RAPHEX 2001 T1. All of the following are true regarding Percentage Depth Dose (PDD) except: A. Increases with increasing energy. B. Depends on field size. C. Is the dose at depth expressed as a % of the dose at dmax. D. Decreases with increasing SSD. E. Decreases as depth increases. T2 For a 10 MV photon beam at 10 cm depth, the TMR will be _______ the PDD at 100 cm, for the same collimator setting. A. Greater than B. Less than C. Approx. equal to DATA FOR NEXT 3 QUESTIONS Depth (cm) TMR 8x8 TMR 20x20 PDD 8x8 PDD 20x20 5 0.924 0.937 86.2 87.9 7.5 0.847 0.880 76.0 79.0 10 0.772 0.818 66.4 70.3 15 0.632 0.697 50.2 55.4 OUTPUT at dmax in tissue 8x8 20x20

At 100cm SSD 0.985 1.055 At 100cm SAD 1.018 1.089

T3. The MU setting to deliver 180 cGy at a depth of 5 cm, at 100 cm SSD, for a field of equivalent square 8x8 cm is _____ MU. A. 106 B. 153

C. 198 D. 204 E. 212 T4. For the field in the previous question the dose at 10 cm depth is _____ cGy. A. 115 B. 139 C. 143 D. 170 E. 185 T5. The MU setting per beam to deliver 250 cGy total to the isocenter via parallel opposed, 23x18 cm whole brain fields, for a patient separation of 15 cm is _____ MU. A. 130 B. 156 C. 182 D. 208 E. 265 T6. If 150 cGy is delivered at a depth of 7 cm, at 100 cm SSD, using a field for which the PDD is 65%, the dose from that field at dmax is _____ cGy. A. 65 B. 130 C. 195 D. 215 E. 23l T7. Parallel opposed AP/PA mantle fields are treated isocentrically. Using the data below, for a total dose of 4000 cGy at the isocenter, the dose to the neck at midplane is _____ cGy. (Ignore off-axis factors.)

AP PA Beam axis SSD =92 SSD =92 thickness = 16 cm Neck SSD = 93 SSD =97 thickness = 10 cm Depth (cm) 3 5 7 8 TMR 0.987 0.945 0.901 0.879 A. 3721 B. 3950 C. 4000 D. 4300 E. 4613 T8. For 24x24 cm, parallel opposed isocentric 6 MV photon beams, the patient’s AP thickness can be up to _____ cm, if the maximum tissue dose is not to exceed 110% of the dose at the isocenter. A. 10 B. l5 C. 25 D. 32 E. 37 T9. If 4000 cGy is delivered to a patient’s mediastinum via AP/PA fields, the lowest total dose at dmax is obtained using: A. Co-60, 80 cm SSD. B. 6 MV photons, isocentric fields. C. 6 MV photons, 100 cm SSD. D. 18 MV photons, 100cm SSD. E. 18 MV photons, isocentric fields. T10. If the SSD of a photon beam is increased from 100 cm to 150 cm, the PDD at 10 cm depth will: A. Increase less than 2%. B. Increase more than 2%. C. Remain the same. D. Decrease less than 2%.

E. Decrease more than 2%. T11. The equivalent square of a 6x30 cm photon beam is a square CxC cm, where C is: A. Closer to 30 than 6. B. Equal to the square root of(6x30). C. The side of the square field which has the same PDD as the rectangular field. D. Is approximately equal to the (area / perimeter) for the rectangle. T12. From the data below, calculate the MU setting for the lateral fields of the three-field plan shown. Wedge factor (WF) = 0.58 TMR=0.782 Output = 0.90 cGy/MU 180 cGy / fraction at isocenter Beams are weighted to deliver equal Doses at the isocenter. A. 49MU B. 85MU C. 120 MU D. 147 MU E. 162 MU

T13. When photon beams are wedged, the “wedge angle” is: A. The angle between the isodose curve at d = 10 cm and the beam axis. B. The angle of the metal wedge itself. C. The angle through which the isodose curve at d = 10 cm is turned by the wedge. D. Half the hinge angle. T14.

During treatment of the wedged pair in the diagram, the wedge for field 2 was accidentally omitted, but the planned MU were delivered. The consequence is: Point A Isocenter Point B A. Cold Same Hot B. Hot Hot Hot C. Hot Same Cold D. Cold Cold Cold E. Hot Cold Hot T15. Which of the following treatment techniques would not deliver a homogeneous dose to the volume V shown below?

A. Two posterior oblique wedged fields. B. A posterior 180° arc pinned at the center of the volume. C. A posterior open field and two lateral wedged fields. D. A posterior 180° arc, split into two 90° wedged arcs (thick ends post). E. An anterior field and two posterior oblique wedged fields. T16. A patient’s skin dose is increased in a megavoltage photon beam by all of the following except: A. increasing the beam energy. B. Adding a 1 cm lucite blocking tray. C. Increasing the field size. D. Adding bolus. E. Increasing obliquity. T17. The purpose of a “beam spoiler” in a photon beam is to: A. Increase dose in the build up region. B. Reduce the PDD at 5 cm depth. C. Filter out scattered electrons, to reduce the skin dose. D. Reduce the energy of a photon beam. E. Increase the TMR beyond dmax. T18. Tissue compensating filters used in megavoltage beams: A. Eliminate skin sparing. B. Must be made of material with the same density as tissue. C. Should be placed as close as possible to the patient’s skin. D. Have an advantage over the use of bolus as a missing tissue compensator.

