Chapter 19 - Sections 19.1-19.12 - Exercises - Problems by Topic - Page 946: 54

$2.9\times10^{4}$ years ago.

$R_{0}=15.3\,dis/min\cdot g\,C$ $R=0.48\,dis/min\cdot g\,C$ $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{5730\,yr}=1.21\times10^{-4}/yr$ Recall that $\ln(\frac{R}{R_{0}})=-kt$ $\implies t=-\frac{\ln(\frac{R}{R_{0}})}{k}=-\frac{\ln(\frac{0.48}{15.3})}{1.21\times10^{-4}/yr}=2.9\times10^{4}\,yr$