b
Let \(N\) be the number of nucleiat any time \(t\) then,

\(\frac{d N}{d t}=100-\lambda N\)

or \(\int_0^N {\frac{{dN}}{{(100 - \lambda N)}}}  = \int_0^t d t\)

\(-\frac{1}{\lambda}[\log (100-\lambda N)]_{0}^{N}=t\)

\(\log (100-\lambda N)-\log 100=-\lambda t\)

\(\log \frac{100-\lambda N}{100}=-\lambda t\)

\(\frac{100-\lambda N}{100}=e^{-\lambda t}\)

\(1 - \frac{{\lambda N}}{{100}} = {e^{ - \lambda t}}\)

\(N=\frac{100}{\lambda}\left(1-e^{-\lambda} t\right)\)

As, \(N=50\) and \(\lambda=0.5\,/sec\)

\(\therefore 50=\frac{100}{0.5}\left(1-e^{-0.5}\right)\)

Solving we get,

\(t=2 \ln \left(\frac{4}{3}\right) \sec\)