Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 38 - Photons and Matter Waves - Problems - Page 1184: 54a

Answer

$3.315\times 10^{-24}$ $kg.m/s$

Work Step by Step

The de Broglie wavelength $(\lambda)$ is express by the formula, $\lambda=\frac{h}{p}$ where $h$ is the planck's constant and $p$ is the momentum of the moving particle. $\therefore$ $p=\frac{h}{\lambda}$ Here, the electron has a wavelength of $\lambda=0.20$ $nm$ $\therefore$ The momentum of the electron is given by $p=\frac{6.63\times 10^{-34}}{0.20\times 10^{-9}}$ $kg.m/s$ or, $p=3.315\times 10^{-24}$ $kg.m/s$
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.