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At radioactive equilibrium, the ratio between two atoms of radioactive elements $A$ and $B$ is $3.1 \times 10^9: 1$. If the half-life period of $A$ is $2 \times 10^{10}$ yrs, then the half-life of $B$ is
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6.45 yrs
At radioactive equilibrium,
$A \longrightarrow B$
$\begin{aligned} \frac{N_A}{N_B} & =\frac{K_B}{K_A}=\frac{t_{1 / 2 A}}{t_{1 / 2 B}} \\ \therefore \quad \frac{3.1 \times 10^9}{1} & =\frac{2 \times 10^{10}}{t_{1 / 2 B}} \text { and, } t_{1 / 2 B}=6.45 \mathrm{yrs} .\end{aligned}$
$A \longrightarrow B$
$\begin{aligned} \frac{N_A}{N_B} & =\frac{K_B}{K_A}=\frac{t_{1 / 2 A}}{t_{1 / 2 B}} \\ \therefore \quad \frac{3.1 \times 10^9}{1} & =\frac{2 \times 10^{10}}{t_{1 / 2 B}} \text { and, } t_{1 / 2 B}=6.45 \mathrm{yrs} .\end{aligned}$
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