Question

For a radioactive material, its activity A and rate of change of its activity R are defined as $A=-\frac{dN}{dt}$ and $R=-\frac{dA}{dt}$, where N(t) is the number of nuclei at time t. Two radioactive sources P(mean life $\tau$) and Q (mean life 2$\tau$) have the same activity at t = 0. Their rates of change of activitites at t = 2$\tau$ are R${}_P$ and R${}_Q$, respectively. If $\frac{R_P}{R_Q}=\frac{n}{e}$, then the value of n is

Answer 1

$N=N_o{e}^{-\lambda t}$
$\begin{array}{cc}A& =-\frac{dN}{dt}\\ R& =-\frac{dA}{dt}\\ & =+\frac{d}{dt}\left(\frac{dN}{dt}\right)\\ & =\frac{d^{2}N}{d{t}^2}\end{array}$

$⟹R=\frac{d}{dt}\left[\frac{d}{dt}\left(N_0{e}^{-\lambda t}\right)\right]$

$=N_0(-\lambda )\frac{d}{dt}\left({\overline{e}}^{\lambda +}\right)$
$=\lambda \left(\lambda N_0\right)e^{-\lambda t}$
$=\lambda A_0{e}^{-\lambda t}$----(1)

From (1)
$\frac{R_p}{R_Q}=\frac{{\lambda }_p{A}_o{e}^{-{\lambda }_{p}t}}{{\lambda }_Q{A}_o{e}^{-{\lambda }_{Q}t}}$
(where $\lambda =\frac{1}{T}$)
$=\frac{\frac{1}{\tau }{e}^{-\frac{1}{\tau }2\tau }}{\frac{1}{2\tau }{e}^{-\frac{1}{2\tau }2\tau }}$
$=2e^{-1}$
So,value of $n$ is 2