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Boards Review > W/L Ch 23 Photon Dosimetry Quiz > Flashcards

W/L Ch 23 Photon Dosimetry Quiz Flashcards

1
Q

Which of the following does not apply to “absorbed dose?”

a. Energy deposited at a point
b. Measured in Gray
c. Measured in air
d. Can measure absorbed dose using ionization chambers

A

c. Measured in air

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2
Q

1 Gy is equal to which of the following?

a. 1000 rads
b. 1 J/kg
c. 100 erg/g
d. 10 rads

A

b. 1 J/kg

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3
Q

Which of the following defines Dmax?

a. Build-up region
b. Depth of maximum equilibrium
c. Depth of dose deposited per unit of time
d. Dose scatter to surface

A

b. Depth of maximum equilibrium

Dmax is where the maximum dose is deposited for that beam.

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4
Q

Mayneord F factor is used in which of the following?

a. Isocentric setups when distance is increased
b. Isocentric setups when field size changes
c. SSD setups when field size changes
d. SSD setups when distance increases

A

d. SSD setups when distance increases

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5
Q

The intensity of a radiation beam is measured at 10 mR/hr at a distance of 20 cm. What will be the intensity of this beam at 30 cm?

a. 22.5 mR/hr
b. 4.44 mR/hr
c. 6.67 mR/hr
d. 15 mR/hr

A

b. 4.44 mR/hr

(old dist/new dist)² x dose

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6
Q

The intensity of a radioactive beam is measured at a distance of 100 cm and found to be 250 mR/min. What will the intensity of this beam be at 105 cm?

a. 226.8 mR/min
b. 238.1 mR/min
c. 262.5 mR/min
d. 275.6 mR/min

A

a. 226.8 mR/min

(old dist/new dist)² x dose

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7
Q

Absorbed dose at depth x 100% = absorbed dose at Dmax is the definition of which of the following?

a. TAR
b. TMR
c. TPR
d. PDD

A

d. PDD

PDD is the ratio expressed as a percentage of the absorbed dose at a given depth to the absorbed dose at a fixed reference depth. This factor is used when calculating nonisocentric treatments.

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8
Q

Absorbed dose at a given depth divided by the dose in air dose at a fixed reference is the definition of which of the following?

a. TAR
b. TMR
c. TPR
d. PDD

A

a. TAR

TAR is the ratio of the absorbed dose at a given depth in a phantom to the absorbed dose at the same point in free space (air).

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9
Q

What is a TAR at the depth of Dmax used to correct for scatter of dose called?

a. Scatter-air ratio
b. Tissue-phantom ratio
c. Backscatter factor
d. Mayneord’s F factor

A

c. Backscatter factor

A backscatter factor is the ratio of the dose rate with a scattering medium to the dose rate at the same point without a scattering medium at Dmax.

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10
Q

As a field size increases from the standard established, the output factor for a treatment machine will do what?

a. Increase
b. Decrease
c. Remain the same

A

a. Increase

As the field size increases, so does the scatter, which increases the dose.

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11
Q

As the field size increases, the PDD does what?

a. Increases
b. Decreases
c. Remains the same
d. No effect

A

a. Increases

As the field size increases, scatter also increases, increasing the PDD.

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12
Q

As the energy increases, what happens to the PDD?

a. Increases
b. Decreases
c. Remains the same
d. No effect

A

a. Increases

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13
Q

As the SSD increases, the PDD does what?

a. Increases
b. Decreases
c. Remains the same
d. No effect

A

a. Increases

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14
Q

As the depth increases, what happens to the PDD?

a. Increases
b. Decreases
c. Remains the same
d. No effect

A

b. Decreases

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15
Q

As the field size increases, the tissue-air ratio does what?

a. Increases
b. Decreases
c. Remains the same
d. No effect

A

a. Increases

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16
Q

As the energy increases, the tissue-air ratio does which of the following?

a. Increases
b. Decreases
c. Remains the same
d. No effect

A

a. Increases

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17
Q

As the source-skin distance increases, what happens to the tissue-air ratio?

a. Increases
b. Decreases
c. Remains the same
d. No effect

A

d. No effect

18
Q

Which of the following is the equivalent square of rectangular field of 10 x 15?

