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 $t\rightarrow\infty$ and $\alpha=2\lambda N_{0}$
$N=\frac{\alpha}{\lambda}=\frac{2\lambda N_{0}}{\lambda}=2N_{0}$ or $N=2N_{0}$