If the exposure rate to an individual standing 7.0 feet from a source of radiation is 0.5 mGy/hr, what will be the dose received after 20 minutes at a distance of 3.0 feet from the source?

You can simply use variation to get the answer. if at 7 feet, the exposure rate is 0.5mGy/hr. Then at 1 feet, the exposure rate would be higher because the rate at which you are exposed to radiation is inversely proportional to the distance you are from the source. i.e the smaller the distance the higher the rate of exposure.

At 1 feet, rate of exposure = 7 x 0.5mGy/hr = 3.5mGy/hr
At 3 feets, rate of exposure = [tex]\frac{3.5mGy/hr}{3}[/tex] = 1.167mGy/hr. The dosage received is 1.167mGy in 1hour at 3feet

But the question is what will be the dose exposed to after 20 minutes at a distance of 3.0 feet from the source?

To calculate the dosage a person is exposed to in 20minutes, we have to know the relationship between exposure time and rate of exposure, which is directly proportional i.e the more exposure time, the higher the rate of exposure .

we calculate as follow

[tex]\frac{1.167*20 }{60} = 0.389mGy[/tex]