Question

Two radioactive substances X and Y emit $\alpha$ and $\beta$ particles respectively. Their disintegration constants are in the ratio 2:3. To have equal probabilities of getting emission of $\alpha$ and $\beta$ particles, find the ratio of number of atoms of X to that of Y at any time instant.

Answer 1

Given,

Two radioactive substance $X$ and $Y$ emit $\alpha$ and $\beta$ particles respectively.

Let, their disintegration constant are $k_1$ and $k_2$ and their ratio is given as,

$\frac{k_1}{k_2}=\frac{2}{3}\dots .\left(1\right)$

Let, at any time the number of atoms of $X$ be $N_1$ and number of atoms of $Y$ is $N_2$.

Now, Decay rate is given by,

$\frac{d{N}_1}{dt}=-k_{1}N\dots .\left(2\right)$

and

$\frac{d{N}_2}{dt}=-k_{2}N\dots .\left(3\right)$

for having equal probability of getting emission of $\alpha$ and $\beta$ particles, the rate of decay must be same for both the substance.

Hence, equating equation (2) and (3) we have

$-k_1{N}_1=-k_2{N}_2$

$\frac{N_1}{N_2}=\frac{k_2}{k_1}$

from equation (1) we have

$\frac{k_2}{k_1}=\frac{3}{2}$

Hence,

$\frac{N_1}{N_2}=\frac{3}{2}$

Hence, the ratio of number of $X$ to that of $Y$ at any time instant is

$N_1:N_2=3:2$