Question

If a radiographer receives an exposure of 0.05 millisievert (msv) at a distance of 1.5 feet from the tube of a portable x-ray unit, what will the exposure be at a distance of 6 feet from the tube?

Answer 1

Answer:

0.003125 msv

Explanation:

$I_1$ = Initial Exposure = 0.05 msv

$I_2$ = Final Exposure

$D_1$ = Initial distance = 1.5 ft

$D_2$ = Final distance = 6 ft

Exposure is inversely related to the distance squared (inverse square law)

$I\propto \dfrac{1}{D}$

So,

$\dfrac{I_1}{I_2}=\dfrac{D_2^2}{D_1^2}\\\Rightarrow I_2=\dfrac{I_1D_1^2}{D_2^2}\\\Rightarrow I_2=\dfrac{0.05\times 1.5^2}{6^2}\\\Rightarrow I_2=0.003125\ msv$

The exposure at the given distance is 0.003125 msv