MOSBY ONLINE - TREATMENT PLANNING

Question 1

Tissue compensators are used to:

A. provide uniform dose distribution in tissue deficit areas
B. delineate field arrangements in soft tissue
C. register patient position during treatment
D. shape the treatment field around the tumor

Answer 1

A. provide uniform dose distribution in tissue deficit areas

Rationale:
Tissue compensators are used in areas where there are tissue deficits. The compensator helps provide uniform dose distribution in this area.

Question 2

A dose of 300 cGy was to be delivered to the thoracic spine at 95 cm SSD. An error caused treatment to be given at an SSD of 92 cm. The actual dose delivered was:

A. 281
B. 291
C. 320
D. 341

Answer 2

C. 320

Rationale:
The inverse square law applies here. 300(95/92)subscript2 = 320.

Question 3

While treating lateral fields in the treatment of the craniospinal axis, to avoid having the caudal margin of the lateral fields diverge into the posterior spine field, the couch should be rotated:

A. way from the collimator
B. toward the collimator
C. until the table is at 5 degrees
D. until the table is at 10 degrees

Answer 3

B. toward the collimator

Rationale:
To prevent divergence of the lateral fields into the posterior field, the couch should be rotated in toward the collimator and degrees calculated by considering the field size and distance.

Question 4

Perforated thermoplastics are better than nonperforated thermoplastics because they:

A. reduce the average skin dose
B. increase the average skin dose
C. are much more sturdy D. make it easy to reproduce the setup

Answer 4

A. reduce the average skin dose

Rationale:
Perforated thermoplastics reduce the average skin dose compared to nonperforated thermoplastics.

Question 5

When using lateral and anterior supraclavicular fields to treat a tumor in the oropharynx, the field junction should be placed:

A. cephalad of the thyroid notch
B. caudal of the thyroid notch
C. caudal of the larynx
D. cephalad of the mandible

Answer 5

A. cephalad of the thyroid notch

Rationale:
If field matching is done while treating the oropharynx, the junction can reasonably be placed cephalad, or superior to, the thyroid notch to avoid overdosing the larynx.

Question 6

“cephalad” means what?

Answer 6

superior

Question 7

“caudad” means what?

Answer 7

inferior

Question 8

Increasing the field size from 5 x 5 to 20 x 20, with all other factors remaining the same, the percent depth dose will:

A. increase
B. decrease
C. remain the same
D. move the Dmax point

Answer 8

A. increase

Rationale:
Increasing the field size will increase the percent depth dose.

FS and PDD are directly related!!!!!

Question 9

The volume of lung within an anterior supraclavicular field can be reduced when treating breast carcinoma by:

A. elevating the arm
B. treating the supraclavicular field every other day
C. placing the angle board under the chest
D. moving the breast down on the chest wall and making the match line as superior as possible

Answer 9

D. moving the breast down on the chest wall and making the match line as superior as possible

Rationale:
To keep the volume of lung as low as possible, the match line should be as superior as possible. This may be achieved in women with larger breasts by using a position or aid that will shift the breast tissue down.

Question 10

The tolerance dose (TD 5/5) for the whole brain is:

A. 45 Gy in 25 fractions
B. 60 Gy in 30 fractions
C. 60 Gy in 15 fractions
D. 45 Gy in 15 fractions

Answer 10

A. 45 Gy in 25 fractions

Rationale:
The tolerance dose for the whole brain is 45 Gy in fractions of 1.8 to 2.0 Gy.

Question 11

Where is the depth of the 80% isodose line for 12 MeV electron beam through water?

A. 3 cm
B. 4 cm
C. 6 cm
D. 10 cm

Answer 11

B. 4 cm

Rationale:
Use the 80% rule. 1/3 of the electron energy is the approximate depth in centimeters of the 80% isodose line.

Question 12

An excerpt from an equivalent square table shows: Using the table, the equivalent square for a field size 9.5 x 13 would be:

A. 10.6
B. 10.9
C. 11.3
D. 11.45

Answer 12

B. 10.9

Rationale:
Using interpolation between the values 10.6 and 11.3. Take the average to get 10.95.

Question 13

The closest blocked equivalent square for a 10 x 20 rectangular field with a 4 x 6 block in the field is:

A. 8.6
B. 11.4
C. 13.3
D. 15.6

Answer 13

C. 13.3

Rationale:
Subtract the area of the block from the area of the open field, and then take the square root. 10 x 20 = 200, 4 x 6 = 24. 200-24 = 176. The square root of 176 = 13.3.

Question 14

Bolus should be composed of a material:

A. having higher absorption than tissue
B. having lower absorption than tissue
C. easy to clean with disinfectant solutions
D. equivalent to tissue in absorption and scattering properties

Answer 14

D. equivalent to tissue in absorption and scattering properties

Rationale:
Bolus should be made of tissue-equivalent material.

Question 15

The dose under a 5 HVL block would be approximately:

A. 3% to 5% of the given dose
B. 5% to 10% of the given dose
C. 10% to 20% of the given dose
D. 20% to 50% of the given dose

Answer 15

A. 3% to 5% of the given dose

Rationale:
A block 5 half-value layers thick should allow about 3% of the dose applied.

Question 16

When a missing tissue compensator is used, it should be placed:

A. directly on the skin surface
B. at some distance from the skin surface
C. on the end of an electron cone
D. on top of a bolus material at isocenter

Answer 16

B. at some distance from the skin surface

Rationale:
To preserve the skin-sparing effect, tissue compensators should be placed at least 15 cm from the patient’s skin surface.

Question 17

The field size of an X-ray or gamma ray beam is defined as the area enclosed within the boundary of the __________% depth dose line.

A. 50
B. 60
C. 80
D. 100

Answer 17

A. 50

Rationale:
The 50% isodose line defines the field size of X-ray or gamma ray beams.

Question 18

When using rotational arc therapy, monitor units or treatment time should be calculated using which of the following methods?

A. Mayneord's factor
B. Paterson-Parker
C. tissue air ratio
D. Clarkson's

Answer 18

C. tissue air ratio

Rationale:
Rotational arc treatment time/monitor units are calculated using the TAR method.

Question 19

Field gaps or abutting junctions are often shifted (feathered) a few times during the course or treatment in order to:

A. reduce the risk of nausea and vomiting
B. reduce the risk of overdosing critical structures
C. make the gap larger
D. change the exit point

Answer 19

B. reduce the risk of overdosing critical structures

Rationale:
Feathering matching junctions, or gaps, are intended to decrease the risk of overdosing or underdosing critical structures.

Question 20

Total dose to the lumpectomy site following accelerated partial breast irradiation utilizing high dose rate brachytherapy will be approximately:

A. 20 Gy
B. 35 Gy
C. 45 Gy
D. 60 Gy

Answer 20

C. 45 Gy

Rationale:
Using APBI techniques, dose to the lumpectomy site typically ranges from 40-50 Gy

Question 21

Posterior cervical nodes need to be treated and the patient’s shoulder will only allow a treatment distance of 115 cm SSD on the 100 cm SAD linear accelerator. The inverse square factor needed in the monitor unit calculation would be:

A. (100/115)2
B. (115/100)2
C. (100/15)2
D. (15/100)2

Answer 21

A. (100/115)2

Rationale:
The inverse square factor in extended distance dose calculations is computed by the relationship: (SAD/SSD + Dmax)superscript2.

Since the energy is not specified here, Dmax can not be accounted for in this exercise.

Question 22

The attenuation of Cerrobend is less dense than lead by about:

A. 5%
B. 10%
C. 15%
D. 25%

Answer 22

C. 15%

Rationale:
Cerrobend or Lipowitz metal is about 15% less dense than lead.

Question 23

Cerrobend or Lipowitz metal is about ____% less dense than lead.

Answer 23

15%

Question 24

The tumor lethal dose of 50 cGy will need to be given to a midline abdominal tumor through a single anterior port using the SSD technique. The abdominal tumor is located at the depth of the 80% dose line, and the spinal cord is at the depth of the 75% dose line. Dose to the cord will be:

A. 66 Gy
B. 50 Gy
C. 47 Gy
D. 37.5 Gy

Answer 24

C. 47 Gy

Rationale:
The spinal cord will receive 47 Gy.

You can compute the applied dose by using TD/%DD. 50/.80 = 62.5 Gy and then taking 75% of the applied dose- 62.5(0.75) = 47 Gy.
Or use direct proportion; 50 Gy:80% as x:75%.

Question 25

The focus-to-film distance on a conventional simulator is 140 cm. A 5 cm wire is placed on the patient’s skin. The focus-to-skin distance is 100 cm. The magnification factor is given by:

A. 100/5
B. 140/100
C. 100/140
D. 5/100

Answer 25

B. 140/100

Rationale:
The magnification formula is FFD/FOD = magnification. Therefore, 140/100 = 1.4.

Question 26

A treatment designed to be given at 120 cm SSD is mistakenly given at 100 cm SSD. What is the error in dose delivered?

A. 20% overdose
B. 20% underdose
C. 44% overdose
D. 44% underdose

Answer 26

C. 44% overdose

Rationale:
Use the inverse square theory. The distance has changed by 20%, so the dose must change in the opposite direction by a square of the distance changed. Check your work by substituting a number for an initial intensity.

Question 27

A dose of 55 Gy will be given through anterior and lateral ports to the maxillary sinus. The lens is included in the treatment fields. The therapeutic ratio is approximately:

A. 0.9
B. 0.18
C. 5.5
D. 11

Answer 27

B. 0.18

Rationale:
The tolerance dose of the lens ranges from 5 Gy to 10 Gy depending on the source. Use the upper limit of the range and the therapeutic ratio formula: 1000/5500 = 0.18.

Question 28

The purpose of an immobilization device is to:

A. make the patient’s position comfortable
B. facilitate reproducing and maintaining the position
C. prevent the patient from falling off the PSA
D. help the therapist set up the treatment faster

Answer 28

B. facilitate reproducing and maintaining the position

Rationale:
Immobilization devices should help in reproducing position and maintaining position.

