ROphex Old Q1-1000 Flashcards

Question 1

Q

Q2. The f-factor is all of the following except:

A. The roentgen-to-rad conversion factor.

B. Generally greater for high-Z materials.

C. Generally greater for low energy.

D. Has the value 0.871 in air.

E. Is 1.0 for water at 1 Mev.

Answer

A

E. Is 1.0 for water at 1 Mev.

The f-factor, also known as the “roentgen-to-rad” conversion factor is 0.871 in air, under conditions for electronic equilibrium.

For media other than air, it varies with photon energy and effective Z. For water, at 1 MeV, f=0.970.

Question 2

Q

Q6. In order to convert exposure (R) to absorbed dose (mGy), the factor for diagnostic x-rays and muscle tissue by which exposure is multiplied is closest to:

A. 0.1

B. 5

C. 9

D. 20

E. 90

Answer

A

C. 9

The f factor (roentgen-to-rad conversion factor) for muscle tissue ranges from 0.921 at 10 keV to 0.960 at 150 keV.

1 rad is equal to 10 mGy.

Question 3

Q

Q9. An exposure of 100 R:

A. Results in the same absorbed dose to muscle, bone or fat.

B. Is a measure of the ability of a photon beam to ionize air.

C. Applies to photons and particulate radiation.

D. Is a unit used in the SI system.

Answer

A

Answer: B. Is a measure of the ability of a photon beam to ionize air.

Other choices explanation:

A. Results in the same absorbed dose to muscle, bone or fat.

–The absorbed dose resulting from an exposure of 100R depends on the photon energy, the effect Z of the absorbing material, which is higher for bone than muscle.

C. Applies to photons and particulate radiation.

–Exposure, the ability of photons to ionize a mass of air, is only defined for photons below 3 MeV, and not for particulate radiation.

D. Is a unit used in the SI system.

–The SI unit is C/kg.

Question 4

Q

Q11. Match the following units with the quantities below:

Absorbed dose •Activity •Exposure •Dose equivalent

A. Bq

B. Sv

C. C/kg

D. Gy

E. J

Answer

A

Answer:

Absorbed dose (D)

–1 Gy= 1 J/kg

Activity (A)

–1 Bq = 1 Dis/sec

Exposure (C)

–1 R = 2.58 x 10-4 C/kg in air

Dose equivalent (B)

–1 Sv = 1 Gy x Q.

–Q depends on energy/RBE.

Question 5

Q

Q13. Which of the following is not equal to one Gray?

A. 1.0 Joule/kg.

B. 100 rads.

C. 1.0 Sv/QF.

D. (100 Roentgen) (f-factor).

E. 100 ergs/ gm.

Answer

A

E. 100 ergs/ gm.

–1 rad = 100 ergs/gm

–1 Gy = 100 rad

Question 6

Q

Q15. Match the unit with the quantity it measures. (Answers may be used more than once or not at all.)

Electron volt - Hertz - Joule - Gray

A. Frequency.

B. Wavelength.

C. Power.

D. Absorbed dose.

E. Energy.

Answer

A

Electron volt = Energy (E)

Hertz = Frequency (A)

Joule = Energy (E)

Gray = Absorbed dose (D)

Question 7

Q

Q17. Match the unit with the quantity it measures. (Answers may be used more than once or not at all.)

Electron volt, Hertz, Joule, Watt

A. Frequency.

B. Wavelength.

C. Power.

D. Absorbed dose.

E. Energy.

Answer

A

Electron volt = Energy (E)

Hertz = Frequency (A)

Joule = Energy (E)

Watt = Power (C)

Question 8

Q

Q19. 5 rem is equivalent to ___ mSv

A. 0.05

B. 0.5

C. 5

D. 50

E. 500

Answer

A

Answer: D

5 rem = 5 cSv = 50 mSv

Remember 1 rem = 1 rad, 1 Gy = 1 Sv, 1 Gy = 100 rads

Question 9

Q

Q21. Which of the following is not equal to a centigray?

A. 100 erg/gram

B. 1 rad

C. 1/0.873 roentgen absorbed in air

D. 1/0.873 roentgen aborbed in tissue

Answer

A

D. 1/0.873 roentgen aborbed in tissue

The F-factor for tissue is about 0.96

Question 10

Q

Q23. 1 mSv is equivalent to:

A. 1 mrem

B. 10 mrad

C. 100 mroentgen

D. 10 mCi

E. 100 mrem

Answer

A

E. 100 mrem

Question 11

Q

Q25. Matching the following with the answer choices:

Rad, Rem, Gray, Sievert

A. 100 gray

B. 0.01 gray

C. 100 rad

D. rad x Q

E. gray x Q

Answer

A

Rad = 0.01 gray (B)

Rem = rad x Q (D)

Gray = 100 rad (C)

Sievert = gray x Q (E)

Question 12

Q

Q27. The quality factor (Q) in radiation protection is most closely related to:

A. roentgen to cGy conversion factor

B. half-value layer

C. electron equilibrium

D. mass attenuation coefficient

E. relative biological effectiveness

Answer

A

E. relative biological effectiveness

The Quality factor (Q) is an approximation for RBE

Question 13

Q

Q29. Match the quality factor (Q) used in radiation protection with the type of radiation:

1.25 MeV gammas, 100 keV x-rays, Fast neutrons, ⁹⁹Mo betas

A. 10

B. 2

C. 1

D. 0.693

E. 20

Answer

A

1.25 MeV gammas = 1 (C)

100 keV x-rays = 1 (C)

Fast neutrons = 20 (E)

⁹⁹Mo betas = 1 (C)

All x-rays and gamma rays used to diagnostic/therapy purposeds have Q of 1. For neutrons the RBE varies with energy, but for protection a “worst case” value of 20 is used.

Question 14

Q

Q31. The difference between exposure and dose is the difference between:

A. the rad and the gray

B. absorption of ionizing radiation and biologic effect

C. photons and charged particles

D. ionization in air and absorption in a medium

E. ionizing and non-ionizing radiation

Answer

A

D. ionization in air and absorption in a medium

Exposure (X) is defined as ionization in air per unit mass produced by x- or gamma-rays. Exposure is limited to ionizing photons. Ioniziation producted by high energy particles is excluded.
Dose (D) is defined as the energy absorbed per unit mass by ionizing radiation. Dose is limited only to ionizing radiation. It may be produced by photons, energetic charged or uncharged particles.

Question 15

Q

Q33. The energy absorbed by a mass of air from x or y rays per roentgen is:

A. greatest in the photoelectric region

B. lowest in the Compton region

C. greatest in the pair production region

D. dependent on the temperature and pressure

E. the same, regardless of the x-ray or y-ray energy

Answer

A

E. the same, regardless of the x-ray or y-ray energy

The energy absorbed by a mass of air from x- or gamma- rays be roentgen is constant, 0.87 cGy/R. It can be calculated by multiplying the number of ion-pairs produced per R per kg by the average energy necessary to produced one ion pair (w) and converting the product from eV to joules.

W = 33 eV/ion pair

Question 16

Q

Q35. Dose equivalent is greater than absorbed dose for:

A. x-rays above 10 MeV

B. superficial x-rays

C. electrons

D. neutrons

E. all charged particles

Answer

A

D. neutrons (Q=20)

Question 17

Q

Q37. Which of the following is not true?

A. 100 MHx = 10^8 cycles/sec

B. 1 curie = 3.7 x 10^10 Bq

C. The speed of light is 3 x 10^8 m/s

D. c = wavelength/ frequency

Answer

A

D. c = wavelength x frequency

Question 18

Q

Q39. A given exposure:

A. always results in the same absorbed dose to muscle or bone

B. is a measure of the ability of a photon beam to ionize air

C. is a measure of the ability of a particle beam to ionize air

D. can be measured in roentgen in the SI system

E. all of the above

Answer

A

B. is a measure of the ability of a photon beam to ionize air

The SI unit is C/kg.

At diagnostic energies, the absorbed dose to bone is greater than that to muscle for the same exposure, because of the predominance of the photoelectric effect.

