Question

A radioactive sample is undergoing $\alpha -$decay. At any time $t_1$, its activity is $A$ and at another time $t_2$, the activity is $A/5$. What is the average life-time for the sample?

• A. $\frac{t_2-t_1}{\ln 5}$
• B. $\frac{\ln (t_2-t_1)}{2}$
• C. $\frac{t_1-t_2}{\ln 5}$
• D. $\frac{\ln 5}{t_2-t_1}$

Answer 1

The correct option is A $\frac{t_2-t_1}{\ln 5}$

For activity of radioactive sample,

$A=A_o{e}^{-\lambda t_1}...\left(1\right)$, and,

$\frac{A}{5}=A_o{e}^{-\lambda t_2}...\left(2\right)$

Dividing equation $\left(1\right)$ by $\left(2\right)$, we get,

$5=e^{\lambda (t_2-t_1)}$

$\Rightarrow \ln 5=\lambda (t_2-t_1)$

$\Rightarrow \lambda =\frac{\ln 5}{t_2-t_1}$

Now, average life-time,

$t_{\text{av}}=\frac{1}{\lambda }=\frac{t_2-t_1}{\ln 5}$