University Physics IV · Nuclear Physics · 13.2
Binding Energy & the Mass Formula
Nuclear masses are measured to eight figures, and almost everything nuclear physics predicts about energy comes from subtracting them. This topic turns that subtraction into binding energy, reads the B/A curve, and fits it with five liquid-drop terms you can defend and one residual you cannot.
Build the model
Connect the measurement to the mechanism.
A bound nucleus weighs less than the nucleons inside it, and the entire energetics of the nucleus is that shortfall. Multiply the mass defect by c² and you have the binding energy B, the work needed to disassemble the nucleus: 492.3 MeV for ⁵⁶Fe, which is 0.94% of its rest mass and some ten million times the fractional binding of a molecule. Because B climbs almost linearly with A, the quantity that carries the physics is B/A, and the measured curve has a shape worth explaining — a steep rise through the light nuclei, a broad maximum near 8.8 MeV per nucleon around A ≈ 60, and a slow decline to 7.57 MeV at ²³⁸U.
The semi-empirical mass formula explains that shape by treating the nucleus as a charged, incompressible drop: a volume term proportional to A because the strong force saturates, a surface correction, a Coulomb repulsion that does not saturate, an asymmetry penalty from the Pauli principle, and a pairing term. Four constants fitted to some two thousand measured masses then give B anywhere on the chart to within about half a percent. What it costs is everything quantum about the individual nucleon: the drop has no shells, so the extra binding at 2, 8, 20, 28, 50, 82 and 126 survives only as a residual — and that residual is the evidence for the shell model.
- Simple definition
- The binding energy of a nucleus is the energy needed to take it apart into free nucleons, and it equals the amount by which the nucleus's mass falls short of the summed masses of those nucleons, multiplied by c².
- Example
- ⁵⁶Fe weighs 55.934936 u; 26 hydrogen atoms plus 30 neutrons come to 56.463400 u. The 0.528464 u shortfall is 0.528464 × 931.494 = 492.3 MeV of binding, or 8.79 MeV per nucleon.
The Z electron masses cancel between the Z hydrogen atoms and the neutral atom; only electron binding survives, ~0.5 MeV in 1802 for uranium.
masses in u, c² = 931.494 MeV/u, M(¹H) = 1.007825 u, mₙ = 1.008665 u; atomic masses, never nuclear
The nucleus's ionisation energy. Its collapse just past N = 50, 82 or 126 is the shell signal the smooth formula cannot carry.
Usually 6–8 MeV; it falls to 3.94 MeV for ²⁰⁹Pb, one neutron past the N = 126 closure
¹⁴N(α, p)¹⁷O has Q = −1.19 MeV, yet a 1.19 MeV alpha cannot open it: momentum conservation lifts the threshold to 1.53 MeV.
Q > 0 releases energy; the threshold applies to endothermic reactions on a target X at rest
Five terms and four constants give B for any nuclide to within about 0.5%, including nuclides nobody has ever weighed.
aV = 15.8, aS = 18.3, aC = 0.714, aA = 23.2 MeV, fitted to some 2000 measured masses
It splits every even-A isobar chain into two parabolas 2δ apart, which is why just four odd-odd nuclides are stable.
±1.5 MeV at A = 64 and ±0.78 MeV at A = 238 — negligible in B, decisive between neighbouring isobars
Gives 12.6 at A = 27 (aluminium is Z = 13) and 91.9 at A = 238 (uranium is Z = 92): Coulomb drags the floor below A/2.
from ∂M/∂Z = 0 at fixed A; including (mₙ − M(¹H))c² = 0.782 MeV raises it a further 0.8%
The mass defect is the binding energy
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, a few hundred keV for uranium against a nuclear binding of 1801.7 MeV, three 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.
B grows with A, so read B/A
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.
Q values and thresholds come off the same table
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/mX) = 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.
Five terms, and the job each one does
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 aV⋅A. Surface: nucleons in the skin have fewer neighbours, costing aS⋅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 aA(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.
At fixed A the mass is a parabola in Z
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 + (aC/4aA)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.
The residual is where the shell model starts
Four 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.
Change one variable at a time
Make the relationship visible.
Leave A at 64: the smooth parabola bottoms at Z = 29, but pairing lifts odd-odd ⁶⁴Cu 1.5 MeV above both even-even neighbours, so it decays both ways. Step A to 65 and the two curves merge into one. Then drag aC down and watch the floor climb back toward Z = A/2.
Z AT VALLEY FLOOR29
PICKED ISOBAR Z29
PAIRING SPLIT 2δ3.00 MeV
PARABOLA CURVATURE3.26 MeV per Z²
Live interpretationZ AT VALLEY FLOOR: 29. PICKED ISOBAR Z: 29. PAIRING SPLIT 2δ: 3.00 MeV. PARABOLA CURVATURE: 3.26 MeV per Z²
Catch the common trap
Explain before calculating.
⁵⁶Fe has B/A = 8.79 MeV and ²³⁸U has B/A = 7.57 MeV. Which statement about these two nuclides is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyThe atomic mass of ⁴He is 4.002602 u. With M(¹H) = 1.007825 u and mₙ = 1.008665 u, find the binding energy of ⁴He and its binding energy per nucleon. Then see what the mass formula predicts, using aV = 15.8, aS = 18.3, aC = 0.714, aA = 23.2 and aP = 12 MeV.
