﻿ The Separation of the Components of Nuclear Binding Energies
San José State University

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Thayer Watkins
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 The Separation of the Components of Nuclear Binding Energies

The binding energy of a nucleus has two sources: One that is due to the formation of substructures and another that is due to the strong force interaction of its nucleons. The substructures formed are primarily the three types of spin pairs; i.e., neutron-neutron, proton-proton and neutron-proton.

By taking the differences in binding energies estimates of the effect of an additional neutron or additional proton on binding enerergy can be obtained. Specifically

#### Effect of an Additional Neutron = EAN(p, n) = BE(p, n+1) − BE(p, n) and Effect of an Additional Proton = EAP(p, n) = BE(p+1, n) − BE(p, n)

If the number of neutrons n is an even number then no additional neutron-neutron spin pair can be formed when a neutron is added and likewise it the number of protons is even then no additional proton-proton spin pair can be formed when a proton is added. Furthermore if the number neutrons is equal to the number of protons then no additional neutron-proton pair can be formed when either a neutron or a proton is added. Thus the change in binding energy when a nucleon is added is due, at least in part, to the the strong force interaction of the rest of the nucleons with the added nucleon. The positioning and/or arrangement of the other nucleons might also change as a result of the presence of the added nucleon.

Here are the data for the nuclides with equal even numbers of neutrons and protons.

The Effect on Binding Energy
n=p EAN
(MeV)
EAP
(MeV)
2 -0.885674 -1.965674
4 1.66539 -0.18511
6 4.946335 1.943572
8 4.143324 0.600274
10 6.761111 2.431341
12 7.33067 2.27131
14 8.47355 2.74811
16 8.64156 2.27654
18 8.7889 1.8578
20 8.3627 1.085
22 9.53 1.6143
24 10.582 2.085
26 10.6832 1.599
28 10.247 0.694
30 10.231 0.458
32 10.06 -0.05
34 10.22 -0.5
36 10.83 -0.6
38 11.64 -0.2
40 11.5 -0.6
42 11.4 -0.9
44 12 -1.1
46 12.2
48 12.6
50 11.3

Here is the graph of the above data. Let SFN(p, n) be the strong force interaction of the p protons and n neutrons with the additional neutron and let REN(p, n) be the change in binding energy due to any rearrangement of the p protons and n neutrons due to the additional neutron. SFP(p, n) and REP(p, n) have corresponding definitions for the effects of an additional proton. Thus

## The Relationship Between the Effects on Binding Energy due to the Strong Force Interaction of Additional Nucleons

The force between two particle with strong force charges of Q and q is presumed to be of the form

#### F = −Qqf(s)

where s is their separation distance and f( ) is a decreasing function of s, such as exp(−s/s0)/s². The effect of that force on potential energy is then of the form QqF(s). For one particle interacting with a structure of other particles this latter formula becomes

#### ΔE = ΣjQqjF(sj) = Q[ΣjQqjF(sj)]

The effect on binding energy is similarly proportional to the strong force charge of the affected particle.

Elsewhere it is found that if the strong force charge of the proton is taken to be +1 then the strong force charge of the neutron is −2/3. This then means that

#### SFN(p, n) = −(2/3)SFP(p, n)

Now suppose that REN(p, n) is equal to REP(p, n). The two equations for EAN and EAP take the form, with the functional dependence (p, n) suppressed,

#### EAN = −(2/3)SFP(p, n) + REPand EAP = SFP + REP

Their solution is immediate. Subtracting EAP from EAN gives

#### −(5/3)SFP = (EAN − EAP) and hence SFP = −(3/5)(EAN − EAP) = (3/5)(EAP − EAN) and SFN = (2/5)(EAP − EAN)

Adding two thirds of the second equation to the first equation gives

#### EAN + (2/3)EAP = (5/3)REPand hence REP = (3/5)[EAN + (2/3)EAP] = (3/5)EAN + (2/5)EAP

Here is a graph of (EAN − EAP) and (EAP + (2/3)EAN)11 The remarkable feature of the graph is the near linearity of (EAN − EAP) as a function of n=p, which leads to SFP and SFNP=REN.

The degree of linearity can quantified with regression analysis. The best fitting linear function of the number of neutrons (or the number of protons) is

#### EAN − EAP = 1.51980 + 0.273027n

The coefficient of determination (R²) for this equation is 0.9854. The t-ratio for the coefficient of n is 36.8. (The t-ratio is the ratio of the value of the coefficient to the standard deviation of its estimate.)

The above is predicated on the assumption that REN being equal to REP, but the linearity of the variables as functions of n=p also follows from REN being equal to REP plus a constant.

It is notable that the variable n=p with both even is none other than the number of alpha modules in the nuclide. The significance then of the linearity of the SFN and SFP is that the binding energies due to the additional neutron or proton per additional alpha module in the nucleus is a constant 0.273028 MeV independent of the number of alpha modules in the nucleus. The linearity of (EAN−EAP) or any multiple of it as a function of n=p could be just a coincidence but it is more likely that it represents a physical characteristic of nuclei. This would mean that REN is equal to REP plus a constant.

(To be continued.)