Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning
ISBN 10: 1285199049
ISBN 13: 978-1-28519-904-7

Chapter 5 - Electron Configurations and the Periodic Table - Questions for Review and Thought - Topical Questions - Page 240a,: 17

Answer

$\nu = 1.1*10^{-15} s^{-1}$ $E=7.4*10^{-19}J$

Work Step by Step

Energy is given by: $E = \frac{hc}{ \lambda}$ $E = \frac{6.625*10^{-34}*3*10^8}{ 270*10^{-9}}$ $E=7.4*10^{-19}J$ Frequency is given by: $ \lambda = \frac{c}{ \nu}$ Substituting gives: $1.1*10^{-15} s^{-1}$
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