- ⁴He has Z = 2 and N = 2, so the constituents are 2 hydrogen atoms and 2 neutrons: 2(1.007825) + 2(1.008665) = 2.015650 + 2.017330 = 4.032980 u. Using ¹H rather than a bare proton lets the two atomic electrons cancel against the two in the helium atom.
- Mass defect: Δm = 4.032980 − 4.002602 = 0.030378 u.
- B = Δm c² = 0.030378 × 931.494 = 28.30 MeV, so B/A = 28.30/4 = 7.07 MeV per nucleon.
- Now the formula. Volume 15.8 × 4 = 63.2; surface 18.3 × 4²⁄³ = 18.3 × 2.5198 = 46.11; Coulomb 0.714 × 2 × 1/4¹⁄³ = 1.428/1.5874 = 0.90; asymmetry zero, since N = Z; pairing +12/√4 = +6.0. So B = 63.2 − 46.11 − 0.90 + 6.0 = 22.2 MeV.
- The formula is 22% low. At A = 4 the surface term is 46 MeV against a volume term of 63 MeV — a drop of four molecules is essentially all surface, and the fit is meant for A ≳ 20.
AnswerB = 28.30 MeV and B/A = 7.07 MeV per nucleon, against 22.2 MeV from the mass formula. ⁴He is far more tightly bound than a liquid drop allows: it is doubly magic, with N = Z = 2.
MediumRutherford's 1919 reaction is ⁴He + ¹⁴N → ¹H + ¹⁷O. Atomic masses: ⁴He 4.002602 u, ¹⁴N 14.003074 u, ¹H 1.007825 u, ¹⁷O 16.999132 u. Find Q, say whether the reaction releases or absorbs energy, and find the smallest alpha kinetic energy that can open it against a nitrogen nucleus at rest.
- Count electrons first: 2 + 7 = 9 on the left and 1 + 8 = 9 on the right, so atomic masses may be used throughout and the electron masses cancel.
- Initial: 4.002602 + 14.003074 = 18.005676 u. Final: 1.007825 + 16.999132 = 18.006957 u.
- Δm = 18.005676 − 18.006957 = −0.001281 u, so Q = −0.001281 × 931.494 = −1.193 MeV. Q is negative: the products are less tightly bound than the reactants, and the reaction absorbs energy.
- An alpha carrying exactly 1.193 MeV still cannot do it, because the products must carry the incoming momentum and so must keep some kinetic energy. For a target at rest, Kₜₕ ≈ |Q|(1 + mₐ/mX) = 1.193 × (1 + 4.0026/14.0031) = 1.193 × 1.2858.
- Kₜₕ = 1.534 MeV. The extra 0.341 MeV is centre-of-mass motion that can never be spent on the reaction.
AnswerQ = −1.19 MeV, so the reaction is endothermic, and the fixed-target threshold is Kₜₕ = 1.53 MeV — about 29% above |Q|.
HardUse the mass formula (aV = 15.8, aS = 18.3, aC = 0.714, aA = 23.2, aP = 12 MeV) to find B for the A = 64 isobars ⁶⁴Ni (Z = 28), ⁶⁴Cu (Z = 29) and ⁶⁴Zn (Z = 30). Then predict how ⁶⁴Cu decays, with the Q value of each branch. Take (mₙ − M(¹H))c² = 0.782 MeV.
- A = 64 gives A¹⁄³ = 4 and A²⁄³ = 16, so volume and surface are common to all three: 15.8 × 64 = 1011.2 MeV and 18.3 × 16 = 292.8 MeV. Pairing is ±12/√64 = ±1.5 MeV, positive for the two even-even isobars and negative for odd-odd ⁶⁴Cu.
- ⁶⁴Ni: Coulomb 0.714 × 28 × 27/4 = 134.95; asymmetry 23.2 × 8²/64 = 23.20; B = 1011.2 − 292.8 − 134.95 − 23.20 + 1.5 = 561.75 MeV.
- ⁶⁴Cu: Coulomb 0.714 × 29 × 28/4 = 144.94; asymmetry 23.2 × 6²/64 = 13.05; B = 1011.2 − 292.8 − 144.94 − 13.05 − 1.5 = 558.91 MeV.
- ⁶⁴Zn: Coulomb 0.714 × 30 × 29/4 = 155.29; asymmetry 23.2 × 4²/64 = 5.80; B = 1011.2 − 292.8 − 155.29 − 5.80 + 1.5 = 558.81 MeV. Without pairing ⁶⁴Cu would be the most bound of the three; the 3.0 MeV split turns it over.
- β⁻ to ⁶⁴Zn: Q = (mₙ − M(¹H))c² + B(Zn) − B(Cu) = 0.782 + 558.81 − 558.91 = 0.68 MeV. Electron capture to ⁶⁴Ni: Q = −0.782 + B(Ni) − B(Cu) = −0.782 + 561.75 − 558.91 = 2.06 MeV, and since that clears 2mec² = 1.022 MeV, β⁺ is open too with Q = 1.04 MeV.
- Measurement: Q(β⁻) = 0.579 MeV and Q(EC) = 1.675 MeV, and ⁶⁴Cu really does decay both ways, 61% by EC and β⁺ and 39% by β⁻. The formula gets the topology exactly right and each Q to a few hundred keV — the residual left when shells are ignored.
AnswerB = 561.75, 558.91 and 558.81 MeV for ⁶⁴Ni, ⁶⁴Cu and ⁶⁴Zn. ⁶⁴Cu is less bound than both neighbours, so it decays in both directions: β⁻ with Q = 0.68 MeV and EC with Q = 2.06 MeV.