Question

Nuclei of a radioactive element $A$ are being produced at a constant rate $\alpha$. The element has a decay constant $\lambda$. At $t=0$, 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 $t\to \infty$.

• A. $\frac{4N_0}{2},2N_0$.
• B. $\frac{3N_0}{2},2N_0$.
• C. $\frac{5N_0}{2},2N_0$.
• D. $\frac{6N_0}{2},2N_0$.

Answer 1

Answer: (B)

$\frac{3N_0}{2},2N_0$.

$\text{Solution}:$

If $\alpha =2\lambda N_0$, t = half life $=\frac{ln\left(2\right)}{\lambda }$

$\therefore N=\frac{1}{\lambda }[2\lambda N_0-(2\lambda N_0-\lambda N_0)e^{-\lambda t}]$

or $N=\frac{\lambda N_0}{\lambda }[2-e^{-ln\left(2\right)}]$ [Here $e^{-ln\left(2\right)}=2^{-1}=\frac{1}{2}]$

or $N=\frac{\lambda N_0}{\lambda }[2-\frac{1}{2}]=\frac{3N_0}{2}$ or

$N=\frac{3}{2}{N}_0$

When $t\to \infty$ and $\alpha =2\lambda N_0$

$N=\frac{\alpha }{\lambda }=\frac{2\lambda N_0}{\lambda }=2N_0$ or $N=2N_0$

$\text{Hence B is the correct option}$