## Chemistry: Principles and Practice (3rd Edition)

Half-life $t_{1/2}$=28.1 years Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{28.1\,y}=0.02466\,y^{-1}$ $t=(2015-1945)\,y=70\,y$ Recall that $\ln(\frac{A}{A_{0}})=-k t$, where $A_{0}$ is the amount of sample at the beginning and $A$ is the amount of sample after time $t$. $\implies \ln(\frac{A}{A_{0}})=-(0.02466\,y^{-1})(70\,y)=-1.7262$ Taking the inverse $\ln$ of both the sides, we have $\frac{A}{A_{0}}=e^{-1.7262}=0.178$