RAPHEX XI

Question 1

A beam of electrons is traveling in a straight line. If an external magnetic field is introduced per-pendicular 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

Answer 1

be deflected to travel perpendicular to the magnetic field and perpendicular to their ini-tial direction of motion
Charged particles traveling in a magnetic field experience the Lorentz force, which will act to curve the path of the electrons in the plane perpendicular to the magnetic field There is no effect on the electron speed. As combined MR-linacs get introduced into the clinic, the effects of the Lorentz force on the Compton electrons need to be considered in order to calculate the correct dose distribution.

Question 2

Which particle(s) cannot undergo bremsstrahlung production? A. electron B. positron C. proton D. neutron E. All of the above can undergo bremsstrahlung,

Answer 2

neutron

Question 3

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
ANSWER
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

Answer 3

the nuclear attractive force can only overcome coulomb repulsion between protons if there is an excess of neutrons

Question 4

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 4

13.5 d

remember I/Io is 0.95

Question 5

Given a physical half-life of 3 hours and an effective half-life of 1.5 hours, what is the biological half-life?

Answer 5

3 h
1/1.5 - 1/3 = 1/3

Question 6

The mass of a nucleus is _____ the sum of the masses of the free individual neutrons and protons contained within the nucleus

Answer 6

less than

Question 7

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.

Answer 7

reduced dose to the periphery of the patien

Question 8

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

Answer 8

B. bending magnets and the exit window
Electrons from the electron gun are accelerated through the waveguide, transported through bending magnets and incident upon the target. The photons from the interactions in the target then exit through the exit window and are collimated by the primary collimators, traverse the flattening filter and ionization chamber, and then are further collimated by the upper and lower jaws.

Question 9

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

Answer 9

high; different
A high atomic number will increase the x-ray production frequency due to increased bremsstrahlung cross section. Different target thicknesses are used for each desired photon energy due to the fact that the energy of the incident electrons will be different.

Question 10

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

Answer 10

. decrease; increase
As the x-ray beam energy increases, the fluence distribution will be more forward peaked. So, higher-energy beams need thicker flattening filters. Therefore, if a 6-MV flattening filter is used for the 15-MV beam, the magnitude of the horns will decrease due to insufficient attenuation along the CAX relative to off-axis points, and the dose rate will increase due to a decreased filter thickness.

Question 11

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

Answer 11

6 MV, 15 MV, 9 MeV
For electron beam therapy, all of the electrons from the accelerating cavity are used for therapy. For photon therapy, however, electrons must be converted into photons in the target via bremsstrahlung interactions, and only a fraction of the electron energy is so converted. Therefore, much higher accelerator currents are needed for photon therapy than for electron therapy. The efficiency of bremsstrahlung x-ray production increases with both atomic number of the target and energy of the electron beam. So a 15-MV photon beam requires lower electron current than does a 6-MV beam. The conversion efficiency for bremsstrahlung production is approximately 1% at 100 kVp and 70% at 20 MeV

Question 12

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

Answer 12

A. small and irregular-shaped tumors at shallow depth
For small, irregular tumors more precise field shaping is clearly more important than for larger tumors, but at larger depths beam scattering largely negates the sharper field definition of small MLC leaves.

Question 13

Which charged particle beam has the steepest distal dose gradient? A. positron B. electron C. proton D. alpha E. carbon

Answer 13

carbon
According to the Bethe mass stopping power equation, energy deposition for charged particles increases as the square of charge and inversely squared with speed as the particle’s energy decreases. The massive rise in stopping power for carbon ions give it a much sharper Bragg peak falloff than the other listed particle

Question 14

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

Answer 14

E. >1; ~1; >1
The mass attenuation coefficient is large for low energies and high-atomic-number media because of the predominance of photoelectric interactions under these conditions. One MeV is in the Compton range, the (μ /ρ) of lead and water do not differ greatly, since Compton interaction is independent of atomic number. At 10 MeV, the dominance of pair production occurs, especially for high-atomic-number medi

Question 15

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

Answer 15

B. ionization created in air and only defined for photons, in C/Kg Exposure refers specifically to ionization in air caused by photons less than 3 MeV

Question 16

As the size of the active volume of the ionization chamber increases, the amount of charge generated for a given dose _____. A. increases B. decreases C. remains the same D. cannot be determined

Answer 16

increases
A larger-volume chamber will generate more charge for the same given dose since there is a larger volume of gas being irradiated.

