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A radioactive sample consists of two distinct species having equal number of atoms $N_0$ initially. The mean-life of one species is $\tau $ and of the other is $5\tau $. The decay products in both cases is stable. The total number of radioactive nuclei at $t = 5\tau $ is
${N_0}\left( {\frac{{{e^5} + 1}}{{{e^5}}}} \right)$
${N_0}\left( {\frac{{{e^4} + 1}}{{{e^5}}}} \right)$
${N_0}\left( {\frac{{e + {e^5}}}{{{e^5}}}} \right)$
$N_0e^{-3}$
Solution
$\mathrm{N}_{1}=\mathrm{N}_{0} \mathrm{e}^{-\mathrm{t} / \tau}=\mathrm{N}_{0} \mathrm{e}^{-\mathrm{s} \tau / \tau}=\mathrm{N}_{0} \mathrm{e}^{-5}$
$\mathrm{N}_{2}=\mathrm{N}_{0} \mathrm{e}^{-\mathrm{t} / 5 \tau}=\mathrm{N}_{0} \mathrm{e}^{-5 \tau / 5 \tau}=\mathrm{N}_{0} \mathrm{e}^{-1}$
$\therefore $ Total number of nuclei
$=\mathrm{N}_{1}+\mathrm{N}_{2}=\frac{\mathrm{N}_{0}}{\mathrm{e}^{5}}+\frac{\mathrm{N}_{0}}{\mathrm{e}}$
$=\mathrm{N}_{0}\left(\frac{\mathrm{e}^{4}+1}{\mathrm{e}^{5}}\right)$
