Question

Nuclei of a radio-active element $A$ are being product at a constant rate $\alpha$. The element has a decay constant $\lambda$. At time $t=0$ there are $N_0$ nuclei of the element present.
(a) Calculate the number of nuclei $A$ as a function of time $t$.
(b) If $\alpha =2N_0\lambda$, calculate the number of nuclei of $A$ after one half-life of $A$, and also the limiting value of $N$ as $t\to \infty$.

Answer 1

(a) $X⟶A⟶B$
Net rate of formulation of $A$ is $\frac{dN}{dt}=\alpha -\lambda N$
$\int \frac{dN}{\alpha -\lambda N}=\int dt+C$ Where $C$ is a constant of integration.
$-\frac{1}{\lambda }in(\alpha -\lambda N)=t+C$
At $t=0N=N_0;C=-\frac{1}{\lambda }in(\alpha -\lambda N_0)$
or $t=\frac{1}{\lambda }in\left(\frac{\alpha -\lambda N_0}{\alpha -\lambda N}\right)$
or $N=\frac{\alpha -(\alpha -\lambda N_0)e^{-\lambda t}}{\lambda }$
(b) When $\alpha =2N_0\lambda$ and $t=T_{1/2}=\frac{In2}{\lambda }$
$\Rightarrow N=\frac{2N_0\lambda -(2N_0\lambda -\lambda N_0)e^{-In2}}{}=\frac{3N_0}{2}$
Also $t\to \infty ;N=2N_0$