Question 19

Q

Q41. The unit of exposure was originally defined as:

A. 1 R = 1 esu of charge per 0.001293 cm^3 of air at STP

B. 1 R = 1 esu of charge per m^3 of air at STP

C. 1 R = 1 C of charge per cm^3 of air at STP

D. 1 R = 1 C of charge per 1.293 kg of air

E. 1 R = 2.58 x 10^-4 of charge per kg of air

Answer

A

A. 1 R = 1 esu of charge per 0.001293 cm^3 of air at STP

Question 20

Q

Q43. Which of the following is true? For x or gamma radiation at standard temperature and pressure, a roentgen is equal to:

A. one electrostatic unit (esu) of charge per cc of air

B. 2.58 x 10^-4 coulombs/ kg in iar

C. an absorbed dose of 0.873 cGy in air

D. A, B, C

E. B, C only

Answer

A

Answer: D

A) the original def of roentgen

B) The SI unit.

C) The f-factor

Question 21

Q

Q45. One patient recieved an exposure of 4 roentgens to an area 10 x 10 cm while a second recieved 1 roentgen to an area 20 x 20 cm. The absorbed dose to the second patient would be:

A. less

B. the same

C. more

Answer

A

Answer: less

While there can be increased dose from scatter with a larger field, there is such a neglible increase in comparison to the increase in exposure that you can essentially ignore it.

Question 22

Q

Q47. Match answer choice with the below options (multiple answers may be present or none of the choices may apply)

Gray, 2. Bq, 3. Rem
Rad, 5. Sv, 6. Curie, 7. Roentgen

A. Dose equivalent

B. Exposure

C. Absorbed dose

D. Activity

E. Energy

Answer

A

A. Dose equivalent - Sv, Rem

B. Exposure - Roentgen

C. Absorbed dose - Gray, Rad

D. Activity - Curie, Bq

E. Energy - None

Question 23

Q

Q54. Exposure is:

A. The amount of energy in joules/kg transferred from a photon beam to a medium

B. Only defined for charged particles below 3 MeV

C. The charge liberated by photons in a given mass of air

D. The absorbed dose multiplied by the quality factor

E. None of the above

Answer

A

C. The charge liberated by photons in a given mass of air

Question 24

Q

Q58. Exposure is:

A. The energy absorbed in a given mass of a medium

B. The air kerma of a photon beam

C. Measured in Sv

D. The ionization in a given mass of air

Answer

A

D. The ionization in a given mass of air

Question 25

Q

Q60. The kerma is the:

A. energy per unit mass absorbed or retained along the path of a charged particle

B. energy per unit mass transferred from charged particles

C. energy per unit mass transferred from photons or uncharged particles to charged particles

D. charge released by photons as they pass through a specified amount of air

Answer

A

C. energy per unit mass transferred from photons or uncharged particles to charged particles

Kerma is defined as the sum of the initial kinetic energies of all the charged particles liberated by indirectly ionizing radiation in a volume element.

However, all of the energy may not be deposited within the volume element if the particle range is large compared to the size of the element, because some fraction of energy can be radiated as bremmsstrahlung.

Therefore Kerma can differ from the absorbed dose at the point of interest.

Question 26

Q

Q62. 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

Answer

A

D. Neutrons

–The dose equivalents for x-rays, gamma-rays, and electrons is the same as the absorbed dose.

–For neutrons, however, it between 5-20 greater depending on the neutron energy, because of higher LET and greater potential for biological damage per Gy.

Question 27

Q

Q64. 100 μSv is equal to ___mrem

A. 100

B. 10

C. 1

D. 0.1

E. 0.01

Answer

A

B. 10

Knowing that 1 Sv = 100 rem

100 uSv = 100 x 10^-6 Sv = 10^-4 Sv = 10^-2 rem = 10 mrem

Question 28

Q

Q66. If 5 coulombs flow through a wire in 2 seconds, the current is:

A. 2 amps

B. 2.5 amps

C. 5 amps

D. 10 amps

Answer

A

B. 2.5 amps

Question 29

Q

Q68. Which of the following is not an SI unit?

A. meter

B. kg

C. second

D. rad

E. bequerel

Answer

A

D. rad

1 rad = 100 erg/gm. The SI unit for dose is the gray = joule/kg.

Question 30

Q

Q70. The rest mass of an electron is:

A. 981 MeV

B. 1.02 MeV

C. 0.51 MeV

D. 1.02 keV

E. 0.51 keV

Answer

A

C. 0.51 MeV

Question 31

Q

Q81. Rank the following in terms of increasing LET: (all have 5 MeV kinetic energy)

A. Neutrons, alphas, electrons

B. Alphas, electrons, neutrons

C. Electrons, neutrons, alphas

D. Neutrons, electrons, alphas

E. Electrons, alphas, neutrons

Answer

A

C. Electrons, neutrons, alphas

–LET depends on both mass and energy. Greater mass, lower energy leads to higher LET

Question 32

Q

Q83. Match the following answers (A-E) with the questions below:

A. +1, 0.51 MeV

B. +1, 930 MeV

C. +1, 140 MeV

D. 0, 930 MeV

E. -1, 0.51 MeV

Is the nucleus of a hydrogen atom.
Is emitted during beta decay when Z decreases by 1.
Is responsible for x-ray production.
When this particle combines with an electron, annihilation photons are emitted.
There are 2 of these in a Tritium nucleus.

Answer

A

Is the nucleus of a hydrogen atom - (B) +1, 930 MeV
Is emitted during beta decay when Z decreases by 1. - (A) +1, 0.51 MeV
Is responsible for x-ray production - (E) -1, 0.51 MeV
When this particle combines with an electron, annihilation photons are emitted - (A) +1, 0.51 MeV
There are 2 of these in a Tritium nucleus - (D) 0, 930 MeV

Question 33

Q

Q86. If a muon has a mass 207x that of an electron, its mass is equivalent to:

A. 560

B. 414

C. 335

D. 207

E. 106

Answer

A

E. 106

Rest mass of electron is 0.51 MeV

Question 34

Q

Q88. A 10 MeV ___ travels at the greatest speed in a vacuum.

A. Alpha particle

B. Neutron

C. Proton

D. Electron

Answer

A

D. Electron

–Lightest particles move the fastest.

Question 35

Q

Q90. Match the following (A-E) to the questions below:

A. Electron

B. Positron

C. Neutron

D. Alpha particle

E. Proton

Emitted by the cathode of the x-ray tube
The particle responsible for MR imaging
Assuming A-E have the same energy, the ____ has the shortest path length in water
Is indirectly ionizing

Answer

A

Emitted by the cathode of the x-ray tube - (A. Electron)
The particle responsible for MR imaging - (E. Proton)
Assuming A-E have the same energy, the ____ has the shortest path length in water - (D. Alpha particle)
Is indirectly ionizing - (C. Neutron)

Question 36

Q

Q83. Match the following answers (A-E) with the questions below:

A. +1, 0.51 MeV

B. +1, 930 MeV

C. 0, 0

D. 0, 930 MeV

E. -1, 0.51 MeV

Loses the most energy per unit path length
Results from pair production and eventually undergoes annihilation

Answer

A

Loses the most energy per unit path length; (B. +1, 930 MeV)
Results from pair production and eventually undergoes annihilation; (A. +1, 0.51 MeV)

Question 37

Q

Q98. Match the particle with the description (A-E)

A. neutron, B. proton, C. antineutrino

D. positron, E. alpha

Nucleus of a hydrogen atom
Is emitted during pure beta minus decay
Is created during pair production
Initiates fission in U235

Answer

A

Nucleus of a hydrogen atom (B. proton)
Is emitted during pure beta minus decay (C. antineutrino)
Is created during pair production (D. positron)
Initiates fission in U235 (A. neutron)

Question 38

Q

Q101. Directly ionizing radiations do not include:

A. electrons

B. positrons

C. neutrons

D. alpha particles

E. beta-rays

Answer

A

C. Neutrons

Neutrons are not charged particles, and generally interact with matter by transferring their energy to protons or other light nuclei, which then produce dense ionization tracts.

Question 39

Q

Q102. Match the following (A-E) with the questions below. Note that answers may be used more than once.

