A radiographer is exposed to a dose rate of 10 millisieverts per hour (mSv/hr) while standing 3 feet from the patient during c-arm procedures. What is the approximate dose rate if the radiographer stands 4 feet from the patient?

• A. 7.5 mSv/hr
• B. 5.6 mSv/hr
• C. 13.3 mSv/hr
• D. 17.7 mSv/hr

Answer: (B)

The concept is the inverse square law: radiation intensity falls off with the square of the distance from the source. When you move away from a source, the dose rate decreases by (initial distance / new distance) squared. Here the distance increases from 3 ft to 4 ft, so the dose rate becomes 10 mSv/hr × (3/4)^2 = 10 × 9/16 = 5.625 mSv/hr, about 5.6 mSv/hr. This is the approximate rate you’d expect, noting that real-world factors like exact geometry and scattered radiation can cause small variations, but the inverse square relationship governs the trend.