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Physics · Ch 8 — Atomic and Nuclear Physics

Mass defect and binding energy

8.4.5

Mass defect and binding energy

A remarkable experimental fact underlies the entire concept of nuclear binding energy: the measured mass of any nucleus is always slightly less than the sum of the masses of its individual, separated constituent nucleons.

Worked case: carbon-12. The carbon-12 nucleus contains 6 protons and 6 neutrons. Adding up their individual masses separately: 6 neutrons contribute 6×1.00866=6.051966\times1.00866=6.05196 u, and 6 protons contribute 6×1.00727=6.043626\times1.00727=6.04362 u, for a combined constituent mass of 12.0955812.09558 u. But mass spectroscopy shows the actual measured mass of the carbon-12 atom is exactly 12 u; subtracting the mass of the 6 orbiting electrons (6×0.00055=0.00336\times0.00055=0.0033 u) gives the carbon-12 nuclear mass as 11.996711.9967 u - which is indeed less than the summed constituent mass of 12.0955812.09558 u, by a difference of Δm=0.09888\Delta m=0.09888 u.

Mass defect. In general, for any nucleus ZAX^{A}_{Z}X with measured mass MM, built from ZZ protons (mass mpm_p) and NN neutrons (mass mnm_n), this shortfall is called the mass defect:

Δm=(Zmp+Nmn)−M(8.20)\Delta m=(Zm_p+Nm_n)-M\qquad (8.20)

Binding energy. Einstein's mass-energy relation E=mc2E=mc^2 explains where the missing mass goes: when the separate nucleons combine to form the nucleus, an amount of mass equal to Δm\Delta m effectively converts into energy that is released, called the binding energy of the nucleus, BE=(Δm)c2BE=(\Delta m)c^2. Equivalently, to pull the nucleus back apart into its separate free nucleons, exactly this much energy would need to be supplied:

BE=[Zmp+Nmn−M]c2(8.21)BE=[Zm_p+Nm_n-M]c^2\qquad (8.21)

It is more convenient in practice to work with tabulated atomic masses (which already include the electrons) rather than bare nuclear masses; adding and subtracting the mass of the ZZ atomic electrons converts this formula into

BE=[ZmH+Nmn−MA]c2(8.24)BE=[Zm_H+Nm_n-M_A]c^2\qquad (8.24) …

Misc Example 8.9Binding energy of the helium-4 nucleus

Worked out. This worked example computes the binding energy of the 24He^{4}_{2}\text{He} nucleus from the given atomic mass of helium (4.00260 u) and of hydrogen (1.00785 u), using the formula BE = [Z m_H + N m_n - M_A] c^2 with Z = 2 protons and N = A - Z = 2 neutrons. The mass defect works out to Δm = [2(1.00785) + 2(1.008665)] - 4.00260 = 0.03038 u, and converting this to energy using 1 u = 931 MeV/c^2 gives a binding energy of 0.03038 times 931, which is approximately 28 MeV - the energy that would have to be supplied to completely separate a helium-4 nucleus …