The t_{1/2} of a radioactive element is 20 m | vedclass

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(B) Radioactive decay follows first-order kinetics. The rate constant $K$ is given by $K = \frac{0.693}{t_{1/2}} = \frac{0.693}{20} \ min^{-1}$.For a

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$20$

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(B) Radioactive decay follows first-order kinetics. The rate constant $K$ is given by $K = \frac{0.693}{t_{1/2}} = \frac{0.693}{20} \ min^{-1}$.For a first-order reaction,the time $t$ required for decay from $x_1\%$ to $x_2\%$ is given by $t = \frac{2.303}{K} \log \frac{100 - x_1}{100 - x_2}$.Here,$x_1 = 33$ and $x_2 = 67$,so the remaining amounts are $100 - 33 = 67$ and $100 - 67 = 33$.Substituting the values: $t = \frac{2.303}{(0.693/20)} \log \frac{67}{33}$.Since $0.693 \approx 2.303 	imes \log_{10}(2)$,we have $t = 20 	imes \frac{\log(67/33)}{\log(2)} \approx 20 	imes \frac{0.307}{0.301} \approx 20 \ min$.

The $t_{1/2}$ of a radioactive element is $20 \ min$. The time interval between the stages of its $33\%$ and $67\%$ decay is ....... $min$.

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