T19. All of the following are techniques used to improve dose homogeneity at the junction between adjacent or orthogonal photon fields except: A. Using half-blocked beams to remove divergence. B. Calculating a gap so that the beams match at a defined depth. C. Using gantry, couch or collimator rotation to remove divergence. D. Shifting the junction during a course of treatment. E. Overlapping coplanar fields on the skin. T20. Adjacent pairs of isocentrie 10 MV AP/PA fields are to be matched at a patient’s midline, at 9 cm depth. The fields have collimator lengths of 20 cm and 24 cm respectively. The gap on the skin is _____ cm. A. 1.2 B. 2.0 C. 2.8 D, 3.3 E. 4.0 T21. In treatment of the craniospinal axis, the upper spine field length is 25 cm at 100 cm SSD. To align with this diverging spine field, the lateral brain fields should have a collimator rotation of: [data: tan 25° = 0.47; tan 14° = 0.25; tan 7° = 0.125; tan 2.5° = 0.044] A. 25 B. 14 C. 7 D. 2.5 E. None of the above. T22. A patient is treated with parallel opposed isocentric 15x15 em 6 MV photon fields, calculated for a separation of 20 cm. The patient loses weight and the separation changes to 16 cm. If no correction is made to the MU, the result will be approximately: A. Less than 1% underdose.

B. Less than 1% overdose. C. 3% underdose. D. 3% overdose. E. 6% overdose. T23. The depth of dmax for a 6 MV photon beam is approximately: A. 0.5 cm. B. 1.0cm. C. 1.5 cm. D. 2.0 cm. E. 2.5 cm. T24.

The readings above are obtained for a 6 MV beam in a water tank. Depth “d” is _____ cm. A. 1.5 B. 5.0 C. 10.0 D. 15.0 E. 20.0

T25. Regarding geometric penumbra: A. It is smaller for a Co-60 unit than a 6 MV linac. B. It decreases with increasing effective focal spot diameter. C. It decreases with increasing SSD. D. It decreases with increasing source-collimator distance. T26. If the dose rate is 250 cGy/mm at 100 cm, at what distance will the dose

rate be 100 cGy/mm? A. 40 cm. B. 158 cm. C. 200 cm. D. 250 cm. E. 285 cm. T27. A craniospinal axis field calculated for treatment at 120 cm SSD, d = 5cm, requires 320 MU. On the first day, it is decided to treat the patient at 140 cm SSD. The new MU setting is approximately: A. 431 MU B. 480 MU C. 510 MU. D. 556 MU. - E. 564 MV. T28. All of the following are required to calculate the timer setting to deliver a prescribed surface dose on a superficial x-ray machine, except: A. Filter to be used. B. BSF for the field size on the skin. C. Dose rate in air at the end of the applicator to be used. D. Air gap, if any, between the applicator and the patient’s skin. E. PDD tables for the appropriate HVL and SSD. T29. When CT images are used in treatment planning: A. Tissue inside the outer contour is always assumed to have a density of 1.0.

B. The CT number of each pixel is proportional to the linear attenuation of the treatment beam. C. Pixel-by-pixel heterogeneity corrections are mandatory. D. Heterogeneity corrections are based on the electron density of the tissues. T30. A lung tumor is treated with oblique 10 MV photon beams. The depth of each field is 12 cm, of which 6 cm is normal lung. If no lung corrections are made, the actual dose to the isocenter will be about _____ compared with the calculated dose. A. 10% high B. 5% high C. The same D. 5% low E. 10%low T31.

For the breast plan shown in the diagram above, the use of a CT scan allowing lung corrections would tend to have the following effect on the wedge angle chosen for optimal homogeneity, compared to a plan with no lung corrections: A. A decreased wedge angle. B. An increased wedge angle. C. No change in the wedge angle.

T32. A patient is treated with 4 MV photons to the anterior neck field shown.

The dose to the cord at 8 cm depth under the cord block due to scattered radiation only is: A. Proportional to SAR (10x20,d = 8 cm) — SAR (10x3, d = 8 cm). B. Proportional to SAR (10x3, d = dmax). C. About 3% regardless of the size of the block. D. Much less than the primary transmission through the block. T33. If a cerrobend block has an HVL of 1.5 cm in a 6 MV photon beam, then a 7.5 cm thick block will transmit: A. 1.5%. B. 3%. C. 4.5%. D. 7.5%. E. 9%. T34. A simulator film is taken at a source-to-film distance of 140 cm. If an area

measuring 15x15 cm is drawn on the film, this is equivalent to a collimator setting of:

A. 7.7 x 7.7 cm. B. 10.7x10.7cm. C. 12.4x12.4cm. D. 13.6x13.6cm. E. 21.0x21.0cm.

T35. A patient is treated with tangential breast fields. The dose to the patient’s

ovaries, 20 cm inferior to the fields, is approximately _____% of the prescribed dose.

