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Unformatted text preview: 2.5 Nuclear Binding Energy Clear evidence for the massenergy equivalence relation is given in the study of mass defects in atomic nuclei. The mass defect is the deficit in the mass of a nucleus, compared with the sum of the masses of its constituent nucleons, and corresponds to the binding energy of the nucleus. I follows from equation 151 that Δ E = Δ mc 2 (152) (since we have established a linear relationship between changes in energy and changes in mass), and also E = m c 2 (153) for the rest energy. In the context of nuclear reactions, we can state the binding energy of a nucleus A Z X N : B = Zm p + Nm n m ( A X) c 2 (154) where I have neglected the energy associated with the binding of the electrons to the nucleus. In using this equation, it often helps to give the masses in terms of mass units U, and to include the unit conversion factor in c 2 : thus c 2 = 931 . 50 MeV /U (155) example: The mass of a helium nucleus is 4.0026 U, whereas the mass of its constituents (two protons and two neutrons) in the free state is 4.0319 U. What is the binding energy of helium? solution: The mass defect is 0.0293 U. Multiplying by c 2 in the appropriate units gives a binding energy of B = 27 . 3MeV (156) this is the energy that would be released in the fusion of the constituent nuclear particles into a single helium nucleus, as might happen in the center of the sun. More practical calculation can be made if nuclei with a given mass defect are transformed via some reaction to nuclei with a different mass defect. In a nuclear reaction of this type, the net energy released would be equivalent to the difference between the total mass defect...
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This note was uploaded on 09/20/2010 for the course AMATH 261 taught by Professor Rogermelko during the Spring '10 term at Waterloo.
 Spring '10
 RogerMelko

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