Question 29

Dose rates in air express the dose measured:

A. at 10 cm depth in a phantom
B. without a phantom, with a buildup cap at a certain distance
C. with a phantom, without a buildup cap at a certain distance
D. without a build up cap or phantom with a 10 x 10 cm field size

Answer 29

B. without a phantom, with a buildup cap at a certain distance

Rationale:
Dose rates in air are measured without a phantom and with a buildup cap to achieve electronic equilibrium at a certain distance from the source.

Question 30

The angle between two beams is known as the :

A. arc angle
B. hinge angle
C. wedge angle
D. divergent angle

Answer 30

B. hinge angle

Rationale:
The degree of separation between two beams is the hinge angle.

Question 31

If the exposure rate at 1 meter is 53.7 R/minute, at what distance is the exposure rate 35 R/minute?

A. 0.5 meters
B. 0.75 meters
C. 1.53 meters
D. 1.24 meters

Answer 31

D. 1.24 meters

Rationale:
Using the inverse square formula. 53.7/35 = 12 /x2, 1.53 = x2; x = 1.24.

Question 32

Rotational arc therapy is elected for treatment to the pituitary gland. The arc is 240 degrees and the calculated monitor units are 196 MU. The MU/degree setting will be:

A. 0.817
B. 0.667
C. 1.22
D. 1.5

Answer 32

A. 0.817

Rationale:
To calculate the MU/degree: 196 MU/240 degrees = 0.817 MU/degree.

Question 33

While using adjacent photon fields for treatment:

A. It doesn’t mater where the hot spot occurs.
B. It doesn’t matter where the cold spot occurs.
C. Dose nonuniformity due to error in abutting can be minimized by increasing the lengths every other day.
D. Dose uniformity can be improved by moving the junction or gap at regular intervals during treatment.

Answer 33

D. Dose uniformity can be improved by moving the junction or gap at regular intervals during treatment.

Rationale:
Adjacent photon fields should have a skin gap, and/or the junction should be feathered periodically.

Question 34

When preparing to assist with the administration of iodine based contrast, the therapist should inspect the contrast container and use only when the content is:

A. expired
B. clear
C. viscous
D. warm

Answer 34

B. clear

Rationale:
Before use, contrast should be inspected and be confirmed as clear, not discolored or cloudy, and having a current date.

Question 35

The physical size of the treatment field is defined at the interception of the central axis at the specific isocentric distance and which isodose line?

A. 10%
B. 20%
C. 50%
D. 80%

Answer 35

C. 50%

Rationale:
The geometric field size is defined at isocenter at the interception of the 50% isodose line.

Question 36

A 6 MV photon beam on the linear accelerator has an output at isocenter of 1.2 cGy/MU. What is the output at an extended distance of 120 cm?

A. 0.8124
B. 0.9876
C. 1.45
D. 1.776

Answer 36

A. 0.8124

Rationale:
The new output can be found by multiplying the old output by the inverse square factor 0.677. Use the inverse square formula.

Question 37

The tolerance dose (TD 5/5) for the whole lung in the adult is:

A. 1800 cGy in 5 fractions
B. 1500 cGy in 5 fractions
C. 1800 cGy in 10 fractions
D. 2400 cGy in 10 fractions

Answer 37

C. 1800 cGy in 10 fractions

Rationale:
The tolerance dose of the whole lung in the adult is about 17.5 Gy when standard fractionation is used:180 to 200 cGy/fraction.

Question 38

Indian club needles have:

A. gold seed
B. one hot end
C. two hot ends
D. even activity

Answer 38

B. one hot end

Question 39

The physician asks you to add a 5 x 4 corner block to a 12 x 14 open lung field. The blocked equivalent square is:

A. 7 x 7
B. 7 x 10
C. 12 x 12
D. 13 x 13

Answer 39

C. 12 x 12

Rationale:
To find the blocked equivalent square, subtract the blocked area from the open area. 12(14)-5(4) = 148. The square root of 148 = 12.1.

Question 40

Diagnostic CT machines are often not suitable for therapy planning because (select two):

1. The table top is concave.
2. The bore is not large enough to accommodate various positions and devices.
3. The slice thickness cannot be varied.
4. The mode can not be switched from axial to helical.

Answer 40

1. The table top is concave.
2. The bore is not large enough to accommodate various positions and devices.

Rationale:
Diagnostic CT scanners do not have bores large enough to accommodate positions and positioning devices used in therapy. Although the table top is concave, the diagnostic CT table often has an insert to make the table flat.

Question 41

An advantage to using custom compensators for missing tissue instead of bolus in megavoltage therapy is:

A. reduced penumbra
B. preservation of skin sparing
C. increased surface dose
D. increased exit dose

Answer 41

B. preservation of skin sparing

Rationale:
Custom compensators are placed near the collimators at a substantial distance away from the patient. This preserved the skin sparing effect of the megavoltage beam.

Question 42

The energy loss of megavoltage (MeV) electron beams in water is approximately_____MeV/cm.

A. 10
B. 5
C. 2
D. 3

Answer 42

C. 2

Rationale:
Electron beams lose about 2 MeV per centimeter of tissue.

Question 43

One disadvantage of using a vertex field to treat primary brain malignancies is that the field exits through the:

A. brain and mediastinum
B. pharynx and spinal cord
C. orbits
D. chest

Answer 43

B. pharynx and spinal cord

Rationale:
When treating the vertex with the patient in neutral position, the vertex field would exit through the pharynx. If the patient is treated with chin flexed, the vertex field exits through the pharynx and spinal cord.

Question 44

A patient was to have received 250 cGy each fraction for 10 fractions at a distance of 100 cm SSD. A readout error demonstrated that the patient was actually treated at a distance of 105 cm SSD. What was the actual total dose delivered?

A. 2267 cGy
B. 2381 cGy
C. 2756 cGy
D. 2625 cGy

Answer 44

A. 2267 cGy

Rationale:
Use the inverse square. 2500(100/105)2 = 2267.

Question 45

The posterior surface of the patient should be flat when treating the entire spine so that:

A. Skin gaps are smaller.
B. Dose is more uniform.
C. Position is reproducible.
D. Patients are most comfortable.

Answer 45

B. Dose is more uniform.

Rationale:
When treating the entire spine, the posterior surface should be as flat as possible so that dose distribution is uniform.

Question 46

A TBI patient will be treated on a 100 cm isocentric accelerator at a distance of 450 cm. The calibrated dose rate at isocenter is 250 cGy/minute. The dose rate at the treatment distance is:

A. 55.5 cGy/minute
B. 1125 cGy/minute
C. 62.5 cGy/minute
D. 12 cGy/minute

Answer 46

D. 12 cGy/minute

Rationale:
Use the inverse square formula. 250(100/450)2 = 12.

Question 47

A patient is receiving 3000 cGy total dose to a spine port using the SSD technique. The total given dose is 3636 cGy. What is the percent depth dose?

A. 100%
B. 120%
C. 82.5%
D. 75%

Answer 47

C. 82.5%

Rationale:
Use the applied dose formula. Given dose = TD/%DD. 3636 = 3000/x; 3000/3636 = 0.825.

Question 48

The CTV is larger than the GTV because it includes the:

A. normal tissue
B. margins for daily uncertainties
C. areas of potential positive disease
D. lymphatic drainage

Answer 48

C. areas of potential positive disease

Rationale:
The CTV includes the tumor volume and areas of potentially positive disease.

Question 49

The output of a 7 MeV electron beam, 12 x 12 cm field is 1.01 cGy/MU. Calculate the monitor units required to deliver 250 cGy to the depth of the 90% dose line.

A. 275 MU
B. 223 MU
C. 227 MU
D. none of the above

Answer 49

A. 275 MU

Rationale:
Use the monitor unit formula for SSD technique.

Question 50

The ratio of dose at depth in tissue to the dose at electronic equilibrium on the beam axis is known as the:

A. tissue air ratio
B. percent depth dose
C. scatter air ratio
D. field size correction factor

Answer 50

B. percent depth dose

Rationale:
The percent depth dose is the ratio of dose at Dmax to dose at a certain depth in tissue.

Question 51

ratio of dose at Dmax to dose at a certain depth in tissue = ______________

Answer 51

PDD

Question 52

In order to reduce the potential for electron contamination, the preferred material for custom block and compensator holders is:

A. foil
B. Lucite
C. aluminum
D. tungsten

Answer 52

B. Lucite

Rationale:
Lucite is a sturdy plastic and is preferred over the other listed materials to reduce the potential for secondary, particle contamination.

Question 53

An excerpt from an equivalent square table shows:

Using the table, the equivalent square for a field size 8 x 13.5 would be:

A. 9.3
B. 9.9
C. 10
D. 10.25

Answer 53

C. 10

Rationale:
Using interpolation between the values 9.9 and 10.1, the equivalent square for 8 x 13.5 would be 10 cm.

Question 54

When treating adjacent, multiple fields on a patient:

A. Patient position should remain the same for all fields.
B. Field borders should abut on the patient’s skin.
C. Fields should be treated on alternate days.
D. Patients should be positioned supine.

Answer 54

A. Patient position should remain the same for all fields.

Rationale:
For multiple, adjacent fields, the patient’s position should remain the same to avoid hot and cold spots.

Question 55

Calculate the equivalent square for a rectangular field 10 x 15.

A. 12 cm
B. 12.5 cm
C. 13.5 cm
D. 13 cm

Answer 55

A. 12 cm

Rationale:
Use Sterling’s formula.

Question 56

The clinical target volume (CTV) includes:

A. gross target and planning target volume
B. gross target and irradiation volume
C. gross target and subclinical disease
D. planning target and irradiation target volume

Answer 56

C. gross target and subclinical disease

Rationale:
The CTV is the gross tumor plus subclinical disease.

Question 57

Total body irradiation must be administered with the patient at an extended distance of 450 cm SSD on our 100 cm SAD linear accelerator. The dose rate at isocenter is 250 cGy/min. The dose rate at the new treatment distance will be:

A. 12.3 cGy/min
B. 810 cGy/min
C. 55.5 cGy/min
D. 506 cGy/min

Answer 57

A. 12.3 cGy/min

Rationale:
The inverse square law applies. As distance changes, the dose rate (intensity) changes inversely, by a square of the distance.

Question 58

Which of the following radioisotopes is sometimes administered into the peritoneal cavity in the treatment of gynecological malignancies?

