RADIATION THERAPY EXAMS SCRIPT 2026/2027
QUESTIONS AND SOLUTIONS RATED A+
✔✔Compared to water, the density of bone is:
A. 1.00
B. 1.65-1.85
C. 0.33-0.35
D. 0.2-0.26 - ✔✔B. 1.65-1.85
✔✔If the width of the vertebral body is 5.3 cm and measures 7.5 cm on the portal
verification film, the magnification factor is:
A. 0.706
B. 1.0
C. 1.41
D. 2.00 - ✔✔C. 1.41
7.5/5.3 = 1.41
✔✔A patient's treatment will be 2.54 minutes. How many seconds is this?
A. 174
B. 152.4
C. 114
D. 254.0 - ✔✔B. 152.4
(60/1)(x/2.54)
1x = 60(2.54)
X = 152.4
✔✔How many cobalt sources are contained in the gamma knife unit?
A. 25
B. 105
C. 201
D. 280 - ✔✔C. 201
✔✔Which of the following is a nonmalignant condition treated by stereotactic
radiosurgery?
A. Nasopharyngeal tumors
B. Medulloblastoma
,C. Astrocytoma
D. Arteriovenous malformation - ✔✔D. Arteriovenous malformation
✔✔The distance between the entrance and exit points for parallel opposed fields is
called the:
A. Source to skin distance
B. Source to axis distance
C. Interfield distance
D. Source to tray distance - ✔✔C. Interfield distance
✔✔A lateral portal verification shows the spinal cord to be at a measured depth of 9 cm
from the skin surface. If the film magnification is 1.5, the actual depth of the cord is:
A. 4 cm
B. 6 cm
D. 9 cm
E. Not enough information - ✔✔B. 6 cm
9/1.5 = 6
✔✔In craniospinal irradiation, using 100 cm SSD, the spinal cord field length is 30 cm.
The skull field length is 15 cm. To align the skull field with the divergent spinal field, the
amount of collimator rotation required for the skull field is:
A. 4.3 degrees
B. 10 degrees
C. 5 degrees
D. 8.53 degrees - ✔✔D. 8.53 degrees
* the rotation of the collimator for the skull field is given by applying the divergence
formula:
Tan⁻¹ ((A/2)(SSD))
The divergence of the spine field is what needs to be known so the collimator angle is
parallel.
Therefore,
Tan⁻¹ ((30/2)(100)) = 8.53
✔✔In craniospinal irradiation, using 100 cm SSD, the spinal cord field length is 30 cm.
The skull field length is 15 cm. To eliminate divergence of the the brain portal into the
,spinal portal while treating the lateral brain fields, the couch should be rotated ____
degrees toward the gantry.
A. 8.53 degrees
B. 6.3 degrees
C. 5.3 degrees
D. 4.3 degrees - ✔✔D. 4.3 degrees
✔✔When treating a left lateral field, the gantry is 90 degrees and the collimator angle is
25 degrees. If an opposing right lateral port is planned with a gantry angle of 270
degrees, the collimator angle should be:
A. 25 degrees
B. 335 degrees
C. 245 degrees
D. 0 degrees - ✔✔B. 335 degrees
90 - 25 = 65
270 + 65 = 335
✔✔If a 1.0 cm Bolus is applied to an area being treated with a 6 MV beam. The
maximum dose point will be located at:
A. The skin surface
B. 0.5 cm below the skin surface
C. 1.0 cm below the skin surface
D. 1.5 cm below the skin surface - ✔✔B. 0.5 cm below the skin surface
*dose maximum is located at 1.5 cm below the skin for a 6 MV beam. If a 1.0 cm Bolus
is then used, the dose maximum is raised to 0.5 cm below the skin surface.
✔✔A patient is mistakenly treated at 82 cm SSD instead of 8o cm SSD due to a
misalignment of the ODI. The planned dose was 300 cGy. What was the approximate
dose delivered?
A. 285 cGy
B. 292 cGy
C. 308 cGy
D. 315 cGy - ✔✔A. 285 cGy
300/x = 82^/80^2
82^2 = 6724
, 80^2 = 6400
300 x 6400 = 1,920,000
X = 1,920,000/6724
X = 285.54
✔✔A 360 degree rotation has an MU/degree of 0.72 MU/degree. How many monitor
units will be delivered?
A. 180
B. 200
C. 259
D. 360 - ✔✔C. 259
360 x 0.72 = 259
✔✔An air filled balloon positioned with a catheter in the bladder is an example of _____
contrast.
A. Soluble
B. Insoluble
C. Negative
D. Positive - ✔✔C. Negative
✔✔Contrast media may be administered:
1. IV
2. Oral
3. Catheter
A. 1 & 2
B. 1 & 3
C. 2 & 3
D. All of the above - ✔✔D. All of the above
✔✔In CT simulation, therapists may alter kVp to lower dose.
