Chemistry: Principles and Practice (3rd Edition)

Published by Cengage Learning

Chapter 7 - Electronic Structure - Questions and Exercises - Exercises - Page 285: 7.32

100. kJ

Work Step by Step

E = hν h = $6.626\times10^{-34}Js$ ν = $2.50\times10^{14}s^{-1}$ E = $(6.626\times10^{-34}Js)(2.50\times10^{14}s^{-1})$ = $1.66\times10^{-19}J$ Energy of one mole of photons =$(1.66\times10^{-19}J)(6.022\times10^{23}) \approx 100. kJ$

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