Chapter 19 - Nuclear Chemistry - Worked Example - Page 831: 20

$1.53\times10^{4}\,y$

Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{5715\,y}=1.2126\times10^{-4}\,y^{-1}$ $R_{0}=15.3\,dpm$ $R=2.4\,dpm$ Recall that $\ln(\frac{R_{0}}{R})=kt$ where $t$ is the age. $\implies \ln(\frac{15.3}{2.4})=1.8524=1.2126\times10^{-4}\,y^{-1}(t)$ $\implies t=\frac{1.8524}{1.2126\times10^{-4}\,y^{-1}}=1.53\times10^{4}\,y$