To achieve a reduction of radiation exposure to half its value, what distance must a radiologic technologist position the patient if the output is 70 uGya/mAs at 75 cm?

To determine the distance a radiologic technologist must position the patient to achieve a reduction of radiation exposure to half its original value, the concept of the inverse square law in radiation physics is applied. This law states that radiation intensity is inversely proportional to the square of the distance from the source.

Given that the original exposure is 70 µGy/mAs at a distance of 75 cm, we need to identify the new distance at which the exposure would be reduced to 35 µGy/mAs (which is half of 70 µGy/mAs).

Using the formula from the inverse square law, we can set up the relationship as follows:

Original intensity / New intensity = (Original distance / New distance)²

Substituting in the values we have:

70 µGy/mAs / 35 µGy/mAs = (75 cm / New distance)²

This simplifies to:

2 = (75 cm / New distance)²

Taking the square root of both sides gives us:

√2 = 75 cm / New distance

Rearranging the equation results in:

New distance = 75 cm / √2 ≈ 75 cm / 1.414 = approximately 53 cm

However, since we