A. 0.005 B. 0.05 C. 0.5 D. 5 E. 10 T36. A TBI patient is treated at 400 cm SSD. If the linac is calibrated to deliver

1.0 cGy/MU at dmax, 100 cm SSD, approximately how many MU are required to deliver 100 cGy from one field at dmax to the TBI patient?

A. 4000 B. 1600 C. 800 D. 400 E. 200 T37. Regarding treatment planning for stereotactic radiosurgery of the brain, which statement is false? A. CT images generally exhibit less geometrical distortion than MRI images. B. MR1 is generally better than CT for imaging brain lesions. C. CT and MRI can be used together to provide optimal tumor localization. D. Both CT and MRI provide electron density information needed to make inhomogeneity corrections in the treatment plan. T38. Currently, the most common MLC leaf width is _____ cm at the isocenter. A. 0.1 B. 0.3 C. 1.0 D. 1.5 E. 2.0

T39. For a single field, stepping at an MLC defined edge is most dramatic at the

_____% isodose level (normalized to the dose on the beam axis at the same depth).

A. 95 B. 80 C. 50 D. 20 T40. The figure shows integral dose-volume histograms for the lung, for plans A and B. Which of the following is false?

A. Approximately the same lung volume receives 90% of the prescribed dose for both plans. B. Plan B gives a dose 30% greater than the prescribed dose to less than 10% of the lung volume. C. In plan B all the lung receives at least 50% of the prescribed dose. D. In plan A 80% of the lung receives 80% of the prescribed dose. T41. Which factors influence the construction of a PTV from a CTV? 1. Patient set-up uncertainty. 2. Organ motion. 3. Proximity of critical structure. 4. Extent of microscopic disease.

A. l,2,3 B. 1,3 C. 2,4 D. 4only E. 1,2,3,4 T42. Advantages of 3-D over 2-D treatment planning systems include all of the following except: A. Use of dose-volume histograms. B. Ability to perform pixel-by-pixel heterogeneity corrections. C. Use of non-coplanar beams. D. Visualization of beam’s-eye-view of target and critical structures. T43. A patient is treated with a field set up at 100 cm SSD. If the patient

must be treated on a stretcher at 120 cm SSD, which of the following is/are true?

1. The collimator setting will increase. 2. The exit dose will decrease slightly. 3. The POD will decrease slightly. 4. The MU will increase. A. 1,3 B. 2,4 C. 4 only D. All of the above. T44. The approximate range of a 6 MeV electron beam passing through 1 cm tissue,

overlying lung (density 0.25 g/cm3) is _____ cm. A. l B. 2 C. 3 D. 6 E. 9

T45. The part of the electron depth dose curve labeled A on the diagram is mainly due to:

A. Electrons with the highest energy which have the greatest range. B. Bremsstrahlung x-rays created by electron interactions with tissue. C. Characteristic x-rays generated by electrons striking the collimators. D. Bremsstrahlung x-rays created by electron interactions with high Z components in the head. T46. Compared with 6 MeV electrons, 16 McV electrons have: 1. A greater surface dose. 2. A lower bremsstrahlung tail. 3. A broader plateau region. 4. A sharper fall off between the 80% and the 20% isodose levels. A. 1,3 B. 2,4 C. 1,2,3 D. 4only E. 1,2,3,4

T47. Cervical neck nodes treated with electrons often require extended SSD, in order to avoid the shoulders with the applicator. If field shaping is achieved

only with inserts in the end of the collimator, the effect will be: A. Increased penumbra width. B. Increased PDD. C. Increased output. D. Decreased beam energy. T48. A 12 MeV electron field delivers 200 cGy at the 90% isodose level. The output of the cone and insert is 1.02 cGy/MU. The MU setting is _____ MU. A. 176 B. 184 C. 218 D. 227 E. 250 T49. A lesion of maximum depth 3 cm is treated with 12 MeV electrons. (PDD data given below). 1 cm of bolus is placed on the skin. If 200 cGy is

delivered to the distal edge of the tumor, the skin dose and the maximum tissue dose are:

Depth 0 1 2 3 4 5 6 PDD 90 95 98 100 80 40 5

Skin Tissue max A. 180 200 B. 211 250 C. 200 238 D. 238 250 E. 195 200 T50. When electron shaping is placed on the skin instead of at the end of the

applicator, this results in: A. A sharper penumbra at the field edge. B. Less thickness of cerrobend required. C. Easier set-up. D. Reduced beam energy. - T51. To reduce hot and cold spots at the junction between adjacent electron fields: A, A gap of at least 1cm should be left on the skin. B. The fields should be overlapped by at least 1cm. C. Only one field should be treated per day. D. The junction should be moved at least once during the course of treatment. T52. Which of the following can be used in total skin electron beam treatment? 1. Low energy electrons 2. Multiple patient positions 3. Beam scatterer or diffuser 4. Boost fields 5. Eye shields A. 1,2,3 B. 1,3 C. 2,4 D. 1,3,5 E. 1,2,3,4,5 T53. A Fletcher applicator is loaded with Cs-137 sources. The activity reported to

the Radiation Safety Officer is 160 mg Ra equ instead of the true value of 160 mCi. The dose rate he measures at 1 meter from the patient will be _____ he expects.

