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Question

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 radioact

Vedclass · Accepted Answer

$\frac{1}{{7\lambda }}$

Vedclass · Answer

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: $\lambda_{A}=8 \lambda, \lambda_{B}=\lambda, N_{0 A}=N_{0 B}=N_{0}$

$ 	herefore \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 $t=\frac{-1}{7 \lambda}$

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 $\frac{1}{e}$ ?

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