Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 7 - Sections 7.1-7.6 - Exercises - Problems by Topic - Page 331: 69d

Answer

434 nm, visible region.

Work Step by Step

$\Delta E_{atom}=-2.18\times10^{-18}\,J(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}})$ $=-2.18\times10^{-18}\,J(\frac{1}{2^{2}}-\frac{1}{5^{2}})$ $=-4.578\times10^{-19}\,J$ $E_{photon}=-\Delta E_{atom}=4.578\times10^{-19}\,J$ $E=\frac{hc}{\lambda}\implies \lambda=\frac{hc}{E}$ $=\frac{(6.626\times10^{-34}\,J\cdot s)(3.00\times10^{8}\,m/s)}{4.578\times10^{-19}\,J}$ $=4.34\times10^{-7}\,m=434\,nm$ This light is found in the visible region (In the visible region, wavelength ranges from 400 nm to 700 nm).
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