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262 Cards in this Set

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G1 -7. Match the unit with the quantity it measures. (Answers may be used more than once or not at all.)
Dose Equivalent
Exposure
Absorbed dose
Activity
Energy

Gl. Gray
G2. Becquerel
G3. Rem
G4. Rad
G5. Sievert
G6. Curie
G7. Roentgen
Gl. C
G2. D
G3. A
G4. C
G5. A
G6. D
G7. B
A.          G8. Exposure is .
The amount of energy in joules/kg transferred from a photon beam to a medium
Only defined for charged particles below 3 MeV
The charge liberated by photons in a given mass of air
The absorbed dose multiplied by the quality factor
None of the above
G8. C "A" is absorbed dose (Gy); it is only defined for photons below 3 MeV; "D" is dose equiv¬alent (Sv).
G9. Dose equivalent is greater than absorbed dose for
A. X-rays above 10 MeV
B. Kilovoltage x-rays
C. Electrons
D. Neutrons
E. All charged particles
G9. D The dose equivalent for x-rays, gamma-rays and electrons is the same as the absorbed dose.
For neutrons, however, it is between 5 and 20 times as great, depending on neutron energy, because of the higher LET and greater potential for biological damage per Gy.
A.          G10. 100 |0.Sv is equal to mrem.
100
10
1
0.1
0.01
G10. B 1.0 Sv = 100 rem. Thus 100 |XSv = 0.0001 Sv = 0.01 rem = 10 mrem.
G11. If a muon has a mass 207 times that of an electron, its mass is equivalent to MeV
A. 560
B. 414
C. 335
D. 207
E. 106
G1l. E The rest mass of an electron is 0.51 MeV Thus, the mass of the muon is 0.51 x 207 = 106 MeV
G12. A 10 MeV travels at the greatest speed in a vacuum.
A. Alpha particle
B. Neutron
C. Proton
D. Electron
G12. D 10 MeV is the kinetic energy of the particle. The lightest particle travels fastest.
G13. Differaxf.t isotopes of the same element will have equal numbers of
A. Electrons
B. Protons and electrons
C. Neutrons
D. Electrons and neutrons
E. Protons and neutrons
G13. B An element is defined by the number of protons in its nucleus. Protons and electrons must be equal in number in a neutral atom, so it is the number of neutrons that differs between isotopes of the same element.
G14. The atom 6C contains electrons, protons, and neutrons.
A. 6 6 8
B. 6 8 8
C. 14 6 8
D. 14 8 20
E. 8 6 14
G14. A In the symbol, the subscript is the atomic number, Z, which is the number of electrons. The number of protons must be equal to Z to balance positive and negative charges. The super­script is the mass number, A, which is the total number of protons and neutrons, N, in the nucleus. Thus N = A- Z= 14-6 = 8
G15. Electron binding energy is …………
A. Greater in the L shell than the K shell of an atom
B. Greater for the K shell of hydrogen than the K shell of barium
C. Increases with Z
D. All of the above
G15. C Binding energy is the energy needed to remove an electron from its orbit, and increases with both Z and proximity to the nucleus (i.e., K > L > M, etc.).
A.      G16. In heavy nuclei such as 235U:
There are more protons than neutrons.
Protons and neutrons are equal in number.
There are more neutrons than protons.
Cannot tell from information given.
G16. C As the mass number increases, more neutrons are needed to balance the attraction of all masses (nucleons) with the repulsion between positively charged protons.
G17. The number of electrons in the outer shell of an atom:
A. Is always 2n2.
B. Is greater for radioactive isotopes than for stable isotopes of the same element.
C. Determines the chemical properties of the atom.
D. Is always between 8 and 16.
E. Is 1 for all inert gases.
G17. C The maximum number of electrons allowed in any shell is 2n2, but the maximum in the outer shell is 8. All isotopes of a given element always have the same electron configura¬tion. Inert gases have the maximum number of electrons in the outer shell.
G18. A radioactive source with a half-life of 6 hours has an activity of 10.0 mCi at noon on Monday. The activity at noon on Tuesday is mCi.
A. 9.375
B. 6.25
C. 2.5
D. 0.625
E. 0.31
G18. D Activity after time t = Initial Activity x exp-t(0.693/half-life)
= 10 x 24(0.693/6) = 0.625 mCi.
Alternatively, since the time is 4 half-lives, the activity can be found by multiplying the ini¬tial activity by l/(24).
G19. The half-life of a radionuclide is:
A. Influenced by temperature and pressure.
B. Directly proportional to the decay constant.
C. Less than the average life.
D. Usually shorter for beta-minus than beta-plus emitters.
E. All of the above are true.
G19. C The average life is 1.44 x half-life. The half-life is inversely proportional to X, and is unaf¬fected by temperature and pressure.
G20. In the expression A = A0 X is .
A. The number of atoms decaying per unit time
B. The fraction of atoms decaying per unit time
C. The fraction of atoms decaying in time t
D. The linear attenuation coefficient
E. The mass attenuation coefficient
G20. B
G21. If the half-life of a radionuclide is 74 days, the decay constant is
A. 3.7 days
B. 37 days
C. 106.8 days
D. 0.0094 per day
E. 0.027 per day
G2I. D T1/2 x A, = 0.693.
G22. If the physical half-life Tp of a radionuclide is much smaller than its biological half-life Tb, the effective half-life will be closest to .
A. Tp
B. Tb
C. (Tp + Tb)/2
D. (TpTb)1/2
E. 1.44 Tp
G22. A Since —!— = — + — If Tp « Tb, this is approximately equal to Tp.
Tiff Tp Tb
G23. 226Ra decays to 222Rn by decay.
A. Beta minus
B. Alpha
C. Gamma
D. Beta plus
E. Both A and C
G23. B The mass number decreases by 4 (226 to 222), so this must be alpha decay.
G24. In the decay of ^Co to ^Ni are emitted.
A. Only monoenergetic electrons
B. Only monoenergetic positrons
C. A spectrum of electrons and several monoenergetic photons
D. A spectrum of positrons and several monoenergetic photons
E. Monoenergetic electrons and several monoenergetic photons
G24. C Since Z increases by 1, this is an example of beta minus decay, which always emits a spec¬trum of betas. In this case, they are accompanied by two gammas.
G25. A positron loses energy passing through matter, and eventually combines with an electron resulting in .
A. A particle with equal mass but no charge
B. A photon with energy equal to the rest masses of both particles
C. Two particles emitted in opposite directions
D. Two photons of equal energy emited in opposite directions
G25. D The rest masses of the two particles combine to yield two 0.51 MeV gammas, emitted in opposite directions. In a PET scanner, they are detected by coincidence counters.
A.      G26. Compared to "Tc, the mass-energy equivalent of 99mTc is
Larger
Smaller
The same
G26. A 99mjc decayS to 99emjtting a 140 keV gamma, so its mass equivalent is greater by this amount.
G27. 5 mCi of 99mTc (140 keV) are placed inside a lead container. A photon detected outside the container could .
A. Have an energy of 145 keV
B. Be a characteristic x-ray
C. Be Cerenkov radiation
D. Be an Auger x-ray
E. Be annihilation radiation
G27. B The 140 keV x-rays could interact with lead to emit lead characteristic x-rays. No photons of energy greater than 140 keV are present. There is no such thing as an Auger x-ray. Annihilation radiation is emitted when a positron and electron combine. Cerenkov radiation is emitted when charged particles travel very fast in a medium such as water.
G28. The following radioactive transformation represents .
zX^z-iY + y + v
A. Alpha
B. Beta minus
C. Beta plus
D. Electron capture
E. Isomeric transition
G28. D As Z decreases by 1, it must be either beta plus or electron capture. However, no positron is created, so beta plus is ruled out.
G29. When internal conversion occurs:
A. Z and A remain the same.
B. Z increases by 1, A remains the same.
C. Z decreases by 1, A remains the same.
D. Z and A decrease by 1.
E. Z and A increase by 1
G29. A Energy is transferred directly to an inner shell electron, which is then ejected.
G30. Which of the following is not a naturally occurring radionuclide?
A. 235Uranium
B. 40Potassium
C. 14Carbon
D. 226Radon
E. 18Fluorine
G30. E 18F is a cyclotron produced positron emitter used in PET.
G31. Which kind of radioactive equilibrium can occur when a very long-lived radionuclide decays to a short-lived daughter?
A. Thermal
B. Secular
C. Transient
D. Non-stable
E. Temporary
G31. B An example is 226Ra decaying to radon gas.
G32. An 192Ir source has an activity of 5.0 x 109 Bq. The activity is mCi.
A. 1.85
B. 13.5
C. 18.5
D. 135
E. 185
G32. D 1.0 Ci = 3.7 x 1010Bq.
Thus 5.0 x 109 Bq = (5.0 x 109)/(3.7 x 1010) = 0.135 Ci or 135 mCi.
G33. A 4.0 mCi source of 192Ir has an exposure rate of mR/hr at 1 m. (Exposure rate
constant = 4.6 R.cm2/mCi-hr.)
A. 1.84
B. 2.56
C. 19.02
D. 25.56
E. 33.82
G33. A Exposure rate = Activity x Exposure rate const, x 1/d2
= 4.0 x 4.6 x 1/(100)2 = 1.84 x 10"3 R/hr = 1.84 mR/hr.
G34. Which of the following does not improve the heat capacity of an x-ray tube.
A. Rotating anode
B. Small target angle
C. Large focal spot
D. Thermionic emission
G34. D Thermionic emission is the emission of electrons from the heated filament.
G35. Two filaments are found in some x-ray tubes. The purpose is to:
A. Function as a spare in case one filament burns out.
B. Produce higher tube currents by using both filaments simultaneously.
C. Double the number of heat units that the target can accept.
D. Enable the smallest focal spot to be used, consistent with the kVp/mA setting.
G35. D Resolution is improved by the use of the small focal spot. However, if a high kVp/mA set¬ting is required, this might overheat a small area of the target; a larger focal spot helps with heat dissipation.
36. The characteristic x-rays emitted from a tungsten target when 100 keV electrons are fired at it:
Have a continuous spectrum of energies up to 100 kV.
Are about equal in intensity to the bremsstrahlung.
Have energies equal to differences in the electron binding energies.
Do not contribute significantly to the imaging process.
G36. C
G37-39. A target material has the following binding energies: K = 30.0 keV L = 4.0 keV M = 0.7 keV
If 40 keV electrons are fired at the target, which of the following types of x-ray can be emitted?
A. Characteristic only
B. Bremsstrahlung only
C. Both A and B
D. Neither A or B
G37. 34 kV x-rays G38. 26 kV x-rays G39. 40.7 kV x-rays
G37. B Bremsstrahlung is emitted in a continuous spectrum up to a maximum energy equal to that of the incident electron.
G38. C In order to emit characteristic x-rays, the incident electron must have an energy equal to or greater than that of an electron shell, so that an orbital electron is ejected. The shell is then filled with an electron from a shell farther out, and the characteristic x-ray is emitted with energy equal to the difference between the two shells. The characteristic x-ray energies possible from this atom are: 29.3, 26.0, and 3.3 kV
G39. D This is greater than the incident energy.
G40. The effective energy of an x-ray beam:
A. Is proportional to the atomic number (Z) of the target material.
B. Is proportional to the mAs.
C. Is not affected by added filtration.
D. Is equal to the kVp.
E. Affects subject contrast.
G40. E The effective energy of an x-ray beam is about 1/2 to 1/3 of the kVp, depending
on filtration. It can be increased by adding additional filtration. Z does not affect the bremsstrahlung spectrum, and mAs has no effect.
G41. The quality of an x-ray beam cannot be characterized only in terms of the peak kV, because beams with the same kV may have different .
A. Filtration
B. Half-value layers
C. Maximum wavelengths
D. Target materials
E. All of the above
G4I. E
G42. The second half-value layer (HVL) of a photon beam is the same as the first HVL:
A. 'For all x-ray tube generated photon beams.
B. Only if the energy is below 100 kVp.
C. Only if the beam is monoenergetic (e.g., gamma rays).
D. Never; it is always less.
G42. C For polyenergetic beams, passing through the first HVL hardens the beam, making the second HVL larger than the first.
G43. Which of the following is the electromagnetic radiation with the highest frequency?
A. Infrared
B. Gamma
C. Radio waves
D. Ultrasound
E. Ultraviolet
G43. B Frequency is proportional to energy. Ultrasound is not electromagnetic radiation.
G44. Which of the following is not ionizing radiation?
A. 2 MHz ultrasound
B. 60Co gammas
C. 90Sr betas
D. 15 MeV photons
E. Neutron leakage from a linear accelerator
G44. A See answer G43 above.
G45. The energy of a photon increases as the increases.
A. Amplitude
B. Wavelength
C. Frequency
D. Speed
G45. C See answer G43 above.
G46. If the exposure at 50 cm from an x-ray target is 10 mR, the exposure at 75 cm is mR.
A. 0.7
B. 4.4
C. 6.7
D. 15.0
E. 22.5
G46. B Using the inverse square law: I75/I5o = (50/75)2.
G47. In an electromagnetic wave the electric and magnetic waves are oriented at ° to each
other, and ° to the direction of propagation.
A. 90, 90
B. 90, 180
C. 180, 90
D. 90, 0
E. 0, 90
G47. A
A.      G48. A photon of frequency 100 MHz has a wavelength of .
3 mm
9 mm
3 cm
9 cm
3 m
G48. E c = Xv.
c = speed of light = 3 x 108 m/s. v = 100 x 106 s"1.
G49. A monoenergetic photon beam whose linear attenuation coefficient is 0.0693 cm 1 traverses 10 cm of a medium. The fraction of the beam transmitted is .
