Three-fourth of the active nuclei present in a radioactive sample decay in $\frac{3}{4}\text{s}$. The half-life of the sample is

The correct option is A $\frac{3}{8}\text{s}$
Given that, at $t=\frac{3}{4}\text{s}$, ${\left(\frac{3}{4}\right)}^{th}$ of the sample decayed, then the fraction of remaining nuclei is,

$\frac{N}{N_0}=\frac{1}{4}$

$\Rightarrow \frac{N}{N_0}=\frac{1}{4}=\frac{1}{2^n}[\because \frac{N}{N_0}=\frac{1}{2^n}]$

$\Rightarrow 2^n=4=2^2$

$\Rightarrow n=2$

Using, $t=n{T}_{1/2}$

$\Rightarrow \frac{3}{4}=2T_{1/2}$

$\therefore T_{1/2}=\frac{3}{8}\text{s}$

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

Why this question? To understand the decay law and apply it to solve the problem related to half life period.