Aftcr $n$ half-life, the numbcrs of atom left undecayed is given by,
$N=N_{0}\left(\frac{1}{2}\right)^{n} $
$n=\frac{t}{T_{\frac{1}{2}}}$ or $N=N_{0}\left(\frac{1}{2}\right)^{\frac{t}{T_{1 / 2}}}$
For $20 \%$ decay of radioactive substance,
$N=N_{0}-\frac{20}{100} N_{0}=0.8\, N_{0} $
or 0.8 $N_{0}=N_{0}\left(\frac{1}{2}\right)^{t_{1 / 18}}\,\,\,...(i)$
For $80 \%$ decay of radioactive substance,
$N=N_{0}-\frac{80}{100} N_{0}=0.2 \,N_{0} $ or $ 0.2 \,N_{0}=N_{0}\left(\frac{1}{2}\right)_{\ldots}^{t_{2 / 18}}\,\,\,...(ii)$
From Eqs. (i) and (ii), we get
$\Rightarrow 4=\left(\frac{1}{2}\right)^{\frac{t_{1}-t_{2}}{18}}$
Taking log on the both sides, we get
$\log 4 =\log \left(\frac{1}{2}\right)^{\frac{t_{1}-t_{2}}{18}} $
$\Rightarrow \log 2^{2}=\log 2^{\left(\frac{t_{2}-t_{1}}{18}\right)} $
$\Rightarrow 2 =\frac{t_{2}-t_{1}}{18} $
$ \Rightarrow t_{2}-t_{1}=36$ min