A. 0.01
B. 0.37
C. 0.50
D. 0.69
E. 0.90
G49. C The fraction transmitted = I/Io = e"Hx.
= 0.0693 cm-1, x = 10 cm. I/Io = e"0693 = 0.5.
G50. The mass attenuation coefficients for most materials (except hydrogen) are similar when interactions predominate.
A. Photoelectric
B. Compton
C. Pair production
D. Photonuclear disintegration
G50. B The mass attenuation coefficient is similar for most materials in the Compton region,
except those containing hydrogen. This is because most materials have approximately one electron per two nucleons (one proton and one neutron), while hydrogen has one electron per nucleon, or twice the number of electrons per unit mass as most other elements.
G51. The photoelectric mass attenuation coefficient is proportional to .
A. Z.E
B. Z2.E2
C. Z3.E3
D. Z3.E"3
E. Z2.E~2
G51. D The probability increases as Z3, and decreases approximately as 1/E3.
G52. The process whereby energy is transferred from a photon beam to electrons in the medium is called:
A. Electron capture.
B. Absorption.
C. Bremsstrahlung.
D. Scatter.
G52. B In the interaction of radiation with matter, energy is generally transferred to electrons and photons. The energy reemitted as photons is called "scatter". The energy transferred to electrons is called "absorbed dose", and is dissipated locally in collisions with atoms.
G53. Regarding photoelectric interactions, all of the following are true except:
A. K, L, and M characteristic x-rays may be emitted if the photon energy is greater than the K shell binding energy.
B. The photoelectron's energy is the energy of the incident photon less the binding energy of the emitted electron.
C. The probability is greatest when the photon energy is a little less than the electron binding energy.
D. In tissue, most of the released energy is locally absorbed.
G53. C The probability is greatest when the photon energy is just greater than the electron binding energy.
G54. A 70 keV photon beam interacts with an atom whose K-shell binding energy is 5 keV An elec¬tron is emitted with a kinetic energy of 65 keV All of the following are true except?
A. Characteristic radiation will be emitted.
B. The 70 keV photon disappears.
C. The electron will be absorbed within 1 cm of its origin in tissue.
D. This is an example of Compton scattering.
G54. D This is an example of photoelectric absorption since the product is a 65 keV photoelectron.
Characteristic radiation will be emitted as a result of an electron from an outer shell falling into the K-shell vacancy. Electrons interact because of their charge and mass by ionization and excitation along their paths. In general, the path of an electron in any medium, except air, is very short.
G55. In Compton interactions, which of the following is true?
A. The photon changes direction but does not lose energy.
B. The electron may acquire any energy from zero up to the energy of the incident photon.
C. Electrons can be emitted between 0° and 90° to the direction of the incident photon.
D. A neutrino is emitted.
G55. C The energy of the incident photon is divided between the scattered photon and the recoil electron. The scattered photon may travel in any direction. The minimum scattered photon energy occurs when the photon is scattered at 180° (backwards). In this case, the electron acquires its maximum kinetic energy and travels in the forward direction. The electron acquires the least energy when it is emitted at 90°.
G56. If a technologist were to stand 2 m away from a patient during fluoroscopy (outside the primary beam) the dose received by the technologist would be mainly due to:
A. Compton electrons.
B. Photoelectrons.
C. Compton scattered photons.
D. Characteristic x-rays generated in the patient.
E. Coherent scatter.
G56. C Even at low KY coherent scatter contributes only a small part of the total scatter. The
characteristic x-rays created by photoelectric interactions within the patient are of very low energy (because of the low Z of tissue) and have an extremely small range. Compton and photoelectrons also have a short range and are unlikely to leave the patient's body.
G57. In Compton scattering, the energy difference between the incident and scattered photons is:
A. Shared equally between the recoiling nucleus and the ejected electron.
B. Maximized when the photon is scattered at 180°.
C. Equal to the binding energy of the ejected electron.
D. Always greater than 0.51 MeV
G57. B The energy lost by the incident photon is used to eject the electron (binding energy), with the remainder given as kinetic energy to the electron. The maximum energy transfer occurs when the photon is backscattered. Nuclear recoil accounts for a very small amount of energy.
A.      G58. The probability, per gram, of a Compton interaction:
Increases as energy increases.
Is independent of energy.
Is proportional to E2Z2.
Is proportional to Z3E"3.
None of the above.
G58. E Compton interactions, per unit mass, are approximately independent of Z, and decrease with increasing energy.
G59. The most probable interaction in soft tissue for a 50 keV photon is:
A. Coherent scatter
B. Photoelectric
C. Compton
D. Pair production
E. Photonuclear disintegration
G59. C Compton is the most probable interaction in soft tissue between 25 keV and 25 MeV
G60. A 2.3 MeV photon undergoes pair production in tissue. The energy deposited locally is MeV
A. 2.3
B. 1.79
C. 1.28
D. 0.51
E. None
G60. C It takes 2 x 0.51 MeV to create the electron-positron pair. The remaining 1.28 MeV is
divided between the two particles as kinetic energy. This kinetic energy is deposited locally, and when the positron and an electron combine and annihilate, two 0.51 MeV photons are emitted.
G61. After a 1.5 MeV photon undergoes pair production, production of will always occur:
A. A pair of 0.51 MeV photons
B. A pair of 0.51 MeV positrons
C. A pair of 0.51 MeV electrons
D. A single 0.51 MeV photon
E. An electron and a positron, each with kinetic energy 0.51 MeV
G61. A See answer G60 above.
G62. The threshold for pair production is MeV
A. 0.51
B. 1.02
C. 1.53
D. 2.04
G62. B See answer G60 above.
G63. In diagnostic x-ray systems, filters are used to "harden" the beam. This process is mainly due to:
A. Coherent scattering.
B. Photoelectric effect.
C. Compton effect.
D. Pair production.
E. A, B, and C only.
G63. B With the exception of the K edges, photoelectric interactions are more likely at low energy than at high energy. After passing through a filter, the total beam intensity is reduced, but the beam contains a relatively greater number of high-energy photons than before filtration.
G64. CT or Hounsfield numbers are linearly related to:
A. Mass density.
B. Electron density
C. Linear attenuation coefficient.
D. Mass absorption coefficient.
E. Effective atomic number.
G64. C CT number = 1000 x [(Hmateriai - M-waterVM-watcr] where |J. is the linear attenuation coefficient.
G65. Which of the following loses the greatest amount of energy per unit path length? A 5 MeV .
A. Electron
B. Alpha particle
C. Proton
D. Neutron
G65. B Energy loss is greatest for the particle with the greatest mass and charge.
G66. Directly ionizing radiation includes all of the following except:
A. Electrons.
B. Positrons.
C. Neutrons.
D. Alpha particles.
G66. C Neutrons are not charged particles; they generally interact by transferring their energy to protons or other light nuclei, which then produce dense ionization tracks.
G67. An electron, a proton, and an alpha particle each have 20 MeV kinetic energy. Which of the following statements is true?
A. The alpha particle travels at almost the speed of light.
B. The alpha particle has the least total energy.
C. The proton has the highest total energy.
D. The electron travels almost at the speed of light.
E. None of the above.
G67. D The electron's rest mass is low compared to its kinetic energy so it must be traveling at nearly the speed of light. The alpha particle has the greatest total energy.
A.      G68. When protons interact with soft tissue, all of the following are true except:
Linear energy transfer (LET) increases towards the end of the proton track.
Protons have a finite range.
Protons undergo exponential attenuation.
The proton track ends in a "Bragg Peak".
G68. C Protons, like all charged particles, have a finite range; photons are attenuated exponentially.
G69. Regarding neutrons, which of the following is false?
A. The threshold for neutron emission as a product of photon interaction is about 8 MeV
B. 1 mGy of neutrons is radiobiologically equivalent to 1 mGy of x-rays.
C. Neutron interactions with matter can cause gamma-ray emission.
D. Neutron interactions with matter can cause proton emission.
G69. B Depending on their energy, neutrons cause up to 20 times as much damage as x-rays. For radiation protection purposes, a factor of 20 is used for all neutrons.
G70. In a diagnostic radiograph, the process mostly responsible for differential attenuation is:
A. Coherent scatter.
B. Compton interactions.
C. Photoelectric interactions.
D. Pair production.
E. Bremsstrahlung.
G70. C The probability of a photoelectric interaction is proportional to Z3, so small differences in Z between, say, bone and muscle result in a large difference in the number of x-rays trans¬mitted to the image receptor.
G71. In diagnostic radiology the greatest source of scatter to the film is from the
A. Film holder
B. Collimators
C. Patient's body
D. Floor and walls
E. Table supporting the patient
G71. C The direct beam interacts with the patient's body, and generates Compton scattered photons.
G72. Which of the following does not reduce patient dose (for the same optical density on the film)?
A. Use of screens
B. Using a high kVp
C. Using a high ratio grid
D. Collimation
G72. C Collimating the beam decreases the area of exposure. Using screens and a high kV tech­nique both reduce patient dose. Grids improve contrast by cleaning up scatter, but require a somewhat higher dose to compensate for attenuation by the grid.
G73. The slope or gradient of a film's characteristic (H&D) curve is also known as its
A. Density
B. Transmittance
C. Opacity
D. Lambda
E. Gamma
G73. E
G74. A radioactive sample is counted for 1 minute and produces 900 counts. The background is
counted for 10 minutes and produces 100 counts. The net count rate and net standard deviation are about , counts.
A. 800, 28
B. 800, 30
C. 890, 28
D. 890, 30
E. 899, 30
G74. D The net count rate is: [(Ns/ts) - (Nb/tb)] = [(900/1) - (100/10)] = 890 counts/min.
The net standard deviation, a, = [(Ns/ts2) + (Nb/tb2)]1/2 - [(900) + (1)]1/2 = 30.
G75. In a chi-square test, looking for a statistically significant difference between two experimental results, claims of such a difference with a p value of 0.01:
A. Means there is unquestionably a difference between the two results.
B. Allows the experimenter a wider latitude of error than would a p value of 0.05.
C. Means there is a 99% chance that the claim is true.
D. Means there is a 99% chance that the claim is incorrect.
G75. C The "p-value" represents the probability of error in accepting the conclusion of the statisti¬cal analysis; i.e., there is a 1% chance that the two results are not different.
G76. The number of binary bits that are required to represent all CT numbers from -1024 to 3096 is bits.
A. 8
B. 9
C. 10
D. 11
E. 12
G76. E Each number in the binary system represents 2 to some power. The first number to the right represents 2° = 1. The second number from the right represents 21 = 2; the third is 22 = 4, and so on. The ones and zeros indicate whether a number is present or not. E.g., 101 would be 4 + 0 + 1 = 5. The range of-1024 to 3096 Hounsfield numbers requires 12 bits.
G77. The computational speed of a computer is measured in units of
A. MB
B. MIPS
C. RVU
D. BAUD
E. BPI
G77. B MIPS stands for Millions of Instructions Per Second.
A.      G78. ROM is memory that can be .
Used and changed freely
Freely read, but not written to
Repeatedly used to store output from an input device
Randomly accessed
G78. B ROM stands for Read Only Memory. Programs and/or data once put into ROM may be used, but not revised.
G79. Parallel processing refers to:
A. Running multiple tasks simultaneously.
B. Using multiple processors to increase speed.
C. Computer networking.
D. Sharing peripheral devices between computers.
G79. B Parallel processors divide a task into pieces, with each piece being solved on a separate processor to increase speed.
G80. There are approximately bits in a megabyte.
A. 1024
B. 2048
C. 8000
D. 2,000,000
E. 8,400,000
G80. E 1 byte = 8 bits. 1 Mbyte = 1024 x 1024 - 1,044,448,575 bytes = 8,388,608 bits.
G81. For protection purposes, the absorbed dose is multiplied by a "radiation weighting factor" wR to obtain the equivalent dose H. For fast neutrons wR is .
A. 1
B. 5
C. 10
D. 20
E. 50
G81. D See NCRP Report 116.
G82. The principal hazard from indoor radon involves:
A. Whole body dose from gamma rays.
B. Skin dose from betas.
C. Lung dose from alpha emission.
D. Bone dose from deposited radionuclides.
G82. C Radon decays through a chain of daughter products that may be inhaled, some of which decay by alpha emission.
G83. The average natural background dose to members of the general public is due to cosmic radiation, terrestrial radiation, and .
A. Fallout
B. Medical x-rays
C. Nuclear plant releases
D. Contamination from radioactive waste disposal
E. Internal radiation
G83. E The average natural background radiation is about 100 mrem/year (excluding radon).
About 40 mrem/yr is contributed by radionuclides within the body, mostly 40K. (The other choices are artificially produced.)
G84. About half the average effective dose equivalent received by the U.S. population is attributable to:
A. Radon
B. Medical procedures
C. Fallout
D. Cosmic radiation
E. Internal radionuclide
G84. A About 55% (2 mSv per year) is due to radon, 11% from internal radionuclides, 8% from cosmic radiation, 8% from terrestrial radiation, and 15% from medical procedures.
G85. For risk-benefit calculation purposes, NCRP Report 116 (1993) estimates a probability of
developing fatal breast cancer from one 2-view mammographic exam to be about %.
(Assume a mean glandular dose of 4 mGy to each breast.)
A. 0.001
B. 0.1
C. 1
D. 10
G85. A NCRP Report 116 gives a probability of fatal cancer of 0.05 per Sv x breast organ weight¬ing factor of 0.05 for a total of 0.0025 per Sv. For x-rays 1 Sv = 1 Gy. Thus the probability is 0.0025 x 10"3 per mGy, or 0.00025% per mGy. For 4 mGy the probability is 0.001%. This is typical of the increased risk for many radiographic exams.
