Explanation:
Fraction left, \(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f = 1 - \frac{{20}}{{100}} = \frac{8}{{10}}\)
Hence, \(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,N = \frac{8}{{10}}{N_0},t = 10{\text{ days }}\)
Now, \(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,N = N_0^{ - \lambda t}\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{8}{{10}}{N_0} = {N_0}{e^{ - \lambda t}}\)
\( \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{e^{\lambda t}} = \frac{{10}}{8}\)
\(\therefore \lambda = \frac{{{{\log }_e}(10/8)}}{{10}} = \frac{{2.303\log (10/8)}}{{10}}\)
\(\,\,\lambda = 0.022\)
If \({T_{1/2}}\) is half-life period,
\(\,\,\,\,\,{T_{1/2}} = \frac{{0.693}}{\lambda } = \frac{{0.693}}{{0.022}}\)
\(\,\,\,\,\,{T_{1/2}} = 31.5{\text{ days}}\)
Again, \(N = {N_0}{\left( {\frac{1}{2}} \right)^n} = {N_0}{\left( {\frac{1}{2}} \right)^{31.5/30}}\)
\(\,\,\,\,\,\, \simeq 51.2\% \)