One of the standard graphs of nuclear physics is a display of the binding energy per nucleon as a function of the number of nucleons in the nuclide.

Beyond the datum for iron the relationship is downward sloping at an almost constant rate. This suggests that the heavier nuclides, such as the isotopes of uranium, would be less stable than the isotopes of iron.

Here is a derivation of the downward sloping of that relation. Let n and p be the numbers of neutrons and protons in a nuclide. A nucleus is held together by the spin pairing of nucleons and the interactions between nucleons due to another type of force.

Spin Pairing of Nucleons

The spin pairing is exclusive in the sense that a neutron can only pair with one other neutron and with one proton. And likewise for a proton. The number of neutron-neutron spin pairs in a nuclide is [n/2]. The number of proton-proton spin pairs is [p/2] and the number of neutron-proton is min(n, p). Therefore the total number of spin pairs is

[n/2] +[p/2] + min(n, p)

The Other Force Between Nucleons

This force is based upon nucleons having a nucleonic charge. It is nonexclusive and is a function of the separation distance of the nucleons. Conventional theory inappropriately combines this force with the spin pairing and calls it the nuclear strong force. A better name for this force is the nucleonic force, the force between nucleons as nucleons. The action associated with spin pairing is not a force in the same sense as gravitation and the electrostatic force are. They have a force field that pervades space; spin pairing does not. They have a variable charge; spin pairing does not.

Let the nucleonic charge of a proton is taken to be unity and the nucleonic charge of a neutron be represented as q. The quantity q is estimated elsewhere to be −2/3. But for now let the analysis continue to be general.

The binding energy for neutron-neutron interactions is proportional to q² .and for neutron-proton interactions proportional to q. If the electrostatic repulsion between protons is left out of the analysis the binding energy for proton-proton interactions would be proportional to 1. The electrostatic interaction will be treated later. The number of neutron-neutron interactions is ½n(n-1). For proton-proton interactions it is ½p(p-1). For neutron-proton interactions it is np. Thus the total binding energy due to nucleonic interactions is proportional to

(½n(n-1))q² + (½p(p-1)) + (np)q

For q being negative this says that like nucleons repulse each othe and unlike ones attract. Here negative values of the above quantity correspond to higher values of binding energy.

It was found elsewhere that the above binding energy is minimized if n is equal to (1/|q|)p. This corresonds to the most stable isotope for a given value of p. From this point let q denote the absolute value of the nucleonic charge ratio. Let n be replaced by (p/q) in the above expression to yield

BE = −[(½(p/q)((p/q)-1))q² + (½p(p-1)) + ((p/q)p)q]
which reduces to
BE = −[(½p((p/q)-1))q + (½p(p-1)) + p²]
and further to
BE = (1 + (1/q))p − 2p²

When n=(p/q) the number of spin pairs is approximately equal to

(p/q)/2 + p/2 + p = (3/2 + 1/(2q))p

Let the binding energy for spin pairing relative to that due to nucleonic interaction be expressed in terms of a constant β so the total binding energy of the stable isotopes for p is given by

BE = β((3/2 + 1/(2q))p + (1 + (1/q))p − 2p²

Now let us compute the binding energy per nucleon, BE/(n+p); i.e.,

BE/(n+p) = [β((3/2 + 1/(2q))p + (1 + (1/q))p − 2p² ]/(1+3/2)p
= (2/5)[β((3/2 + 1/(2q)) + (1 + (1/q)) − 2p ]
= (2/5)[β((3/2 + 1/(q)) + (1 + (1/q)) − 2p ]
= (2/5)[(3β/2 − 1) + (β − 1)/q − 2p]

The significant result is that this form of the theory predicts that BE/(n+p) decreases with p. Here again is the graph of the data for BE/(n+p)

The analysis captures the physical reality for large nuclides but not that of small nuclides. This is in part due to the fact that the n=p/|q| rule fits well for large nuclides but not so well for smaller ones.