: 149 Fundamentals of Radiation Protection Transportation of all types of packages including those which have radioactive contents fall under which federal agency? A) EPA B) NBS C) DOT D) OSHA E) NRC

The correct answer is: C It is useful to know the abbreviations for the agencies above and something about scope of authority and enforcement. EPA - Environmental Protection Agency NBS - National Bureau of Standards DOT - Department of Transportation OSHA - Occupational Safety and Health Administration NRC - Nuclear Regulatory Commission

Question Number: 254 Fundamentals of Radiation Protection After four half-lives, the original activity of a radionuclide will be reduced by a factor of: A) 4 B) 8 C) 16 D) 32 E) 64

The correct answer is: C Use A = Ao (1/2)^n and solve for A o/A Ao/A = 1/(1/2)^n = 1/(1/2)^4 = 16

Question Number: 55 Fundamentals of Radiation Protection What is the decay constant for Carbon-14 given that the half life is 5770 years? A) 0.120/year B) 0.0012/year C) 1.2 x 10^-4/year D) 1.2/year E) 12/year

The correct answer is: C k, the decay constant, is related to half life via the equation: k = 0.693/half life k = 0.693/5770 years k = 0.00012/year equivalent to 1.2 x 10^-4/year

Question Number: 167 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) 2200 dpm D) 4440 dpm E) 7770 dpm

The correct answer is: D 1 curie = 3.7 x 10 10 disintegrations per second. There are 60 seconds in one minute. 1 picocurie is equal to 1 x 10 -12 curie. Therefore: (3.7 x 1010) x (1 x 10 -12) (60) (2000) = 4440

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

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: 14 Fundamentals of Radiation Protection 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 E) 0.5 mm

The correct answer is: 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.

Question Number: 146 Fundamentals of Radiation Protection The Department of Transportation (DOT) label Yellow III allows what surface radiation dose rate, when the package is shipped non-exclusive use? A) None above background B) Less than 5 µSv per hr C) Between 5 µSv per hr and 100 µSv per hr D) Between 500 µSv per hr and 2000 µSv per hr E) Between 500 µSv per hr and 5000 µSv per hr

The correct answer is: D 49 CFR Part 173 describes the labels. Radioactive Yellow III has three vertical red stripes. The surface radiation rate allowed is 500-2000 µSv/hr (50-200 mrem/hr). The upper limit at 1 meter, if shipped non-exclusive use, is 500 µSv/hr (50 mrem/hr).

Question Number: 361 Fundamentals of Radiation Protection Alpha particles are commonly referred to as High LET radiations. A typical value for the electrons set in motion by a 2 MeV alpha particle would be: A) 0.25 keV per micrometer. B) 2.5 keV per micrometer. C) 25 keV per micrometer. D) about 1000 times that of a 1.25 MeV Co-60 gamma ray. E) 1000 keV per micrometer.

The correct answer is: D A Co-60 gamma ray (1.25 MeV average) would be about 0.25 keV per micrometer, alpha of 2 MeV = 250 keV per micrometer. (1990). BEIR V. Washington, DC., pg.11.

Question Number: 30 Fundamentals of Radiation Protection Three half value layers of lead will produce a fold reduction in x-rays. A) 4 B) 5 C) 6 D) 8 E) 16

The correct answer is: D A half value layer produces a reduction of 1/2. Therefore: 1/2 * 1/2 * 1/2 = 1/8 for a reduction factor of 8.

Question Number: 377 Fundamentals of Radiation Protection According to ICRP 30 (and the current regulations in the USA, 10 CFR 20) the radiation dose limit for the skin the depth at which it is measured is: A) 5 rems, 1 cm B) 15 rems, 0.3 cm C) 15 rems, 1 cm D) 50 rems, 0.007 cm E) 50 rems, 1 cm

The correct answer is: D A shallow-dose equivalent of 50 rem (0.5 Sv) to the skin of the whole body or to the skin of any extremity ( 10 CFR Part 20.1201.). Shallow-dose equivalent (Hs), which applies to the external exposure of the skin of the whole body or the skin of an extremity, is taken as the dose equivalent at a tissue depth of 0.007 centimeter (7 mg/cm 2). See: 10 CFR Part 20.1003

Question Number: 312 Fundamentals of Radiation Protection Which of the following naturally occurring radionuclides contributes the most to the annual average background dose equivalent to a member of the U.S. due to ingestion? A) Th-232 B) Rn-222 C) Ra-226 D) K-40 E) C-14

The correct answer is: D About 80% of our internal dose from ingestion of naturally occurring radionuclides is due to K-40. Johnson and Birky, Health Physics and Radiological Health Handbook.

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.

The correct answer is: D All sealed sources should be checked for leaks semiannually under NRC RAM licenses.

Question Number: 284 Fundamentals of Radiation Protection Uptake and retention of a radionuclide is influenced by all of the following EXCEPT: A) Portal of entry B) Chemical property of radionuclide C) Solubility D) Radioactive half-life E) Particle size

The correct answer is: D All the factors are a component of the biological half-life for the radionuclide NCRP Report No. 65

Question Number: 379 Fundamentals of Radiation Protection Which of the following are NOT critical inputs into x-ray shielding calculations? A) Occupancy factor B) Use factor C) Allowable air kerma rate D) Number of doctors allowed to use the machine E) Number of patients

The correct answer is: D All the other answers are inputs into the NCRP 147 shielding calculations. See also Cember, Chapter 10.

Question Number: 398 Fundamentals of Radiation Protection The effective half life of I-131 is 7.6 days, and the physical (radiological) half life is 8.05 days. The biological half life of I-131 is: A) 9 days B) 7.6 days C) 1 year D) 136 days E) 76 days

The correct answer is: D Calculate: (1/radiological half life) + (1/biological half life) = 1/effective half life. Using algebra (1/effective half life) -(1/radiological half life) =(1/biological half life) (1/7.6) -(1/8.05) = 1/136 Note that you may add together decay constants and removal constants to determine an effective removal rate.

Question Number: 170 Fundamentals of Radiation Protection Which of the following is not a naturally occurring radionuclide? A) Carbon-14 B) Potassium-40 C) Radium-226 D) Xenon-133 E) Vanadium-50

The correct answer is: D Carbon-14, Potassium-40, Radium-226, and Vanadium-50 are naturally occurring radionuclides. Xenon-133 is manmade, as a fission product.

Question Number: 386 Fundamentals of Radiation Protection A Co-60 check source was 5 microCuries 11 years ago. What is the activity of the source today? A) Zero -it has all decayed away B) 5 microCuries C) 2.5 microCuries D) 1.25 microCuries E) Approximately 3 microCuries

The correct answer is: D Co-60 has a 5.27 year half life, and approximately 2 half lives have elapsed, so the activity would be half of half, or approximately 2.5 microCuries after 5.27 years, and then 1.25 microCuries after another 5.27 years. Answer D is the closest answer.

