Binding energy and the mass formula

Weigh a nucleus and it comes out lighter than the nucleons inside it. That shortfall, the mass defect, times c², is the work needed to take the nucleus apart: B = [Z⋅M(¹H) + N⋅mₙ − M(A, Z)]c². Work with atomic masses throughout and the Z electrons on the left, one per hydrogen atom, cancel the Z in the neutral atom on the right; only the electrons' own binding energies fail to cancel, about 0.76 MeV for uranium against a nuclear binding of 1801.7 MeV, four parts in ten thousand. The effect is large by nuclear standards and invisible by chemical ones. ⁵⁶Fe falls short of its 26 hydrogens and 30 neutrons by 0.528464 u out of 55.934936 u — 0.94% of its mass — while a water molecule's chemical binding is about six parts in ten billion of its mass. That is why nuclear mass tables are quoted to six decimal places in u: the physics lives in the fifth and sixth.

Total binding energy rises almost linearly with mass number: 2.22 MeV for ²H, 28.30 for ⁴He, 92.16 for ¹²C, 492.3 for ⁵⁶Fe, 1801.7 for ²³⁸U. That near-linearity is itself the first result — it says each nucleon binds only to the handful of neighbours within the roughly 1.4 fm range of the strong force, not to all A − 1 others. Were the force long-ranged like the Coulomb force, B would go as A(A−1)/2 and B/A would climb without limit. So the informative quantity is B/A, and its curve has three regions: a steep rise through the light nuclei, where too large a fraction of nucleons sits on the surface; a broad maximum near 8.8 MeV per nucleon around A ≈ 60, with ⁶²Ni at 8.795 MeV holding the record and ⁵⁶Fe at 8.790 just behind; and a slow decline to 7.57 MeV at ²³⁸U as Coulomb repulsion accumulates. Moving toward that maximum from either side releases energy: fusing deuterium and tritium gives 17.6 MeV, and splitting ²³⁸U into two A ≈ 119 fragments about 200 MeV.

A reaction's energy release is that same subtraction done twice: Q = (Σmᵢₙ − Σmₒᵤₜ)c², equivalently ΣBₒᵤₜ − ΣBᵢₙ. Atomic masses again do the work, provided the electron count balances. In ⁴He + ¹⁴N → ¹H + ¹⁷O there are 2 + 7 = 9 electrons on the left and 1 + 8 = 9 on the right, so Q = (4.002602 + 14.003074 − 1.007825 − 16.999132) × 931.494 = −1.19 MeV: endothermic. A negative Q is not the whole story in the laboratory. Fire the alpha at a stationary nitrogen nucleus and the products must carry off the incoming momentum, so part of the beam energy can never be spent on the reaction; the threshold is Kₜₕ ≈ |Q|(1 + mₐ/m_X) = 1.19 × 1.286 = 1.53 MeV. Decays are the exothermic case of the same accounting: ²³⁸U → ²³⁴Th + ⁴He has Q = (238.050788 − 234.043601 − 4.002602) × 931.494 = +4.27 MeV. That is where alpha decay gets its energy — though nothing in this arithmetic says how long the wait will be.

Treat the nucleus as a charged drop of incompressible nuclear fluid. Volume: saturation gives almost every nucleon the same number of neighbours, so B gains a_V⋅A. Surface: nucleons in the skin have fewer neighbours, costing a_S⋅A²⁄³, since area goes as R² and R = r₀A¹⁄³. Coulomb: every proton pair repels and this term does not saturate — for a uniformly charged sphere it is (3/5)(e²/4πε₀)Z(Z−1)/R, and with r₀ = 1.2 fm the coefficient comes out at 0.72 MeV, calculated rather than fitted. Asymmetry: neutrons and protons fill separate ladders of levels, so converting one kind into the other lifts nucleons up an already occupied ladder, costing a_A(A−2Z)²/A. Pairing: identical nucleons couple in pairs to zero angular momentum, so even-even nuclei gain and odd-odd nuclei lose. For ⁵⁶Fe the five terms are +884.8, −267.9, −121.3, −6.6 and +1.6 MeV, giving B = 490.6 against 492.3 measured, 0.34% low; for ²³⁸U they are +3760.4, −702.8, −964.6, −284.2 and +0.8, giving 1809.5 against 1801.7, this time 0.43% high.

Fix A and walk along an isobar. The volume and surface terms do not move, the Coulomb term climbs as Z(Z−1), and the asymmetry term falls to zero at Z = A/2, so the atomic mass is a parabola in Z. Setting ∂M/∂Z = 0 gives Zₘᵢₙ ≈ (A/2)/[1 + (a_C/4a_A)A²⁄³], which is 12.6 at A = 27 and 91.9 at A = 238: the Coulomb term drags the valley floor below A/2, so heavy stable nuclei need N > Z. For odd A the pairing term vanishes, there is a single parabola, and normally exactly one stable isobar sits at the bottom while every other member β-decays toward it — β⁻ from the neutron-rich arm, β⁺ or electron capture from the proton-rich one. For even A the pairing term splits the parabola in two, even-even below odd-odd by 2δ = 24/√A, which is 3.0 MeV at A = 64. That is why only four odd-odd nuclides are stable (²H, ⁶Li, ¹⁰B and ¹⁴N) and why even-A chains often carry two or three stable isobars with an odd-odd member between them decaying both ways.

Five fitted constants reproduce two thousand measured masses to a few MeV in several hundred, but the leftovers are not noise. They peak wherever a nucleon number reaches 2, 8, 20, 28, 50, 82 or 126: those nuclides are bound several MeV more tightly than the drop allows, and the drop has no way to know it, containing no shells, no spins and no parities. Separation energies make the point sharply. Sₙ is 7.37 MeV for ²⁰⁸Pb and 3.94 MeV for ²⁰⁹Pb, because the 127th neutron has to start a new shell — a cliff the smooth formula renders as a gentle slope. The fit also fails outright for light nuclei: at A = 4 the surface term reaches 46.1 MeV against a volume term of only 63.2 MeV, leaving B = 22.2 MeV against 28.30 measured, 22% low, and a drop of four molecules is all surface anyway. Deformed rare-earth and actinide nuclei add a systematic residual of their own. Read the right way, that residual map is not a failure of the model but the experimental signal for the next one.