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. The number $N$ of nuclei of $A$ at time $t$ is, 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

• A. $2N_0$, $\frac{5}{2}{N}_0$
• B. $3N_0$, $2N_0$
• C. $4N_0$, $2N_0$
• D. $\frac{3}{2}{N}_0,2N_0$

Answer 1

Answer: (D)

If $\alpha =2\lambda N_0$, t=half life $=\frac{ln\left(2\right)}{\lambda }$
$\therefore N=\frac{1}{\lambda }\left[2\lambda N_0-\left(2\lambda N_0-\lambda N_0\right)e^{-\lambda t}\right]$
or $N=\frac{\lambda N_0}{\lambda }\left[2-e^{-ln\left(2\right)}\right]$[Here $e^{-ln(2)}=2^{-1}=\frac{1}{2}]$
or $N=\frac{\lambda N_0}{\lambda }\left[2-\frac{1}{2}\right]=\frac{3N_0}{2}$ 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 $N=2N_0$