Question 17

Which of the following is particularly important to consider when using an ion chamber for a flattening-filter-free photon beam measurement compared to measurement of a flattened beam? A. temperature and pressure B. polarity correction C. stem effect D. leakage current E. ion recombination

Answer 17

ion recombination
According to the addendum to the AAPM Task Group 51 calibration protocol, the ion-recombination correction is larger for flattening-filter-free photon beams compared to flattened beams

Question 18

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

Answer 18

directly proportional to
As the pressure increases, the density of the air in the chamber will increase, resulting in a corresponding increase in the charge measured.

Question 19

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. C. D. E.
4.5 3.5 2.4 1.0

Answer 19

1.0
The optical density is a measure of the amount of light that can pass through exposed film, which can be related to the dose received by the film. The optical density is defined as OD = log (I0 / It), where I0 and It are the transmission values measured before and after an exposure, respectively. For a 250-cGy exposure, OD = log (30,000/3000) = log (10) = 1.0.

Question 20

A single AP 6-MV photon beam traverses 4 cm of lung tissue (density = 0.25 g/cc). The MU are calculated without homogeneity corrections to a point beyond lung. The delivered dose to that point will be _____ than calculated by approximately _____% A. less; 1–4 B. less; 5–8 C. greater; 1–4 D. greater; 5–8 E. greater; 10–12

Answer 20

greater; 10–12
The heterogeneity correction increases by approximately 3%/cm in healthy lung tissue for a 6-MV x-ray beam. Traversing 4 cm of lung would increase dose beyond lung by approximately 10 to 12% if the MUs were calculated, assuming homogeneous water equivalence.

Question 21

A patient is being treated AP/PA isocentrically with 6-MV x-rays. The actual patient separation is 1 cm smaller than the separation used for the MU calc. What is the approximate dose error to midplane depth? A. There is no dose error. B. 1.5% C. 5% D. 7.5% E. 10%

Answer 21

1.5%
A rule of thumb for photon attenuation is about 3%/cm for 6-MV photons in tissue. Decreasing the separation with the same MU will result in an overdose.

0.5 cm on either side

Question 22

Two beam arrangements are being considered for partial breast irradiation. The advantage of four non-coplanar photon fields over a technique using two coplanar photon mini-tangents with an en-face electron beam is _____. A. less volume of breast tissue irradiated B. decreased risk of collision with the patient C. improved coverage for targets deep within the breast D. All of the above are true. E. None of the above is true.

Answer 22

improved coverage for targets deep within the breast
The photon mini-tangents with en-face electron beam will generally irradiate less normal breast tissue and is more likely to avoid collision issues. The non-coplanar beams could collide with patient arms, head, or the couch. However, for deep targets, the photon-only technique should result in better target coverage and less irradiation of the heart and lungs.

Question 23

Defined from the perspective of a person standing at the end of the couch looking toward the gantry, the gantry angle increases from 0 to 360 degrees as the gantry rotates clockwise, while the couch angle increases from 0 to 90 degrees as the couch rotates to the left and decreases from 360 to 270 degrees as the couch rotates to the right. Which of the following gantry and couch angle combinations is likely to pose a collision issue for a brain patient? A. Gantry = 0; Couch = 90 B. Gantry = 50; Couch = 90 C. Gantry = 5; Couch = 270 D. Gantry = 50; Couch = 270 E. Gantry = 310; Couch = 270

Answer 23

Gantry = 50; Couch = 270
Usually when the couch kick is toward the gantry rotation and the gantry is beyond 30 degrees, the field falls into a collision zon

Question 24

Beam quality, as measured by %dd(10) of flattened beams, is approximately _____ compared to flattening-filter-free (FFF) beams. A. 20% higher B. 5% higher C. the same D. 5% lower E. 20% low

Answer 24

B. 5% higher
Flattened beams are slightly harder than flattening-filter-free beams due to beam hardening in the flattening filter

