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: 3


$\lambda = 223~nm$

Work Step by Step

We can find the wavelength of the light: $K_{max} = \frac{h~c}{\lambda} - E_0$ $K_{max}+E_0 = \frac{h~c}{\lambda}$ $\lambda = \frac{h~c}{K_{max}+E_0}$ $\lambda = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(1.30~eV+4.28~eV)(1.6\times 10^{-19}~J/eV)}$ $\lambda = 2.23\times 10^{-7}~m$ $\lambda = 223~nm$
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