Define the average life of a radioactive sample and obtain its relation to decay constant and half life.
"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 $t=\infty$
After $3$ hours, only $0.25 \,mg$ of a pure radioactive material is left. If initial mass was $2 \,mg$ then the half life of the substance is ...... $hr$
If half life of a radioactive element is $3\, hours$, after $9\, hours$ its activity becomes
Curie is a unit of
Obtain the amount of $_{27}^{60} Co$ necessary to provide a radioactive source of $8.0\; mCi$ strength. The half-life of $^{60}_{27} Co$ is $5.3$ years.
The activity of a sample reduces from $A_0$ to ${A_0} / \sqrt{3}$ in one hour. The activity after $3$ hours more will be