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

The activity of radioactive substance is given as:
$A=\frac{dN}{dt}=\lambda N=\lambda N_{(}t=0)e^{-\lambda t}$...(i)

Mean life time $\tau$ is related to $\lambda$ as:
$\lambda =\frac{1}{\tau }$...(ii)

Given activity of $P$ and $Q$ are equal at time $t$ :
${\lambda }_{P}N{P}_{(}t=0)e^{-{\lambda }_{P}t}={\lambda }_{Q}N{Q}_{(}t=0)e^{-{\lambda }_{Q}t}$...(iii)

The rate of change of activity can be found by differentiating (i)
$\frac{dA}{dt}=\lambda N_{(}t=0)e^{-\lambda t}$...(iv)

Calculating $R_P$ and $R_Q$:
$R_P={\lambda }_P^{2}N{P}_{(}t=0)e^{-{\lambda }_P(t+2\tau )}$
$R_Q={\lambda }_Q^{2}N{Q}_{(}t=0)e^{-{\lambda }_P(t+2\tau )}$

$\frac{R_P}{R_Q}=\frac{{\lambda }_P^{2}N{P}_{(}t=0)e^{-{\lambda }_{P}t}}{{\lambda }_Q^{2}N{Q}_{(}t=0)e^{-{\lambda }_Q(t+2\tau )}}$...(v)

Equation (v) can be written as:
$\frac{R_P}{R_Q}=\frac{{\lambda }_P{\lambda }_{P}N{P}_{(}t=0)e^{-{\lambda }_{P}t}{e}^{-{\lambda }_{P}2\tau }}{{\lambda }_Q{\lambda }_{Q}N{Q}_{(}t=0)e^{-{\lambda }_{Q}t}{e}^{-{\lambda }_{Q}2\tau }}$

From equation (iii):
$\frac{R_P}{R_Q}=\frac{{\lambda }_P{e}^{-{\lambda }_{P}2\tau }}{{\lambda }_Q{e}^{-{\lambda }_{Q}2\tau }}$
$\frac{R_P}{R_Q}=\frac{{\lambda }_P{e}^{({\lambda }_Q-{\lambda }_P)2\tau }}{{\lambda }_Q}$

From equation (i):
$\frac{{\lambda }_P}{{\lambda }_Q}=\frac{\frac{1}{\tau }}{\frac{1}{2\tau }}=2$

$\frac{R_P}{R_Q}=2e^{({\lambda }_Q-{\lambda }_P)2\tau }$

$\frac{R_P}{R_Q}=2e^{(\frac{1}{2\tau }-\frac{1}{\tau })2\tau }$

$\frac{R_P}{R_Q}=2e^{-1}=\frac{2}{e}$

$\Rightarrow n=2$