A. electrons, B. protons, C. neutrons

D. neutrinos, E. gamma-rays

Most responsible for nuclear medicine imaging
Most responsible for MR imaging
Most difficult to detect
Emitted in beta minus decay

Answer

A

Most responsible for nuclear medicine imaging - (E. gamma-rays)
Most responsible for MR imaging - (B. protons)
Most difficult to detect - (C. neutrons)
Emitted in beta minus decay - (A. electrons)

Question 40

Q

Q104. A neutron is heavier than an electron. The ratio of their masses is approximately:

A. 10:1

B. 500:1

C. 1000:1

D. 1400:1

E. 1800:1

Answer

A

E. 1800:1

The exact value = 939.55 MeV / 0.511 MeV = 1839

Question 41

Q

Q106. An electron, proton, and photon each have 1000 MeV total energy (kinetic energy + rest mass energy). Which of the following statements is true?

A. The electron travels at almost the speed of light

B. The proton travels at almost the speed of light

C. The photon travels at almost the speed of light

D. the photon has the most kinetic energy

E. the electron and the proton have the same rest mass

Answer

A

A. The electron travels at almost the speed of light

IF the kinetic energy of a particle is >> than the rest mass, then the velocity is very close to the speed of light.

Photons always travel at the speed of light.

Question 42

Q

Q108. Which of the following is never emitted during radioactive decay?

A. alpha particle

B. proton

C. positron

D. gamma-ray

E. neutrino

Answer

A

B. Proton

Question 43

Q

Q110. A radionuclide can decay by either beta minus or positron emission. The 2 daughter nuclei are:

A. Isomers

B. Isobars

C. Isotones

D. Isotopes

Answer

A

B. Isobars.

Isobars have equal A, but different number of protons and neutrons.

In beta minus n🡪p + B-
In beta plus, p🡪n + B+

Thus A is constant, but Z differs between the daughters.

Question 44

Q

Q113. Which of the following is true?

A. Elements can only have one stable isotope.

B. Elements can only have one radioactive isotope.

C. Isotopes of the same element have the same number of protons.

D. Isotopes of the same element have the same number of neutrons.

E. Stable elements have equal numbers of protons and electrons.

Answer

A

C. Isotopes of the same element have the same number of protons.

Question 45

Q

Q115. The product of an (n;γ) reaction is an __ of the target atom.

A. Isotone

B. Isobar

C. Isotope

D. Isomer

Answer

A

C. Isotope

–N increases, but Z remains the same, so the product is an isotope.

Question 46

Q

Q119. 131-Iodine and 125-Iodine have:

A. Have different chemical properties.

B. Have different Z values.

C. Occupy different columns on the periodic table.

D. Have the same number of neutrons.

E. None of the above.

Answer

A

E. None of the above.

Question 47

Q

Q123. All of the following are true except: Tritium 3H ___ hydrogen 1H.

A. is an isotope of

B. has more neutrons than

C. has more electrons than

D. is chemically identical to

Answer

A

C. has more electrons than

Question 48

Q

Q127. Elements which have the same Z but different A are called:

A. isobars

B. isomers

C. isotones

D. isotopes

Answer

A

D. isotopes

Question 49

Q

Q133. The number of neutrons in a U-238 atom (Z = 92) is:

A. 330

B. 238

C. 146

D. 92

E. Cannot tell from information given.

Answer

A

C. 146

–A(238) minus Z(92)

Question 50

Q

Q141. When an electron is removed from an atom, the atom is said to be ___ .

A. Radioactive

B. Ionized

C. Inert

D. Excited

E. Metastable

Answer

A

B. Ionized

Question 51

Q

Q145. In heavy nuclei such as 235Uranium:

A. There are more protons than neutrons.

B. Protons and neutrons are equal in number.

C. There are more neutrons than protons.

D. Cannot tell from information given.

Answer

A

C. There are more neutrons than protons.

–As the mass number increases, more neutrons are needed to balance the attraction of all masses (nuelons) with the repulsion between positively charged particles.

Question 52

Q

Q154. Tritium is an isotope of hydrogen with the symbol 3H.

A. its atomic number is 3

B. its mass number is 3

C. its atomic number is 2

D. its mass number is 1

Answer

A

B. its mass number is 3

Question 53

Q

Q158. The energy equivalent of one atomic mass unit is approximately

A. 10^0 eV

B. 10^3 eV

C. 10^6 eV

D. 10^9 eV

E. 10^12 eV

Answer

A

D. 10^9 eV

1 amu = 931.5 MeV, or approximately 109 eV

Question 54

Q

Q160. The number of electrons per gram of material is approximately equal to (N = Avogadro’s number)

A. NZ

B. NA

C. NA/Z

D. NZ/A

E. N

Answer

A

D. NZ/A

= (atoms/mole x (electrons/atom) / (grams/mole) = electrons / gram

Question 55

Q

Q162. The binding energy of an electron in the K shell of an atom is the energy needed:

A. To remove an electron from the K shell.

B. For the electron to jump from the K to the L shell.

C. To keep an electron in the K shell.

D. For the electron to jump from the L to the K shell.

Answer

A

A. To remove an electron from the K shell.

–This is the definition of electron binding energy.

Question 56

Q

Q164. Electron binding energy:

A. is greater in the K-shell than the L-shell

B. is greater for the K-shell of barium than the K-shell of hydrogen

C. increase with increasing Z

D. all of the above

E. none of the above

Answer

A

D. all of the above

Question 57

Q

Q168. The K-shell binding energy of tungsten (Z=74) is approximately 69.5 keV. Therefore, one would expect the K-shell binding energy of oxygen (Z=8) to be about:

A. 92 keV

B. 9.2 keV

C. 0.81 keV

D. 3 eV

Answer

A

C. 0.81 keV

The binding energy is proportional to Z2.

Thus: E/69.5= 82/742

Question 58

Q

Q170. Which of the following is not true? The electron binding energy:

A. decreases with increasing distance from the nucleus

B. decreases with increasing nuclear charge

C. is a few electron volts for the outer electrons of an atom

D. must be overcome if ionization is to take place

Answer

A

B. decreases with increasing nuclear charge

It increases approximately as the square of the atomic number.

Question 59

Q

Q174. From the following data, calculate the minimum energy required to separate a deuteron into its component parts (1 amu = 931 MeV)

Proton = 1.00727 amu

Neutron = 1.00866 amu

Deuteron = 2.01355 amu

A. 1.875 MeV

B. 2.02 MeV

C. 2.22 MeV

D. 2.38 MeV

E. 4.03 MeV

Answer

A

C. 2.22 MeV

The difference in Mass between a deuteron (2.10355amu) and its components, a proton (1.00727 amu) and a neutron (1.00866 amu) is 0.00238 amu. This is equivalent to 0.00238 x 931.2 MeV = 2.22 MeV.

((Proton) 1.00727 amu + (neutron) 1.00866 amu - (deuteron) 2.01355 amu) x 931 MeV

Question 60

Q

Q178. An atom with an ionization potential of 13.4 eV is bombarded with 7 eV photons. The minimum number of photons needed to ionize the atom is:

A. 1

B. 1.91

C. 2

D. 3

E. none of the above

Answer

A

E. none of the above

If the energy of the photon is less than the binding energy of the electron, ionization is impossible.

Question 61

Q

Q180. Tungsten has a K-shell binding energy of 69.5 keV. Which of the following is true?

A. The L-shell has a higher binding energy.

B. Carbon has a higher K-shell binding energy.

C. Two successive 35 keV photons could remove an electron from the K-shell.

D. A 69 keV photon could not remove the K-shell electron, but could remove an L-shell electron.

Answer

A

D. A 69 keV photon could not remove the K-shell electron, but could remove an L-shell electron.

–An electron can only be removed from an atom by absorbing a single photon of energy greater than or equal to the BE.

Question 62

Q

Q182. The binding energy per nucleon usually:

A. does not change in beta decay

B. increases after radioactive decay to the ground state

C. is independent of Z

D. is minimum for intermediate values of Z

Answer

A

B. increases after radioactive decay to the ground state

Decay to ground state usually very stable.

High binding energy means to be stable.

Question 63

Q

Q184. 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

Answer

A

C. Increases with Z

–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.)

Question 64

Q

Q191. The binding energy per nucleon is:

A. equal to the atomic mass

B. larger for stable nuclei than for radioactive nuclei

C. equal for different isotopes of the same element

D. decreased when an atom is ionized

Answer

A

B. larger for stable nuclei than for radioactive nuclei

Binding energy, or mass defect of a nucleus, is a measure of stability.

Answer 65

A

B. 1

The nth shell can contain a maximum of 2n^2 electrons, but no shell can contain more than 8 if it is the outer shell. In this case, it is Argon.