Question 25

A patient is treated with two adjacent fields on a linac (100 cm SAD). Both fields are SAD setups to a depth 10 cm (90 cm SSD). The first and second fields have symmetric collimator settings of 1616 cm2 and 2020 cm2, respectively. What is the gap (in cm) required on the skin for these fields to intersect at a depth of 5 cm? A. 0.70 B. 0.75 C. 0.80 D. 0.90 E. 1.00

Answer 25

0.9

remember that depth given is to SAD and needs to be corrected to depth to surface

Question 26

In the right lateral DRR below, hash marks represent 1 cm at the isocenter, which is placed at midplane depth (100 cm SAD). To avoid beam divergence at point P, which would be near the edge of the field, approximate gantry angle offsets of _____ degrees from lateral and couch offsets of _____ degrees from center are needed

Answer 26

C. 3; 1.7
The gantry angle and couch angle offsets will be determined by the anterior-posterior and superior-inferior distances between the isocenter and divergence point in the plane of the isocenter, respectively. The angles are calculated as:
gantry angle offset = tan–1 (5/100) = 2.9 degrees, and couch angle offset = tan–1 (3/100) = 1.7 degrees.

Question 27

A 1010 cm2 6-MV photon field is incident upon a flat phantom. On the plane perpendicular to the beam central axis at dmax, the dose to a point 1 cm outside of the edge of the collimated field is of the order of _____% of the field maximum dose. A. 0.05% B. 0.5% C. 5% D. 50% E. 90%

Answer 27

5%

Question 28

Standard linear accelerator MLCs are NOT typically used for IMRT electron therapy primarily because _____. A. the leaves are not thick enough to attenuate the electrons B. electron collimators should be close to the patient’s skin surface in order to decrease the dose penumbra
ANSWER
C. electron dose rates are typically too high for use with MLCs D. there would be too much electron dose leakage between MLC leaves E. All of the above are true.

Answer 28

A 1010 cm2 6-MV photon field is incident upon a flat phantom. On the plane perpendicular to the beam central axis at dmax, the dose to a point 1 cm outside of the edge of the collimated field is of the order of _____% of the field maximum dose. A. 0.05% B. 0.5% C. 5% D. 50% E. 90%

Question 29

A patient is set to 100 cm SSD for treatment with an electron beam. One cm of bolus is then added to cover the entire field. The SSD that should be used to calculate the MUs to deliver the desired dose is: _____. A. 99 cm B.
99.5 cm ANSWER
C. 100 cm D. 101 cm E. 101.5 cm

Answer 29

99 cm
The electrons first encounter tissue-equivalent material at an SSD of 99 cm. By the time they reach the skin, they have already traversed a centimeter of tissue-equivalent material.

Question 30

What are the four solid square sections in the corners of the phantom testing?

Answer 30

contrast
These sections are testing contrast using different thicknesses of material to attenuate the imaging beam and creating varying contrast.

Question 31

Under what conditions are temperature-pressure corrections most important when doing ion chamber dosimetry? A. in the summer when an air conditioning system is malfunctioning B. in the winter when a heating system is malfunctioning C. during extreme weather conditions, such as stormy weather D. at extremely high altitudes, such as in Denver, Colorado

Answer 31

at extremely high altitudes, such as in Denver, Colorado
Room temperature is rarely more than 10 degrees above or below normal, which is only about a 3% correction. Similarly, barometric pressure, even during hurricane conditions, is rarely more than 3% low. But at 5000 ft altitude, the pressure can easily be >20% below sea level values

Question 32

According to the AAPM Task Group 142 report, all of the following are daily QA checks except _____. A. x-ray output constancy B. light field/radiation field coincidence C. room lasers D. door interlock E. collimator size indicator

Answer 32

light field/radiation field coincidence
According to AAPM Task Group 142, light field and radiation field coincidence is a monthly, not daily, check.

Question 33

the monthly QA check should be within _____ of the baseline value. A. 0.5% B. 1% C. 2% D. 3% E. 5%

Answer 33

2%
The x-ray output constancy measured by a physicist during monthly linac QA should be within 2% of the baseline value. The daily x-ray output constancy, often measured by a therapist with a less precise device, should be within 3% of the baseline value.

