Q. There are two radioactive nuclei $A$ and $B$ . $A$ is an alpha emitter and $B$ is a beta emitter. Their disintegration constants are in the ratio of $1:2$ . What should be the ratio of the number of atoms of $A$ and $B$ at any time $t$ so that the probabilities of getting alpha and beta particles are the same at that instant?
NTA AbhyasNTA Abhyas 2022
Solution:
Given here, $\frac{\lambda _{A}}{\lambda _{B}}=\frac{1}{2}$
Probability of getting $\alpha $ and $\beta $ are same.
Thus, rate of disintegration are equal.
$\therefore \lambda _{A}N_{A}=\lambda _{B}N_{B}=\left|\frac{dN}{dt}\right|$
Hence, the ratio of atoms is $\frac{N_{A}}{N_{B}}=\frac{\lambda _{B}}{\lambda _{A}}=\frac{2}{1}$
Probability of getting $\alpha $ and $\beta $ are same.
Thus, rate of disintegration are equal.
$\therefore \lambda _{A}N_{A}=\lambda _{B}N_{B}=\left|\frac{dN}{dt}\right|$
Hence, the ratio of atoms is $\frac{N_{A}}{N_{B}}=\frac{\lambda _{B}}{\lambda _{A}}=\frac{2}{1}$