IIT JEE Physics (1978 to 2018: 41 Years) Topic-wise Complete Solutions

Problem (IIT JEE 2015): For a radioactive material, its activity $A$ and rate of change of its activity $R$ are defined as $A=-\mathrm{d}N/\mathrm{d}t$ and $R=-\mathrm{d}A/\mathrm{d}t$, 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 activities at $t=2\tau$ are $R_\text{P}$ and $R_\text{Q}$, respectively. If $R_\text{P}/R_\text{Q}=n/e$, then the value of $n$ is _____.
Solution In a radioactive decay, the number of nuclei at a time $t$ are given by, \begin{align} \label{byb:eqn:1} N(t)=N_0e^{-\lambda t}, \end{align} where $N_0$ is number of nuclei at $t=0$ and the decay constant $\lambda$ is related to the mean life $\tau$ by $\lambda=1/\tau$. Differentiate above equation to get the activity $A(t)$ and the rate of change of activity $R(t)$, \begin{align} \label{byb:eqn:2} &A(t)=-\mathrm{d}N/\mathrm{d}t=\lambda N_0 e^{-\lambda t} \\ \label{byb:eqn:3} &R(t)=-\mathrm{d}A/\mathrm{d}t=\lambda^2 N_0 e^{-\lambda t}. \end{align} Let $N_\text{0,P}$ and $N_\text{0,Q}$ be the number of nuclei of P and Q at $t=0$, respectively. Given, $\lambda_\text{P}=1/\tau$ and $\lambda_\text{Q}=1/(2\tau)$. The activities of P and Q at $t=0$ are equal i.e., \begin{align} \label{byb:eqn:4} &\lambda_\text{P} N_\text{0,P}=\lambda_\text{Q} N_\text{0,Q}. \end{align} Use above equations to get the ratio of rate of change of activities of P and Q at time $t=2\tau$ as, \begin{align} \frac{R_\text{P}}{R_\text{Q}}&=\frac{\lambda_\text{P}^2}{\lambda_\text{Q}^2} \frac{N_\text{0,P}}{N_\text{0,Q}} \frac{e^{-2\lambda_\text{P} \tau}}{e^{-2\lambda_\text{Q} \tau}}=\frac{\lambda_\text{P}}{\lambda_\text{Q}} \left(\frac{\lambda_\text{P} N_\text{0,P}}{\lambda_\text{Q} N_\text{0,Q}}\right) \frac{e^{-2\lambda_\text{P} \tau}}{e^{-2\lambda_\text{Q} \tau}}\nonumber\\ &=\frac{1/\tau}{1/(2\tau)} \left(1\right)\frac{e^{-2(1)}}{e^{-2(1/2)}}=\frac{2}{e}. \end{align}