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 288: 7.89

Answer

Frequency (ν) = $6.08\times10^{14}s^{-1}$ Energy (E) = $4.03\times10^{-19} J$

Work Step by Step

ν = c/λ c = $2.997925\times10^{8}ms^{-1}$ λ = 493 nm = $493\times10^{-9}m$ ν = $\frac{2.997925\times10^{8}ms^{-1}}{493\times10^{-9}m}$ = $6.08\times10^{14}s^{-1}$ E = hν h = $6.626\times10^{-34}Js$ ν = $6.08\times10^{14}s^{-1}$ E = $6.626\times10^{-34}Js\times6.08\times10^{14}s^{-1}$ = $4.03\times10^{-19}J$
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