Chapter 25 - Nuclear Chemistry - Example 25-4 - Applying the Integrated Rate Law for Radioactive Decay: Radiocarbon Dating - Page 1122: Practice Example A

$4.7\times10^{3}\,years$

Decay constant $\lambda=\frac{0.693}{t_{1/2}}=\frac{0.693}{5730\,y}=1.21\times10^{-4}\,y^{-1}$ Original activity $R_{0}=15\,dpm$ Activity after time $t$, $R=8.5\,dpm$ Recall that $\ln(\frac{R_{0}}{R})=\lambda t$ where $t$ is the age. $\implies \ln(\frac{15\,dpm}{8.5\,dpm})=0.567984=1.21\times10^{-4}\,y^{-1}(t)$ Or $t=\frac{0.567984}{1.21\times10^{-4}\,y^{-1}}=4.7\times10^{3}\,y$