Nuclear Binding Energy Curve

The energy holding a nucleus together, per nucleon

A nucleus weighs slightly less than the sum of its separate protons and neutrons — the missing mass, converted via E = mc², is the binding energy B, the energy you'd need to supply to pull the nucleus apart into free nucleons. Dividing by the mass number A gives binding energy per nucleon, B/A, the single most informative number in nuclear physics: plot it against A for every known nuclide and you get a curve that explains, in one picture, both fission and fusion.

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The semi-empirical mass formula

The Bethe-Weizsäcker semi-empirical mass formula (SEMF, 1935) models the nucleus as a charged liquid drop and reproduces the B/A curve remarkably well with five physically motivated terms:

B(A,Z) = a_V·A − a_S·A^(2/3) − a_C·Z(Z−1)/A^(1/3) − a_A·(A−2Z)²/A + δ(A,Z)

a_V·A                 volume term: each nucleon binds roughly equally (short-range force)
−a_S·A^(2/3)           surface term: surface nucleons have fewer neighbours, less binding
−a_C·Z(Z−1)/A^(1/3)    Coulomb term: protons repel each other electrostatically
−a_A·(A−2Z)²/A       asymmetry term: penalises N ≠ Z (Pauli exclusion)
δ(A,Z)                pairing term: bonus for even-even nuclei, penalty for odd-odd

Each coefficient (a_V ≈ 15.8 MeV, a_S ≈ 18.3 MeV, a_C ≈ 0.71 MeV, a_A ≈ 23.2 MeV, in one common fit) has a direct physical story, which is exactly why the simulation lets you drag each one independently and watch the predicted curve deform — raise a_C and the whole heavy end of the curve droops, because Coulomb repulsion, growing as Z², is what caps how large a stable nucleus can get.

Why the curve peaks at iron

The volume term grows binding energy roughly linearly with A, but the surface term (a penalty, since surface nucleons are under-bonded) and especially the Coulomb term (a penalty growing with Z²/A^(1/3)) both eat into it as A grows. B/A rises steeply for light nuclei, where surface effects dominate the penalty side, peaks around A ≈ 56 (iron and nickel, the most tightly bound nuclides that exist), and then declines slowly for heavier nuclei as Coulomb repulsion between an ever-growing number of protons wins out. That single hump is the whole story of where nuclear energy comes from.

Fusion and fission are both just climbing toward the peak

Any nuclear reaction that moves a nucleus toward the iron peak on the B/A curve releases energy, because it converts mass into binding energy that gets carried away as kinetic energy of the products. Light nuclei below the peak release energy by fusing into something closer to iron (hydrogen fusing to helium in stars, deuterium-tritium fusion in a reactor); heavy nuclei above the peak release energy by fissioning into two medium-mass fragments closer to the peak (uranium or plutonium splitting in a reactor or bomb). Iron and nickel sit at the summit and release energy from neither direction, which is exactly why iron is the endpoint of fusion inside a massive star's core and the point at which stellar fusion can no longer support the star against collapse.

Frequently asked questions

Why does binding energy per nucleon, not total binding energy, determine stability?

Total binding energy just grows with the size of the nucleus and doesn't tell you how tightly held each nucleon is. Binding energy per nucleon normalises for size, so comparing B/A across nuclides directly shows which arrangement of nucleons is more energetically favourable, and therefore which direction a nuclear reaction releases energy.

Why does the Coulomb term have Z(Z−1) instead of just Z²?

It's counting the number of distinct proton-proton pairs that repel each other, and a proton doesn't repel itself, so Z protons form Z(Z−1)/2 pairs, not Z²/2. The formula uses Z(Z−1) as the standard convention; the effect is negligible for large Z but keeps the formula exact for very light nuclei.

Why is iron the peak instead of some other element?

Iron-56 and nickel-62 sit at the point where the volume term's near-linear growth in binding energy is exactly balanced against the combined penalties from the surface and Coulomb terms. Below that point, adding nucleons (fusion) still nets a binding-energy gain; above it, the Coulomb repulsion penalty from adding more protons outweighs the volume gain, so splitting (fission) nets a gain instead.

Try it live

Everything above runs in your browser — open Nuclear Binding Energy Curve and change the parameters while it is running. Nothing is installed, nothing is uploaded, the whole model lives in one tab.

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The Nuclear Binding Energy Curve: Why Fusion and Fission Both Release Energy