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

$\ln(\frac{A_{0}}{A})=kt$ where $A_{0}$ is the initial activity, $A$ is the present activity and $k$ is the decay constant. $\implies \ln (\frac{4.3\times10^{6}}{2.6\times10^{5}})=2.8057=k\times 21.2\,y$ $\implies k=\frac{2.8057}{21.2\,y}=0.132344\,y^{-1}$ Half-life $t_{1/2}=\frac{0.693}{k}=\frac{0.693}{0.132344\,y^{-1}}=5.2\,y$