San José State University
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Why Are There So Many
More Neutrons than Protons
in Heavier Nuclei?

The stable isotopes of the heavier element have a great surplus of neutrons compared to the number of protons. For example the stable isotope of uranium, U_238, has 92 protons and 146 neutrons, nearly a (3/2) to 1 ratio. This may be a matter of that combination involves a lower energy than other combinations of the same number of nucleons.

The conventional explanation of the preponderance of neutrons in nuclei is that because protons repel each other through the electrostatic force they must be diluted by neutrons. The problem with that explanation is that it implies that the more neutrons the better whereas there is clearly a limit to the number of neutrons relative to the number of protons in a nucleus. There has to be a balance between the number of neutrons and the number of protons. That proper balance involves approximately 50 percent more neutrons than protons.

The conventional theory of nuclear structure maintains that all nucleons are attracted to each other through the so-called strong force that drops off with distance more rapidly than does the electrostatic force. This theory is nothing more than giving a name to what holds a nucleus together. An empirical analysis based on the binding energies of nuclides reveals that the conventional theory is wrong.

What primarily holds a nucleus together is the formation of spin pairs of nucleons; proton-proton, neutron-neutron and proton-neutron spin pairs. However this spin pairing is exclusive in the sense that a neutron can pair with only one other neutron and with one proton. The same goes for a proton.

This spin pairing leads to the formation of chains made up of modules of the form -n-p-p-n- or equivalently -p-n-n-p-. These chains of modules can close forming rings. Such rings constitute nuclear shells. The smallest such ring is the alpha particle consisting of two protons and two neutrons. The above theory can be appropriately called the alpha module rings model of nuclear structure.

The Interactive Force Between Nucleons

In addition to the force associated with spin pairing there is a nonexclusive force of interaction between nucleons. The analysis of the binding energies of nuclides implies that nucleons possess a nucleonic charge that is different for neutrons and protons.

Let the nucleonic charge of a proton be taken to 1 and that of a neutron be denoted as q. Then the force due a neutron-neutron interaction is proportional to q². The forces between neutron-proton and proton-proton interactions are proportional to q·1 and 1·1, respectively. If q is a negative value the force between a neutron and a proton is an attraction, whereas the forces for the other two interactions are necessarily repulsions. The energies involved for these interactions are in the same proportions: q², q and 1. Ignoring the electrostatic repulsion between protons for the moment, the energy for a nucleus of p protons and n neutrons is proportional to the sum of the products of the numbers of the various interactions and the proportionality factors; i.e.,

E = ½(p²−p) + pnq + ½(n²−n)q²

Consider the minimization of E subject to the constraint that (p+n)=a. This can be achieved using the Lagrangian multiplier method. This involves minimizing

½(p²−p) − pnq + ½(n²−n)q² +λ(a − (p+n))

The first order equations for a constrained minimum are therefore:

(p−½) + nq − λ = 0
(n−½)q² + pq − λ = 0

These can reduced to

(n−½)q² + pq = (p−½) + nq
and further to
(n−½)q² − nq = (p−½) − pq
n(q²−q) −½)q² = p(1−q) −½
and still further to
n(q²−q) = p(1−q) − ½(1 − q²)

Dividing through by (q−1) gives

nq = −p + ½(1+;q)
and finally
n = (−1/q)p + ½(1/q+1)

For q=−2/3 this evaluates to

n = (3/2)p − (1/4)

The minimum energy combination of p and n would tend to be the most stable compositional arrangement. For p=92 the formula gives n=229.75 rather than 238, but it is the right order of magnitude.

The Effect of the Electrostatic
Repulsion between Protons

But as noted the above ignores the electrostatic repulsion between protons. That can be represented by making the energy of the interaction between protons as (1+d) rather that 1, where d represents the relative magnitude of the electrostatic force compared to that of the nucleonic force at the distances involved in the nuclear interactions. This makes the quantity to be minimized equal to

½(p²−p)(1+d) − pnq + ½(n²−n)q² +λ(a − (p+n))

The first order equations for a constrained minimum are then:

(p−½)(1+d) + nq − λ = 0
(n−½)q² + pq − λ = 0

These can reduced to

(n−½)q² + pq = (p−½)(1+d) + nq
and further to
(n−½)q² − nq = (p−½)(1+d) − pq
n(q²−q) −½q² = p((1+d)−q) −½(1+d)
and still further to
n(q²−q) = p((1+d)−q) − ½((1+d) − q²)
and finally to
n(q²−q) = p(1−q +d) − ½(1− q² + d)

Dividing through by (1−q) gives

−nq = p(1+d/(1-q)) −½(1+q + d/(1-q))
and finally
n = (−1/q)p(1+d/(1-q)) + ½[(1/q+1) + d/(q(1-q) ]

For q=−2/3 and d=1/10 this evaluates to

n = (3/2)p(1+3/50) −(1/2) [(1/2) + (9/100)]

For p=92 the formula gives (p+n)=238.11 rather than 238, a deviation of about 0.05 of 1%. This 0.1 is a plausible value for d, the ratio of the electrostatic repulsion between protons to the nucleonic repulsion.

A regression of n on p for the nuclides with p≥26 (iron) gives

n = 1.57054p − 10.83610

The coefficient of determination (R²) for this equation is 0.92345 and the t-ratio for the coefficient of p is 172.4. The ratio of that coefficient to 3/2 is 1.0470273. The previous analysis says that this ratio should be (1+3d/5). This means that d, the ratio of the electrostatic repulsion to the nucleonic force repulsion, should be 0.078.


The Alpha Module Rings model of nuclear structure explains why there must be a preponderance of neutrons in heavier nuclides. It even gives an explanation for quantitative magnitude there being roughly 50 percent more neutrons than protons in the heavier nuclides.

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