Question

In a radioactive equilibrium $A\to B\to C\to$ the ratio of number of atoms of $A$ to $B$ is $1:2.8\times {10}^6.$ If half- life of $A$ is $1620yr$ (years).
what is half - life of $B$ ?

• A. $4.536\times {10}^{9}Yr$
• B. $5.785\times {10}^{7}Yr$
• C. $4.536\times {10}^{6}Yr$
• D. $5.785\times {10}^{-3}Yr$

Answer 1

The correct option is A $4.536\times {10}^{9}Yr$

$\frac{d{N}_A}{dt}={\lambda }_A{N}_A=,\frac{-d{N}_B}{dt}={\lambda }_B{N}_B$

$\frac{N_A}{N_B}=\frac{{\lambda }_B}{{\lambda }_A}=\frac{(t_{1/2}{)}_A}{(t_{1/2}{)}_B}$

$\therefore \frac{N_A}{N_B}=\frac{(t_{1/2}{)}_A}{(t_{1/2}{)}_B}$

$\frac{1}{2.8\times {10}^6}=\frac{1620}{t_{1/2}B}$

$t_{1/2}B=1620\times 2.8\times {10}^6$

$=4536\times {10}^6$

$=4.536\times {10}^9$