G86. The radiation protection quantity which has been used in attempts to estimate the cancer risk from x-ray irradiation of personnel is .
A. Exposure (X)
B. Air Kerma (K)
C. Absorbed dose (D)
D. Dose equivalent (H)
E. Effective dose (E)
G86. E In NCRP Report 116 the Effective Dose (E) attempts to weight the radiation dose to different organs by the relative cancer risk of each organ.
G87. Deterministic or non-stochastic effects of radiation include all of the following except:
A. Bone marrow damage
B. Skin damage
C. Cataract induction
D. Leukaemia
E. Infertility due to gonadal irradiation
G87. D Cancer is a probabilistic or stochastic effect; the chance of getting cancer depends on the dose, but the severity is not dose related.
A.      G88. According to NCRP there is a negligible increase in the risk of adverse effects to the fetus, compared with other risks of pregnancy, up to a total dose of mGy.
5
20
100
500
1000
G88. C
G89. The NRC requires personnel to wear a radiation monitor if they are likely to receive %
of the annual dose limit.
A. 90
B. 50
C. 25
D. 10
E. 1
G89. D Prior to 1994 it was 25%, but since this date it has been set at 10%. G90. C
G90. Regulations in the U.S. limit the total dose equivalent to the fetus of a declared pregnant radiation worker to mSv.
A. 50
B. 25
C. 5
D. 2.5
E. 0.5
G90. C
G91. The recommended annual effective dose limits for members of the public (NCRP 116) are mSv for infrequent exposure, and mSv for continuous exposure.
A. 5, 1
B. 1, 5
C. 10, 5
D. 5, 5
E. 10, 20
G9I. A
G92. Which of the following has the greater shielding value?
A. 2 Tenth Value Layers (TVLs).
B. 6 Half Value layers (HVLs).
C. They are equal.
D. It depends on the material.
G92. A 2 TVLs = 1/102 = 0.01.
6 HVLs = 1/26 = 0.16.
G93. The exposure rate at 1 m from a radioactive source is 100 mR/hr. A 0.06 mm lead
shield is placed around the source. The exposure rate is now mR/hr.
(HVL in lead is 0.02 mm.)
A. 75
B. 50
C. 25
D. 12.5
E. 5
G93. D The shield is 3 HVLs thick, and therefore reduces the dose rate by a factor of l/23 = 1/8 = 0.125.
G94. In the event of a 137Cs "dirty bomb" explosion, regarding the use of potassium iodide pills, which of the following is true?
A. KI should be taken as soon as possible.
B. KI should be taken if the thyroid has the potential of receiving a dose greater than 15 rem.
C. A dose of 130 mg per day is suggested for adults.
D. KI will offer no protection.
G94. D Sodium iodide is only used to protect the thyroid from the radioactive iodine released in a nuclear explosion. 13'I has a relatively short half-life, and would not be used in a "dirty bomb".
G95-98. Match the most appropriate instrument to the procedure in each question.
A. Liquid scintillation counter
B. Nal well counter
C. Geiger-Mueller (GM) counter
D. Thermoluminescent dosimeter (TLD)
E. Ionization chamber survey meter
G95. Gamma ray sealed source wipe test
G96. Contamination survey for 99mTc.
G97. Radiation survey of a diagnostic x-ray installation
G98. Personnel monitoring
G95. B A Nal well counter is an efficient device for measuring low-level gammas. It can also pro¬vide energy discrimination.
G96. C A GM counter has a fast response and the ability to detect low levels of gamma rays.
G97. E An ionization chamber survey meter is capable of accurate x-ray dose rate measurements with minimal energy dependence.
G98. D The small size and relative energy independence of TLD make it useful as a personnel monitoring device.
G99. The basic consideration in setting limits for disposal of radioactive materials into the sewer system is:
A. Contamination of the sewer.
B. Risk to swimmers.
C. Fish death.
D. Entrance into the food and fresh water chains.
E. Evaporation into the air.
G99. D Very low concentrations of radioactive materials, when ingested, can produce high, local¬ized radiation doses to internal organs.
G100. "ALARA" stands for:
A. As Long As Reasonably Allowable.
B. As Low As Responsibly Attainable.
C. As Low As Reasonably Achievable.
D. As Little As possible RadioActivity
G100. C ALARA is a basic tenet of radiation protection. Obviously, radiation levels could be
reduced to negligible levels with huge amounts of shielding that would be prohibitively expensive and unwieldy. The ALARA concept seeks to strike a reasonable balance between safety and practicality
DI-D3. Concerning radiation interaction with matter, select the answer (A-E) that is most closely associated with the question. (Answers may be used more than once).
Compton scattering
Photodisintegration
Photoelectric effect
Pair Production
Bremsstrahlung
Dl. The primary interaction of 99mTc photons in soft tissue. D2. The primary interaction of 30 kVp x-rays with breast tissue. D3. The primary interaction of 201T1 photons in bone.
D1. A 99mTc photons have an energy of 140 keV
D2. C Depending on filtration type and thickness, a 30 kVp spectrum is dominated by relatively low energy photons, which interact predominantly by the photoelectric effect in breast tissue.
D3. A The main photon energy of 201T1 is 70 keV The photoelectric effect accounts for less than 25% of the interactions at this energy in bone.
D4. Changing the focal spot from 0.6 mm to 1.0 mm has the following effect:
A. Heat loading capacity of the anode is decreased.
B. Geometric unsharpness of the image is improved.
C. High contrast spatial resolution of the image is improved.
D. Available exposure times at a given mA station are increased.
E. Patient dose is decreased.
D4. D Increasing in the focal spot size increases the area over which the instantaneous heat is
distributed. This increases the heat loading capacity of the anode track, and thus the avail¬able exposure times for a given mA. Unsharpness and resolution are degraded. The dose remains the same, since the total mAs is unchanged.
D5. Which of the following will increase the image contrast that is due to the screen-film image
receptor?
A. Decreasing the grid ratio
B. Decreasing the kVp
C. Increasing the developer temperature
D. Increasing the focal spot size
E. Increasing x-ray beam filtration
D5. C This is the only answer that refers to properties of the image receptor.
D6. Which of the following will increase subject contrast in a screen-film imaging system?
A. Decreasing the grid ratio
B. Decreasing the kVp
C. Increasing the developer temperature
D. Increasing the focal spot size
E. Increasing x-ray beam filtration
D6. B Due to increased attenuation coefficients with reduced kVp, A and E will reduce subject contrast. Focal spot size affects resolution, not contrast.
D7. Low contrast detectability refers to the ability of a system to distinguish:
A. A calcified lung nodule
B. A non-calcified lung nodule
C. Between overlying and underlying tissues
D. The size of a small fracture
E. Vessels during the arterial phase of a normal angiogram
D7. B Low contrast detectability represents the ability of a system to reproduce an object whose linear attenuation coefficient does not vary greatly from the surrounding material.
D8. For a screen-film image, the degree of mottle is:
A. Usually determined by the number of primary x-ray photons absorbed in the film.
B. Independent of the mAs.
C. An inherent property of the AgBr structure in the film emulsion.
D. Increased by increasing film speed.
E. Measured in line pairs per mm (lp/mm).
D8. D Increasing film speed reduces the need for primary x-ray photons and hence increases
mottle (quantum noise). The vast majority of the x-ray photons that contribute to the image are absorbed in the intensifying screen(s), not in the film.
D9. The modulation transfer function (MTF) is a tool for describing the of an imaging
system.
A. Properties of the characteristic (H&D) curve
B. Sharpness
C. Noise content
D. Latitude
D9. B
D10. Increasing the mAs will the x-ray beam quantity, and will the x-ray beam
quality.
A. increase, increase
B. decrease, increase
C. increase, decrease
D. decrease, decrease
E. increase, not change
D10. E Increasing the mAs increases the number of x-rays, but the shape of the spectrum does not change.
D11. Most of the energy transferred from incident electrons to the target of a diagnostic x-ray tube goes to the production of .
A. Characteristic radiation
B. Bremsstrahlung
C. Heat
D. Compton scatter
D11. C Typically, 99% or more of the energy deposited by the electrons in the target of an x-ray tube is dissipated as heat.
D12. The effect of adding 0.1 mm Cu filtration to an x-ray beam is:
A. Lower patient skin exposure.
B. Increased HVL.
C. Reduced contrast at the same kVp.
D. Longer exposure times.
E. All of the above.
D12. E Increasing filtration increases the HVL by removing more of the lower energy photons than the penetrating higher energy photons. This in turn reduces contrast. The more penetrating filtered beam will then require less exposure at the skin entrance to properly expose the film. However, since the filter reduces the overall number of photons (both high and low energy) in the beam, the exposure time must be increased to compensate
D13-D 17. Match the following target and filter materials with the radiographic examinations most closely associated with their use. (Use each answer only once).
Copper, aluminum
Molybdenum
Aluminum
Tin, copper, aluminum
Leaded acrylic shaped wedges
D13. Filter used for airway studies
D14. Filter used for routine angiography
D15. Target used for screen-film mammography
D16. Filter used for chest radiography
D17. Filter used for scoliosis radiography
D13. D High filtration is required to reduce the contrast of the spinal column compared to soft tissue.
D14. C A1 filters are used in routine radiography to selectively filter out the lower energy photons.
D15. B In mammography low-energy photons are required to improve contrast.
D16. A Filtration is used to increase latitude and reduce patient dose.
D17. E These shaped filters are attached to the bottom of the collimators, and are used to compen¬sate for density differences, thus providing a more uniform exposure over the film.
D18. The maximum wavelength of the photons in an x-ray spectrum is primarily determined by the…………..
A. kVp
B. Anode material
C. Filtration
D. Anode angle
E. Ripple
D18. C The maximum wavelength corresponds to the lowest photon energy, which is determined by the filtration.
A.      D19. In order to decrease the optical density of an over-exposed radiograph from 2.0 to 1.2,
the mAs should be reduced by approximately %. (Assume a slope of the characteristic
curve of 3.0.)
5-10
10-20
20^10
40-60
E. Greater than 95
D19. D OD2 - OD1 = y log(E2/El) where E is exposure, proportional to mAs. -0.8 = 3.0 log (E2/E1) E2/E1 = 0.54 or 46% reduction
D20. For a fixed mAs and kVp, increasing the exposure time will significantly affect the:
A. Overall film density.
B. Overall film latitude.
C. Speed of film/screen combination.
D. Motion unsharpness.
E. Patient exposure.
D20. D Increasing exposure time increases motion unsharpness. For a fixed kVp and mAs, the film and patient exposures, latitude and speed are unaffected. No change in kVp means no change in system latitude. The screen/film system usually is not significantly affected by exposure time variations, though if the exposure time becomes quite long, the speed is reduced by the reciprocity failure effect.
D21. A high-ratio grid is used to eliminate a significant portion of the scattered photons. As a result, the radiation dose to the patient will and the radiographic contrast will……...
A. increase, decrease
B. increase, increase
C. decrease, increase
D. decrease, decrease
E. be unchanged, increase
D21. B Reducing scatter improves contrast. As the grid ratio increases, the patient dose increases, as more radiation is required to replace the radiation absorbed in the grid.
D22. A radiographic chest unit with a focused grid produces films with optical densities much lighter towards the edges. A possible cause is:
A. Upside-down grid.
B. Lateral misalignment of the focal spot and the grid center.
C. Poor film-screen contact.
D. Patient motion.
D22. A An upside-down grid will produce a film that is dark in the center and almost blank toward the edges.Patient motion produces a blurred image; lateral misalignment produces an over¬all lighter image; poor screen-film contact produces a spotted image.
D23. The tomographic unit with the highest spatial resolution is:
A. MRI.
B. CT.
C. Stereo fluoroscopy.
D. 3-D ultrasound.
E. Laminar section tomography.
D23. E MRI, CT, and ultrasound have resolutions of about 1 lp/mm or less. Stereo fluoroscopy is limited to the TV resolution of about 2 lp/mm. Laminar section tomography is limited to the film-screen resolution of 4 to 8 lp/mm
D24. Compared with a single-screen, single-emulsion film system, double-screen double-emulsion films in radiography .
A. Are less sensitive to processor artifacts.
B. Are associated with a higher patient dose.
C. Produce lower image contrast.
D. Produce lower spatial resolution.
E. Are associated with lower speed.
D24. D Double-emulsion films may produce higher contrast than single-emulsion films, and are
faster, thus reducing patient dose. Using two screens reduces spatial resolution due to more light diffusion in the thicker screens, and some possible crossover of light from the far screens to the emulsions.
D25. A radiograph with an optical density (OD) of 3.0:
A. Transmits 1/3 times as much light as a film with an OD of 1.0.
B. Has a combined OD of 6.0 when placed over another film of OD 2.0.
C. Looks "light gray" on a standard view box.
D. Transmits 0.1% of the light incident upon it.
D25. D Optical densities are additive, so the net OD of both films is 5.0. An OD of 3.0 looks black on a standard view box.
OD = log10(I/T), where I and T are the incident and transmitted light intensities. For OD = 3.0, T = 0.001, and for OD = 1.0, T = 0.1.
D26-D28. For a fixed mAs and kVp, decreasing the exposure time will
A. Increase
B. Decrease
C. Not affect
D26. Film density D27. Film latitude
D28. Speed of the screen-film combination
D26. C Film density is a function of exposure. Decreasing exposure time does not affect the total exposure. (This assumes reciprocity law is valid.)
D27. C Latitude is a function of kVp, which was not changed.
