The mass of a nuclide is less than the sum of the masses of its constituent nucleons (neutrons and protons). When this mass deficit is expressed in energy units via the Einstein formula of E=mc² it is called the binding energy of the nuclide. This webpage is an attempt to explain the binding energies of four nuclides; the deuteron (H 2 nuclide), the triteron (H 3 nuclide), the He 3 nuclide and the alpha particle (He 4 nuclide).

The Underestimation of the Mass of the Neutron

The masses of charged particles can be measured. The mass of the neutron, a neutral particle, must be deduced. It is deduced from data concerning the deutron, the Hydrogen 2 isotope. When a deuteron is formed from a neutron and a proton a gamma photon is emitted. The energy of this photon is 2.224573 million electron volts (MeV). If deuterons are subjected to gamma rays of energy at least 2.224573 MeV they disassociate. The conventional estimate of the mass of the neutron is based upon the unwarranted assumption that the binding energy of the deuteron is exactly equal to the energy of the photon involved in its formation or disassociation. The binding energy of a nuclide is basically the change in the potential energy of the constituent nucleons. When the potential energy of a system decreases the energy can go into an increased kinetic energy or the energy of an emitted photon. In the case of the electrons in an atom the decrease in potential energy is equally divided between the increase in kinetic energy and the energy of the emitted photon. This follows from the electrostatic force being strictly inversely proportional to the square of the separation distance of the charges. The nuclear strong force has a different functionality so the division of the change in potential energy between kinetic energy and the energy of the photon is not equal, but the kinetic energy component is not zero. An analysis based upon the Virial Theorem and a plausible formula for the strong force indicates that the binding energy of the deuteron is 3.210953 MeV instead of 2.224573 MeV. This means that the conventional estimate of the mass of the neutron is 0.98386 MeV too low. Thus the correct binding energy of a nuclide is the conventional value plus 0.98386 MeV times the number of its neutrons.

The Enhancement of Binding Energies
Due to the Formation of Nucleonic Spin Pairs

The binding energies for a set of nuclides with the same number of protons and successive numbers of neutrons can be compiled. The difference in the binding energy of a nuclide with #n neutrons and one with #n-1 neutrons is the incremental binding energy (IBE) of the #n neutron. The IBE's for neutrons in the isotopes of tin (50 protons) are shown below.

The sawtooth pattern is due, in part, to the formation of neutron-neutron spin pairs. When the enhancements are computed the results are as follows.

The best estimate for the binding energy due to the formation of a neutron-neutron spin pair is the average value in the highest shell. For the display above that would be 1.53667 MeV.

The models of the H 3 (triteron) and He 3 nuclides are as shown.

      

The strong force nucleonic charge of a neutron is −2/3 when that of the proton is defined as +1. The neutron pair has a nucleonic charge of −4/3 and the product of this with the nucleonic charge of the proton is −4/3. The nucleonic charge of the proton pair in the He 3 nuclide is +2 and the product of this with the nucleonic −2/3 charge of the neutron is also −4/3. Thus the forces of attraction in the triteron and He 3 nuclide are exactly the same.

Thus, according to the models presented above, the difference in the binding energy of the Helium 3 nuclide and the triteron should be solely due to the difference in the binding energy due to the formation of a neutron-neutron pair compared to that of a proton-proton pair. The difference in the uncorrected binding energies of the two nuclides is 0.763763 MeV. For the corrected values it is 1.750143 MeV.

For the binding energy due to the formation of a neutron-proton spin pair the value to be used is 1.936573625 MeV. This is not too different from the value of 1.98671 MeV derived in a previous study. If the binding energy due to the formation of a neutron-proton pair is subtracted from the corrected binding energy of a deuteron the result is the binding energy due the strong force. That value is (3.210953−1.936573625)=1.265379375 MeV. The values for the other nuclides are shown in the table below.

The binding energy due to the formation of a neutron-neutron pair is 1.51991625 MeV and for a proton-proton pair is 0.75615325 MeV. The difference between the value for a neutron-neutron pair and a proton-proton pair is 0.763763 MeV, exactly the same as the difference in the uncorrected binding energies of the triteron and the Helium 3 nuclide.

The quantum theory of cluster deuterons suggests that the binding energy due to the strong force should be roughly proportional to the fourth power of the geometric mean of the sizes of the clusters. For the deuteron in which both clusters are of size 1 the mean is 1 and its fourth power is 1. For both the triteron and the Helium 3 nuclide the geometric mean cluster size is √2 and its fourth power is 4. For the alpha particle both clusters are of size 2 and its fourth power is 16. In the above table the binding energies for both the triteron and the Helium 3 nuclide is 5.0615175 MeV. The ratio of this value to the binding energy for the strong force in the deuteron (1.265379375 MeV) is 4. The ratio of the binding energy due to the strong force in the alpha particle is 20.24607 MeV. The ratio of this value to the value for the deuteron is 16.