There is abundant evidence that within stable nuclei that neutron-neutron and proton-proton pairs exist, but there is no evidence of the existence of such pairs outside of nuclei. However free neutron-proton pairs do exist and are called deuterons, the nuclei of heavy hydrogen. This material is an attempt explain these phenomena. It makes use of results found through the analysis of the binding energies of nuclei. Those results are:

• Nuclei are held together primarily by the formation of spin pairs, neutron-neutron, proton-proton and neutron-proton.
• Such pairing is exclusive in the sense that a neutron can form a spin pair only with one other neutron and with one proton. The same applies for a proton.
• The formation of such spin pairs results in chains of modules of the form -n-p-p-n- (or equivalently, -p-n-n-p-).
• These chains close forming rings which rotate and spin creating nuclear shells.
• In addition to the exclusive spin pairing there is a nonexclusive nucleonic force such that like nucleons are repelled from each other and unlike nucleons are attracted.
• This nucleonic force can be explained by neutrons and protons having a nucleonic charge of opposite signs.
• If the nucleonic charge of a proton is taken as +1 then the nucleonic charge of a neutron is −2/3.
• For more on the Alpha Module Model of nuclear structure see Nucleus.

These results are in contrast to the conventional theory of nuclear forces in which there is a nuclear strong force which at small separation distances is more powerful than the electrostatic repulsion between protons. This hypothetical strong force is supposedly an equal attraction between all nucleons. The conventional strong force apparently tries to combine the effects of spin pairing and the nucleonic force into one force.

Nucleonic Pairing

The binding energy for the spin pairing of a neutron with neutron is on the order of 3 million electron volts (MeV) whereas the binding energy due to the repulsive nucleonic force between two neutrons is only about 1/3 MeV. The same applies for protons. The binding energy for a neutron-proton pair is likewise about 3 MeV but the nucleonic force between a neutron and a proton is an attraction creating a binding energy of about 1/3 MeV. From these figures it would seem that all types of free nucleonic pairs should be possible.

But consider the energy relationships between a neutron-neutron pair and a neutron-proton pair. The energies in the spin pairing would be the same. The nucleonic repulsion between two neutrons involves a positive amount of potential energy, but the attraction between a neutron and a proton constitutes a negative amount of potential energy. Thus the conversion of a neutron in a neutron-neutron pair to a proton would release energy. In a free neutron-neutron pair this conversion would take place. Within a nucleus a neutron in neutron-neutron pair could already be part of a neutron-proton pair so its conversion to a proton would create a disallowed -p-n-p- linkage so the conversion can't happen.

If the neutron-neutron pair in a nucleus is not linked to a proton then the conversion might tke place. All of the beta electron emitters neutrons that are not linked to a proton.

The same prevails for a proton-proton pair. Because of the repulsion between protons due both to the nucleonic force and the electrostatic force the conversion of a proton to a neutron would release energy. Such energy is required to convert a proton to a neutron because a neutron has a larger mass than a proton. In a free proton-proton pair such a conversion would take place. But within a nucleus such a conversion would create a disallowed -n-p-n-linkage and hence be prohibited.

Thus free neutron-neutron and free proton-proton cannot exist because upon their creation they would immediately convert to a neutron-proton pair. Within a stable nucleus that conversion cannot take place because it would create a violation the exclusity of the spin pair formation.

Under the conventional theory of the strong force attraction between all nucleons there is no energy advantage of a free neutron-proton pair over a neutron-neutron pair. The only energy advantage of a neutron-proton pair over that of a proton-proton pair is the disappearance of the electrostatic repulsion between the two protons.

Quantitative Relationships

Although the above accomplishes to the purpose of this webpage more can be said. If the force between two particles is proportional to the product of their charges then the potential energy associated with their combining into a system is also proportional to the product of their charges. If the nucleonic charge of a proton is taken as +1 and that of a neutron is q. The potntial energy in a neutron-proton due the nucleonic force is proportional to 1·q=q and in a neutron-neutron pair it is proportional to q·q=q². The potential energy in a proton-proton pair is a bit more complicated because of the electrostatic repulsion. Let d represent the force between two protons due the electrostatic repulsion relative to the nucleonic force. Then potential energy in a proton-proton pair is proportional to (1+d).

The energy released when one proton in a proton-proton pairs converts into a neutron is then (1+d−q). The energy released when a neutron in a neutron-neutron pair converts into a proton is proportional to q²−q. There is also additional energy released because the mass of the neutron is greater than the combined masses of a proton and an electron.

The best estimates of q and d are −2/3 and +1/5, respectively. This means the energy released when a proton in a proton-proton pair converts to a neuton is proportional to 1.87. Some of this goes to cover the mass difference between a neutron and a proton. The energy released by the conversion of a neutron into a proton and an electron is proportional to 10/9=1.11.

Conclusion

The results of the analysis of nuclear binding energy provide a cogent explanation of why free neutron-neutron or free proton-proton pairs do not exist but free neutron-proton pairs do exist and yet all such nucleon pairs do exist within stable nuclei. The conventional theory of the nuclear strong force provides no such explanations.