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 An Estimate of the Nuclear Strong Force Chargeof a Neutron Relative to that of Proton Based Upon the Pattern of the Scatter Diagram of the Proton and Neutron Numbers of Nuclides

A Previous study have developed evidence that the nucleonic (strong force) charge of a neutron is of the opposite sign and smaller in magnitude from that of a proton. Let ν denote the ratio of the nucleonic charge of a neutron to that of a proton. The actual value of ν is undoubtedly a simple fraction. Previous work indicated that the relative magnitude of the neutron charge could be 2/3 or 3/4. Furthermore such a difference in charge of the nucleons can account for the scatter diagram of the values of the proton and neutron numbers of the known nuclides, shown below.

Another study demonstrated that the binding energy increments experienced by additional nucleons to a nuclide is a function of two components. One is simply the difference in the number of protons and neutrons in the nuclide. This component has to do with the formation of a neutron-proton spin pair. The other component has to do with the interaction of nucleons through the strong force and it is a function of the net nucleonic charge of the nuclide. If p and n are the numbers of protons and neutrons, respectively, of the nuclide then the net nucleonic charge ζ is

#### ζ = p − νn

where ν is the magnitude of the nucleonic charge of the neutron relative to that of a proton.

The binding energy associated with the interaction nucleons through the strong force is a nonlinear function of ζ, but for small values of ζ to a reasonable approximation it is kζ, where k is a constant. Nucleons also interact through the formation of spin pairs. For example, the addition of another neutron to a nuclide with an odd number of neutrons would result in the formation of a neutron-neutron spin. Let Enn be the binding energy associated with the formation of a neutron-neutron spin pair. If there are unpaired protons in the nuclide the addition of another neutron would result in the formation of a neutron-proton spin pair with a binding energy of Enp. The binding energies associated with the the formation of spin pairs are not really constants independent of the levels of n and p but for the present they are assumed to be constants.

The energy change associated with the addition of another neutron to a nuclide with p protons and n neutrons in which n is odd and less than p is

#### IBEn = kζ + Enn + Enp

The maximum number of neutrons for a nuclide with p protons is reached when IBEn≤0. This means that

#### k(p − νnmax) + Enn + Enp = 0 and hence nmax = (1/ν)p + Enn/kν + Enp/kν

The binding energy of an additional proton to a nuclide with p protons and n neutrons in which p is odd and less than n is

#### IBEp = −kζ + Epp + Enpand thus IBEp = kνn - kp + Epp + Enp

For IBEp to be positive requires a minimum n of

#### nmin = (1/ν)p − Epp/kν − Enp/kν

Thus the slope of the relationship between nmax and p should be (1/ν) the same as the slope of the relationship between nmin and p.

In the graph below the relationships between the number of protons and the maximum and minimum number of neutrons are plotted.

The relationships are not linear over the whole range of values of p but over some intervals they are reasonably close to being linear. For the maximums an approximating line goes from the nuclide with 15 protons and 31 neutrons to the nuclide with 102 protons and 160 neutrons. The difference in the proton numbers for this line is 87 protons and 129 neutrons. The ratio of these two quantities is 1.48276. This should be 1/ν and thus ν is equal to 0.6744.

For the minimums the approximating line runs from the nuclide with 15 protons and 31 neutrons to the nuclide with 80 protons and 95 neutrons. The difference in the proton numbers is 65 and in the neutron numbers 86. Their ratio is 1.32308 and its reciprocal is 0.75581. Thus the maximum neutron line gives an estimate of of ν of (2/3) and the minimum neutron line an estimate of (3/4).

## The Range of Neutron Numbers

According to the formulation above the difference between the maximum and minimum number of neutrons should be

#### Δ = (Epp + Enn + 2Enp)/kν

The graph below of the actual differences shows that the difference is not constant but it is approximately so over a minrange of proton numbers. This difference is the number of isotopes for an element with an atomic number p.

Roughly from p=60 to p=80 the level of the difference is constant. At p=60 the value of nmin is 66. At p=80 the value of nmin is 95. Thus the increase in the number of neutrons for the increase in the number of protons of 20 is 29. The ratio is 1.45 and its reciprocal, the charge of the neutron relative to that of the proton, is 0.68966, or roughly 2/3.

The maximum number of neutrons at p=60 is 101. At p=80 it is 128 for a change of 27 over an interval of 20. The ratio is 1.35 and its reciprocal is 0.74074, or roughly 3/4.

## The Electrostatic Repulsion of Protons

In addition to the factors which affect neutrons, protons are subject to electrostatic repulsion. This factor is thought to only effective in large nuclides where the average distance between proton becomes large. The effect of this repulsion on the binding energy for an additional proton is negative and proportional to the number of protons in the nuclide, say −qp. The binding energy of an additional proton to a nuclide with p protons and n neutrons in which p is odd and less than n is

#### IBEp = −kζ −qp + Epp + Enpand thus IBEp = kνn - (k+q)p + Epp + Enp

For IBEp to be positive requires a minimum n of

#### nmin = (1/ν)((k+q)/k)p − Epp/kν − Enp/kν

Thus the slope of the relationship between nmax and p should be (1/ν) whereas the slope of the relationship between nmin and p should be the larger value of (1/ν)(k+q)/k.

An examination of the data reveals there is one interval in which the slope of the relation between nmin and p it larger than that between nmax and p. That interval is between p=80 and p=100. At p=100 nmin is equal to 142 and nmax is 159. The changes from nmin=95 and nmax=128 at p=80 are 47 and 31, respectively. The ratio for nmax implies that ν equal 0.64516, or approximately 2/3. The slope of the relation between nmin and p is 2.35. If ν=2/3 then (k+q)/k is equal to 2.35*(2/3), which is equal to 1.56667. Thus 1+q/k=1.56667 a and hence q/k=0.56667.

## Conclusions

Not only is the strong force charge of the neutron opposite in sign to that of the proton but it is smaller in magnitude. This difference in magnitude accounts for nuclides, especially heavier ones, having many more neutrons than protons. A proton is attracted to the neutrons of a nucleus but repelled by the protons. It takes about four neutrons to counterbalance the effect of three protons.