Chemistry: The Molecular Science (5th Edition)

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

Chapter 18 - Nuclear Chemistry - Questions for Review and Thought - Topical Questions - Page 817b: 38



Work Step by Step

If Original activity $N_{0}=100$, Present activity $N=1$ Rate constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{2.4\times10^{4}\,y}=2.8875\times10^{-5}\,y^{-1}$ $\ln(\frac{N_{0}}{N})=kt$ where $t$ is the time required. $\implies \ln(\frac{100}{1})=4.605=(2.8875\times10^{-5}\,y^{-1})\times t$ Or $t= \frac{4.605}{2.8875\times10^{-5}\,y^{-1}}= 1.6\times10^{5}\,y$ $=1.6\times10^{5}\times3600\times24\times365\,s$ $=5.05\times10^{12}\,s$
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