Question

a radiograph acquired at 40 inches results in a receptor exposure of 10 milligray (mgy). what is the new receptor exposure if the radiograph is acquired with the same technical factors and 72 inches?
3.1 mgy
5.6 mgy
18 mgy
32.4 mgy

Explanation:

Step1: Recall the inverse - square law

The inverse - square law for radiation exposure states that the intensity of radiation (receptor exposure) is inversely proportional to the square of the distance. Let $E_1$ be the initial exposure, $d_1$ be the initial distance, $E_2$ be the new exposure, and $d_2$ be the new distance. The formula is $E_1d_1^{2}=E_2d_2^{2}$.

Step2: Identify the given values

We are given that $E_1 = 10$ mGy, $d_1 = 40$ inches, and $d_2=72$ inches. We need to solve for $E_2$.

Step3: Rearrange the formula to solve for $E_2$

From $E_1d_1^{2}=E_2d_2^{2}$, we can get $E_2=\frac{E_1d_1^{2}}{d_2^{2}}$.

Step4: Substitute the values into the formula

$E_2=\frac{10\times40^{2}}{72^{2}}=\frac{10\times1600}{5184}=\frac{16000}{5184}\approx 3.1$ mGy

Answer:

3.1 mGy