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If $50 \%$ of a radioactive substance dissociates in 15 min , then the time taken by substance to dissociate $99 \%$ will be
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Verified Answer
The correct answer is:
99 min
Radioactive disintegration is a Ist order reaction, hence
$\begin{aligned}
k & =\frac{2.303}{t} \log \frac{a}{a-x} \\
\therefore \quad k & =\frac{2.303}{15} \log \frac{100}{100-50} \\
\text { also } \quad k & =\frac{2.303}{t} \log \frac{100}{100-99} \\
\therefore \frac{2.303}{15} \log \frac{100}{50} & =\frac{2.303}{t} \log \frac{100}{1} \\
\frac{1}{15} \log 2 & =\frac{1}{t} \log 100 \\
\frac{0.3010}{15} & =\frac{2}{t} \\
t & =\frac{2 \times 15}{0.3010}=99 \mathrm{~min}
\end{aligned}$
$\begin{aligned}
k & =\frac{2.303}{t} \log \frac{a}{a-x} \\
\therefore \quad k & =\frac{2.303}{15} \log \frac{100}{100-50} \\
\text { also } \quad k & =\frac{2.303}{t} \log \frac{100}{100-99} \\
\therefore \frac{2.303}{15} \log \frac{100}{50} & =\frac{2.303}{t} \log \frac{100}{1} \\
\frac{1}{15} \log 2 & =\frac{1}{t} \log 100 \\
\frac{0.3010}{15} & =\frac{2}{t} \\
t & =\frac{2 \times 15}{0.3010}=99 \mathrm{~min}
\end{aligned}$
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