Question

A sample of a radioactive substance undergoes $80 \%$ decomposition in $345$ minutes. Its half life is ___ minutes.

• A. $\frac{\ln 2}{\ln 5} \times 345$
• B. $\frac{\ln 5}{\ln 2} \times 345$
• C. $\frac{\ln 5}{\ln 4} \times 345$
• D. $\frac{\ln 4}{\ln 5} \times 345$

Answer 1

Answer: (A)

Radioactive decomposition is first order reaction
So, $K=\frac{2.303}{t} \log \frac{a}{a-x}=\frac{2.303}{t} \log \frac{100}{100-x}$
$\therefore t_{1 / 2}=\frac{0.693}{ K }$
$\frac{0.693}{t_{1 / 2}}=\frac{2.303}{345} \log \frac{100}{100-80}$
$\frac{0.693}{t_{1 / 2}}=\frac{2.303}{345} \log \frac{100}{20}$
$\frac{0.693}{t_{1 / 2}}=\frac{2.303}{345} \log 5$
$\therefore [0.693=2.303 \times \log 2]$
So, $\frac{2.303 \times \log 2}{ t _{1 / 2}}=\frac{2.303}{345} \log 5$
$t_{1 / 2}=345 \times \frac{\log 2}{\log 5}$