Question

At radioactive equilibrium, the ratio between the number of the atoms of two radioactive elements $\left(X\right)$ and $\left(Y\right)$ was found to be $3.2\times {10}^9:1$ respectively. If half-life of the element $\left(X\right)$ is $1.6\times {10}^{10}$ years, then half-life of the element $\left(Y\right)$ (in years) is

Answer 1

$X\stackrel{{\lambda }_X}{\to }Y\stackrel{{\lambda }_Y}{\to }Z$

At radioactive equilibrium,

$\left({\lambda }_X\right)\times \left(N_X\right)=\left({\lambda }_Y\right)\times \left(N_Y\right)$

$\frac{{\lambda }_x}{{\lambda }_y}=\frac{N_Y}{N_X}$

$\frac{{\left(t_{\frac{1}{2}}\right)}_Y}{{\left(t_{\frac{1}{2}}\right)}_X}=\frac{N_Y}{N_X}$ or $\frac{{\left(\frac{t_1}{2}\right)}_Y}{1.6\times {10}^{10}}=\frac{1}{3.2\times {10}^9}$

${\left(t_{\frac{1}{2}}\right)}_Y=5$ years