A. Cesium-137
B. Iridium-192
C. Iodine-131
D. Phosphorus-32

Answer 58

D. Phosphorus-32

Rationale:
Phosphorus-32 can be suspended in liquid and administered intracavitarily for the management of gynecologic malignancies that have spread to the abdominal cavity.

Question 59

Rotational arc therapy is elected for treatment to the pituitary gland. The arc is 240 degrees and the calculated monitor units are 196 MU. The MU/degree setting will be:

A. 0.817
B. 0.667
C. 1.22
D. 1.5

Answer 59

A. 0.817

Rationale:
To calculate the MU/degree: 196 MU/240 degrees = 0.817 MU/degree.

Question 60

Adjacent fields with respective lengths of 40 cm and 36 cm should intersect at a depth of 10 cm. The patient will be treated on a 100 cm isocentric accelerator using the SSD technique. Calculate the skin gap.

A. 4.4 cm
B. 3.4 cm
C. 3.8 cm
D. 2.2 cm

Answer 60

C. 3.8 cm

Rationale:
Gap formula is L/2 (d/SSD) for field one + L/2 (d/SSD) for field two.

Question 61

A good surface landmark to approximate the apex of the lung is the:

A. clavicle
B. sternal angle
C. thyroid notch
D. cricoid notch

Answer 61

A. clavicle

Rationale:
The apex of the lung is approximately 3 cm superior to the clavicle.

Question 62

The linear accelerator dose rate is 250 MU/minute. What will be the treatment time for a treatment field receiving 100 MU?

A. 0.4 seconds
B. 2.5 seconds
C. 4.0 seconds
D. 24 seconds

Answer 62

D. 24 seconds

Rationale:
At 250 MU/minute. Using a direct proportion, it will take 0.4 minutes to deliver 100 MU. Convert the 0.4 minutes to 24 seconds also using direct proportion.

Question 63

Assuming ideal patient contour, two fields utilizing 30 degree wedges have an optimal hinge angle of:

A. 20 degrees
B. 180 degrees
C. 120 degrees
D. 60 degrees

Answer 63

C. 120 degrees

Rationale:
Use the hinge angle formula.

Question 64

A tumor that can be completely eradicated by radiation is known to be:

A. radioresistant
B. radiocurable
C. reoxygenated
D. recurrent

Answer 64

B. radiocurable

Rationale:
A tumor that can be completely eradicated by radiation is called radiocurable.

Question 65

Surface mold brachytherapy is appropriate for all of the following except:

A. pelvic sidewall tumors
B. hard palate tumors
C. orbit tumors
D. superficial skin cancer

Answer 65

A. pelvic sidewall tumors

Rationale:
Mold therapy is appropriate in superficially located tumors in the oral cavity, skin, or eye.

Question 66

The 50% isodose line for a 12 MeV electron beam would be at an approximate depth of:

A. 3 cm
B. 4 cm
C. 5 cm
D. 6 cm

Answer 66

C. 5 cm

Rationale:
Use the rules for electron beam therapy: 4(E) = depth of 50% isodose in millimeters. 4(12) = 48 mm or E = 2.5 x depth of 50% isodose in centimeters. 12 = 2.5(x).

Question 67

Percent depth dose varies with:

1. depth
2. fractionation
3. field size
4. radiation qualityA. 1, 3
   B. 2, 4
   C. 1, 3, 4
   D. 1, 2, 3, 4

Answer 67

C. 1, 3, 4

Rationale:
Percent depth dose varies with the depth in tissue, the field size, and beam energy.

Question 68

Kidney tolerance (TD 5/5) in the adult would be:

A. 2000 cGy delivered in 5 fractions
B. 1500 cGy delivered in 5 fractions
C. 2000 cGy delivered in 10 fractions
D. 2400 cGy delivered in 10 fractions

Answer 68

C. 2000 cGy delivered in 10 fractions

Rationale:
The tolerance dose of the kidney in the adult is about 1800 to 2300 cGy using standard fractionation of 180 to 200 each fraction.

Question 69

Calculate the monitor units necessary to deliver 150 cGy to a depth of 5 cm using a single field SAD technique. The reference dose rate is 0.998 cGy/MU. The field size factor is 1.02 and the TAR is .856.

A. 172 MU
B. 147 MU
C. 176 MU
D. 182 MU

Answer 69

A. 172 MU

Rationale:
Use the monitor unit formula. TD/dose rate x field size factor x TAR.

Question 70

Radiation therapy to treat medullary thyroid cancer requires large treatment portals that include:
1. the lesion
2. bilateral supraclavicular nodes
3. superior mediastinum

A. 1, 2
B. 1, 3
C. 2, 3
D. 1, 2, 3

Answer 70

D. 1, 2, 3

Rationale:
If medullary thyroid cancer is treated with radiation, the field will include the lesion, bilateral supraclavicular nodes, and superior mediastinal nodes.

Question 71

Because electrons scatter so readily in the air, they are:

A. removed before striking the target
B. focused by cones extending close to isocenter
C. bent by 90 or 270 degrees toward the collimator
D. used only at distances shorter than the SAD

Answer 71

B. focused by cones extending close to isocenter

Rationale:
Cones are used to help keep electrons focused due to their potential to scatter in air.

Question 72

In the usual orientation of the three planes with respect to the patient:

A. The transverse plane lies across the patient.
B. The sagittal plane lies across the patient.
C. The coronal plane lies parallel with the x axis.
D. The sagittal plane lies parallel with the z axis.

Answer 72

A. The transverse plane lies across the patient.

Rationale:
The transverse plane corresponds to the axial plane and lies across the patient, dividing the patient into upper and lower halves

Question 73

Using a 100 cm SAD machine, an anterior SSD reading of 86.5 cm, and posterior SSD reading of 85.5 cm are given. The total anterior to posterior separation along the central axis is:

A. 14.5 cm
B. 24 cm
C. 28 cm
D. 14 cm

Answer 73

C. 28 cm

Rationale:
The separation can be found by adding the depths of isocenter in parallel opposed fields. 13.5 + 14.5 = 28.

Question 74

Hypogastric lymph nodes would likely be included in radiation field for the treatment of the:

A. base of tongue
B. stomach
C. esophagus
D. rectum

Answer 74

D. rectum

Rationale:
The hypogastric lymph nodes are part of external iliac chain of nodes in the pelvis and would likely be included in fields for rectal carcinoma.

Question 75

An entire spine needs treatment with field dimensions of 9 x 45. The 100 cm SAD linear accelerator to be used has a collimator limit of 40 x 40. What treatment distance will be required?

A. 89 cm SSD
B. 140 cm SSD
C. 105 cm SSD
D. 112.5 cm SSD

Answer 75

D. 112.5 cm SSD

Rationale:
Using direct proportion. Take the largest dimension needed. 4:100 as 45:x. 40(x) = 45(100). X = 112.5.

Question 76

The dose rate on a linear accelerator is 200 MU/minute. How long is a treatment requiring 240 MU?

A. 0.694 minutes
B. 0.833 minutes
C. 1.2 minutes
D. 1.44 minutes

Answer 76

C. 1.2 minutes

Question 77

Wedge angle refers to:

A. the angle of the actual wedge filter
B. the angle at which an isodose curve at a specific depth is tilted as a result of the wedge being inserted in the beam
C. the angle at which an isodose curve at a specified depth is tilted with respect to the central axis of the beam

Answer 77

B. the angle at which an isodose curve at a specific depth is tilted as a result of the wedge being inserted in the beam

Rationale:
The tilt of the isodose curve at a specified depth when a wedge filter is inserted is how the wedge angle is specified.

Question 78

If the output of a machine is 130 rad/min at 80 cm, what is the output at 60 cm?

A. 65 rad/min
B. 231 rad/min
C. 110 rad/min
D. 150 rad/min

Answer 78

B. 231 rad/min

Rationale:
Use the inverse square. 130 (80/60)2 = 231.

Question 79

Assuming ideal patient contour, a suitable wedge angle for two fields separated by 90 degrees would be:

A. 30 degrees
B. 45 degrees
C. 60 degrees
D. 90 degrees

Answer 79

B. 45 degrees

Rationale:
Use the hinge angle formula. Hinge = 180 - 2(w). 90 = 180 - 2(x).

Question 80

An entire spine is treated by matching two fields both treated at 100 cm SSD. Field number 1 measures 8 x 20 and field number 2 measures 8 x 25. In order to match these fields at a depth of 5 cm, a_______cm skin gap would be required.

A. 1.1 cm
B. 2.0 cm
C. 5.0 cm
D. 3.2 cm

Answer 80

A. 1.1 cm

Rationale:
Use the gap formula. L/2(d/SSD) + L/2 (d/SSD) = skin gap
20/2 (5/100) + 25/2 (5/100) = 1.1 cm.

Question 81

What is the equivalent square for a 15 x 6 field?

A. 9.5 x 9.5
B. 8.6 x 8.6
C. 10 x 10
D. 2.1 x 2.1

Answer 81

B. 8.6 x 8.6

Rationale:
Use Sterling’s formula. 2(w x L)/ w + L. 2(15 x 6)/ 15 + 6 = 8.6.

Question 82

A TBI patient will be treated on a 100 cm isocentric accelerator at a distance of 450 cm. The calibrated dose rate at isocenter is 250 cGy/minute. The dose rate at the treatment distance is:

A. 55.5 cGy/minute
B. 1125 cGy/minute
C. 62.5 cGy/minute
D. 12 cGy/minute

Answer 82

D. 12 cGy/minute

Rationale:
Use the inverse square formula. 250(100/450)2 = 12.

Question 83

A patient is treated using the SAD technique on a 100 cm isocentric linear accelerator. The field size is 24 x 20. The isocenter is located 5 cm below the patient’s skin surface. What does the field measure on the patient’s skin?

A. 24 x 20
B. 19 x 15
C. 23 x 19
D. 22 x 22

Answer 83

C. 23 x 19

Rationale:
Use direct proportion.

Question 84

Custom tissue compensators are used to:

A. filter scattered electrons
B. reduce the beam energy
C. compensate for surface irregularity
D. none of the above

Answer 84

C. compensate for surface irregularity

Rationale:
Custom compensators compensate for surface irregularities.

