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 lifetime for the sample?

• A. $(t_2–t_1 )/\ln \hspace{0.17em}5$
• B. $\ln \hspace{0.17em}(t_2+t_1 )/2$
• C. $(t_1–t_2 )/\ln \hspace{0.17em}5$
• D. $ln\hspace{0.17em}5/(t_2–t_1 )$

Answer 1

The correct option is A

$(t_2–t_1 )/\ln \hspace{0.17em}5$

Step 1: Given data

Decay $=\alpha$ decay

Activity at time $t_1$ $=A$

Activity at time $t_2$ $=A/5$

Initial activity$=A_0$

Step 2: Find average life

Let Average life $=\frac{1}{\lambda }$

For activity of radioactive sample,

$$\begin{gathered} A=A_0{e}^{-\lambda t_1}\left(1\right) \\ A/5=A_0{e}^{-\lambda t_2}\left(2\right) \end{gathered}$$

On dividing $\left(1\right)$ by $\left(2\right)$, we get

$$\begin{gathered} 5=e^{-\lambda (t_1-t_2)} \\ \ln \left(5\right)=(t_2-t_1)\lambda \\ \Rightarrow \lambda =\ln \left(5\right)/t_2–t_1 \end{gathered}$$

Average life $=\frac{1}{\lambda }=(t_2-t_1)/\ln \left(5\right)$

Hence, option A is correct.