## Chemistry: Principles and Practice (3rd Edition)

(a) Decay rate= $1238\,dpm=kN=\frac{0.693}{t_{1/2}}\times2.64\times10^{10}$ ($k$ is the decay constant, $N$ is the number of particles) $\implies$Half-life $t_{1/2}=\frac{0.693}{1238/min}\times2.64\times10^{10}$ $=1.4778\times10^{7}\,min=1.4778\times10^{7}\,min\times\frac{1\,h}{60\,min}\times\frac{1\,d}{24\,h}\times\frac{1\,y}{365\,d}$ $=28.1\,y$ (b) Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{28.1\,y}=0.02466\,y^{-1}$ Original rate $R_{0}=1238\,dpm$ Rate after time $t$, $R=1000\,dpm$ Recall that $\ln(\frac{R_{0}}{R})=kt$, where $t$ is the time required. $\implies \ln(\frac{1238\,dpm}{1000\,dpm})=0.2135=0.02466\,y^{-1}(t)$ $\implies t=\frac{0.2135}{0.02466\,y^{-1}}=8.66\,y$