Question

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

• A. $3.2\times {10}^9$ years
• B. $5\times {10}^9$ years
• C. 10 years
• D. 5 years

Answer 1

The correct option is D

5 years

$X\stackrel{{\lambda }_1}{\to }Y\stackrel{{\lambda }_2}{\to }Z$

At radioactive equilibrium,

$\left({\lambda }_x\right)\times \left(N_X\right)=\left({\lambda }_y\right)\times \left(N_Y\right)$

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

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

${\left(\frac{t_1}{2}\right)}_Y=5$ year