D28. C Speed depends on the thickness and sensitivity of the screen material, which was not changed.
D29. The following film processor parameters all affect mammography image quality except:
Nitrate depletion.
Developer concentration.
Developer temperature.
Replenishment rates.
Developer immersion time.
D29. A B-E directly influence the density and speed of the various film/screen systems. Although bromine concentration in the developer affects both speed and contrast, nitrate depletion has no effect on the development of mammography films.
D30. In order to pass ACR accreditation, using the approved phantom, a minimum of fibers
must be visible on the x-ray image.
A. 1
B. 2
C. 3
D. 4
E. 5
D30. D To pass ACR at least 10 objects must be visible in the image of the phantom. These must include four fibers, three speck groups, and three masses.
D31. In mammography, average glandular tissue dose depends on:
A. Breast compression.
B. Breast thickness
C. kVp.
D. mAs.
E. All of the above.
D31. E Average dose to the glandular tissue is an indicator of the risk of carcinogenesis. The dose depends on the energy distribution in the incident x-ray beam (kVp, HVL), mAs, and com¬pressed breast thickness.
D32. Calcifications are seen on mammograms primarily because of their
A. Atomic number
B. Density
C. Electrons/gram
D. Size
E. Location
D32. A At the low x-ray energy used in mammography, the interactions are predominantly photo¬electric. Thus attenuation (and hence contrast) is proportional to Z3. The Z of calcium is 20, while the effective Z of fat and breast tissue is between 6 and 8.
D33. In mammography the ACR recommends a compression force of 25 to 40 pounds. This will result in:
A. Reduced geometric unsharpness.
B. Contrast improvement.
C. Reduced radiation dose.
D. Diminished motion unsharpness.
E. All of the above.
D33. E Good compression moves the breast tissue closer to the film, reducing geometric unsharp- ness from the focal spot. It also reduces the breast thickness near the chest wall, resulting in less scatter and better contrast, and a lower radiation dose. Compression also prevents movement, reducing motion unsharpness.
D34. In order to limit loss in spatial resolution due to geometric unsharpness, the measured focal spot size for screening mammography should be less than………mm.
A. 2.0
B. 1.0
C. 0.6
D. 0.3
E. 0.15
D34. C See NCRP Report 85: "Mammography - A User's Guide".
D35. In the diagram of the image intensifier shown below, all of the numbered arrows represent photons, except:
A. I.
B. II.
C. III.
D. IV
E. V
D35. D IV represents electrons produced in the photocathode. I and II are x-ray photons, and III and V are light photons.
D36. A 9-in. multi-mode image intensifier (II) is switched to the 6-in. mode. As a result, the image
will be , and the automatic brightness control system (ABC) will the exposure
to the II and the patient.
A. magnified, decrease
B. magnified, increase
C. minified, increase
D. magnified, not change
E. minified, decrease
D36. B Switching to the 6-in. mode spreads out 6-in. of input information over the entire output
phosphor, resulting in a magnified image. This also reduces the light intensity at the output phosphor because only the x-ray photons from the center 6-in. portion of the input con¬tribute to the light. Thus, the ABC system will usually drive the patient dose higher to maintain constant brightness.
D37. For fluoroscopy performed in the 6-in. II mode and displayed on a 525 line TV monitor, spatial resolution is most limited by the …
A. Scatter from the patient
B. Grid
C. II tube
D. Optical system
E. TV system
D37. E The TV system has a resolution of only about 1.8 lp/mm. The II has a resolution of about 4 lp/mm. The grid and optical lenses do little to degrade the spatial resolution.
D38. In a digital subtraction angiography (DSA) a plumbicon tube is used to obtain the video image. If the tube is driven at standard broadcast TV rates, it takes…………|isec to scan one line.
A. 5.3
B. 31.7
C. 63.5
D. 90.1
D38. C Standard broadcast TV consists of 30 frames/sec with 525 lines/frame. It is actually made up of two interleaved half frames of 262.5 lines lasting l/60tb of a second. Time for one line = (1/30 sec)/525 lines = 63.5 sec.
D39. In cardiac catheter procedures, all of the following changes would reduce the patient's dose except:
Reducing the frame rate.
Increasing the f number of the lens.
Increasing the aperture of the lens.
Increasing the conversion gain of the II.
Increasing the source-to-image receptor distance (SID).
D39. B Dose is directly proportional to frame rate. Large f-number lenses are less efficient at
gathering light. Larger apertures permit more light to reach the cine film. A higher conver¬sion gain produces more light for less radiation. Larger SIDs utilize geometry to reduce the patient's radiation dose.
D40. is the persistence of an image on the fluoroscopic TV monitor. With motion, the image
becomes blurred because of this persistence.
A. Lag
B. Vignetting
C. Flare
D. Veiling glare
E. Blooming
D40. A Vignetting is the brightness change from the center to the edge of the II. The center is
brighter than the edge. Flare and veiling glare are related parameters that indicate a degra¬dation in contrast due to light scattering in the image intensifier. Blooming is the excessive brightness that occurs at the outside edge of a highly attenuating object.
D41. In CT, all of the following will decrease noise except:
A. Increased scan time
B. Decreased patient thickness
C. Increased slice thickness
D. Decreased window width
E. Increased detector efficiency
D41. D Increased scan time results in a higher dose; therefore, more photons are detected resulting in less noise. A thinner patient causes less attenuation, resulting in more photons at the detector, and less noise. A thicker slice increases the total number of photons detected, reducing noise. Window width has no effect on the number of photons detected. Higher detector efficiency increases the number of photons detected, thus reducing noise.
42. For an increase from 200mA to 400mA in a CT exam the result would be to
A. Increase high contrast (spatial) resolution
B. Increase low contrast resolution (decrease noise)
C. Change neither the high or low contrast resolution
D42. B Increasing mA results in an increased number of photons, which will lower the noise (quantum mottle), and increase low contrast resolution
D43. The CT value of white matter is 40 HU, and that of gray matter is 45 HU. The approximate subject contrast between white and gray matter is……..
A. 0.12
B. 0.5
C. 1.2
D. 5.0
E. 12.0
D43. B For low CT numbers (-200 to 200) the percent contrast can be approximated by: % contrast = (CT number difference)/10 = (45-40)/10 = 0.5
D44. Spiral/helical CT scanning consists of continuously moving the couch while rotating the x-ray beam continuously through the use of "slip ring" electrical contactors. This results in effective¬ly thicker slices, but allows a complete region of anatomy to be scanned in a single breath hold. Which of the following is false?
A. For a given beam width, partial volume effects will decrease.
B. Registration from slice to slice of organs that move during breathing will improve.
C. An x-ray tube with a higher heat capacity is needed.
D44. A Partial volume effects depend on slice width. Helical CT scanning effectively gives thicker slices for a given beam width. The main advantage of helical scanning is better registration of organs; breathing between each slice, as in conventional CT scans, does not ensure that the organ will be in the same position on each slice, as the depth of inspiration may change. As the beam is on continuously, with no opportunity to cool the target between slices (as with axial scanning), a tube with a higher heat capacity is required.
D45. If a CT bone window is set at a width of 1000 with the center at 500, the range of CT numbers that will be displayed as black is .
A. Greater than 500
B. Less than 500
C. Less than-500
D. Less than 0
E. Less than 1000
D45. D The center is at 500, and the width is 1000, i.e., 0-1000. CT numbers below 0 are outside the window, and are displayed as black.
D46. A digital image is acquired on a 9-in. receptor. In order to preserve 3 lp/mm and provide 256 shades of gray, the amount of storage capacity required is .
A. 1 MB
B. 2MB
C. 4MB
D. 8MB
E. 1.7 GB
D46. B A 9-in. (or 229 mm) image receptor digitized to provide 3 lp/mm requires
(229 mm x 6 lines/mm), i.e., a 1370 square matrix. The total number of pixels is 13702 = 1.88 x 106. Since 256 shades of gray are provided, each pixel includes 1 byte (=8 bits) of information. Thus, 1.88 x 106 bytes, or approximately 2 MB are required.
D47. All of the following are image processing filters except:
A. Recursive.
B. Spatial low pass.
C. Scanned projection.
D. TID.
E. Matched.
D47. C Scanned projection radiography is a type of digital imaging system, not a type of software filter.
D48. The principle advantage of digital subtraction angiographic studies, compared with cine, is the…….
A. Reduction in patient dose
B. Improved high contrast spatial resolution
C. Improved low contrast detectability
D. Ability to digitally store data
E. Reduction in overall scatter
D48. C Subtracting structural background improves the visualization of low contrast objects.
On a frame-by-frame basis, digital subtraction angiography delivers about 50 times more radiation than cine, and therefore produces more scatter. Digital storage is an advantage, but is not the principle reason for using digital radiography. The digital system has lower resolution than a cine film.
D49. If the matrix size of a CT image is 512 and the CT number range is 4096 numbers wide, the number of bits required for each image is .
512 x4096
512 x 512 x4096
512x512x12
512x64
D49. C The image consists of 512 x 512 pixels with 12 bits per pixel = 4096 = 212. Note that you would actually need to store each pixel in 2 bytes since you cannot use a fraction of a byte. Thus, in reality, you end up using 16 bits per pixel (a byte is 8 bits).
D50. A DICOM standard for cine specifies that the image matrix is 512 x 512 with a depth of 8 bits per pixel. You have the need to store high-speed, high-resolution DSA raw data (1024 x 1024 x 16 bits). The relative disk storage space needed to store an individual image for the latter is times as large as the original.
A. 32
B. 16
C. 8
D. 4
E. 2
D50. C The matrix height and width are each twice as large as the DICOM standard
(1024 vs. 512); thus four times as many pixels are required. In addition, each pixel needs twice as many bits (16 vs. 8), so the total storage is 4 x 2 = 8 times as large.
D51. A flat panel digital radiographic detector has a square 20 x 20 cm image receptor field. The full field of the detector is coupled to a nominal 2048 x 2048 CCD array. The relative spatial resolution (expressed as line pairs per mm) when going from a 20 x 20 to a 10 x 10 field of view is .
A. Four times better
B. Twice as good
C. The same
D. Half as good
E. One fourth as good
D51. C Digital radiographic (DR) detectors use a fixed number of pixels to image the entire field of view (2048 x 2048 for 20 x 20 cm in this example). Only some of the pixels are used when the image is coned down (1024 x 1024 for 10 x 10 cm in this example). Since the pixel size is fixed, spatial resolution does not change with collimation.
D52. An imaging system is equipped with variable frame rate pulsed fluoroscopy in place of contin¬uous fluoroscopy. The mAs is programmed to yield equivalent noise perception to the old system. (No other changes are made). This system is used to observe the swallowing reflex. Which of the following is true:
A. The sharpness of the clinical images is unchanged.
B. The contrast of the clinical images is unchanged.
C. Pulsed video at 30 frames/second always results in less dose than continuous video.
D. Pulsed video never results in less dose than continuous video.
D52. B Continuous video implies continuous x-ray production during the entire acquisition of a fluoroscopic frame. Pulsed fluoroscopy typically uses the same kVp and about the same mAs, but the mAs is delivered in a short pulse. The resulting stroboscopic effect yields less motion blur than continuous fluoroscopy, thus improving the sharpness of a rapidly moving structure. Contrast is related to kVp, and is unchanged. Since the mAs can be programmed, the dose per second can be programmed. In some cases lower frame rates are programmed for dose reduction, while in other cases the dose rate may actually be increased.
D53. For hydrogen imaging in a 1.0 T MRI unit, the frequency of the RF signal is about
A. 40 Hz
B. 40 kHz
C. 40 MHz
D. 400 MHz
E. 4 GHz
D53. C The frequency for hydrogen is 42 MHz x magnetic field strength in tesla.
D54. In MRI the signal-to-noise ratio can be increased by all of the following except:
A. Switching from a volume to a surface coil.
B. Increasing the number of acquisitions.
C. Increasing the static magnetic field strength.
D. Decreasing the slice thickness.
E. Increasing TR.
D54. D
D55. In MRI contrast is created by all of the following except:
A. Differences in hydrogen content.
B. Differences in T1 time of tissues.
C. Differences in T2 time of tissues.
D. Administration of a contrast agent.
E. Differences in atomic number.
D55. E Unlike x-ray images, atomic number differences have no effect on MRI images.
D56. In MRI, pure water will have a T1 and a T2.
A. long, long
B. long, short
C. short, long
D. short, short
D56. A
D57-D59. Match the following MRI terms. (Answers may be used more than once.)
Shim coils
RF
Gradient fields
T1
T2
D57. Used to localize MR signal
D58. Used to tip the net magnetization of spins
D59. Used to adjust magnetic field uniformity
D57. C Gradient fields are required in three planes for localization.
D58. B The RF is used to excite (or tip) the hydrogen nuclei out of equilibrium with the main field.
D59. A Shim coils and sometimes small blocks of metal are used to improve the uniformity of the main magnetic field.
D60. The ultrasound propagation velocity through is at least twice the velocity of propaga¬tion through the other human tissues listed.
A. Brain
B. Fat
C. Muscle
D. Soft tissue
E. Skull (bone)
D60. E Ultrasound velocity is inversely related to the compressibility of the conducting material.
All body tissues have approximately the same compressibility and velocity (1540 m/s) except bone, which is much less compressible than the other tissues. The velocity in bone is about 4080 m/s.
D61-D62. Match the reflected ultrasound intensity with the interface in the question:
0.999
0.412
0.050
0.008
0.0002
D61. Air-muscle interface D62. Bone-muscle interface
D6I. A Reflected intensity = [(Z2 - Z,)/(Z2 + Z,)]2-
D62. B
D63. The wavelength of a 2 MHz ultrasound beam is mm.
