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: 39

Answer

6700 yr

Work Step by Step

Rate constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{5730\,y}=1.21\times10^{-4}\,y^{-1}$ $\ln(\frac{N_{0}}{N})=kt$ where $t$ is the age, $N_{0}$ is the original activity and $N$ is the present activity. $\implies \ln(\frac{9200}{4100})=0.8082=(1.21\times10^{-4}\,y^{-1})\times t$ Or $t=\frac{0.8082}{1.21\times10^{-4}y^{-1}}=6700\,yr$
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