Answer 66

A

D. 8

–The outer shell contains a maximum of 8 electrons, hence 8 groups in the periodic table.

–When it is filled, the atom is very stable and unlikely to interact.

–Exception is helium which has 2 electrons in outer shell, but is very stable.

Answer 67

A

C. is 2n^2 where n is the principle quantum number

The maximum number in the outer shell is 8, but in general the maximum number of electrons in a shell of quantum number n is 2n2.

Answer 68

A

C. Determines the chemical properties of the atom.

–The maximum number of electrons in any shell is 2n2, but the maximum in the outer shell is 8.

Answer 69

A

D. It is the number of decays per unit time.

Answer 70

A

B. 4.05 hrs.

–1/Teff = 1/Tp + 1/Tb

–Note that Teff is shorter than Tp or Tb

Answer 71

A

B. 600

0.001=[-(0.693xT/60)]

T=598 days

Answer 72

A

A. 4.64 mCi

A/A0 = e^ (-0.693xT/14) = e^(-0.693x3/14) = e^-0.1485 = 0.862

A= 4 mCi, so A0 = A/0.862 = 4/0.862 = 4.64 mCi

Answer 73

A

D. 0.0625

–(1/2)^4=1/16=0.0625

Answer 74

A

C. 2.83

*note: the image of the formula has an error. it is Ao=2/(e^-0.693x37/74)

Answer 75

A

B. Close to Tb

–The effective half-life is (Tp x Tb)/(Tp + Tb).

–IF Tp + Tb= Tp this reduces to Tb

Answer 76

A

A. dN/dt

Activity is defined as the number of nuclear transformations occurring in a given time.

dN/dt. Activity also equals λN.

Answer 77

A

D. mass defect

Answer 78

A

C. a radioisotope with a 14 day half life will have more than 75% of its initial activity left after 7 days

After 0.5 T1/2, more than 25% of the activity has decayed.

The activity remaining is 70.7%.

Answer 79

A

D. 0.625

A = 10 mCi x exp ^ –(0.693/6 hrs x (24 hrs)

Answer 80

A

C. Less than the average life.

–The average life is 1.44 x half-life.

–The half-life is inversely proportional to λ.

Answer 81

A

B. The fraction of atoms decaying per unit time

Answer 82

A

D. 0.0094 per day

–T1/2 X λ = 0.693

Answer 83

A

A - linear on a log based scale

Answer 84

A

D. 70 hrs

The decay constant, λ, is the fraction of radioactive atoms will will decay per unit time.

Here λ = 0.01/hr.

Therefore T1/2 = 0.693/0.01= 70

Answer 85

A

C. give a lower patient dose

Although the initial dose rates are similar, the total dose will be less.

Total dose = dose rate x average life

The average life = 1.44 x half-life

Answer 86

A

B. 115 days

T1/2 = 0.693/lambda = 0.693/0.006 = 115 days

Answer 87

A

C. 1 hour

1/T_eff = 1/T_phy + 1/T_bio

Answer 88

A

A. Alpha decay

Mass number decreases by 4

Answer 89

A

B. Beta minus decay.

–The neutron changes to a proton plus electron and the electron is ejected as β-.

Answer 90

A

D. Beta decay.

–In beta decay the energy is shared between the electron and the anti-neutrino. The sum of their energies is the same, but each may have any energy from zero to the maximum energy available.

–Alphas and gammas are emitted with discrete energies

An isomeric transition is a nuclear process in which a nucleus with excess energy following the emission of an alpha particle or a beta particle emits energy without changing its number of protons or neutrons.

Answer 91

A

E. C or D

EC and B+ always compete each other.

EC happens when proton rich nucleus capture e (from any shell, usually K)

Beta plus decay will produce a positron though, which EC does not

Answer 92

A

E. All of the above

–In beta decay and electron capture, the daughter nucleus is in an excited state. Decay to the ground state is then accompanied by emission of the excitation energy in the form of one or more gammas. If the daughter has a measurable lifetime (Tc-99m), this is called an isomeric transition.

Answer 93

A

D. Betas.

–Internal conversion is an isomeric transition. Energy can be emitted as a gamma, or transferred directly to an inner shell electron.

•In the later case, the electron is emitted leaving a vacancy, which is filled by an outer electron. This is accompanied by characteristic radiation and auger electrons.

Answer 94

A

A. Increases after radioactive decay.

–Radionuclides decay in order to gain greater stability, and increase the BE per nucleon.

–The value is highest for intermediate values of Z.

Answer 95

A

D. Annihilation photons.

–Positrons combine with electrons in the patient and emit two annihilation photons of 0.511 MeV in oposite directions.

•These are detected by the coincidence counters.

Answer 96

A

A. 1, 3

Positrons, like betas, have a spectrum of energies.

Two 511 keV annihilation photons are emitted when the positron loses its kinetic energy and combines with an electron.

Answer 97

A

B. Beta.

–In beta decay the emitted particle (electron or positron for beta minus and beta plus, respectively) is shared with a neutrino. Thus the beta particle can have any energy between zero and the maximum available energy.

Answer 98

A

C. A spectrum of electrons and several monoenergetic photons

–Since Z increase by 1, this is an example of beta minus decay, which always emits a spectrum of betas.

In this case, they are accompanied by 2 gammas.

Answer 99

A

A. Larger

–99mTc decays to 99Tc emitting a 140 keV gamma. So its mass equivalent is greater by this amount.

Answer 100

A

B. Be a characteristic x-ray

–The 140 keV x-rays could interact with lead to emit lead characteristic x-rays.

Answer 101

A

A. Z and A remain the same.

–Energy is transferred directly to an inner shell electron, which is then ejected.

Internal conversion occurs in the nucleus and competes with gamma emission. Sometimes the electric fields of the nucleus interact with orbital electrons with enough energy to eject them from the atom.

This process is not the same as emitting a gamma ray which knocks an electron out of the atom.

It is also not the same as beta decay, since the emitted electron was previously one of the orbital electrons, whereas the electron in beta decay is produced by the decay of a neutron.

An example is 203Hg, which decays to 203Tl by beta emission, leaving the 203Tl in an electromagnetically excited state. It can proceed to the ground state by emitting a 279.190 keV gamma ray, or by internal conversion

Answer 102

A

D. 1, 2, 3, 4

Positron emission and electron capture often compete as a decay process in the same radionuclide with an excess of protons.

In electron capture, an electron, usually from the K shell, combines with a proton to form a neutron and an emitted neutrino.

The K shell vacancy causes characteristic x-rays and auger electrons to be emitted.

Answer 103

A

E. this is most likely in atoms with Z >82

Heavy nuclei tend to decay by alpha particle emission. Z decreases by 2 and A decreases by 4.

Radium is an example.

Answer 104

A

E. 1.0

B1 + Y3 = 0.2 + 0.8 MeV

Answer 105

A

C. one gamma ray, one beta ray, and one antineutrino

Answer 106

A

The mass number of Y is A+1 (B)
The atomic number of Y is Z-1 (B)
The decay of X is accompanied by the emission of antineutrinos (A)
0.511 MeV beta rays are emitted (A)
7.6 MeV beta rays are emitted (B)
In any one disintegration a total of 4.86 MeV will be emitted (A)

Answer 107

A

C. monoenergetic photons and electrons with a spectrum of energies

In beta decay, electrons are always emitted with a spectrum of energies.

Answer 108

A

1.59 MeV gamma ray (B. sometimes occurs)
40 keV characteristic x-ray (D. cannot be determined from the diagram)
1.12 MeV beta minus (B. sometimes occurs)
4.8 MeV antineutrino (B. sometimes occurs)
3.75 MeV gamma ray (C. never occurs)

Answer 109

A

A. The maximum energy of the positrons is 1.19 MeV

In positron emission, the energy available for the positron and neutrino is the difference in energy levels minus 2x the rest mass of an electron.

i.e. 2.21-1.02=1.19 MeV

Answer 110

A

C. is followed by the creation of 511 keV photons

Positron decay is always accompanied by neutrino emission, and the positrons have a spectrum of emissions.

When the positrons stop, it annihilates with another electron, producing a pair of 511 keV photons.

