Chapter 18 - Nuclear Chemistry - Questions for Review and Thought - More Challenging Questions - Page 817f: 93

Decay constant $k=0.1806/day$ Half-life $t_{1/2}=3.84\,days$

$\ln(\frac{A_{0}}{A})=kt$ where $k$ is the decay constant. $\implies \ln (\frac{7.00\times10^{4}\,Bq}{1.15\times10^{4}\,Bq})=1.806=k\times 10.0\,d$ $\implies k=\frac{1.806}{10.0\,d}=0.1806\,d^{-1}$ Half-life $t_{1/2}=\frac{0.693}{k}=\frac{0.693}{0.1806\,d^{-1}}=3.84\,d$