Question

Two radioactive samples $\text{1}$ and $\text{2}$ have half lives $T_1$ and $T_2(T_1>T_2)$ respectively. At $t=0$, the activity of $\text{2}$ was twice the activity of $\text{1}$. Their activity will become equal after a time

• A. $\frac{T_1{T}_2}{T_1-T_2}$
• B. $\frac{T_1-T_2}{2}$
• C. $\frac{T_1+T_2}{2}$
• D. $\frac{T_1{T}_2}{T_1+T_2}$

Answer 1

The correct option is A $\frac{T_1{T}_2}{T_1-T_2}$

Given that, $2R_1=R_2(\text{At}t=0)$

$\Rightarrow 2{\lambda }_1{N}_1={\lambda }_2{N}_2---\left(1\right)$

Let the activity be same after time $t$.
By the law of radioactivity,

${\lambda }_1{N}_1{e}^{-{\lambda }_{1}t}={\lambda }_2{N}_2{e}^{-{\lambda }_{2}t}---\left(2\right)$

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

$2e^{{\lambda }_{1}t}=e^{{\lambda }_{2}t}$

$\Rightarrow 2=e^{({\lambda }_2-{\lambda }_1)t}$

Taking $\ln$ on both sides

$\ln 2=({\lambda }_2-{\lambda }_1)t$

$\Rightarrow \ln 2=(\frac{\ln 2}{T_2}-\frac{\ln 2}{T_1})t$

$\Rightarrow 1=(\frac{1}{T_2}-\frac{1}{T_1})t$

$\Rightarrow t=\frac{T_1{T}_2}{T_1-T_2}$

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