In a prostate cancer treatment patients were given six injections of radium-223 chloride, an \(\alpha\)-emitter of RBE 20 at a dose of 50 kBq per kilogram of body weight intravenously in each injection. What was the dosage in (a) sievert (b) rem. Assume all radiation from radium-223 was absorbed and \(\alpha\) particle energy = 5.8 MeV.
Solution.
(a) Each decay produces one alpha particle of energy 5.8 MeV, which is \(9.7 \times 10^{-13}\) J. The dose of 50 kBq per kilogram of body weight will deliver the absorbed dosage to be
\begin{equation*}
\textrm{Absorbed dosage/injection } = 50\times 10^{3} \times 9.7 \times 10^{-13}\:\textrm{ J}\:\textrm{ per kg} = 4.8\times 10^{-8}\:\textrm{Gy}.
\end{equation*}
The total in six injections
\begin{equation*}
\textrm{Absorbed dosage} = 6\times 4.8\times 10^{-8}\:\textrm{Gy} = 2.9 \times 10^{-7}\:\textrm{Gy} .
\end{equation*}
Now, we can multiply this by the RBE for alpha particles to obtain the effective dosage in Sv.
\begin{equation*}
\textrm{Effective Dosage } = 2.9 \times 10^{-7}\:\textrm{Gy} \time 20\:\textrm{Sv/Gy} = 5.8 \times 10^{-6}\:\textrm{Sv}.
\end{equation*}
(b) to convert this into rem unit we just multiply this by 100.
\begin{equation*}
\textrm{Effective Dosage } =5.8 \times 10^{-6}\:\textrm{Sv} = 5.2 \times 10^{-4}\:\textrm{rem}.
\end{equation*}

