Chapter 17 - Radioactivity and Nuclear Chemistry - Exercises - Problems - Page 638: 71

18 hours.

Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{6.0\,h}=0.1155\,h^{-1}$ Original amount $A_{0}=0.050\,mg$ Amount remaining $A=6.3\times10^{-3}\,mg$ Recall that $\ln(\frac{A_{0}}{A})=kt$ where $t$ is the time required. $\implies \ln(\frac{0.050\,mg}{6.3\times10^{-3}\,mg})=2.07147=0.1155\,h^{-1}(t)$ $\implies t=\frac{2.07147}{0.1155\,h^{-1}}=18\,h$