Question Number: 328 Fundamentals of Radiation Protection Which of the following types of decay "competes" with positron decay? A) Auger electrons B) Internal conversion electrons C) Beta decay D) Electron capture E) X-ray emission

The correct answer is: D Each of these decay modes occurs when the proton - to - neutron ratio in the nucleus is too great for nuclear stability. Turner, J., Atoms, Radiation and Radiation Protection. Chapter 3.

Question Number: 352 Fundamentals of Radiation Protection Which of the following types of decay "competes" with positron decay? A) Auger electrons B) Internal conversion electrons C) Beta decay D) Electron capture E) X-ray emission

The correct answer is: D Each of these decay modes occurs when the proton - to - neutron ratio in the nucleus is too great for nuclear stability. Turner, J., Atoms, Radiation and Radiation Protection. Chapter 3.

Question Number: 210 Fundamentals of Radiation Protection All of the following statements about atomic structure are true EXCEPT: A) The nucleus carries a positive charge. B) Protons and neutrons are collectively known as "nucleons". C) The diameter of the nucleus is relatively small compared to the diameter of the atom. D) Electrons, protons, and neutrons are all of about equal mass. E) Electrons carry a negative charge.

The correct answer is: D Electrons are about 1/1837 the mass of a proton. The rest mass energy equivalence of an electron is .511 MeV, for 1 atomic mass unit (amu) it is 931.5 MeV. Protons have a mass energy equivalence of 938.256 MeV, neutrons 939.550 MeV.

Question Number: 147 Fundamentals of Radiation Protection Medical use of byproduct material falls under which part of 10 CFR? A) Part 19 B) Part 20 C) Part 30 D) Part 35 E) Part 50

The correct answer is: D Four parts of the Code that are important to know: Part 19--Notices, Instructions, Reports to Workers Part 20--Standards for Protection Against Radiation Part 30--Licensing of By-Product Material Part 35--Medical Use of Byproduct Material

Question Number: 338 Fundamentals of Radiation Protection Which laser wavelengths are generally considered retina hazards for safety purposes? A) 100 to 340 nm B) 340 to 1200 nm C) 400 to 700 nm D) 400 to 1400 nm E) 340 to 1500 nm

The correct answer is: D From 400 to 700 nm is the visible region, but near infrared is also focused on the retina from 700 to 1400 nm. ANSI Z136.1.

Question Number: 165 Fundamentals of Radiation Protection When calculating for structural shielding, the occupancy factor can have a value of: A) 1 B) 2 C) 3 D) 4 E) 5

The correct answer is: D Full occupancy, T = 1. Other occupancy factors are summarized in "Introduction to Health Physics", chapter 10 and are listed in NCRP report No. 147.

Question Number: 53 Fundamentals of Radiation Protection When ionizing energy ejects an electron from a water molecule, it makes an H2O+ ion. When an electron is added to a water molecule it makes H2O-. H2O - then decomposes into: A) Carbonium ions B) Superoxide ions C) Acetyl CoA D) Free radicals E) Electrically neutral hydrogen

The correct answer is: D H2O- decomposes and releases a hydrogen radical. A radical contains an unpaired electron. They are short lived and very reactive. H2O+ decomposes and releases a hydroxyl radical. These radicals can interact with biological molecules. This is one of the theories concerning the possible mechanisms of damage for ionizing radiation.

Question Number: 120 Fundamentals of Radiation Protection What is the approximate efficiency for the radioiodines (I-125, I-131) on an activated charcoal filter at the appropriate flow rate for that instrument? A) 5% B) 25% C) 75% D) 95% E) 99%

The correct answer is: D If the flow rate approaches 5 liters per minute, the efficiency can reach 98%.

Question Number: 226 Fundamentals of Radiation Protection Why are the dose equivalent limits for occupationally exposed persons so much higher than those set for members of the general public? A) Radiation workers are paid enough to compensate for the risk. B) Radiation workers have continuous monitoring of their exposure and the work environment is continuously monitored. C) Radioactive materials licensees only hire workers with a limited knowledge of radiological principles. D) The number of radiation workers is very low compared to the population, so the total risk is very small. E) Radiation workers develop a higher tolerance for the effects of ionizing radiation than that of the general public.

The correct answer is: D In Cost-Benefit analysis in radiation protection, the benefit of an implemented process, technology, etc. must outweigh the detriment. For a benefit to an entire population, the total radiation detriment is small, since radiation workers comprise such a small percentage of the total population.

Question Number: 95 Fundamentals of Radiation Protection 100 pCi is equivalent to: A) 1 x 10 -6 Ci B) 100 dpm C) 1 x 1014 Bq D) 1 x 10 -10 Ci E) 3.7 x 10-10 Ci

The correct answer is: D In environmental counting, disintegrations per minute (dpm) are used more frequently than dps. Picocurie is 1 x 10^-10 curie. Units are as follows: Curie Millicurie Microcurie Nanocurie Picocurie

Question Number: 322 Fundamentals of Radiation Protection If the half life of a radionuclide is 10 days, the tenth life would be about: A) 3 days. B) 7 days. C) 14 days. D) 33 days. E) 100 days.

The correct answer is: D In order to calculate the 1/10 life, multiply the half by 3.32. 3.32 is the ratio of ln 10/1n 2 2.3/.693 = 3.32

Question Number: 334 Fundamentals of Radiation Protection A criticality accident in Japan in 1999 resulted in doses of over 5 Gy to an individual. Doses of over 5 Gy generally result in all the following EXCEPT: A) Depression of bone marrow B) Bruising, bleeding around the gums C) Loss of intestinal lining D) Increased capillary flow to the stratum granulosum E) Vomiting, nausea

The correct answer is: D Increased blood flow to the skin is not an effect specific to radiation exposure.

Question Number: 355 Fundamentals of Radiation Protection Low energy neutrons (thermal and near thermal) can deposit energy in human tissue through different reactions. The PRIMARY contributor of absorbed energy from thermal neutrons in the body to tissue samples with dimensions less than 1 cm is: A) Inelastic scattering B) O-16 (n, alpha) C-13 C) H-1 (n, gamma) H-2 D) N-14 (n, proton) C-14 E) Spallation

The correct answer is: D Inelastic scattering is important above 10 MeV. O-16 (n, alpha) C-13 is important above 5 MeV. H-1 (n, gamma) H-2 is a major contributor for the whole human body from its 2.2 MeV gamma. N-14 (n, proton) C-14 since the proton is charged, most energy is deposited locally into a small mass. Spallation is only important above ~ 20 MeV. (1990). BEIR V. Washington, DC., pg.17.

Question Number: 138 Fundamentals of Radiation Protection Human excreta containing medically administered radioactive materials: A) must be collected and held. B) must be excreted into special containers. C) must be monitored. D) can be disposed of in ordinary sewage systems. E) is limited to 1 curie per year.

