Fundamentals of Physics Extended (10th Edition)

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

Chapter 42 - Nuclear Physics - Problems - Page 1303: 16

Answer

Binding Energy per nucleon, $\Delta E_{be/n} = 8.23 MeV$

Work Step by Step

The equation of binding energy is $\Delta E_{be} = \Sigma mc^2 - Mc^2 $ Where $ \Sigma mc^2 = [Zm_H + (A-Z) m_n] $ Substitute into $ \Delta E_{be}$ equation $\Delta E_{be} = [Zm_H + (A-Z) m_n - M_{Eu}] c^2 $ For Europium $^{152}_{63}Eu$, $ M_{Eu} = 151.921 742 u $ $A = 152$ $Z = 63$ and using $m_H = 1.007 825 u$ $m_n = 1.008 665 u $ Now we can substitute all values into the equation $\Delta E_{be} = [(63)(1.007 825 u) + (152-63) (1.008 665 u) - 151.921 742 u] c^2 $ $\Delta E_{be} = (1.342418 u)c^2$ But $1 \space u = 931.494013 \space MeV/c^2$ So $\Delta E_{be} = (1.342418 u) (931.494013 \space MeV/c^2) c^2$ $\Delta E_{be} = 1250.454 MeV$ Now the question wants binding energy per nucleon, and there is 152 nucleons in Europium, so $\Delta E_{be/n} = \frac{\Delta E_{be}}{A}$ $\Delta E_{be/n} = \frac{1250.454 MeV}{152}$ $\Delta E_{be/n} = 8.23 MeV$
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