ROphex Old Q1-1000 Flashcards

Question

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 1

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.

Answer 2

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.

Answer 3

B. 10 Knowing that 1 Sv = 100 rem 100 uSv = 100 x 10^-6 Sv = 10^-4 Sv = 10^-2 rem = 10 mrem

Answer 4

B. 2.5 amps

Answer 5

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

Answer 6

C. 0.51 MeV

Answer 7

C. Electrons, neutrons, alphas –LET depends on both mass and energy. Greater mass, lower energy leads to higher LET

Answer 8

1. Is the nucleus of a hydrogen atom - (B) +1, 930 MeV 2. Is emitted during beta decay when Z decreases by 1. - (A) +1, 0.51 MeV 3. Is responsible for x-ray production - (E) -1, 0.51 MeV 4. When this particle combines with an electron, annihilation photons are emitted - (A) +1, 0.51 MeV 5. There are 2 of these in a Tritium nucleus - (D) 0, 930 MeV

Answer 9

E. 106 Rest mass of electron is 0.51 MeV

Answer 10

D. Electron –Lightest particles move the fastest.

Answer 11

1. Emitted by the cathode of the x-ray tube - (A. Electron) 2. The particle responsible for MR imaging - (E. Proton) 3. Assuming A-E have the same energy, the ____ has the shortest path length in water - (D. Alpha particle) 4. Is indirectly ionizing - (C. Neutron)

Answer 12

1. Loses the most energy per unit path length; (B. +1, 930 MeV) 2. Results from pair production and eventually undergoes annihilation; (A. +1, 0.51 MeV)

Answer 13

1. Nucleus of a hydrogen atom (B. proton) 2. Is emitted during pure beta minus decay (C. antineutrino) 3. Is created during pair production (D. positron) 4. Initiates fission in U235 (A. neutron)

Answer 14

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.

Answer 15

1. Most responsible for nuclear medicine imaging - (E. gamma-rays) 2. Most responsible for MR imaging - (B. protons) 3. Most difficult to detect - (C. neutrons) 4. Emitted in beta minus decay - (A. electrons)

Answer 16

E. 1800:1 The exact value = 939.55 MeV / 0.511 MeV = 1839

Answer 17

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.

Answer 18

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.

Answer 19

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

Answer 20

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

Answer 21

E. None of the above.

Answer 22

C. has more electrons than

Answer 23

D. isotopes

Answer 24

C. 146 –A(238) minus Z(92)

Answer 25

B. Ionized

Answer 26

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.

Answer 27

B. its mass number is 3

Answer 28

D. 10^9 eV 1 amu = 931.5 MeV, or approximately 109 eV

Answer 29

D. NZ/A = (atoms/mole x (electrons/atom) / (grams/mole) = electrons / gram

Answer 30

A. To remove an electron from the K shell. –This is the definition of electron binding energy.

Answer 31

D. all of the above

Answer 32

C. 0.81 keV The binding energy is proportional to Z2. Thus: E/69.5= 82/742

Answer 33

B. decreases with increasing nuclear charge It increases approximately as the square of the atomic number.

Answer 34

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

Answer 35

E. none of the above If the energy of the photon is less than the binding energy of the electron, ionization is impossible.

Answer 36

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.

Answer 37

B. increases after radioactive decay to the ground state Decay to ground state usually very stable. High binding energy means to be stable.

Answer 38

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.)

Answer 39

B. larger for stable nuclei than for radioactive nuclei Binding energy, or mass defect of a nucleus, is a measure of stability.

Answer 40

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 41

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 42

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 43

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 44

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

Answer 45

B. 4.05 hrs. –1/Teff = 1/Tp + 1/Tb –Note that Teff is shorter than Tp or Tb

Answer 46

B. 600 0.001=[-(0.693xT/60)] T=598 days

Answer 47

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 48

D. 0.0625 –(1/2)^4=1/16=0.0625

Answer 49

C. 2.83 \*note: the image of the formula has an error. it is Ao=2/(e^-0.693x37/74)

Answer 50

B. Close to Tb –The effective half-life is (Tp x Tb)/(Tp + Tb). –IF Tp + Tb= Tp this reduces to Tb

Answer 51

A. dN/dt Activity is defined as the number of nuclear transformations occurring in a given time. dN/dt. Activity also equals λN.

