b
(b) \(\lambda = \frac{{0.693}}{{{T_{1/2}}}} = \frac{{0.693}}{{20}} = 0.03465\)

Now time of decay \(t = \frac{{2.303}}{\lambda }\log \frac{{{N_0}}}{N}\)

\( \Rightarrow {t_1} = \frac{{2.303}}{{0.03465}}\log \frac{{100}}{{67}} = 11.6\, min \)

and \({t_2} = \frac{{2.303}}{{0.03465}}\log \frac{{100}}{{33}} = 32\,min\)

Thus time difference between points of time

\(= t_1 -t_2 =32 -11.6 = 20.4\, min \, \approx \, 20\, min.\)