﻿ Nuclear Basic Models: SOM

# SOM Basic Nuclear Models

Author: Phil Fraley

Six models are basic: proton (already covered), neutron, deuteron, triton, helion and alpha. Larger nuclei contain mixtures of these.

Equation (3) will be the nuclear radius:

r = r0 A1/3 = 0.853 A1/3    (3)

This is a lower bound on nuclear radius calibrations, just as the proton radius was the lowest value that could reasonably fit the SOM idea.

## Alpha Model

Figure 4
Alpha Model The radius of the black circle is the average proton radius. Axis 3 is perpendicular as indicated by the small red bullseye.

In SOM, pairs of neutrons are replaced by a p p- pair. Therefore, the helium-4 nucleus, alpha, becomes 3 protons and one antiproton as shown in Fig. (4).

### Spacings

The radius (and p p- distance) is 0.853 41/3 = 1.354 fm It is represented by the vertical light blue line in the figure. The p p distance is √3 1.354 fm = 2.345 fm and one (of three) is the darker blue horizontal line in the figure. The black circles were scaled to give a 0.546 fm average proton radius.

### Energy

p p-: -65.5/1.354
p p- overlap (Fig. (3)): +10.0
p p: +65.5/2.345
p p overlap: -2.4

There are 3 of each of these for a total of -38.38 MeV. A positive energy of 12.78 MeV is needed to give the binding energy of -25.60 MeV. The BE in SOM is 2.6 MeV less than normally calculated because 2 neutrons are replaced by p p-. The energies for axes 1, 2 and 3 are 0.75 M v2, M v2 and 1.5 M v2. v/c for axes 1, 2 and 3 are 0.1167c, 0.1348c and 0.09529c (equating to 12.78 MeV and dividing both sides of equation by M c2 = 938.259 MeV). M is a proton mass.

### Angular Momentum

The angular momentum for axes 1, 2 and 3 are: 1.5 M r v = 1.303ћ, √3 M r v = 1.303ћ and 3 M r v = 1.843ћ. The average is 1.483ћ. r is 1.354 fm and M is a proton mass. Multiply and divide by M rp c = ћ. Example axis-1: ћ 1.5 1.354 0.1167/0.210 where rp is 0.210 fm.

### Energy-Equivalent ћ Turning

What fraction h' of ћ pointing and turning with the particles gives the same energy? E = h' ћ v/r
Replace ћ by M rp c and divide both side of equation by M c2. The totals of the two or three proton masses for axes 1,2 and 3 are 0.6515ћ, 0.6515ћ and 0.9215ћ. The average is 0.7415ћ. Example axis-1: h'1 = 2/3 (12.78 1.354/(938.259 0.1348)) = 2/3 x 0.6515ћ while the other 2 masses each give 1/6 x 0.6515ћ

### Summary

Scale, 65.5 MeV, and r0 = 0.853 multiplying A1/3 were the parameters selected. Secondarily, the selected radius of the proton, 0.546 fm, shifts the scale of the overlap correction. The whole result is non-linear and I did not study it as function of the proton radius.

Because of symmetry, the alpha is a keystone of SOM. The simplest model is given: overlapping bonds plus rotation of a semi-rigid structure. An angular momentum close to √2 ћ results (or, from an SOM perspective, √2/2 ћ total component in the direction of motion). The QM angular momentum, spin and orbital, are zero. One has to imagine the orbits as continuously and randomly changing. The cause is the randomness of the underlying partics and particles. Think of the outer layers of a ball of yarn - different colors for each particle. An analogy to the model here is the n=1 hydrogen electron orbit where the angular momentum is ћ (or a component of 1/2ћ in the direction of the velocity) even though the QM value is zero. The yarn’s outer layer can be thought of as looser in the hydrogen case since the radii are not precise by comparison to the alpha model. The average in each case is zero because the (instantaneous) vectors point in random directions.

Figure 4
Triton Model The radius of the black circle is the average proton radius. Axis 3 is perpendicular as indicated by the small red bullseye.

## Triton Model

For the triton, 2 neutrons are replaced by p p- as shown in the figure.

The p p distance is allowed to vary, but the average radius is set by Eqn. (3) to about 1.230 fm. The angular momentum is constrained to be about √2ћ. Since the formulas and methods are similar to the alpha case, only the results are given.
p p: 65.5/1.7
p p overlap: -5.216 (Fig 3)
p p-: -65.5/2.3 (2 times)
p p- overlap:1.951 (2 times)
PE = -19.741 MeV
BE = -5.90 MeV (again, p p-, not 2 neutrons: so 2.6 MeV less negative)
KE: 13.842 MeV
Axis-3 energy is different: KE = M vp+2 + 1.6989/2 M vp+2
(1.6989 is the square of the radius ratios: 1.4606/1.1206)

p p distance: 1.70 fm (horiz. darker blue line in Fig. 4)
p p- distance: 2.30 fm
rp+: 1.1206 fm
rp-: 1.4606 fm (vertical light blue line in Fig. 4)
rave.: 1.234 fm

angular momentum, axes 1, 2 and 3: 1.463ћ, 0.983ћ and 1.763ћ with average 1.403ћ
velocities, axes 1, 2 and 3: 0.1402c, 0.1215c and 0.08931c (p+)
(axis-3: multiply p+ velocity by the radius ratio above to get the p- velocity)

Figure 5
Neutron Model The radius of the black circle is the average proton radius.

## Neutron Model

Figure 6
Deuteron Model The radius of the black circle is the average proton radius.

## Deuteron Model

The deuteron is the only nucleon whose measured value is much greater than Eqn. (3) predicts. Calculating the angular momentum for 0.853 21/3 and the scale and soe+ radius values just above, yields 1.12ћ. Adjusting the radius to get 1.41ћ yields 1.75 fm for the radius, closer to the measured value of 1.6 to 1.7 fm. At 2 r = 3.5 fm, 0.2776 is the fraction multiplying -65.5 MeV for the p p- potential energy and 0.2820 is the fraction multiplying +32.7 MeV for the soe+ p potential energy. Since 1/(2r) is 0.2857, there is overlap for both cases and the difference for the p p- overlap of (0.2857 - 0.2776) times 65.5 can be seen (by extrapolation) in Fig. 3. The overlap calculation for soe+ p is very similar, but only the result is given. BE = 2.22 MeV was subtracted from the PE and KE = M v2. 2 M v r is the angular momentum. Figure 6 gives the geometry.

Figure 7
Helion Model The radius of the black circle is the average proton radius.

## Helion Model

Figure 7 shows the model. A complication arises because the p- mass is centered in the middle of the p p interaction. It is not obvious how to model this. Using 0.853 31/3 for the radius and 21/2ћ angular momentum gives an estimate that 10 MeV of 23.5 MeV of the p p interaction was blocked. So the helion is not useful for verifying any of the models and scaling factors.