College Physics (4th Edition)

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

Chapter 29 - Problems - Page 1117: 10


The binding energy is $2.84\times 10^7~eV$

Work Step by Step

The mass of an $\alpha$ particle is $4.00151~u$ We can find the mass of two neutrons and two protons: $M = 2(1.008665~u)+2(1.007276~u)$ $M = 4.031882~u$ We can find the missing mass: $\Delta m = 4.031882~u-4.00151~u$ $\Delta m = 0.030372~u$ We can assume that the energy of the missing mass is the binding energy: $E = \Delta m~c^2$ $E = (0.030372~u)(3.0\times 10^8~m/s)^2$ $E = (0.030372)(1.66054\times 10^{-27}~kg)(3.0\times 10^8~m/s)^2$ $E = (4.539\times 10^{-12}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$ $E = 2.84\times 10^7~eV$ The binding energy is $2.84\times 10^7~eV$
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