$N(t)=N_{0} e^{-\lambda t}$
where $N_{0}$ is number of radioactive nuclei in the sample at some aribitrary time $t=0$ and $\lambda$ is the radioactive decay constant.
Given: $\lambda_{A}=8 \lambda, \lambda_{B}=\lambda, N_{0 A}=N_{0 B}=N_{0}$
$ \therefore \quad \frac{N_{B}}{N_{A}} =\frac{e^{-\lambda t}}{e^{-8 \lambda t}} $
$\frac{1}{e}=e^{-\lambda t} e^{8 \lambda t}=e^{7 \lambda t}$
$\Rightarrow-1=7 \lambda t$ or $t=\frac{-1}{7 \lambda}$
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$\left(h=6.6 \times 10^{34} \;J \cdot sec\right)$,
