Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 38 - Quantization - Exercises and Problems - Page 1154: 64

Answer

$v = 4.425\times 10^4~m/s$

Work Step by Step

We can find the energy of each photon: $E = \frac{h~c}{\lambda}$ $E = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{121.6\times 10^{-9}~m}$ $E = 1.635\times 10^{-18}~J$ By conservation of momentum, since both atoms come to a halt, both atoms must have had the same initial speed, and thus the same kinetic energy. By conservation of energy, the total initial kinetic energy must be equal to the total energy of both photons. Therefore, the initial kinetic energy of each atom must be equal to the energy of one photon. We can find the initial speed of each atom: $K = \frac{1}{2}mv^2 = E$ $v^2 = \frac{2E}{m}$ $v = \sqrt{\frac{2E}{m}}$ $v = \sqrt{\frac{(2)(1.635\times 10^{-18}~J)}{1.67\times 10^{-27}~kg}}$ $v = 4.425\times 10^4~m/s$
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