San José State University

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Thayer Watkins
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 The Net Force Between Alpha Particles and other Substructures of Nuclei

one model of nuclei is 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.

A nucleus is held together by the force between these substructures and the forces are proportional to the number of interactions of the various types. The number of interactions between alpha particles is #a(#a-1)/2 and likewise for the number of interaction between neutron pairs (#nn(#nn-1)/2) and proton pairs (#pp(#pp-1)/2). The number of interactions of unlike types of substructures is just the product of their numbers. For example, the number of interactions between alpha particles and neutron pairs is #a#nn.

## The Forces Within Nuclei

Generally the force is determined by a formula of the form

#### F = Z1Z2f(s)

where 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 K/s², where K is a constant and for the nuclear strong force f(s) is most likely H*exp(−s/0)/s², where K and s0 are constants. The potential energies then are of the form

#### V(s) = Z1Z2∫s∞f(z)dz

For the electrostatic force the charges are simple. The charge of the proton is +1 and that of the neutron is 0. For the nuclear strong force the situation is more complicated. Let the 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.

The force and the potential energy due to an interaction of two alpha particles through the strong force should be proportional to

#### Z1Z2 = [2sn+2sp]² = 4(sn² + 2snsn + sp²)

On the other hand, the force between an alpha particle and a deuteron is proportional to

#### Z1Z2 = (2sn+2sp)(sn+sp) which reduces to Z1Z2 = 2(sn² + 2snsn + sp²)

This is exactly half of the value for an alpha alpha interaction. When a deuteron is added to the nuclide containing only alpha particles the potential energies lost due to the interaction with the alpha particles should be one half of the potential energy lost when an alpha particle is added to such a nuclide. This proposition might be tested empirically.

The force between a neutron pair and a deuteron is proportional to 2sn(sn+sp). Between a proton pair and a deuteron the force is proportional to 2sp(sn+sp). The forces between an alpha particle and the nucleon pairs are then proportional to 4sn(sn+sp) and 2sp(sn+sp).

The forces and potential energies also depend upon the separation distances between the particles. Separation distances are largely a matter of shell structures. Therefore any quantitative analysis must be carried out separately for each shell.

## Testing of the Implications of the Model

According to the model the effect of an additional single neutron on the structural binding energy of a nuclide should be one half of the effect of an additional neutron 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 and SBEa are the effects on structural binding energy of a deuteron and an alpha particle, respectively, then

#### 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 displayed above is

#### BEn = −2.1443 + 0.53483BEnn

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 4.2886 MeV.

The coefficient of determination for this equation is 0.99025. The t-ratio for the regression coefficient is 48.3 so it is highly significantly different from zero. It is also significantly different statistically from 0.5.

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 = −1.14569 + 0.47098BEpp

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 2.29137 MeV.

The coefficient of determination for this equation is 0.9347. The t-ratio for the regression coefficient is 15.1 so it is highly significantly different from zero. In this case the regression coefficient is not significantly different statistically from 0.5.

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 + SBEpwhich implies BEd = Ed + BEn + BEp

The graph for testing this proposition is as follows: The regression equation is

#### BEd = 6.54302 + 0.66130(BEn+BEp)

The coefficient of determination for this equation is 0.933. The t-ratio for the regression coefficient is 16.7 so it is highly significantly different from zero. It is also statistically different from the model value of 1.0. 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.

A similar sort of relationship should prevail between the nucleonic charge of the deuteron and the sum of the nucleonic charges of neutron pairs and proton pairs. The sum of the effects for nucleon pairs should be twice that of the deuteron; i.e.,

#### SBEd = 0.5(SBEn + SBEp) which implies BEd = [Ed−0.5(Enn + Epp)] + 0.5(BEnn + BEpp)

The graph for the testing of this proposition is: The regression equation is

#### BEd = 3.83573 + 0.37290(BEnn+BEpp)

The coefficient of determination for this equation is 0.977. The t-ratio for the regression coefficient is 25.8 so it is highly significantly different from zero. It is also statistically different from the model value of 0.5.

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 = 0.5SBEaand 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 partical 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 to the first approximation, the errors being on the order of 10 percent or less.

(To be continued.)