a. 12 x 12
b. 3.6 x 3.6
c. 11.5 x 11.5
d. 13 x 13

A

a. 12 x 12

2(W x L) / W + L

19
Q

Which of the following is the equivalent square of a rectangular field of 20 x 10?

a. 12 x 12
b. 15 x 15
c. 11.5 x 11.5
d. 13.3 x 13.3

A

d. 13.3 x 13.3

2(W x L) / W + L

20
Q

A wedge filter is used to alter the isodose distribution on an anterior chest field. The monitor units setting must be ___________ to account for the wedge filter in the field.

a. increased
b. decreased
c. unchanged

A

a. increased

21
Q

A tray holding custom blocks is measured as 97 cGy, and the dose without the tray is 100 cGy. What is the tray transmission factor?

a. 1.03
b. 0.97
c. 1.00
d. Not enough information to determine

A

b. 0.97

97 / 100

22
Q

What is the patient separation if depth from the anterior posterior (AP) is 9 cm and depth from posterior anterior (PA) is 11 cm?

a. 18 cm
b. 19 cm
c. 20 cm
d. 21 cm

A

c. 20 cm

9 + 11 = 20

23
Q

What is the patient’s depth from the AP if the source-axis distance (SAD) is 100 cm and source-skin distance (SSD) is 91 cm?

a. 8 cm
b. 9 cm
c. 10 cm
d. 11 cm

A

b. 9 cm

SAD - SSD
100 - 91 = 9

24
Q

What is the depth from the PA if the SAD is 100 cm and the SSD is 89 cm?

a. 8 cm
b. 9 cm
c. 10 cm
d. 11 cm

A

d. 11 cm

SAD - SSD
100 - 89 = 11

25
Q

Find the PDD at 10 cm of a 6-MV beam of radiation that has a 5 x 5 cm treatment field.

a. 68%
b. 64.1%
c. 65.1%
d. 62.8%

A

b. 64.1%

p. 495, table 23-6

26
Q

Find the PDD at 19 cm of an 18-MV beam of radiation that has a 12 x 12 cm treatment field.

a. 56%
b. 55.6%
c. 53.7%
d. 51.7%

A

a. 56%

p. 496, table 23-7

27
Q

Find the PDD at 3.5 cm of an 18-MV beam of radiation that has a 23 x 15 cm treatment field.

a. 99.8%
b. 98%
c. 100%
d. 98.8%

A

c. 100%

The PDD for the 18-MV beam at a depth of 3.5 cm is 100% at any field size because this is Dmax.

28
Q

A patient is treated on a 6-MV accelerator; field size is 10 x 12 cm and depth of tumor is 6 cm. What is the PDD?

a. 83.02
b. 83.18
c. 83.0
d. 83.2

A

b. 83.18

The equivalent square is 10.9. Using interpolation, the PDD is 83.18.

p. 495, table 23-6

29
Q

A patient is treated on a 6-MV machine at 100 cm SSD. The collimator setting is 15 x 15. There is no blocking used for this treatment. The prescription states that a dose of 3000 cGy is to be delivered to a depth of 3 cm in 10 fractions using a posterior treatment field using the nonisocentric method of calculation. The patient central axis separation is 20 cm. What is the MU setting for the treatment? Reference dose rate (RDR) for a 15 cm equivalent square is 0.993 cGy/MU.

a. 315
b. 321
c. 319
d. 317

A

d. 317

daily dose / (PDD)( RDR)

30
Q

The physician prescribes the AP/PA field to be weighted 3:1 for the posterior field. How much dose is coming from the AP and the PA if total dose prescribed is 200 cGy?

a. AP = 50 cGy, PA = 150 cGy
b. AP = 75 cGy, PA = 125 cGy
c. AP = 150 cGy, PA = 50 cGy
d. AP = 175 cGy, PA = 25 cGy

A

a. AP = 50 cGy, PA = 150 cGy

dose / fields
200 / 4
= 50

3:1 for the posterior field, so 150 cGy for PA, 50 cGy for AP

31
Q

Calculate the PDD of depth (d) if the dose at Dmax is 180 cGy and the dose at d is 127 cGy.

