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

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

COMEDKCOMEDK 2011Chemical Kinetics

Solution:

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\times1.204}{480}=5.78\times10^{-3} min^{-1} $      $\left(\log 16 =1.204\right) $
$t_{\frac{1}{2}} = \frac{0.693}{k} = \frac{0.693}{5.78 \times10^{-3}} $
$= 119.89\approx120 min $