Question 85

In the maintenance phase of managing ALL, whole brain irradiation may be given with the inferior border of the treatment field at the:

A. base of skull
B. mastoid tip
C. cervical spine #2
D. cervical spine #7

Answer 85

C. cervical spine #2

Rationale:
In the maintenance phase of ALL, whole brain irradiation may be administered using the helmet technique where the inferior border is at C2.

Question 86

An advantage of using the wedged pair technique is that beyond the point of beam intersection, there is:

A. rapid dose buildup
B. rapid dose falloff
C. isodose uniformity
D. heel-effect dose

Answer 86

B. rapid dose falloff

Rationale:
When using the wedged pair, the highest dose concentration is near the hinge, and then there is rapid dose falloff beyond the hinge.

Question 87

A patient is receiving 3000 cGy total dose to a spine port using the SSD technique. The total given dose is 3636 cGy. What is the percent depth dose?

A. 100%
B. 120%
C. 82.5%
D. 75%

Answer 87

C. 82.5%

Rationale:
Use the applied dose formula. Given dose = TD/%DD. 3636 = 3000/x; 3000/3636 = 0.825.

Question 88

Monitoring setup reproducibility:

A. is usually accomplished by weekly port films
B. is always accomplished by using electronic portal imaging
C. can be accomplished by visual comparison of simulation films/DRRs and port films, or by electronic portal imaging
D. is not necessary in most patients

Answer 88

C. can be accomplished by visual comparison of simulation films/DRRs and port films, or by electronic portal imaging

Rationale:
Periodic comparison of treatment verification ports to DRRs or conventional simulation films is a means to monitor setup reproducibility.

Question 89

The isodose chart shown is for an electron beam energy of approximately:

A. 5 MeV
B. 7 MeV
C. 9 MeV
D. 12 MeV

Answer 89

C. 9 MeV

Rationale:
Using the rules for electron energy and isodose depths, you can apply the 80% rule: E/3 = 3 cm (depth of the 80% dose line).

Question 90

What thickness of cerrobend would be equivalent to 5 cm of lead?

A. 4.2
B. 5
C. 6
D. 7.2

Answer 90

C. 6

Rationale:
Cerrobend is about 15% less dense than lead. To figure the equivalent attenuation of cerrobend to lead, use the factor 1.2. 1 cm lead = 1.2 cm cerrobend.

Question 91

Bone metastasis to the thoracic spine is often managed using________treatment field(s).

A. parallel opposed
B. 90 degree hinged
C. single posterior
D. IMRT

Answer 91

C. single posterior

Rationale:
The thoracic spine is usually treated using a posterior port since the thoracic spine is located near the posterior surface.

Question 92

The _____% isodose line defines the field size of X-ray or gamma ray beams.

Answer 92

50

Question 93

A treatment designed to be given at 120 cm SSD is mistakenly given at 100 cm SSD. What is the error in dose delivered?

A. 20% overdose
B. 20% underdose
C. 44% overdose
D. 44% underdose

Answer 93

C. 44% overdose

Rationale:
Use the inverse square theory. The distance has changed by 20%, so the dose must change in the opposite direction by a square of the distance changed. Check your work by substituting a number for an initial intensity.

Question 94

Digitally reconstructed radiographs (DRRs) can be generated from:

A. conventional simulation fluoroscopic images
B. ultrasound images
C. conventional X-ray images
D. CT images

Answer 94

D. CT images

Rationale:
DRRs are generated from CT images.

Question 95

The depth of the 80% dose line for a clinical electron beam energy is given by:

A. 1/4 of the energy
B. 1/3 of the energy
C. 1/2 of the energy
D. 1/10 of the energy

Answer 95

B. 1/3 of the energy

Rationale:
The depth of the 80% dose line is found by dividing the energy by 3.

Question 96

The factors that determine the number of treatment fields to be used includes:

1. beam quality
2. adjacent normal tissue
3. tumor depth
4. tumor shapeA. 1, 3
   B. 1, 4
   C. 1, 2, 3
   D. 1, 2, 3, 4

Answer 96

D. 1, 2, 3, 4

Rationale:
The number of treatment fields to be used depends on the depth of the tumor, adjacent tissue, shape of the tumor and beam quality.

Question 97

If a patient is to be treated at 140 cm SSD on a linear accelerator and the treatment field size is 22 x 40 at this extended distance, what would be the required collimator setting?

A. 22 x 40
B. 16 x 29
C. 11 x 20
D. 8 x 16

Answer 97

B. 16 x 29

Rationale:
Use direct proportion: 22:140 as x:100. Use this for both dimensions. The collimator setting is the measured field size at isocenter.

Question 98

In order to make custom Cerrobend blocks, the following information is needed:
1. target-to-film distance
2. beam energy
3. patient thickness

A. 1, 2
B. 2, 3
C. 1, 3
D. 1, 2, 3

Answer 98

A. 1, 2

Rationale:
In order to make custom blocks, the cutter must know the energy of the machine, the target-to-film distance, and the target-to-tray distance.

Question 99

The use of special blocks or multileaf collimators to create a treatment beam area of proportions other than those capable of the primary collimator system is known as:

A. beam shaping
B. beam spoiling
C. beam shifting
D. beam flattening

Answer 99

A. beam shaping

Rationale:
Beam shaping is accomplished with customized blocks or collimators.

Question 100

The average life of a radioactive source with a half-life of 32 minutes is:

A. 64 minutes
B. 320 minutes
C. 46 minutes
D. 96 minutes

Answer 100

C. 46 minutes

Rationale:
Use the average life formula. Ta = 1.44(half-life).

Question 101

The collimator setting on a 100 cm SAD linear accelerator is 10 x 20. When treating a patient at 120 cm SSD, what does this field measure on the patient’s skin?

A. 8.3 x 16.7
B. 12 x 24
C. 10 x 20
D. 20 x 40

Answer 101

B. 12 x 24

Rationale:
Field size and distance have a direct relationship. Direct proportion can be used here. As distance increases, field size increases.

Question 102

The Dmax for a 6 MV photon beam is approximately:

A. 1 cm
B. 1.5 cm
C. 2 cm
D. 2.5 cm

Answer 102

B. 1.5 cm

Rationale:
The point of electronic equilibrium for a 6 MV photon beam is 1.5 cm.

Question 103

Increasing the distance from 80 cm to 90 cm causes the dose rate to change by:

A. 0.888
B. 0.79
C. 1.125
D. 1.265

Answer 103

B. 0.79

Rationale:
Increasing the distance from 80 to 90 will change the dose rate by a factor of 0.790. Use the inverse square factor (80/90)superscript2.

Question 104

Radiation therapy using a solid or sealed radioisotope for the body surface or tissue implantation is:

A. brachytherapy
B. total body irradiation
C. large field irradiation
D. cobalt therapy

Answer 104

A. brachytherapy

Question 105

When treating an extremity, the limb should be positioned:

A. parallel to the horizontal axis of the beam
B. perpendicular to the table
C. resting on a sponge
D. parallel to the torso

Answer 105

A. parallel to the horizontal axis of the beam

Rationale:
For dose homogeneity, limbs should be positioned parallel to the table or to the horizontal axis of the beam.

Question 106

During verification placement imaging for radioactive needles, the image receptor is located 120 cm from the source. The 3-inch needles measure 4.0 cm on the film. How far away were the needles from the image receptor?

A. 130 cm
B. 120 cm
C. 90 cm
D. 30 cm

Answer 106

D. 30 cm

Rationale:
Using direct proportion and magnification rules: 4/3 = magnification and 120/SAD = magnification. By substitution 4/3 = 120/SAD. SAD = 90 cm therefore, the needles had to be 30 cm from the image receptor. Draw a diagram to visualize.

Question 107

If the cGy/MU is 1.0 at 100 cm SSD, what is the cGy/MU at 200 cm SSD?

A. 0.25 cGy/MU
B. 2.0 cGy/MU
C. 0.5 cGy/MU
D. 4.0 cGy/MU

Answer 107

A. 0.25 cGy/MU

Rationale:
Use the inverse square formula.

Question 108

Treating a patient with a 4 MV photon beam delivers maximum dose below the skin surface at a depth of:

A. 0.5 cm
B. 1.0 cm
C. 1.5 cm
D. 2.0 cm

Answer 108

B. 1.0 cm

Rationale:
Dmax dose for a 4 MV beam is at 1.0 cm below the skin surface.

Question 109

When comparing an isodose profile from a linear accelerator with one from an orthovoltage machine, the linear accelerator isodose will have the following characteristics:
1. greater penetration
2. greater side scatter
3. higher % DD at depth
4. flatter isodose lines

A. 1, 2, 3
B. 2, 3, 4
C. 1, 3, 4
D. 1, 2

Answer 109

C. 1, 3, 4

Rationale:
The linear accelerator beam will be of greater energy so it will show greater penetration, less side scatter, higher percent depth dose and flatter isodose lines.

Question 110

A dose of 300 cGy was to be delivered to the thoracic spine at 95 cm SSD. An error caused treatment to be given at an SSD of 92 cm. The actual dose delivered was:

A. 281
B. 291
C. 320
D. 341

Answer 110

C. 320

Rationale:
The inverse square law applies here. 300(95/92)2 = 320.

Question 111

When treating the maxillary antrum, the organ at the greatest risk for damage is the:

A. pituitary
B. brain
C. lens
D. sphenoid

Answer 111

C. lens

Rationale:
The floor of the orbit is the superior edge of the maxillary sinus. Treatment fields are in close proximity to the radiosensitive lens.

Question 112

A source of I-131 has an activity of 2.5 mCi. What is the activity after 12 days?

A. 0.9 mCi
B. 30 mCi
C. 0.21 mCi
D. 2.5 mCi

Answer 112

A. 0.9 mCi

Rationale:
Recall that the half-life of iodine 131 is 8 days. Use the activity formula.

Question 113

A patient treated with two fields separated by 120 degrees, likely needs a ________degree wedge inserted for even dose distribution.

A. 30
B. 45
C. 60
D. 15

Answer 113

A. 30

Rationale:
Use the hinge angle formula.

Question 114

The best position for treating the vertex field is:

A. chin extended as much as possible
B. chin flexed
C. chin turned to one side
D. chin extension at 12 cm

Answer 114

B. chin flexed

Rationale:
The vertex is best treated with the chin flexed. This position makes the field easy to image.

