The Binding Energy Curve

A fair measure of stability

The last lesson found binding energies of \(2.22\,\text{MeV}\) for the deuteron, \(28.3\,\text{MeV}\) for helium-4 and \(492\,\text{MeV}\) for iron-56. A big nucleus has a big binding energy simply because it has many nucleons. To compare nuclei fairly, divide by the number of nucleons.

The deuteron has \(B/A = 2.22/2 = 1.11\,\text{MeV}\): it is loosely held. Helium-4 has \(28.3/4 = 7.07\,\text{MeV}\) and iron-56 has \(492/56 = 8.79\,\text{MeV}\). The larger \(B/A\), the harder it is, on average, to pull a nucleon out. So a larger \(B/A\) means a more stable nucleus.

The curve

Work out \(B/A\) for every stable nucleus from its measured mass, and plot it against \(A\). The result is one of the most important graphs in physics, the binding energy curve of Fig 3.1.

Read the curve from left to right. It rises very steeply through the lightest nuclei, with a sharp spike at helium-4 and smaller ones at carbon-12 and oxygen-16. From about \(A = 30\) it flattens near \(8\,\text{MeV}\). It reaches a broad top of \(8.8\,\text{MeV}\) near iron-56 and nickel-62, the most tightly bound nuclei. Beyond them it falls slowly, to \(7.6\,\text{MeV}\) for uranium.

Why the curve has this shape

Three effects shape the curve, and each comes from the nuclear force or the electric force.

1. The flat middle. The nuclear force saturates: each nucleon is held only by its few neighbours. So each nucleon adds about the same binding, and \(B\) grows in proportion to \(A\). That keeps \(B/A\) nearly constant, about 8 MeV, from \(A = 30\) to \(A = 170\).
2. The low start. In a small nucleus, most nucleons sit on the surface. A surface nucleon has fewer neighbours to hold it, so light nuclei have a small \(B/A\).
3. The slow fall. Every proton repels every other proton, since the electric force has a long range. In a heavy nucleus this repulsion grows faster than the short-range binding, and \(B/A\) falls.

Which nuclei are stable

Not every pair of \(Z\) and \(N\) makes a stable nucleus. Plot the stable nuclei on a graph of \(N\) against \(Z\), and they lie in a narrow band, the line of stability of Fig 3.2.

For light nuclei the band follows \(N = Z\). Helium-4, carbon-12, oxygen-16 and calcium-40 each have equal numbers of protons and neutrons. For heavier nuclei the band bends upwards: iron-56 has 30 neutrons for 26 protons, tin-120 has 70 for 50, and lead-208 has 126 for 82. The ratio \(N/Z\) climbs from 1 to about 1.5.

The reason is the same competition that bends the binding energy curve. A proton adds nuclear binding but also repulsion with every other proton. A neutron adds nuclear binding and no repulsion at all. In a big nucleus the repulsion is large, so extra neutrons are needed to hold it together. They do not screen the protons from one another; they simply add more attraction.

The neutrons cannot be added without limit either. Too many neutrons also make a nucleus unstable, as the chapter on radioactivity explains. And beyond lead, \(Z = 82\), no amount of neutrons is enough. Every nucleus with \(Z\) above 82 is radioactive, although bismuth-209 lives so long that it was long thought stable.

Climbing towards the peak

The curve now answers the puzzle of the two fires. Energy is released whenever nucleons end up more tightly bound than they started. A tighter binding means a larger mass defect, so the products are lighter than the reactants. The mass lost comes out as energy.

Heavy nuclei sit on the falling right-hand side. Split one into two middle-sized pieces, and each nucleon moves up the curve. Uranium has \(B/A = 7.6\,\text{MeV}\) and nuclei near \(A = 118\) have about \(8.5\,\text{MeV}\). Each of the 235 nucleons gains about \(0.9\,\text{MeV}\). This is nuclear fission, and it releases about \(235 \times 0.9 \approx 210\,\text{MeV}\), the famous "200 MeV per fission".

Light nuclei sit on the steep left-hand side. Join two of them, and the product lies far higher. Two deuterons have \(4 \times 1.1 = 4.4\,\text{MeV}\) of binding in all, and helium-4 has \(4 \times 7.1 = 28.4\,\text{MeV}\). Joining them into helium-4 releases about \(24\,\text{MeV}\). This is nuclear fusion. The exact masses give \(23.8\,\text{MeV}\), the value in Fig 3.1.

So splitting and joining do not contradict each other. Both move nucleons towards the peak near iron, from opposite sides. Iron itself can only lose. Split iron-56 into two silicon-28 nuclei, at \(8.45\,\text{MeV}\), and the products are less bound. About \(56 \times (8.79 - 8.45) \approx 19\,\text{MeV}\) must be supplied. Iron is nuclear ash: it can be neither split nor fused for energy.

Nuclear energies dwarf chemical ones. Burning one carbon atom in oxygen gives about \(4\,\text{eV}\). One fission gives about \(200\,\text{MeV}\), some fifty million times more. The difference is the difference between an atom's electrons, bound by electron volts, and its nucleus, bound by millions of electron volts.