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 952: 20.63


65.3 years

Work Step by Step

Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{28.1\,y}=0.02466\,y^{-1}$ Original amount $A_{0}=1.00\,g$ Amount remaining $A=0.200\,g$ Recall that $\ln(\frac{A_{0}}{A})=kt$ where $t$ is the time required. $\implies \ln(\frac{1.00\,g}{0.200\,g})=1.60944=0.02466\,y^{-1}(t)$ $\implies t=\frac{1.60944}{0.02466\,y^{-1}}=65.3\,y$
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