﻿ A Revision of the Estimates of the Parameters of the Nuclear Force Formula based upon the Difference in the Nucleonic Charge of the Neutron Compared to the Proton With the Results Applied to the Estimation of the Binding Energy of the formation of a Neutron-Proton Spin Pair

San José State University

applet-magic.com
Thayer Watkins
Silicon Valley
USA

 Revision of the Estimates of the Parameters of the Nuclear Force Formula based upon the Difference in the Nucleonic Charge of the Neutron Compared to the Proton With the Results Applied to the Estimation of the Binding Energy of the formation of a Neutron-Proton Spin Pair

## Derivation of Estimates of the Parameters of the Nuclear Strong Force

The nuclear strong force between two particles of nucleonic charge Z1 and Z2 is assumed to be given by the formula

#### F = HZ1Z2e−s/s0/s²

where H and s0 are parameters to be estimated. A value of s0 can be derived from the Yukawa Relation and the mass of the pi mesons. Its value is 1.522 fermi. The sign of F is given by the sign of Z1Z2. If it is negative the force is an attraction; if positive the force is a repulsion. The sign is negative only if Z1 and Z2 are of opposite signs. If Z1 and Z2 are both of the same sign, either positive or negative, then the sign of F is positive and it is a repulsion.

Previous analysis was based upon the presumption that the nucleonic charges of a proton and neutron were equal. Now it is known that they are of opposite sign and the value for the neutron is two-thirds of that of a proton. The nucleonic charge of the proton can be defined as +1. This make the nucleonic charge of a neutron −2/3 and the product of the charges is also −2/3.

What was found in the previous analysis was, in effect, the value of H|Z1Z2| where |Z1Z2|=2/3. Therefore

## The Potential Energy Function

For a force F which is a function of distance s the potential energy function V(s) is

#### V(s) = −∫s∞F(x)dx

For the force formula given previously this evaluates to

#### V(s) = −∫s∞(HZ1Z2*exp(−x/s0)/x²)dx

If the variable of integration x is replaced by z=x/s0 then the above expression becomes

#### V(s/s0) = −(HZ1Z2/s0)∫s/s0∞exp(−z)(dz/z²) or, equivalently V(s/s0) = −(H/s0)W(s/s0)

where W(s/s0)=∫s/s0exp(−z)(dz/z²), a dimensionless function.

Thus the parameter which is crucial for the potential energy of a deuteron is (2/3)(H/s0). Its value is 1.9906×10-12 joules or in units of millions of electron volts 12.2424 MeV.

## The Separation Distance of Nucleons in a Deuteron

A group of physicists under the editorship of Savely G. Karshenboin published in 2008 a book devoted to the compilation of the best estimates of physical properties of particles, simple atoms and simple molecules (Precision Physics of Simple Atoms and Molecules, Springer-Verlag).

The best estimate of the root-mean-square (rms)-charge diameter of a deuteron from page 70 of the above mentioned work is 4.260 fermi with a margin of error of ±0.02 fermi. The recommended estimate of the rms-charge radius of the proton, given on page 49 of the above work, is 0.895 fermi. Precision Physics of Simple Atoms and Molecules does not give an estimate for the radius of the neutron. Another source gives the rms-radius of the neutron as 1.11 fermi.

Thus the separation distance of the centers of the nucleons is

#### s = 4.260−0.895−1.113=2.252 fermi.

The ratio of this distance to the scale parameter s0 is 2.252/1.522=1.48. The potential energy due to the strong force in a deuteron is then

#### V(1.48) = −(2/3)H*W(1.48)

The value of W(1.48) is approximately 0.100714, a pure number.

## The Binding Energies of Nuclides

The mass of nuclide is less than the sum of the masses of its constituent protons and neutrons. This mass deficit when expressed in energy units is called its binding energy. When a deuteron is formed there is an emission of a gamma ray with energy of 2.224573 MeV. The conventional estimate of the mass of a neutron is based upon the assumption that this is the binding energy of the deuteron. The mass of the neutron is deduced from the measured mass of a proton and a deuteron. This however involves a conceptual error. Generally when a quantum-mechanical system loses energy part of that loss goes into increased kinestic energy and part into the energy of an emitted photon. For the electrons in an atom there is an exactly equal division of the potential energy loss into increased kinetic energy and the photon energy. For the nuclear strong force the division will not necessarily be equal, as it is for the electrostatic force. A previous study, using the Virial Theorem, concludes that the binding energy of the deuteron is underestimated by 0.98638 MeV and consequently the mass of the neutron is underestimated by the same amount.

Nuclei are held together, in part, by the strong force attraction between neutrons and protons. The strong force between neutrons is a repulsion. Likewise the strong force between protons is a repulsion as well as the electrostatic force. Nuclei are also held together by the formation of spin pairs of nucleons. For small nuclides the energies involved in the spin pair formation is larger than that due to the strong force.

The binding energies of nuclides could depend upon the binding energies for the three types of pairs and the parameters of the potential energy function. Let Pnp denote the binding energy created by the formation of a neutron-proton pair. The binding energies may be error due to an error in the mass of the neutron. Let Δ denote the error in the mass of the neutron expressed in MeV.

For a deuteron the equation to be satisfied is

## Conclusion

The values of the parameters H and s0 in the formula for the nuclear strong force are

#### H = 4.5446×10-27 kg m³/sec²and s0 = 1.522×10-15 m

The relevant parameter in the potential energy function for the force between particles of nuclear charges Z1 and Z2 is (H/s0)|Z1Z2|. For the deuteron |Z1Z2| is equal to (2/3) and

#### (H/s0)|Z1Z2| = 12.2424 MeV

The enhancement in binding energy due to the formation of a neutron-proton spin pair is 1.98671 MeV. In contrast, the binding energy in the deuteron due to the strong force is only 1.22424 MeV.