The correct answer is: D Licensees for medical radiopharmaceuticals calculate sewage flow volume per year. Excreta radioactivity is exempt; non-excreta licensed material can be put into the sewage system at a rate of 1 curie per year per institution. An additional 1 curie C-14 and 5 curies H-3 is also allowed in the sewer per year under 10 CFR Part 20 .

Question Number: 208 Fundamentals of Radiation Protection For an isotope having an allowed concentration of 1 x 10 -5 µCi /cc, what is the minimum discharge time to not exceed the concentration when 1.5 curies of activity are discharged into a stream having a flow rate of 140,000 gallons/hour? (1 gallon = 3785.6 cc). A) 1.5 hours B) 10.7 hours C) 140 hours D) 283 hours E) 1072 horus

The correct answer is: D Look at units given and units in the answer to determine how to attack this problem. We are given allowable µCi/cc, total radioactivity to be released in Ci, and discharge rate in gal/hr. If we set up the equation to divide µCi/cc/hr into µCi/cc, we will be left with units of hours, which is the unit that the answer is expressed in. Therefore: [(1.5 Ci * 1 x 10 6 µCi/Ci)/(140,000gal * 3785.6 cc/gal)/1hr]/ (1x 10 -5 µCi/cc) = 283.02 hr

Question Number: 150 Fundamentals of Radiation Protection How does the NRC regulate x-ray machines? A) Any medical x-ray machine B) Any x-ray machine except medical x-ray machines C) All x-ray machines D) NRC does not regulate x-ray machines E) Only through agreement states

The correct answer is: D NRC regulates reactor-produced radionuclides and not x-ray machines. The state generally regulates them according to their own regulations, regardless of whether the state is an NRC Agreement State. OSHA has authority to enforce its standards to protect hospital workers.

Question Number: 60 Fundamentals of Radiation Protection Radon-222 has a half life of 3.8 days. What will the activity be of 1.0 mCi of radon 222 after 15.2 days? A) 0.05 mCi B) 0.5 mCi C) 0.62 mCi D) 0.062 mCi E) 0.125 mCi

The correct answer is: D One solution: One half life results in decrease of 1/2 in the activity of the radon 222. How many half lives is 15.2 days? 15.2/3.8 = 4 half lives 4 half lives = 1/2 * 1/2 * 1/2 * 1/2 = 1/16 1/16 of the original activity will remain. Therefore: 1/16 * 1.0 mCi = 0.062 mCi.

Question Number: 383 Fundamentals of Radiation Protection How often must radiation workers be notified of their radiation dose according to 10 CFR 20? A) After every work package is completed B) After any re-fueling outage C) At least once during and immediately follow a re-fueling outage D) Annually E) Twice a year, regardless of the number of outages

The correct answer is: D Only once a year is required by regulation, other requirements may be required based on license conditions specific to your site. See 10 CFR Part 20

Question Number: 140 Fundamentals of Radiation Protection What is a "phantom" in reference to radiation exposure studies? A) Calculation of ideal point in radiation formula B) Type of radiation detection system C) Non-portable survey or monitoring unit D) Material to simulate tissue E) Unanticipated double beta decays

The correct answer is: D Phantoms are tissue-equivalent material, i.e., materials that simulate absorption of radiation in the same manner as tissue.

Question Number: 38 Fundamentals of Radiation Protection Which of the following radionuclides in soil is a major contributor to genetically significant radiation exposure? A) Carbon-14 B) Tritium C) Radium-226 D) Potassium-40 E) Argon-41

The correct answer is: D Potassium-40 is present in significant quantities soil.

Question Number: 293 Fundamentals of Radiation Protection All of the following are naturally occurring radionuclides EXCEPT: A) K-40 B) Sm-147 C) Re-187 D) Pu-241 E) Th-232

The correct answer is: D Pu-241 is part of the Neptunium series which is artificially produced. Moe, H.J. (1992) Operational Health Physics Training .

Question Number: 151 Fundamentals of Radiation Protection The NRC regulates radionuclides produced in a linear accelerator by which criterion? A) Greater than a minimum activity B) Greater than a minimum specific activity C) Each radionuclide falls into its own category D) EPAct 2005 E) Through agreement states only

The correct answer is: D Radioactive materials (RAM) produced by a reactor are called by-product materials. These are regulated by the NRC. Accelerator produced radionuclides became regulated under the NRC via changes in definition in the Energy Policy Act of 2005.

Question Number: 392 Fundamentals of Radiation Protection Radiosensitive cells tend to be: A) larger than other cells. B) red in color. C) filled with ectoplasm. D) those dividing rapidly. E) those dividing slowly.

The correct answer is: D Radiosensitivity of cells is best explained by The Law of Bergoine and Tribondeau. It has three generalizations: Cells have a high division rate. Cells have a long dividing future. Cells are of an unspecialized type. See Gollnick, Chapter 4 on Radiobiology.

Question Number: 321 Fundamentals of Radiation Protection Radon gas can be found in which of the following decay chains? A) Thorium B) Uranium C) Actinium D) All of the above E) None of the above

The correct answer is: D Radon 220, 222, 219 are found in the Thorium, Uranium and Actinium decay chains respectively.

Question Number: 45 Fundamentals of Radiation Protection Two cost effective means available to test for radon gas in a building are which of the following? A) Geiger-Mueller Counter, alpha track detector B) Scintillation counter, charcoal canister C) Proportional counter, alpha track detector D) Alpha track detector, charcoal canister E) E-Perm, proportional counter

The correct answer is: D Radon gas is an alpha emitter. Since ambient levels are extremely low, traditional grab sampling techniques are very expensive. The two most cost effective methods for measuring radon in buildings are charcoal canister and alpha track detector. Both devices are left exposed to the ambient air in the building for 3-7 days and 2-4 weeks respectively.

Question Number: 192 Fundamentals of Radiation Protection If an atom is assembled from its constituent parts (electrons, protons, and neutrons), the mass of the assembled atom is less than the sum of the mass of the parts, largely because: A) free electrons are less massive than bound electrons. B) coulombic repulsion among protons tends to decrease the effective mass of the protons. C) the requirements of quantized angular momentum demand a reduction in mass associated with an increase in total system angular momentum. D) of the transformation of constituent mass to binding energy. E) the gravitons required to complete the assembly uses up mass from the constituent parts.

The correct answer is: D Since E = mc2, energy can be converted to mass, and vice versa. The difference in mass between the component parts and the assembled atom is termed "mass defect".

Question Number: 218 Fundamentals of Radiation Protection A technician surveys a shielded pure beta-emitting source and obtains a measurement of 0.2 R/hr. If all beta particles are shielded, the reading is caused by: A) Compton scattering in the shield material. B) Pair production in the shield material. C) Photoelectric effect in the shield material. D) Bremsstrahlung. E) Cerenkov photons.

