Using Co-60, if the radiation intensity at a certain distance is 600 mR/hr, what thickness of lead is required to reduce this intensity to 75 mR/hr?

To determine the thickness of lead required to reduce the radiation intensity from 600 mR/hr to 75 mR/hr using Co-60, we first need to understand the relationship between the thickness of a shield and the intensity of the radiation passing through it. This is typically modeled using the concept of half-value thickness (HVT), which is the thickness of a particular material required to reduce the intensity of radiation by half.

The formula relating the initial intensity (I0), the transmitted intensity (I), the number of half-values (n), and the half-value thickness (HVT) is:

[ I = I_0 \times (1/2)^n ]

To reduce the intensity from 600 mR/hr to 75 mR/hr, we can calculate the number of half-values needed.

1. Start by determining the reduction factor:

[ \frac{600 \text{ mR/hr}}{75 \text{ mR/hr}} = 8 ]

2. This indicates that the intensity needs to be reduced by a factor of 8. Each half-value thickness reduces the intensity by a factor of 2.

3. Calculate the number of half-values required to achieve this reduction:

To find how