A. 0.02
B. 0.55
C. 0.77
D. 2.0
E. 5.0
D63. C Wavelength X = velocity v/frequency f. The average velocity in tissue is 1540 m/s. Frequency = 2x 106 s"1.
D64. An ultrasound transducer that is used to examine tissue receives a reflection echo 64 |is after the signal is sent. The depth of the structure that caused the echo is cm.
A. 2
B. 5
C. 7
D. 10
E. 15
D64. B Half the time is required for the pulse to go forward, and half for it to return. Distance = velocity x time.
D = (1540 m/s) x (32 x 10"6 s) = 4.9 cm.
D65. In ultrasound, the intensity of a reflected echo depends on all of the following except:
A. Depth to interface.
B. Transducer diameter.
C. Transducer frequency.
D. Difference in acoustical impedance at the interface.
D65. B Transducer diameter affects only lateral resolution.
D66. The intensity of a reflected 5 MHz sound beam is 40 dB. The incident intensity is 20 mW/cm2. The echo intensity is about mW/cm2.
A. 20 x 104
B. 4 x 102
C. 4 x 10"2
D. 20 x 10'2
E. 20 x 10-4
D66. E The reflected intensity (I) is lower than the incident intensity (I0). The relative power in decibels (dB) is:
dB = 10 log10 (Io /I).
40 = log10 (20/1).
(20/I)4 = 10.
I = 20 x lO^4.
D67. Modern ultrasound units have spatial resolution limits of about in both axial and lateral
directions.
A. 5 mm
B. 1 mm
C. 0.5 mm
D. 0.1 mm
E. 0.05 mm
D67. B The spatial resolution of ultrasound units is similar to that of CT scanners, about 1 mm.
D68. A shadowing artifact in ultrasound may be due to the reduction of reflected intensity behind all of the following except:
A. A strong attenuator.
B. A highly reflective interface.
C. Gas or air.
D. Water.
D68. D A shadowing artifact is caused by a lack of reflection from an area. This can be caused by an incident beam being highly attenuated, or if the beam is strongly reflected from an overlying interface, such as between air or gas and tissue. Water has a low absorption coefficient, and acts as a window, generally producing no shadowing.
D69. The resonant frequency of an ultrasound transducer is determined by
A. Crystal thickness
B. The critical angle
C. The Q factor
D. Acoustical impedance
E. TGC
D69. A The resonant frequency (f) is determined by the speed of sound in the medium (c) and the crystal thickness.
f = c/(2 x thickness).
D70. Color Doppler ultrasound measures blood flow based upon information from all of the follow¬ing factors except:
A. Impedance
B. Wavelength
C. Ultrasound speed
D. Angle of incidence
E. Frequency shift
D70. A
D71. A pre-calibrated 13'I sodium iodide "therapy" dose is delivered to the hospital on Monday at noon. If the dose to be administered on Thursday at noon is 10 mCi, the activity at noon on Monday is mCi. (T1/2 = 8.04 days.)
A. 5.4
B. 7.7
C. 13.0
D. 19.2
E. 25.5
D71. C At = A0 exp-(0.693t/T1/2) where A0 is the initial activity, and A, is the final activity after time t.
Substituting A, = 10 mCi, t = 3.0 days, Ti/2 = 8.04 days, gives A0 = 12.95 mCi.
D72. An unshielded syringe contains 20 mCi of 99mTc. The exposure rate at 20 cm from the syringe is . <T = 0.6 R-cm2/mCi-hr.)
0.03 mR/hr
30 mR/hr
0.6 R/hr
600 R/hr
240 R/hr
D72. B Assuming the syringe to be a point source,
Exposure rate = T A (1/d2) = 0.6 x 20 x (1/202) = 0.03 R/hr = 30 mR/hr.
D73. The United States Pharmacopia sets the standard for "Mo breakthrough as of "Mo per 1.0 mCi of 99mTc.
A. <0.1 mCi
B. <0.01 mCi
C. <0.15 (J-Ci
D. <1.0 jaCi
D73. C
D74. Flourine-18 used in FDG imaging is produced by:
A. Neutron bombardment of 17F in a nuclear reactor.
B. Elution from a fluorine generator.
C. A (p,n) reaction in a cyclotron.
D. High-energy electron bombardment of a sulfur target using a betatron.
D74. C 18F is produced by a (p,n) reaction on lsO.
D75. To check the activity of a vial containing a pure beta emitter in a well counter, would
be required.
A. A calibrated beta source from the manufacturer
B. The exact volume of the well counter
C. A calibrated 99mTc source
D. The gamma factor and the exposure rate constant for the beta source.
D75. A
D76. When comparing studies using the same activity of l31I and 99mTc, all of the following state¬ments are true except:
A. The detector counts are higher with the higher energy isotope, 131I.
B. The resolution of the 131I image is lower.
C. The 13'I image is noisier.
D. The radiation dose to the patient is higher for the 13'I study.
E. The exposure to the technologist performing the scan is higher for the 13'I study.
D76. A Detector counts decrease with increasing energy. Although the fraction of photons escaping the patient is higher for 131I, the efficiency of the Nal crystal is about 25% for 131I (364 keV) compared with about 80% for "mTc (140 keV).
D77. While imaging a patient with 99mTc, the multichannel analyzer (MCA) displays a small peak at 280 KeV The most likely explanation is:
A. Lead (Pb) x-rays.
B. Backscatter.
C. Coincidence.
D. Iodine escape peak.
E. Compton scatter.
D77. C Two 140 keV photons entering a scintillation crystal simultaneously would create the same light pulse as a 280 keV photon.
D78. The main purpose of the multi-channel analyzer and energy discriminator on a gamma camera is .
A. To reduce the camera dead time
B. For quality assurance
C. To increase camera sensitivity
D. To reduce counting of scattered photons which possess no useful positional information
D78. D Scattered photons have less energy than the photopeak. They degrade the image contrast because most do not contribute to the accurate positional information of the activity distri¬bution in the body. Energy discrimination allows most of these scattered photons to be eliminated.
D79. Most diagnostic nuclear medicine scans are performed with an image acquisition setting of 8
bits per pixel. This means that the maximum number of counts per pixel is 28 or 256. The mini¬mum detectable count difference which is significant at the 2a level is about %.
A. 15
B. 12.5
C. 9
D. 6
E. 3
D79. B The standard deviation o = VN where N is the average number of counts per pixel.
The minimum detectable count difference would occur when the counts are highest, i.e., 256. 2a = 2a/256 = 32.
Expressed as a percent of 256 this is 32/256 x 100% = 12.5%. According to Poisson statis¬tics, a pixel with a number of counts that differs by more than 12.5% from the average occurs less than 5% of the time.
D80. If the background is 100 counts and the sample is 900 counts, the net standard deviation is .
A. 50.0
B. 40.0
C. 31.6
D. 28.3
E. 24.1
D80. C The net standard deviation is ^(crb2ackgroun#crs2aitple) = V(i00+90<) = 316.
D81. A cold spot artifact appears in a scintillation camera image. The artifact could be caused by all of the following except:
A. The camera is incorrectly peaked for the radionuclide in the study.
B. The photomultiplier tube is defective.
C. The patient is wearing metallic jewelry.
D. An out-dated uniformity correction is used.
E. The wrong collimator was used.
D81. E The wrong collimator would increase septal penetration and increase or decrease camera sensitivity, but could not produce a cold spot in the image.
A.          D82. The crystal thickness routinely used in an Anger camera for 99mTc imaging is inches.
0.25-0.5
1-2
2-3
3-4
D82. A
D83. In a gamma camera, positional information is obtained by:
A. Using an array of small detectors.
B. Moving a crystal over the patient.
C. Marking the region of interest on the patient.
D. Analyzing the light output from an array of photomultiplier tubes
D83. D
D84. A patient is to be injected with 740 MBq (20 mCi) of 150 water (T1/2 = 124 seconds).
If there is a delay of 10 seconds in administering the injection, the actual activity injected is MBq.
A. 760
B. 740
C. 735
D. 710
E. 699
D84. E 740 exp(-0.693 x 10/124) = 699 MBq.
D85. All of the following are true statements about PET scanning, except:
A. Radioisotopes are cyclotron produced.
B. Positrons are not detected directly.
C. Coincident detection at 180° is required.
D. Images are generally axial tomograms.
E. The detector photopeak is centered at 1.02 MeV
D85. E The detector photopeak must be centered around 512 keV, the energy of each annihilation photon. Positron emitters are proton rich, and are made in cyclotrons. The positrons are not detected directly, but rather the annihilation photons are detected in coincidence at 180° from each other.
D86. In PET, scatter and background are greatly reduced because of
A. The thickness of the parallel hole collimator
B. Pulse height analysis
C. Coincidence detection
D. Off peak imaging
D86. C Coincidence detection, i.e., detection of simultaneous anti-parallel photons, eliminates most noise.
D87. In a SPECT image the assigned values in each pixel in the reconstructed image represent .
A. Densities
B. Absorption factors
C. Attenuation factors
D. Radioisotope concentrations
E. Both B and C.
D87. D In nuclear medicine scintigrams, the measured values are representative of the radioisotope distribution. In the case of SPECT, where the reconstructed image is of a known slice thickness, each pixel in the image is associated with a unit volume; hence the values repre¬sent isotope concentrations. However, they are affected by the attenuation of the radiation in the body.
D88. The Mammography Quality Standards Act (MQSA) requires all facilities to have all of the fol¬lowing except .
A. Annual physics testing
B. Daily processor sensitometry
C. Minimum image quality standards
D. FDA certification
E. Special 15:1 high ratio grids
D88. E
D89. In daily gamma camera QA testing, a lead bar phantom is used to check
A. That the energy window is centered
B. Resolution
C. Uniformity
D. Collimator transmission
D89. B "Peaking" consists of tuning the energy window so that it is centered about the photopeak of the isotope used in the study. The bar phantom is used to check resolution.
D90. Daily QA of an x-ray film processor includes tests of all of the following except
A. Base and fog density
B. Speed
C. Contrast
D. Replenishment rate
E. Developer temperature
D90. D
D91-D92. Match the following maximum permissible fluoroscopic tabletop exposure rate with the type of II exposure control system (under-table x-ray tube):
A. 5 mR/min
B. 10 mR/min
C. 5 R/min
D. 10 R/min
E. No limit
D91. Manual control
D92. Automatic brightness control
D91. C With manual control, where the operator sets the mA and kVp, and it remains at those settings, the federal performance standard limits table top exposure rate to 5 R/min for under-table x-ray tubes.
D92. D With auto-brightness control systems, the table top exposure rate can be 10 R/min. The philosophy is that in this mode it is allowable to go to higher exposure rates for the thick patients.
D93. In 131I therapy for thyroid cancer, the whole body clearance curve is commonly plotted versus time. The radiation absorbed dose to the patient is proportional to the .
Administered activity of I-131
Administered activity per unit body surface area
Administered activity per unit body weight
Peak counts in the clearance curve
Area under the clearance curve normalized to per unit body weight
D93. E The absorbed dose depends on the patient specific clearance kinetics. The same activity
administered to two different patients of the same weight could result in different absorbed doses, if they metabolized and cleared the 131I at different rates.
D94. The dose to personnel from scattered radiation during a fluoroscopic procedure depends on .
A. kVp
B. mAs
C. Distance
D. Field size and patient thickness
E. All of the above
D94. E
D95. Well-collimated AP and lateral chest radiographs are taken on a patient. She later discovers that she was pregnant at the time of the study. The expected fetal radiation dose would be about mSv.
A. 0.005
B. 0.5
C. 5
D. 50
E. 100
D95. A According to HEW (FDA) 76-8231 the embryo would receive about 0.002 to 0.004 of
the primary beam. Since AP chest radiographs usually deliver less than 30 mR at the skin surface, and the LAT is about 2.5 times greater, the dose to the fetus would be less than 0.004 x (30 x 3.5) = 0.42 mR or 0.0042 mGy, which is about 0.0042 mSv.
D96-D98. Match the following exposure conditions with the appropriate dose.
A. 1 mSv
B. 0.1 mSv
C. 2 mSv
D. 2 (xGy
E. 50 mSv
D96. The maximum organ dose for patients undergoing nuclear medicine procedures
D97. The regulatory weekly dose limit in controlled areas
D98. The annual effective dose limit for a nuclear medicine technologist
D96. E
D97. A
D98. E
D99. Regulations limit the radiation dose equivalent to patients undergoing radiological procedures to mSv/year.
A. 500
B. 50
C. 5
D. 1
E. None of the above
D99. E There are no regulations limiting the radiation dose to patients undergoing radiological procedures. Physicians are expected to weigh the risks and benefits to determine whether a procedure should be performed.
D100. Patient entrance exposure increases approximately with
A. (kVp x mAs)2
B. kVp2 x mAs
C. kVpxmA
D. kVp2 x mA
E. kVp x mAs
D100. B Patient exposure or x-ray output is approximately proportional to kVp2 and the product of the tube current and exposure time (mAs).
T1. The depth of maximum dose for a photon beam is approximately equal to:
A. The depth at which dose and kerma are equal.
B. The maximum range of the secondary electrons.
C. The depth at which electronic equilibrium occurs.
D. All of the above.
E. None of the above.
Tl. D
T2. The TMR for a 10 x 10 cm2 4 MV photon field at 100 cm SAD is 0.73 at 10 cm depth. If the
SSD were changed from 90 cm to 95 cm, the expected change in the TMR would be:
A. Increase of 5%.
B. Increase of 7.5 %.
C. Decrease of 5%.
D. Decrease of 7.5 %.
E. Zero.
T2. E Over the normal clinical range of SSD, TMR is independent of SSD, since it represents the attenuation of a given thickness of tissue. (Percent depth PDD, on the other hand, has an inverse square component, and does depend on SSD.)