The correct answer is: D Since all betas are shielded, it is apparent that the reading is due to photons. This could be confirmed by the fact that the measurement is given in units of Roentgen, which only applies to photons. Since the source is a "pure beta-emitter", the photons are being produced by Bremsstrahlung.

Question Number: 173 Fundamentals of Radiation Protection If a point source of radiation had 1000 units of gamma radiation at 1 ft, what would the radiation level be at 10 ft? A) 10000 units B) 1000 units C) 100 units D) 10 units E) 1 unit

The correct answer is: D Since radiation obeys the inverse square law, the radiation would be 1/102 of that at 1 foot. Hence, 1/100 * 1000 = 10 units.

Question Number: 348 Fundamentals of Radiation Protection Radio-frequency radiation can induce currents in the human body at low frequencies and can be a hazard above exposure limits. Induced currents at lower frequencies typically manifest as: A) burns to the eyes and face. B) burns to the hands. C) heating of the joints, especially the knees. D) heating of the ankles. E) burning sensation on the skin.

The correct answer is: D Since the ankles are the narrowest part of the body in contact with the ground , as currents pass through them, there is heating.

Question Number: 351 Fundamentals of Radiation Protection A person at a football game is "lazed" by a red laser. He starts rubbing his eye and gets concerned, as it starts becoming more and more painful. He visits his family physician, who finds some abnormal marks on the cornea that he cannot identify. The MOST LIKELY cause of the marks on the cornea is which of the following? A) Laser energy absorption in the cornea, resulting in injury B) Reflection within the eye, causing injury to the cornea C) Filtration by the glasses the victim was wearing with subsequent phase shifting the red light so that it is absorbed by the cornea and causing injury D) A foreign body co-incidentally entered the eye at the time of the lasing, resulting in corneal abrasions. E) The corneal abnormalities were pre-existing.

The correct answer is: D Since the laser was visible, it was not absorbed by the cornea. There is the possibility of pre-existing abnormalities, but the development of the pain over time would indicate that the injury was progressing due to the foreign body being "ground" into the cornea.

Question Number: 214 Fundamentals of Radiation Protection Which of the following has the highest specific activity? A) One gram of U-238 B) One gram of Cs-137 C) One gram of Co-60 D) One gram of P-32 E) One gram of Ra-226

The correct answer is: D Specific activity is activity per unit mass. Since the mass is constant for each choice (one gram), it must be determined which has the highest decay rate. Since the decay rate has an inverse relationship to the half-life, the nuclide with the shortest half-life will have the highest decay rate and, therefore, the highest activity. The half-lives are as follows: U-238 half-life: 4.47 x 109 years Cs-137 half-life: 30.17 years Co-60 half-life: 5.27 years P-32 half-life: 14.82 days Ra-226 half-life: 1.6 x 103 years

Question Number: 36 Fundamentals of Radiation Protection Which instrument would you use to locate a lost, small radium source? A) Cutie pie B) Condenser R meter C) Geiger-Mueller meter D) NaI scintillation counter E) Proportional counter

The correct answer is: D The GM meter would be effective but since the radium source would be producing gamma ray, the most sensitive meter would be the scintillation counter.

Question Number: 307 Fundamentals of Radiation Protection The Quality Factor (Q) in H = DQN is an assigned factor for radiation protection applications that denotes the modification of the effectiveness of a given absorbed dose. The numerical value of Q is based partly on: A) biological effects and experimental results. B) observed effects and legal requirements. C) experimental results and observed effects. D) biological effects and judgment. E) observed effects and judgment.

The correct answer is: D The Quality Factor is not experimentally determined. It is an assigned value based on dose modifying factors; Q is for radiation protection applications; that is, for relatively low dose rates, not for use in radiobiology or experiments, and not for acute accidental exposures. Moe, H.J. (1992) Operational Health Physics Training .

Question Number: 246 Fundamentals of Radiation Protection The Roentgen is a measure of: A) absorbed exposure to gamma rays. B) absorbed exposure to gamma and x-rays. C) exposure to gamma rays only. D) exposure of air to gamma and x-rays. E) absorbed dose in air.

The correct answer is: D The Roentgen is a measure of exposure, i.e. the ionization of air by photons. The Roentgen is only defined for photons (x or gamma) in air below 3 MeV. The Roentgen may be quantified as 2.58 x 10^-4 Coulombs/kilogram air or 1 esu/cc air.

Question Number: 209 Fundamentals of Radiation Protection The energy absorbed by 1 gram of air exposed to 1 Roentgen of gamma rays is equivalent to: A) 100 ergs. B) 100 rads. C) 32.5 eV. D) 87 ergs. E) Dependent upon the radiation energy

The correct answer is: D The Roentgen is a measurement of the ionization of air by photons. Air, as used when quantifying the Roentgen, is dry air at STP (0 ° C, 760 mm Hg). Representations of the Roentgen which are useful to remember: 1R = 1 esu/cc air 1R = 2.58 x 10-4 coulombs/kilogram air 1R = 87 ergs/gm air 1R = 98 ergs/gm tissue

Question Number: 203 Fundamentals of Radiation Protection The three naturally-occurring radioactive decay chains are: A) Thorium, Neptunium, Uranium B) Polonium, Plutonium, Neptunium C) Hydrogen, Oxygen, Carbon D) Thorium, Actinium, Uranium E) Neptunium, Actinium, Polonium

The correct answer is: D The Thorium, Actinium, and Uranium series begin with the nuclides Th-232, U-235, and U-238, respectively. The fourth decay series is Neptunium, which no longer occurs naturally and begins at Np-237.

Question Number: 324 Fundamentals of Radiation Protection Which of the following is the BEST definition of photon attenuation? A) Attenuation is the reduction in photon intensity by the processes of ionization and excitation interactions. B) Attenuation is the reduction in photon intensity by geometry and shielding configuration. C) Attenuation is the reduction in photon intensity by Compton interactions. D) Attenuation is the reduction in photon intensity by scattering and absorption processes. E) Attenuation is the reduction in photon intensity by photoelectric interactions.

The correct answer is: D The absorption occurs via the photoelectron, Compton electron and the pair production electron. Scattering results from the Compton photon and the pair production annihilation photon.

Question Number: 255 Fundamentals of Radiation Protection If Gallium has two naturally occurring isotopes, Gallium-68.926 with a 60.11% abundance and Gallium-70.925 with a 39.89% abundance, what is the atomic weight of natural Gallium? A) 33.53 B) 34.86 C) 34.96 D) 69.72 E) 69.93

The correct answer is: D The atomic weight of natural Gallium will be equal to the sum of the ratios of each naturally occurring isotope of Gallium. So: A = (68.926 * .6011) + (70.925 * .3989) = 69.72

Question Number: 20 Fundamentals of Radiation Protection The becquerel is a measure of the: A) rate of disintegration of a radioisotope. B) ionizing power of a radioisotope. C) quantity of a radioisotope. D) activity of a radioisotope. E) dose equivalent from radiation.

