RAD 230 - Week 3

Grays, or milligrays, in air are the measurement for x-ray _________ .

• quantity and intensity

• exposure and quantity

• intensity and exposure

• quantity, exposure, and intensity

Standard x-ray machines produce about _____ µGya/mAs at 70 kVp measured at 100 cm SID.

• 20

• 50

• 100

• 200

X-ray intensity increases in direct proportion to increases in __________.

• mAs

• kVp

• distance

• filtration

Quantity refers to:

• kVp

• mAs

• HVL

• Shades of gray

If the distance from the source to the image (SID) is reduced by half, how is the x-ray

intensity at the image affected?

• It is increased 4 times.

• It is doubled.

• It is reduced by 1/2.

• It is reduced by 1/4.

Answer explanation

If the SID is reduced by 1/2, the intensity at the image increases 4 times, because of the

inverse square law. We can even make our own problem to demonstrate this. If we start with an SID of 200 cm and our intensity is 8 Rads, what would our intensity be if we reduced to 100 cm?

I₁ / I₂ = (SID₂/SID₁)², so let's plug in our numbers.

I₁ = 8 Rads

I₂ = x

D₁ = 200 cm

D₂ = 100 cm

After we square our distances, cross multiply, and divide we get 32 Rads. 32 is 4 times more than 8. This makes sense because we're decreasing the distance from the x-ray source which always increases the exposure.

If filter thickness is _______, then x-ray intensity is ______.

• reduced; reduced

• increased; reduced

• reduced; increased

• increased; reduced and reduced; increased

Answer explanation

If filter thickness is increased, then x-ray intensity is decreased and if filter thickness is

decreased, then x-ray intensity is increased.

Example: Let's say we're shooting BBs at a pane of glass, and let's say we're shooting some at 5 miles per hour, some at 30 mph, and some at 80 mph (these are our kVps). If we shoot any of these at a single thin sheet of glass, they'll probably all get through, right? So what can we do if we only want the 80 mph BBs to get through? We can add another pane of glass (add filtration). That probably won't work either, BUT if we keep thickening the pane of glass (increasing filtration) eventually the ONLY BBs that will get through will be the 80 mph BBs - the highly energetic ones.

Filtration removes low energy x-ray photons from the primary beam so that only the useful higher energy photons hit the patient for image production.

A 10% increase in kVp has _______ effect on x-ray intensity than/as a 10% increase in mAs.

• the same

• much greater

• less

• much less

Answer explanation

A 10% increase in kVp has much greater effect on x-ray intensity than a 10% increase in

mAs. Remember, changing kVp has a geometric effect on intensity whereas changing mAs has a direct effect i.e. a 10% increase in kVp would be much greater than a 10% increase in mAs.

If the quantity of electrons hitting the target is doubled, the x-ray intensity is _________.

• reduced by half

• just slightly increased

• increased by a factor of two

• increased by a factor of four

Answer explanation

If x-ray quantity is doubled, the optical density on the finished radiograph will be increased by

a factor of two. A change in mAs has a direct effect on x-ray beam quantity/intensity/exposure. A doubling of mAs would equal a doubling of exposure on our film.

If a technologist changes the technique from 70 kVp @ 200 mAs to 70 kVp @ 400 mAs, the

x-ray intensity will _______.

• double

• remain the same

• decrease by half

• increase by 1/4

Answer explanation

You can of course use the inverse square law to calculate this OR you can go with what we already know about intensity and mA/mAs. The problem says we go from 200 to 400 mAs. If we double the mAs (all other factors being constant, such as the 70 kVp in the problem), as we do in the problem, then the intensity must double.

If the intensity of a 70 kVp exposure at 20 mAs is 1.0 mGy_a, what would it be at 5 mAs?

• 0.25 mGy_a

• 0.5 mGy_a

• 2.0 mGy_a

• 4.0 mGy_a

Answer explanation

You can use the inverse square law to solve for this because we're changing intensity by changing mAs. We can also just remember what we know about the relationship between intensity and mAs. If we change from 20 mAs to 5 mAs, that's a reduction in mAs by a factor of 4. Because intensity is directly related to mAs, we know then that if the intensity of radiation at 20 mAs is 1 mGy_a, then the exposure at 4 times less mAs must also be 4 times less, so the exposure is 0.25 mGy_a.

If the intensity of a 70 kVp exposure at 20 mAs is 1.0 mGya, it would be 0.25 mGya at 5 mAs

because exposure is proportional to mAs.