Answer 52

D. mass defect

Answer 53

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 54

D. 0.625 A = 10 mCi x exp ^ –(0.693/6 hrs x (24 hrs)

Answer 55

C. Less than the average life. –The average life is 1.44 x half-life. –The half-life is inversely proportional to λ.

Answer 56

B. The fraction of atoms decaying per unit time

Answer 57

D. 0.0094 per day –T1/2 X λ = 0.693

Answer 58

A - linear on a log based scale

Answer 59

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 60

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 61

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

Answer 62

C. 1 hour 1/T_eff = 1/T_phy + 1/T_bio

Answer 63

A. Alpha decay Mass number decreases by 4

Answer 64

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

Answer 65

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 66

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 67

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 68

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 69

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 70

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 71

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 72

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 73

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 74

A. Larger –_99mTc decays to 99Tc_ emitting a 140 keV gamma. So its mass equivalent is greater by this amount.

Answer 75

B. Be a characteristic x-ray –The 140 keV x-rays could interact with lead to emit lead characteristic x-rays.

Answer 76

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 77

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 78

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 79

E. 1.0 B1 + Y3 = 0.2 + 0.8 MeV

Answer 80

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

Answer 81

1. The mass number of Y is A+1 (B) 2. The atomic number of Y is Z-1 (B) 3. The decay of X is accompanied by the emission of antineutrinos (A) 4. 0.511 MeV beta rays are emitted (A) 5. 7.6 MeV beta rays are emitted (B) 6. In any one disintegration a total of 4.86 MeV will be emitted (A)

Answer 82

C. monoenergetic photons and electrons with a spectrum of energies In beta decay, electrons are always emitted with a spectrum of energies.

Answer 83

1. 1.59 MeV gamma ray (B. sometimes occurs) 2. 40 keV characteristic x-ray (D. cannot be determined from the diagram) 3. 1.12 MeV beta minus (B. sometimes occurs) 4. 4.8 MeV antineutrino (B. sometimes occurs) 5. 3.75 MeV gamma ray (C. never occurs)

Answer 84

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 85

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 86

B. 1, 2 1. 99mTc emits 140 keV x-rays. So 130 kev scatter photon is possible. 2. Characteristic x-rays resulting form a photoelectic interaction with the lead container could be detected. 3. 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. 4. Auger electrons are emitted when characteristic x-rays are reabsorbed by the same atom. There is no such thing as a auger x-ray. 5. Annihaltion radiation occurs when a positron ane electron combine, producing two 0.511 MeV photons.

Answer 87

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 88

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 89

E. 1, 2, 3

Answer 90

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 91

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 92

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 93

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

Answer 94

1. 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. 2. 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 95

E. 18Fluorine –Is a cyclotron produced positron emitter used in PET.

Answer 96

B. Beta particle

Answer 97

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 98

1. Oxygen-15 (A. sample bombarded with charged particles in a cyclotron) 2. Radium-226 (B. naturaly occuring in uranium ore) 3. Technetium-99m (E. elution from the column of a generator) 4. Strontium-90 (D. separated from used reactor fuel rods) 5. Cobalt-60 (C. sample placed in neutron flux of a reactor)

Answer 99

1. Bombarding a sample with neutrons in a reactor (B. beta minus emission) 2. Bombarding with protons in a cyclotron (C. beta plus emission) 3. Bombarding with deuterons in a cyclotron (C. beta plus emission) 4. 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 100

A. Much longer than

Answer 101

B. Somewhat longer than

Answer 102

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 103

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 104

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

Answer 105

1. Secular equilibrium occurs when the decay constant of the daughter is slightly greater than the decay constant of the parent (B) 2. Transient equilibrium requires decay to a metastable state of the daughter (B) 3. In transient equilibrium the activity of the daughter is always less than that of the parent (B) 4. Equilibrium may exist if the half life of the daughter is shorter than that of the parent (A) 5. 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 106

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 107

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 108

C. Equilibrium has not yet occured Needs 4 half lives (in this case about 24 hours)

Answer 109

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 110

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

Answer 111

E. AMU; atomic mass unit

Answer 112

D. 135 5.0 x 10^9 Bq = 5x10^3 MBq 1 mCi = 37 MBq

Answer 113

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 114

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 115

B. 2 –Exp Rate= 6.29, 1 HVL and Exp Rate=3.15, 2 HVL and Exp Rate=1.57 12. 9 Rcm^2/(mCi-hr) x (1 m/ 100 cm)^2 x 19.5 mCi = 251.55 x 10^-4 Rm^2/hr 251. 55 x 10^-4 Rm^2/hr x (1/2m)^2 = 62.9 x 10^-4 R/hr = 6.29 mR/hr

Answer 116

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 117

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 118

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 119

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

Answer 120

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

Answer 121

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 122

A. 7 is the anode, 8 is the cathode The anode is the target (positive) and the filament is the cathode (negative).

Answer 123

C. full-wave rectified Four rectifiers are required for full-wave rectification.

Answer 124

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 125

1. Transformer (B. increases or decreases voltage) - used to step up or step down, (volts to kilovolts) 2. Miliammeter (E. measures tube current) - Measure the current in mA. It is connected in series with the x-ray tube itself. 3. 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 126

C. (N2/N1) x V1 N1/N2=V1/V2 V2 = N2/N1 x V1

Answer 127

E. none of the above Step-up increases the voltage, but decreases the current and power P=IV

Answer 128

A. small effective focus and large heat loading capacity

Answer 129

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 130

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 131

D. all of the above Three phase units have nearly constant voltage output resulting in higher beam current, average voltage, and dose rate.

