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: 21.65


4300 Bq

Work Step by Step

Number of moles n=$\frac{140\,g}{40\,g/mol}=3.5\,mol$ Number of particles N=$n\times N_{A}$ $=3.5\times6.022\times10^{23}=2.1077\times10^{24}$ Among these, only 0.0117% is radioactive. Therefore, $N=2.1077\times10^{24}\times0.000117$ $=2.466\times10^{20}$ 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=$kN=1.744\times10^{-17}\,s^{-1}\times2.466\times10^{20}$ $=4300\,decays/s=4300\,Bq$
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