Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 32 - Elementary Particles - Problems - Page 944: 2

Answer

$4.4\times10^{-17}m$

Work Step by Step

The kinetic energy of the electrons is many thousands of times greater than their rest mass energy, so their energy is basically their KE, and E = pc. Use the equation for the de Broglie wavelength, $p=\frac{h}{\lambda}$. $$E=pc=\frac{hc}{\lambda}$$ $$\lambda=\frac{hc}{E}$$ $$=\frac{(6.63\times10^{-34} J \cdot s)(3.00\times10^8m/s)}{(28\times10^9eV)(1.60\times10^{-19}J/eV)}$$ $$=4.4\times10^{-17}m$$
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