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The Testing of the Nucleonic Charge Hypothesis

Nuclei are held together by the net effect of the strong force and the electrostatic force. There may be substructures such as alpha particles existing in nuclei. The electrostatic force clearly involves charges. The force between two nuclear structures having charges of e1 and e2 is given by

F = Ke1e2/s²

where s is the distance between the centers of the two structures.

For a single proton e=1 and for a single neutron e=0.

Generally a force that is carried by particles the way the electrostatic force is carried by photons and the nuclear strong force is carried by π mesons is determined by a formula of the form

F = HZ1Z2f(s)/s²

where H is a constant, Z1 and Z2 are the charges on the substructures and s is the distance between their centers. For the electrostatic force f(s) is equal to 1 and for the nuclear strong force f(s) is most likely exp(−s/s0). The potential energies then are of the form

V(s) = HZ1Z2sf(z)dz

Let the strong force charges of the neutron and the proton be denoted as sn and sp, respectively. Either of these can be set to an arbitrary amount, say 1, and the value of the other and the force constant H adjusted accordingly. For now it is convenient to keep them arbitrary.

There is evidence that the nucleons of a nucleus combine whenever possible into substructures such as alpha particles and nucleon pairs. According to this model a nucleus is made up of a number of alpha particles (denoted as #a), one or none deuterons (denoted as #d), and numbers of neutron pairs or proton pairs (denoted as #nn and #pp, respectively). There may also a singleton neutron or proton left over (denoted as *n and *p, respectively). There are restrictions on these quantities. The number of deuterons can only be zero or one because two deuterons would form an alpha particle. If #nn>0 then #pp=0, and vice versa. Likewise if *n>0 then *p=0 and vice versa.

There are six types of substructures, including the singleton nucleons, but only three can have numbers beyond 1. The number of interactions of unlike particles is 6*5/2=15. The number of different types of interactions is potentially 3+15=18, but of these 18 at least two must be zero. There cannot be simultaneously singleton neutrons and protons (*n*p=0) and likewise #nn#pp=0.

According to the nucleonic charge hypothesis here are the charges of the various substructures of nuclei:

Nuclear
Substructure
ChargeEquivalency
neutron pair2snTwo singleton
neutrons
proton pair2spTwo singleton
protons
deuteronsn+spA singleton
neutron and
a singleton
proton
alpha
particle
2(sn+sp)neutron pair
and proton pair
or
two deuterons

Effect of adding one neutron can be determined by tabulating the binding energies of all nuclides which could contain all alpha particles then all that contain all alpha particles plus one neutron. Such effects so measured may be dependent up on which shell the added particle goes into.

The alpha particle shells contain 1, 2, 4, 7, 11, 16, 22 alpha particles. These correspond to shell occupancies for the nuclear magic numbers for neutrons and protons of 2, 6, 14, 28, 50, 82, 126; i.e., 2, 4, 8, 14, 22, 32, 44.

The Effect of a Deuteron and the Combined
Effects of a Neutron Pair and a Proton Pair

According to the nucleonic charge hypothesis the effect of an added deuteron on the structural binding energy of a nuclide should be one half of the sum of the effects of an additional neutron pair and an additional proton pair on the structural binding energy. The effect on total binding energy is the effect on structural binding energy plus the binding energy resulting from the formation of the substructure.

If SBEd, SBEnn and SBEpp are the effects on structural binding energy of a deuteron, a neutron pair and a proton pair, respectively, then

SBEd = 0.5S(SBEnn+SBEpp)

But SBEx is equal to the effect on binding energy of an additional substructure x, BEx minus the binding energy stemming from the formation of the substructure x, Ex. Therefore

BEd = Ed + 0.5(BEnn−Enn + BEpp−Epp)
which reduces to
BEd = [Ed−0.5(Enn+Epp)] + 0.5(BEnn+BEpp)

The graph of the data is as follows.

The first four data points on the left correspond to the first, second and part of the third alpha particle shells. These and the rest of the data for the third alpha particle shell are not relevant for testing the above proposition. When the data for the fourth shell are regressed the result is as follows:

BEd = 0.76146 + 0.48376(BEnn+BEpp)

The coefficient of determination for this equation is 0.99681 and the t-ratio for the regression coefficient is 39.5. The deviation of the regression coefficient from the predicted value of 0.5 is not significantly different from 0 at the 95 percent level of confidence. There are however only 5 degrees of freedom for the regression.

