College Physics (4th Edition)

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

Chapter 29 - Problems - Page 1117: 11


The binding energy is $2.22~MeV$

Work Step by Step

The mass of a deuteron is $2.013553~u$ We can find the mass of one neutron and one proton: $M = (1.008665~u)+(1.007276~u)$ $M = 2.015941~u$ We can find the missing mass: $\Delta m = 2.015941~u-2.013553~u$ $\Delta m = 0.002388~u$ We can assume that the energy of the missing mass is the binding energy: $E = \Delta m~c^2$ $E = (0.002388~u)~c^2$ $E = (0.002388~u)(\frac{931.5~MeV/c^2}{1~u})~c^2$ $E = 2.22~MeV$ The binding energy is $2.22~MeV$
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