Beta decay cannot occur unless the energy levels of the parent and daughter nuclei differ by 1.02 MeV ( 2x 0.511 MeV)

Answer 111

A

B. 1, 2

99mTc emits 140 keV x-rays. So 130 kev scatter photon is possible.
Characteristic x-rays resulting form a photoelectic interaction with the lead container could be detected.
Cerenkov radiation is emitted when charged particles travel faster than the speed of light in a medium such as water. Relativity states that the ultimate speed limit is the speed of light in vacuum. It follows that Cerenkov radiation never occurs in vacuum. But, propagation of light can be slowed down considerably in materials due to interactions between light (photons) and particles of the material. Thus, it becomes possible for a particle moving at relativistic speeds to actually exceed the speed of light in that medium. When that happens, the particle emits radiation in the form of a ‘shock wave’, widely known as Cerenkov radiation.
Auger electrons are emitted when characteristic x-rays are reabsorbed by the same atom. There is no such thing as a auger x-ray.
Annihaltion radiation occurs when a positron ane electron combine, producing two 0.511 MeV photons.

Answer 112

A

C. A. charge: 0, rest mass: 0

Characteristic x-ray (photons) (0, 0 MeV); in electron capture the Z decreases by 1 but A is the same as a P -> N. See attached image.

charge: +1, rest mass: 0.511 MeV = positron
charge: +1, rest mass: 930 MeV = proton
charge: 0, rest mass: 930 MeV = neutron
charge: -1, rest mass: 0.511 MeV = electron

Answer 113

A

A. electron capture

Nuclide decays in an isobaric transition by either electron capture or beta plus.

Remember that internal conversion is a process where the excited nucleus interacts electromagnetically with one of the orbital electrons of the atom. This causes the electron to be emitted (ejected) from the atom. Since an electron is lost from the atom, a hole appears in an electron shell which is subsequently filled by other electrons that descend to that empty, lower energy level, and in the process emit characteristic X-ray(s), Auger electron(s), or both.

Answer 114

A

E. 1, 2, 3

Answer 115

A

C. emission of an Auger electron

In the Auger process, one may think of a characteristic x-ray as being emitted, but immediately undergoing a photoelectic interaction with another orbital electron, which is then emitted from the atom.

The energy of the Auger electron is equal to the energy of the characteristic x-ray less the binding energy of the electron.

The Auger process is similar to internal conversion, except that in the latter case the fictitious photon precipitating the interaction is generated from the nucleus.

Answer 116

A

B. Beta minus decay.

–Adding neutrons to the nucleus results in too many neutrons for stability. The neutron changes to a proton and a β−, which is emitted.

–The daughter nuclide may be in an excited state, and emit gammas.

Answer 117

A

E. bombarding with neutrons in a cyclotron

A. 137Cs, 90Sr

B, C. Various PET isotopes.

D. 99mTc

Cyclotrons can accelerate only charged particles; radioisotopes can be created by bombarding samples placed in the neutron flux of a reactor (Cobalt).

Answer 118

A

C. A fission product, which is obtained from used reactor fuel rods.

Answer 119

A

Separation from reactor-generated fission products. (B. Beta minus emitter)

–Cs-137 is an example of a radioisotope separated from reactor fuel. It decays by beta minus, followed by gamma emission.

Cyclotron-produced radionuclides. (A. Beta plus emitter)

–Cyclotrons accelerate protons or deuterons, increasing the Z of the target, and hence creating nuclides that will reduce Z by positron emission.

In fusion, a neutron is generated

In fission, a neutron bombards the target (think of it like the white ball when starting a game of pool)

Answer 120

A

E. 18Fluorine

–Is a cyclotron produced positron emitter used in PET.

Answer 121

A

B. Beta particle

Answer 122

A

A. 32/15P

The addition of one neutron and the loss of an alpha particle (2 protons and 2 neutrons) means that there is one less neutron and there are two less protons. i.e. three less nucleons.

Answer 123

A

Oxygen-15 (A. sample bombarded with charged particles in a cyclotron)
Radium-226 (B. naturaly occuring in uranium ore)
Technetium-99m (E. elution from the column of a generator)
Strontium-90 (D. separated from used reactor fuel rods)
Cobalt-60 (C. sample placed in neutron flux of a reactor)

Answer 124

A

Bombarding a sample with neutrons in a reactor (B. beta minus emission)
Bombarding with protons in a cyclotron (C. beta plus emission)
Bombarding with deuterons in a cyclotron (C. beta plus emission)
Separating fission products from reactor fuel rods (B. beta minus emission). In general, the higher the atomic number, the larger the neutron-proton ratio for stability. Therefore decay by beta minus.

Answer 125

A

A. Much longer than

Answer 126

A

B. Somewhat longer than

Answer 127

A

C. The same

–The nuclide is radon. It must be the daughter of a long-lived parent, or it would not be found at all. As it is constantly being replaced by the parent, it appears to decay with the half-life of the parent. In 6 days, this decay is negligible.

Answer 128

A

B. Secular

–Example is radon.

In transient:

–Half-life of the parent is not much longer then the daughter.

–Example is milking of 99mTc

Answer 129

A

B. Secular equilibrium

Takes 4 half-lifes to establish, and then the daughter and parent decay with the half-life of the parent.

Answer 130

A

Secular equilibrium occurs when the decay constant of the daughter is slightly greater than the decay constant of the parent (B)
Transient equilibrium requires decay to a metastable state of the daughter (B)
In transient equilibrium the activity of the daughter is always less than that of the parent (B)
Equilibrium may exist if the half life of the daughter is shorter than that of the parent (A)
Transient equilibrium exists if the half life of the parent is somewhat greater than that of the daughter (A)

–

Secular equilibrium is when parent half life >> daughter half life. The activity of the daughter therefore will be quickly approximately equal to the activity of the parent.

Real life example of secular equilibrium is Ra226 -> Rn222 -> Po218

Transient equilibrium occurs when half life of daughter is similar to that of the parent (parent half life = daughter half life) this means that the relative activity of the daugher will increase to a max and then decline at the same rate as the parent. Ex. Mo99 -> Tc99m.

When daughter’s half life is >> parent, no equilibrium will occur

Answer 131

A

C. 4 half lives

The activity of the sample increases exponentially with time (the curve is the inverse of the decay curve). After one half-life, half the maximum activity is achieved.

After n half-lives, the activity is (1-0.5^n) A_max.

Thus 4 half-lives gives 94% A_max.

Answer 132

A

D. The daughter decays with the half-life of the parent.

EDIT: This contradicts answer from Q431:

When equilibrium is achieved, whether transient or secular, the activity of the daughter is slightly greater than the activity of the parent (except when there is more than one decay mode whereby less than 100% of the parent decays yield the daughter).

Answer 133

A

C. Equilibrium has not yet occured

Needs 4 half lives (in this case about 24 hours)

Answer 134

A

E. None of the above

When equilibrium is achieved, whether transient or secular, the activity of the daughter is slightly greater than the activity of the parent (except when there is more than one decay mode whereby less than 100% of the parent decays yield the daughter).

Answer 135

A

B. 3.7 X 10^2

1mCi = 37 MBq, so 10 mCi = 3.7 x 10^2 MBq

Answer 136

A

E. AMU; atomic mass unit

Answer 137

A

D. 135

5.0 x 10^9 Bq = 5x10^3 MBq

1 mCi = 37 MBq

Answer 138

A

C. 3.3 mR/hr

10 mCi x 3.3 Rcm^2/(mCi.hr)) = 33 Rcm^2/hr

33 Rcm^2/hr x (1 m/ 100 cm)^2 = 33 x 10^-4 Rm^2/hr

33 x 10^-4 Rm^2/hr = 3.3 mRm^2/ hr x (1/1m)^2 = 3.3 mR/hr

–Exposure Rate= Exp. Rate Const. x Activity x 1/d^2

Answer 139

A

C. Air kerma rate

–Air kerma strength SK specifies the strength of a source in terms of its kerma rate in air at 1 m, and has units of μGy m^2 h-1.

