If an x-ray exposure at 40 inches results in a beam intensity of 200 microgray (μGy), what is the expected intensity at 50 inches?

If an x-ray exposure at 40 inches results in a beam intensity of 200 microgray (μGy), what is the expected intensity at 50 inches?

• A. 128 μGy
• B. 160 μGy
• C. 250 μGy
• D. 312 μGy

Answer: (A)

To determine the expected intensity of the x-ray beam at a new distance, the inverse square law is applied. This law states that the intensity of radiation is inversely proportional to the square of the distance from the source. Mathematically, it can be represented as: \[ I_1 \times D_1^2 = I_2 \times D_2^2 \] Where: - \( I_1 \) is the initial intensity, - \( D_1 \) is the initial distance, - \( I_2 \) is the final intensity, - \( D_2 \) is the final distance. In this scenario, the initial intensity (\( I_1 \)) is 200 μGy at a distance (\( D_1 \)) of 40 inches, and we are trying to find \( I_2 \) at \( D_2 \) of 50 inches. First, substitute the known values into the formula: \[ 200 \, \text{μGy} \times (40 \, \text{inches})^2 = I_2 \times (50 \, \text{inches})^2 \] Calculating the squares gives: \[ 200

To determine the expected intensity of the x-ray beam at a new distance, the inverse square law is applied. This law states that the intensity of radiation is inversely proportional to the square of the distance from the source. Mathematically, it can be represented as:

[ I_1 \times D_1^2 = I_2 \times D_2^2 ]

Where:

• ( I_1 ) is the initial intensity,

• ( D_1 ) is the initial distance,

• ( I_2 ) is the final intensity,

• ( D_2 ) is the final distance.

In this scenario, the initial intensity (( I_1 )) is 200 μGy at a distance (( D_1 )) of 40 inches, and we are trying to find ( I_2 ) at ( D_2 ) of 50 inches.

First, substitute the known values into the formula:

[ 200 , \text{μGy} \times (40 , \text{inches})^2 = I_2 \times (50 , \text{inches})^2 ]

Calculating the squares gives:

[ 200