Question

A radioactive substance with decay constant of $0.5s^{-1}$ is being produced at a constant rate of 50 nuclei per second. If there are no nuclei present initially, the time (in second) after which 25 nuclei will be present is :

• A. 1
• B. $\text{ln}2$
• C. $\text{ln}2\left(\frac{4}{3}\right)$
• D. $2\text{ln}\left(\frac{4}{3}\right)$
• E. None of the above

Answer 1

The correct option is D $2\text{ln}\left(\frac{4}{3}\right)$

Let N be the number of nuclei at any-time t then
$\frac{dN}{dt}=50-\lambda N$
${\int }_0^N\frac{dN}{(50-\lambda N)}={\int }_0^{t}dt$
$-\frac{1}{\lambda }\left[\text{ln}\right(50-\lambda N){]}_0^N=t$
$\text{ln}(50-\lambda N)-\text{ln}50=-\lambda t$
$\text{ln}\left[\frac{50-\lambda N}{50}\right]=-\lambda t$
$\frac{50-\lambda N}{50}=e^{-\lambda t}$
$1-\frac{\lambda N}{50}=e^{-\lambda t}$
$N=\frac{50}{\lambda }[1-e^{-\lambda t}]$
As $N=25\text{and}\lambda =0.5/\text{sec}$
$25=\frac{50}{0.5}[1-e^{-0.5t}]$
on solving we get
$t=2\text{ln}\left[\frac{4}{3}\right]\text{sec}$
Hence option (D) is correct.