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 - Cumulative Exercises - Page 935: 21.63


18.4 Bq

Work Step by Step

Number of moles n=$\frac{6.00\times10^{-1}\,g}{40\,g/mol}=0.0150\,mol$ Number of particles=$n\times N_{A}$ $=0.0150\times6.022\times10^{23}=9.033\times10^{21}$ Among these, only 0.0117% is radioactive. Therefore, $N=9.033\times10^{21}\times0.000117$ $=1.05686\times10^{18}$ Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{1.26\times10^{9}\times365\times24\times3600\,s}$ $=1.744\times10^{-17}\,s^{-1}$ Radioactivity=$kN=1.744\times10^{-17}\,s^{-1}\times1.05686\times10^{18}$ $=18.4\,decays/s=18.4\,Bq$
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