If the data for the fourth and fifth alpha particle shells are combined the regression equation is

BEd = 0.26206 + 0.50093(BEnn+BEpp)

The coefficient of determination for this equation is 0.99207 and the t-ratio for the regression coefficient is 33.5 Needless to say, this regression coefficient is not statistically significantly different from 0.5 at the 95 percent level of confidence. This regression has 9 degrees of freedom.

The Effects of Neutrons and the Effect of Neutron Pairs

The nucleonic charge of a neutron pair is twice that of a neutron. Therefore the effect on the structural binding energies of nuclides of a neutron should be one half that of a neutron pair. Thus

SBEn = 0.5SBEnn

But SBEnn is equal to the effect on binding energy of an additional neutron pair BEnn minus the binding energy stemming from the formation of the neutron pair, Enn. For the singleton neutron there is no formation of a substructure. Therefore

BEn = 0.5(BEnn−Enn)
which reduces to
BEn = −0.5Enn + 0.5BEnn

The graph of the data is as follows.

The regression equation for the data in the fourth alpha particle is

BEn = −2.75815 + 0.56085BEnn

The model predicts the negative value of the constant and thus indicates that the binding energy involved in the formation of a neutron pair is 5.516302619 MeV.

The coefficient of determination for this equation is 0.92151. The t-ratio for the regression coefficient is 7.6 so it is significantly different from zero, but it is not significantly different statistically from 0.5 at the 95 percent level of confidence.

The Effects of Protons and the Effect of Proton Pairs

The analogous equation, graph and regression equation for the effect of an additional proton versus a proton pair are as follows.

BEp = −0.5Epp + 0.5BEpp

BEp = −-0.51847 + 0.37703BEpp

The model predicts the negative value of the constant and this is confirmed. Thus indicates that the binding energy involved in the formation of a proton pair is 1.03694 MeV.

The coefficient of determination for this equation is 0.85849. The t-ratio for the regression coefficient is 5.5 so it is significantly different from zero, but in this case the regression coefficient is not significantly different statistically from 0.5.

The Effect of a Deuteron and the Sum of the Effects
of a Singleton Neutron and a Singleton Proton

The charge of a deuteron is (sn+sp). This is the same as the sum of the charges of the singleton neutron and the singleton proton. Thus the effect of a deuteron should be the same as the sum of the effects of the singleton neutron and proton; i.e.,

SBEd = SBEn + SBEp
which implies
BEd = Ed + BEn + BEp

The graph for testing this proposition is as follows:

The regression equation for the data for the fourth alpha particle shell is

BEd = 5.05486 + 0.80947(BEn+BEp)

The coefficient of determination for this equation is 0.76344. The t-ratio for the regression coefficient is 4.0 so it is significantly different from zero, but it is not statistically different from the model value of 1.0 at the 95 percent level of confidence. It is notable that the constant term, which represents the effect on binding energy of the formation of a deuteron, is positive. It is also substantially larger than the binding energy of 2.22457 MeV for a deuteron.

The Effect of an Alpha Particle and the Effect of a Deuteron

The final comparison to be considered is that of the relation between the effects of a deuteron and an alpha particle. The nucleonic charge for the deuteron is (sn+sp) whereas that of an alpha particle is 2(sn+sp). Thus the effect of the addition of a deuteron on structural binding energy should be one half of that of an alpha particle.

SBEd = 0.5SBEa
and hence
BEd − Ed = 0.5(BEa−Ea)
which reduces to
BEd = (Ed−0.5Ea) + 0.5BEa

The graph of the data is as follows.

In the above display the yellow line represents one half of the effect of an additional alpha particle and the blue line the difference between the effect of an additional deuteron and one half of the effect of an additional alpha particle. There is a shell structure involved in the above display. For more than 14 alpha particles (28 neutrons and 28 protons) the effects are nearly constant. There is very little variation in the effects of an additional deuteron or an additional alpha particle within this shell so it is not possible to obtain a meaningful regression coefficient to compare with the theoretical 0.5.

If the comparison is extended back from the shell of more than 14 alpha particles to 11 alpha particles the scatter diagram is as follows.

The regression equation for the data displayed above is

BEd = 1.8856 + 0.62093BEa

The coefficient of determination for this equation is 0.943. The t-ratio for the regression coefficient is 14.1 so it is highly significantly different from zero. It is also significantly different from 0.5.

Conclusions

While the predictions concerning the relationships of the effects of various constituents of nuclei are not fully confirmed there is generally confirmation of the relationships. In almost all cases the difference between the regression coefficient and the predicted value is not significantly different from zero at the 95 percent level of confidence.

(To be continued.)


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