Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning
ISBN 10: 1285199049
ISBN 13: 978-1-28519-904-7

Chapter 18 - Nuclear Chemistry - Summary Problem - Page 816: 5d



Work Step by Step

If original activity $A_{0}=100$, then present activity $A=10.0$. Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{24100\,y}$ $\ln(\frac{A_{0}}{A})=kt$ $\implies t=\frac{\ln(\frac{A_{0}}{A})}{k}$ $=\frac{\ln(\frac{100}{10.0})}{\frac{0.693}{24100\,y}}=8.00755\times10^{4}\,y$ $=8.00755\times10^{4}\,y\times\frac{365\,d}{1\,y}\times\frac{24\,h}{1\,d}=\frac{3600\,s}{1\,h}$ $=2.53\times10^{12}\,s$
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