At any instant, two elements $X_1$ and $X_2$ have same number of radioactive atoms. If the decay constant of $X_1$ and $X_2$ are $10 \lambda$ and $\lambda$ respectively. Then the time when the ratio of their atoms becomes $\frac{1}{e}$ respectively will be:

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Aicharya · Accepted Answer

By radioactivity law, we have,
$\mathrm{N}=\mathrm{N}_o e^{-\lambda t}$
(where, the symbols have their usual meanings) Hence, for two elements,
$\frac{\mathrm{N}_1}{\mathrm{~N}_2}=\frac{\mathrm{N}_o e^{-10 \lambda t}}{\mathrm{~N}_o e^{-\lambda t}}=\frac{1}{e}$
Hence, $e^{-9 \lambda t}=e$
$t=\frac{1}{9 \lambda}$

At any instant, two elements $X_1$ and $X_2$ have same number of radioactive atoms. If the decay constant of…

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