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, $\frac{2.303\times \log 2}{t_{1/2}}=\frac{2.303}{345}\log 5$
$t_{1/2}=345\times \frac{\log 2}{\log 5}$