Question

Some nuclei of a radioactive material are undergoing radioactive decay. The time gap between the instances when a quarter of the nuclei have decayed and when half of the nuclei have decayed is given as :
(where $\lambda$ is the decay constant)

• A. $\frac{ln\left(2\right)}{\lambda }$
• B. $\frac{ln\left(\frac{3}{2}\right)}{\lambda }$
• C. $\frac{2ln\left(2\right)}{\lambda }$
• D. $\frac{1}{2}\frac{ln\left(2\right)}{\lambda }$

Answer 1

The correct option is B $\frac{ln\left(\frac{3}{2}\right)}{\lambda }$

The number of undecayed nuclei present in a radioactive sample is given by,
$N=N_0{e}^{-\lambda t}$

When a quarter of nuclei have decayed,
$\frac{3N_0}{4}=N_0{e}^{-\lambda t_1}$
$ln\left(\frac{3}{4}\right)=-\lambda t_1...\left(i\right)$

When half the nuclei have decayed,
$\frac{N_0}{2}=N_0{e}^{-\lambda t_2}$
$ln\left(\frac{1}{2}\right)=-\lambda t_2...\left(ii\right)$


$ln\left(\frac{3}{4}\right)-ln\left(\frac{1}{2}\right)=-\lambda \left(t_1\right)+\lambda t_2$

$ln\left(\frac{3}{2}\right)=\lambda (t_2-t_1)$

$\therefore \Delta t=\frac{ln\left(\frac{3}{2}\right)}{\lambda }$

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