The half life period of a radioactive substance is 69.3 minutes. In how much time will it be disintegrated 80% of its original amount?

For radio active disintegration $t_{1/2}=69.3$  min.  $N_t=\frac{20}{100}{N}_0=\frac{N_0}{5}$
$\therefore \lambda =\frac{\ln 2}{t_{1/2}}=\frac{0.693}{69.3}=0.01{\min }^{-1}$
Applying 1-order reaction
$t=\frac{1}{\left(0.01\right)}\ln \left(\frac{5N_0}{N_0}\right)=\frac{\ln 5}{\left(0.01\right)}$
 Time taken (t) =  $\frac{2.303\times 0.7}{\left(0.01\right)}$    $\Rightarrow t=161$   minutes