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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 |
The nuclear strong force between two particles of nucleonic charge Z_{1} and Z_{2} is assumed to be given by the formula
where H and s_{0} are parameters to be estimated. A value of s_{0} 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 Z_{1}Z_{2}. If it is negative the force is an attraction; if positive the force is a repulsion. The sign is negative only if Z_{1} and Z_{2} are of opposite signs. If Z_{1} and Z_{2} are both of the same sign, either positive or negative, then the sign of F is positive and it is a repulsion.
What was found in the previous analysis was, in effect, the value of H|Z_{1}Z_{2}| where |Z_{1}Z_{2}|=2/3. Therefore
For a force F which is a function of distance s the potential energy function V(s) is
For the force formula given previously this evaluates to
If the variable of integration x is replaced by z=x/s_{0} then the above expression becomes
where W(s/s_{0})=∫_{s/s0}^{∞}exp(−z)(dz/z²), a dimensionless function.
Thus the parameter which is crucial for the potential energy of a deuteron is (2/3)(H/s_{0}). Its value is 1.9906×10^{-12} joules or in units of millions of electron volts 12.2424 MeV.
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
The ratio of this distance to the scale parameter s_{0} is 2.252/1.522=1.48. The potential energy due to the strong force in a deuteron is then
The value of W(1.48) is approximately 0.100714, a pure number.
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 P_{np} 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
The values of the parameters H and s_{0} in the formula for the nuclear strong force are
The relevant parameter in the potential energy function for the force between particles of nuclear charges Z_{1} and Z_{2} is (H/s_{0})|Z_{1}Z_{2}|. For the deuteron |Z_{1}Z_{2}| is equal to (2/3) and
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.
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