﻿ The Maximum, the Minimum and the Optimum Number of Neutrons for each Element
San José State University

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The Maximum, the Minimum
and the Optimum Number of
Neutrons for each Element

Nuclei are held together by the spin pairing of their nucleons (protons and neutrons). Such spin pairing is exclusive in the sense that one neutron can spin pair with one other neutron and one proton, no more. The same applies to a proton.

There is another force involved due to the interactions of nucleons. It is nonexclusive, but it is such that like nucleons repel each other and unlike ones attract. It should be called the interactive nucleonic force, the force between nucleons. It is based upon protons and neutrons having nucleonic charges of different magnitudes.. The conventional theory of the so-called nuclear strong force conflates spin pairing and the interactive nucleonic force and is just flat wrong. It only explains the simple fact that there are stable nuclei containing multiple positive charges. When it is used to try to statistically explain the binding energies of nuclei all of the regression coefficients except one have the wrong sign or relative magnitude. See Test of Conventional Theory.

## Objective

This material is an investigation of the minimum energy combinations of p and n. Such combinations might tend to be the most stable compositional arrangements of protons and neutrons.

## Optimization

The effect of the interactive nucleonic force on the binding energy is proportional to the number of the different types of interactions, proton-proton, proton-neutron and neutron-neutron. Let p and n be the numbers of protons and neutrons in a nucleus, respectively. The numbers of the proton-proton, neutron-neutron and proton-neutron interactions are ½p(p-1), ½n(n-1) and np, respectively. The amount of binding energy involved in the different types of interactions differs because the nucleonic charges of the proton and neutron differ. Empirical analysis indicates that the nucleonic charge of a neutron differs in sign and is smaller in magnitude from that of a proton. If the nucleonic charge of a proton is taken to be 1 and the nucleonic charge of a neutron is denoted as q then the effects of nucleonic charge on binding energies of the proton-proton, proton-neutron and neutron-neutron interactions are in the ratios of 1, q and q².

Thus the energy EI due to the nucleonic interactions, ignoring the electrostatic repulsion between protons, can be expressed as

#### EI = b[½p(p-1) + npq + ½n(n-1)q²]

where b is a constant.

Consider the minimization of EI/b 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 be reduced to

#### (n−½)q² + pq = (p−½) + nq and further to (n−½)q² − nq = (p−½) − pq and 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)

This relationship shows promise because in heavier nuclides there are about 50 percent more neutrons than protons.

But proton-proton interactions involve their electrostatic repulsion. Let d be the ratio of the effect of the electrostatic repulsion on energy. In effect, the nucleonic charge of a proton is (1+d) rather than 1. The ratios of the binding energies due to proton-proton, proton-neutron and neutron-neutron are then (1+d)², q and q² respective ly. In effect the charge of a proton in proton-proton intersctions is (1+d) rather than 1.

Thus the energy EI due to the nucleonic interactions can be expressed as

#### EI = b[½p(p-1)(1+d)² + npq + ½n(n-1)q²]

where b is a constant.

Consider EI/b. This makes the quantity to be minimized equal to

#### ½(p²−p)(1+d)² − pnq + ½(n²−n)q² +λ(a − (p+n))

The first order equations for a constrained minimum are then:

#### ( p−½)(1+d)² + nq − λ = 0 (n−½)q² + pq − λ = 0

These can be reduced to

#### (n−½)q² + pq = (p−½)(1+d)² + nq and (n−½)q² −nq = (p−½)(1+d)² −pq n(q²−q) −½q² = p((1+d)² −q) −½)(1+d)² nq(q−1) = p((1−q+2d +d² ) + ½(q²−(1+d)²)

After dividing through by q(1−q)

#### n = (−1/q)p(1+(2d +d²)/(1−q) − ½(1/q + 1 + ((2d+ d²) /q(1-q))

For q=−2/3 and d=0.04 the above formula evaluates to

#### n = 1.57344p −0.57344

This value of n is the optimal number of neutrons for a given number of protons. The value of (p+n) is a prediction of the most stable isotope for an element given by p.

Here is a plot of the optimal n as a function of p, along with the maximum n and minium n as a function of p.

This value of n is too close to the maximum to be a reasonable estimate of the most stable isotope for a given value of p.

## The Inclusion of Spin Pairing

To be physically relevant the energy to be minimized must include that due to the spin pairing of nucleons. The numbers of proton-proton, neutron-proton and neutron-neutron spin pairs are equal to floor(p/2), min(p, n), and floor(n/2), respectively.

Let the binding energy due spin pairing be denoted as S where

#### S = cPP(p%2) + cNPmin(p, n) + cNN(n%2)

and where (p%2) and (n%2) denote the number of pairs in p and q, respectively. The optimal values of p and n are the values that minimize

#### E/b = S/b + ½(p²−p)(1+d) − pnq + ½(n²−n)q² +λ(a − (p+n))

S is a discontinuous function of p and n. It can be approximated by (p%2)≅p/2 and (n%2)≅n/2. Let this approximation of S be denoted as S*. The the energy based upon S* is then E*.

