After 2 hours the amount of a certain radioactive substance reduces to 1 / 16^th of the original amount (the decay process follows first-order kinetics). The half-life of the radioactive substance is

Correct answer: (B) 30 ~min. frac0.693t_1 / 2=2.303/(2 × 60) log a/(a / 16) t_1 / 2=30 ~min

Q18 / 60 · KVPY 2011 (SA) · Chemistry

After 2 hours the amount of a certain radioactive substance reduces to $1 / 16^{\text {th }}$ of the original amount (the decay process follows first-order kinetics). The half-life of the radioactive substance is

$\frac{0.693}{t_{1 / 2}}=\frac{2.303}{2 \times 60} \log \frac{a}{a / 16}$ $t_{1 / 2}=30 \mathrm{~min}$

Hence, the correct answer is option: 'B'.