## Chemistry: The Molecular Science (5th Edition)

If Original activity $N_{0}=100$, then present activity $N=10.0$. Rate constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{3.82\,day}=0.1814/day$ $\ln(\frac{N_{0}}{N})=kt$ where $t$ is the time required. $\implies \ln(\frac{100}{10.0})=2.30=(0.1814/day)\times t$ Or $t= \frac{2.30}{0.1814/day}= 12.7\,days$