Question 115

Applying a radioactive material inside or in close proximity to the patient is known as:

A. external beam radiation therapy
B. proton beam radiation therapy
C. tomotherapy
D. brachytherapy

Answer 115

D. brachytherapy

Question 116

The tolerance dose (TD 5/5) for the whole brain is:

A. 45 Gy in 25 fractions
B. 60 Gy in 30 fractions
C. 60 Gy in 15 fractions
D. 45 Gy in 15 fractions

Answer 116

A. 45 Gy in 25 fractions

Rationale:
The tolerance dose for the whole brain is 45 Gy in fractions of 1.8 to 2.0 Gy.

Question 117

The use of the wedge filter helps to:
1. achieve a more homogenous dose distribution
2. compensate for the absence or presence of tissue
3. avoid hot spots

A. 3
B. 2
C. 1, 2, 3
D. 2, 3

Answer 117

C. 1, 2, 3

Rationale:
The wedge compensates for the absence or presence of tissue to achieve more homogenous dose distribution and avoid hot and cold spots.

Question 118

The width of penumbra increases with increasing:

A. SSD
B. source to collimator distance
C. energy
D. SAD

Answer 118

A. SSD

Rationale:
Penumbra increases with increasing SSD and source size.

Question 119

If a tumor receives 225 cGy at a point where the percent depth dose is 75%, the dose at maximum buildup is:

A. 150 cGy
B. 170 cGy
C. 250 cGy
D. 300 cGy

Answer 119

D. 300 cGy

Rationale:
Using the applied dose formula. Applied dose = TD/%DD. 225/75% = 300.

Question 120

Calculate the magnification factor for an object measuring 6 cm on a film. The SOD was 70 cm and the SFD was 100 cm.

A. 1.43
B. 0.7
C. 0.08
D. 4.6

Answer 120

A. 1.43

Rationale:
Use magnification formula. SFD/SOD = magnification.

Question 121

A patient is treated on a linear accelerator. The treatment distance is 94 cm SSD. The TAR at the treatment depth is 0.888 and the output is 1.02 cGy/monitor unit. How many monitor units are required to deliver 100 cGy?

A. 98 MU
B. 110 MU
C. 113 MU
D. 100 MU

Answer 121

B. 110 MU

Rationale:
Use the monitor unit formula. TD/output x TAR. Use the factors given.

Question 122

For patients with bladder cancer, the preoperative radiation treatment fields should be large enough to include the bladder and regional nodes such as:

A. obturator and inguinal
B. obturator, external iliac, and internal iliac
C. hypogastric and external iliac
D. external iliac, hypogastric, and inguinal

Answer 122

B. obturator, external iliac, and internal iliac

Rationale:
Treatment fields for bladder carcinoma should include the tumor volume, external and internal iliacs, and obturator lymph nodes.

Question 123

Which of the following formulas represents the activity formula?

A. A = mcsuperscript2

B. superscript A = Aoe^(λ)(t)

C. A = A/ λ

D. T 1/2 = 1/ λ

Answer 123

B. superscript A = Aoe^(λ)(t)

Rationale:
The activity formula should has an ending activity (A) and original activity (A0). The formula should show the relationship between activity and decay constant and the elapsing of time.

Question 124

Treatment planning involves selection of:
1. daily dose
2. beam arrangement
3. fractionation
4. tumor dose

A. 2, 3
B. 1, 4
C. 2, 3, 4
D. 1, 2, 3, 4

Answer 124

D. 1, 2, 3, 4
Rationale:
Treatment planning involves prescription of dose and fractionation and beam arrangement.

Question 125

A 6 MV photon beam on the linear accelerator has an output at isocenter of 1.2 cGy/MU. What is the output at an extended distance of 120 cm?

A. 0.8124
B. 0.9876
C. 1.45
D. 1.776

Answer 125

A. 0.8124

Rationale:
The new output can be found by multiplying the old output by the inverse square factor 0.677. Use the inverse square formula.

Question 126

A large area must be treated measuring 45 x 60 cm. On a 100 cm SAD linac with a limited collimator setting of 40 x 40, a distance of ______ would be required to accommodate this field size.

A. 145 cm
B. 160 cm
C. 95 cm
D. 150 cm

Answer 126

D. 150 cm

Rationale:
Field size and distance have a direct relationship. Direct proportion can be used here. Since one field dimension is larger than the other, the largest field dimension required should be used in the direct proportion formula

Question 127

The expected total dose to point A for treatment of the cervix from both external radiation and brachytherapy is approximately:

A. 35 Gy
B. 70 Gy
C. 90 Gy
D. 140 Gy

Answer 127

C. 90 Gy

Rationale:
A patient with bulky disease can receive 40 to 45 Gy with external beam and an additional 40 to 50 Gy with low-dose or high-high dose brachytherapy.

Question 128

A fixed point in air around which the treatment machine rotates is called the:

A. isocenter
B. surface dose
C. isodose curve
D. gantry

Answer 128

A. isocenter

Rationale:
The isocenter is a fixed point in air around which the gantry, collimator, and table rotate.

Question 129

A wedge transmission factor of 0.895 indicates a wedge attenuation of the beam by:

A. 89.5%
B. 10.5%
C. 100%
D. 3% to 5%

Answer 129

B. 10.5%

Rationale:
The transmission factor indicates the percent of beam coming through. 100%-89.5% = 10.5% (attenuated).

Question 130

All thermoplastic materials need some kind of base plate support on the couch top. It is preferable to have these plates made of:

A. high-density material
B. nontissue equivalent material
C. carbon fiber material
D. Lucite

Answer 130

C. carbon fiber material

Rationale:
The base plate support for thermoplastic immobilization devices should preferably be made of carbon fiber due to its low attenuation.

Question 131

The dose under a 5 HVL block would be approximately:

A. 3% to 5% of the given dose
B. 5% to 10% of the given dose
C. 10% to 20% of the given dose
D. 20% to 50% of the given dose

Answer 131

A. 3% to 5% of the given dose

Rationale:
A block 5 half-value layers thick should allow about 3% of the dose applied.

Question 132

The equivalent square field size must be determined in order to calculate the:

A. dose to be delivered to the central axis of an irregular field
B. relative dose values from a reference beam
C. spread of the beam from a certain distance from the target
D. air gap on the patient’s external contour

Answer 132

A. dose to be delivered to the central axis of an irregular field

Rationale:
The equivalent square is necessary to calculate the dose to be delivered to the central axis of a nonsquare field.

Question 133

As SSD increases, depth dose:

A. does not change
B. increases
C. decreases

Answer 133

B. increases

Rationale:
The dose at depth increases as the source to skin distances increases.

Question 134

Overall, most intracranial malignancies are:

A. glioblastoma multiforme
B. astrocytoma
C. metastatic
D. pituitary adenoma

Answer 134

C. metastatic

Rationale:
Overall, the majority of intracranial malignancies are metastatic.

Question 135

The percent depth dose is 91.8% for a 10 MV beam with a field size of 12 x 12 at 5 cm depth and 80 cm SSD. Calculate the percent depth dose for the same field size and depth for 100 cm SSD.

A. 81.5
B. 83.8
C. 87.4
D. 92.9

Answer 135

D. 92.9

Rationale:
Use Mayneord’s formula. Use the F factor then multiply by the old percent depth dose.

Question 136

What is the range in soft tissue of a 12 MeV electron beam?

A. 3 cm
B. 4 cm
C. 6 cm
D. 10 cm

Answer 136

C. 6 cm

Rationale:
The range, in centimeters, of an electron beam in tissue/water is 1/2 of the beam energy.

Question 137

The tumor lethal dose of 50 cGy will need to be given to a midline abdominal tumor through a single anterior port using the SSD technique. The abdominal tumor is located at the depth of the 80% dose line, and the spinal cord is at the depth of the 75% dose line. Dose to the cord will be:

A. 66 Gy
B. 50 Gy
C. 47 Gy
D. 37.5 Gy

Answer 137

C. 47 Gy

Rationale:
The spinal cord will receive 47 Gy. You can compute the applied dose by using TD/%DD. 50/.80 = 62.5 Gy and then taking 75% of the applied dose- 62.5(0.75) = 47 Gy. Or use direct proportion; 50 Gy:80% as x:75%.

Question 138

The physician asks you to add a 4 x 4 cm corner block to a 15 x 15 cm open lung field. The blocked equivalent square is:

A. 11 x 11
B. 14.4 x 14.4
C. 19 x 19
D. 12.6 x 12.6

Answer 138

B. 14.4 x 14.4

Rationale:
To find the blocked equivalent square, subtract the blocked area from the open area. 15(15) - 4(4) = 209. The square root of 209 = 14.4

Question 139

What would be the finishing angle for a clockwise rotational arc starting at 280 degrees with an MU setting of 205 and MU/degree of 1.21?

A. 94 degrees
B. 101 degrees
C. 89 degrees
D. 132 degrees

Answer 139

C. 89 degrees

Rationale:
Draw the gantry dial. Compute the number of degrees rotated when administering 205 MU. 205/1.21 Mu/dgree = 169 degrees. Then, count degrees from the starting angle.

Question 140

The maximum, acceptable amount of dose transmitted through the multileaf collimators is:

A. 1%
B. 5%
C. 10%
D. 15%

Answer 140

B. 5%

Rationale:
The acceptable amount of transmission through the MLC, or fabricated blocks, is 5%.

Question 141

Anatomical data for CT imaging and reconstruction is acquired through the:

A. coronal plane
B. transverse plane
C. sagittal plane
D. none of the above

Answer 141

B. transverse plane

Rationale:
Anatomical data is acquired through the transverse plane during scanning.

Question 142

What type of image would yield the poorest contrast quality when visualizing soft tissue?

A. CT image
B. MRI image
C. mega-voltage portal image
D. diagnostic X-ray image

Answer 142

C. mega-voltage portal image

Rationale:
Mega-voltage photons are highly penetrating, so there is little tissue absorption on a portal image compared to a CT or diagnostic X-ray image.

