Question

Nuclei of radioactive element A are produced at rate '$t^2$' at any time t. The element A has decay constant $\lambda$. Let N be the number of nuclei of element A at any time t. At time $t=t_0,\frac{dN}{dt}$ is minimum. N is :

• A. $\frac{2t_0-\lambda t_0^2}{{\lambda }^2}$
• B. $\frac{t_0-\lambda t_0^2}{{\lambda }^2}$
• C. $\frac{2t_0-\lambda t_0^2}{\lambda }$
• D. $\frac{t_0-\lambda t_0^2}{\lambda }$

Answer 1

The correct option is A $\frac{2t_0-\lambda t_0^2}{{\lambda }^2}$

$\frac{dN}{dt}=t^2-\lambda N$
for $\frac{dN}{dt}$ to be minimum;$\frac{d^{2}N}{d{t}^2}=0$
$\Rightarrow \frac{d^{2}N}{d{t}^2}=2t-\lambda \frac{dN}{dt}=2t-\lambda (t^2-\lambda N)=0$
or $N=\frac{2t_0-\lambda t_0^2}{{\lambda }^2}$