Questions Flashcards
Why is mA the better controlling factor to increase the exposure to the IR?
Because mAs is a linear relationship to intensity. Increase the mAs and you know the exact number of photons that will be produced. kVp is not linear
Give an example of when kVp might be used instead of mAs to increase exposure to the IR?
To penetrate more tissue - increase kVp, decrease mAs. Also to reduce patient dose
What does a change in kVp do to the electrons within the x-ray tube? How does this influence the intensity of the x-ray beam?
Electrons have more kinetic energy, accelerate faster. More photons hitting anode. kVp - highest possible energy a photon can have
56” SID 40 mAs - 33” SID what is new mAs
40 x (33/56)^2 = 14 mAs
40” SID, 40 mAs - 72” SID what is new mAs
40 x (72/40)^2 = 129.6 mAs
72” SID, 20 mAs - 10mAs, what is new SID
72^2 x (10/20) = square root of 2592 = 51 “
40” SID to 60”SID, 70 kVp to 90 kVp 10 mAs to _____ mAs
10 x (60/40)^2 x (70/90)^4 = 8.2 mAs
Calculate the anticipated primary beam intensity change due to a change from 70 kVp at 13.6 mR to 80 kVp.
13.6 x (80/70)^2 = 17.8 mR
primary beam is power of 2.
What new kVp would be required to double density on a radiograph if the original kVp was 80?
80 x 1.15 = 92 kVp. 15% rule.
