Question

State the law of radioactive decay. Hence derive the expression $N=N_0{e}^{-\lambda t}$ where symbols have their usual meaning

Answer 1

Law of radioactive decay:

1. It states that “For a particular time, the rate of radioactive decay of an atom is directly proportional to the number of nuclei of the elements present at that time.”
2. This can be written as, $\frac{dN}{dt}\propto N$, where N is the number of nuclei.
3. This can be further expressed as $\frac{dN}{dt}=-\lambda N$, where $\mathrm{\lambda }$ is the proportionality constant.

Derivation of the expression:

Step 1: Given

Number of nuclei present initially is $N_0$.

Number of nuclei present at time $t$ is $N$.

Step 2: Formula Used

${\log }_{e}a-{\log }_{e}b={\log }_e\left(\frac{a}{b}\right)$, where $a$ and $b$ are constants.

Step 3: Solution

Rearrange the expression $\frac{dN}{t}=-\lambda N$ and integrate it.

$$\begin{gathered} \frac{dN}{t}=-\lambda N \\ \Rightarrow \frac{dN}{N}=-\lambda t \\ \Rightarrow {\int }_{N_0}^N\frac{dN}{N}=-{\int }_0^t\lambda tdt \\ \Rightarrow {\int }_{N_0}^N\frac{dN}{N}=-\lambda {\int }_0^{t}tdt \end{gathered}$$

Solve the expression further using logarithm.

$$\begin{gathered} {\log }_{e}N-{\log }_e{N}_0=-\lambda t \\ \Rightarrow {\log }_e\left(\frac{N}{N_0}\right)=-\lambda t \\ \Rightarrow \frac{N}{N_0}=e^{-\lambda t} \\ \Rightarrow N=N_0{e}^{-\lambda t} \end{gathered}$$

Hence, $N=N_0{e}^{-\lambda t}$ is derived.