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The Differential Impact of Neutrons
and Protons on the Binding Energy of Nuclides

When the number of neutrons in a nucleus is increased the binding energy increases substantially, typically by 7 to 8 million electron volts (MeV). In contrast when a proton is added the binding energy of the nuclide may or may not go up. If the binding energy does go up it is only for a fraction of the amount which results when a neutron is added, typically less than 3 MeV. In may cases the binding energy goes down when a proton is added.

The material is a quantitative analysis of the differential effects of additional protons and neutrons on the binding energy of nuclides. In order to isolate the the effects and eliminate any increment in binding energy that results from the formation of an alpha particle in a nucleus the analysis is carried out in terms of the addition of protons or neutrons is a nuclide that contains or could contain an integral number of alpha particles. Hereafter nuclides that contain or could contain an integral number of alpha particles will be referred to as alpha nuclides.

Many more neutrons than protons may be added to a nuclide and still have a stable nuclide. The case of the Oxygen 16 nuclide is a typical case.

As many as ten neutrons can be added to the O16 nuclide and the result is still a stable nuclide. On the other hand only one through six protons can be added to the O16 nuclide and still have stability. The situation becomes much more extreme for large nuclides. Scores of neutrons can be added to the Sn50 (tin) nuclide but not one proton can be added without creating instability.

Although the effect of additional neutrons is quite different for the effect of additional protons there is a simple relationship between their effects. Below is a graph of the difference in binding energies due to adding k neutrons compared to adding k neutrons.

The relationship is nearly linear; there being only a slight upward curvature to the line. This same relationship prevails starting from other alpha nuclides. For example, here is the case for the Magnesium24 alpha nuclide.

Let us call the binding energy for a nuclide that is k alpha particles and n neutrons BN(k,n) and the the binding energy for a nucide that is k alpha particles and n protons difference in the binding energy of BP(k,n). The difference Φ(k,n) is then

Φ(k,n) = BN(k,n) − BN(k,n)

For a given nuclide such as the O16 the relationship can be approximated by a quadratic function (with no constant term) obtained by regression. For the O16 nuclide, which has k=4, the quadratic function found is

Φ(4,n) = 3.39272n + 0.20168n²

The quadratic regression was carried out for the data on differences in binding energy for all of the alpha nuclides for which it could be applied. Here are the results.

Regression Coefficients for Difference Function Φ(k,n)
of Alphas
of points
He4 1 0.78285 0.21009 5
Be8 2 1.84575 0.17692 5
C12 3 2.96481 0.08577 5
O16 4 3.39272 0.20168 7
Ne20 5 4.37950 0.11777 7
Mg24 6 5.13126 0.08459 7
Si28 7 5.78401 0.09303 7
S32 8 6.47674 0.05657 7
Ar36 9 6.84907 0.10923 7
Ca40 10 7.12983 0.15901 7
Ti44 11 7.81775 0.15064 7
Cr48 12 8.42226 0.13378 7
Fe52 13 8.77401 0.21462 6
Ni56 14 9.49577 0.10697 6
Zn60 15 9.73595 0.18723 4
Ge64 16 10.07115 0.24493 4
Ga68 17 10.87158 0.08928 4
Kr72 18 11.32 0.11 3
Sr76 19 11.84 0 3
Zr80 20 12.1 0 2
Mo84 21 12.3 0 2

The graphs of the linear and quadratic regression coefficients as a function of the number of alpha particles in the starting nuclide are as follows.

For the linear coefficients a quadratic function should give a reasonable approximation. The regression equation for the linear coefficients as a function of the number of alpha particles in the starting nuclide.

c1 = 0.91927k − 0.01657k²

The coefficient of determination for this equation, R², is 0.9943.

The regression of the quadratic coefficient on the number of alpha particles is less successful.

c2 = 0.17412 − 0.0105k

The coefficient of determination for this equation is only 0.066 and the t-ratio for the linear coefficient is less than 1. This means that the average value of 0.14068 is just as appropriate as the regression equation; i.e.,

c2 = 0.14068

Thus the difference function can be expressed as

Φ(k,n) = (0.91927k − 0.01657k²)n + 0.14068n²

(To be continued.)

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