College Physics (4th Edition)

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

Chapter 26 - Problems - Page 1012: 47

Answer

The amount of energy that is released is $5.58~MeV$ which is $8.93\times 10^{-13}~J$

Work Step by Step

We can find the mass $M$ that is lost: $221.97039~u = 217.96289~u+4.00151~u+M$ $M = 221.97039~u - 217.96289~u - 4.00151~u$ $M = 0.00599~u$ $M = (0.00599~u)(\frac{931.494~MeV/c^2}{1~u})$ $M = 5.58~MeV$ We can assume that the missing mass is converted into energy. We can find this energy: $E = Mc^2$ $E = (5.58~MeV/c^2)(c^2)$ $E = 5.58~MeV$ $E = (5.58~MeV)(\frac{1.60\times 10^{-13}~J}{1~MeV})$ $E = 8.93\times 10^{-13}~J$ The amount of energy that is released is $5.58~MeV$ which is $8.93\times 10^{-13}~J$.
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.