The activity of a radioactive sample is $A_1$ at time $t_1$ and $A_2$ at time $t_2$. If the mean lif

Question

The activity of a radioactive sample is $A_1$ at time $t_1$ and $A_2$ at time $t_2$. If the mean life of the sample is $	au$, then the number of nuclei decayed in time $\left(t_2-t_1ight)$ is

Professor PC Thomas Classes · Answer

$
A=\left|\frac{d N}{d t}ight|=\lambda N_0 e^{-\lambda t}=\frac{N_0}{	au} e^{-t / 	au}
$
At
$
\begin{aligned}
& t=t_1 ; A_1=\frac{N_0}{	au} e^{-t_1 / 	au} \
& t=t_2 ; A_2=\frac{N_0}{	au} e^{-t_2 / 	au}
\end{aligned}
$
The number of nuclei decayed in time $\left(t_2-t_1ight)$ is or
$
\begin{aligned}
& N_1-N_2=N_0 e^{-t_1 / 	au}-N_0 e^{-t_2 / 	au}=A_1 	au-A_2 	au \
& N_1-N_2=\left(A_1-A_2ight) 	au
\end{aligned}
$