[exposure rate constant for Ra = 8.25 R-cm2/mg-hr; for Cs-137 = 3.26 R-cm2/mCi-hr]. A. Higher than

B. Lower than C. The same as T54. A Cs-137 source had an activity of 15.5 mg Ra equ in June 1992. Its activity in December 1998 is _____ mg Ra equ. A. 8.8 B. 12.5 C. 13.3 D. 14.7 E. 17.8 T55. The exposure rate at 4 m from a patient containing 60 mg Ra equ of Cs-137 is about _____ mR/hr. Assume a point source, and that the patient absorbs 30% of the radiation. A. l B. 2 C. 3 D. 4.5 E. 16 T56. The dose distribution in tissue up to 5 cm from an Ir-192 seed can be predicted from inverse square alone because: A. No other factors are involved. B. At short distances tissue attenuation is negligible at this energy. C. Dose build-up from betas emitted by the source cancels out attenuation. D. Scattered radiation is of very low energy and does not contribute to absorbed dose. E. Scatter dose build-up and attenuation cancel each other out. T57. On the plan of a permanent Au-l98 seed implant, the initial dose rate of 50 cGy/hr covers the volume. The total dose to this volume, after decay, will be _____ cGy. (The half-life of Au-198 is 2.7 days.) A. 9330

B. 5360 C. 5000 D. 4666 E. 3240 T58. The diagram below represents a lateral film of a Fletcher-type gynecological applicator. “Point A” is represented by which point on the diagram?

A. Point1 B. Point2 C. Point 3 D. Point4 E. Point 5 T59. The dose rate at 1 m from a brachytherapy safe is 64 mR/hr. How many

HVLs of shielding are required to reduce this to 2 mR/hr? A. 3

B. 5 C. 7 D. 9 E. 11

T60. The dose rate is 40 mSv/hr at I m from a brachytherapy patient. How many minutes can a person remain at 2 m before receiving a dose of! mSv? A. 6 B. 12 C. 24 D. 30 E. 60 T61. When patients are treated with Cs-137 brachytherapy sources, the dose deposited at 1 cm in tissue is from: A. Gamma and beta minus. B. Beta minus only. C. Gamma and alpha. D. Gamma only. E. Beta plus and x-rays. T62—65. Match the photon energy with the radionuclide: A. 30 keV average B. 141 keV C. 380 keV average D. 662 keV E. 1.25 MeV average T62. Cs-137 T63. 1-125 T64. Ir-192 T65. Co-60 T66. Cs-l37 sources used for brachytherapy have a similar distribution to radium sources because: A. They have similar half-lives. B. They have almost the same maximum energy. C. The inverse square law dominates the dose distribution. D. The same activity (in mCi) can be used.

T67. Pd-103 is sometimes favored over 1-125 for permanent prostate implants because: A. The half-life is shorter. B. The time taken to deliver 90% of the total dose is longer. C. The initial dose rate is lower. D. The initial activity is lower. E. All of the above. T68. For an “ideal” Fletcher tandem and ovoids, with the loading shown below, the typical dose rate at point A is _____ cGy/hr. Tandem: 15-10-10mg Ra equ Ovoids: 15 each A. 90 B. 70 C. 55 D. 35 E. 20 T69. When an HDR planning system uses optimization to create a homogeneous

dose along the surface of a cylindrical applicator, the dwell times will be: A. Higher in the center than at the ends. B. All approximately equal. C. Alternately high and low. D. Higher at the ends than in the center. T70. An HDR treatment takes 366 seconds on the first treatment day. If the

source is Ir-l92 (half-life 74 days) the treatment time exactly one week later will be _____ sec.

A. 343 B. 355 C.372 - D.387 E. 391

T71. In linear accelerators, flattening filters are required in order to A. Reduce dose rates to a safe level for treatment. B. Compensate for the pulsing of the radiation. C. Increase the maximum energy of the beam. D. Attenuate the peak and create a broad, flat beam. E. None of the above. T72. A monitor unit is: A. Part of the hardware of a record and verify system. B. A length of time (fraction of a minute) that the linac beam is turned on. C. A unit used to describe the electron energy of a linac. D. Proportional to the dose delivered at a reference point. T73. Bending magnets are required in linear accelerators for energies greater than 6 MV because: 1. The accelerator waveguide is too long to be mounted parallel to the beam axis. 2. Higher beam currents are required in the electron mode. 3. They are needed to select the correct energy. 4. They are used to boost the electron energy. A. 1,3 B. 2,4 C. 1,2,3 D. 4only B. 1,2,3,4 T74. Most of the neutrons produced by a high-energy linac are produced: A. In the head of the linac. B. In the patient. C. In the floor and walls of the room. D. In the electron mode rather than the photon mode.