Question 34

2 parts of DICOM

Answer 34

header and data

Question 35

If your treatment planning system and treatment delivery system are from different vendors, what DICOM-RT files need to be transferred out of the treatment planning system to effectively treat a patient? A. RT-Plan B. RT-Dose C. RT-Registration Object D. All of the above need to be transferred. C. None of the above need to be transferred.

Answer 35

RT-Plan
A DICOM RT-Plan file needs to be exported from the treatment planning system in order to treat the patient, as it holds information about the planned beam geometry and MU. Other files that would be helpful include RT-Image (e.g., DRRs), RT-Structure Set (contours), and the planning CT scan (for planning CT to CBCT registration for setup verification

Question 36

ICRP and NCRP reports contain estimates of excess cancer risk for radiation (e.g., 5% excess lifetime risk per 1 Sv) for the general population. Are these coefficients suitable for estimating the radiation-induced cancer risk of radiotherapy patients? A. Yes, they are. B. No, the average age of radiotherapy patients is significantly higher than in the general population and lifetime risk is, therefore, lower.
ANSWER
C. No, radiotherapy patients are known to have significantly more mutations that make them more radiosensitive and lifetime risk is, therefore, higher.
D. No, the coefficients only work for low x-ray energies. E. No, the coefficients were estimated from data on foreign populations, whose average weight is significantly different from the U.S. population.

Answer 36

B. No, the average age of radiotherapy patients is significantly higher than in the general population and lifetime risk is, therefore, lower.
The lifetime cancer risk decreases significantly with age at exposure.

Question 37

The instantaneous dose rate behind a secondary barrier is measured to be 3.0 mR/hr. How many TVLs of shielding are needed to decrease the dose rate to 1.0 mR/hr? A. 0.3 B. 0.5 C. 1.0 D. 2.0 E. 3.0

Answer 37

0.5 The dose rate needs to decrease by a factor of one third. 1/3 = (1/10)n or n = –log(1/3) = 0.48.

Question 38

The fraction of ultrasound waves reflected at the interface between two different media depends on the difference in the _____ and the _____ between the two media. A. density; atomic number B. density; speed of sound C. density; temperature D. atomic number; speed of sound E. atomic number; temperature

Answer 38

density; speed of sound
Ultrasound wave reflections or echoes are caused by variations in acoustic impedance of materials on opposite sides of the interfaces. The acoustic impedance (Z) of a medium is defined as the product of the density of the medium and the speed of ultrasound waves in the medium.

Question 39

Which of the following is FALSE regarding an electronic portal imaging device (EPID)? A. An EPID makes it possible to view the portal images instantaneously. B. Portal images can be stored on computers for later viewing and archiving. C. Current EPIDs provide submillimeter image resolution. D. An EPID can be used for patient-specific IMRT QA. E. Amorphous silicon has low resistance to radiation damage.

Answer 39

Amorphous silicon has low resistance to radiation damage.
EPID’s image resolution is less than 0.5 mm. EPID dosimetry can be used for IMRT QA patient-specific dose distribution verification. Amorphous silicon is used because of its high resistance to radiation damage.

Question 40

A disadvantage of using filtered backprojection algorithms in the presence of high-atomic-number materials, such as dental artifacts, for CT reconstruction is _____. A. streak artifacts B. aliasing artifacts C. susceptibility artifacts D. All of the above are true. E. None of the above is true.

Answer 40

streak artifacts

Question 41

What is the major advantage of using kV cone-beam CT rather than gantry mounted kV orthogonal planar radiographs? A. Image acquisition and registration is faster. B. Images can be acquired at large couch angles. C. Images can be aligned to a soft tissue target. D. All of the above are true. E. None of the above is true.

Answer 41

Images can be aligned to a soft tissue target

Question 42

What is/are the benefit/s of using optical surface imaging (OSI) over orthogonal kV x-rays for setup verification of a supine breast patient? A. The patient surface is a better surrogate than bones for breast treatment. B. OSI provides a larger field of view so that the patient’s ipsilateral arm position can be checked and aligned.
ANSWER
C. OSI does not utilize ionizing radiation. D. All of the above are true. E. None of the above is t

Answer 42

D. All of the above are true.
OSI uses light, not ionization radiation, gives information about the topology of the breast, and has a large field of view that will allow you to focus on the arm position, where tangent field portal images would not.

