Question

Let $T$ be the mean life of a radioactive sample. $75\%$ of the active nuclei present in the sample initially will decay in time

• A. $2T$
• B. $4T$
• C. $\frac{1}{2}\left(\ln 2\right)T$
• D. $2\left(\ln 2\right)T$

Answer 1

The correct option is D $2\left(\ln 2\right)T$

$75\%$ active nuclei decayed.

The remaining nuclei is $25\%\text{of}{N}_0$.

$\therefore N=\frac{25}{100}{N}_0$

$\Rightarrow \frac{N}{N_0}=\frac{25}{100}=\frac{1}{4}$

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

$\Rightarrow \frac{1}{4}=e^{-\lambda t}$

$\Rightarrow 2^{-2}=e^{-\lambda t}$

$\Rightarrow 2\ln 2=\lambda t$

$\therefore t=\frac{2\ln \left(2\right)}{\lambda }=2T\ln \left(2\right)[\because T=\frac{1}{\lambda }]$

Hence, $\left(D\right)$ is the correct answer.