T3. A Sievert (Sv) is .
A. 100 ergs per gram
B. The SI unit of dose equivalent
C. Absorbed dose/RBE
D. The SI unit of exposure
T3. B Dose equivalent, measured in Sv (formerly rem) takes into account the relative biological effectiveness (RBE) of different types of radiation. For example, 1 cGy of neutrons causes more biological damage than 1 cGy of photons, and thus has a higher value in Sv.
100 ergs/gm = 1 cGy, formerly 1 rad.
T4. All of the following are true regarding Percent Depth Dose (PDD) except:
A. The value can be adjusted for extended SSD using Mayneord's F factor.
B. It comprises an attenuation component and an inverse square component.
C. At a given depth it increases as field size increases.
D. For a given field size and depth it increases as energy increases.
E. It is defined as: (dose rate at dmax/dose rate at depth) x 100%.
T4. E The formula is inverted.
A.          T5. Collecting all the negative ions produced by a beam of photons in a small volume of air, under
conditions of electronic equilibrium, is a direct measure of:
Dose equivalent
LET
Absorbed dose
Exposure
Specific ionization
T5. D This is a measure of exposure in air. To convert to absorbed dose in a small mass of material, eg tissue, exposure is multiplied by the "f factor".
T6. The Monitor Unit (MU) setting to deliver 180 cGy at 5 cm depth to a 6 x 28 cm2 field at 100 cm SSD is .
A. 180
B. 191
C. 199
D. 207
E. 215
T6. D The equivalent square of 6 x 28, using 4A/P, is 9.9; use 10. MU = [dose/fx]/[(cGy/MU at SSD) x (PDD/100)] - 180/(1.0x0.871) = 207 MU.
T7. The maximum tissue dose in the previous question is cGy.
A. 180
B. 195
C. 207
D. 215
E. 233
T7. C Dose at dl/dose at d2 = PDD1/PDD2. Dose at dmax = 180 x (100/87.1) = 207 cGy.
T8. The exit dose at 20 cm depth in the previous question is cGy.
A. 175
B. 130
C. 116
D. 99
E. 81
T8. E Dose at d20= 180 x (39.2/87.1) = 81.0 cGy.
T9. Parallel opposed 22 x 18 cm2 whole brain fields are treated isocentrically. The SSD is 92 cm on each side. A small corner block is added, and the tray factor is 0.96. The MU setting to deliver a total of 250 cGy at midline is MU per field.
A. 138
B. 132
C. 129
D. 124
E. 119
T9. A Equivalent square of 18 x 22 = 20.
MU = [dose/fx]/[(cGy/MU at SAD) x TMR x TF] = 125/(1.089 x 0.869 x 0.96) = 138 MU.
T10. The total dose at dmax in the previous question is cGy.
A. 233
B. 245
C. 255
D. 275
E. 288
T10. C At 102% of the M/L dose, this is the only reasonable answer. The dose at dmax is always greater than the M/L dose, so A and B cannot be correct, and the other answers are too large for this energy and separation. To calculate the actual dose at this point:
Depth to M/L = 8 cm. Depth to dmax = 1.6 cm.
Depth to exit point (16 - dmax) = 14.4 cm.
Distance to M/L = 100 cm. Dist. to dmax = 93.6 cm. Dist. to exit point = 106.4 cm. For SAD: dose at dl/dose at d2 = [TMR1/TMR2] x [dist2/distl]2. Entrance dose at d^ = 125 x [(1.0/.869) x (100/93.6)2] = 164.2 cGy. Exit dose = 125 x [(.711/.869) x (100/106.4)2] = 90.3 cGy. Total dose at this point = 164.2 + 90.3 = 254.5 cGy.
The total dose at dmax for parallel opposed fields is always greater than the dose at the isocenter; the lower the energy, and the larger the thickness, the greater this difference will be.
T11. A patient's pelvis measures 28 cm AP, and is to be treated with AP/PA photon beams. 18 MV would be preferable to 6 MV because:
A. Dose in the build-up region would be greater.
B. Dose variation throughout the treated volume would be less.
C. The beam has a sharper penumbra.
D. Skin dose would be greater.
E. All of the above.
T11. B As photon energy increases, the increased dose at a depth of dmax for parallel opposed beams decreases. For a separation of 28 cm, it would typically be 115% for 6 MV, and 107% for 18 My for typical pelvic fields.
T12. The total dose delivered at depth dmax from a pair of parallel opposed fields, expressed as a per¬centage of the total dose at midplane:
A. Decreases as photon energy increases.
B. Decreases as field size increases.
C. Increases as patient thickness increases.
D. Is slightly less for an SSD setup than for an SAD setup.
E. All of the above.
T12. E Any factor that increases the PDD will decrease the total dose at dmax, compared with the total dose at midplane. Treating at SSD rather than SAD gives slightly higher PDDs.
T13. Changing parallel opposed fields from 100 cm SAD to 100 cm SSD while treating to the same prescription dose at patient midline would cause which of the following changes in MU and
MU Total dose at dmax
A. decrease increase
B. decrease decrease
C. increase increase
D. increase no change
E. increase decrease
T13. E As the prescription point moves further away from the source, the MU must be increased.
Increasing the SSD slightly increases the PDD (due to a smaller inverse square factor), making the total dose at dmax smaller.
T14. A patient is to be treated with 6 MV photons to a sarcoma of the leg, requiring an extended SSD of 130 cm. Compared with the PDD at 100 cm SSD, the PDD at 8 cm depth at 130 cm SSD will be approximately than that at 100 cm SSD.
A. 0.5% greater
B. 3% greater
C. 10% greater
D. 5% less
E. 1.5% less
T14. B PDD increases with increasing SSD because of the change in the inverse square factor.
PDD (130 cm SSD, d8) = PDD(100 cm SSD, d8) x [(100 + 8)/100 + dmax) x (130 +dmax)/130 + 8)]2
For 6 MV dmax is 1.6 cm.
T15. The side of the equivalent square of a 5 x 25 cm2 field is:
A. Approximately equal to V (5 x 25)
B. Closer to 25 than to 5
C. The square field, which has the same PDD as the rectangle
D. Approximated by the (area/perimeter) of the rectangle
T15. C The equivalent square CxC of a rectangular field has the same PDD and TMR as the
rectangle. It is smaller in area than the rectangle (i.e., CxC < 5x25 in this case), since it is the field with the same scatter contribution on the beam axis. A useful rule of thumb is that C = 4x (area / perimeter). The use of "equivalent square" enables PDD and TMR tables to be simplified to only square fields rather than tabulating the many rectangular fields in use.
T16. The difference in gantry angle position for a wedged pair of photon fields is 120°. Assuming the surface is perpendicular to the beam axes, the most appropriate wedge angle, to achieve a uniform distribution, is:
15°
30°
45°
T16. B The rule is: Wedge angle = 90 - (Hinge angle/2), where the hinge angle is the angle
between the beam axes. This wedge will cause the isodose lines from each field to run par¬allel to each other, so that the fall-off from one beam across the volume is matched by a fall-off in the opposite direction from the other beam.
A.          T17. In the diagram below, the wedge angle is


a
b
c
d
e
T17. C The wedge angle is the angle through which the isodose at 10 cm depth is rotated from its position in the open beam.
T18. In the diagram shown below, three 6 MV beams deliver equal doses at the isocenter. The dose gradient across the volume is 110% to 95%. Dose homogeneity could be improved by:
T18. B A larger wedge angle will reduce the dose at the anterior side of the volume, increase the dose at the opposite side, and thus create a more homogeneous distribution. Renormalization has no effect on dose homogeneity.
T19. Which of the following delivers the greatest scatter dose to the contralateral breast from a medial tangential breast field?
A. A universal wedge
B. A dynamic wedge
C. A physical (metal) wedge
D. All of the above are about equal
T19. C The metal wedge delivers a greater scatter dose to the contralateral breast than a universal wedge (which is smaller, and situated higher up in the head), or a dynamic wedge (which is achieved with a closing collimator).
20. During treatment of a wedged field, the wedge is accidentally omitted, but the planned MU are delivered. The wedge transmission factor is 0.75. The consequence is that the dose at the fol­lowing points will be………..times that planned.
T20. A The dose on the axis is increased by 1/0.75 = 1.33. The thick end of the wedge is increased by a greater amount, and the thin end by a smaller amount, relative to the beam axis, because of the shape of the wedge.
T21. Wedges are used for all of the following except:
A. To increase dose homogeneity for breast tangent fields.
B. To increase the anterior hot spot for a larynx treated with lateral opposed fields.
C. To increase dose homogeneity for orthogonal fields in partial-brain treatments.
D. As a missing-tissue compensator.
E. To increase dose homogeneity in a three-field plan (anterior and two opposing laterals).
T21. B Wedges are used to reduce the anterior hot spot in a larynx plan treated with lateral opposed fields. If a hot spot is required, the wedges may be omitted.
T22. If a single arc 360° rotation is performed on a pelvis, the shape of the isodose which is 90% of the dose at the isocenter is:
A. Circular.
B. Oval, wider in the AP dimension than the lateral dimension.
C. Oval, wider in the lateral dimension than the AP dimension.
D. Star shaped.
T22. B Since the depth to the isocenter is less in the AP dimension than the lateral dimension, a greater dose is delivered per degree of arc anteriorly and posteriorly, creating an oval iso¬dose. Lower isodose values may appear star-shaped on the plan if the treatment-planning system approximates continuous rotation with fixed beams every 10°.
T23. "Skin sparing" is reduced in photon beams by all of the following except:
A. Using bolus.
B. Treating through a plaster cast.
C. Using a beam spoiler.
D. Treating tissue under a skin fold.
E. Increasing photon energy.
T23. E As photon energy increases, skin dose decreases.
T24. The surface dose for a 10 x 10 cm2 field on a 6 MV linac at 100 cm SSD is about:
A. 95-100%.
B. 80-95%.
C. 5-15%
D. 2-5%.
E. None of the above.
T24. E The surface dose is about 15% to 40%, depending on field size.
T25. A 1 cm thick Lucite™ sheet is placed above a patient's surface in a photon beam to act as a beam spoiler. (Monitor unit calculations include a correction for attenuation by the beam spoiler.) Which of the following is true?
A. As the spoiler-to-surface distance increases, the surface dose decreases.
B. As the spoiler thickness increases, the surface dose decreases.
C. The spoiler has the same effect on the surface dose as bolus.
D. All of the above are true.
T25. A Spoilers are designed to increase the dose in the build-up region (to simulate a lower ener¬gy photon field) without raising the skin dose to 100%, as with bolus.
T26. Which of the following could be used as a tissue compensator?
A. Shaped bolus
B. A wedge
C. Cerrobend
D. Dynamic MLC
E. All of the above
T26. E Bolus can be used with an electron beam to achieve a desired isodose distribution at depth.
Wedges are often used as simple compensators, (e.g., thick ends up in AP/PA treatment of the thorax, in tangential breast treatment, etc.) but have the disadvantage of having uniform thickness in height. Two-dimensional "missing tissue" compensators can be fabricated using a device that mills the shape in styrofoam (based on the patient's surface contour), which is then filled with cerrobend. Dynamic MLC can be used to create any desired dose pattern; this is one technique used in IMRT.
T27. The oblique fields in the diagram are angled so that their posterior borders are aligned. LAO gantry angle = 60°. Field width = 18 cm (symmetrical) at 100 cm SAD. (Gantry angles are defined as 0° = anterior, 90° = patient left, 180° = posterior, 270° = patient right). RPO Gantry angle = .

A. 240°+ 5°
B. 240°+10°
C. 240°-5°
D. 240°- 10°
E. 240°-20°
T27. D Divergence = tan 1 (9/100) = 5° for each field. To eliminate divergence, the RPO gantry angle = 60° + 180° - (2 x divergence) = 230°.
A.          T28. Adjacent single direct spine fields are "matched" at 5 cm depth. The field lengths are 20 cm and 28 cm respectively, at 100 cm SSD. The gap to be left on the skin between the light field edges is……..cm.
1.2
1.6
2.0
2.2
2.4
T28. A Gap = [(CI + C2)/2] x d/SAD
= [(20 + 28)/2] x 5/100 - 1.2 cm.
T29. All of the following are true for a 6 MV photon beam except:
A. The attenuation is approximately 6% per cm.
B. The PDD for a 10 x 10 cm2 beam at 10 cm depth is approximately 68%
C. The TMR for a 10 x 10 cm2 beam at 10 cm depth is approximately 0.784.
D. dmax is approximately 1.6 cm
T29. A Attenuation in a 6 MV photon beam is about 3.5% per cm.
T30. Which of the following is true regarding geometric penumbra?
A. It increases as source diameter increases.
B. It decreases as SSD increases.
C. It is independent of source-collimator distance.
D. It decreases as depth increases.
T30. A This is why a 60Co unit has a wider penumbra than a linac.
T31-T35. Match the depth dose curves A-E in the diagram shown with the type of beam.
T31. 20 MeV electrons
T32. 3 mm A1 HVL x-rays
T33. 60Co photons
T34. 18 MV photons
T35. 10 MV photons
T3I. D
T32. E
T33. C
T34. A
T35. B
T36. Flattening filters in megavoltage photon beams are designed to achieve a flat beam at 10 cm depth. This means that at d^, the profile of a 30 x 30 cm2 6 MV beam is
A. Higher at the edges than the center
B. Lower at the edges than the center
C. The same as at 10 cm depth
T36. A The flattening filter "overflattens" at shallow depths, and "underflattens" at greater depths.
