Chapter 21 - Nuclear Chemistry - Questions and Exercises - Exercises - Page 934: 21.40

12.3 hours

Initial rate $R_{0}=1245\,dps$ Rate after time $t$, $R=350\,dps$ Time $t=\text{22 hours and 32 minutes}=22.533\,h$ Recall that $\ln(\frac{R_{0}}{R})=kt=\frac{0.693}{t_{1/2}}\times t$ where $k$ is the decay constant and $t_{1/2}$ is the half-life. $\implies \ln(\frac{1245\,dps}{350\,dps})=1.26896=\frac{0.693}{t_{1/2}}\times22.533\,h$ Or $t_{1/2}=\frac{0.693\times22.533\,h}{1.26896}=12.3\,h$