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 817d: 68


5.2 years

Work Step by Step

$\ln(\frac{A_{0}}{A})=kt$ where $A_{0}$ is the initial activity, $A$ is the present activity and $k$ is the decay constant. $\implies \ln (\frac{4.3\times10^{6}}{2.6\times10^{5}})=2.8057=k\times 21.2\,y$ $\implies k=\frac{2.8057}{21.2\,y}=0.132344\,y^{-1}$ Half-life $t_{1/2}=\frac{0.693}{k}=\frac{0.693}{0.132344\,y^{-1}}=5.2\,y$
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