Chapter 29 - The Nucleus - Learning Path Questions and Exercises - Exercises - Page 998: 22

0.993 mCi

Initial activity $R_{0}=10.0\,mCi$ Decay constant $\lambda =\frac{0.693}{t_{1/2}}=\frac{0.693}{18.0\,min}=0.0385\,min^{-1}$ $t=1\,h=60\,min$ Recall that $\ln(\frac{R_{0}}{R})=\lambda t$ where $R$ is the current activity. $\implies\ln(\frac{10.0\,mCi}{R})=0.0385\,min^{-1}\times60\,min=2.31$ Taking the inverse $\ln$ of both sides, we have $\frac{10.0\,mCi}{R}=e^{2.31}=10.0744$ Or $R= \frac{10.0\,mCi}{10.0744}=0.993\,mCi$