Question 43

Precise placement of the infrared reflective box is critical for _____ respiratory gating. A. amplitude-based B. phase-based C. time-based D. image-based E. All of the above are true.

Answer 43

amplitude-based
Amplitude-based respiratory gating requires precise placement of the respiratory gating surrogate because the distance from the infrared camera affects the amplitude of the surrogate signal.

Question 44

_____ would be expected to have the smallest amount of modulation in the fluence. A. An IMRT plan to a target overlapping with a previously irradiated target B. An IMRT prostate plan for a patient with bilateral hip prostheses C. Single brain metastasis treated with dynamic conformal arcs (DCA) D. Paraspinal SBRT treated with VMAT E. Conventionally fractionated VMAT head and neck irradiation

Answer 44

Single brain metastasis treated with dynamic conformal arcs (DCA)
Plans with concave dose distributions, sharp dose gradients adjacent to organs at risk, or limited beam arrangements tend to have higher levels of modulation. DCA plans have no modulation, as the leaves move to conform to the projection of the PTV as the gantry moves

Question 45

You cannot verify the profile across an IMRT field using a single ionization chamber at a fixed position in a 3D water tank because the _____. A. typical IMRT dose rate is too high B. typical MLC openings are too small C. beam quality in an IMRT field is different from that in an open field D. IMRT dose rate to any point in the field is time-dependent E. None of the above is true.

Answer 45

IMRT dose rate to any point in the field is time-dependent
The IMRT field would have to be delivered once for each measurement point, which is too time consuming (a detector array inside the tank would work).

Question 46

Which of the following is FALSE when using electron beams for intraoperative radiotherapy (IOERT)? A. minimal shielding is required B. high dose rate C. tumor beds are easily accessible to electron cones D. All of the above are false. E. None of the above is false

Answer 46

tumor beds are easily accessible to electron cone
Mobile IOERT units provide electron beams at energies of 6, 9, and 12 MeV at high dose rates (about 10 Gy/min). These units treat tumors that are a few cm deep with short treatment time (2 to 5 minutes). They require minimal operating room (OR) shielding for electron beams. Cone and linac can limit access to tumor bed

Question 47

_____ isotope is produced in a cyclotron. A. F-18 B. Rn-222 C. Tc-99m D. Ir-192 E. Co-60

Answer 47

Medical cyclotrons are used to produce typically short-lived radioisotopes that are proton-rich. F-18 can be produced in a cyclotron by bombarding O-18 with protons. Rn-222 is part of the naturally occurring uranium decay series. Tc-99m is the decay product of molybdenum-99, which is produced in nuclear reactions. Co-60 and Ir-192 are also reactor-produced, created from the bombardment of Co-59 and Ir-191, respectively, with neutrons.

Question 48

A patient is 160 cm tall. What is the minimum distance from the linac source that the patient needs to be positioned for a total body irradiation (TBI) treatment with a collimator angle of 0 degrees and a maximum jaw setting of 4040 cm2 at 100 cm SAD? A. 100 cm B. 160 cm C. 300 cm D. 400 cm E. 450 cm

Answer 48

The minimum distance can be determined using similar triangles, 40/100 = 160/x, so x = 400 cm. Typically a 45-degree collimator is used, in which case the necessary distance would be reduced by approximately 2.

Question 49

Which detector can be used for small-field data collection required for IMRT field sizes  33 cm2? A. G-M counter B. Large-volume ion chamber C. well chamber D. diode E. 0.6-cc Farmer chamber

Answer 49

diode
G-M counters and large-volume ion chambers are used mainly for radiation surveys, not for linac beam output. A well chamber is used to measure the dose rate from sealed sources. A Farmer chamber is used to measure linac beam output, but the size of the sensitive volume (0.6 cc) is too large for small field dosimetry. A diode can be used for such a measurement.

