Chemistry: Molecular Approach (4th Edition)

Published by Pearson
ISBN 10: 0134112830
ISBN 13: 978-0-13411-283-1

Chapter 7 - Quantum Mechanics and The Atom - Exercises - Page 330: 46

Answer

(a) $8.87\times10^{5}\,kJ/mol$ (b) $5.57\times10^{9}\,kJ/mol$

Work Step by Step

Recall: Energy of one photon= $\frac{hc}{\lambda}$ Energy of one mole of photons= $\frac{hc}{\lambda}\times6.022\times10^{23}/mol$ (a) $E=\frac{(6.626\times10^{-34}\,J\cdot s)(3.00\times10^{8}\,m/s)}{0.135\times10^{-9}\,m}\times6.022\times10^{23}/mol$ $=8.87\times10^{5}\times10^{3}\,J/mol=8.87\times10^{5}\,kJ/mol$ (b) $E=\frac{(6.626\times10^{-34}\,J\cdot s)(3.00\times10^{8}\,m/s)}{2.15\times10^{-5}\times10^{-9}\,m}\times6.022\times10^{23}/mol$ $=5.57\times10^{12}\,J/mol=5.57\times10^{9}\,kJ/mol$
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