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

Question #118051
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?
1
Expert's answer
2020-05-27T10:36:08-0400

By definition, the equivalent dose will be:


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

where Dα=EαmD_{\alpha} = \dfrac{E_{\alpha}}{m} and Dγ=EγmD_{\gamma} = \dfrac{E_{\gamma}}{m} are absorbed dose from alpha particles and gamma rays respectively. Thus:


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

Substitute numerical values:


H=203J75kg+110J75kg=70J75kg0.93Sv=93remH = 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.


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