Question

Nuclei of a radioactive element A are being produced at a constant rate
$\alpha$. The element has a decay constant $\lambda$. At time
$t=0$, there are $N_0$ nuclei of the element.
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

For production, $\frac{d{N}_A}{dt}=\alpha$

For decay, $\frac{d{N}_A}{dt}=-\lambda N_A$

Net rate $=\frac{d{N}_A}{dt}=\alpha -\lambda N_A$

$\frac{d{N}_A}{\alpha -\lambda N_A}=dt$

Integraing, ${\int }_{N_0}^N\frac{d{N}_A}{\alpha -\lambda N_A}={\int }_0^{t}dt$

$-\frac{1}{\lambda }ln\left(\frac{\alpha -\lambda N}{\alpha -\lambda N_0}\right)=t$

$ln\left(\frac{\alpha -\lambda N}{\alpha -\lambda N_0}\right)=-\lambda t$

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

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

$N=\frac{1}{\lambda }[\alpha -(\alpha -\lambda N_0\left)e^{-\lambda t}\right]$

If $\alpha =2N_0\lambda$, we get $N=N_0(2-e^{-\lambda t})$

For one half time, $t=t_{1/2}=ln2/\lambda$ then we have,

$N=N_0(2-e^{-ln2})=N_0(2-1/2)=3/4N_0$

When $t\to \infty ,e^{-\lambda t}\to 0$ so, $N=2N_0$