A radioactive nucleus $n_2$ has 3 times the decay constant as compared to the decay constant of another radioactive nucleus $n_1$. If initial number of both nuclei are the same, what is the ratio of number of nuclei of $n_2$ to the number of nuclei of $n_1$, after one half-life of $n_1$ ?

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{N}_2=\mathrm{N}_0 \mathrm{e}^{-3 \lambda t} \\ & \mathrm{~N}_1=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}} \\ & \frac{\mathrm{N}_2}{\mathrm{~N}_1}=\mathrm{e}^{-2 \lambda \mathrm{t}} \\ & \mathrm{t}_{\text {half lifeofN }} \mathrm{t} \\ & \mathrm{t}=\frac{\ln 2}{\lambda}{ }_{\mathrm{n}} 1 \\ & \frac{\mathrm{~N}_0}{2}=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}}\end{aligned}$
$\begin{aligned} & \lambda \mathrm{t}=\ln 2 \\ & \mathrm{t}=\frac{\ln 2}{\lambda} \\ & =\mathrm{e}^{-2 \lambda \frac{\ln 2}{\lambda}} \\ & \frac{\mathrm{~N}_2}{\mathrm{~N}_1}=\frac{1}{4}\end{aligned}$

A radioactive nucleus $n_2$ has 3 times the decay constant as compared to the decay constant of another…

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