College Physics (4th Edition)

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

Chapter 28 - Problems - Page 1074: 5


$\lambda = 1.84\times 10^{-11}~m$

Work Step by Step

We can find the mass of the electron when it is moving at this relativistic speed: $m = \gamma~m_0$ $m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$ $m = \frac{m_0}{\sqrt{1-\frac{(0.6~c)^2}{c^2}}}$ $m = \frac{m_0}{0.8}$ $m = 1.25~m_0$ We can find the de Broglie wavelength: $\lambda = \frac{h}{mv}$ $\lambda = \frac{6.626\times 10^{-34}~J~s}{(1.25)(1.6\times 10^{-31}~kg)(0.6)(3.0\times 10^8~m/s)}$ $\lambda = 1.84\times 10^{-11}~m$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.