Question 50

_____ is a typical dose rate to the periphery of the prostate immediately following an I-125 prostate seed implant with a prescribed total dose of 144 Gy. A. 0.1 cGy/hr B. C.
1 cGy/hr 7 cGy/hr
D. 70 cGy/hr E. 700 cGy/hr

Answer 50

7 cGy/hr
For a I-125 implant with a prescribed dose of 144 Gy, Dtotal = Ḋ0 Tav = Ḋ0 1.44 T1/2. 144 Gy = Ḋ0 1.44  59.4 days  24 hours/day, Ḋ0 = 0.07 Gy/hr.

Question 51

Which of the following statements concerning CT and MRI imaging in HDR treatment planning is FALSE? A. MRI provides better visualization of the soft tissue anatomy. B. The shape of the applicator is more accurate in the MRI scan. C. The CT scan will provide the electron density of the target. D. CT and MRI scans can be registered to one another. E. Only a subset of HDR applicators are considered to be MR-compatible.

Answer 51

The shape of the applicator is more accurate in the MRI scan.
MRI provides superior soft tissue imaging, but MR images have larger geometric distortions than CT images.

Question 52

Given the choice between two isotopes for a prostate implant, the use of the source with a lower energy will provide _____ homogeneity for the same number of seeds per unit volume. A. more B. less C. the same amount of D. Not enough information is given to answer this question.

Answer 52

A lower energy means that the dose will be deposited closer to each particular source.

Question 53

_____ should be performed daily for an HDR afterloader. A. Source positioning accuracy B. Source strength measurement C. Printing new decay tables D. Testing the emergency hand crank E. All of the above are tru

Answer 53

Source position accuracy
HDR QA procedures are specified in the report from AAPM Task Group 56. Daily checks include checks of the audio-visual systems, radiation indicator lights, and interlocks, as well as checks of the source positioning accuracy. Source strength measurements and the printing of new decay tables are done at the time of the source exchange

Question 54

The treatment time for a high-dose-rate brachytherapy plan with an Iridium-192 source was 300 seconds using a nominal source strength of 10 Ci. The source strength at the time of treatment was 6.71 Ci. How long is the treatment? A. 201 seconds B. 300 seconds C. 447 seconds D. 600 seconds E. This cannot be determined

Answer 54

447 seconds 10/0.671  300 s = 447 s

Question 55

A limitation of HDR treatment of superficial lesions with a multi-channel surface applicator is _____. A. treatment depth B. irregular surface contour C. internal scatter to adjacent organs at risk D. sensitivity to patient movement E. All of the above are true.

Answer 55

treatment depth
A multi-channel surface applicator would be placed on the patient’s surface. It is flexible enough to conform to an irregular surface contour and is attached to the patient, so it should have little sensitivity to patient movement. Superficial treatment with HDR is limited by treatment depth, however, as prescribing too deep would cause a high skin dose.

Question 56

A HDR treatment is planned to deliver 600 cGy to 0.5 cm into the vaginal tissue using a 2.5-cm diameter cylinder. If a 3-cm diameter cylinder was used by mistake, _____ is the approximate dose delivered to the prescription depth in the middle of the cylinder length. A. 459 cGy B. 525 cGy C. 600 cGy D. 684 cGy E. 780 cGy

Answer 56

525 cGy
Use of a larger cylinder will decrease the dose to the prescription point. For a line source, the dose falloff goes as 1/r when it is close to the line source. In the plan, the distance from the source to the prescription point was (2.5 cm / 2 + 0.5 cm) = 1.75 cm. In the delivery, the distance from the source to the prescription point was (3 cm / 2 + 0.5 cm) = 2 cm. Therefore, the dose that is actually delivered is 600 cGy  (1.75/2) = 525 cGy

Question 57

A given permanent implant delivers a total prescribed dose of 100 Gy. How much dose is delivered if the implant is removed after two half-lives? A. 25 Gy B. 50 Gy C. 63 Gy D. 75 Gy E. 82 Gy

Answer 57

75 Gy
For a permanent implant, the total (prescribed) dose, Dtotal, is delivered after complete decay of the sources (i.e., t ≫ T1/2 ). Permanent implant dose can be calculated:
Dtotal = 1.44 Ḋ0 T1/2 or Dtotal = Ḋ0 Tav , where Ḋ0 is the initial dose rate and Tav is the average life (1.44T1/2 ).
The Dtotal (Ḋ0 Tav) is 100 Gy for this permanent implant.
After two half-lives the dose rate is decreased to 1⁄4 Ḋ0 , and the undelivered permanent dose is 1⁄4 Ḋ0 Tav (25 Gy). The delivered dose is 75 Gy.