Answer 132

1. Maximum kVp is 50 kVp (A. spectrum I) 2. Higher exposure rate (B. spectrum II) 3. Higher HVL (B. spectrum II) 4. Produced by 3 phase or constant potential generator (E. cannot be determined) 5. Will produce relatively less scatter in soft tissue (A. spectrum I). scatter increases with kVp in the diagnostic range.

Answer 133

E. 99:1 For diagnostic machines the efficiency is low, but for therapeutic is better at around 15-20%

Answer 134

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 135

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 136

1. An incoming 80 keV projectile electron (A) 2. An incoming 75 keV gamma ray (A) 3. An incoming 66 keV x-ray (B) 4. A 58 keV projectile electron (B) Have to have enough to get over the binding energy

Answer 137

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 138

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

Answer 139

1. 34 kV x-rays (B. Bremmstrahlung is emitted in a continous spectrum up to a maximum energy equal to that of the incident electrons.) 2. 26 kV x-rays (C. The characteristic x-ray energies possible from this atom are 29.3, 26.0 and 3.3 kV) 3. 40.7 kV x-rays (D. Greater than incident energy.)

Answer 140

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 141

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 142

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 143

C. kV across the tube A, B, and E affect the shape of the spectrum, but not the kVp

Answer 144

A. Inherent and added filtration

Answer 145

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 146

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 147

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 148

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 149

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 150

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 151

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 152

E. All of the above

Answer 153

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 154

E. patient skin dose

Answer 155

1. 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.) 2. The quality of the beam is independent of the tube current (mA) (A. TRUE) 3. Small changes in filament current affect the beam quality (B. FALSE) 4. 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 156

D. inversely proportional to the linear attenuation coefficient HVL = 0.693/ (linear attenuation coefficient)

Answer 157

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 158

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 159

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 160

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 161

E. have a lower effective energy

Answer 162

1. Betas (A. Ionizing elementary particles) 2. Heat radiation (D. Non-ionizing photons) 3. Visible light (D. Non-ionizing photons) 4. X-rays (C. Ionizing photons) 5. Ultrasound (E. Other)

Answer 163

A. 2 MHz ultrasound

Answer 164

B. 1&2 The radiation must have sufficient energy to ionize atoms in the film emulsion.

Answer 165

C. 2 eV Eλ=Eλ

Answer 166

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

Answer 167

C. Frequency

Answer 168

E. 3m c=λυ

Answer 169

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

Answer 170

D. is deflected by a magnetic field Charged particles are deflected by magnetic fields.

Answer 171

C. energy is directly proportional to frequency Velocity = frequency x wavelength Energy = planck’s constand x frequency

Answer 172

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 173

C. 0.693/x X = HVL, therefore μ=0.693/HVL

Answer 174

B. 6.93 cm HVL = 0.693/0.1

Answer 175

C. 0.50 I/Io = e ^ (-μx) or HVL = 0.693/0.0693 = 10 cm

Answer 176

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 177

B. Absorption. –The energy transferred to electrons, aka “absorbed dose” Scatter is the energy remitted as photons

Answer 178

B. Linear attenuation coefficient.

Answer 179

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

Answer 180

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 181

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

Answer 182

D. the linear attenuation coefficient

Answer 183

D. Z^3 E^-3

Answer 184

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 185

E. 13.86 cm 0.693/0.05 = HVL

Answer 186

E. 1 - e^ (-0.693n)

Answer 187

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

Answer 188

B. 10 TVL = 3.3 HVL (1/2)^3.3 = 1/10

Answer 189

B. 10 1% = 2 TVLs so at midline is 1 TVL

Answer 190

C. 7 The activity is increased by a factor of 4. Therefore 2 additional HVL’s are required.

Answer 191

E. 3.1 6/ 1.2 = 5 HVLs so 0.5^5 = 0.03 of I_o

Answer 192

D. bone Photoelectric process; Z^3/ E^3

Answer 193

1. Fat, muscle (A) 2. Muscle, bone (A) 3. Iodine, bone (B) 4. Fat, air (A) 5. Muscle, air (B) Note - it is for photoelectric that fat counterintuitively absorbs LESS than air. Please see attached graph

