The activity of sample of radioactive materials is $A_1$ at time $t_1$ and $A_2$ at time $t_2(t_2>t_1)$. It's mean life is $T$, then

The correct option is A $A_2=A_1{e}^{\left(\frac{t_1-t_2}{T}\right)}$
According to the radioactive decay law,

$A_1=A_0{e}^{-\lambda t_1}...\left(1\right)$

$A_2=A_0{e}^{-\lambda t_2}...\left(2\right)$

Using equations $\left(1\right)$ and $\left(2\right)$,

$\frac{A_1}{A_2}=\frac{A_0{e}^{-\lambda t_1}}{A_0{e}^{-\lambda t_2}}=e^{-\lambda (t_1-t_2)}$

$\Rightarrow \frac{A_1}{A_2}=e^{-\frac{(t_1-t_2)}{T}}[\because T=\frac{1}{\lambda }]$

$\Rightarrow \frac{A_2}{A_1}=e^{\left(\frac{t_1-t_2}{T}\right)}$

$\therefore A_2=A_1{e}^{\left(\frac{t_1-t_2}{T}\right)}$

Hence, $\left(\text{A}\right)$ is the correct answer.