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One simple and plausible approach to explaining the binding energies of nuclides is in terms of the energy of bonds between the nucleons in a nuclide. The presumption is that the mass deficits of nuclides have something to do with the potential energy and the potential energy of a combination of nucleons is determined by the sum of the energies of the nucleon pairs in the combination. An examination of the binding energy per bond reveals great differences among the nuclides. For examples, consider the data for the simplest nuclides.
| Nuclide | Symbol | Binding Energy MeV | Bonds | Binding Energy per Bond MeV |
| deuteron | H² | 2.224573 | 1 | 2.22457 |
| tritium | H³ | 8.481821 | 3 | 2.827274 |
| helium 3 | He³ | 7.718058 | 3 | 2.572689 |
| alpha particle | He4 | 28.295674 | 6 | 4.715949 |
The difference in binding energy between the H² nuclide and the He³ nuclide might possibly be explained entirely by the electrostatic repulsion between the two protons in He³. If the difference between the binding energies corresponds to a difference in potential energy then the separation distance between the two protons can be computed. It is 1.884 fermi. See Structure of Tritium. This is a surprisingly low value given that the separation distance between the nucleons in a deuteron is between 3 and 4 fermi. This means that the difference in binding energy between the triteron and He3 is not due only to the repulsion of the protons in He3. In addition to that factor there is likely a significant difference in structure.
The masses of charged particles can be measured by ejecting them into a magnetic field and measuring the radius of curvature of their trajectoris. A deuteron disintergrates into a proton and a neutron when subjected to a gamma ray of energy at least 2.224573 MeV. Gamma rays of this energy are emitted when protons and deuterons combine to form deuterons. This value is taken to be the mass deficit of the deuteron. From this value and the masses of the deuteron and proton the mass of the neutron is deduced. This value may however be in error.
For transitions of electrons in atoms the loss of potential energy is evenly divided between the energy of the emitted photon and the increase in kinetic energy of the electron. The nuclear force is different from the electrostatic force so the division between the change in kinetic energy and the energy of the photon may not be even. However if it were the mass deficit of the deuteron would be on the order of 4.45 MeV and the mass of the neutron would be about 2.225 MeV higher than the accepted value. The binding energies of nuclides would higher by 2.225 MeV times the number of neutrons they contain.
Suppose the mass of the neutron is Δ higher than the accepted value. The binding energies per bond for the deuteron, the tritium nuclide and the alpha particle would then be increased as follows:
We may now ask what Δ has to be in order for the binding energy per bond to be the same for the two nuclides. The graph shows the relationships.
The answer is (4.715949-2.224573)/(2/3)=3.737059 MeV. The equal binding energy per bond for the two nuclides is then 5.961632 MeV.
Using the above value for Δ the binding energy per bond for tritium is then 5.318646333 MeV. Close but off by about 10 percent. However the structure of tritium may not be that of an equilateral triangle. The tritium nuclide may involve a linear form with the proton position halfway between the two neutron. If the force were electrostatic then the weighted number of bonds would be 2.5. Using this value for the number of bonds the binding energy per bond, using the value of Δ as 3.737059 the binding energy per bond for tritium would be 6.3823756 MeV this means that for some value for the number of bonds between 2.5 and 3.0 the binding energy per bond for tritium woud be the same as for the deuteron and the alpha particle.
If the mass of the neutron is higher by 3.737059 MeV than the accepted value then one anomaly in binding energy is solved; i.e., Be5 has a negative binding energy. The binding energy for Be5 would then be 3.737059−0.75=2.987059 MeV.
Just for the record the value of Δ which makes the binding energy per bond equal for the deutron and the tritron is 1.808112 MeV. And for the alpha particle and the triteron it is 3.996648 MeV. The average of these two values is 2.90238 MeV. The average of the three solution values for Δ is 3.18060633 MeV.
But because the alpha particle contains two repelling protons the binding energy per bond should not be the same as for the deuteron which does not. The difference in binding energy between the triteron and He3 as a function of the error in the mass of the neutron is then
If this same decrement prevailed for the alpha particle then the binding energy of the alpha particle without the effect of the repulsion between its two protons would have to 29.066915 MeV. The corresponding value as a function of Δ would be
To equate the binding energies per bond with that of the deuteron the value of Δ has to satisfy the equation
This would mean that the binding energy per bond is 7.46191 MeV. This is based upon the presumption that the effect of the repulsion on the structure of the alpha particle is the same as that for He3. It is likely that the effect for the alpha particle is much less than for the He3.
(To be continued.)
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