The Binding Energy per Nucleon Curve

Introduction

Total binding energy grows with the size of a nucleus, so comparing ⁴He with ²³⁸U directly says little. Dividing by the number of nucleons gives the average binding energy per nucleon, B/A, a measure of how tightly each nucleon is held. Plotted against mass number, B/A traces one of the most important graphs in science. Its single broad peak near iron explains why stars shine by fusion, why reactors run on fission, and why iron is so abundant in the universe.

Core explanation

Shape of the curve. B/A rises steeply from 1.11 MeV for ²H through 2.57 MeV for ³He and 2.83 MeV for ³H to 7.07 MeV for ⁴He. After a few irregularities among the light nuclei it climbs more gently — 7.68 MeV for ¹²C, 7.98 MeV for ¹⁶O, 8.55 MeV for ⁴⁰Ca — reaching a broad maximum of about 8.8 MeV for A ≈ 56–62. Beyond the peak it falls slowly, to about 8.0 MeV at A = 150 and 7.57 MeV for ²³⁸U.

The top of the curve. The three most tightly bound nuclides per nucleon are:

Nuclide B/A / MeV --- --- ⁶²Ni 8.795 ⁵⁸Fe 8.792 ⁵⁶Fe 8.790

Nickel-62 has the highest binding energy per nucleon. Iron-56, however, has the lowest mass per nucleon, because it has a slightly larger proportion of protons, and a hydrogen atom is lighter than a neutron by 0.00084 u. Both statements appear in textbooks; they answer slightly different questions. The differences are tiny — a few keV per nucleon — and for most purposes the region around iron and nickel is simply "the peak".

Why the curve rises. The strong force is short-range and saturates: a nucleon attracts only its immediate neighbours. In a small nucleus a large fraction of nucleons sit on the surface with fewer neighbours, so the average binding is low. As A increases, the surface fraction falls roughly as A^(−1/3) and B/A rises towards the value for a nucleon fully surrounded.

Why the curve falls. Every proton repels every other proton through the long-range Coulomb force. The total repulsion grows approximately as Z² ÷ A^(1/3), faster than the strong-force binding, which grows as A. Beyond about A = 60 the added repulsion outweighs the reduced surface effect, and B/A declines. The extra neutrons that heavy nuclei need to limit this repulsion also cost energy, through the asymmetry effect described on the page about the mass formula.

Energy release. A nuclear change releases energy if the products have greater total binding energy than the reactants. Because B/A rises towards the peak from both sides, two routes release energy: joining light nuclei (fusion) and splitting heavy ones (fission). Nothing can release energy by fusing beyond the peak or splitting below it.

Special features. ⁴He sits well above its neighbours, which is why alpha particles are emitted as a unit and why ⁸Be, which would split into two ⁴He nuclei, is unbound. Nuclei built from alpha-like units, such as ¹²C and ¹⁶O, also sit slightly above the smooth trend.

Step-by-step reasoning

To estimate the energy released in a nuclear change using the curve:

1. Read B/A for each reactant and product from the curve or a table. 2. Multiply each by its mass number to obtain total binding energies. 3. Sum the binding energies of the products and of the reactants. 4. Energy released ≈ total B (products) − total B (reactants); a positive result means energy is released.

Visual explanation

Sketch B/A (vertical, 0–9 MeV) against A (horizontal, 0–250). Draw a sharp rise from the origin with a spike at ⁴He, a rounded summit near A = 60, and a long gentle slope down to uranium. Add a right-pointing arrow on the left labelled "fusion" and a left-pointing arrow on the right labelled "fission", both climbing towards the summit.

Real-world analogy

Think of the curve as a hill with its summit at iron. A ball on either side rolls towards the summit when you turn the picture upside down: in the energy view, the peak of B/A is the bottom of an energy valley. Light nuclei roll in from one side by fusing; heavy nuclei roll in from the other side by splitting.

Real-world example

Massive stars fuse progressively heavier nuclei — hydrogen to helium, helium to carbon and oxygen, and finally silicon to nickel-56 and neighbouring nuclei — releasing energy at each stage. Once the core is iron-group material, further fusion absorbs energy, the core collapses and a supernova follows. Radioactive ⁵⁶Ni produced in the explosion decays through ⁵⁶Co to ⁵⁶Fe, one reason iron is so abundant.

Why?

Why does fission of uranium release about 200 MeV? The fragments, with A around 90–145, have B/A of roughly 8.4–8.6 MeV compared with 7.6 MeV for uranium. Gaining about 0.8–0.9 MeV for each of roughly 236 nucleons gives close to 200 MeV per fission.

Common misconception

"Iron-56 has the highest binding energy of all nuclei." Iron-56 has a high binding energy per nucleon, but its total binding energy (about 492 MeV) is far smaller than that of uranium-238 (about 1802 MeV). Total and per-nucleon values answer different questions.

Worked example

Question: Estimate the energy released when ²³⁵U plus a neutron splits into ¹⁴¹Ba, ⁹²Kr and three neutrons. Take B/A = 7.59 MeV for ²³⁵U, 8.33 MeV for ¹⁴¹Ba and 8.51 MeV for ⁹²Kr.

Reasoning: B(reactants) = 235 × 7.59 = 1784 MeV (free neutrons contribute nothing). B(products) = 141 × 8.33 + 92 × 8.51 = 1174.5 + 782.9 = 1957 MeV. Energy released ≈ 1957 − 1784 = 173 MeV.

Answer: About 170 MeV, consistent with the roughly 200 MeV per fission observed once the later decays of the fragments are included.

Quick check

1. Why can fusing two iron-56 nuclei never be used as an energy source? Answer: Iron sits at the peak of the B/A curve, so any heavier product has lower binding energy per nucleon and energy must be supplied.

Exam focus

Be able to sketch the curve with correct values at ⁴He, the peak and ²³⁸U, and to explain its rise (surface effect) and fall (Coulomb repulsion). Calculations use total binding energies, never B/A values subtracted directly.

Advanced insight

The irregularities along the curve are signatures of nuclear structure. Nuclei with even numbers of both protons and neutrons are more tightly bound than their odd neighbours, and nuclei with magic numbers of nucleons stand above the smooth trend. A smooth liquid-drop model reproduces the overall shape, and the deviations from it are the evidence for the nuclear shell model.

Summary

B/A rises steeply for light nuclei, peaks at about 8.8 MeV around A = 56–62 and falls gently to 7.6 MeV for uranium. The rise reflects the shrinking surface fraction; the fall reflects growing Coulomb repulsion. ⁶²Ni has the highest B/A and ⁵⁶Fe the lowest mass per nucleon. Energy is released by fusion below the peak and fission above it.

Practice questions

1. State the approximate binding energy per nucleon of ⁴He, the peak nuclides and ²³⁸U. Answer: About 7.1 MeV for ⁴He, 8.8 MeV at the peak and 7.6 MeV for ²³⁸U. 2. Explain why B/A increases from ²H to about A = 60. Answer: The strong force saturates, and as the nucleus grows a smaller fraction of nucleons lie on the surface with fewer neighbours, so the average binding increases. 3. Using B/A = 1.11 MeV for ²H, 2.83 MeV for ³H and 7.07 MeV for ⁴He, estimate the energy released in ²H + ³H → ⁴He + n. Answer: B(reactants) = 2.22 + 8.49 = 10.71 MeV; B(products) = 28.28 MeV; energy released ≈ 17.6 MeV. 4. Why do both fission and fusion release energy? Answer: Both move nucleons towards the peak of the B/A curve, so the products are more tightly bound in total than the reactants.