Question

an x - ray exposure is made using 80 kilovoltage peak (kvp) and 100 milliampere - seconds (mas). at which of the following distances would the image receptor (ir) receive the most exposure? 30 inches (76 cm) source - to - image distance (sid) 40 inches (102 cm) source - to - image distance (sid) 48 inches (122 cm) source - to - image distance (sid) 72 inches (183 cm) source - to - image distance (sid)

Explanation:

Step1: Recall the inverse - square law

The intensity of radiation at the image receptor is inversely proportional to the square of the source - to - image distance (SID). Mathematically, $I\propto\frac{1}{SID^{2}}$, where $I$ is the intensity of radiation (exposure).

Step2: Analyze the relationship between SID and exposure

As the SID decreases, the exposure at the image receptor increases.

Step3: Compare the given SID values

We have SID values: 30 inches, 40 inches, 48 inches, and 72 inches. The smallest SID value among them is 30 inches.

Answer:

30 inches (76 cm) source - to - image distance (SID)