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The Effect of the Electrostatic
Repulsion of Protons in Nuclear Structure

A quantitative estimate of the effect of the repulsion of protons for each other would be very valuable. The potential energy due the electrostatic force is a simple known function of the separation of protons. Therefore an estimate of the effect on binding energy of the repulsion of protons could be converted into information of the separation distance of protons.

One plausible method of isolating the effect of the electrostatic repulsion between protons is to consider two nuclides that differ only in that one has a proton where the other has a neutron. For example, consider the triteron (one proton and two neutrons) and the He3 nuclide (two protons and one neutron). The binding energy of the triteron is 8.48 million electron volts (MeV), whereas that of He3 is 7.76 MeV. The difference might be ascribed to the charge of the second proton. The situation is a bit more complicated. He3 has a proton-proton spin pair where the triteron has a neutron-neutron spin pair. The binding energies of the various spin pairs might not be the same.

In order to avoid the problem of possible differences in the binding energy effects of pair formation consider a proton or a neutron being added to a core nuclide in which all of the nucleons are paired up. Such a core nuclide is one that could be made up of alpha particles or something like alpha particles. For examples, the nuclides with four neutrons and four protons, the Be8 nuclide. Such nuclides will be called alpha nuclides.

The alpha nuclides may contain actual alpha particles or chains of neutrons and protons in which there are modules involving two neutrons and two protons.

Such chains arise because any nucleon can form only one pair with a nucleon of the same type and only one pair with a nucleon of the opposite type. An alpha module is of the form:

An alpha particle is just a special case of an alpha module:

For more on this see Quasi Alpha Particles.

The following table illustrates such a construction.

Number of
Alpha Modules
Energy of
Alpha +
Energy of
Alpha +
Difference Increment
(MeV) (MeV) (MeV) (MeV)
0 0 0 0
1 26.33 27.41 1.08 1.08
2 56.3144 58.1649 1.8505 0.7705
3 94.1053 97.108063 3.002763 1.152263
4 128.21961 131.76266 3.54305 0.540287
5 163.0762 167.40597 4.32977 0.78672
6 200.5282 205.58756 5.05936 0.72959
7 239.285 245.01044 5.72544 0.66608
8 274.0572 280.42222 6.36502 0.63958
9 308.5735 315.5046 6.9311 0.56608
10 343.137 350.4147 7.2777 0.3466
11 377.089 385.0047 7.9157 0.638
12 413.547 422.044 8.497 0.5813
13 449.296 458.3802 9.0842 0.5872
14 484.682 494.235 9.553 0.4688
15 515.45 525.223 9.773 0.22
16 545.9 556.01 10.11 0.337
17 575.9 586.62 10.72 0.61
18 606.5 617.93 11.43 0.71
19 637.9 649.74 11.84 0.41
20 669.2 681.3 12.1 0.26
21 700 712.3 12.3 0.2
22 730.3 743.4 13.1 0.8

The plot of the difference in binding energies plotted versus the number of alpha modules is shown below.

The number of alpha modules serves as a proxy for the size of the nuclide. There seems to be a decreasing marginal effect of additional alpha modules. This can be illustrated by plotting the increment in the difference in binding energy versus the number of alphas, as shown below.

There appears to a general decline in the incremental effect of an additional alpha module with the size of the nuclide, but with some sort of cycle.

The same construction was carried out for two neutrons and two protons; and likewise for three and four of the nucleons. The results are shown below.

The relationships are roughly linear with respect to the number of alpha modules, with slightly declining slopes. The relationships in terms of the number of exchanged nucleons also seem to be linear. This is tested below.

Interpretation of the Differences in Binding Energies

Although the effect of a possible difference in the binding energy due to the formation of neutron and proton spin pairs is eliminated for K=1 it could show up in the results for K>1.

The linearity of the relation between the binding energy difference and K then is evidence that there is negligible difference for the effect of the formation of neutron-neutron and proton-proton spin pairs.

There is however another consideration in the interpretation of the differences. Previous studies indicate that neutrons repel neutrons and protons repel protons through the strong force as well as the electrostatic force. A neutron is attracted to a proton and it is this force that holds a nucleus together.

The forces between nucleons can be accounted for as a result of neutrons and protons having a nucleonic (strong force) charge. The nucleonic charge of a neutron is not only opposite in sign from that of a proton but smaller in magnitude.

Let the nucleonic of a proton be designated as +1 and that of a neutron as −q. Previous studies have found q to be 2/3 but for now q will be left general.

Suppose two particles have nucleonic charges of q1 and q2, where these can be positive or negative. The force between the particles is then proportional to q1q2. If this product is positive then the force is a repulsion and if it is negative the force is an attraction. Likewise the potential energy for the two particles is proportional to the product of the nucleonic charges. This then applies to the binding energy.

