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

0.003125 msv

Step-by-step 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 (1)/(D)`

So,

`(I_1)/(I_2)=(D_2^2)/(D_1^2)\\\Rightarrow I_2=(I_1D_1^2)/(D_2^2)\\\Rightarrow I_2=(0.05* 1.5^2)/(6^2)\\\Rightarrow I_2=0.003125\ msv`

The exposure at the given distance is 0.003125 msv