ROPhex 2020 Flashcards
- A beam of electrons is traveling in a straight line. If an external magnetic field is introduced perpendicular to the direction of travel of the electrons, the electrons will _____.
A. speed up
B. slow down
C. be deflected to travel parallel to the magnetic field and perpendicular to their initial
direction of motion
D. be deflected to travel perpendicular to the magnetic field and perpendicular to their initial
direction of motion
E. be unaffected by the magnetic field
Key: D
- Which particle(s) cannot undergo bremsstrahlung production?
A. electron
B. positron
C. proton
D. neutron
E. All of the above can undergo bremsstrahlung,
Key: D
- Assume that 3.12 x 1010 6-MeV electrons are needed to treat a superficial tumor to 2 Gy. Given
that the charge of an electron is 1.602 x 10–19 C, what is the current of the beam hitting the patient
if the dose is delivered in 10 s? (1 A = 1 x 1012 pA,)
A. 0.5 pA
B. 5 pA
C. 50 pA
D. 500 pA
E. 5000 pA
Answer: D
I (amps) = Q (Coloumbs) / t (seconds)
= (3.12 x 1010 electrons) x (1.602 x 10–19 C/ electron) / (10 s)
= (5 x 10-9 C) / (10 s)
= 5 x 10-10 Amps or 5 x 102 pA
- In order to be stable, elements with high atomic numbers (Z > 20) must have more neutrons than
protons in the nucleus because _____.
A. the nuclear attractive force can only overcome coulomb repulsion between protons if there
is an excess of neutrons
B. neutron-proton coulomb attraction helps stabilize the nucleus
C. gravitational attraction between neutrons and protons stabilized the nucleus
D. the magnetic moments of neutrons and protons act as an attractive force
E. without an excess of neutrons, high-atomic-number elements would all undergo beta
minus decay
Key: A
- An isotope that decreases in activity by 5% per day has an approximate half-life of _____.
A. 1.5 d
B. 5 d
C. 10 d
D. 13.5 d
E. 20 d
Answer: D
A = A0e−λt
A/A0 = 0.95 = e−λt
ln (0.95) = -0.05 = -λ(1 day)
A/A0 = 0.5 = e−0.05 x t1/2
ln (0.5) = -0.693 = -(0.05/ day) x t1/2
t1/2 = 13.86
Alternatively; the concept is:
(1-0.05)n = (0.95)n =0.5
where n = the time in days
-
103; 46 Pd decays to 103; 45 Rh. What is its primary mode of radioactive decay?
A. alpha decay
B. beta minus decay
C. electron capture
D. gamma emission
E. photonuclear disintegration
Key: C
Atomic number Z decreased by 1 but mass number A is unchanged; meaning 1 proton -> 1 neutron.
Electron capture competes with beta plus decay, grabs a low lying K shell electron into the nucleus and results in P -> N, releases a neutrino and a gamma ray but NO POSITRON
Beta minus (-) decay results in an electron (-) and an ANTI-neutrino to balance the gain of a charge from N -> P
- Given a physical half-life of 3 hours and an effective half-life of 1.5 hours, what is the biological
half-life?
A. 1 hour
B. 1.5 hours
C. 2.25 hours
D. 3 hours
E. 4.5 hours
Key: D
1/Teff = 1/ Tphysical + 1/ Tbiologic
1/1.5 = 1/3 + 1/ Tbiologic
1/3 = 1/ Tbiologic
- The mass of a nucleus is _____ the sum of the masses of the free individual neutrons and protons
contained within the nucleus.
A. greater than
B. less than
C. equal to
D. There is not enough information to answer this question.
Answer: B
- _____ are common to both diagnostic x-ray tubes and megavoltage linear accelerators.
A. Focusing magnets and monitor chambers
B. A heated cathode and a metal target
C. A rotating anode and a transmission target
D. A high-voltage source and a glass enclosure tube
Key: B
- The advantage of using a bowtie filter in CT imaging is _____.
A. it absorbs low-energy x-rays
B. reduced scatter-to-primary ratio
C. reduced dose to the periphery of the patient
D. All of the above are true.
E. None of the above is true.
Key: D
- The output of an x-ray tube increases _____.
A. linearly with the tube voltage
B. linearly with the tube current
C. linearly with the thickness of x-ray target
D. with the square of the tube current
E. with the square of the filament current
Key: B
- An x-ray tube is engineered with _____ to manage the heating of the anode.
