Question

Nuclei of a radioactive element is produced at constant rate $\alpha$ and they decay with decay constant $\lambda$. At $t=0$, the number of nuclei is 0. The number of active nuclei at time $t$ is

• A. $\frac{\alpha }{\lambda }(1-e^{-\lambda t})$
• B. $\alpha (1-e^{-\lambda t})$
• C. $\frac{\alpha }{\lambda }-e^{-\lambda t}$
• D. $\alpha e^{-\lambda t}$

Answer 1

The correct option is A $\frac{\alpha }{\lambda }(1-e^{-\lambda t})$

According to the information,
$\frac{dN}{dt}=-\lambda N+\alpha$
Rearranging and integrating,
${\int }_0^N\frac{dN}{-\lambda N+\alpha }={\int }_0^{t}dt$

Let $-\lambda N+\alpha =p$
$\Rightarrow -\lambda dN=dp$
$\Rightarrow dN=-\frac{dp}{\lambda }$

Substituting back in integral,
${\int }_{\alpha }^p\frac{dp}{p}={\int }_0^t-\lambda dt$
This gives $p=\alpha e^{-\lambda t}$
Putting the expression for $p$, we get
$p=-\lambda N+\alpha =\alpha e^{-\lambda t}$
This gives
$N=\frac{\alpha }{\lambda }(1-e^{-\lambda t})$