= Solution
\b[The idealized isotope mass is $1.63\times10^{-2}\ \mathrm{kg}$, about $16.3\ \mathrm g$.] Both the <alpha particle> and daughter recoil energy are recovered in the stipulated <heat> balance. Thus the recovered energy per transformation is
$$
E_h=4.6007\ \mathrm{MeV}=7.3712\times10^{-13}\ \mathrm J,
$$
and the <nuclear decay constant> is
$$
\lambda=\frac{\ln2}{138.4(86400)\ \mathrm s}=5.7966\times10^{-8}\ \mathrm{s}^{-1}.
$$
With $N_0=(m/M_{\rm mol})N_A$, thermal <power> is $P_0=\lambda N_0E_h$. Consequently
$$
\boxed{m=\frac{P_0M_{\rm mol}}{\lambda N_AE_h}
=\frac{2000(0.210)}{(5.7966\times10^{-8})(6.02214\times10^{23})(7.3712\times10^{-13})}
\simeq0.0163\ \mathrm{kg}.}
$$
The conversion of $210\ \mathrm{kg\,kmol^{-1}}$ to $0.210\ \mathrm{kg\,mol^{-1}}$ is essential. This is an isotope-only energy-accounting result under the paper's prescribed branch and perfect heat-recovery assumptions; it is not the mass or design of a complete generator.
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