Every atom in your body heavier than hydrogen was cooked inside a star, and the currency the star paid in was nuclear binding energy. When four hydrogen nuclei fuse into helium in the Sun's core, the helium comes out lighter than the ingredients, and the missing mass is poured into space as sunlight. Run the same accountancy across the whole periodic table and one number governs everything — which nuclei give up energy when they fuse, which give it up when they split, and why the ash left at the heart of a dying star is iron.

You already know that a nucleus weighs less than its separate nucleons, and that this mass defect is the binding energy B = \Delta m\,c^2. This page answers the next question: can we predict B for any nucleus from first principles? Remarkably, a five-term formula built by picturing the nucleus as a charged drop of liquid — the semi-empirical mass formula (SEMF), or Bethe–Weizsäcker formula — gets it right to a fraction of a percent across hundreds of nuclei. We will build it term by term, and watch the famous binding-energy-per-nucleon curve fall straight out of it.

Binding energy and the mass defect, in one line

A nucleus {}^{A}_{Z}X holds Z protons and N = A - Z neutrons. Weigh the free constituents, weigh the assembled nucleus, and the difference is the mass defect:

B is the energy you must supply to tear the nucleus into free, infinitely-separated nucleons — equivalently, the energy released when they snap together. A big B means a well-glued nucleus. But raw B grows with size (more nucleons, more glue), so the honest measure of stability is the binding energy per nucleon, B/A — the average glue holding each nucleon in place. Everything below is a machine for predicting B, and hence B/A, from just A and Z.

The liquid-drop idea

Here is the leap. The strong nuclear force is short-ranged and attracts every nucleon to its immediate neighbours, and nuclear matter has a nearly constant density regardless of size — exactly like the molecules in a drop of incompressible liquid. So model the nucleus as a tiny charged droplet and ask what would set its energy. Four mechanical effects and one quantum effect suggest themselves, and each becomes a term in the SEMF:

• a volume gain: every nucleon is glued to its neighbours — bulk attraction;
• a surface penalty: nucleons at the skin have fewer neighbours — like surface tension;
• a Coulomb penalty: the protons all repel one another electrically;
• an asymmetry penalty: a quantum (Pauli) cost when the proton and neutron counts are unequal;
• a pairing nudge: nucleons like to couple up spin-to-spin.

The word semi-empirical is the honest label: the form of each term is argued from physics, but the five coefficients are fitted to measured masses. Let's build them one at a time.

Term 1 — Volume: the bulk reward +a_V A

The strong force saturates: each nucleon bonds only to the handful of nucleons it actually touches, not to the whole nucleus. So the total binding from the bulk is simply proportional to the number of nucleons — double the nucleons, double the bonds. That is a term +a_V A, and because it is the only term growing linearly with A it dominates: on its own it would give a constant B/A = a_V \approx 15.75\ \text{MeV} for every nucleus. Everything else is a correction that eats into this reward.

Term 2 — Surface: the skin penalty -a_S A^{2/3}

The volume term overcounts, because it assumed every nucleon has a full set of neighbours. A nucleon at the surface has neighbours only on the inward side, so it is under-bound — exactly the physics of surface tension, which is why raindrops are round. We must subtract an amount proportional to the surface area. A drop of A nucleons has radius R \propto A^{1/3} (volume \propto A), so its area goes as R^2 \propto A^{2/3}:

This term matters most for small nuclei, where the surface is a large fraction of the whole — it is the reason the very light nuclei sit low on the B/A curve. As A grows, A^{2/3}/A = A^{-1/3} shrinks, so the surface penalty per nucleon fades away, letting B/A climb.

Term 3 — Coulomb: proton repulsion -a_C\,\dfrac{Z(Z-1)}{A^{1/3}}

The strong force binds; the electrostatic force fights back. Every one of the Z protons repels every other proton, and there are \tfrac{1}{2}Z(Z-1) such pairs. Spread that charge through a uniform sphere of radius R \propto A^{1/3} and the electrostatic energy scales as Z(Z-1)/R \propto Z(Z-1)/A^{1/3}. Since repulsion reduces the binding, it enters with a minus sign:

This is the term that grows relentlessly for heavy nuclei — it climbs roughly as Z^2/A^{1/3} \sim A^{5/3}, faster than the linear volume reward. It is why stability demands ever more neutrons (which feel no Coulomb push) as Z rises, and ultimately why there is a heaviest possible nucleus at all.

