Chemistry: Principles and Practice (3rd Edition)

Published by Cengage Learning
ISBN 10: 0534420125
ISBN 13: 978-0-53442-012-3

Chapter 21 - Nuclear Chemistry - Questions and Exercises - Exercises - Page 934: 21.37

Answer

$1.29\times10^{5}\,years$

Work Step by Step

Rate= $\frac{382\,atoms}{60\,s}=6.37\,\frac{atoms}{s}$ $N=3.75\times10^{13}\,atoms$ $Rate=kN \implies k=\frac{Rate}{N}=\frac{6.37\,\frac{atoms}{s}}{3.75\times10^{13}\,atoms}=1.70\times10^{-13}\,s^{-1}$ $t_{1/2}=\frac{0.693}{k}=\frac{0.693}{1.70\times10^{-13}\,s^{-1}}=4.08\times10^{12}\,s$ $=4.08\times10^{12}\,s\times\frac{1\,hr}{3600\,s}\times\frac{1\,day}{24\,hr}\times\frac{1\,year}{365.25\,day}=1.29\times10^{5}\,years$
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