During a fluoroscopic exam lasting seven minutes, if the exposure rate to the operator is calculated to be 15mR/minute at a distance of three feet, what will be the technologist's total exposure standing at five feet from the table?

To determine the technologist's total exposure standing at five feet from the table during a seven-minute fluoroscopic exam, it is essential first to calculate the exposure at this new distance using the inverse square law, which explains how radiation intensity decreases with distance from the source.

Initially, the exposure rate is provided as 15 mR/minute at three feet. The exposure at five feet can be calculated based on the fact that intensity varies inversely with the square of the distance. The inverse square law states that if you double the distance from a point source of radiation, the intensity is reduced to one-fourth.

To apply this, you first need to establish how the distances relate:

• From three feet to five feet, the increase in distance is a factor of:

(5 feet)^2 / (3 feet)^2 = 25 / 9

This means the exposure rate at five feet would be reduced by a factor of 25/9. Therefore, the exposed rate at five feet is:

Exposure rate at five feet = (15 mR/min) × (9/25) = 5.4 mR/min

Now, with an exposure rate of 5.4 mR/min for a duration of seven minutes