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${ }^{227} \mathrm{Ac}$ has a half-life of 22 years with respect to radioactive decay . The decay follows two parallel paths : ${ }^{227} \mathrm{Ac} \rightarrow{ }^{227} \mathrm{Th}$ and ${ }^{227} \mathrm{Ac} \rightarrow{ }^{223} \mathrm{Fr}$. If the percentage of the two daughter nuclides are $2.0$ and $98.0$, respectively, the decay constant (in year $^{-1}$ ) for ${ }^{227} \mathrm{Ac} \rightarrow{ }^{227}$ Th path is closest to
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$6.3 \times 10^{-4}$
$\mathrm{Ac}^{\mathrm{R}_{1}} / \mathrm{R}_{2} \backslash \mathrm{Ac}$
$\% \mathrm{Th}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{\mathrm{T}}}=\frac{2}{100}$
$\% \mathrm{Ac}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{\mathrm{T}}}=\frac{98}{100}$
$\therefore \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{2}{98}$
$\mathrm{R}_{\mathrm{T}}=\mathrm{R}_{1}+\mathrm{R}_{2}$
$\frac{0.693}{22}=\mathrm{R}_{1}+\frac{98}{2} \mathrm{R}_{1}$
$\therefore \mathbf{R}_{1}=\mathbf{6 . 3} \times \mathbf{1 0}^{-4}$
$\% \mathrm{Th}=\frac{\mathrm{R}_{1}}{\mathrm{R}_{\mathrm{T}}}=\frac{2}{100}$
$\% \mathrm{Ac}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{\mathrm{T}}}=\frac{98}{100}$
$\therefore \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{2}{98}$
$\mathrm{R}_{\mathrm{T}}=\mathrm{R}_{1}+\mathrm{R}_{2}$
$\frac{0.693}{22}=\mathrm{R}_{1}+\frac{98}{2} \mathrm{R}_{1}$
$\therefore \mathbf{R}_{1}=\mathbf{6 . 3} \times \mathbf{1 0}^{-4}$
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