{ }^{227}Ac has a half-life of 22 , years wit | vedclass

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(D) Radioactive decay follows first-order kinetics. The overall decay constant $k$ is related to the half-life $t_{1/2}$ by $k = \frac{0.693}{t_{1/2}}

Vedclass · Accepted Answer

$6.3 	imes 10^{-4}$

Vedclass · Answer

(D) Radioactive decay follows first-order kinetics. The overall decay constant $k$ is related to the half-life $t_{1/2}$ by $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 22 \, years$,so $k = \frac{0.693}{22} \, year^{-1}$.The decay occurs via two parallel paths with rate constants $k_1$ and $k_2$. The total rate constant is $k = k_1 + k_2$.The branching fractions are given as percentages: $\frac{k_1}{k} = \frac{2.0}{100} = 0.02$ and $\frac{k_2}{k} = \frac{98.0}{100} = 0.98$.We need to find $k_1$. From the branching fraction,$k_1 = 0.02 	imes k$.Substituting the value of $k$: $k_1 = 0.02 	imes \frac{0.693}{22} = 0.02 	imes 0.0315 = 0.00063 = 6.3 	imes 10^{-4} \, year^{-1}$.Thus,the correct option is $(D)$.

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