A. True
B. False - ✔✔B. False
✔✔The identification of treatment fields relative to stable landmarks is:
A. Localization
QUESTIONS AND SOLUTIONS RATED A+
✔✔Compared to water, the density of bone is:
A. 1.00
B. 1.65-1.85
C. 0.33-0.35
D. 0.2-0.26 - ✔✔B. 1.65-1.85
✔✔If the width of the vertebral body is 5.3 cm and measures 7.5 cm on the portal
verification film, the magnification factor is:
A. 0.706
B. 1.0
C. 1.41
D. 2.00 - ✔✔C. 1.41
7.5/5.3 = 1.41
✔✔A patient's treatment will be 2.54 minutes. How many seconds is this?
A. 174
B. 152.4
C. 114
D. 254.0 - ✔✔B. 152.4
(60/1)(x/2.54)
1x = 60(2.54)
X = 152.4
✔✔How many cobalt sources are contained in the gamma knife unit?
A. 25
B. 105
C. 201
D. 280 - ✔✔C. 201
✔✔Which of the following is a nonmalignant condition treated by stereotactic
radiosurgery?
A. Nasopharyngeal tumors
B. Medulloblastoma
,C. Astrocytoma
D. Arteriovenous malformation - ✔✔D. Arteriovenous malformation
✔✔The distance between the entrance and exit points for parallel opposed fields is
called the:
A. Source to skin distance
B. Source to axis distance
C. Interfield distance
D. Source to tray distance - ✔✔C. Interfield distance
✔✔A lateral portal verification shows the spinal cord to be at a measured depth of 9 cm
from the skin surface. If the film magnification is 1.5, the actual depth of the cord is:
A. 4 cm
B. 6 cm
D. 9 cm
E. Not enough information - ✔✔B. 6 cm
9/1.5 = 6
✔✔In craniospinal irradiation, using 100 cm SSD, the spinal cord field length is 30 cm.
The skull field length is 15 cm. To align the skull field with the divergent spinal field, the
amount of collimator rotation required for the skull field is:
A. 4.3 degrees
B. 10 degrees
C. 5 degrees
D. 8.53 degrees - ✔✔D. 8.53 degrees
* the rotation of the collimator for the skull field is given by applying the divergence
formula:
Tan⁻¹ ((A/2)(SSD))
The divergence of the spine field is what needs to be known so the collimator angle is
parallel.
Therefore,
Tan⁻¹ ((30/2)(100)) = 8.53
✔✔In craniospinal irradiation, using 100 cm SSD, the spinal cord field length is 30 cm.
The skull field length is 15 cm. To eliminate divergence of the the brain portal into the
,spinal portal while treating the lateral brain fields, the couch should be rotated ____
degrees toward the gantry.
A. 8.53 degrees
B. 6.3 degrees
C. 5.3 degrees
D. 4.3 degrees - ✔✔D. 4.3 degrees
✔✔When treating a left lateral field, the gantry is 90 degrees and the collimator angle is
25 degrees. If an opposing right lateral port is planned with a gantry angle of 270
degrees, the collimator angle should be:
A. 25 degrees
B. 335 degrees
C. 245 degrees
D. 0 degrees - ✔✔B. 335 degrees
90 - 25 = 65
270 + 65 = 335
✔✔If a 1.0 cm Bolus is applied to an area being treated with a 6 MV beam. The
maximum dose point will be located at:
A. The skin surface
B. 0.5 cm below the skin surface
C. 1.0 cm below the skin surface
D. 1.5 cm below the skin surface - ✔✔B. 0.5 cm below the skin surface
*dose maximum is located at 1.5 cm below the skin for a 6 MV beam. If a 1.0 cm Bolus
is then used, the dose maximum is raised to 0.5 cm below the skin surface.
✔✔A patient is mistakenly treated at 82 cm SSD instead of 8o cm SSD due to a
misalignment of the ODI. The planned dose was 300 cGy. What was the approximate
dose delivered?
A. 285 cGy
B. 292 cGy
C. 308 cGy
D. 315 cGy - ✔✔A. 285 cGy
300/x = 82^/80^2
82^2 = 6724
, 80^2 = 6400
300 x 6400 = 1,920,000
X = 1,920,000/6724
X = 285.54
✔✔A 360 degree rotation has an MU/degree of 0.72 MU/degree. How many monitor
units will be delivered?
A. 180
B. 200
C. 259
D. 360 - ✔✔C. 259
360 x 0.72 = 259
✔✔An air filled balloon positioned with a catheter in the bladder is an example of _____
contrast.
A. Soluble
B. Insoluble
C. Negative
D. Positive - ✔✔C. Negative
✔✔Contrast media may be administered:
1. IV
2. Oral
3. Catheter
A. 1 & 2
B. 1 & 3
C. 2 & 3
D. All of the above - ✔✔D. All of the above
✔✔In CT simulation, therapists may alter kVp to lower dose.
A. True
B. False - ✔✔B. False
✔✔The identification of treatment fields relative to stable landmarks is:
A. Localization