Answer 140

A

B. 2

–Exp Rate= 6.29, 1 HVL and Exp Rate=3.15, 2 HVL and Exp Rate=1.57

9 Rcm^2/(mCi-hr) x (1 m/ 100 cm)^2 x 19.5 mCi = 251.55 x 10^-4 Rm^2/hr
55 x 10^-4 Rm^2/hr x (1/2m)^2 = 62.9 x 10^-4 R/hr = 6.29 mR/hr

Answer 141

A

C. 3

5 mCi x 3 hrs x 2 Rcm^2/(mCi x hr) = 30 Rcm^2

30 Rcm^2 x (1 m/ 100 cm)^2 x (1/1m)^2 = 30x10^-4 R = 3 mR

Answer 142

A

C. 8.25 mR/hr

The exposure rate at r cm for A mg of Ra is 8.25 x A x (1/r)^2 x R/hr

Answer 143

A

C. Co-60 emits more than one gamma per disintegration

Cobalt emits two gammas per disintegration (99.8% of the time). Half life has no influence on gamma factor, and although cesium has a lower energy gamma than cobalt, this does not have a marked effect on the gamma factor.

Answer 144

A

C. Create thermionic emission.

–The process where by a filament is heated to a sufficient temperature to emit electrons is called “thermionic emission”.

Answer 145

A

D. Thermionic emission

–Thermionic emission is the emission of electrons from the heated filament.

Answer 146

A

D. Enable the smallest focal spot to be used, consistent with the kVp/mA setting.

–Resolution is improved by the use of the small focal spot.

–However, if a high kVp/mA setting is required, this might overheat a small area of the target

–A Larger focal spot helps with heat dissipation.

One filament is used for low current exposures when a small focal spot will give a sharper image. At high currents, heat dissipation in the target is a problem if the focal spot is too small, so a second filament with a larger focal spot is used.

NOTE: This is a feature usually found in diagnostic, not therapeutic machines

Answer 147

A

A. 7 is the anode, 8 is the cathode

The anode is the target (positive) and the filament is the cathode (negative).

Answer 148

A

C. full-wave rectified

Four rectifiers are required for full-wave rectification.

Answer 149

A

E. all of the above

A transformer is an electrical apparatus designed to convert alternating current from one voltage to another. It can be designed to “step up” or “step down” voltages and works on the magnetic induction principle.

Answer 150

A

Transformer (B. increases or decreases voltage)
- used to step up or step down, (volts to kilovolts)
Miliammeter (E. measures tube current)
- Measure the current in mA. It is connected in series with the x-ray tube itself.
Rectifier (A. allows current to flow in one direction only)
- Rectifiers allow current to pass in only one direction; x-ray tubes act as a rectifier in a “self-rectified” circuit.

Answer 151

A

C. (N2/N1) x V1

N1/N2=V1/V2

V2 = N2/N1 x V1

Answer 152

A

E. none of the above

Step-up increases the voltage, but decreases the current and power

P=IV

Answer 153

A

A. small effective focus and large heat loading capacity

Answer 154

A

E. is always less than the kVp

The effective energy of an x-ray beam is approximately 1/3 to ½ of the kVp. This may be increased by added filtration (beam hardening). For a given KV, two factors that can alter the spectrum are the amount of filtration in the beam and the high voltage waveform used to produce the x-rays.

Contrast is a function of x-ray energy.

Not affected by mAs or Z.

Answer 155

A

C. delivers the same exposure in half the time

Four rectifiers are required for full-wave rectification.

The voltage ripple is 100% for both full and half-wave rectification.

The x-ray spectrum depends only on the kVp, filtration, and the use of a single- or three-phase power.

Answer 156

A

D. all of the above

Three phase units have nearly constant voltage output resulting in higher beam current, average voltage, and dose rate.

Answer 157

A

Maximum kVp is 50 kVp (A. spectrum I)
Higher exposure rate (B. spectrum II)
Higher HVL (B. spectrum II)
Produced by 3 phase or constant potential generator (E. cannot be determined)
Will produce relatively less scatter in soft tissue (A. spectrum I). scatter increases with kVp in the diagnostic range.

Answer 158

A

E. 99:1

For diagnostic machines the efficiency is low, but for therapeutic is better at around 15-20%

Answer 159

A

D. The energy of the characteristic x-rays.

–At 100kVp, the highest characteristic x-rays (69 kV) are already present, so the characteristic peaks will remain the same, althought the intensity will increase.

Answer 160

A

C. 2,4

Brem x-rays of any energy up to that of the projectile electron can be created.

For characteristic x-ray emission, the electrons must have an energy greater than or equal to the binding energy of an orbital electron.

–In this case, only the L or M shell electrons could be ejected. An electron from an outer shell then fills the vacancy, and a characteristic x-ray of energy equal to the difference in the electron binding energies is emitted.

12-2=10.

Answer 161

A

An incoming 80 keV projectile electron (A)
An incoming 75 keV gamma ray (A)
An incoming 66 keV x-ray (B)
A 58 keV projectile electron (B)

Have to have enough to get over the binding energy

Answer 162

A

A. 0-20

–Below 69 kV, K characteristic x-rays cannot be emitted, and L x-rays are generally filtered out.

–Above 68 kV, the proportion of K x-rays increases with increasing energy, but bremsstrahlung always accounts for the greater part of the spectrum.

Answer 163

A

C. Have energies equal to differences in the electron binding energies.

Answer 164

A

34 kV x-rays (B. Bremmstrahlung is emitted in a continous spectrum up to a maximum energy equal to that of the incident electrons.)
26 kV x-rays (C. The characteristic x-ray energies possible from this atom are 29.3, 26.0 and 3.3 kV)
40.7 kV x-rays (D. Greater than incident energy.)

Answer 165

A

C. produces characteristic x-rays less than 20 keV

Characteristic K x-rays at about 17.5 and 19 keV are desirable for maximum calcium and soft tissue contrast.

Answer 166

A

B. 4

K-shell = 30 keV. M-shell = 0.7 keV.

Electron ejected is 25.3 keV is ejected from the M-shell

L-shell electron binding energy is thus is 30 - 25.3 - 0.7 = 4 keV.

Answer 167

A

E.cba

C) heat loss > B) production of bremsstrahlung radiation > A) production of characteristic radiation

90% of energy lost as heat.

Of the 10% converted into x-rays

90% is bremsstrahlung
10% is characteristic X-rays.

This can depend on Z of material.

Answer 168

A

C. kV across the tube

A, B, and E affect the shape of the spectrum, but not the kVp

Answer 169

A

A. Inherent and added filtration

Answer 170

A

B. Somewhat more than 25%.

–For Monochromatic beam two HVLs would reduce the intensity to 25%.

–However, for a broad spectrum the beam becomes hardened as it passes through the filtration, making the 2nd HVL greater than the first.

Answer 171

A

D. 3.5

–The first HVL reduces output from 200 to 100 and is 2.5 mm. The second HVL reduces output from 100 to 50 and is 6.0 – 2.5 = 3.5 mm Al.

Answer 172

A

B. A broad beam could introduce scattered x-rays, giving a false reading.

–For HVL measurement, only the primary beam should be measured. With a broad beam the detector will record scattered radiation also, which will give an incorrect HVL value.

Answer 173

A

E. The half-value layer in Al.

–The HVL, or thickness of Al (in mm) required to reduce the intensity of the beam to 50%, is the method used to quantify the quality (i.e., penetrability) of the beam.

–Different combinations of kVp and filtration can result in the same HVL.

Answer 174

A

D. Increasing the tube voltage.

The effective photon energy is approximately equal to between 1/3-1/2 of the maximum photon energy.

Factors influencing x-ray quality include:

peak voltage (kVp)
voltage waveform: reducing ripple increases quality
beam filtration: increasing filtration increases quality through beam hardening
anode material: photon energy depends on the binding energies of shells in the anode material

On a side note, do not confuse this with output.

X-ray output is proportional the current in mA, and the square of the voltage in kVp; (kVp)^n x mA; where n is usually 2 for an x-ray tube

Answer 175

A

B. Greater Less

–The added filtration filters out more low- than high-energy photons, so the resulting beam has a greater effective energy.

–However the intensity is reduced.

Answer 176

A

E. Affects subject contrast.

The effective photon energy is approximately equal to between 1/3-1/2 of the maximum photon energy.

Factors influencing x-ray quality include:

peak voltage (kVp)
voltage waveform: reducing ripple increases quality
beam filtration: increasing filtration increases quality through beam hardening
anode material: photon energy depends on the binding energies of shells in the anode material

On a side note, do not confuse this with output.

X-ray output is proportional the current in mA, and the square of the voltage in kVp; (kVp)^n x mA; where n is usually 2 for an x-ray tube

Answer 177

A

E. All of the above

Answer 178

A

C. Only if the beam is monoenergetic (e.g., gamma rays).

–For polyenergetic beams, passing through the first HVL hardens the beam, making the second HVL larger than the first.

