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The law of radioactive decay states that the rate of decay of a sample of radioactive nuclei is proportional to the number of nuclei present. The rate of decay is typically measured by the **activity** of the sample, which is the number of radioactive decays that occur per unit time. The activity expressed in units of becquerels (Bq) or curies (Ci).

Let $N$ be the population of radioactive nuclei at time $t$. The population decreases with time with a decay rate \begin{align} \frac{\mathrm{d}N}{\mathrm{d}t}=-\lambda N, \end{align} where $\lambda$ is the decay constant. The negative sign indicates that the number of radioactive nuclei decreases over time. The decay constant is a characteristic of the specific type of radioactive decay and is related to the half-life of the sample.

The population at time $t$ is given by \begin{align} N=N_0e^{-\lambda t}. \end{align}

The population is reduced to half in time \begin{align} t_{1/2}=\frac{0.693}{\lambda}. \end{align} The time $t_{1/2}$ is called half-life of the nuclei.

The average life of the nuclei is given by \begin{align} t_\text{av}=\frac{1}{\lambda}. \end{align}

The population after $n$ half lives is given by \begin{align} N=N_0/2^n. \end{align}

**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}
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}
A(t)&=-\mathrm{d}N/\mathrm{d}t=\lambda N_0 e^{-\lambda t} \\
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}
&\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}

**Problem (IIT JEE 2000): **
Two radioactive materials $\mathrm{X_1}$ and $\mathrm{X_2}$ have decay constants $10\lambda$ and $\lambda$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $\mathrm{X_1}$ to that of $\mathrm{X_2}$ will be $1/e$ after a time

- $1/(10\lambda)$
- $1/(11\lambda)$
- $11/(10\lambda)$
- $1/(9\lambda)$

**Solution: **
The populations of $\mathrm{X}_1$ and $\mathrm{X}_2$ after time $t$ are given by,
\begin{align}
&N_1=N_0e^{-10\lambda t},\\
&N_2=N_0e^{-\lambda t}.
\end{align}
Divide first equation by second and substitute the values to get,
\begin{align}
\frac{N_1}{N_2} & =\frac{1}{e}=\frac{e^{-10\lambda t}}{e^{-\lambda t}} \\
&=e^{-9\lambda t}.
\end{align}
Take logarithm of above equation and solve to get $t={1}/{(9\lambda)}$.

**Problem (IIT JEE 1989): **
The decay constant of a radioactive sample is $\lambda$. The half-life and mean-life of the sample are respectively given by

- $1/\lambda$ and $(\ln 2)/\lambda$
- $(\ln 2)/\lambda$ and $1/\lambda$
- $\lambda(\ln 2)$ and $1/\lambda$
- $\lambda/(\ln 2)$ and $1/\lambda$

**Solution: **
The half life and average life of a radioactive sample with decay constant $\lambda$ are given by,
\begin{align}
T_{1/2}=\ln(2)/\lambda=0.693/\lambda
\end{align}
and $\tau=1/\lambda$.