Mass Defect And Binding Energy

Question

A helium-4 nucleus is assembled from two protons and two neutrons. Using mass of hydrogen atom 1.00783 u, neutron 1.00867 u, and helium-4 atom 4.00260 u, what is the total binding energy of the helium nucleus?

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The binding energy of a nucleus is the energy equivalent of its mass defect, the difference between the summed masses of the free constituents and the actual nuclear mass. Working with atomic masses (electron masses cancel), the mass defect is $\Delta m = [2(1.00783) + 2(1.00867)] - 4.00260 = (2.01566 + 2.01734) - 4.00260 = 4.03300 - 4.00260 = 0.03040$ u. Converting with $1\,	ext{u} = 931.5$ MeV gives $BE = 0.03040 	imes 931.5 \approx 28.3$ MeV. The value 7.07 MeV is the binding energy per nucleon ($28.3 \div 4$), not the total. The value 14.2 MeV mistakenly uses only half the nucleons. The value 4.03 MeV multiplies the mass defect by the wrong factor, confusing mass-number units with energy. The factor 931.5 MeV per atomic mass unit follows directly from $E = mc^2$ applied to one unified mass unit, and using atomic rather than bare nuclear masses is valid here because the two electron masses on each side of the balance cancel exactly. This applies the standard NCERT mass-defect method. A plausibility check confirms a per-nucleon binding of about 7 MeV, which is consistent with helium-4 being an unusually tightly bound, highly stable light nucleus.

Mass Defect And Binding Energy

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