Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 38 - Quantization - Exercises and Problems - Page 1114: 8

Answer

(a) $E = 2.26~eV$ (b) $\lambda = 1.66\times 10^{-10}~m$

Work Step by Step

(a) We can find the energy: $E = \frac{h~c}{\lambda}$ $E = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{550\times 10^{-9}~m}$ $E = 3.614\times 10^{-19}~J$ $E = (3.614\times 10^{-19}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$ $E = 2.26~eV$ (b) We can find the wavelength: $\lambda = \frac{h~c}{E}$ $\lambda = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(7500~eV)(1.6\times 10^{-19}~J/eV)}$ $\lambda = 1.66\times 10^{-10}~m$
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