The first order equations for a constrained minimum of E* are then:

#### (1/b)(∂S*/∂p) + (p−½)(1+d) + nq − λ = 0 (1/b)(∂S*/∂n) + (n−½)q² + pq − λ = 0

These can be reduced to

#### (n−½)q² + pq + (1/b)(∂S*/∂n) = (p−½)(1+d) + nq + (1/b)(∂S*/∂p) and finally to n = (−1/q)p(1+d) + ½(1/q + 1 + d/q(1-q)) + [(1/(b(1−q))( (∂S*/∂p) − (∂S*/∂n))]

And now for the evaluation of the partial derivatives of S*.

When n>p

## The Testing of the Model

For the 2929 nuclides the following variables were computed which represent the formation of substructures.

• The number of alpha modules
• The number of proton-proton spin pairs not included in an alpha module
• The number of neutron-proton spin pairs not included in an alpha module
• The number of neutron-neutron spin pairs not included in an alpha module

To represent the interactions between nucleons the following variables were computed.

• The interactions among the p protons: ½p(p-1)
• The interactions among the p protons and n neutrons: np
• The interactions among the n neutrons: ½n(n-1)

The effect of the formation of a substructure on binding energy is not a constant amount applying under all conditions. For x formations it is in the nature of (a − b/x) where a and b are coefficients. When this is multiplied by x the result is (ax − b). This means that the regression equation for binding energy should include a constant that represents the cumulative total of the b coefficients for all of the substructures.

Here are the regression equation coefficients and their t-ratios (the ratios of the coefficients to their standard deviations).

The Results of Regression Analysis
VariableCoefficient
(MeV)
t-Ratio
Number of
Alpha Modules
42.64120923.0
Number of
Proton-Proton
Spin Pairs
13.8423452.0
Number of
Neutron-Proton
Spin Pairs
12.77668165.5
Number of
Neutron-Neutron
Spin Pairs
13.6987565.3
Proton-Proton
Interactions
−0.58936−113.8
Neutron-Proton
Interactions
0.3183195.8
Neutron-Neutron
Interactions
−0.21367−96.6
Constant−49.37556−112.7
0.99988

It should be noted that the nuclides of a single proton and a single neutron were left out of the analysis. Also that there is a great difference among the frequencies of the extra spin pairs. There are 2919 with an alpha module and only 10 without. There are 2668 nuclides with extra neutron-neutron spin pairs, but only 164 out of the 2929 nuclides which have one or more extra proton-proton spin pairs. There are 1466 with an extra neutron-proton spin pair.

## Results and Conclusions

The coefficient of determination (R²) for this equation is 0.9998825 and the standard error of the estimate is 5.47 MeV. The average binding energy for the nuclides included in the analysis is 1072.6 MeV so the coefficient of variation for the regression equation is 5.47/1072.6=0.0051. Most impressive are the t-ratios. A t-ratio of about 2 is considered statistically significant at the 95 percent level of confidence. The level of confidence for a t-ratio of 923 is beyond imagining.

It is notable that the coefficients for all three of the spin pair formations are roughly equal. They all are larger from what one would expect from the binding energies of small nuclides. But the model is not that the binding energy per spin pair is some constant a; instead the model presumes the binding energy per spin pair is (a−b/#x) where #x is the number of the spin pairs of a particular type. When the binding energy per spin pair is multiplied times #x the result is (a#x−b). All of the b's get added together to form the constant of the regression equation. Thus a comparison cannot be made, for example, of the coefficient for the number of alpha modules with the the binding energy of an alpha particle.

## Regression Results

The results of regressing binding energy on these six variables of spin pairing and nucleon interactions are:

Below is a graph of the actual binding energies of nuclides plotted against their estimates based upon the regression equation.

The ratio of the regression coefficients of nn and np should be q; i.e.,

#### cnn/cnn = bq²/bq = q = −0.21367/0.31831=0.671264.

The actual value of q is undoubtedly q ratio of small whole numbers. Thus the best estimate of q from empirical analysis is −2/3.

When the regression coefficients are used

#### (∂S*/∂p) = ½(13.84234) + 12.77668 = 19.69785 (∂S*/∂n) = ½(13.69875 ) = 6.84938 and thus (∂S*/∂p) − (∂S*/∂n) = 12.84847

Then for q=-2/3 and (1+d)²=1.1

#### n = 1.5(1.06)p − 0.09 +(3/5)12.84847/0.31831 n = 1.59p + 24.21879

This is much too high to be a formula for the stable nuclides.

## Conclusions

The minimization of energy (maximization of binding energy) is not relevant for determining the most stable isotopes for each element.