The dose at d,^ is therefore greater towards the edges of the beam, and this effect increas¬es with field size.
T37. During routine quality assurance checks, the optical distance indicator (ODI) is found to
read 100 cm when the distance is actually 101.5 cm. If uncorrected, the effect on treatments set up at 100 cm (as indicated by the erroneous ODI) would be:
A. An overdose of 3%
B. An overdose of 1.5%
C. An overdose of 1 %
D. An underdose ofl.5%
E. An underdose of 3%
T37. E The patient would be further away from the isocenter, and the dose would be lower by the inverse square factor of (100/101.5)2.
T38. A TBI patient is treated at 400 cm SSD. If the linac is calibrated to deliver 1.0 cGy/MU at dmax, 100 cm SSD, approximately how many MU are required to deliver 100 cGy from one field at dmax to the TBI patient?
4000
B. 1600
C. 800
D. 400
E. 200
T38. B Using the inverse square law, the MU will be: 100 x (400/100)2 = 1600 MU.
T39. The dose distribution shown for breast tangents could be improved by .

A. Using a larger wedge angle
B. Inverting the direction of the wedges
C. Using a smaller wedge angle
D. Using a lower photon energy
E. Both A and D
T39. C The plan is over-wedged.
T40. A superficial x-ray treatment is prescribed with a 2 mm A1 filter. By mistake, a 1 mm filter is used. All of the following will occur except:
A. Increased dose rate.
B. Increased dose for the same timer setting.
C. Decreased HVL.
D. Increased PDD.
T40. D The thinner filter will increase dose rate, but will not harden the beam as much as the 2 mm filter, so the HVL and PDD will both decrease.
T41. For a superficial x-ray unit, if there is no measured data, all of the following factors are neces¬sary to select the correct PDD table from published data, except:
A. Field size.
B. kVp.
C. SSD.
D. HVL.
T41. B The HVL (in A1 or Cu) defines the penetrability of a low-energy x-ray beam. Different
combinations of kVp and filtration can produce beams with the same HVL. The SSD also affects the PDD and is important for machines that treat at short SSDs.
T42. Which of the following is not true for CT images of the torso used directly for computerized treatment planning?
A. The patient must be scanned in the treatment position.
B. A flat top is required for the CT table.
C. The CT image is a gray scale representation of the linear attenuation coefficient of each pixel.
D. CT numbers must be converted into electron densities before pixel-by-pixel inhomo- geneity corrections can be made.
E. Triangulation points or surface marks are unnecessary since the isocenter can be related to internal organs.
T42. E Although the isocenter of a plan can be related to internal structures, it must also be related to surface landmarks (preferably triangulation points) in order to position the patient cor¬rectly for treatment.
A.          T43. Regarding inhomogeneity corrections, which of the following are true?
1. At high energy, build-up and build-down effects at tissue-lung interfaces could under¬dose a tumor adjacent to normal lung.
2. 10 cm of lung in a 6 MV beam will increase the dose beyond the lung by about 24%.
3. The density of lung is about 1/4 to 1/3 that of soft tissue.
4. Attenuation corrections for transmission are greater for 15 MV photons than for 6 MV photons.
1,2,3
1,3
2,4
4 only
All of the above.
T43. A 10 cm of lung is approximately equivalent to 3 cm tissue, and 10 - 3 = 7 cm of "missing" tissue. Thus the dose beyond the lung is increased by approximately 24% (ignoring scatter effects).
Transmission corrections are greater at lower energy, where the attenuation per cm of tissue is greater.
T44. In irregular field calculations the dose can be divided into TMR^ and SMR. The part represent¬ed by TMRo is:
A. The scatter component of the dose on the central beam axis.
B. The primary component of the dose on the central beam axis.
C. The backscatter factor for the blocked field.
D. The TMR for the equivalent square of the blocked field.
E. The tissue-maximum ratio for the open, unblocked field.
T44. B TMRq is the primary component and SMR is the scatter component for the blocked field. TMRQ is found by extrapolating TMR(R) to zero field radius.
T45. For megavoltage beams the scatter component of dose at a point in tissue, as a percent of the total dose at that point, increases with all of the following except:
A. Depth
B. Field size
C. Photon energy
D. Dose rate
T45. C In the megavoltage range, the lower the energy, the greater the contribution of scatter. For this reason, irregular field calculations are more important for lower energy beams, since blocking reduces scatter dose from under the blocked area.
T46. A field with a collimator setting of 10 x 10 cm2 is set up at 100 cm SSD. The field projects to cm at 7 cm depth.
A. 9.3x9.3
B. 10x10
C. 10.7x10.7
D. 11.5x11.5
E. 17x17
T46. C By similar triangle geometry: wl/w2 = dist 1/dist 2. W(d7)/10 = 107/100.
T47. The collimator setting required to treat a 40 cm field at 130 cm on a linac (for which the SAD is 100 cm) is cm.
A. 23.7
B. 25.2
C. 26.7
D. 30.8
E. 37.0
T47. D By similar triangle geometry: coll/40 = 100/130.
T48. A patient is simulated at 100 cm SAD with 15 x 15 cm AP/PA fields. The AP thickness is 22 cm, and the isocenter is at midplane depth. If the treatment is now changed to 100 cm SSD, the appropriate collimator setting is cm.
A. 12.2 x 12.2
B. 13.5x13.5
C. 15x15
D. 16.7 x 16.7
E. 22.6x22.6
T48. B By similar triangle geometry:
coll/15 = 100/111. The field at 100 cm SSD must still project to 15 cm at d = 11 cm, i.e., at 111 cm.
T49. It is recommended that the dose to a pacemaker be kept below 2.0 Gy. In a lung treatment of 40 Gy with 10 MV photons, the fields should be no closer than to the pacemaker.
A. 5 mm
B. 2 cm
C. 7 cm
D. 10 cm
T49. B 2 Gy out of 40 Gy is 5%. This occurs at about 2 cm from the field edge.
T50. The main cause of dose inhomogeneity in TBI treatment is
A. Inverse square effects
B. Use of compensators
C. Use of high energy photons
D. Use of a beam spoiler
E. Differences in patient thickness
T50. E Inhomogeneity is mainly due to differences in thickness (e.g., forearms vs. trunk). B, C, and D tend to improve dose homogeneity.
T51. Advantages of multileaf collimators (MLCs), compared to conventional cerrobend blocks include:
A. Sharper penumbra.
B. Lower leakage radiation.
C. Can accommodate larger field sizes.
D. Permits Intensity Modulated Radiation Therapy (IMRT).
E. All of the above.
T5 I. D Multileaf collimators (MLC) have leaf widths on the order of 3 to 10 mm, meaning that field edges have checkerboard shapes. Cerrobend blocks can be cut to precisely match the desired shape of the treatment field. MLC field sizes may be limited, whereas cerrobend can be used with the largest field available. There is some radiation leakage between leaves, and for most MLCs the ends of the leaves do not exactly match beam divergence. In fact, other than for IMRT, MLCs have no dosimetric advantages over cerrobend blocks. Their main advantage is ease of use.
T52. The tongue-and-groove effect is related to which of the following:
A. An increase in dose between two adjacent leaves of a multileaf collimator.
B. A most pronounced field-size effect on the output factor.
C. An decrease in the overall radiation fluence by about 1%.
D. It may be absent in some multileaf design.
E. None of the above.
T52. E The "tongue-and-groove" effect may decrease (not increase) interleaf dose, is present in all current MLC models, and has no effect on output.
A.          T53. According to the dose-volume histogram shown, all of the following are true except:

The minimum dose to this volume is approximately 30%.
50% of the volume receives at least 75% of the dose.
75% of the volume receives at least 50% of the dose.
25% of the volume receives at least 75% of the dose.
Part of the volume receives 100% of the dose.
T53. B The graph shown is a cumulative dose volume histogram (as opposed to a differential dose volume histogram). It depicts the fraction of the volume (y-axis) that receives more than the dose values given on the x-axis. For example, 100% of the volume receives at least 30% of the prescribed dose; 75% of the volume receives less than 50% of the dose; 25% of the volume receives at least 75% of the dose.
T54. Which of the following is (are) not included in the CTV (clinical target volume) as defined by ICRU Reports 50 and 62:
A. Gross target volume (GTV).
B. Internal margin (IM).
C. Set-up margin (SM).
D. Lymphatic spread.
T54. C Setup margin is included in creation of the PTV from the CTV
T55. The fundamental difference between intensity-modulated radiation therapy (IMRT) and 3-D conformal radiation therapy (3-DCRT) is:
A. In IMRT, the intensity distribution of each beam is optimized and is usually non-uniform. In 3-DCRT, the intensity distribution is either uniform (as in open fields) or linear (as in wedged fields).
B. The number of beams used in a plan is usually more in IMRT than that in 3-DCRT.
C. The beam directions are non-coplanar for IMRT and coplanar for 3-DCRT.
D. The delineation of targets and critical organs are different between IMRT and 3-DCRT.
T55. A Intensity distribution is the only fundamental difference between IMRT and 3-DCRT. The number and direction of beams are not affected by the planning method, nor is the delin¬eation of targets or critical organs.
T56. Intensity-modulated radiation therapy (IMRT) can be delivered using either "Step and Shoot" (segmental IMRT) or "Sliding Window" (dynamic IMRT) techniques. The major difference between these two methods is:
A. "Step and Shoot" does not require a multileaf collimator (MLC).
B. "Sliding Window" requires fewer monitor units.
C. Only "Step and Shoot" requires inverse treatment planning.
D. Only "Sliding Window" can deliver continuously variable dose intensities.
E. "Step and Shoot" produces more neutron contamination.
T56. D " Step and Shoot" IMRT is delivered using a fixed number of MLC shaped sub-fields,
each with a fixed number of delivered monitor units, or dose. After each sub-field is treat¬ed, the beam is turned off, the MLC field shape reset, the beam turned on again, and so on. Because there are a fixed number of sub-fields, the number of intensity levels possible must be finite. Both methods require more monitor units than conventional treatment, with "Sliding Window" usually requiring slightly more MU than "Step and Shoot."
T57. Inverse treatment planning usually does not incorporate:
A. Specification of dose volume constraints.
B. Beam's eye view (BEV) computer display for design of field sizes.
C. Intensity modulated beam delivery.
D. Monte Carlo dose calculations.
E. CT simulation.
T57. D Monte Carlo dose calculations are still too time consuming to be used for most individual patient treatment planning, although they are used as a basis of comparison to determine the accuracy of analytical type dose calculations which are still used in virtually all treat¬ment planning systems including IMRT.
T58. The usual settings for a portal film are 2 MU (treatment field) plus 4 MU (full field), with a
source-film distance of 130 cm. For treatment at extended SSD, where the source-film distance is now 160 cm, the appropriate settings would be MU plus MU.
A. 2, 4
B. 3, 6
C. 4, 8
D. 6, 8
E. 8, 16
T58. B The MU need to be increased by the inverse square law: (160/130)2 = 1.5.
T59. Compared with a conventional simulator film, a digitally reconstructed radiograph (DRR) of a thorax from CTs with 5 mm slice separation will have resolution.
A. worse
B. better
C. about the same
T59. A The DRR has noticeably worse resolution. A smaller slice separation will improve resolu¬tion, but the data file will be larger. However, DRRs have the advantage in that the data can be manipulated to emphasize different features such as bone, lung, etc., which may yield a more useful image for treatment planning in spite of poorer resolution.
T60-T62. Match the imaging modality with its approximate spatial resolution in the axial plane (choices may be used more than once):
A. 0.5-1.0 mm
B. 1-2 mm
C. 3-4 mm
D. 5-6 mm
T60. CT
T61. MRI
T62. PET
T60. A See remarks following answer T62.
T61. A See remarks following answer T62.
T62. C Both CT and MRI have pixel sizes less than 1 mm. PET scan resolution is limited to a few millimeters mainly because of the finite range of the positrons, and also because the 511 keV annihilation gammas are not exactly anti-parallel.
T63. Advantages of EPID images (Electronic Portal Imaging Devices), compared to conventional portal films include all of the following except:
A. Improved spatial resolution.
B. Improved latitude.
C. Improved contrast.
D. Real-time image display.
E. Digital image storage.
T63. A Most EPIDs have on the order of 256-512 rows of pixels, and 256-512 pixels per row, giving them a spatial resolution no better than about 1 mm, which is vastly inferior to radiographic film.
T64-T67. Match the electron beam energy with the feature described below:
A. 6 MeV
B. 9 MeV
C. 12 MeV
D. 16 MeV
E. 20 MeV
T64. Has the highest surface dose
T65. Has a range of 6 cm in tissue
T66. Has a 90% depth dose at 2.7 cm depth
T67. Has the sharpest fall-off between 80% and 20%
T64. E
T65. C Electrons lose 2 MeV per cm in tissue.
T66. B Rule of thumb: 90% depth dose = Energy(MeV)/3.
T67. A
T68. The dose measured beyond the practical range of a therapy electron beam is due to:
Very low energy electrons.
The highest energy electrons in the spectrum.
Characteristic x-rays generated in tissue.
Bremsstrahlung.
T68. D Electron interactions with high Z materials in the head of the linac generate bremsstrahlung x-rays.
T69. When a custom electron insert has dimensions smaller than the range of the electrons, all of the following are likely to occur except:
A. The output (cGy/MU) will be reduced.
B. The surface dose will decrease.
C. The PDD will decrease beyond dmax.
D. Dmax will shift to a shallower depth.
T69. B Surface dose will generally increase, due to scatter from the insert.