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: 277 Fundamentals of Radiation Protection Natural Uranium consists of three primary isotopes whose natural abundances are: 99.2739, 0.7204, and 0.0057. Those isotopes from high abundance to low are: A) U-238, U-235, and U-230 B) U-239, U-238, and U-235 C) U-238, U-235, and U-232 D) U-238, U-235, and U-234 E) U-239, U-238, and U-232

The correct answer is: D The decay products of these uranium isotopes consist of long decay chains that decay by both alpha and beta radiation. Bevlacqua, Contemporary Health Physics: Problems and Solutions. 1995.

Question Number: 346 Fundamentals of Radiation Protection Ionizing radiation can produce deleterious effects that depend on the type of radiation, the tissue and the proximity of the radiation, among other things. The tissue(s) that is/are most likely to be impacted by 30 to 300 MHz radio-frequency radiation is/are which of the following? A) Hands B) Brain C) Gastro-intestinal tract epithelial lining D) Eyes E) Ears

The correct answer is: D The eyes are mostly avascular, and as such are not easily cooled. The testes are also of concern.

Question Number: 261 Fundamentals of Radiation Protection Ionizing radiation has been directly associated with cataract formation. Select the statement which is INCORRECT. A) The cataractogenic dose response is considered a threshold effect. B) Fast neutrons are more effective at producing cataracts than other forms of radiation. C) The cataract effect is dependent on age at time of irradiation. D) Occupational exposure to x-rays accounts for approximately 1% of the cataracts observed in x-ray technicians. E) Radiogenic cataracts are distinct in that they originate on the anterior epithelium of the lens.

The correct answer is: D The lens is rarely exposed in occupational medical x-ray applications. Even when it is, it is far less than the annual limit of 15 rem (.15 Sv) on the average. Since the limit is set to allow much less of an incidence of cataracts than 1%, this statement is incorrect. All other distractor data are valid.

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

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: 262 Fundamentals of Radiation Protection Which one of the following lists the skin response to acute radiation exposure in correct chronological order? A) Dry desquamation, moist desquamation, erythema, recovery B) Moist desquamation, dry desquamation, erythema, recovery C) Dry desquamation, moist desquamation, recovery, erythema D) Erythema, dry desquamation, moist desquamation, recovery E) Erythema, moist desquamation, dry desquamation, recovery

The correct answer is: D The order may also be termed: reddening, peeling, blistering, recovery.

Question Number: 118 Fundamentals of Radiation Protection The primary function of a radiation detecting instrument that operates in the proportional region is to measure which of the following? A) Alpha, beta, gamma B) Gamma only C) Beta distinguished from gamma D) Alpha distinguished from beta E) Alpha distinguished from gamma

The correct answer is: D The reason that proportional counters are used in counting is that they can separate the alpha from the beta due to differences in pulse height. Since alphas have a higher specific ionization, they produce a larger pulse than beta at a given voltage.

Question Number: 29 Fundamentals of Radiation Protection A very thin windowed Geiger-Mueller detector can detect which of the following radiations? A) Alpha B) Beta C) Gamma D) Alpha, beta, gamma E) Beta, gamma

The correct answer is: D The thin windowed GM counter can detect all three types of ionizing radiation.

Question Number: 220 Fundamentals of Radiation Protection In the interaction of thermal neutrons with tissue, the major dose is from: A) elastic scattering by hydrogen atoms. B) the N-14(n,p)C-14 reaction. C) inelastic scattering by hydrogen atoms. D) the H-1(n,gamma)H-2 reaction. E) the H-2(n,gamma)H-3 reaction.

The correct answer is: D Thermal neutrons interact by absorption (also called capture). Dose is produced by the interactions described in both Answers B and D. Even though the gamma has a much lower LET than the proton, the dose is greater in Answer D due to the great amount of hydrogen present in the body.

Question Number: 221 Fundamentals of Radiation Protection The average energy required to produce an ion pair in air by x or gamma radiation is which of the following? A) 16 eV B) 32 eV C) 32.5 eV D) 33.7 eV E) 35 eV

The correct answer is: D This is also called the "W" value of air. All gases can be assigned a W value. It is of particular importance in radiation measurements using gas-filled detectors.

Question Number: 362 Fundamentals of Radiation Protection The analysis of Japanese bomb survivors was limited to those persons who had doses: A) > 1 microGy B) > 1 milliGy C) > 0.0001 milliGy D) < 4 Gy E) < 10 Gy

The correct answer is: D This is because the analyses were used to determine stochastic risk coefficients more so than non-stochastic thresholds, and at 0.4 Gray LD 50/60 had been reached. (1990). BEIR V. Washington, DC., pg.183.

Question Number: 86 Fundamentals of Radiation Protection Name the initial and final member of the neptunium series. A) Th-232; Pb-208 B) U-238; Pb-206 C) U-235; Pb-207 D) Np-237; Bi-209 E) Np-239; Pb-207

The correct answer is: D This series has essentially decayed away due to Neptunium's relative short half-life compared to that of the earth. It is not important for dose determination from natural sources. The most investigated series is that of U-238 since it contains Rn-222 and Ra-226. Its relative abundance is used to determine the age of earth.

Question Number: 114 Fundamentals of Radiation Protection A Geiger counter should never have enough voltage to cause: A) primary electrons. B) secondary electrons. C) an avalanche of electrons. D) continuous discharge. E) recombination.

The correct answer is: D To operate properly in the Geiger region of the six-region curve for gas-filled detectors, the voltage must be great enough to produce secondary electrons with an avalanche range extending the entire length of the anode but not exceed a voltage to cause breakdown of the potential field, i.e., continuous discharge.

Question Number: 194 Fundamentals of Radiation Protection You have 400 millicuries of a radionuclide on June 15th. On June 20th, you have 350 millicuries of the material. How much of the material will be present on July 15th of the same year? A) .02 curies B) .08 curies C) .15 curies D) .18 curies E) .45 curies

The correct answer is: D Use A = Ao e^- ((ln2/halflife)*time) First establish the decay constant : 350 = 400 e^- ((ln2/x)*5) = 25.95/day Then plug in new time: A = 400 e^- ( 25.95/day * 30 days) A = 179.5 mCi

Question Number: 168 Fundamentals of Radiation Protection Given a reading of 100 mr/hr (gamma) at 10 feet, what would be the reading at 2 feet assuming a point source geometry? A) 500 mr/hr B) 1000 mr/hr C) 2200 mr/hr D) 2500 mr/hr E) 5000 mr/hr

The correct answer is: D Use inverse square law. r2 = (100)(10/2)2 . The radiation level drops off as one over the square of the distance. Answer = 2500 mr/hr

Question Number: 231 Fundamentals of Radiation Protection Of the following cell types, which is LEAST radiosensitive? A) Lymphocytes B) Endothelial cells C) Epithelial cells D) Nerve cells E) Erythrocytes

The correct answer is: D Using the Law of Bergonie and Tribondeau and the heirarchy of effects from the Acute Radiation Syndrome, it becomes easy to select cells in their order of radiosensitivity. CNS syndrome occurs at the highest dose during acute exposure. Also, nerve cells are non-dividing, highly specialized, and very mature.

