San José State University

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 The Limits of Nuclear Stability in Terms of Alpha Particles and Excess Neutrons

There are 2931 nuclides stable enough to have had their masses measured and their binding energies computed. There is evidence that at least some of the protons and neutrons are combined into alpha particles. The nuclides which could contain only alpha particles and additional (hereafter called excess) neutrons were selected. Within this group for each number of alpha particles the minimum number and the maximum number of excess neutrons were compiled. The results are displayed in the following graph. 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. Another study indicated that the value of ν is 2/3. However yet another study found a value for ν of 0.73.

If the value of ν is 2/3 then the net nucleonic charge of an alpha particle is 2−2(2/3)=+2/3. This is just the negative of the nucleon charge of −2/3 for a neutron.

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. For a nuclide made up entirely of a alpha particles and xsn excess neutrons the net nucleonic charge would be of the form

#### η = a − μxsn

where μ is the nucleonic charge of a neutron relative to that of an alpha particle. If ν=2/3 then μ is equal to 1. If ν=3/4 then the nucleonic charge of an alpha particle would be 1/2 and μ would be 1.5.

The binding energy associated with the interaction of alpha particles and excess neutrons through the strong force is a nonlinear function of η, but for small values of η to a reasonable approximation it is Kη, 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 excess neutrons would result in the formation of a neutron-neutron spin pair. Let Enn be the binding energy associated with the formation of a neutron-neutron spin pair. The binding energies associated with the formation of spin pairs is not really a constant independent of the levels of a and xsn but for the present it is assumed to be a constant.

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 a alpha particles and xsn excess neutrons in which xsn is odd it is

#### IBEn = Kη + Ennor, expanded IBEn = K(a−μxsn) + Enn

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

#### K(a − μxsnmax) + Enn = 0 and hence xsnmax = (1/μ)a + Enn/K

There is also an incremental binding energy for an additional alpha particle, IBEa. Its value is given by

#### IBEa = −Kη = −K(a − μxsn) = Kμxsn − Ka

There are no additional spin pairs formed by the addition of another alpha particle. The minimum number of excess neutrons is where IBEa=0 and hence

#### xsnmin = (1/μ)a

According to this simplified theory the slope of the relationships between xsnmax and xsnmin should both be (1/μ). The actual relationships display a number of piecewise linear sections with different slopes. The initial section for xsnmin is determined by the constraint that xsn cannot be negative. There is however one section which best fits the model. It is the initial section of the relationship for xsnmax and a between a=3 and a=23. The regression equation for this section of the data is

#### xsnmax = 1.05584a + 6.36926

This means that (1/μ)=1.05584 and hence μ=0.94711. The standard deviation of the slope is 0.04328 so the z-ratio for the deviation of the slope from 1.00000 is (1.05584−1)/0.04328=1.29. Therefore the slope of the relationship between xsnmax and a is not statistically significantly different from 1.0 at the 95 percent level of confidence. Thus μ is not significantly different from 1.0 and thus the magnitude of the nucleonic charge of a neutron is two-thirds that of a proton.

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

#### xsnmax = (1/μ)a + Enn/(Kμ) xsnmin = (1/μ)a

If these two equations are averaged the result is

#### xsnmid = (1/μ)a + Enn/(2Kμ)

Here is the display for xsnmid as function of the number of alpha particles. If the two equations are subtracted the result is

#### xsnrange = (Enn/(Kμ)

Thus if Enn were constant then xsnrange would be constant. But over the initial levels of alpha particles, the number of excess neutrons cannot be less than zero. At the limit of the number of alpha particles the range of excess neutrons goes to zero. The actual range displays this sort of pattern but with more complications.