Question

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

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

Answer 1

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

Given, $2R_A=R_B$

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

Radio activity of sample $\text{A and B}$ becomes equal after some time $t$.

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

Dividing (1) by (2) we get,

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

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

Taking log on both sides we get,

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

$\because \ln 2=\lambda T$

$\Rightarrow 0.693=0.693[\frac{1}{T_2}-\frac{1}{T_1}]t$

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

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.