﻿ Explanation of Why Neutrons in a Nucleus are Stable but Free Neutrons are not Stable
San José State University

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Thayer Watkins
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Explanation of Why Neutrons
in a Nucleus are Stable but
Free Neutrons are not Stable

When the neutron was discovered in 1933 it was widely welcomed as giving an explanation for the discrepancy between the structure of the periodic table and the ordering of the elements based upon their atomic weight. The neutron was hailed as a fundamental particle of the universe on par with the proton and electron. Imagine the consternation of the physicists who found that free neutrons are unstable and decay into pairs of protons and electrons with a half-life of 18 minutes. Yet neutrons within nuclei appeared to be perfectly stable, or perhaps not quite perfectly stable because some nuclides are beta emitters.

The stability of neutrons within nuclei is explained by the energies which would be involved in the decay of a neutron which is spin paired with a proton. Although the correct explanation is not in terms of electrostatic force it is worthwhile to look at this force for a deuteron. A deuteron is a proton and neutron spinning about their center of mass. The proton and neutron are held together by the spin pairing force and the nuclear strong force. The potential energy of the deuteron is negative and it requires an input of energy to split a deuteron apart. But if the neutron of a deuteron decayed into a proton the deuteron would become a pair of protons a small distance apart. The potential energy of that arrangement can be easily computed. It is based upon the formula

#### V = ke²/s

where k is a constant, e is the charge of a proton and s is separation distance of the nucleon centers. Thus

#### V = 9×109(1.6×10-19)²/(2.25×10-15) V = 1.024×10-13 joules V = (1.024×10-13)(6.24×1018) electron volts V = 6.4×105 eV V = 0.64 MeV (million electron volts)

Thus in order to shift the nucleons of a deuteron into a proton spin pair would require the input of 0.64 MeV of energy. There is no place for this energy to come from so a neutron paired with a proton would be stable. However this is not the answer to the matter of the stability of neutrons within nuclei. The reason the above is not the answer to neutron stability is because the electron was left out of the computation. The creation of a proton-electron pair at that separation would create negative potential energy of −0.64 MeV. So the decay of a neutron is not prevented by the electrostatic force.

It is important to recognize that although neutron is electrically neutral it has a charge distribution. There is positive electrical charge in the interior of the neutron which is counter balanced by a negative charge toward the surface. The proton's charge distribution is all positive. The binding energies of the three types of spin pairing are approximately equal the shift from a neutron-proton spin pairing to a proton-proton spin pairing would not prevent the decay of a neutron.

A possible alternative is the so-called nuclear strong force. According to the conventional theory of nuclear structure the interaction force between all combinations of neutrons and protons are attractions of the same magnitudes. Thus, according to the conventional theory, there is nothing to prevent the decay of a neutron within a nucleus.

There is an alternative to the conventional called the Alpha Module Model of nuclear structure. According that theory protons and neutrons have a strong force charge. If the strong force charge of a proton is designated as +1 then that of a neutron is −2/3. Thus unlike nucleons are attracted to each other and like one are repelled. A nucleus is held together by the spin pairing of the nucleons and the attraction of neutrons and protons for each other. Therefore if the neutron in a deuteron were to decay the potential energy due to the strong force would switch from one based upon an attraction proportional to 2/3 to one based upon a repulsion proportional to 1. This is not counter balanced by an effect on the electron produced by the decay. Electrons are not affected by the nuclear strong force. Without a source for this additional energy the neutron cannot decay. Hence the neutron in a deuteron is stable.

In a more complex nucleus such as an alpha particle the same would apply but with the additional complicating factor that the neutrons are spin paired with each other as well as with protons. Spin pairing is exclusive in the sense that a nucleon can form a spin pair with one and only one nucleon of the same type and one and only nucleon of the opposite type. Thus in an alpha particle if a neutron decays a violation of the exclusivity rule is created. Same applies in more complex nuclei because nucleons, wherever possible, are formed into modules of the form -n-p-p-n-, or equivalently -p-n-n-p-.

The decay of free neutrons is energy feasible because the mass of a neutron is greater than the sum of the masses of the proton and electron it decays into. But where a neutron is paired with a proton its decay is not energy feasible and thus such neutrons within nuclei are stable. This is according to the Alpha Module Model of nuclear structure. The conventional model of nuclear structure has nothing to say on this issue.