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

$5.05\times10^{12}\,s$
If Original activity $N_{0}=100$, Present activity $N=1$ Rate constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{2.4\times10^{4}\,y}=2.8875\times10^{-5}\,y^{-1}$ $\ln(\frac{N_{0}}{N})=kt$ where $t$ is the time required. $\implies \ln(\frac{100}{1})=4.605=(2.8875\times10^{-5}\,y^{-1})\times t$ Or $t= \frac{4.605}{2.8875\times10^{-5}\,y^{-1}}= 1.6\times10^{5}\,y$ $=1.6\times10^{5}\times3600\times24\times365\,s$ $=5.05\times10^{12}\,s$