San José State University

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 The Role of the Range in Nuclear Stability in the Estimate of the Ratio of the Nucleon Charge of a Neutron to that of a Proton

There are 2931 nuclides stable enough to have had their masses measured and their binding energies computed. For each number of protons the minimum number and the maximum number of neutrons were compiled. The results are displayed in the following graph. The difference between the maximum and the minimum is called the range. The range of the number of protons for each level of the number of neutrons is displayed below. A Previous study 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 limits to the values of the proton numbers of the known nuclides, shown above.

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 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.

## An Additional Neutron

The incremental binding energy of a neutron, IBEn, is the binding energy change associated with the addition of another neutron to a nuclide. For a nuclide with p protons and n neutrons in which n is odd and less than p it is

#### IBEn = kζ + Enn + Enpor, expanded IBEn = k(p−νn) + Enn + Enp

For stability the values of n and p must be such that IBEn≤0. The minimum number of protons for a nuclide with p protons is reached when IBEn is driven to zero. This means that

#### k(pmin − νn) + Enn + Enp = 0 and hence pmin = νn − Enn/k − Enp/k

The maximum number of neutrons is also where IBEn=0 and hence

## An Additional Proton

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 = 0

For IBEp to be greater than or equal to zero requires a minimum n of

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

The maximum p is the value such that IBEp=0; i.e.,

#### pmax = νn + Epp/k + Enp/k

Now the equations for the maximum and minimum number of neutrons can be displayed together; i.e.,

#### nmax = (1/ν)n + Enn/(kν)+ Enp/(kν) nmin = (1/ν)p − Epp/(kν) − Enp/(kν)

If these two equations are subtracted the result is

#### nrange = (Enn+Epp+2Enp)/(kν)

Thus if Enn, Epp and Enp are constant then nrange should be constant. But over the initial levels of protons, the number of neutrons cannot be less than one. At the limit of the number of protons the range of neutrons goes to zero. The actual range displays this sort of pattern but more complicated. The equations for the maximum and minimum number of protons are

#### pmax = νn + Epp/k + Enp/k pmin = νn − Enn/k − Enp/k

The difference of these two equations is

#### prange = (Epp + Enn + 2Enp)/k

This is essentially the same as for nrange except that nrange is also divided by ν. Thus

#### nrange = prange/ν or, equivalently ν = prange/nrange Since prange and nrange are not constant it is problematical which levels to use in applying the above formula. The maximum ranges seem appropriate. The maximum range of the number of protons is 26; the maximum range of the number of neutrons is 39. The ratio of these two values is exactly two-thirds.