= Solution
\b[The idealized external photon estimate is approximately $0.557\ \mathrm{mSv\,h^{-1}}$ at one metre.] First apply <radioactive decay> to the certified <radioactive activity>:
$$
A=2.4\times10^9\,2^{-3/5.27}
=1.61751\times10^9\ \mathrm{Bq}.
$$
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 is
$$
E_\gamma=(1.17+1.33)\ \mathrm{MeV}
=4.00544\times10^{-13}\ \mathrm J.
$$
The incident photon energy flow per area at distance $r$ is therefore
$$
\Psi_E(r)=\frac{AE_\gamma}{4\pi r^2}.
$$
Interpret the given $3\ \mathrm{m^{-1}}$ as the <linear energy-absorption coefficient> $\mu_{\rm en}$, rather than a total photon-removal coefficient. For an absorbing medium of <mass density> $\rho$, a thin volume $S\,dx$ absorbs energy per time $\Psi_E S\mu_{\rm en}\,dx$ and has <mass> $\rho S\,dx$. Its <absorbed dose> rate is consequently
$$
\dot D=\Psi_E\frac{\mu_{\rm en}}{\rho}.
$$
For the estimate assume a water-like $\rho=10^3\ \mathrm{kg\,m^{-3}}$ and local deposition of the secondary charged-particle energy. At $r=1\ \mathrm m$,
$$
\boxed{
\dot D=
\frac{(1.61751\times10^9)(4.00544\times10^{-13})}{4\pi}
\frac{3}{1000}
=1.54671\times10^{-7}\ \mathrm{Gy\,s^{-1}}
=0.556815\ \mathrm{mGy\,h^{-1}}.
}
$$
For <gamma rays>, the <radiation weighting factor> is one, giving \b[$\dot H=0.556815\ \mathrm{mSv\,h^{-1}}$] under these assumptions.<Absorbed, Equivalent, and Effective Dose by International Commission on Radiological Protection>[More Details: photon radiation weighting factor one and the distinction between absorbed and equivalent dose]{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.
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