Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 30 - Nuclear Physics and Radioactivity - Problems - Page 882: 47

Answer

$0.91g$

Work Step by Step

$\frac{\Delta N}{\Delta t}=\lambda N$ $\lambda=\frac{\ln(2)}{T_{0.5}}=1.75998\times10^{-17}s^{-1}$ $N=\frac{2.4\times10^5s^{-1}}{1.75998\times10^{-17}s^{-1}}=1.36365\times10^{22}$nuclei $m=1.36365\times10^{22}nuclei\times\frac{1 a}{nucleus}\times\frac{1mol}{6.02\times10^{23}a}\times39.963998\frac{g}{mol}=0.91g$
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