Write and derive the law of radioactivity decay.
Write and derive the law of radioactivity decay.
"In any radioactive sample, which undergoes $\alpha, \beta$ or $\gamma$-decay, it is found that the number of nuclei undergoing the decay per unit time is proportional to the total number of nuclei in the sample".
If $\mathrm{N}$ is the number of nuclei in the sample and $\Delta \mathrm{N}$ undergo decay in time $\Delta t$ then $\frac{\Delta \mathrm{N}}{\Delta t} \propto \mathrm{N}$
The number $\Delta \mathrm{N}$ of the decaying nucleus is always positive, $\therefore \frac{\Delta \mathrm{N}}{\Delta t}=\lambda \mathrm{N}$
where $\lambda$ is called decay constant or disintegration constant. If the period $\Delta t$ corresponds to zero,
$\lim _{\Delta t \rightarrow 0} \frac{\Delta \mathrm{N}}{\Delta t}=-\lambda \mathrm{N}$
$\therefore-\frac{d \mathrm{~N}}{d t}=\lambda \mathrm{N}$$...(1)$
where $d \mathrm{~N}$ is the change in $\mathrm{N}$, which may be positive or negative. Here it is negative because as time goes by the number of nucleus remaining will decreases.
Writing equation $(1)$ as follows,
$\frac{d \mathrm{~N}}{\mathrm{~N}}=-\lambda d t$
“In any radioactive sample, which undergoes $\alpha, \beta$ or $\gamma$-decay, it is found that the number of nuclei undergoing the decay per unit time is proportional to the total number of nuclei in the sample".
If $\mathrm{N}$ is the number of nuclei in the sample and $\Delta \mathrm{N}$ undergo decay in time $\Delta t$ then $\frac{\Delta \mathrm{N}}{\Delta t} \propto \mathrm{N}$
The number $\Delta \mathrm{N}$ of the decaying nucleus is always positive,
$\therefore \frac{\Delta \mathrm{N}}{\Delta t}=\lambda \mathrm{N}$
where $\lambda$ is called decay constant or disintegration constant.
If the period $\Delta t$ corresponds to zero,
$\lim _{\Delta t \rightarrow 0} \frac{\Delta \mathrm{N}}{\Delta t}=-\lambda \mathrm{N}$
$\therefore-\frac{d \mathrm{~N}}{d t}=\lambda \mathrm{N}$
where $d \mathrm{~N}$ is the change in $\mathrm{N}$, which may be positive or negative. Here it is negative because as time goes by the number of nucleus remaining will decreases.
Writing equation $(1)$ as follows,
$\frac{d \mathrm{~N}}{\mathrm{~N}}=-\lambda d t$
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$(C)$ Slope of the graph of $\log _{e}$ (no. of undecayed nuclei) $Vs$. time represents the reciprocal of mean life time $(\tau)$.
$(D)$ Product of decay constant ( $\lambda$ ) and half-life time $\left(T_{1 / 2}\right)$ is not constant.
Choose the most appropriate answer from the options given below
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