Question 143

The actual physical dimensions of the tumor, including regions of presumed occult spread, is the:

A. absorbed dose volume
B. irradiation volume
C. treatment volume
D. tumor volume

Answer 143

D. tumor volume

Rationale:
The tumor volume shows the physical dimensions of the tumor, including regions of presumed spread.

Question 144

Opposed tangential fields would likely be used in the treatment of the:

A. breast
B. spleen
C. femur
D. esophagus

Answer 144

A. breast

Rationale:
Opposing tangential fields are used in the treatment of the breast.

Question 145

Which of the following therapeutic ratios would be expected to provide the highest probability of cure?

A. 0.85
B. 1
C. 1.45
D. 1.75

Answer 145

D. 1.75

Rationale:
The highest probability of cure is expected with the highest therapeutic ratio. Normal tissue tolerance/tumor lethal dose = therapeutic ratio.

Question 146

When performing manual contouring, it is important to:

A. contour the patient in treatment position
B. verify the beam energy to be used.
C. establish the beam arrangement planned
D. all of the above

Answer 146

A. contour the patient in treatment position

Rationale:
When contouring, patient data must be acquired while the patient is in the treatment position. It is not necessary to know energy and beam arrangement at the time of contouring. However, the therapist should acquire contour data in every possible aspect of anatomy that might be encountered by the beam after planning.

Question 147

The half-life of I-125 is 60 days. In 6 months, the activity of the source will be ____ of the original.

Question 148

The half-life of I-125 is 60 days. In 6 months, the activity of the source will be ____ of the original.

A. 70%
B. 50%
C. 1/8
D. 5%
E. 1/64

Answer 148

C. 1/8

Rationale:
Follow the principle of half-lives. One half-life is 60 days. There are 3 half-lives in 6 months. Three half-lives decrease activity to 1/8 of the original.

Question 149

In preparation for Fletcher’s suite brachytherapy, an orthogonal set of films was acquired on the conventional simulator. The hinge angle for the set of films was:

A. 45 degrees
B. 90 degrees
C. 180 degrees
D. not enough information given

Answer 149

B. 90 degrees

Rationale:
An orthogonal set of films are films taken 90 degrees apart.

Question 150

With all other factors remaining the same, increasing the field size from 6 x 6 to 10 x 10, backscatter would:

A. increase
B. decrease
C. remain the same
D. fall away

Answer 150

A. increase

Rationale:
Increasing the field size increases the backscatter.

Question 151

A device used to measure patient diameter and body thickness is called a:

A. wedge
B. compensator
C. bolus
D. caliper

Answer 151

D. caliper

Question 152

The ratio of dose at depth in tissue to the dose at electronic equilibrium on the beam axis is known as the:

A. tissue air ratio
B. percent depth dose
C. scatter air ratio
D. field size correction factor

Answer 152

B. percent depth dose

Rationale:
The percent depth dose is the ratio of dose at Dmax to dose at a certain depth in tissue.

Question 153

An advantage to using custom compensators for missing tissue instead of bolus in megavoltage therapy is:

A. reduced penumbra
B. preservation of skin sparing
C. increased surface dose
D. increased exit dose

Answer 153

B. preservation of skin sparing

Rationale:
Custom compensators are placed near the collimators at a substantial distance away from the patient. This preserved the skin sparing effect of the megavoltage beam.

Question 154

Target volume delineation for lung tumors may involve cardiac or respiratory gating procedures. The rationale for this is that lung tumors:

A. often metastasize
B. are difficult to image
C. move with cardiac and respiratory motion
D. change shape with cardiac and respiratory motion

Answer 154

C. move with cardiac and respiratory motion

Rationale:
Due to cardiac and respiratory motion, lung tumors move as the heart beats and also on inspiration and expiration. To decrease the risk of target miss, cardiac or respiratory gating procedures may be utilized.

Question 155

What is the dose at Dmax for a patient receiving 200 cGy to a depth of 3 cm where the percent depth dose is 97%?

A. 206 cGy
B. 394 cGy
C. 400 cGy
D. 300 cGy

Answer 155

A. 206 cGy

Rationale:
Use the applied dose formula. 200/97% = 206.

Question 156

One disadvantage of using a vertex field to treat primary brain malignancies is that the field exits through the:

A. brain and mediastinum
B. pharynx and spinal cord
C. orbits
D. chest

Answer 156

B. pharynx and spinal cord

Rationale:
When treating the vertex with the patient in neutral position, the vertex field would exit through the pharynx.
If the patient is treated with chin flexed, the vertex field exits through the pharynx and spinal cord.

Question 157

The range in water of a 20 MeV electron beam is approximately:

A. 10 cm
B. 3.33 cm
C. 5 cm
D. 2 cm

Answer 157

A. 10 cm

Rationale:
The range of an electron beam is found by dividing the energy by 2.

Question 158

When treating the whole brain with opposing laterals, one way to minimize beam divergence into the eye opposite the entrance port would be to:

A. rotate the collimator
B. center the axis near the lateral canthus
C. turn the patient’s head slightly away from the beam
D. rotate the couch away from the gantry

Answer 158

B. center the axis near the lateral canthus

Rationale:
Since there is no beam divergence at the central axis, placing the central axis near the lateral canthus and then using beam shaping would minimize beam divergence to the opposite eye.

Question 159

Which radioisotope is commonly used for permanent, interstitial implants of the prostate gland?

A. Cesium-137
B. Cobalt-60
C. Gold-198
D. Iodine-131

Answer 159

C. Gold-198

Rationale:
Gold-198, Iodine-125, or Palladium-103 seeds are commonly used for permanent prostate implants.

Question 160

A magnification ring measures 5.0 cm on a simulation film. The actual size of the ring is 3.0 cm. The ring is located 85 cm from the target. What is the target to film distance?

A. 85 cm
B. 100 cm
C. 142 cm
D. 167 cm

Answer 160

C. 142 cm

Rationale:
Use the magnification formula. Image size/object size = mag factor and target to film distance/target to object distance = mag factor. 5/3 = 1.67; x/85 = 1.67; 85(1.67) = 142 cm.

Question 161

The term used to define the area on the surface of the patient on which radiation is incident is the:

A. irradiated volume
B. central axis
C. target volume
D. entrance point

Answer 161

D. entrance point

Rationale:
The entrance point is the area on the surface of the patient on which radiation is incident.

Question 162

A volume that consists of the demonstrated tumor and any other tissue presumed to contain tumor is called the:

A. integral volume
B. treatment volume
C. target volume
D. irradiated volume

Answer 162

C. target volume

Rationale:
The tumor and any other tissue presumed to contain tumor is the target volume.

Question 163

The shape of the isodose curves depends largely upon:
1. field size
2. depth dose
3. flattening filters
4. SSD

A. 1, 4
B. 1, 3
C. 2, 4
D. 2, 3

Answer 163

B. 1, 3

Rationale:
The shape of the isodose curve depends on the field size, filters and accessories such as blocks, wedges, cones and multi-leaf collimators.

Question 164

Tissue inhomogeneites that may alter dose distribution would include:

1. air cavities
2. lung
3. fat
4. bone

A. 1, 2
B. 3, 4
C. 1, 3, 4
D. 1, 2, 3, 4

Answer 164

D. 1, 2, 3, 4

Rationale:
During treatment planning, any tissues without the same density as water may alter dose distribution.

Question 165

The dose rate used for low dose rate (LDR) brachytherapy procedures is typically:

A. 20 Gy/minute
B. 10 Gy/minute
C. 2 Gy/minute
D. 0.05 Gy/minute

Answer 165

C. 2 Gy/minute

Rationale:
The normal range for LDR brachytherapy is 0.5 to 2.0 Gy/minute.

Question 166

A dose of 4500 cGy is delivered at a depth of 10 cm where the percent depth dose is 62%. What is the dose to an organ anterior to the target where the percent depth dose is 72%?

A. 4950 cGy
B. 4050 cGy
C. 5226 cGy
D. 5676 cGy

Answer 166

C. 5226 cGy

Rationale:
You may use direct proportion or solve for the applied dose and multiply by 72%. 4500/62% = 7258. 7258(72%) = 5226.

Question 167

For patients with pancreatic cancer, extra care is taken to limit dose to the radiosensitive:

A. heart, stomach, and small bowel
B. stomach, kidney, and small bowel
C. stomach, rectum, and bladder
D. liver, kidney, and spinal cord

Answer 167

D. liver, kidney, and spinal cord

Rationale:
Beam arrangement and field shaping should keep doses low to the low tolerance liver and kidneys and spinal cord during radiation to the pancreas.

Question 168

An 8 MeV electron beam is used to deliver 300 cGy to the 90% isodose line. The cone size is 15 x 15 and the SSD is 100 cm SSD. The cone factor is .872 for the specified cone, and the reference dose rate is 1.0 cGy/MU. What MU setting should be used to deliver this treatment?

A. 344 MU
B. 300 MU
C. 382 MU
D. 333 MU

Answer 168

C. 382 MU

Rationale:
Use the monitor unit formula for SSD technique. TD/cone factor x dose rate x %DD.

Question 169

What type of tissue will appear darkest on a CT image?

A. bone
B. water
C. muscle
D. lung

Answer 169

D. lung

Rationale:
Lung tissue has the lowest density compared to the other tissues, so it will absorb the least amount of radiation, resulting in the appearance of a darker image.

Question 170

A left-sided brain lesion is to receive 3000 cGy using parallel opposed lateral fields with 1.5:1.0 (left: right) weighting. The total dose delivered to the left lateral field will be:

A. 1800 cGy
B. 4500 cGy
C. 1200 cGy
D. 2000 cGy

Answer 170

A. 1800 cGy

Rationale:
For unequally weighted fields, add the weights: 1.5+1 = 2.5. Then divide the total dose by the total weights to figure one portion: 3000/2.5 = 1200. The left lateral receives 1.5 portions: 1200 (1.5) = 1800

Question 171

The dose at the point where the axis of the beam emerges from the patient is known as the:

A. entrance dose
B. exit dose
C. integral dose
D. cumulative dose

Answer 171

B. exit dose

Rationale:
The dose at the point where the beam comes out of the patient is the exit dose.

