Question

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

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

Answer 1

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.