Answer 179

A

E. patient skin dose

Answer 180

A

The quality of the beam depends on the wave form (A. TRUE. The beam is described by it quality and quantity. The Quantity, or intensity is best described by the exposure rate. The quality is best described by the graph of intensity as a function of photon energy. The quality depends on the wave form, fitration, and kVp.)
The quality of the beam is independent of the tube current (mA) (A. TRUE)
Small changes in filament current affect the beam quality (B. FALSE)
A graph of intensity vs photon energy is the best description of the x-ray tube output (A. TRUE. HVL is a crude measure of quality.

Answer 181

A

D. inversely proportional to the linear attenuation coefficient

HVL = 0.693/ (linear attenuation coefficient)

Answer 182

A

D. radiation intensity

A, B, and C determine the spectrum reaching the absorbers.

For E, readings to done so as to allow to much scatter to reach detector will give incorrect readings.

Answer 183

A

D. 13%

Two HVLs reduce the intensity to 25%.

But the reduction due to inverse square also applies. 25% x (50/70)^2 = 13%

Answer 184

A

E. the half value layer in aluminum

The value of the HVL depends on both the kVp and the amount of filtration, and is more easily measured than either of those.

Answer 185

A

D. increasing the tube voltage

Full wave rectification doubles the effective tube current (as compared to half-wave) without changing the effective photon energy.

Decreasing filtration increases dose rate, but decreases the effective photon energy.

Answer 186

A

E. have a lower effective energy

Answer 187

A

Betas (A. Ionizing elementary particles)
Heat radiation (D. Non-ionizing photons)
Visible light (D. Non-ionizing photons)
X-rays (C. Ionizing photons)
Ultrasound (E. Other)

Answer 188

A

A. 2 MHz ultrasound

Answer 189

A

B. 1&2

The radiation must have sufficient energy to ionize atoms in the film emulsion.

Answer 190

A

D. Origin

Answer 191

A

C. 2 eV

Eλ=Eλ

Answer 192

A

E. 4.4

–By inverse square law: I75 = I50 x (50/75)2 = 4.4 mR/min

Answer 193

A

C. Frequency

Answer 194

A

A. 90, 90

Answer 195

A

E. 3m

c=λυ

Answer 196

A

B. 12.4

(4.14 x 10^-15 eV s) x (3 x 10 ^8 m/s) / (10^-7 m) = 12.4 eV

Answer 197

A

D. is deflected by a magnetic field

Charged particles are deflected by magnetic fields.

Answer 198

A

C. energy is directly proportional to frequency

Velocity = frequency x wavelength

Energy = planck’s constand x frequency

Answer 199

A

A Linear attenuation coefficient

The mass energy absorption coefficient is the linear attenuation coefficient divided by the density (i.e., attenuation per unit mass rather than per cm.)

The mean attenuation length is the linear attenuation coefficient x 1.44.

Answer 200

A

C. 0.693/x

X = HVL, therefore μ=0.693/HVL

Answer 201

A

B. 6.93 cm

HVL = 0.693/0.1

Answer 202

A

C. 0.50

I/Io = e ^ (-μx)

or HVL = 0.693/0.0693 = 10 cm

Answer 203

A

B. Compton

–The mass attenuation coefficient is similar for most materials in the compton region, except this 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.

Answer 204

A

B. Absorption.

–The energy transferred to electrons, aka “absorbed dose”

Scatter is the energy remitted as photons

Answer 205

A

B. Linear attenuation coefficient.

Answer 206

A

D. increases with increasing photon energy and decreasing atomic number

Absorption normally decreases with increasing photon energy and increases with increasing atomic number.

Answer 207

A

A. has a minimum value for energies between 1 and 10 MeV

From energies of 1-10 MeV, only the compton effect is important and total cross section is at a minimum.

Answer 208

A

B. fraction of the primary photons transmitted by x cm of attenuator

Answer 209

A

D. the linear attenuation coefficient

Answer 210

A

D. Z^3 E^-3

Answer 211

A

C. only Compton interactions occur

The mass attenuation coefficient is the linear attenuation coefficient divided by the density.

Pair production increases with Z.

Photoelectic increases with Z^3.

Compton is independent of mass/Z.

Answer 212

A

E. 13.86 cm

0.693/0.05 = HVL

Answer 213

A

E. 1 - e^ (-0.693n)

Answer 214

A

B. False

99mTc is a monoergetic gamma emitter. The first and second HVL would then be the same.

Answer 215

A

B. 10

TVL = 3.3 HVL

(1/2)^3.3 = 1/10

Answer 216

A

B. 10

1% = 2 TVLs so at midline is 1 TVL

Answer 217

A

C. 7

The activity is increased by a factor of 4. Therefore 2 additional HVL’s are required.

Answer 218

A

E. 3.1

6/ 1.2 = 5 HVLs

so 0.5^5 = 0.03 of I_o

Answer 219

A

D. bone

Photoelectric process; Z^3/ E^3

Answer 220

A

Fat, muscle (A)
Muscle, bone (A)
Iodine, bone (B)
Fat, air (A)
Muscle, air (B)

Note - it is for photoelectric that fat counterintuitively absorbs LESS than air. Please see attached graph

Answer 221

A

E. 1000

Z^3 so (60/6)^3 = 1000

Answer 222

A

A. K

The photon is most likely to undergo a photoelectric interaction when its energy is just above the binding energy of the electron.

Answer 223

A

C. 4.4

–The energy of the photon is totally absorbed by the electron, which uses 4.4 keV to overcome its binding energy, leaving 3.6 keV as the energy of the emitted photoelectron.

Answer 224

A

B. 8

Photoelectric is related to Z^3; so (14/7)^3

Answer 225

A

C. The probability increases rapidly with increasing energy.

Answer 226

A

D. A characteristic x-ray

Answer 227

A

D. scattered photon following photoelectric interaction

Answer 228

A

D. 16 to 1

The photoelectic component of the linear attenuation coefficient is proportional to Z^3 and the physical density (ρ).

(t_B/t_A)=(14/7)^3 x (2/1) = 2^3 x 2 =2^4 = 16

Answer 229

A

E. 30 keV and iodine

Calculate out Z^3/E^3

Answer 230

A

B. The recoil electron generally acquires most of the initial photon energy.

–In Compton interactions, interaction at any angle other than grazing incidences leads to a significant transfer of energy to the electron. That is why the energy of the scattered radiation from a megavoltage beam is significantly lower than that of the primary beam.

Answer 231

A

C. Compton electrons can be backscattered.

–Electrons cannot be backscattered, they can only be scattered at angles between 0 and 90o to the direction of the incident photon.

Answer 232

A

B. 1:1

–The number of compton interactions depends on the number of electrons. Most materials have about the same number of electrons per gram.

–THE EXCEPTION IS HYDROGEN. It has twice as many electrons as it has no neutrons in the neucleus.

Answer 233

A

D. Photoelectric, Compton

–After a compton interaction, the compton scatter photon can undergo any type of interaction (even pair production, if its energy is greater than 1.02 MeV).

–However, after a photoelectric interaction, the photon is absorbed.

Answer 234

A

C. Electrons can be emitted between 0° and 90° to the direction of the incident photon.

Answer 235

A

C. Compton scattered photons.

Answer 236

A

B. Maximized when the photon is scattered at 180°.

–The maximum energy transfer occurs when the photon is backscattered. 0.256 MeV. At 90 degrees it is 0.51 MeV.

Answer 237

A

E. None of the above.

–Independent of Z

–Decrease with increasing energy

Answer 238

A

C. Compton

0.24 MeV is 50% probability cutoff for Photoelectric/ Compton. 24 MeV is the 50% probability cutoff for Compton/ Pair production.

Answer 239

A

B. 0-90 degrees with the incident photon

Answer 240

A

A. zero

No energy is lost in classical scattering.

Answer 241

A

A photon may undergo a Compton interaction followed by a pair production interaction (A, TRUE)
A photon may undergo 2 consecutive Compton interactions (A, TRUE)
A photon may undergo a Compton interaction followed by a photoelectric interaction (A, TRUE)
A photon may scatter from an atom without losing energy (A, TRUE)
The probability of an 80 keV photon being scattered is greater in Sn (Z=50) than in Al (Z=13) (B, FALSE - Compton is not dependent on Z; remember it is asking about photon SCATTERED, not absorbed which would be photoelectric)

Answer 242

A

D. Photons above 1 MeV

In a compton interaction, the angle at which the scattered photon has the lowest energy is always 180o, i.e. backscatter.

