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: 23

Answer

$L = 0.426~nm$

Work Step by Step

We can find the energy of the photon: $E = \frac{h~c}{\lambda}$ $E = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{600\times 10^{-9}~m}$ $E = 3.313\times 10^{-19}~J$ We can find the length of the box: $E_n = \frac{n^2~h^2}{8~m~L^2}$ $E_1 = \frac{1^2~h^2}{8~m~L^2}$ $L^2 = \frac{h^2}{8~m~E_1}$ $L = \frac{h}{\sqrt{8~m~E_1}}$ $L = \frac{6.626\times 10^{-34}~J~s}{\sqrt{(8)~(9.109\times 10^{-31}~kg)~(3.313\times 10^{-19}~J)}}$ $L = 4.26\times 10^{-10}~m$ $L = 0.426~nm$
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