College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 28 - Problems - Page 1074: 7


The kinetic energy of the electrons should be $~250~eV$

Work Step by Step

Let $E_x = 16,000~eV$ Note that the momentum of the electron $p_e = \sqrt{2~m_e~E_e}$ To produce the same diffraction pattern, the de Broglie wavelength of the electron should be the same as the wavelength of the x-ray photons. We can find the kinetic energy of the electrons: $\lambda = \frac{hc}{E_x} = \frac{h}{p_e}$ $\frac{hc}{E_x} = \frac{h}{\sqrt{2~m_e~E_e}}$ $\sqrt{2~m_e~E_e} = \frac{E_x}{c}$ $2~m_e~E_e = \frac{E_x^2}{c^2}$ $E_e = \frac{E_x^2}{2~m_e~c^2}$ $E_e = \frac{[(16,000~eV)(1.6\times 10^{-19}~J/eV)]^2}{(2)~(9.1\times 10^{-31}~kg)~(3.0\times 10^8~m/s)^2}$ $E_e = (4.00\times 10^{-17}~J)(\frac{1~eV}{1.6\times 10^{-19}~eV})$ $E_e = 250~eV$ The kinetic energy of the electrons should be $~250~eV$
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