T75. Borated polyethylene is used in entrance doors and mazes of high-energy linac rooms. Its function is to: A. Attenuate high-energy photons. B. Thermalize high-energy neutrons, and absorb thermal neutrons. C. Make the door lighter than one composed entirely of lead. D. Absorb low-energy scattered photons. T76. After a superficial x-ray treatment, it was found that a 3 mmAl filter had been used instead of the intended 2 mm Al filter. The effect on the patient is: Skin dose PDD A. Increase No change B. Increase Increase C. Increase Decrease D. Decrease Decrease E. Decrease Increase T71. According to AAPM’s TG-40 report, linear accelerator output should be: 1. Spot checked daily before treatment. 2. Spot checked bi-weekly by the physicist 3. Calibrated monthly by the physicist. 4. Calibrated annually by the physicist in a water tank. 5. Calibrated bi-annually by the physicist in a water tank. A. 1,3,4 B. 1,2,4 C. 1,2,5 D. 2,4 E. 2,5 T78. The edge of the light field on a linac coincides with the isodose line which is

____ of the dose on the axis in the plane of the film. A. 100% B. 95%

C. 90% D. 80% E. 50% T79. According to AAPM’s TG-40 report, the coincidence between the light and radiation fields should be checked: A. Daily B. Weekly C. Bi-weekly D. Monthly E. Annually T80. Ionization chambers that are not sealed require temperature and pressure corrections to account for: A. Variations in the mass of gas in the collection volume. B. The expansion of the gas at high pressure. C. The contraction of the gas at high temperature. D. Variations in the probability of ionization at different temperatures. E. None of the above. T81. If a linac dose rate is 500 cGy/minute at the isocenter, the shielding

required to reduce this to under 2 mrem/hour at 4 meters is _____TVLs. A. 1. B. 2 C. 6 D. 9 E. 20 . T82. Which of the following would not be used in therapy room shielding calculations? if A. Beam energy. B. Use factor. C. Inverse square law. D. Scatter-air ratio.

E. Occupancy factor. T83. The dose to a resident’s hands from a brachytherapy procedure is 25 mSv (2.5 rem). The number of procedures that the resident can perform per year without exceeding maximum permissible dose is: A. 1 B. 2 C. 5 D. 10 E. 20 T84. All of the following are true regarding film badges worn by radiation oncology residents, except: A. They record the whole body dose. B. They should be removed for brachytherapy cases, and a ring badge worn instead. C. They can distinguish between different types of radiation. D. They should be worn on the trunk. E. They should be changed every month. T85. The instrument best suited for finding a dropped Ir-192 seed is: A. Thimble Farmer chamber B. Large volume ionization survey meter C. Geiger counter D. Thermoluminescent dosimeter E. Diode T86. When surveying a patient after removal of radioactive sources, all of the following procedures are necessary except: A. Check detector batteries are functioning. B. Check detector function against a check source. C. Use a detector that has been calibrated within the last year. D. Move container of removed sources away from patient during measurement. E. Calibrate detector against a source of the same type before each use.

T87. Orthogonal films of a gynecological applicator are required for dosimetry planning. AP and lateral films are taken, but the lateral film has very poor contrast. All of the following solutions may provide films with acceptable contrast except: A. Retake lateral, and reduce collimator setting to the minimum area possible. B. Retake lateral using a higher ratio grid. C. Retake lateral with increased mAs. D. Take orthogonals at 45° and 315° instead of 0° and 90°. T88. Bone-soft tissue contrast is better in a simulator film than in a megavoltage “port” film because: A. Of the type of cassette used. B. Of the type of film used. C. Of the absence of Compton interactions in the simulator beam. D. Of the absence of photoelectric interactions in the megavoltage beam. E. Of the predominance of pair production in the megavoltage beam. T89. What type of MRI study best helps differentiate CSF from surrounding edema? A. TI with gadolinium. B. T2. C. Fluid attenuated inversion recovery (FLAIR). D. Perfusion. T90. In MRI, when gadolinium is given for a T1 acquisition, which of the following is

true? A. Nasal mucosa will appear bright. B. CSF will appear bright. C. Vitreous fluid will appear bright. D. Bone is well delineated. T91. A 6 MeV photon incident on water will most probably interact by: A. Photoelectric absorption. B. Compton scatter.

C. Pair production. D. Photonuclear disintegration. E. Coherent scatter. T92—95. Match the PDD at 10 cm depth with the beam energy. A. 85% B. 64% C. 56% D. 33% E 5% T92. Co-60 T93. 6 MV photons T94. 20 MeV electrons T95. 18 MV photons T96—98. Match the following annual dose equivalent with the correct value. A. 0.4mSv B 1mSv C. 20mSv D. 50mSv E. 5000 mSv T96. Average dose to a member of the U.S. population from natural background radiation (excluding radon). T97. Average dose to a member of the U.S. population from medical x-rays. T98. MPD for a radiation worker.