Term 4 — Asymmetry: the Pauli cost of N \ne Z

Protons and neutrons are fermions, each filling its own ladder of quantum energy levels, two per rung. The most economical way to pack A nucleons is to keep the two ladders level — N = Z. Force a surplus of one species and the extra particles must be stacked onto higher rungs (the Pauli exclusion principle forbids doubling up), which costs binding energy. The cost turns out to grow as the square of the imbalance and to be diluted by size:

For light nuclei this pushes stability towards N = Z (carbon-12, oxygen-16). For heavy nuclei it is in a permanent tug-of-war with the Coulomb term, which wants extra neutrons — the compromise between the two is exactly what carves the curved "valley of stability".

Term 5 — Pairing: the even–odd nudge \delta(A,Z)

The last term is a small quantum bonus for pairing up. Two identical nucleons in the same orbital with opposite spins overlap especially well and bind a little extra. So a nucleus does best when both its proton count and its neutron count are even (all nucleons paired), worst when both are odd (two unpaired), and in between when one is odd:

This tiny term punches above its weight: it explains why even–even nuclei are far more abundant in nature, why there are only four stable odd–odd nuclei in the whole chart, and the sawtooth wobble superimposed on the smooth B/A curve.

The whole formula

Assemble the five pieces and you have the semi-empirical mass formula in its full glory. Read it as a story: a bulk reward, chipped at by the surface skin, by proton repulsion, by proton–neutron imbalance, and finally tweaked by pairing.

Worked example — iron-56, term by term

Let's grind the formula for the icon of nuclear stability, {}^{56}_{26}\text{Fe}: A = 56, Z = 26, N = 30. Both counts are even, so it is an even–even nucleus and pairing is positive. First the geometric factors: A^{1/3} = 56^{1/3} \approx 3.826 and A^{2/3} \approx 14.64.

• Volume: a_V A = 15.75 \times 56 = 882.0\ \text{MeV}.
• Surface: -a_S A^{2/3} = -17.8 \times 14.64 \approx -260.6\ \text{MeV}.
• Coulomb: -a_C\,\dfrac{Z(Z-1)}{A^{1/3}} = -0.711 \times \dfrac{26\times 25}{3.826} \approx -120.8\ \text{MeV}.
• Asymmetry: -a_A\,\dfrac{(A-2Z)^2}{A} = -23.7 \times \dfrac{(56-52)^2}{56} = -23.7 \times \dfrac{16}{56} \approx -6.8\ \text{MeV}.
• Pairing: +\dfrac{a_P}{\sqrt{A}} = +\dfrac{11.18}{\sqrt{56}} \approx +1.5\ \text{MeV}.

Add them up:

The measured value for iron-56 is B \approx 492\ \text{MeV}, B/A \approx 8.79\ \text{MeV} — a five-term back-of-envelope estimate landing within 1\% of reality. Notice the shape of the sum: the volume reward is huge, but surface and Coulomb between them claw back nearly 380\ \text{MeV}. The balance of those two competitors is the whole story of the curve we plot next.

The payoff: the B/A curve

Now feed the SEMF every mass number. For each A we pick the most stable proton number using Z \approx A / \big(2 + 0.0155\,A^{2/3}\big) (the value that maximises B, balancing Coulomb against asymmetry), compute B, and plot B/A. Out comes one of the most important curves in physics — with no experimental input beyond five fitted numbers.

Read the curve from left to right. It rises steeply from the light nuclei, because the surface penalty per nucleon (\propto A^{-1/3}) is dying away. It flattens into a broad peak near A \approx 56\text{–}62 at about 8.8\ \text{MeV per nucleon}. Then it declines gently, because the Coulomb term (\propto Z^2/A^{1/3}) is now growing faster than the volume reward. Try the slider: dial the Coulomb strength down and the peak flattens and slides to the right — proof that the summit's position is set by the surface-versus-Coulomb contest.

Why iron/nickel sits at the top — and what it means

The peak is a compromise between two opposing trends per nucleon:

• the surface cost per nucleon, a_S A^{-1/3}, keeps falling as A grows — so light nuclei are always improved by getting bigger;
• the Coulomb cost per nucleon, roughly \propto A^{2/3}, keeps rising — so heavy nuclei are always worsened by getting bigger.

Somewhere in the middle the two effects cross, and B/A is maximised: right around iron-56 and nickel-62. Because these nuclei sit at the summit, you cannot extract energy from them either way. Everything lighter can climb towards the peak by fusing (join two light nuclei, move up and to the right → energy out). Everything heavier can climb by fissioning (split a heavy nucleus, move up and to the left → energy out). Both arrows point at iron. That single fact is why stars burn light elements by fusion, why reactors burn heavy elements by fission, and why the core of a dying massive star ends as inert iron ash — there is simply no nuclear energy left to extract at the peak. (The quantitative energy released in each reaction is the subject of a sibling page; here we keep it qualitative.)