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

$2.53\times10^{12}\,s$
If original activity $A_{0}=100$, then present activity $A=10.0$. Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{24100\,y}$ $\ln(\frac{A_{0}}{A})=kt$ $\implies t=\frac{\ln(\frac{A_{0}}{A})}{k}$ $=\frac{\ln(\frac{100}{10.0})}{\frac{0.693}{24100\,y}}=8.00755\times10^{4}\,y$ $=8.00755\times10^{4}\,y\times\frac{365\,d}{1\,y}\times\frac{24\,h}{1\,d}=\frac{3600\,s}{1\,h}$ $=2.53\times10^{12}\,s$