Answers

T1. D PDD increases with increasing SSD because it has two components, attenuation and inverse square. The inverse square component decreases as distance increases. T2 A TMRs are a measure of attenuation only, whereas PDDs comprise attenuation and inverse square components. T3. E

T4. B Dose at d2=(PDD2/PDD1) x Dose at dl =(66.4/86.2)x 180= 139 cGy. T5. A

T6. E The dose at dmax = the dose at depth! PDD = 150 / 0.65 = 231 cGy. T7. D Dneck = Daxis x (TMR d5 / TMR d8) = 4000 x (0.945 / 0.879) = 4300 cGy. The mid point of the neck is at 98 cm (AP) and 102 cm (PA). However, the inverse square corrections from ant and post cancel each other

out, so they can be omitted. T8. C The maximum dose occurs at dmax. This effect is field size dependent; for example, it would be 22 cm for 8x8 cm fields. T9. D Dose at dmax, expressed as a percent of dose at midplane for parallel opposed fields decreases with increasing energy and increasing SSD. T10. B PDDs increase with increasing SSD, because the inverse square

component decreases. The correction (known as Maynard’s F factor) in this case is:

PDD(150 SSD)=PDD (100 SSD) x [(1101(100 + dmax)]2 x [(150+dmax)1160]2

Dmax is a small factor, and can be ignored. Thus, the factor is 1.063, i.e., a 6.3% increase. T11. C The equivalent square of a rectangular field has the same PDD and

TMR as the rectangle. It is smaller in area than the rectangle (i.e., Cxc c 6x30 in this case), since it is the field with the same scatter

contribution on the beam axis. A useful rule of thumb is that C = 4x (area! perimeter). The use of “equivalent square” enables PDD and TMR tables to be simplified to only square fields rather than tabulating the many rectangular fields in use. T12. D Dose perfield=180/3=60cGy.

T13. C This is also the angle between the isodose curve at 10 cm depth and the line perpendicular to the beam axis. T14. B Omitting the wedge increases the dose at all points. The isocenter

dose is increased by 1/(wedge transmission factor), B more than this, and A less than this.

T15. B A 180° posterior arc gives a distribution with a hot spot posteriorly, and doses which fall off across the volume. Wedges are needed for homogeneity. TI6. A Skin dose decreases as photon energy increases. T17. A If only high-energy photons are available and superficial structures would be underdosed,spoilers may be used. The ideal is to maintain a low skin dose and increase dose in the build-up region, to emulate a

lower energy beam. However, while it is impossible to exactly mimic a lower energy beam with a spoiler, the build-up characteristics may be

preferable to using bolus. T18. D Adding bolus to a patient’s surface to correct for “missing tissue” eliminates skin sparing in megavoltage photon beams. Tissue compensators, by moving the absorber away from the skin, preserve skin sparing. Wedges are sometimes used as crude tissue compensators. T19. B A thru D are all techniques used to improve homogeneity at photon field junctions. Overlapping fields on the skin will always give a hot spot.

T20. B By similar triangle geometry: gap = d x (C/2)/SAD for each beam = (9x10/l00) + (9x12/l00) = 2.0 cm.

T21. C The divergence of the spine field is tan-1 (25/2 x 100) = tan-1 0,125 = 7° T22. E The depth to midline changes from 10 cm to 8 cm. The attenuation in a 6 MV beam is about 3% per cm. Thus, a change of 2 cm will cause an overdose of 6%, if no MU correction is made. T23. C T24. C The PDD is 92.0/1 36.0 = 0.676. This is the PDD at 10cm depth for 6 MV photons. T25. D Geometric penumbra increases with increasing source size or effective focal spot size, with increasing SSD, and with decreasing source-

collimator distance.

T26. B By the inverse square law: Dose rate Dose rate1/Dose rate2 = [dist2]2/[dist1]2

Thus, dist 22 = 1002 x (250/100) = 25000 and dist 2 = 158 cm. T27. A The effective TMR will not change, and the collimator output (Sc) will change minimally, so this is essentially an inverse square problem. MU at 140 SSD,d5 = 320 x (145/125)2 = 431 MU. T28. E For a skin dose, PDDs are not required. T29. D Heterogeneity corrections, which may or may not be used at the planner’s discretion, are based on the electron density of the tissues

traversed. The CT number is related to the linear attenuation coefficient of the CT scanner beam (about 140 kV).

T30. A The path length is 6 cm tissue + 6 cm lung. Assuming a density of 0.3g/cm3 6 cm of lung is equivalent to 2 cm of tissue, giving a total effective path length of 6+2 = 8 cm. This is 4 cm less than 12 cm. At 10 MV the attenuation is about 2.5% per cm.

Thus, not using lung corrections will lead to an overdose of about 4x2.5 = 10%. T31. A The increased transmission through lung will create higher doses at the chest wall, and thus a slightly thinner wedge will give a more

homogeneous dose distribution. T32. A SAR (scatter-air ratio) represents the portion of the dose deposited due to scatter. The dose under the cord block is made up of primary

dose (attenuated to about 3% by the block),and scatter from the unblocked portion of the beam. This is represented by the difference

in the SARs for the whole field and the blocked area. T33. B 7.5 cm is 5 HVLs. The transmission is (1/2)5 = 0.03 or 3%. T34. B By similar triangle geometry the collimator setting C is:

C= 15x(l00/140)=10.7cm. T35. C The dose is due to internal scatter, head leakage, and scatter from the collimators and wedge. T36. B Using the inverse square law, the MU will be: 100 x (400/100)2 = 1600 MU. T37. D Only CT provides a map of electron densities. MRI can map hydrogen density. T38. C While newer units with 0.5 cm leaves are being developed, currently 1.0 cm is the most common. T39. C Stepping is most dramatic at the 50% isodose in the penumbra, and becomes somewhat less defined at the 80% and 20% levels. However,

over a course of treatment the stepping becomes blurred out by patient motion and set-up variation, so a beam film presents the “worst case” scenario.

