= Solution
The number of radioactive atoms is determined by the supplied isotope mass, rather than by the total body mass:
$$
N=\frac{0.0164\,\mathrm g}{40\,\mathrm{g\,mol^{-1}}}
(6.02214076\times10^{23}\,\mathrm{mol^{-1}})
=2.4691\times10^{20}.
$$
The <radioactive activity> is $\mathcal A=\lambda N$, with <nuclear decay constant> $\lambda=\ln2/T_{1/2}$. Using $1\,\mathrm{yr}=365.25\times86400\,\mathrm s$,
$$
\lambda=\frac{\ln2}{(1.3\times10^9)(3.15576\times10^7)\,\mathrm s}
=1.6895\times10^{-17}\,\mathrm{s^{-1}}.
$$
Therefore
$$
\boxed{\mathcal A=4.17\times10^3\,\mathrm{Bq}\simeq4.2\,\mathrm{kBq}.}
$$
A <becquerel> is one decay per second; this is total activity over all decay branches. Multiplying by the argon-producing <branching fraction> would give that branch's activity, not the total asked for. Activity alone is also distinct from an absorbed radiation dose.
Back to article page