Answer 194

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

Answer 195

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 196

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 197

B. 8 Photoelectric is related to Z^3; so (14/7)^3

Answer 198

C. The probability increases rapidly with increasing energy.

Answer 199

D. A characteristic x-ray

Answer 200

D. scattered photon following photoelectric interaction

Answer 201

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 202

E. 30 keV and iodine Calculate out Z^3/E^3

Answer 203

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 204

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 205

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 206

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 207

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

Answer 208

C. Compton scattered photons.

Answer 209

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 210

E. None of the above. –Independent of Z –Decrease with increasing energy

Answer 211

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

Answer 212

B. 0-90 degrees with the incident photon

Answer 213

A. zero No energy is lost in classical scattering.

Answer 214

1. A photon may undergo a Compton interaction followed by a pair production interaction (A, TRUE) 2. A photon may undergo 2 consecutive Compton interactions (A, TRUE) 3. A photon may undergo a Compton interaction followed by a photoelectric interaction (A, TRUE) 4. A photon may scatter from an atom without losing energy (A, TRUE) 5. 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 215

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 216

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 217

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 218

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 219

B. Ei - 1.02 MeV.

Answer 220

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

Answer 221

C. 1.28 –Takes 2 x 0.51 MeV to create the electron-positron pair. –Thus the remaining 1.28 is deposited.

Answer 222

A. Two 0.51 MeV photons will be emitted

Answer 223

1. Chiefly responsible for loss of contrast in a diagnostic radiograph. (C. Compton) 2. Most probable at photon energies between 100 keV and 2MeV. (C. Compton) 3. No energy is transferred or locally absorbed. (A. Coherent scatter) 4. Probability of interaction increases as energy increases (D. Pair production)

Answer 224

1. Pair production (in Pb) - D 2. Compton (in H2O) - C 3. Photoelectric (in H2O) - B “A” is the total mass attenuation coefficient in water.

Answer 225

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

Answer 226

B. Annihilation photons

Answer 227

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 228

1. A photon may undergo a photoelectric interaction followed by a pair production interaction (B, FALSE) 2. A photon may undergo 2 consecutive photoelectric interactions (B, FALSE) 3. A photon may undergo a photoelectric interaction followed by a compton interaction (B, FALSE) 4. A photon may be totally absorbed by an atom (A, TRUE) 5. 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 229

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

Answer 230

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 231

1. A 35 keV x-ray incident on a vessel filled with iodine contrast material (A, photoelectric) 2. A 100 keV photon incident on muscle tissue (B, compton) 3. A 15 MeV photon incident on muscle tissue (B, compton) 4. A 18 keV photon incident on fatty breast tissue (A, photoelectric) 5. A 90 keV photon incident on bone (B, compton) 6. A 90 keV photon incident on a lead barrier (A, photoelectric) ; K edge of lead is 88 keV.

Answer 232

C. Linear attenuation coefficient. –CT number = 1000 x [(μmaterial – μwater)/μwater]

Answer 233

1, This is an example of Compton scattering (B, FALSE - this is photoelectric) 2. Characteristic radiation will be emitted (A, TRUE) 3. Auger electron emission is possible (A, TRUE) 4. The electron will be absorbed within 1 cm of its origin (A, TRUE)

Answer 234

C. 4 τ/ρ is proportional to Z3. 0.004 x (82/8)3 is approximately 4

Answer 235

B. decreases to about 3 MeV then rises again

Answer 236

B. 1 is greater than 2

Answer 237

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 238

B. a 2 MeV beta particle

Answer 239

C. 1, 3 & 4 only Photoelectric is a photon interaction.

Answer 240

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 241

C. Protons undergo exponential attenuation. –_Photons_ , NOT protons, are attenuated exponetially.

Answer 242

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 243

E. 1, 2, 3, 4

Answer 244

B. Interacts primarily with the oxygen in H2O. –A neutron is most likely to interact with hydrogen, which has a similar mass.

Answer 245

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

Answer 246

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 247

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 248

C. 1 –Optical Density = log (incident intensity/transmitted intensity) –OD = log(1/0.1)=log 10=1

Answer 249

1. D; H&D curves are plots of film density (log of opacity) versus the log of exposure 2. A, radioactive decay is a straight line on a semi-log scale.

Answer 250

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 251

A. Inversely proportional to focal spot size. –Geometric sharpness or edge gradient is reduced by minimizing magnification.

Answer 252

C. 1.5 –OD is additive –It is the log of (incident/transmitted) light intensity. 10^(0.75+0.75) = 10^1.5

Answer 253

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 254

C. Photoelectric interactions. –Remember, need to see bones.

Answer 255

C. Patient's body

Answer 256

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 257

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

Answer 258

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 259

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 260

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.

ROphex Old Q1-1000 Flashcards

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