Question 172

An excerpt from an equivalent square table shows: Using the table, the equivalent square for a field size 9.5 x 13 would be:

A. 10.6
B. 10.9
C. 11.3
D. 11.45

Answer 172

B. 10.9

Rationale:
Using interpolation between the values 10.6 and 11.3. Take the average to get 10.95.

Question 173

If the pituitary gland is treated using a three-field technique, including laterals and vertex fields, wedges will be used in:

A. the lateral fields
B. all three fields
C. the vertex field
D. none of the fields

Answer 173

A. the lateral fields

Rationale:
Due to the contour of the skull in the coronal plane, wedges would be used in the lateral fields only for a midline tumor such as one located in the pituitary gland.

Question 174

Assuming ideal patient contour, the optimal hinge angle for a 45 degree wedged pair is:

A. 30 degrees
B. 45 degrees
C. 60 degrees
D. 90 degrees

Answer 174

D. 90 degrees

Rationale:
Use the formula: hinge = 180 - 2(wedge angle).

Question 175

A tumor lethal dose is 5000 cGy. The spinal cord is adjacent to the tumor. The approximate therapeutic ratio is:

A. 0.9
B. 1
C. 1.11
D. 2.1

Answer 175

A. 0.9

Rationale:
Therapeutic ratio is normal tissue tolerance/tumor lethal dose. Without considering the volume of cord irradiated, the accepted tolerance is 45 Gy. 45/50 = 0.9.

Question 176

The most likely photon beam energy used to treat areas in the head and neck would be:

A. 6 MV
B. 10 MV
C. 18 MV
D. 25 MV

Answer 176

A. 6 MV

Rationale:
The anatomy in the head and neck area is thin so the photon energy most likely used would be 6 MV.

Question 177

A 6 MV photon beam on the linear accelerator has an output at isocenter of 1.2 cGy/MU. What would be the inverse square factor used in the monitor unit calculation for an extended distance treatment at 120 cm SSD?

A. 0.677
B. 0.823
C. 1.21
D. 1.48

Answer 177

A. 0.677

Rationale:
The inverse square factor is (SAD/SSD+Dmax)2. Using this formula, the inverse square factor would be 0.677.

Question 178

A dose of 3000 cGy is to be delivered at a depth of 5 cm with a %DD of 85%. What is the applied dose?

A. 3529 cGy
B. 2550 cGy
C. 5550 cGy
D. 6000 cGy

Answer 178

A. 3529 cGy

Rationale:
Use the applied dose formula. Applied dose = TD/%DD. 3000/85% = 3529.

Question 179

An excerpt from an 18 MV TAR chart shows:

The TAR for a field size 11 x 11 at a depth of 4.5 is:

A. 1.009
B. 1.011
C. 2.02
D. 1.006

Answer 179

D. 1.006

Rationale:
Interpolate-take the average-between 10 cm square at 4 cm depth and 12 cm square at 5 cm depth. 1.009 + 1.002 = 2.011/2 = 1.055.

Question 180

A dose of 55 Gy will be given through anterior and lateral ports to the maxillary sinus. The lens is included in the treatment fields. The therapeutic ratio is approximately:

A. 0.9
B. 0.18
C. 5.5
D. 11

Answer 180

B. 0.18

Rationale:
The tolerance dose of the lens ranges from 5 Gy to 10 Gy depending on the source. Use the upper limit of the range and the therapeutic ratio formula: 1000/5500 = 0.18.

Question 181

When planning portal arrangements, it is important to keep in mind that the treatment couch can be rotated:

A. 360 degrees and moved right to left
B. to any position and moved up and down
C. up to 90 degrees and moved up and down and right to left
D. about 180 degrees, moved up and down and right to left

Answer 181

D. about 180 degrees, moved up and down and right to left

Rationale:
Due to the immobile gantry, the treatment couch can be rotated approximately 90 degrees to the right and left of neutral/0 degrees, up and down and right to left, according to safe limits set by the manufacturer.

Question 182

The physical size of the treatment field is defined at the interception of the central axis at the specific isocentric distance and which isodose line?

A. 10%
B. 20%
C. 50%
D. 80%

Answer 182

C. 50%

Rationale:
The geometric field size is defined at isocenter at the interception of the 50% isodose line.

Question 183

A magnification ring measures 5.0 cm on a simulation film. The actual size of the ring is 3.0 cm. The ring was located 85 cm from the target. What is the magnification factor?

A. 0.6
B. 0.85
C. 1.17
D. 1.67

Answer 183

D. 1.67

Rationale:
Using the magnification formula image size/object size = magnification factor. 5/3 = 1.67.

Question 184

Positron Emission Tomography (PET) is based on imaging with photons produced in:

A. pair production
B. annihilation of positrons by electrons
C. fluorescence caused by positrons in certain screens
D. Compton scattering of photons

Answer 184

B. annihilation of positrons by electrons

Rationale:
PET scan images are a result of positron annihilation.

Question 185

When treating opposed oblique fields for a lung tumor:

A. SSDs for each field should be documented and verified.
B. SSDs for each field should be the same and verified.
C. Isocenter should be at the mid point of the tumor.
D. Isocenter should be at midline of the patient.

Answer 185

A. SSDs for each field should be documented and verified

Rationale:
Oblique fields for the treatment of lung tumors will not necessarily be at the midpoint of the tumor or the midline of the patient’s measured separation. Oblique fields should be verified against the treatment plan and SSD reading documented

Question 186

What thickness of cerrobend would be equivalent to 8 cm of lead?

A. 6.7
B. 7
C. 9.6
D. 12

Answer 186

C. 9.6

Rationale:
Cerrobend is about 15% less dense than lead. To figure the equivalent attenuation of cerrobend to lead, use the factor 1.2. 1 cm lead = 1.2 cm cerrobend.

Question 187

For a sloping skin surface, even dose distribution and maintenance of skin sparing using a high energy beam can be achieved by using a:

A. wedge
B. bolus
C. laser
D. wider beam

Answer 187

A. wedge

Rationale:
A sloping skin surface will cause inhomogeniety of dose lines. A wedge can be inserted to even dose distribution while, at the same time, maintaining skin sparing.

Question 188

If the exposure rate at 1 meter is 53.7 R/minute, at what distance is the exposure rate 35 R/minute?

A. 0.5 meters
B. 0.75 meters
C. 1.53 meters
D. 1.24 meters

Answer 188

D. 1.24 meters

Rationale:
Using the inverse square formula. 53.7/35 = 12 /x2, 1.53 = x2; x = 1.24.

Question 189

The backscatter factor depends on (select two):

1. field size
2. distance
3. energy
4. time

Answer 189

1. field size
2. energy

Rationale:
The backscatter or peak scatter factor is dependent on the field area and energy of the beam.

Question 190

Total dose to the lumpectomy site following accelerated partial breast irradiation utilizing high dose rate brachytherapy will be approximately:

A. 20 Gy
B. 35 Gy
C. 45 Gy
D. 60 Gy

Answer 190

C. 45 Gy

Rationale:
Using APBI techniques, dose to the lumpectomy site typically ranges from 40-50 Gy.

Question 191

The dDmax dose for a 250 cGy tumor dose if the depth dose percentage is 96% would be:

A. 240
B. 260
C. 300
D. 350

Answer 191

B. 260

Rationale:
Using the applied dose formula. Applied dose = TD/%DD. 250/96% = 260.

Question 192

An 8 MeV electron beam will be used to deliver 300 cGy to Dmax. The SSD is 100 cm, dose rate for 10 x 10 is 1.0 cGy/MU, an 8 x 8 cm cone will be used and the cone factor is 0.872. What MU setting should be used?

A. 344 MU
B. 430 MU
C. 172 MU
D. 3.44 MU

Answer 192

A. 344 MU

Rationale:
Using the monitor unit formula: 300/1 x .872 x 1.00 = 344 MU.

Question 193

For a field treated with the isocenter at a depth of 12.5 cm on a 100 cm SAD linear accelerator, the collimator setting is 20 x 25. The field would measure ______ on the patient’s skin surface.

A. 20 x 25
B. 7.5 x 12.5
C. 17.5 x 21.9
D. 22.9 x 28.6

Answer 193

C. 17.5 x 21.9

Rationale:
Use direct proportion. Figure the SSD: 100-12.5 = 87.5 cm. The collimator setting is the measured field size at isocenter. 20:100 as x:87.5. Do the same for the length.

Question 194

Clarkson’s method is used to determine the scatter produced in:

A. irregular fields
B. rectangular fields
C. square fields
D. extended fields

Answer 194

A. irregular fields

Rationale:
Clarkson’s method is for use in irregular fields.

Question 195

The closest blocked equivalent square for a 10 x 20 rectangular field with a 4 x 6 block in the field is:
A. 8.6
B. 11.4
C. 13.3
D. 15.6

Answer 195

C. 13.3

Rationale:
Subtract the area of the block from the area of the open field, and then take the square root. 10 x 20 = 200, 4 x 6 = 24. 200-24 = 176. The square root of 176 = 13.3.

Question 196

The closest blocked equivalent square for a 10 x 20 rectangular field with a 4 x 6 block in the field is:
A. 8.6
B. 11.4
C. 13.3
D. 15.6

Answer 196

C. 13.3

Rationale:
Subtract the area of the block from the area of the open field, and then take the square root. 10 x 20 = 200, 4 x 6 = 24. 200-24 = 176. The square root of 176 = 13.3.

Question 197

What is the CT number of water on a Hounsfield scale?

A. 1000
B. -1000
C. 0
D. -3000

Answer 197

C. 0

Rationale:
The CT number of water is 0.

Question 198

The skin dose, in the case of electron beams, is about:
A. 80% to 90%
B. 30% to 40%
C. 100%
D. 10%

Answer 198

A. 80% to 90%

Rationale:
Electron beams do have some skin sparing. Skin dose is usually 80% to 90% of dose.

Question 199

The treatment plan for an esophagus calls for a total dose of 3000 cGy to the AP field, 1500 cGy to the RPO field, and 1500 cGy to the LPO field. The prescription is for a total of 6000 cGy to be delivered in 40 fractions. What will the daily dose be to each field AP:RPO:LPO?

