A radioactive substance takes 20 , text{min} | vedclass

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(A) For a first-order radioactive decay,the rate constant $K$ is given by $K = \frac{2.303}{t} \log \frac{A_0}{A_t}$.For $25 \%$ decay,the remaining a

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$96.4$

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(A) For a first-order radioactive decay,the rate constant $K$ is given by $K = \frac{2.303}{t} \log \frac{A_0}{A_t}$.For $25 \%$ decay,the remaining amount is $75 \%$ of the initial amount $(A_t = 0.75 A_0)$:$K = \frac{2.303}{20} \log \frac{100}{75} = \frac{2.303}{20} 	imes 0.1249 = 0.01438 \, 	ext{min}^{-1}$.For $75 \%$ decay,the remaining amount is $25 \%$ of the initial amount $(A_t = 0.25 A_0)$:$t = \frac{2.303}{K} \log \frac{100}{25} = \frac{2.303}{0.01438} \log 4 = \frac{2.303 	imes 0.6020}{0.01438} \approx 96.4 \, 	ext{min}$.

$A$ radioactive substance takes $20 \, \text{min}$ to decay $25 \%$. How many minutes will it take to decay $75 \%$?

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