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?

0.003125 msv

Explanation:

= Initial Exposure = 0.05 msv

= Final Exposure

= Initial distance = 1.5 ft

= Final distance = 6 ft

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

So,

The exposure at the given distance is 0.003125 msv

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