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What wholebody dose is received by a $\SI{70}{kg}$ laboratory worker exposed to a $\SI{40}{mCi}$ (\SI{1}{Ci}=\SI{3.7e10}{Bq}) \isotope[60][27]{Co} source, assuming the person's body has a crosssectional area $\SI{1.5}{\meter\squared}$ and is normally about $\SI{4.0}{m}$ from the source for $\SI{4.0}{h}$ per day? \isotope[60][27]{Co} emits $\upgamma$ rays of energy $\SI{1.33}{MeV}$ and $\SI{1.17}{MeV}$ in quick succession. Approximately 50\% of the $\upgamma$ rays interact in the body and deposit all their energy. The rest passes through. $\star$
$\SI{0.45}{mSv}$
The Cobald source has an activity of \begin{align} A &= \SI{40}{mCi}\\ &= 40 \cdot \numprint{e3} \cdot \SI{3.7e10}{Bq}\\ &= \SI{1.48e9}{Bq}. \end{align} This activity is assumed to be radiated uniformely distributed in all directions  i.e. into a sphere. In a distance of $\SI{4}{m}$ to the source, this sphere has an area of \begin{align} \hat A &= 4\pi r^2\\ &= \SI{201}{\meter\squared}. \end{align} Since the body has a crosssectional area of $\SI{1.5}{\meter\squared}$, he is hit by the activity \begin{align} \tilde A &= \SI{1.48e9}{Bq} \cdot \frac{\SI{1.5}{\meter\squared}}{\SI{201}{\meter\squared}}\\ &= \SI{1.10e7}{Bq}, \end{align} of which only 50\% interact with the body, i.e. \begin{align} A' &= \SI{5.52e6}{Bq}. \end{align} During $\SI{4}{h}$ of work, the body of the worker absorbes the energy \begin{align} E &= \SI{4}{h} \cdot \SI{5.52e6}{Bq} \cdot (\SI{1.33}{MeV}+\SI{1.17}{MeV})\\ &= \SI{1.98e11}{MeV}\\ &= \SI{0.0318}{J}. \end{align} The absorbed dose therefore is \begin{align} D &= \frac{E}{m}\\ &= \SI{0.0004548}{Gy}\\ &= \SI{0.455}{mGy}. \end{align} Since $\upgamma$rays have a radiation quality factor of $q=1$, the wholebody dose is \begin{align} H &= qD\\ &= \SI{1}{\sievert\per\gray} \cdot \SI{0.455}{mGy}\\ &= \SI{0.455}{mSv}. \end{align}
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