Unit 2 PRI

The original technique is 300 mA at 50 ms. If the mA is increased to 500, what exposure time will maintain the same exposure to the IR?

• 0.030 s

• 0.30 s

• 3 ms

• 15 ms

which technique would give the highest patient dose?

• 74 kVp/400 mA @ 0..04 sec

• 90 kVp/ 200 mA @ 0.02 sec

• 65 kVp/300 mA @ 0.1 sec

• 87 kVp/400 mA @ 0.02 sec

Once the part has been penetrated, a doubling of the mAs results in __________ the exposure to the IR.

• one fourth

• one half

• two times

• four times

How much change in kVp will change the radiation exposure to the IR by a factor of 2

• 10%

• 15%

• 20%

• 30%

As mAs increases, the quantity of radiation reaching the IR ________.

• increases

• decreases

• squared

• in not affected

In order to reduce patient exposure, what exposure factors should be considered?

• lower kVp, higher mAs

• higher kVp, higher mAs

• higher kVp, lower mAs

• it makes no difference

which of the following mA and exposure time combinations does not produce 12 mAs?

• 200 mA at 0.06 s

• 400 mA at 0.03 s

• 600 mA at 20 ms

• 800 mA at 0.15 s

As mAs increases, scatter noise ______.

• increases

• decreases

• no change

To reduce patient radiation exposure, it is best to use a lower kVp and higher mAs.

• True

• False

Using twice as much mAs as needed will result in:

• adequate image brightness and increased image brightness

• adequate image brightness and unnecessary patient radiation exposure

• increased image brightness and unnecessary patient radiation exposure

• all options are correct