Question

The activity of a sample of a radio active material is $A_1$ at time $t_1$ and $A_2$ at time $t_2$ ($t_2>t_1$). If its mean life is $T_1$, then

• A. $A_1{t}_1=A_2{t}_2$
• B. $A_1{A}_2=t_2-t_1$
• C. $A_2=A_1{e}^{\frac{t_1-t_2}{T}}$
• D. $A_2=A_1{e}^{t_1/t_2}T$

Answer 1

Answer: (C)

$A_2=A_1{e}^{\frac{t_1-t_2}{T}}$

$\begin{array}{c}\text{Given-}\ast \text{Activity at}t=t_1\text{is}{A}_1\\ \ast \text{Activity at}t=t_2\text{is}{A}_2\\ \ast \text{Mean life}=T\\ \text{we}\text{know that}\text{mean life,}\\ T=\frac{1}{\lambda },\lambda =\text{disintegration}\\ \Rightarrow \lambda =\frac{1}{T}-\left(1\right)\end{array}$

$\begin{array}{c}\text{By radioactive law of disintegration-}\\ \text{activity}A=A_0\cdot e^{-\lambda t},A_0=\text{initial activity}\\ \Rightarrow A_1=A_0\cdot e^{-\lambda t_1}\text{and}{A}_2=A_0\cdot e^{-\lambda t_2}\end{array}$

$\begin{array}{c}\Rightarrow \frac{A_2}{A_1}=e^{\lambda t_1-\lambda t_2}\\ \Rightarrow A_2=A_1\cdot e^{\lambda (t_1-t_2)}\\ \text{From eq}1,\text{we get -}\\ A_2=A_1\cdot e^{(t_1-t_2)/T}\\ \text{Henu option (C) is correct.}\end{array}$