Question

Two radioactive samples of different elements having half lives $t_1$ and $t_2$ respectively, have the same number of nuclei at $t=0$. The time after which their activities are the same is

• A. $\frac{t_1{t}_2}{0.693}\ln \left(\frac{t_2}{t_1}\right)$
• B. $\frac{t_1{t}_2}{0.693(t_2-t_1)}\ln \left(\frac{t_2}{t_1}\right)$
• C. $\frac{t_1{t}_2}{0.693(t_2+t_1)}\ln \left(\frac{t_2}{t_1}\right)$
• D. $\frac{0.693}{t_1{t}_2}\ln \left(\frac{t_2}{t_1}\right)$

Answer 1

The correct option is B $\frac{t_1{t}_2}{0.693(t_2-t_1)}\ln \left(\frac{t_2}{t_1}\right)$

The activity is given by,

$A=-\lambda N$

According to the law of radioactivity,

$N=N_0{e}^{-\lambda t}$

$\Rightarrow A=-\lambda N_0{e}^{-\lambda t}$

As the two samples have the same number of nuclei at $t=0$,

$A_1=-{\lambda }_1{N}_0{e}^{-{\lambda }_{1}t}$

$A_2=-{\lambda }_2{N}_0{e}^{-{\lambda }_{2}t}$

Let the activities of the two samples be the same at time $t$,

$\Rightarrow A_1=A_2$

$\Rightarrow -{\lambda }_1{N}_0{e}^{-{\lambda }_{1}t}=-{\lambda }_2{N}_0{e}^{-{\lambda }_{2}t}$

$\because \lambda =\frac{0.693}{T_{\frac{1}{2}}}$

$\Rightarrow \frac{0.693}{t_1}{e}^{\left(\frac{-0.693}{t_1}\right)t}=\frac{0.693}{t_2}{e}^{\left(\frac{-0.693}{t_2}\right)t}$

Taking $\ln$ both sides,

$\Rightarrow -0.693t[\frac{1}{t_1}-\frac{1}{t_2}]=\ln \left(\frac{t_1}{t_2}\right)$

$\Rightarrow 0.693t\left[\frac{t_2-t_1}{t_1{t}_2}\right]=\ln \left(\frac{t_2}{t_1}\right)$

$\Rightarrow t=\frac{t_1{t}_2}{0.693(t_2-t_1)}\ln \left(\frac{t_2}{t_1}\right)$

Hence, option $\left(B\right)$ is correct.

Why this question? To improve the skill of solving the problem related to activity.