A radioactive sample (Z = 22) decreases by 90 | vedclass

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(C) For a first-order reaction, the rate constant $k$ is given by: $k = \frac{2.303}{t} \log \frac{a}{a - x}$.Given that the sample decreases by $90\%

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(C) For a first-order reaction, the rate constant $k$ is given by: $k = \frac{2.303}{t} \log \frac{a}{a - x}$.Given that the sample decreases by $90\%$, the remaining amount $(a - x)$ is $10\%$ of the initial amount $a$, so $(a - x) = 0.1a$.Substituting the values: $k = \frac{2.303}{10} \log \frac{a}{0.1a} = \frac{2.303}{10} \log 10 = \frac{2.303}{10} 	imes 1 = 0.2303 \ 	ext{year}^{-1}$.The half-life $t_{1/2}$ is calculated as: $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.2303} \approx 3 \ 	ext{years}$.

$A$ radioactive sample $(Z = 22)$ decreases by $90\%$ in $10 \ \text{years}$. What will be the half-life of the sample in $\text{years}$?

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