Question Number: 216 Fundamentals of Radiation Protection Following electron capture, the following processes may occur: 1. Z increases by 1 2. Z decreases by 1 3. Characteristic x-rays are emitted 4. Auger electron emission A) 1,3 B) 1,3,4 C) 2,3 D) 2,3,4 E) 3,4

The correct answer is: D We know that selections 1 and 2 cannot both be correct, however, this does not help since 1 and 2 do not both appear in the same answer. Good try. During electron capture, a high proton-to-neutron ratio causes the nucleus to capture a negatively-charged electron from the K-shell, transforming a proton to a neutron. Z has decreased by 1. Then, a higher energy level electron fills the vacancy in the K-shell. The difference in electron energies is emitted as a characteristic x-ray, which if it ionizes the "host" atom, will produce Auger electrons. Additional characteristic x-rays are produced as other higher energy level electrons fall to the lowest available energy sites in a cascade fashion.

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

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: 257 Fundamentals of Radiation Protection Assuming a normal cancer fatality rate of 20%, what would be the total probability of developing a fatal cancer for a group of occupationally exposed workers with a 3 in 1,000 probability of contracting a radiation induced fatal cancer? A) 3 in 1,000 B) 20 in 1,000 C) 230 in 1,000 D) 203 in 1,000 E) 200 in 1,000

The correct answer is: D total = normal probability + occupational probability = (200 in 1000) + (3 in 1000) = 203 in 1000

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

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: 175 Fundamentals of Radiation Protection Radium-226 decays by alpha to Radon-: A) 226 B) 225 C) 224 D) 222 E) 220

The correct answer is: D As an alpha particle is a helium nucleus, it contains 2 neutrons and 2 protons for an atomic mass of 4. Losing 4 atomic mass units converts Radium 226 to Radon 222.

Question Number: 289 Fundamentals of Radiation Protection A gram-atomic weight of an element contains which of the following? A) 3.011 x 1013 atoms B) 3.7 x 1023 molecules C) 1.24 x 1010 atoms D) 6.022 x 1023 atoms E) 2.22 x 1012atoms

The correct answer is: D Avogadro number: 6.022 x 1023 or a mole Moe, H.J. (1992) Operational Health Physics Training .

Question Number: 298 Fundamentals of Radiation Protection A radionuclide with a parent half-life longer than that of the daughter is in transient equilibrium. The ratio of the number of parents atoms to the number of daughter atoms is: A) doubled. B) reduced by half. C) increases by a factor of ten. D) constant. E) decreasing by the natural log of 2.

The correct answer is: D In this case, the daughter activity decays at the same rate as the parent activity. "The formation rate of daughter atoms equals the decay rate of daughter atoms." Moe, H.J. (1992) Operational Health Physics Training .

Question Number: 382 Fundamentals of Radiation Protection According to 10 CRF 20, whole body dosimeters must be worn: A) by all radiation workers at all times. B) whenever a reactor is in operation. C) on the turbine floor of a PWR, but only during operations. D) on the abdomen, upper arms, upper legs, or head, at the point where the highest dose is expected. E) on the abdomen only.

The correct answer is: D Note this is the definition of the whole body from 10 CFR Part 20 and includes the head.

Question Number: 237 Fundamentals of Radiation Protection The NCRP has reported which of the following to be the largest contributor to radiation exposure of the general population? A) Radiation from nuclear power plants B) Medical x-rays C) Cosmic and ultraviolet radiation D) Radon and thoron daughters E) Consumer products

The correct answer is: D Radon/thoron daughters comprise 200 mrem of the 360 mrem annual average radiation exposure from all sources in the United States. Data is from NCRP Report No.160.

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.

The correct answer is: D The measure of ionization produced in air by X-ray or gamma radiation is called the Roentgen (R).

Question Number: 266 Fundamentals of Radiation Protection The roentgen is equal to: A) 1.0 coulomb/kg B) 1.00 x 10^-3 coulomb/kg C) 5.28 x 10^-3 coulomb/kg D) 2.58 x 10^-4 coulomb/kg E) 5.28 x 10^-4 coulomb/kg

The correct answer is: D It is helpful to memorize the following values for the Roentgen: 1R = 2.58 x 10^-4 coulomb/kg air = 1 esu/cc air = 87 ergs/gm any absorber = 98 ergs/gm tissue

Question Number: 229 Fundamentals of Radiation Protection The Systeme Internationale units Gray and Sievert represent the same respective quantities in which of the following? A) Rad and curie B) Rem and curie C) Curie and rad D) Rad and rem E) Rem and rad

The correct answer is: D Special Unit SI Unit 100 rad 1 Gray 100 rem 1 Sievert 1 dps (2.7 x 10 -11 Ci) 1 Becquerel

Question Number: 271 Fundamentals of Radiation Protection 10 CFR Part 20 requires Very High Radiation Areas to be posted at 500 Rad/hr at 1 meter. All other radiation postings in 10 CFR Part 20 are based on mrem/hr dose rates. The Very High Radiation Area posting requirement is in Rad/hr because: A) the type of radiation is unknown so mrem cannot be reported. B) radiation detection instruments are unreliable at high dose rates. C) neutron detection instruments read out in mrem/hr eliminating the need to calculate dose. D) the type of radiation is unknown so mrem must be calculated for the type of instrument used. E) quality factors do not apply to dose rates this high.

The correct answer is: E "The values of Q have been selected on the basis of relevant values of relative biological effectiveness but they also take into account the fact that the dose-equivalent limits are based on extrapolations from higher absorbed doses at which deleterious effects in man can be directly assessed. These values of Q are therefore not necessarily representative of values of RBE for other observed effects such as non-stochastic effects in man at high absorbed doses. It is particularly important that dose equivalent should not be used to assess the likely early consequences of severe accidental exposure." ICRP 26 pg. 5

Question Number: 316 Fundamentals of Radiation Protection Which of the following is equivalent to 475 mCi of activity? A) 1.06 x 101 MBq B) 1.76 x 10-2 MBq C) 1.06 x 101 TBq D) 1.76 x 104 TBq E) 1.76 x 10-2 TBq

The correct answer is: E (475 mCi)(Ci/ 1x 103 mCi)(3.7 x 1010 Bq/Ci)(TBq/ 1 x 10 1Z Bq) = 1.76 x 10-2 TBq

Question Number: 369 Fundamentals of Radiation Protection The range of a beta particle in air is roughly: A) 12 inches per MeV. B) 3 to 4 cm for a 4 to 5 MeV particle. C) 12 feet per Merg. D) 500 cm per MeV. E) 12 feet per MeV.

