Chapter 19 - Sections 19.1-19.12 - Exercises - Problems by Topic - Page 946: 49

10.8 hours

$R_{0}=1.5\times10^{5}/s$ $R=2.5\times10^{3}/s$ $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{1.83\,h}=0.379/h$ Recall that $\ln(\frac{R}{R_{0}})=-kt$ $\implies t=-\frac{\ln(\frac{R}{R_{0}})}{k}=-\frac{\ln(\frac{2.5\times10^{3}}{1.5\times10^{5}})}{0.379/h}=10.8\,h$