A. an angled anode
B. a rotating anode
C. materials with high heat storage capacity
D. All of the above are true.
E. None of the above is true
Key: D
- What characteristic x-ray energies are possible for molybdenum, given energy levels for its K, L,
and M shells of 20 keV, 2.5 keV, and 0.5 keV, respectively?
A. Continuous spectrum from 0.5 keV to 20 keV
B. Continuous spectrum from 2 keV to 19.5 keV
C. 20 keV, 2.5 keV, 0.5 keV
D. 19.5 keV, 17.5 keV, 2 keV
E. 22.5 keV, 20.5 keV, 3 keV
Key: D
- After the replacement of the ionization chamber in a linac, which machine parameters are expected
to change the most?
A. flatness
B. symmetry
C. output
D. beam profile
E. energy
Key: C
- The target in a dual-energy linear accelerator lies between the _____.
A. electron gun and the waveguide
B. bending magnets and the exit window
C. primary collimators and the flattening filter
D. flattening filter and the ion chamber
E. ion chamber and the upper/lower jaws
Key: B
- For a dual-energy linear accelerator, a target should have a _____ atomic number and _____ target thickness(es) for each photon energy.
A. high; the same
B. high; different
C. low; the same
D. low; different
Key: B
- Given the definitions in the figures of a linac couch below, a six-degree-of-freedom couch enables
_____ and _____ corrections that are unavailable on a standard, four-degree-of-freedom couch.
[see attached imaging]
A. Lat, Vrt
B. Lng, Yaw
C. Pitch, Yaw
D. Roll, Yaw
E. Pitch, Roll
Key: E
- The _____ in a medical linac generates the high power and short-duration pulses to run the
electron gun and the klystron.
A. power supply
B. modulator
C. magnetron
D. wave guide
E. accelerator tube
Key: B
The function of the modulator is to provide high voltage pulses to the microwave transmitter (klystron)
The magnetron is an oscillator where electrons are emitted from a hot cathode and whirl past resonant cavities which are part of the anode at speeds that generate microwave energy.
- Modern multileaf collimators utilize a tongue and groove design in order to reduce the _____.
A. transmission through the tip of the leaf
B. transmission between adjacent leaves on the same bank
C. transmission through the center of a leaf
D. collisions between leaves on opposite banks
E. collisions between leaves on the same bank
Key: B
- If a flattening filter designed for a 6-MV beam is accidentally applied to a 15-MV beam, the
magnitude of the horns at dmax will _____ and the dose rate per pulse will _____ compared to the
use of the proper 15-MV flattening filter.
A. increase; increase
B. increase; decrease
C. increase; remain the same
D. decrease; increase
E. decrease; decrease
Key: D
Horns start forming laterally at deeper depths due to scatter of the central beam, since the 6 MV filter is under-filtering the 15 MV beam, the central part of the isodose has more forward energy conserved and less scatter laterally to contribute to the horns.
- Arrange the following MV photon and MeV electron beams in order of greatest to least electron
current in the linac accelerating cavity for the same dose rate.
A. 9 MeV, 6 MV, 15 MV
B. 9 MeV, 15 MV, 6 MV
C. 6 MV, 9 MeV, 15 MV
D. 6 MV, 15 MV, 9 MeV
E. 15 MV, 6 MV, 9 MeV
Key: D
2 rules: First, generating electrons is always more efficient than generating photons. Second, HIGHER energy beams are more efficient to produce than lower energy beams.
- Your department has two MLC linacs, one with 3-mm wide leaves the other with 5-mm leaves. For
which treatment would the 3-mm MLC provide the greatest advantage?
A. small and irregular-shaped tumors at shallow depth
B. small and irregular-shaped tumors at large depth
C. large spherical-shaped tumors at shallow depth
D. large spherical-shaped tumor at large depth
Key: A
Resolution of the light field decreases with depth, so the advantage of the 3 mm MLC will be lost the deeper you go.
The benefit of a smaller MLC is to be more precise in design of the treatment field. This benefit is seen in more irregular tumors.
- _____ will absorb the highest dose when exposed to the same fluence of 100-kV photons.
A. Bone
B. Water
C. Muscle
D. Fat
E. All receive the same dose.
Key: A
At 100 MV, photoelectric interactions dominate with are proprtional to Z3 / E3
- _____ is the minimum thickness of copper needed to effectively block a 20-MeV electron beam,
given that Copper (ρ = 9 g/cm3) has an approximate total mass stopping power of 2.0 MeV cm2/g?
