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The Net Nucleonic Charge Explaining the Incremental
Binding Energy of Protons Due to Their Interaction
Through the Strong Force in Nuclides

Previous studies have developed evidence that the incremental binding energy (IBE) of a proton or neutron is composed of that which is due to the interaction of nucleons through the strong force and that which is due to the formation of spin pairs. By computing the IBE over a change of two nucleons instead of one and considering the evidence only for even-even or odd-odd nuclides the effect of the formation of neutron-neutron pairs and proton-proton pairs can be eliminated. This modified IBE still contains the effect of the formation of neutron-proton pairs. In previous studies it was found that this modified IBE for protons and for neutrons is a function of the difference in the number of protons and neutrons. For example,

IBEp' = f(z)
where z=p-n

What is explored here is the refinement that

IBEp' = g(ζ) + h(z)

where z=p-n but ζ=(p-1)-νn. The component h( ) is due to the formation of neutron-proton pairs and is a function of the difference z, but the component g( ) is due to the interaction through the strong force. The quantity ζ must differ from z because a nucleon cannot interact with itself. The strong force charge of the neutron may be different from the strong force charge of the proton.

A previous study developed an estimate of the strong force charge of the neutron relative to that of a proton as being about 0.7. It seemed obvious that the ratio would have to be a simple fraction such as (2/3) or (3/4). Two thirds seemed the more intuitive choice.

The value of ν can be estimated through the use of a powerful spread sheet such as EXCEL to find the value of ν which maximizes the coefficient of determination (R²) for the regression that explains the IBE's. Using the data for the IBE of protons in the odd-odd nuclides it is found that ν equal to 0.7638 maximizes the R² to 0.977829057. This is in contrast to a value of 1.0 for which the R² is 0.966385. Thus the nucleonic charge of a neutron appears to be three quarters of that of the proton.

For the IBE of protons in the even-even nuclides the R² maximizing value of ν is 0.7885. The value of R² is 0.958364654.

For the IBE' for neutrons in the even-even nuclides the R² maximizing value of ν is 07554 which gives a value of R² of 0.954904957. For the odd-odd nuclides R² is maximized at 0.956170899 when ν is equal to 0.7614.

Estimates of the Nucleonic (Strong Force)
Charge of a Neutron Relative to that of a Proton
 even-even
nuclides
odd-odd
nuclides
IBE' of Protons0.78850.7638
IBE' of Neutrons0.75540.7614

The probable value for ν is then 3/4.

The Preponderance of Neutrons in the Stable Nuclides

The previous results help to explain the preponderance of neutrons in stable nuclides. For example, the stable U238 isotope of uranium has 92 protons and 146 neutrons. If the effects of the formation of nucleonic spin pairs are ignored for the moment one can say that there would be a net attraction of a nucleus for a proton only is the number of neutrons is greater than (4/3)p. For uranium with p=92 this means the number of neutrons would have to be greater than 123.

On the other hand there would not be a be net attraction for a neutron unless the number of protons is greater than (3/4)n. So the line p=(3/4)n+1 would be some sort of dividing line between the different types of instability.

If a proton were to be added to a nucleus of p protons and n neutrons and both p and n are even there would be no neutron pairs or proton pairs formed. If furthermore the value of p is more than n then no additional neutron-proton pair will be formed. Then clearly p>(3/4)n so the addition of another proton would produce an unstable nuclide. Let the binding energy due to ζ be kζ. That is to say k is the binding energy per unit net nucleonic charge. If Enp is the binding energy associated with the formation of a neutron-proton pair then there would be a net attraction of a proton if

k((3/4)n - (p-1)) + Enp > 0

Consequently n would have to satisfy the condition

n > (4/3)[p-1-Enp/k]

If p is an odd number then the addition of another proton would form a proton-proton pair. If Epp is the binding energy of a proton-proton pair then the condition for there being a net attraction for a proton is

k[(3/4)n − (p-1)] + Enp + Epp > 0
which means that
n > (4/3)[p-1-Enp/k-Epp/k]

This constitutes the minimum level for the number of neutrons, unless it is negative. The minimum is then the larger of the above value and zero.

On the other hand there would be a net attraction for another neutron to be added to an odd neutron nuclide in which n<p if

k[p − (3/4)(n-1)] + Enp + Enn > 0
which means that
n < (4/3)[ p + Enp/k + Enn/k] + 1

where Enn is the energy associated with the formation of a neutron-neutron pair. This is the maximum for the neutron number as a function of the proton number.

Thus a range of neutron numbers which allow stability is established based upon the odd or evenness of p and n and the relative values of p and n and the values of Enp, Epp and Enn. Of course the value of ν is also crucial.

Below is a plotting of the maximum and minimum neutron numbers as a function of the proton number.

The scatter diagram of the actual proton and neutron numbers is shown below.

A notable correspondence.

The Nucleonic Charges of Quarks that would
Determine the Charges of the Nucleons

A proton is composed of two up quarks and a down quark; a neutron of one up quark and two down quarks. Let u and d be the nucleonic charges of the up and down quarks, respectively. If the charge of the proton is +1 and that of a neutron −3/4 then

2u + d = +1
u + 2d = −3/4

Adding the two equations together gives 3(u+d)=1/4 and hence u+d=1/12. Subtracting this equation from the first equation gives u=11/12. Subtracting it from the second equation gives d=−5/6. These are surprising values but they are what the data indicates.

A value of −2/3 for the relative charge of a neutron would lead to u=8/9 and d=−7/9, equally surprising values.

If the charge of the neutron is taken as +1 and that of the proton as −4/3 then u=−11/9 and d=10/9.

Conclusions

Not only is the strong force charge of the neutron opposite in sign to that of the proton but it is smaller in magnitude. This difference in magnitude accounts for nuclides, especially heavier ones, having many more neutrons than protons. A proton is attracted to the neutrons of a nucleus but repelled by the protons. It takes four neutrons to counterbalance the effect of three protons.


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