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 - General Questions - Page 817d: 73


$3.6\times10^{9}$ years.

Work Step by Step

Recall that $\ln(\frac{original\, quantity}{present\, quantity})=\frac{0.693}{t_{1/2}}t$ $\implies \ln(\frac{1}{0.951})=\frac{0.693}{4.9\times10^{10}\,y}\times t$ $\implies 0.05024=(1.414\times10^{-11}\,y^{-1})\times t$ Or $t=\frac{0.05024}{1.414\times10^{-11}\,y^{-1}}=3.6\times10^{9}\,y$
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