Question

The 6.25% of radioactive substance is left after 480 min. The half life period is

• A. 60 min
• B. 90 min
• C. 150 min
• D. 120 min

Answer 1

Answer: (D)

Applying first order kinetic equation,
$k=\frac{2.303}{t}\log \frac{a}{\left(a-x\right)}$
$k=\frac{2.303}{480}\log \frac{100}{6.24}=\frac{2.303}{480}\log 16$
  $=\frac{2.303\times 1.204}{480}=5.78\times 10^{-3}mi{n}^{-1}$      $\left(\log 16=1.204\right)$
$t_{\frac{1}{2}}=\frac{0.693}{k}=\frac{0.693}{5.78\times 10^{-3}}$
$=119.89\approx 120min$