College Physics (4th Edition)

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

Chapter 26 - Problems - Page 1012: 55


$p = 6.48~MeV/c$

Work Step by Step

Mote that the rest energy of an electron is: $mc^2 = 0.511~MeV$ We can find the momentum $p$: $E^2 = (pc)^2+(mc^2)^2$ $(pc)^2 = (6.5~MeV)^2 -(0.511~MeV)^2$ $(pc)^2 = 42.0~(MeV)^2$ $pc = \sqrt{42.0~(MeV)^2}$ $pc = 6.48~MeV$ $p = 6.48~MeV/c$
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.