Question

Obtain an expression to deduce the amount of the radioactive substance present at any moment. Obtain the relation between half-life period and decay constant.

Answer 1

Rutherford and Soddy found that the rate of disintegration is independent of physical and chemical conditions. The rate of disintegration at any instant is directly proportional to the number of atoms of the element present at that instant. This is known as radioactive law of disintegration.
Let $N_0$ be the number of radioactive atoms present initially and $N$, the number of atoms at a given instant $t$. Let $dN$ be the number of atoms undergoing disintegration in a small interval of time $dt$. Then, the rate of disintegration is,
$-\frac{dN}{dt}\propto N$
$\frac{dN}{dt}=-\lambda N$ ....(1)
where $\lambda$ is a constant known as decay constant or disintegration constant. The negative sign indicates that $N$ decreases with increase in time.
Equation (1) can be written as,
$\frac{dN}{N}=-\lambda dt$
Integrating, ${\log }_{e}N=-\lambda t+C$ ...(2)
where $C$ is a constant of integration.
At $t=0$, $N=N_0$
$\therefore {\log }_e{N}_0=C$
Substituting for $C$, equation (2) becomes
${\log }_{e}N=-\lambda t+{\log }_e{N}_0$
${\log }_e\left[\frac{N}{N_0}\right]=-\lambda t$
$\frac{N}{N_0}=e^{-\lambda t}$
$N=N_0{e}^{-\lambda t}$ ....(3)
Equation (3) shows that the number of atoms of a radioactive substance decreases exponentially with increase in time (figure).
Initially the disintegration takes place at a faster rate. As time increases, $N$ gradually decreases exponentially. Theoretically, an infinite time is required for the complete disintegration of all the atoms.
Relation between half-life period and decay constant : The half life period of a radioactive element is defined as the time taken for one half of the radioactive element to undergo disintegration.
From the law of disintegration,
$N=N_0{e}^{-\lambda t}$
Let $T_{\frac{1}{2}}$ be the half life period. Then, at $t=T_{\frac{1}{2}}$,
$N=\frac{N_0}{2}$
$\therefore \frac{N_0}{2}=N_0{e}^{-\lambda T_{\frac{1}{2}}}$
${\log }_{e}2=\lambda T_{\frac{1}{2}}$
$T_{\frac{1}{2}}=\frac{{\log }_{e}2}{\lambda }=\frac{{\log }_{10}2\times 2.3026}{\lambda }=\frac{0.6931}{\lambda }$
The half life period is inversely proportional to its decay constant.