Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company
ISBN 10: 0078021529
ISBN 13: 978-0-07802-152-7

Chapter 20 - Questions and Problems - Page 953: 20.95


2770 years.

Work Step by Step

Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{5730\,y}=1.21\times10^{-4}\,y^{-1}$ $\ln(\frac{A_{0}}{A})=kt$ $\implies \ln(\frac{0.260}{0.186})=0.3349=1.21\times10^{-4}\,y^{-1}(t)$ Then, $t=\frac{0.3349}{1.21\times10^{-4}\,y^{-1}}=2.77\times10^{3}\,y$
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