13.Nuclei
medium

Define the average life of a radioactive sample and obtain its relation to decay constant and half life. 

Option A
Option B
Option C
Option D

Solution

"In radioactive sample the average life time of a nucleus for which the nucleus exists. This time is called average or mean life".

OR

"The time interval, during which the number of nuclei of a radioactive element becomes equal to the $e^{\text {th }}$ part of its original number is called the mean life or average life of that element".

The average life is denoted by $\tau$

$\therefore$ Average life time $\tau=\frac{\text { Sum of life time of all nucleus }}{\text { No. of total nucleus }}$

The relation between average life and decay constant : Suppose at $t=0$ time radioactive sample contains $\mathrm{N}_{0}$ nuclei. After $t$ time the number of nuclei decreases to $\mathrm{N}$ and the number of nuclei which decay in the time interval $t$ to $t+\Delta t$ is $d \mathrm{~N}$.

Since the $d t$ time is very small, the life time of each $d \mathrm{~N}$ nucleus can be taken approximate to $t$ $\therefore$ Total life time of $d \mathrm{~N}$ nucleus $=t d \mathrm{~N}$

$\therefore$ Total life time of all $\mathrm{N}_{0}$ nucleus $=\int_{0}^{\mathrm{N}_{0}} t d \mathrm{~N}$

$\therefore$ Average life time $=\frac{\begin{array}{c}\text { Life time of } \\ \text { all } \mathrm{N}_{0} \text { nucleus }\end{array}}{\mathrm{N}_{0}}$

$\therefore \tau=\frac{1}{\mathrm{~N}_{0}} \int_{0}^{\mathrm{N}_{0}} t d \mathrm{~N}$$…(1)$ 

But exponential law $\mathrm{N}=\mathrm{N}_{0} e^{-\lambda t}$

$\therefore d \mathrm{~N}=-\lambda \mathrm{N}_{0} e^{-\lambda t} d t$

$\ldots(2)$

$\therefore \tau=\frac{1}{\mathrm{~N}_{0}} \int_{0}^{\mathrm{N}_{0}}-\lambda t \mathrm{~N}_{0} e^{-\lambda t} d t$

(Putting value of equation

$(2)$ in $(1)$)

Now when $\mathrm{N}=\mathrm{N}_{0}$, then $t=0$ and

$\mathrm{N}=0$, then $t=\infty$

Standard 12
Physics

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