Answer to Question #118051 in Atomic and Nuclear Physics for Jess

Question

An unwary physicist accidentally ingests radioactive sample that is both an alpha emitter (RBEα = 20) and a gamma emitter (RBE = 1).

The 75 kg physicist absorbs 3 J of energy in the form of alpha particles and 10 J of energy in the form of gamma rays.

What is the total dose equivalent in rem that the physicist has been exposed to?

Answer 1

By definition, the equivalent dose will be:

$$H = RBEα\cdot D_{\alpha} + RBE\cdot D_{\gamma}$$

where $D_{\alpha} = \dfrac{E_{\alpha}}{m}$ and $D_{\gamma} = \dfrac{E_{\gamma}}{m}$ are absorbed dose from alpha particles and gamma rays respectively. Thus:

$$H = RBEα\cdot \dfrac{E_{\alpha}}{m} + RBE\cdot \dfrac{E_{\gamma}}{m}$$

Substitute numerical values:

$$H = 20\cdot \dfrac{3J}{75kg} + 1\cdot \dfrac{10J}{75kg} = \dfrac{70J}{75kg} \approx 0.93Sv = 93 rem$$

Answer. The equivalent dose is 93 rem.