The correct answer is: E 12 feet per MeV is the thumb-rule from Chapter 3, Health Physics and Radiological Health Handbook , 4th ed.

Question Number: 132 Fundamentals of Radiation Protection For transportation of radioactive materials, the Department of Transportation (DOT) classifies them as hazardous materials when their specific activity exceeds: A) 0.002 µCi/g. B) 0.005 µCi/g. C) 1000 dpm/100cm2. D) 220 dpm/100cm2. E) certain nuclide-specific values.

The correct answer is: E 49 CFR Part 173 classifies the material as hazardous by radiotoxicity of each individual radionuclide.

Question Number: 393 Fundamentals of Radiation Protection A report of a worker dose of over 15 Gy in a period of 2 minutes is in the news. The expected outcome for the worker's health is: A) no noticeable effects. B) some nausea and vomiting, followed by recovery. C) nausea, vomiting, then illness, followed by a long recovery. D) no vomiting, but reduced red blood cell count. E) death in a few days.

The correct answer is: E A 15 Gy dose is 1500 rads and is most likely fatal in less than 48 hours.

Question Number: 380 Fundamentals of Radiation Protection A technique that would NOT be useful in preventing a criticality is which of the following? A) Reduce enrichment of the uranium B) Change the geometry such that there is more surface area C) Add in poisons D) Limit the amount of fissile materials E) Install additional moderators

The correct answer is: E Additional moderators would cause more neutrons to be thermalized, increasing the probability that they could cause fission. See Cember, Chapter 12.

Question Number: 319 Fundamentals of Radiation Protection An unstable atom with an excess of protons in the nucleus could be expected to decay by which of the following? A) Alpha B) Electron capture C) Beta plus D) Electron capture or beta plus E) Electron capture, beta plus, or alpha

The correct answer is: E All are possible, especially in high Z materials.

Question Number: 387 Fundamentals of Radiation Protection Tc-99m decays to Tc-99 by emitting a gamma ray - this type of decay is called: A) Electron capture B) Internal Conversion C) Photoelectric effect D) Pair Production E) Isomeric Transition

The correct answer is: E All atoms that have a metastable state (as denoted by the "m" ) decay by isomeric transition. Metastable atoms are those which have residual energy left in the nucleus after a particle decay and remain in this state for greater than 10 -6 seconds. Metastable atoms decay by emitting either a gamma ray or undergoing internal conversion.

Question Number: 331 Fundamentals of Radiation Protection An incident 0.667 MeV photon traveling at a speed of 3 x 10 10 cm/sec undergoes a Compton interaction in air, and the scattered photon is reduced in energy to 190 keV. Accounting for the relativistic effects, the speed of the scattered photon is now: A) 1 x 1010 cm/sec B) 1.5 x 1010 cm/sec C) 2 x 1010 cm/sec D) 2.5 x 1010 cm/sec E) 3 x 1010 cm/sec

The correct answer is: E All photons travel at the speed of light. (3 x 10 10 cm/sec)

Question Number: 344 Fundamentals of Radiation Protection Certain persons are known to be genetically sensitive to ionizing radiation. Which of the following conditions are NOT associated with this susceptibility? A) Xeroderma Pigmentosum B) Ataxia Telangeictasia C) Bloom's disease D) Down's Syndrome E) Collet's Syndrome and Cestan-Raymond Syndrome

The correct answer is: E Answers A - D are all diseases or syndromes which involve impaired or limited ability to repair DNA damage. Hall, E., Radiobiology for the Radiologist.

Question Number: 290 Fundamentals of Radiation Protection A mole of U-238 weighs: A) 238 amu. B) 92 amu. C) 92 grams. D) 1 gram. E) 238 grams.

The correct answer is: E By definition, the atomic weight (roughly equal to the mass number) is the number of grams per mole of the element. Moe, H.J. (1992) Operational Health Physics Training .

Question Number: 364 Fundamentals of Radiation Protection The dose response relationship for laboratory animals appears to vary with all the following EXCEPT: A) LET B) Dose rate C) Sex D) Age E) Animal size

The correct answer is: E Dose is measured in ergs/gm (J/kg). The energy absorbed in tissue per unit mass for a given radiation flux would be constant. (1990). BEIR V. Washington, DC., pg. 5.

Question Number: 345 Fundamentals of Radiation Protection The most important radiogenic cancers include all of the following EXCEPT: A) Leukemia B) Breast C) Respiratory D) Digestive E) Prostate

The correct answer is: E In this report it states that prostate cancer is not associated with radiation exposure. BEIR V Report.

Question Number: 375 Fundamentals of Radiation Protection Irradiation of bandages with Cs-137 to doses of 100 Gy and higher: A) can result in the bandages becoming radioactive. B) can cause the bandages to emit neutrons. C) will result in genetic mutations in people using the bandages. D) can cause death to those who use the bandages. E) is done routinely to sterilize medical equipment.

The correct answer is: E It is common to use gamma ray sources to sterilize medical equipment.

Question Number: 299 Fundamentals of Radiation Protection You have counted a radioactive sample every hour for a period of twelve hours. A plot of the count rate versus time on a semi-log graph reveals a curved decay line. This indicates: A) an isotope with a short half-life. B) an isotope with a long half life. C) a counting error has occurred. D) only one radionuclide is present. E) more than one radionuclide is present.

The correct answer is: E On a semi-log graph a plot of a single radionuclide decaying will be a straight line. More than one radionuclide with differing half-lives will give a curved line based on total counts obtained. Moe, H.J. (1992) Operational Health Physics Training .

Question Number: 313 Fundamentals of Radiation Protection One roentgen of gamma radiation deposits approximately how much energy in soft tissue? A) 87 erg/gm B) 33.7 ev/gm C) 1 J/Kg D) 2.58 x 10-4 c/Kg E) 98 erg/gm

The correct answer is: E One R deposits approximately 87 erg/gm in air and creates an electrical charge of 2.58 x 10 -4 c/Kg in air.

Question Number: 311 Fundamentals of Radiation Protection The most predominant constituent of primary cosmic radiation is which of the following? A) Alpha particles B) Neutrons C) Electrons D) Mesons E) Protons

The correct answer is: E Primary cosmic radiation consists of approximately 87% high energy protons, 12% alpha particles, and 1% other nuclei. Johnson and Birky, Health Physics and Radiological Health Handbook.