However, at 100 keV, the backscattered photon retains about 70% of the incident energy, while at 1 MeV the backscattered photon retains only about 20%.

Rule of thumb for megavoltage photons is that backscatter is about 250 keV and side scatter is about 500 keV.

Answer 243

A

C. 2.0

Hydrogen is the exception for Z and Compton non-dependence.

Hydogren has 1 electron per nucleon (proton).

All others have 1 electorn per two nucleon (proton+neutron).

Thus hydrogen has twice the number of electrons per nucleon.

Answer 244

A

B. 0 degrees

At high photon energies backscattered Compton photons always have an energy of 255 keV. When the photon is backscattered the electron must go in the forward direction: 0 degrees

Answer 245

A

B. An electron and a positron are produced.

–When the positron loses its kinetic energy and annihilates with an electron, two 0.51 MeV photons are emitted in opposite directions.

–Threshold energy for pair production is 2 mec2=1.02 MeV.

–The electron and positron share the photon energy less 1.02 Mev.

Answer 246

A

B. Ei - 1.02 MeV.

Answer 247

A

B. Annihilation photons always have an energy of 0.51 MeV each.

Answer 248

A

C. 1.28

–Takes 2 x 0.51 MeV to create the electron-positron pair.

–Thus the remaining 1.28 is deposited.

Answer 249

A

A. Two 0.51 MeV photons will be emitted

Answer 250

A

Chiefly responsible for loss of contrast in a diagnostic radiograph. (C. Compton)
Most probable at photon energies between 100 keV and 2MeV. (C. Compton)
No energy is transferred or locally absorbed. (A. Coherent scatter)
Probability of interaction increases as energy increases (D. Pair production)

Answer 251

A

Pair production (in Pb) - D
Compton (in H2O) - C
Photoelectric (in H2O) - B

“A” is the total mass attenuation coefficient in water.

Answer 252

A

Curve #1 (E. total absorption in Pb)
Curve #2 (D. total absorption in H2O)
Curve #3 (C. photoelectric effect in H2O)
Curve #4 (A. Compton effect)
Curve #5 (B. pair production in Pb)

Answer 253

A

B. Annihilation photons

Answer 254

A

B. photoelectric

Photoelectric is considered total absorption since the characteristic x-rays are self-absorbed in the attenuating material in a very short distance

Answer 255

A

A photon may undergo a photoelectric interaction followed by a pair production interaction (B, FALSE)
A photon may undergo 2 consecutive photoelectric interactions (B, FALSE)
A photon may undergo a photoelectric interaction followed by a compton interaction (B, FALSE)
A photon may be totally absorbed by an atom (A, TRUE)
The probability of a photoelectric interaction increases rapidly with energy (B, FALSE)

A photon disappears following a photoelectric interaction.

The photon also dissappears following pair production.

Following a compton interaction, the scatter photon may interact again.

Answer 256

A

C. 25 keV

Compton is the most probable interaction from 25 keV to 25 Mev.

Answer 257

A

B. photoelectric effect

Photoelectric interactions are more likely at low than high energy; after passing through a filter, the total beam energy is reduced, but the beam contains a relatively greater number of high energy photons than before filtration.

Answer 258

A

A 35 keV x-ray incident on a vessel filled with iodine contrast material (A, photoelectric)
A 100 keV photon incident on muscle tissue (B, compton)
A 15 MeV photon incident on muscle tissue (B, compton)
A 18 keV photon incident on fatty breast tissue (A, photoelectric)
A 90 keV photon incident on bone (B, compton)
A 90 keV photon incident on a lead barrier (A, photoelectric) ; K edge of lead is 88 keV.

Answer 259

A

C. Linear attenuation coefficient.

–CT number = 1000 x [(μmaterial – μwater)/μwater]

Answer 260

A

1, This is an example of Compton scattering (B, FALSE - this is photoelectric)

Characteristic radiation will be emitted (A, TRUE)
Auger electron emission is possible (A, TRUE)
The electron will be absorbed within 1 cm of its origin (A, TRUE)

Answer 261

A

C. 4

τ/ρ is proportional to Z3.

0.004 x (82/8)3 is approximately 4

Answer 262

A

B. decreases to about 3 MeV then rises again

Answer 263

A

B. 1 is greater than 2

Answer 264

A

D Cannot tell from this plot

Cannot tell, because they are Mass absorption coefficient. If they were linear attenuation coefficient, could tell, because higher curve would have greater density.

Answer 265

A

B. a 2 MeV beta particle

Answer 266

A

C. 1, 3 & 4 only

Photoelectric is a photon interaction.

Answer 267

A

D. The electron travels almost at the speed of light.

–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 due to its high mass

Answer 268

A

C. Protons undergo exponential attenuation.

–Photons , NOT protons, are attenuated exponetially.

Answer 269

A

A. 1 cm, 8 m

1/0.0013 = 770,

assuming tissue is about equal to water (which has density of 1 g/ cm^3)

Converting everything to cm;

800 cm/ 1 cm = 800 so it comes out the closest to 770

Answer 270

A

E. 1, 2, 3, 4

Answer 271

A

B. Interacts primarily with the oxygen in H2O.

–A neutron is most likely to interact with hydrogen, which has a similar mass.

Answer 272

A

E. All of the above

–When neutrons are absorbed by nuclei, the nucleus becomes unstable, and any of the emissions listed becomes possible.

Answer 273

A

A. They transfer energy to protons which have a high LET.

–They transfer energy to protons, which have a large mass, and are densely ionizing, especially at the end of their tracks (Bragg peak).

Answer 274

A

B. 1 mGy of neutrons is radiobiologically equivalent to 1 mGy of x-rays.

–Depending on their energy, neutrons cause up to 20 times as much damage for x-rays.

–MUST USE FACTOR of 20 for RADIATION PROTECTION

Answer 275

A

C. 1

–Optical Density = log (incident intensity/transmitted intensity)

–OD = log(1/0.1)=log 10=1

Answer 276

A

D; H&D curves are plots of film density (log of opacity) versus the log of exposure
A, radioactive decay is a straight line on a semi-log scale.

Answer 277

A

B. Compton scattered photons

–Compton-scattered photons travel in random directions, and contain no useful diagnostic information. The grid absorbs most of these photons, and thus improves image contrast.

Answer 278

A

A. Inversely proportional to focal spot size.

–Geometric sharpness or edge gradient is reduced by minimizing magnification.

Answer 279

A

E. 4 only

Answer 280

A

C. 1.5

–OD is additive

–It is the log of (incident/transmitted) light intensity.

10^(0.75+0.75) = 10^1.5

Answer 281

A

C. 0.001

10^ -(1.5+1.5)= 10^-3, so only 0.001 is transmitted through

For a single film:

A = log₁₀ (I_o/ I) = - log₁₀ T, where T is the transmission

1.5 = - log₁₀ T or 10^-1.5 = T

T= 0.031

Answer 282

A

C. Photoelectric interactions.

–Remember, need to see bones.

Answer 283

A

C. Patient’s body

Answer 284

A

C. Using a high ratio grid

–Grids improve contrast by cleaning up scatter, but require a somewhat higher dose to compensate for attenuation by the grid.

–Collimation decreases beam exposure.

Answer 285

A

E. Gamma

(H&D) curve - Plots of film density (log of opacity) versus the log of exposure

Answer 286

A

C. Reduce scatter to the film

Grids tend to increase patient dose (for the same optical density on the film) but they intercept scatter generated by interactions in the patient, which would reduce contrast.

Answer 287

A

D. absorb electrons produced by the photoelectric effect

Compton scattered photons travel in random directions and thus contain no useful diagnostic information. The grid absorbs most of these photons, thus improving image contrast. Photoelectrons rarely travel far enough to leave the patient.

Answer 288

A

C. 2400 2600

–If a large number of measurements (N) are made, approximately 67% will fall between +σ, and 96% between +2σ of the mean.

The standard deviation σ=sqrt(N), or 50 in this case. There for 2500+(2 x σ)= 2400-2600.