10 g of a radioactive element is disintegrat | vedclass

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(D) Radioactive disintegration follows first-order kinetics.The rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{a}{a-x}$.Given $a = 10 \

Vedclass · Accepted Answer

$0.693$

Vedclass · Answer

(D) Radioactive disintegration follows first-order kinetics.The rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{a}{a-x}$.Given $a = 10 \ g$,$a-x = 1 \ g$,and $t = 2.303 \ 	ext{min}$.Substituting the values: $k = \frac{2.303}{2.303} \log \frac{10}{1} = 1 	imes \log(10) = 1 \ 	ext{min}^{-1}$.The half-life $t_{1/2}$ is calculated as $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{1} = 0.693 \ 	ext{min}$.

$10 \ g$ of a radioactive element is disintegrated to $1 \ g$ in $2.303 \ \text{minutes}$. What is the half-life (in minutes) of that radioactive element?

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