College Physics (4th Edition)

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

Chapter 28 - Problems - Page 1074: 16


The change in velocity is $0.078~m/s$

Work Step by Step

We can use the kinetic energy $E$ to write an expression for the speed of a particle: $E = \frac{1}{2}mv^2$ $v = \sqrt{\frac{2~E}{m}}$ We can find the speed of the particle when it absorbs a photon with a wavelength of $660~nm$: $v = \sqrt{\frac{2~E}{m}}$ $v = \sqrt{\frac{2~(\frac{hc}{\lambda})}{m}}$ $v = \sqrt{\frac{2~h~c}{m~\lambda}}$ $v = \sqrt{\frac{(2)(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(1.0\times 10^{-16}~kg)(660\times 10^{-9}~m)}}$ $v = 0.078~m/s$ The change in velocity is $0.078~m/s$
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.