Answer to Question #261132 in Physics for Frimpong

Question

A 60-kg patient about to have a brain scan is injected with 0.02 Ci of the radionuclide Technetium-99. Technetium-99 decays to the ground state with a half-life of 6.0 hours. The Technetium-99 nucleus decays by emitting a 143-keV photon. Assume that half of these photons escape the body without interacting. What biological equivalent dose does the patient receive? The relative biological effectiveness (RBE) for these photons is 0.97. Assume that all of the Technetium-99 decays while in the body.

Answer 1

Biological equivalent dose

$H_T=D_{T,R}\cdot W_R=\frac{E}{m}\cdot W_R$

$H_T$ is the equivalent dose;

$D_{T,R}$ is the absorbed dose;

$W_R$ is the relative biological effectiveness;

$E$ is the radiation energy absorbed by matter of mass $m$.

$E=0.02\cdot3.7\cdot 10^{10}\cdot143000\cdot1.6\cdot10^{-19}/2=8.5\cdot10^{-6}\ (Gy)$

$H_T=D_{T,R}\cdot W_R=\frac{E}{m}\cdot W_R=\frac{8.5\cdot10^{-6}}{60}\cdot0.97=1.4\cdot10^{-7}\ (Sv)$ . Answer