Chemistry: Principles and Practice (3rd Edition)

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

Chapter 7 - Electronic Structure - Questions and Exercises - Exercises - Page 285: 7.31


$1.48\times10^{3} kJmol^{-1}$

Work Step by Step

E= hν h = $6.626\times10^{-34}Js$ ν = $3.70\times10^{15}Hz$ E = $(6.626\times10^{-34}Js)(3.70\times10^{15}s^{-1})$ = $2.45\times10^{-18}J$ Energy of one mole of photons = ($2.45\times10^{-18}J$)($6.022\times10^{23}mol^{-1}$) = $1.48\times10^{3} kJmol^{-1}$.
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