A. 100:50:50
B. 150:150:150
C. 50:100:100
D. 75:37.5:37.5

Answer 199

D. 75:37.5:37.5

Rationale:
This plan calls for unequal weighting. Each fraction should deliver a total of 150 cGy. The AP field is weighted at 50% of the total. 150 x 50% = 75. The remaining 75 will be equally distributed between the oblique fields. 75/2 = 37.5. Check your work. Dose for each field should total 150 cGy.

Question 200

Shoulder retractors are important for:

A. pushing the shoulder into the treatment field
B. pulling the shoulders out of the treatment field
C. improving dose uniformity
D. patient comfort during long procedures

Answer 200

B. pulling the shoulders out of the treatment field

Rationale:
Shoulder retractors are used to pull shoulders down out of the field.

Question 201

In one month, the output of a cobalt machine would reduce by about:

A. 0.1%
B. 1%
C. 2%
D. 5%

Answer 201

B. 1%

Rationale:
Cobalt output decreases at about 1% a month.

Question 202

The natural slope of the chest may cause high dose regions in the upper mediastinum. This area of increased dose may be managed by the use of a:

A. bolus
B. custom compensator
C. bite block
D. wing board

Answer 202

B. custom compensator

Rationale:
The slope of the chest causes uneven dose distribution leading to higher dose regions in the upper mediastinum, where the anatomy is thinner. Custom compensators, standard wedges, or dynamic wedges may be used to produce more even dose distribution.

Question 203

A representation of the dose in an irradiated volume, as a function of spatial position along a single line is known as a/an:

A. isodose curve
B. percent depth dose line
C. dose profile
D. dose volume histogram

Answer 203

C. dose profile

Rationale:
A dose profile shows the dose in a volume as a function of position along a single line.

Question 204

When it is determined that an immobilization device is needed, the therapist should:

A. make the device, then determine beam orientation
B. determine beam orientation, then make the device
C. determine device, patient position, and then beam orientation
D. make the device, then determine patient position

Answer 204

B. determine beam orientation, then make the device

Rationale:
Before fabricating a device, the therapist should consider the most appropriate patient position, likely beam orientation, and then which device.

Question 205

Which of the following is not a necessary step in tumor volume definition?

A. contour of the tumor highlighted throughout the patient volume
B. measurements of anatomical separation
C. staging of the disease
D. identification of adjacent normal tissue

Answer 205

C. staging of the disease

Rationale:
In defining the tumor volume, patient measurements are taken, visible tumor margins are highlighted, and normal adjacent tissue is identified.

Question 206

The calculated treatment time on a cobalt unit is 4.23 minutes. This treatment time in minutes and seconds is:

A. 4 minutes 23 seconds
B. 4 minutes 30 seconds
C. 4 minutes 14 seconds
D. 4 minutes 19 seconds

Answer 206

C. 4 minutes 14 seconds

Rationale:
Use direct proportion. 60 sec 1 minute as x sec: 0.23 minutes.

Question 207

The depth of inguinal lymph nodes :

A. can range from 5 to 10 cm and is not influenced by patient size
B. is at approximately 4 cm from the surface in all patients
C. can range from 2 to 18 cm depending on patient size
D. is always one-third of the patient’s anterior to posterior separation

Answer 207

C. can range from 2 to 18 cm depending on patient size

Rationale:
There are both deep and superficial inguinal lymph nodes located from 2 to 18 cm deep depending on the patient’s size.

Question 208

For breast irradiation, internal mammary lymphatics would be best treated using:

A. an anterior 10 MV photon field
B. widened tangential fields
C. an anterior electron field
D. a posterior electron field

Answer 208

C. an anterior electron field

Rationale:
Internal mammary lymphatics may be included in tangential fields; however this would increase lung volume. An anterior field could be matched to tangential fields using low-energy photons or electron beam. The electron beam would be best so that dose to limit dose to the mediastinum. Internal mammary nodes are located approximately 3 cm below the anterior surface in patients of normal body habitus.

Question 209

The range of a 10 MeV electron beam in tissue is approximately:

A. 10 mm
B. 10 cm
C. 5 cm
D. none of the above

Answer 209

C. 5 cm

Rationale:
The range of an electron beam can be found by dividing the energy by 2.

Question 210

During CT simulation, laser alignment marks should be placed:

A. at the level of anticipated treatment isocenter
B. at the level of the central scan
C. at the first image level
D. at the final image level

Answer 210

A. at the level of anticipated treatment isocenter

Rationale:
Markings should be placed at the level of the anticipated treatment isocenter during CT simulation.

Question 211

External beam radiation for cancers of the midesophagus is best delivered using:

A. anterior field only
B. right and left lateral fields
C. posterior and two posterior obliques
D. anterior and two posterior obliques

Answer 211

D. anterior and two posterior obliques

Rationale:
Feedback: The middle section of the esophagus is best treated with an anterior and two posterior obliques to adequately deliver dose to the volume while sparing the dose-limiting spinal cord.

Question 212

A field area measures 14 x 23 at a distance of 80 cm. What is the measured field size at a distance of 60 cm?

A. 10.5 x 17.25
B. 18.6 x 30.6
C. 25 x 41
D. 7.9 x 12.9

Answer 212

A. 10.5 x 17.25

Rationale:
Use direct proportion.

Question 213

A midline brain tumor will be treated to 4500 cGy, through right and left lateral ports. Of concern is the normal brain tissue. The therapeutic ratio is approximately:

A. 0.75
B. 1
C. 1.33
D. 1.5

Answer 213

C. 1.33

Rationale:
Therapeutic ratio is normal tissue tolerance/tumor lethal dose. The tolerance of brain tissue is accepted as 60 Gy. 60/45 = 1.33.

Question 214

The depth of electronic equilibrium for a 10 MV photon beam is:
A. 0.5 cm
B. 1.0 cm
C. 1.5 cm
D. 2.0 cm

Answer 214

D. 2.0 cm

Rationale:
The depth of electronic equilibrium for a 10 MV beam is 2.0 cm.

Question 215

What is the first objective in radiation therapy treatment planning?

A. identify routes of tumor spread and beam energy
B. define critical structures and beam angles
C. determine tumor boundaries and critical structures
D. choose beam energy and angles

Answer 215

C. determine tumor boundaries and critical structures

Rationale:
The first objective of treatment planning is to define the tumor volume and identify critical structures and where they lie in respect to the tumor volume.

Question 216

A method of radiation therapy in which the source is moved around the patient, but only through certain sectors of a circle is called:

A. step and shoot
B. arc therapy
C. adjacent field technique
D. feathering technique

Answer 216

B. arc therapy

Rationale:
Arc therapy is a method of radiation therapy delivery in which the source is moved around the patient through certain sectors of a circle.

Question 217

Standard wedges are designed for use in high voltage beams and vary in angle from_____to____degrees.
A. 15 to 75
B. 30 to 60
C. 45 to 75
D. 15 to 60

Answer 217

D. 15 to 60

Rationale:
Standard wedges vary in size from 15 to 60 degrees.

Question 218

Isodose curves would shift away from the skin surface for underlying:
1. lung
2. air cavities
3. bone
A. 1, 2
B. 2, 3
C. 1, 3
D. 1, 2, 3

Answer 218

A. 1, 2

Rationale:
Isodose lines would shift away from the skin surface for underlying lung tissue or air-filled cavities.

Question 219

Increasing the field size from 5 x 5 to 20 x 20, with all other factors remaining the same, the percent depth dose will:
A. increase
B. decrease
C. remain the same
D. move the Dmax point

Answer 219

A. increase

Rationale:
Increasing the field size will increase the percent depth dose.

Question 220

The volume larger than the treatment volume receiving some significant dose is the:

A. tumor volume
B. target volume
C. irradiated volume
D. clinical volume

Answer 220

C. irradiated volume

Rationale:
The volume receiving some dose that is larger that the treatment volume is the irradiated volume.

Question 221

When using film for portal imaging, the size of the enlarged image or shadow on the film is dependent on:
1. target-to-object distance
2. object-to-table distance
3. target-to-film distance
A. 1, 2
B. 2, 3
C. 1, 3
D. 1, 2, 3

Answer 221

C. 1, 3

Rationale:
Magnification of objects on film depends on target-to-object distance, target-to-film distance, and object-to-film distance.

Question 222

If a patient has a large pelvic tumor involving the bladder and prostate, and there is a presacral mass, the best arrangement would be:
A. opposed anterior and posterior fields
B. opposed lateral fields
C. four-field box technique
D. 90 degree wedged pair

Answer 222

A. opposed anterior and posterior fields

Rationale:
In this instance, the goal would be to treat all areas of tumor at various depths in the patient’s pelvis. At least, the initial fields would be opposing anterior and posterior fields to give even dose to all points of interest and spare pelvic bones.

Question 223

In radiation therapy planning, the x, y, z coordinate system is used to:

A. align patients daily for treatment
B. describe a point in space relative to isocenter
C. describe the location of isocenter
D. assure all planes are parallel

Answer 223

B. describe a point in space relative to isocenter

Rationale:
The x, y, z coordinate system is used to describe the location of a point in space relative to isocenter.

Question 224

Partial breast irradiation may involve treatment techniques that employ all of the following except:

A. 5-field breast technique
B. brachytherapy
C. 3-D conformal IMRT
D. intraoperative radiation therapy

Answer 224

A. 5-field breast technique

Rationale:
Partial breast irradiation may be accomplished by using brachytherapy, intra-operative therapy, or 3-D conformal IMRT.

Question 225

Electron beams are useful in radiation therapy primary because they:

A. have a greater biological effect that X-rays
B. provide more skin sparing than gamma rays
C. have a definite range in tissue
D. have maximum LET at electronic equilibrium

Answer 225

C. have a definite range in tissue

Rationale:
Electrons have a definite range in tissue, making them useful in the treatment of superficial lesions with critical tissues beyond.

Question 226

An 8 MeV electron beam is used to deliver 200 cGy to the 80% dose line. The field size is 8 x 8 and the SSD is 100 cm. The dose rate is 1.0 cGy/MU for a 10 x 10 field. The cGy/Mu for this cone size is 0.872. What are the calculated monitor units?
A. 229 MU
B. 287 MU
C. 143 MU
D. 115 MU

Answer 226

B. 287 MU

Rationale:
Using the monitor unit formula: 200/1 x 0.872 x 0.80 = 287 MU.