Past exam of the MPhil in nuclear energy course of the University of Cambridge / 2019 / Module NE1 / 1 / c / Solution

The idealized external photon estimate is approximately $0.557\mathrm{mSv}{\mathrm{h}}^{-1}$ at one metre. First apply radioactive decay to the certified radioactive activity:Treat the source as isotropic and small compared with the observation distance. Neglect photon self-absorption, capsule attenuation and attenuation in air. The metal and capsule stop the primary beta particles, so the direct external contribution is taken to be the two gamma rays; secondary bremsstrahlung is neglected. Their total energy per transformation isThe incident photon energy flow per area at distance $r$ is thereforeInterpret the given $3{\mathrm{m}}^{-1}$ as the linear energy-absorption coefficient ${\mu }_{\mathrm{en}}$, rather than a total photon-removal coefficient. For an absorbing medium of mass density $\rho$, a thin volume $Sdx$ absorbs energy per time ${\Psi }_{E}S{\mu }_{\mathrm{en}}dx$ and has mass $\rho Sdx$. Its absorbed dose rate is consequentlyFor the estimate assume a water-like $\rho =10^3\mathrm{kg}{\mathrm{m}}^{-3}$ and local deposition of the secondary charged-particle energy. At $r=1\mathrm{m}$,For gamma rays, the radiation weighting factor is one, giving $\dot{H}=0.556815\mathrm{mSv}{\mathrm{h}}^{-1}$ under these assumptions.^[ref]

The phrase “equivalent absorbed dose” mixes two quantity names: absorbed dose is measured in gray, while equivalent dose is measured in sievert. Their rates are numerically equal here because the weighting factor is one. The estimate is a physical energy-deposition calculation; the assumed mass density is essential, and the stated inverse-length coefficient must not be used as if it already had area-per-mass units.