## Chemistry (4th Edition)

Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{28.1\,y}=0.02466\,y^{-1}$ Original amount $A_{0}=1.00\,g$ Amount remaining $A=0.200\,g$ Recall that $\ln(\frac{A_{0}}{A})=kt$ where $t$ is the time required. $\implies \ln(\frac{1.00\,g}{0.200\,g})=1.60944=0.02466\,y^{-1}(t)$ $\implies t=\frac{1.60944}{0.02466\,y^{-1}}=65.3\,y$