Question 58

The air kerma rate at 4 meters from a brachy source is 0.25 μGy h–1. What is the source strength of this source? A. 0.25 U B. 0.5 U C. 1 U D. 2 U E. 4 U

Answer 58

In this case, Sk = 0.25 μGy/h  (4 m)2 = 4 U.

Question 59

In brachytherapy, high specific activity is a preferred physical property because it affects the _____. A. source size B. photon energy C. HVL D. All of the above are true. E. None of the above is true.

Answer 59

source size

Question 60

ICRU Report 78 on “Prescribing, Recording, and Reporting Proton-Beam Therapy” defines the length of the spread-out Bragg Peak to be the distance in water between the _____% proximal and distal doses on the depth dose curve. A. B. C. D.
10 50 75 90
E. 100

Answer 60

90 Another phrase for the spread-out Bragg Peak length is the range modulation.

Question 61

Compared to proton therapy, carbon ions have all of the following properties EXCEPT _____. A. higher mass and charge B. lower skin and entrance dose C. higher OER (oxygen enhancement ratio) D. higher LET (linear energy transfer) E. sharper lateral dose penumbr

Answer 61

higher OER (oxygen enhancement ratio)
Because of carbon’s higher mass and charge, there is less lateral scatter, higher stopping power (LET), and a more enhanced Bragg peak. Higher LET also results in higher RBE and LOWER OER.

Question 62

Intensity-modulated proton therapy (IMPT) _____. A. uses apertures and compensators to shape the beam B. uses an MLC to modulate the intensity C. uses changing magnetic fields to scan the beam D. All of the above are true. E. None of the above is true.

Answer 62

uses changing magnetic fields to scan the beam
Pencil beam scanning uses magnetic fields to scan the beam while the proton delivery system changes the energy.

Question 63

Range uncertainty in proton therapy arises because _____. A. the calculation algorithms for proton beams are not as accurate as those for x-rays B. the output of the proton accelerator is variable C. patient setups in proton centers lack the precision of volumetric imaging D. Hounsfield units do not map perfectly to proton stopping powers E. range uncertainty is only a problem for photon radiotherapy

Answer 63

D. Hounsfield units do not map perfectly to proton stopping power
Different compounds can result in the same HU but slightly different stopping powers. Variable setups have similar effects, as in IMRT, and similar PTV margins are used. In the beam direction, a perfectly aligned beam will always stop at the same point, but that point is usually a few percent (typically 2 to 4%) of the range in water. The output stability of the proton accelerator is comparable to a linac.

Question 64

Comparing pencil beam scanning (PBS) protons to passive scattering protons, which of the following are true? A. PBS results in sharper penumbra. B. PBS allows more conformality on the proximal side of the target. C. PBS is less sensitive to motion. D. PBS uses compensators. E. None of the above is true.

Answer 64

PBS allows more conformality on the proximal side of the target.
With passive scattering, a narrow proton beam is scattered over a large area in order to cover the target volume. A compensator is used in order to produce conformality on the distal side of the target, and apertures are used to shape the beam in the plane perpendicular to the beam axis. In pencil beam scanning (PBS), the proton beam is magnetically scanned across the target volume, delivering dose one point at a time. Since different energies can be used as the beam scans to deliver dose in different planes, PBS can deliver conformality both on the distal and proximal sides of the target without using compensators, and it can deliver lateral conformality without the use of apertures. However, the lack of an aperture does result in a worse penumbra than that seen for passive scattering. PBS is also more sensitive to patient motion since dose is delivered to the patient one spot at a time, unlike passive scattering, which delivers dose to the entire volume simultaneously.