Fundamentals of Physics Extended (10th Edition)

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

Chapter 37 - Relativity - Problems - Page 1149: 58a

Answer

$\gamma = 1.0019569$

Work Step by Step

We can find $\gamma$: $K = (\gamma-1)~mc^2$ $\gamma-1 = \frac{K}{mc^2}$ $\gamma = 1+\frac{K}{mc^2}$ $\gamma = 1+\frac{(1.0000000\times 10^3~eV)(1.602 176 462\times 10^{-19}~J/eV)}{(9.109 381 88\times 10^{-31}~kg)(2.998\times 10^8~m/s)^2}$ $\gamma = 1.0019569$
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.