Question

The nuclei of a radioactive element is produced at constant rate $\alpha$ and they decay with decay constant $\lambda$. At $t=0$, number of nuclei is zero, then the number of nuclei at time $t$ is

• A. $\frac{\alpha }{\lambda }(1-e^{-\lambda t})$
• B. $\alpha -\frac{\alpha }{\lambda }{e}^{-\lambda t}$
• C. $\frac{\alpha }{\lambda }{e}^{-\lambda t}$
• D. $\alpha (1-e^{-\lambda t})$

Answer 1

The correct option is A $\frac{\alpha }{\lambda }(1-e^{-\lambda t})$

Given nuclei of a radioactive element is produced at constant rate $\alpha$ and they decay with decay constant $\lambda$

$\frac{dN}{dt}=\alpha -\lambda N$

${\int }_0^N\frac{dN}{\alpha -\lambda N}={\int }_0^{t}dt$

$[\frac{ln(\alpha -\lambda N)}{-\lambda }{]}_0^N=t$

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

$N=\frac{\alpha }{\lambda }(1-e^{-\lambda t})$