T70. The thickness of lead required to shield a 6 MeV electron beam is approximately mm.
A. 0.5
B. 3
C. 6
D. 12
E. 24
T70. B 6 MeV electrons are stopped in 3 cm of water, or approximately 3 mm of lead.
T71. In Total Skin Electron Beam Therapy (TSET), multiple oblique beams do all of the following except:
A. Improve the uniformity of the skin dose.
B. Decrease the effective depth of dmax.
C. Increase dose uniformity in the target volume.
D. Decrease the whole body x-ray dose.
T71. D Oblique incidence tends to increase skin dose and decrease depth dose. In general, the more oblique fields used, the higher the x-ray dose to the patient's whole body but the greater the dose uniformity.
T72. An 192Ir HDR unit has an activity of 8.6 Ci on May 2. The treatment time for a vaginal cylinder is 300 seconds. If the treatment time for a similar treatment is not to exceed 10 minutes, the source must be changed on:
A. July 15
B. June 23
C. July 1
D. August 5
T72. A 10 mins = 600 sec. The treatment time will double when the activity has dropped to one half, i.e., after one half-life, or 74 days.
T73. A patient has a temporary implant of the thigh containing 110 mCi of 192Ir seeds. Assuming
there is no lead bed shield, a physician can spend minutes at the bedside (30 cm from
the implant) without exceeding the recommended weekly MPD for a radiation worker. (Exposure rate const for l92Ir = 4.6 R cm2 mCr'-hr"1)
A. 5
B. 10
C. 30
D. 60
E. 90
T73. B The MPD is 5000 mrem/yr or 100 mrem/wk. For protection purposes this is taken to be equal to 100 mR/wk.
The exposure rate is: Activity x Exposure rate constant x 1/d2.
= 110 mCi x 4.6 R.cm2/mCi.hr x l/(30)2 = 0.562 R/hr. or 562/60 = 9.4 mR/min. 100/9.4 = 10.6 minutes. In practice, portable lead shields at the bedside would decrease the whole body dose to personnel.
T74. The half-life of a permanent prostate seed implant is 60 days. 90% of the total dose will have been delivered in about
A. 90 days '
B. 6.5 months
C. 1 year
D. 2 years
E. 5 years
T74. B 90% of the dose will have been delivered when the initial dose rate drops to 10% of its
value, i.e., when It/I0 =10/100 = exp-(0.693 t/60). Thus t = 199 days, or 6.5 months. (For a more intuitive answer: 50% of the dose is delivered after 1 half-life, 75% after 2 half-lives, 87.5% after 3, and 93.75% after 4 half-lives.)
T75-T76. A planar implant is planned using a uniform distribution of sources. The implant consists of a 4 x 6 cm2 plane of sources, with all sources 1 cm apart, and all sources having the same air kerma strength. Consider the possibility of using either ,25I or 192Ir seeds for the implant.

T75. For the same dose rate, the required total activity of l25I will be that needed if
192Ir were to be used.
A. Greater than
B. Less than
C. The same as
D. Cannot tell from the information given
T76. Which of the following is true?
A. Compared to the dose rate in the center of the implant, the dose rate 1 m away from the patient will be higher for 125I than it would be if 192Ir were to be used.
B. The Paterson-Parker tables for planar implants could be used to accurately calculate the dose if 125I were to be used.
C. The Paterson-Parker tables for planar implants could be used to accurately calculate the dose if 192Ir were to be used.
D. Compared to the dose rate in the center of the implant, the dose rate at the edges of the implant will be higher for 125I than it would be if l92Ir were to be used.
T75. A A greater activity of 125I is required because it has a lower exposure rate constant than l92Ir.

T76. D Paterson-Parker tables can only be used for isotopes with energies comparable to 226Ra,
such as 192Ir, and 137Cs; not low-energy isotopes such as 125I. Paterson-Parker also requires a non-uniform distribution of sources to achieve a uniform dose distribution. With uniform source distribution the center of the implant will always be hotter than the periphery, with the degree of non-uniformity increasing with increasing isotope energy (less attenuation of high energy photons).
T77. NRC regulations require that brachytherapy calculations are checked for accuracy before:
A. Any sources are applied to the patient.
B. 10% of the prescribed dose is delivered.
C. 50% of the prescribed dose is delivered.
D. The sources are removed.
T77. C Non-Agreement states follow NRC regulations, and Agreement states, which make their own regulations, must have rules which are at least as stringent.
A.          T78. Radioactive packages for transportation are labeled with White I, Yellow II, or Yellow III. The yellow labels also have a Transport Index (TI) that indicates the dose rate:
On the surface.
At 1 cm from the surface.
At 1 m from the surface.
At 1 m from the center
T78. C
T79. Regarding an 125I prostate seed implant, which of the following is true?
A. All doctors and nurses must wear film badges.
B. Lead shields are required at the patient's bedside.
C. The patient must be in a private room.
D. A dropped seed should be picked up with a pair of tweezers, not with the fingers.
T79. D The photon energy is about 35 keV. The patient's body provides substantial attenuation, and the exposure rate at 1 m from the patient is around background. However, by the ALARA principle, seeds should always be handled with tweezers to reduce exposure.
T80. Unused 125I seeds must be stored for a minimum of before being discarded.
A. 10 months
B. 5 years
C. 10 half-lives
D. 20 half-lives
E. They can never be discarded by the user, but must be returned to the manufacturer.
T80. C After a minimum of 10 half-lives (20 months for 125I) the seeds must be surveyed with an appropriate instrument to ensure that the dose rate is not above background. They can then be discarded in regular trash, provided documentation is kept.
T81. The following is a table of dose rates (cGy/hr) at various calculation points for a tandem and ovoid gynecological applicator loaded with 3 sources in the tandem, and one in each ovoid.

The prescription is 2000 cGy at the average of points A right and left (Aavg). In order to decrease the relative dose to the rectum, the physician decides to change the activity of source #3 to 6.7 mCi. This would have the effect of:

A. Decreasing the total prescribed dose at point Aavg.

B. Decreasing the rectal dose by 6%.

C. Increasing the point Aavg dose by 5%, and decreasing the rectal dose by 10%.

D. Decreasing the bladder and rectal doses by 15%.
T81. B The total dose to point A average is fixed at 2000 cGy—only the relative doses to
other points, and the time taken to deliver 2000 cGy will change. The initial plan delivers 41/61 = 67.2% to the rectum. After the change, the dose rates at A are 56 (R) and 58 (L). The rectal dose rate is 36 cGy/hr. Thus, the new plan delivers 36/57 = 63.2% to the rectum, or 6% less.
T82. Calculate the approximate total treatment time (in minutes) to deliver an HDR dose of 500cGy using a 3 cm diameter vaginal cylinder applicator, given the following informations:


Source activity = 10 Ci 192Ir Treatment distance = 0.5 cm depth
in tissue Total treatment length = 6.0 cm Exposure rate constants
= 4.61 R-cm2/mCi-hr for 192Ir = 8.25 R-cm2/mCi-hr for 226Ra
A. 2.69
B. 4.63
C. 4.82
D. 5.02
E. 22.22
T82. C (1) 10 Ci 192Ir expressed in mgRaEq = 10,000 mCi.
10,000 mCi = 10,000 x 4.61/8.25 = 5588 mgRaEq.
(2) Treatment distance = Cylinder diam/2 + 0.5 cm = 2.0 cm.
(3) Per Patterson-Parker tables: 898 mg-hr per 10 Gy at 2 cm = 449 mg-hr per 5 Gy = (449/5588 hr) x (60 min/hr) = 4.82 min.
T83. The range of 2 MeV betas in air is about (the density of air is 0.00129 gm/cm3)
50 cm
B. 1 m
C. 10 m
D. 20 m
E. 50 m
T83. C Betas have a range of 0.5 cm per MeV in water, so 2 Mev betas would travel 1 cm.
Air has a density of approx. 1/1000 that of water, so 2 MeV betas will travel about 10 m. 90Sr decays to 90Y, which emits 2.2 MeV betas.
T84. In Intravascular Brachytherapy (IVBT) with beta-emitting sources, the maximum total expo¬sure to a physician standing adjacent to the patient occurs during the treatment "dwell" time:
A. When the source is in transit in the femoral artery.
B. When the source is in transit in the aorta.
C. When the source is in transit in "in air."
D. When the source is in the transport/delivery "home" position prior to use.
T84. D IVBT beta sources (90Sr and 32P) have a very short range and are completely stopped by
the source holder (prior to insertion), and by the patient (after insertion). The only exposure is due to a small amount of bremsstrahlung radiation from the holder or the patient. During transit, however, the operator is exposed directly to the unshielded betas.
T85. The principal benefit of gamma versus beta IVBT is:
A. Longer source half-life.
B. Superior dose homogeneity.
C. Lower shielding requirements.
D. Less chance of release of radioactivity.
E. Shorter treatment times.
T85. B For photons, the dose falls off more slowly between the inner and outer arterial walls, than for betas.
T86. Components of a linear accelerator treatment head that are used in the x-ray mode only, and not in the electron mode are:
A. Scattering foils and target.
B. Target and flattening filter.
C. Primary collimators and monitor chamber.
D. Magnetron and bending magnet.
T86. B In the x-ray mode the electron beam strikes the target to produce x-rays. This narrow, for¬ward peaked beam is broadened and flattened by the flattening filter. Scattering foils are used only in the electron beam mode, and the other components are used for both.
T87. A flattening filter:
A. Compensates for patient irregularities.
B. Is designed to decrease depth dose.
C. Is of uniform thickness across the beam.
D. Is used in the electron mode only.
E. None of the above.
T87. E The flattening filter is used only in the photon mode, is thicker in the center than at the edges, and is designed to create a flat photon beam (±3%) at a depth of 10 cm.
T88. Regarding neutrons produced by therapy linacs, all of the following are true except:
They are slowed down by materials containing hydrogen.
They are more effectively shielded with concrete than an equal mass of lead.
More neutrons are generated in the photon mode than the electron mode.
They can be measured with the same survey meter used to measure photon leakage.
T88. D Neutrons are indirectly ionizing. They are therefore measured by detecting the products of their interactions, such as protons or alphas, or by foil activation.
T89. On linear accelerators, the light field coincides with the isodose line at dmax.
A. 95%
B. 90%
C. 80%
D. 75%
E. 50%
T89. E
T90. If the temperature is 22°C and the pressure is 770 mm mercury, the correction to an ionization chamber reading would be .
A. 1.013
B. 1.0
C. 0.994
D. 0.987
E. 0.974
T90. D The temperature/pressure correction is used to correct the ion chamber reading to that which would be obtained at Standard Temperature and Pressure (STP), i.e., 22°C and 760 mm Hg. The correction is [(273 + C)/295] x [760/P]. In this case, 760/770 = 0.987.
T91. Which of the following is true regarding neutron shielding for linacs?
A. The neutron dose is higher in an electron beam than a photon beam of the same energy.
B. When thermal neutrons are captured, "capture gammas" of about 1 MeV are emitted.
C. Lead is an efficient neutron moderator.
D. A 6 MV photon beam can generate neutrons.
T91. B Lead is inefficient as a moderator, but it is placed in the door after the borated polyethylene to attenuate the capture gammas. The neutron dose is about 10 times higher for photons than for electrons. The threshold for neutron production is 8 MeV
T92. How many HVLs of shielding material are required to reduce 250 mR/hr to 2 mR/hr?
A. 11
B. 9
C. 7
D. 5
E. 3
T92. C 250 x(l/27)= 1.95.
T93. When shielding a linear accelerator room to be used for IMRT, compared to conventional RT, all of the following are true except:
A. The primary shielding increases by a factor of about 3, depending on the number of IMRT cases.
B. Secondary wall thickness will increase.
C. Beam-on time increases for the same dose per fraction delivered to the patient.
D. Workload will increase.
E. The patient's whole body dose increases with IMRT.
T93. A MU, workload, and leakage (hence whole body dose) increase with IMRT. However, the PTV receives the same dose per fraction as conventional RT, so the primary beam exiting the patient, and requiring a primary barrier is the same.
T94. Shielding around a linear accelerator must be designed to reduce the exposure to radiation workers to a maximum of per week.
A. 1.0 mSv
B. 0.1 mSv
C. 0.05 mSv
D. 0.001 mSv
T94. A The maximum permissible dose is 5000 mrem/yr (50 mSv), or 100 mrem/wk (1 mSv). In practice, however, 0.1 mSv/week is generally used to comply with the ALARA principle.
95. The MPD for the hands of a radiation worker, according to NCRP Report 116 is mSv
per year.
A. 1
B. 5
C. 10
D. 50
E. 100
T95. D
T96. Megavoltage photons can be detected by all of the following except:
A. An ion chamber.
B. A Geiger counter.
C. A diode.
D. A Thermoluminescent Dosimeter (TLD).
E. A flow meter.
T96. E The flow meter is not a radiation detection device. It is used to measure the flow of air in a fume hood or flow of air in a stack monitoring system in a Cyclotron Facility.
T97. The door of a room containing sealed radioactive sources:
A. Must be posted with a yellow and magenta "Caution Radioactive Materials" sign.
B. Need not be posted unless there are more than 10 mCi present.
C. Must be posted with the exposure rate at 1 m from the sources.
D. Must be surveyed weekly to ensure that the sources are correctly shielded.
E. A and D.
T97. A Rooms containing sealed radioactive sources, such as 137Cs sources, 125I seeds, etc., must be surveyed quarterly. Records must be kept showing inventory and use of the sources.