A radioactive nuclide is produced at the constant rate of $n$ per second. It decays into a stable nucleus with decay constant $\lambda$. If $N_0$ is a number of nuclei at $t=0$, then expected number $N$ of nuclei in existence after $t$ seconds is given by

The correct option is D $N=\frac{n}{\lambda }+(N_0-\frac{n}{\lambda })e^{-\lambda t}$
The population is increasing by $n$ and simultaneously decreasing by $\lambda N$.

$\therefore \frac{dN}{dt}=n-\lambda N$

Integrating the above equation within proper limit,

$\Rightarrow {\int }_{N_0}^N\frac{dN}{n-\lambda N}={\int }_0^{t}dt$

$\Rightarrow -\frac{1}{\lambda }\left[\ln \right(n-\lambda N)-\ln (n-\lambda N_0\left)\right]=t$

$\Rightarrow \ln \left[\frac{n-\lambda N}{n-\lambda N_0}\right]=-\lambda t$

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

$\Rightarrow n-\lambda N=(n-\lambda N_0)e^{-\lambda t}$

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

$\Rightarrow N=\frac{n}{\lambda }[1-e^{-\lambda t}]+N_0{e}^{-\lambda t}$

$\therefore N=\frac{n}{\lambda }+(N_o-\frac{n}{\lambda })e^{-\lambda t}$

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