T40. D In plan A 80% of the lung receives about 45% of the prescribed dose. T41 A See ICRU Report 50 (1993), Prescribing, Recording and Reporting

Photon Beam Therapy. Microscopic disease is included in the CTV. T42. B This can be done with most 2-D systems also. T43. C The collimator setting will decrease. The PDD and exit dose will increase slightly. T44. E The range of 6 MeV electrons is 3 cm in tissue, or 1 cm tissue + 2 cm

tissue equivalent lung. Since lung density is 1/4 that of soft tissue, 2 cm tissue is equivalent to 4x2 = 8 cm in lung. Thus, the total range is 1+8 = 9 cm.

T45. D When electrons interact with high Z components in the head of the linac, bremsstrahlung and a smaller number of characteristic x-rays

are produced. This bremsstrahlung “tail” increases with increasing energy, but is usually between 2% and 5%.

T46. A The opposite is true for 2 and 4. T47. A The most noticeable effect of increased SSD is increased penumbra width. PDD and beam energy are unaffected, and output decreases. T48. C MU = Dose/Output x PDD = 200 /(1.02 x 0.9)= 218 MU. T49. D 200 cGy is delivered to a depth of 4 cm (1 cm bolus +3 cm tissue). The POD is 80% at d =4 cm. Thus, the maximum tissue dose (100%) is 200 x 100/80 = 250 cGy. The skin dose (now at 1 cm depth under the bolus)

is 95% of the maximum, i.e., 250 x 0.95 = 238 cGy. T50. A The reason for placing field shaping on the skin is to achieve a sharper penumbra. The output factor may not be the same as for the same effective field size with an insert in the applicator. T51. D Hot and cold spots are inevitable when adjacent electron fields are abutted. Moving the junction as many times as practical during

treatment will reduce the hot and cold spots accordingly. T52. E All the items noted are used to get as homogeneous a dose as possible over the skin surface. T53. B The dose rate will be lower by the ratio of the exposure rate constants, i.e.,

3.26/8.25 = 0.40. Also, whereas 160 mCi (64 mg Ra equ) is a reasonable loading for a Fletcher type applicator, 160 mg Ra equ would be unusually high.

T54. C The time interval is 6.5 years. ‘~ At=A0exp—(0.693xt/T1/2)= 15.5exp—(0.693x6.5/30)= l3.3mg Ra equ. T55. B Exposure rate= Γ x Ax 1/R2 x n.

where f = exposure rate constant = 8.25 R. cm2 / mg.hr A = activity =60 mg R = distance from source = 4m = 400 cm n = fraction emitted (i.e., not attenuated) by patient = 0.7 Rate=8.25 x60 x 1/(400)2x0.7=0.002R/hr=2mR/hr.

T56. E For Ra-226, Cs-l37, and Ir-l92, scatter and attenuation approximately

cancel each other at distances up to about 5 cm, so that the inverse square law predominates.

T57. D Total dose = Initial dose rate x mean life. Mean life = Half-life x 1.44. Total dose =50 cGy/hr x 2.7 days x 24 hr/day x 1.44 = 4666 cGy. T58. B “Point A” was originally defined as a point 2 cm superior to the mucosa of the lateral fornices, in the plane of the tandem, (i.e ., 2 cm superior to the ovoids) and 2 cm lateral to the tandem. The sup/inf distance is often taken to be 2 cm superior to the flange on the tandem (which abuts the cervix). However, as the relative positions of the tandem and ovoids vary from case to case, care should

be taken to ensure that point A measured from the flange does not lie too close to the colpostats, as this will tend to increase the dose rate and possibly result in under-treatment. A better way to prescribe dose is to the isodose surface encompassing the tumor volume, although this is technically more difficult

to define. T59. B To reduce 64 to 2 requires a reduction factor of 2/64 = 1/32. This is achieved with n HVLs, where 1/2n=1/32.Thus,n=5. T60. A Dose rate at 2 m =40 mSv/hr x 1/22 = 10 mSv/hr. Time to receive 1 mSv = 1/10 hr = 6 mm. T61. D Cs-137 decays by beta minus with subsequent gamma emission, but the beta emissions are absorbed in the source and its encapsulation. T62. D T63. A T 64. C T65. E T66. C This allowed rules developed for Ra sources (e.g., Patterson-Parker) to

be used for Cs sources, provided the activity was expressed in mg Ra equ.

T67. A Because it has a shorter half-life, the clinically relevant initial dose

rate (and initial activity) of Pd-103 is higher, and the total dose is delivered in a shorter time.

T68. C

T69. D Due to the inverse square law, a point near the end of the cylinder is further away from most of the sources than a point at an equal radius near the center. Thus, longer dwell times are required at the ends to compensate for this effect. T70. E Timed = Time0 exp(O.693 x d / T1/2) = 366 x exp(O.693 x 7/74) = 391 sec. T71. D Without a flattening filter the dose rate along the central axis would be substantially greater than that at the edge of a clinically useful (40x40 cm) beam. T72. D A monitor unit (MU) represents the amount of charge collected by the

monitor ionization chamber (located in the path of the beam, in the head), when a specific dose has been delivered at a reference point in a phantom. The beam terminates after the set number of MUs have been delivered. As the dose rate on a linac is not absolutely constant, dose cannot be monitored by time alone. However, linacs have a backup timer in case the monitor (and backup) chambers fail to terminate the beam.

