A first-order nuclear reaction is half-comple | vedclass

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(B) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 45 \ minutes = 0.75 \ hours$.So,$k = \fr

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$7.5$

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(B) For a first-order reaction,the rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}}$.Given $t_{1/2} = 45 \ minutes = 0.75 \ hours$.So,$k = \frac{0.693}{0.75} \ hr^{-1}$.The time $t$ required for $99.9\%$ completion is given by $t = \frac{2.303}{k} \log \frac{a}{a - 0.999a} = \frac{2.303}{k} \log \frac{1}{0.001} = \frac{2.303}{k} \log 10^3$.Substituting the value of $k$: $t = \frac{2.303 	imes 0.75}{0.693} 	imes 3$.Since $\frac{2.303}{0.693} \approx \frac{1}{0.301} \approx 3.32$,we get $t = 3.32 	imes 0.75 	imes 3 = 7.5 \ hours$.

$A$ first-order nuclear reaction is half-completed in $45 \ minutes$. How long does it take for $99.9\%$ of the reaction to be completed in $hours$?

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