Question

If 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 than 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

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

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

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

Solving

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