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 950: 20.34


$5\times10^{8}$ years.

Work Step by Step

Mass of $\,^{238}U$ decayed to $\,^{206}Pb$= $0.08\,mg\,\,^{206}Pb\times\frac{238\,mg\,\,^{238}U}{206\,mg\,\,^{206}Pb}=0.0924\,mg$ Original mass of $^{238}U=$$1.09\,mg+0.0924\,mg$ $=1.1824\,mg$ Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{4.51\times10^{9}\,y}=1.5366\times10^{-10}\,y^{-1}$ Recall that $\ln(\frac{original\,mass}{present\,mass})=kt$ $\implies \ln(\frac{1.1824}{1.09})=0.08137=1.5366\times10^{-10}\,y^{-1}(t)$ $\implies t=\frac{0.08137}{1.5366\times10^{-10}y^{-1}}=5\times10^{8}\,y$
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