Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 19 - Sections 19.1-19.12 - Exercises - Problems by Topic - Page 946: 53

Answer

$2.4\times10^{4}\,yr$

Work Step by Step

$R_{0}=15.3\,dis/min\cdot g\,C$ $R=0.85\,dis/min\cdot g\,C$ $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{5730\,yr}=1.21\times10^{-4}/yr$ Recall that $\ln(\frac{R}{R_{0}})=-kt$ $\implies t=-\frac{\ln(\frac{R}{R_{0}})}{k}=-\frac{\ln(\frac{0.85}{15.3})}{1.21\times10^{-4}/yr}=2.4\times10^{4}\,yr$
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