Question

a portable chest x - ray is to be acquired on an emergency room (er) patient. the technique chart indicates the exposure should be made with 120 kilovoltage peak (kvp), 3.5 millampere - seconds (mas), and a 72 inch source - to - image distance (sid). the room is small and the exposure must be made at 55 inches. how should the technique be changed?
increase mas to 4.1
increase mas to 4.7
decrease mas to 3.0
decrease mas to 2.0

Explanation:

Step1: Recall the inverse - square law for x - ray exposure

The relationship between intensity (or mAs) and distance in x - ray is $I_1/I_2=(D_2/D_1)^2$. Since intensity is proportional to mAs, we can write $\frac{mAs_1}{mAs_2}=(\frac{D_2}{D_1})^2$, where $mAs_1$ and $D_1$ are the initial mAs and distance, and $mAs_2$ and $D_2$ are the new mAs and distance.

Step2: Identify the given values

We have $mAs_1 = 3.5$, $D_1=72$ inches, and $D_2 = 55$ inches. We need to find $mAs_2$.

Step3: Rearrange the formula to solve for $mAs_2$

$mAs_2=mAs_1\times(\frac{D_1}{D_2})^2$.

Step4: Substitute the values into the formula

$mAs_2=3.5\times(\frac{72}{55})^2=3.5\times\frac{72^2}{55^2}=3.5\times\frac{5184}{3025}\approx3.5\times1.714\approx4.7$.

Answer:

Increase mAs to 4.7