T73. A The length of the accelerator waveguide increases as photon energy increases. When it becomes greater than about 30 cm, it is mounted horizontally, and the electron beam is turned through 90° or 270° to point down at the isocenter. Linacs with dual photon energies and electron beams also require bending magnets for energy selection.

T74. A Most neutrons are generated by interactions between high-energy photons and high Z components in the head of the linac. T75. B Polyethylene scatters the high-energy neutrons generated by high- energy linacs. Scattering reduces the neutron energy to thermal levels, and these thermal neutrons can then be absorbed by the polyethylene

and boron. Boron is more efficient at capturing thermal neutrons than polyethylene, and also emits capture gammas of much lower energy. The door must contain sufficient lead to attenuate these gammas.

T76. E A thicker filter will reduce output, and hence skin dose. Because a thicker filter hardens the beam by removing more low-energy photons,

the PDD, as a percent of the skin dose, will increase.

T77. A The output should be spot checked daily by the therapists, and if outside the range specified by the physicist, should be checked by the physicist before treatment. A calibration is part of the monthly physics Q/A. Additionally, a calibration in water is part of the annual Q/A.

T78. E The light field coincides with the 50% isodose line, i.e., the line joining all points which are 50% of the dose on the axis in the plane of the film. T79. D Light/radiation coincidence is part of the monthly Q/A. T80. A The charge collected is proportional to the mass of gas in the chamber.

As temperature increases, or presssure decreases, the mass of gas in the chamber decreases. Chambers are calibrated at standard temperature and pressure (22°C and 760 mm Hg), and the chamber

reading must be corrected to the value that would have been obtained at 22°C, 760 mm. T81. C For photon shielding, 1 cGy is equivalent to 1 rem. The dose rate at I m = 500 rem/min = 500 x 60 rem/hr = 30 x 106 mrem/hr. The dose rate at 4 m = 30/42 x 106 mrem/hr = 1.9 x 106

mrem/hr. To reduce the dose rate by a factor of 106 requires 6 TVLs. T82. D SAR (scatter-air ratio) is used in Clarkson dose calculations for irregular fields. T83. E The annual dose limit to a radiation worker’s hands is 500 mSv (50 rem). (See NCRP report 116). T84. B The ring badge is an additional badge used to measure the dose to the

hands. The whole body dose, which will generally be lower, is still recorded by the film badge. T85. C The dose rate at some distance from a single Ir-192 seed is small, so a sensitive detector

is needed. The Geiger counter uses gas amplification to convert the small signal into a measurcable one. TDLs are too insensitive and do not give an instant reading, and the other detectors are less sensitive than the Geiger counter.

T86. E Annual calibration is required. The detector must be checked before each use to see that its batteries are operational, and that the

expected reading is obtained from a check source. However, the check source does not need to be the same type of source that is being removed from the patient. These detector checks are important because getting a zero

reading from a detector with dead batteries would give the impression that no sources were present, when in fact a source could have been

left behind. T87. C Increasing mAs will make the film darker, but have no effect on contrast. Contrast is improved by reducing scatter. Reducing the

collimator setting to the minimum necessary field of view usually has the greatest effect. If different grids are available, the one with the greatest grid ratio will “clean up” the most scatter. If these two techniques still don’t work, for patients with very large lateral separations one solution is to take orthogonal R and L ant obliques, making use of the somewhat reduced separation.

T88. D At diagnostic energies the probability of photoelectric interactions increases as Z3, which magnifies the difference in attenuation

between bone and tissue. In megavoltage beams the Compton effect predominates, with virtually no photoelectric effect, and shows differences in electron density rather than Z.

T89. C CSF appears dark on a T1 image, and bright on a T2 image. Findings of interest such as transependymal edema or intraventricular tumor can

be obscured by the brightness of the CSF on a T2 image. FLAIR

allows the CSF to appear dark, and thus make T2 changes more conspicuous. T90. A Gadolinium is a ferromagnetic agent which is useful in the imaging of a

variety of lesions. It is always important to verify that adequate gadolinium was delivered and looking at the intensity of uptake (i.e., brightness) of the nasal mucosa can help confirm this.

T91. B Compton interactions are most likely over the range of(monoenergetic) photon energies

from 25 keV to 25 MeV. Even in diagnostic beams, there are generally more Compton than photoelectric interactions.

T92. C T93. B T94. E T95. A T96. B See NCRP report 94. T97. A The average annual dose to members of the public is about 1.6 mSv, excluding the dose from radon. Medical x-rays, cosmic, internal, and

terrestrial radiation each contribute 0.3 to 0.4 mSv, and nuclear medicine contributes about 0.14 mSv.

T98. D (See NCRP report 116.) The maximum permissible whole body dose for a

radiation worker is 5 rem or 50 mSv per year. Most radiation workers receive only a fraction of this, due to the principle of ALARA (as low as reasonably achievable) used when designing shielding.

Copyright© 2001 by RAMPS,Inc. the New York chapter of the AAPM.