a. 47%
b. 70.6%
c. 56%
d. 75%

A

b. 70.6%

direct proportion
180 cGy / 100% = 127 cGy

32
Q

Calculate the PDD of depth (d) if the dose at Dmax is 250 cGy and the dose at d is 231 cGy.

a. 70%
b. 75%
c. 85.2%
d. 92.4%

A

d. 92.4%

direct proportion

33
Q

Using the 6-MV PDD chart, what is the dose at 10 cm if the dose at Dmax is 300 cGy per day on a 6-MV beam for a field size of 15 x 18?

a. 205 cGy
b. 210 cGy
c. 215 cGy
d. 225 cGy

A

b. 210 cGy

Eq Sq
2(15x18) / 15+18
= 16.36

Look up PDD on 6MV chart and interpolate to get a PDD of 70.26%

p. 495, table 23-6

34
Q

A patient is being treated on the 18-MV beam to a PA spine at a depth of 10 cm. The collimator setting is 10 x 13 with no blocks. The dose is 300 cGy per fraction. What is the dose delivered to Dmax?

a. 375 cGy
b. 240 cGy
c. 400 cGy
d. 300 cGy

A

a. 375 cGy

Eq Sq Law
2(10x13) / 10+13
= 11.3

Look up the PDD on 18MV and interpolate to get a PDD of 80.03%

35
Q

Find the skin gap necessary to match two symmetric ports at midline that are 21 and 37 cm in length. Assume the midline depth is 13 cm and both ports are treated at 100 cm SAD.

a. 1.53
b. 3.6
c. 2.5
d. 3.77

A

d. 3.77

Gap = 1/2 d (L1/SSD) + (L2/SSD2)

36
Q

A patient is treated using two superiorly/inferiorly adjacent fields. The collimator setting for the superior field is 10 cm wide by 13 cm long. The collimator setting for the inferior field is 20 cm wide by 35 cm long. Both fields are treated at 100 cm SSD. Calculate the gap at the skin surface if the fields abut at a depth of 5 cm.

a. 1.5
b. 1.2
c. 2.0
d. 1.0

A

b. 1.2

Gap = 1/2 d (L1/SSD) + (L2/SSD2)

37
Q

Which of the following are adjacent fields that match on the surface and have an overlap at depth due to divergence?

a. Separated fields
b. Abutted fields
c. Half beam blocking fields
d. Geometric matching

A

b. Abutted fields

38
Q 
Methods of obtaining dose uniformity across field junctions include which of the following?
I.   Dosimetric isodose matching 
II.  Separated fields 
III. Junction shift 
IV. Half beam blocking 
V.  Geometric matching

a. I, II, III, and V
b. I, III, IV, and V
c. II, III, IV, and V
d. III, IV, and V

A

b. I, III, IV, and V

39
Q

A patient is being treated on the 6-MV beam to an opposed lateral whole brain (midline). The collimator setting is 14 x 18 with no blocks. The dose is 4000 cGy/20 fractions. The separation is 22 cm. What is the dose delivered to Dmax?

a. 200 cGy
b. 212 cGy
c. 150 cGy
d. 61.9 cGy

A

b. 212 cGy

Eq Sq Law
2(14x18) / 14+18
= 15.75

The entrance and exit dose must be calculated. The PDD must be determined at Dmax (entrance dose) and the exit dose.

REF: pp. 502-503

40
Q

A patient is being treated on the 18-MV beam to an AP/PA stomach (midline). The collimator setting is 12  12 with no blocks. The dose is 4400 cGy/22 fractions. The separation is 24 cm. The cord is 4 cm from the PA. What dose is delivered to the spinal cord?

a. 206.4 cGy
b. 180 cGy
c. 220 cGy
d. 133.7 cGy

A

a. 206.4 cGy

First the PDD for 12 cm, 4 cm, and 20 cm must be determined. The 12 cm represents the midline structure where the physician has prescribed 100 cGy per field. The 4 cm represents the spinal cord depth from the PA, and the 20 cm represents the cord depth from the AP. The dose to 4 cm and 20 cm must be determined and then added together.

REF: pp. 506-507

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