Question

A radioactive substance decays 20% in 10 min. If at the start there are $5\times {10}^{10}$ atoms present, the time after which the number of atoms are reduced to ${10}^{18}$ atoms is______ (in hrs).

Answer 1

Radioactive substance decays 20% in 10 min hence the rate constant of first order disintegration is
$k=\frac{2.303}{10}log\left(\frac{100}{80}\right)=0.0223mi{n}^{-1}$
Initially, there are $5\times {10}^{20}$ atoms, which reduce to ${10}^{18}$ atoms after time t min.
Hence, $t=\frac{2.303}{k}\log \frac{N_0}{N}=\frac{2.303}{0.0223}log\frac{5\times {10}^{20}}{{10}^{18}}=278.7min=4.65hr$