The binding energy due to the strong force repulsion of two neutrons is then of the form −Hq²F(s), where H is the constant for the strong force and F(s) is a function of the separation distance s. On the other hand, the binding energy due to the attraction between a neutron and proton is +HqF(s). The binding energy due to the interaction of two protons is then −(HF(s)+J/s), where J is the constant for electrostatic attraction.

The binding energy due to the interaction of a proton with an alpha module is then

BE = −2HF(s) + 2HqF(s) − 2J/s
and hence
BE = −[2H(1−q)F(s) + 2J/s]

The binding energy due to the interaction of a neutron with an alpha module is

BE = −2Hq²F(s) + 2HqF(s)
BE = −2Hq(1−q)F(s)

The difference Δ of the neutron interaction less the proton interaction is then

Δ = 2H(1−q²)F(s) + J/s

This is for one alpha module and one nucleon exchanged. For α alpha modules and K nucleons exchanged the difference would be αKΔ.

The simple difference does not isolate the electrostatic interaction. That requires a more sophisticated manipulation of the data.

The Incremental Binding Energies of Additional Alpha Modules

Consider the increments in binding energy for additional alpha modules. Such increments represents the interaction of an additional alpha module with the other alpha module and with the single nucleon. From the previous table these are:

Number of
Alpha Modules
Energy of
Alpha Modules
in Alpha + 1 proton
Energy of
Alpha Modules
in Alpha + 1 neutron
(MeV) (MeV)(MeV)
1 26.33 27.41 1.08
2 29.9844 30.7549 0.7705
3 37.7909 38.943163 1.152263
4 34.11431 34.654597 0.540287
5 34.85659 35.64331 0.78672
6 37.452 38.18159 0.72959
7 38.7568 39.42288 0.66608
8 34.7722 35.41178 0.63958
9 34.5163 35.08238 0.56608
10 34.5635 34.9101 0.3466
11 33.952 34.59 0.638
12 36.458 37.0393 0.5813
13 35.749 36.3362 0.5872
14 35.386 35.8548 0.4688
15 30.768 30.988 0.22
16 30.45 30.787 0.337
17 30 30.61 0.61
18 30.6 31.31 0.71
19 31.4 31.81 0.41
20 31.3 31.56 0.26
21 30.8 31 0.2
22 30.3 31.1 0.8
23 30.9
24 31.7
25 29.6

The graph of the two series of increments is quite remarkable.

The two curves nearly match but always the value for the additional proton is below the value for the additional neutron. Also note that the numbers of alpha modules after which the incremental binding energy drops off sharply are 3, 7 and 14. These numbers correspond to 6, 14 and 28 nucleons, each being a magic number representing the filling of a shell.

The incremental binding energies may be considered to be of the forms

IBE = M − 2Hq(1−q)F(s)
IBE = M − [2H(1−q)F(s) + 2J/s]

where M is the binding energy resulting from the interaction of an additional alpha modules with the other alpha modules in the nuclide.

If the second equation is multiplied by q and subtracted from the first, the result is:

IBE−q*IBE = (1−q)M + 2qJ/s

Dividing this series by (1-q) gives a series equal to M+2(q/(1-q))J/s. Now the problem is to find the values of M for the various numbers of alpha modules. Those values are available from the incremental binding energies of the alpha modules in the alpha nuclides. They are given in the table below along with the series for M+2(q/(1-q))J/s where q=2/3 and hence q/(1-q)=2.

Number of
Alpha Modules
M M+4J/s 4J/s
(MeV) (MeV) (MeV)
1 28.295674 29.57 1.274326
2 28.203836 32.2959 4.092064
3 35.662218 41.247689 5.585471
4 35.457608 35.735171 0.277563
5 33.025523 37.21675 4.191227
6 37.612031 39.64077 2.028739
7 38.28 40.75504 2.47504
8 35.24377 36.69094 1.44717
9 34.93504 36.21454 1.2795
10 35.3363 35.6033 0.267
11 33.4227 35.866 2.4433
12 35.9873 38.2019 2.2146
13 36.235 37.5106 1.2756
14 36.291 36.7924 0.5014
15 31.004 31.428 0.424
16 30.958 31.461 0.503
17 30.45 31.83 1.38
18 30.7 32.73 2.03
19 31 32.63 1.63
20 31.7 32.08 0.38
21 31.1 31.4 0.3
22 30.5 32.7 2.2
23 30.7
24 31.3
25 31.5

Below is the graph of the incremental binding energies of alpha modules in alpha nuclides and that amount plus a term that is proportional to the electrostatic repulsion between the additional proton and the protons in the alpha modules.

The difference is equal to 4J/s. The relation between separation distance and potential energy is as follows:

s (in fermi) = 1.44/PE (in Mev)

Based on this formula and the binding energies for J/s found the distances between the proton added to the alpha nuclides and the last alpha modules to be added are as shown.

The extremes are implausible and even the lowest levels are surprisingly high, but plausible. Thus it was shown that the theory could be used to obtain estimates of the spacing of nucleons in nuclei.

(To be continued.)

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