-
-
Moving the image intensifier closer to the patient during traditional fluoroscopy
- decreases the SID
- decreases patient dose
- improves image quality
A
1 only
B
1 and 2 only
C
1 and 3 only
D
1, 2, and 3
 -
- D
-
- 1, 2, and 3
-
With milliamperes (mA) increased to maintain output intensity, how is the ESE affected as the source-to-skin distance (SSD) is increased?
A
The ESE increases.
B
The ESE decreases.
C
The ESE remains unchanged.
D
ESE is unrelated to SSD.
-
- B
-
- The ESE decreases.
-
Which of the following will increase patient dose during fluoroscopy?
- Decreasing the SSD
- Using 2.5 mm Al filtration
- Restricting tabletop intensity to less than 10 R/min
A
1 only
B
1 and 2 only
C
2 and 3 only
D
1, 2, and 3
-
- A
-
- 1 only
-
If a patient received 2000 mrad during a 10-minute fluoroscopic examination, what was the dose rate?
A
0.02 rad/min
B
0.2 rad/min
C
2.0 rad/min
D
20 rad/min
-
- B
-
- 0.2 rad/min
- Two thousand mrad is equal to 2 rad. If 2 rad were delivered in 10 minutes, then the dose rate is 2 ÷ 10, or 0.2 rad/min.
-
What quantity of radiation exposure to the reproductive organs is required to cause temporary infertility?
A
100 rad
B
200 rad
C
300 rad
D
400 rad
-
- B
-
- 200 rad
-
Which of the following is (are) associated with Compton scattering?
| 1. |
High-energy incident photons |
| 2. |
Outer-shell electrons |
| 3. |
Characteristic radiation |
A
1 only
B
1 and 2 only
C
2 and 3 only
D
1, 2, and 3
-
- B
-
- 1 and 2 only
-
Aluminum filtration has its greatest effect on
A
low-energy x-ray photons
B
high-energy x-ray photons
C
low-energy scattered photons
D
high-energy scattered photons
-
- A
-
- low-energy x-ray photons
-
Patient exposure can be minimized by using which of the following?
| 1. |
Accurate positioning |
| 2. |
High-kV, low-mAs factors |
| 3. |
Rare earth screens |
A
1 only
B
1 and 2 only
C
1 and 3 only
D
1, 2, and 3
-
- D
-
- 1, 2, and 3
-
If a patient received 1,400 mrad during a 7-minute fluoroscopic examination, what was the dose rate?
A
200 rad/min
B
5 rad/min
C
2.0 rad/min
D
0.2 rad/min
-
- D
-
- 0.2 rad/min
- A measure 1,400 mrad is equal to 1.4 rad. If 1.4 rad were delivered in 7 minutes, then the dose rate would be 0.2 rad/min:
-
If a patient received 4,500 mrad during a 6-minute fluoroscopic examination, what was the dose rate?
A
0.75 rad/min
B
2.7 rad/min
C
7.5 rad/min
D
27 rad/hr
-
- A
-
- 0.75 rad/min
- Since 4,500 mrad is equal to 4.5 rad, if 4.5 rad were delivered in 6 minutes, then the dose rate must be 0.75 rad/min:
- Thus, x = 0.75 rad/min.
-
Which of the following most effectively minimizes radiation exposure to the patient?
A
Small focal spot
B
Low-ratio grids
C
Increased SID
D
High-speed intensifying screens
-
- D
-
- High-speed intensifying screens
-
An increase of 1.0 mm added aluminum filtration of the x-ray beam would have which of the following effects?
- Increase in average energy of the beam
- Increase in patient skin dose
- Increase in milliroentgen output
A
1 only
B
1 and 2 only
C
2 and 3 only
D
1, 2, and 3
-
- A
-
- 1 only
-
A fluoroscopic examination requires 3 minutes of exposure on time. If the exposure rate for the examination is 250 mR/hr, what is the approximate exposure for the three minute procedure?
A
83.3 R
B
83.3 mR
C
12.5 R
D
12.5 mR
-
- D
-
- 12.5 mR
- If the exposure rate for the examination is 250 mR/hour (60 minutes), then a 3-minute examination would be proportionally less—as the equation below illustrates:
- 60 x = 750
- x = 12.5 mR, dose in 3 minutes
-
Patient dose increases as fluoroscopic
A
FOV increases
B
FOV decreases
C
FSS increases
D
FSS decreases
-
- B
-
- FOV decreases
-
If a patient received 2000 mrad during a 10-minute fluoroscopic examination, what was the dose rate?
A
0.2 rad/min
B
2.0 rad/min
C
5 rad/min
D
200 rad/min
-
- A
-
- 0.2 rad/min
-
If the exposure rate to an individual standing 2.0 m from a source of radiation is 15 R/min, what will be the dose received after 2 minutes at a distance of 5 m from the source?
A
1.2 R
B
2.4 R
C
4.8 R
D
9.6 R
-
- C
-
- 4.8 R
- 25 x = 60
- x = 2.4 R/minute at 2 m = 4.8 R after 2 minutes
-
If the entrance dose for a particular radiograph is 320 mR, the radiation exposure at 1 m from the patient will be approximately
A
32 mR.
B
3.2 mR.
C
0.32 mR.
D
0.032 mR.
-
- C
-
- 0.32 mR.
- Therefore, if the entrance dose for this image is 320 mR, the intensity of radiation at 1 m from the patient is 0.1% of that, or 0.32 mR (0.001 × 320 = 0.32).
-
Which of the following result(s) from restriction of the x-ray beam?
- Less scattered radiation production
- Less patient hazard
- Less radiographic contrast
A
1 only
B
1 and 2 only
C
2 and 3 only
D
1, 2, and 3
-
- B
-
- 1 and 2 only
-
- can pose a safety hazard to personnel
- can have a negative impact on image quality
- occurs with low-energy incident photons
A
1 only
B
1 and 2 only
C
2 and 3 only
D
1, 2, and 3
-
- B
-
- 1 and 2 only
-
If a patient received 0.9 rad during a 3-minute fluoroscopic examination, what was the dose rate?
A
3 mrad/min
B
30 mrad/min
C
300 mrad/min
D
3,000 mrad/min
-
- C
-
- 300 mrad/min
- If 0.9 rad were delivered in 3 minutes, then the dose rate would be 0.9/3, or 0.3 rad/min. Three-tenths rad is equal to 300 mrad.
-
Which of the following will reduce patient dose during fluoroscopy?
| 1. |
Decreasing the source-skin distance (SSD) |
| 2. |
Using 2.5 mm Al filtration |
| 3. |
Restricting tabletop intensity to less than 10 R/min |
A
1 only
B
1 and 2 only
C
2 and 3 only
D
1, 2, and 3
-
- C
-
- 2 and 3 only
-
An increase in total filtration of the x-ray beam will increase
A
patient skin dose
B
beam HVL
C
image contrast
D
milliroentgen (mR) output
-
- B
-
- beam HVL
-
Guidelines for the use of protective shielding state that gonadal shielding should be used
- if the patient has reasonable reproductive potential
- when the gonads are within 5 cm of the collimated field
- when tight collimation is not possible
A
1 only
B
1 and 2 only
C
1 and 3 only
D
2 and 3 only
-
- B
-
- 1 and 2 only
|
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