A. 0.06 cm
B. 0.6 cm
C. 1.1 cm
D. 3.6 cm
E. 5.1 cm
Answer: C
Use unit cancellation when in doubt:
20 MeV x g/ (2 MeV x cm2) x cm3 / (9 g) = 1.1 cm
- Which charged particle beam has the steepest distal dose gradient?
A. positron
B. electron
C. proton
D. alpha
E. carbon
Key: E
- _____ is the transmission of a 5-HVL block.
A. 0.3125%
B. 0.625%
C. 1%
D. 3.125%
E. 6.25%
Answer: D
(1/2)5 = 0.03125
- A 10-MeV photon undergoes Compton scatter. The minimum energy of the scattered photon is
equal to _____.
A. 1.022 MeV
B. 0.511 MeV
C. 0.256 MeV
D. 0.128 MeV
E. 0.064 MeV
Answer: C
The minimum energy of the scattered photon is obtained when it is scattered in 180◦. At high photon energies the minimum photon energy is close to 0.25 MeV and thus the maximum electron energy is close to hν − 0.25 MeV.
- _____ is the dominant photon interaction in human tissue from a photon beam in the therapeutic
treatment range of 4 to 25 MeV, and _____ affects the likelihood of an interaction.
A. Photoelectric effect; atomic number
B. Photoelectric effect; electron density
C. Compton scatter; atomic number
D. Compton scatter; electron density
E. Pair production; electron density
Key: D
- The ratio of mass attenuation coefficients (μ/ρ) for lead and water, (μ/ρ)Pb / (μ/ρ)water, is _____ at
0.1 MeV, _____ at 1 MeV, and _____ at 10 MeV, respectively:
A. >1; >1; >1
B. >1; <1; >1
C. >1; >1; <1
D. <1; >1; >1
E. >1; ~1; >1
Key: E
- Exposure (X) refers to _____.
A. ionization created in the patient due to charged particle irradiation, in C/kg
B. ionization created in air and only defined for photons, in C/Kg
C. ionization created in any medium and only defined for photons, in C/Kg
D. energy deposited per unit mass in air, in J/Kg
E. energy deposited per unit mass in any medium, in J/Kg
Key: B
- The radiant energy incident on a sphere of cross-sectional area dA is defined as the _____, and the
SI unit is _____.
A. fluence; l/m2
B. fluence; J/m2
C. fluence; J/m2s
D. energy fluence; 1/m2
E. energy fluence; J/m2
Key: E
- The loss of charged particle equilibrium for MV photon interactions within low-density lung tissue is worse for _____ photon energies and _____ field sizes.
A. higher; larger
B. higher; smaller
C. lower; larger
D. lower; smaller
Key: B
- The factor kQ _____, as defined in the Report of AAPM Task Group 51 on high-energy x-ray and electron beam dosimetry.
A. converts measured charge to dose
B. converts the chamber calibration factor from Cobalt-60 to your given beam energy
C. converts percent depth ionization to percent depth dose
D. corrects the measured charge for temperature/pressure, ion recombination, and polarity
E. corrects the depth from the point of measurement to the effective point of measurement
Key: B
- In the diagram below, which arrow is pointing to where the air is in a cylindrical ionization
chamber? (Refer to attached picture)
A. i
B. ii
C. iii
D. iv
E. None of the above is correct.
Key: B
Picture refers to a farmer chamber
- The charge measured by an ionization chamber is _____ the pressure of the gas in the chamber.
A. unrelated to
B. directly proportional to
C. inversely proportional to
D. proportional to the square of
E. proportional to the inverse square of
Key: B
All chamber readings have to be converted to -a reference temperature (22 C = 295 K) and pressure (760 mm Hg).
The correction factor increases if T increases, decreases if P increases.
- If the raw transmission values for radiochromic films exposed to 0 cGy and 250 cGy are 30,000 and 3000, respectively, what is the net optical density corresponding to 250 cGy?
A. 10.0
B. 4.5
C. 3.5
D. 2.4
E. 1.0
Key: E
Optical density (OD) = -log10 (I<span>1 </span>/ Io)
= -log10 (30,000/3,000)
= -log10 (10) = -1
NET OD = 1
- Treatment with a half field can be achieved with either an asymmetric collimator jaw or MLC.
Which technique is better?
A. asymmetric collimator
B. MLC
C. There is no difference.
D. This cannot be determined from information given.
Key: A
Less leak w/ collimator jaw vs the tongue/groove MLCs
