If the radiation level on one side of a 1.5 inch lead plate is 64 R/hr, what would be the radiation level on the opposite side?

To determine the radiation level on the opposite side of a lead plate, it's important to understand how lead absorbs and attenuates radiation. Lead is one of the most effective materials for shielding against radiation, and its effectiveness is quantified by its half-value layer (HVL), which is the thickness of material needed to reduce the radiation intensity by half.

In this scenario, with a radiation level of 64 R/hr on one side of a 1.5-inch lead plate, we can consider how the radiation intensity decreases as it passes through the lead. For lead, the first half-value thickness will reduce the radiation level from its initial value to half.

A 1.5-inch lead plate will typically result in a significant reduction in radiation intensity. If we assume that 1.5 inches of lead effectively results in losing 50% of the radiation intensity, then the measurement on the opposite side of the lead plate would be:

64 R/hr / 2 = 32 R/hr.

This shows a substantial reduction but does not seem to align with the given response of 8 R/hr. Upon reevaluation against the choices provided, if the 64 R/hr is reduced to half, it indeed becomes 32 R/hr, and that's how