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Radioactive material $'A'$ has decay constant $8 \lambda$ and material $'B'$ has decay constant $ ' \lambda '$. Initially they have same number of nuclei . After what time, the ratio of number of nuclei of material $'B'$ to that $'A'$ will be $\frac{1}{e}$ ?
$\frac{1}{{\lambda }}$
$\frac{1}{{9\lambda }}$
$\frac{1}{{8\lambda }}$
$\frac{1}{{7\lambda }}$
Solution
The number of radioactive nuclei $'N'$ at any time $t$ is given as
$N(t)=N_{0} e^{-\lambda t}$
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 $t=\frac{-1}{7 \lambda}$