Question Number: 349 Fundamentals of Radiation Protection Radio-frequency exposure limits are based on: A) cancer induction. B) hearing loss from resonance with the small bones in the inner ear. C) skin ulceration in the vicinity of the exposure under grounded conditions. D) hair loss. E) heating beyond the ability of the body to cool.

The correct answer is: E RF frequencies, such as microwaves, induce friction between water molecules, thus creating a thermal hazard. See C95.1 for details.

Question Number: 320 Fundamentals of Radiation Protection Which of the following does NOT affect the mode of radioactive decay? A) Neutron to proton ratio is too high B) Neutron to proton ratio is too low C) The amount of excess energy of the parent D) The atomic mass of the parent E) The electron quantum state

The correct answer is: E Radioactive decay is affected only by conditions within the nucleus.

Question Number: 366 Fundamentals of Radiation Protection The human data used in creating the BEIR V report include all the following EXCEPT: A) Atomic bomb survivors B) Ankylosing spondylitis patients C) Cervical cancer patients D) Tinea capitis patients E) Radium dial painters

The correct answer is: E Radium dial painters were exposed to alpha radiation. BEIR V report does not cover alpha. (1990). BEIR V. Washington, DC., pg.2.

Question Number: 381 Fundamentals of Radiation Protection Tritium production in a pressurized water reactor can be due to which of the following? A) Ternary fission B) Absorption of neutrons by Boron C) Absorption of neutrons by deuterium D) Both 1 and 2 only E) 1, 2, and 3

The correct answer is: E Ternary fission is where tritium is produced during fission. Some deuterium is naturally occurring in water and is produced by neutron absorption on hydrogen in water. Although the cross section for the reaction is low, still possible. Boron can absorb neutrons, 10B, and form tritium and two helium nuclei. 2H(n,γ)3H, is

Question Number: 347 Fundamentals of Radiation Protection Reference man is frequently used to ascertain dose from ionizing radiation. In non-ionizing radiation, the resonant frequency for reference man is approximately: A) 70Hz B) 7kHz C) 700kHz D) 7MHz E) 70 MHz

The correct answer is: E The 70 MHz frequency exhibits a wavelength on the order of magnitude of the effective diameter of reference man. See FCC website on hazards from RFR and microwaves.

Question Number: 233 Fundamentals of Radiation Protection According to NCRP Report No. 160, the average annual radiation dose to persons in the United States from natural and man-made sources is: A) 100 mrem. B) 260 mrem. C) 300 mrem. D) 360 mrem. E) 620 mrem.

The correct answer is: E The NCRP 160 report increased previous estimates by about 300 millirem, largely due to medical exposures.

Question Number: 294 Fundamentals of Radiation Protection The nucleus transfers its energy to an orbital electron, which is then ejected at a discrete energy. This decay mode is called: A) isomer. B) positron. C) beta. D) characteristic. E) internal conversion.

The correct answer is: E The ejection of the electron can cause secondary radiation; an electron will transition to the "vacancy" releasing a characteristic x-ray to go to a lower energy state. This x-ray can go on to interact with an electron in the atom causing it to be ejected producing an auger electron. Moe, H.J. (1992) Operational Health Physics Training .

Question Number: 363 Fundamentals of Radiation Protection The excess of mortality in the Japanese bomb survivor study was found to be which of the following? A) Linear for all cancers B) Linear for all cancers in the range below 4 Sv C) Linear-quadratic for all cancers below 4 Sv D) Linear-quadratic for all cancers E) Linear quadratic for leukemia below 4 Sv

The correct answer is: E The linear-quadratic dose/response curve forms a gentle "S" shape. (1990). BEIR V. Washington, DC., pg.5.

Question Number: 274 Fundamentals of Radiation Protection Cigarette smokers who smoke 1.5 packs a day receive approximately 8 Rem/yr to the lungs due to: A) Rn-222 and Po-210 B) Ra-226 and Po-210 C) Bi-214 and Ra-226 D) Po-218 and Pb-214 E) Pb-210 and Po-210

The correct answer is: E The tobacco plant has a signifcant affinity to natural radium decay products. When burned and inhaled, these decay products reach the lungs. NCRP Report No. 56 pg. 29

Question Number: 281 Fundamentals of Radiation Protection The reason radioactive compounds are often mixed with quantities of non-radioactive but chemically identical material is to: A) allow scientists to test the non-radioactive component of the compound. B) increase the compounds specific activity so that small losses of compound will not affect the radioactivity of the compound. C) increase the amount of compound to work with, making measurement easier, decreasing the statistical probability of error. D) ensure that enough of the compound will be available to run a number of analysis before decaying seven half lives. E) ensure that the loss of small amounts of compound will have a small affect on the radioactivity of the compound.

The correct answer is: E The use of a carrier ensures that only a very small fraction of molecules of the compound contain the radioisotope, so that if small amounts of the compound are lost through adsorption and so on, the amount of radioactivity lost will be negligible. Malcolme-Lawes, Introduction to Radiochemistry.

Question Number: 199 Fundamentals of Radiation Protection Secondary radiation following photoelectric interaction can include: 1. Scattered photons 2. Photons of shorter wavelengths 3. Characteristic x-rays 4. Auger electrons 5. Neutrons A) 1,2,3,4,5 B) 1,2,3,4 C) 1,3.5 D) 2,3,4 E) 3,4

The correct answer is: E There are no scattered photons, since the photon is completely absorbed. The characteristic x-rays created would have longer, not shorter wavelengths. The production of neutrons would require a photoneutron interaction, which is not described by photoelectric interaction.

Question Number: 358 Fundamentals of Radiation Protection As it pertains to dosimetry and energy transfer, the interaction of a photon with an atom, causing an electron to be set in motion, is BEST described by: A) Photoelectric effect. B) Compton scatter (if the scattered photon is accounted for). C) Internal conversion. D) Dose. E) Kerma.

The correct answer is: E This is the definition of Kerma. (1990). BEIR V. Washington, DC., pg.12.

Question Number: 305 Fundamentals of Radiation Protection The binding energy of a neutron in most elements is approximately which of the following? A) 33.7 ev B) 100 kev C) 70 kev D) 2 Mev E) 8 Mev

The correct answer is: E This is the energy which holds a neutron in the nucleus except for the lighter nuclei. To remove the neutron, one must supply at least this much energy. Moe, H.J. (1992) Operational Health Physics Training .

Question Number: 278 Fundamentals of Radiation Protection Uranium and Thorium decay products are predominately an internal hazard. Historically, the radiological concerns of uranium have been overshadowed by the chemical toxicity of uranium to the: A) lymph nodes. B) lung. C) bone marrow. D) liver. E) kidney.

The correct answer is: E Uranium, a heavy metal is chemically toxic to the kidneys. Bevlacqua, Contemporary Health Physics: Problems and Solutions. 1995.