Law of Radioactive Decay
By 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}
Problems from IIT JEE
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 $\tau=1/\lambda$.
