San José State University
Thayer Watkins
Silicon Valley
& Tornado Alley

The Statistical Explanation of the Binding
Energies of Nuclides Based Upon a Three-Way
Classification of Nucleon Shell Occupancies
and Allowing for a Possible Error in the
Mass of the Neutron

A regression model which is an outgrowth of the Alpha Module Model of nuclear structure preforms wonderfully well in explaining the binding energies of 2931 nuclides. It explains 99.99434 percent of the variation in binding energy of the nuclides based upon the numbers of the three types of nucleon spin pairs and the three types of nucleonic (strong force) interactions for a three-way shell classification. The coefficient of variatio in error of the regression estimates is 0.355 of 1 percent.

There is a little bit of room for the improvement of the statistical performance of the model. The problem is to modify the regression equation by adding new variables that are consistent with the model. This is not easy to do. However there is one modification that is outside the model i.e., the possibility of an error in the mass of the neutron. The binding energy of a nuclide is its mass deficit expressed in energy units via the Einstein formula of E=mc². The mass deficits of all nuclides except one are the difference between the masses of its constituent nucleons and its measured mass. The one exception is for the deuteron. This case is how the mass of a neutron is computed. When a proton and a neutron form a deuteron a gamma ray of energy 2.224573 million electron volts (MeV) is emitted. When deuterons are irradiated with gamma rays of energy at least 2.224573 MeV deuterons are broken up into their constiuent nucleons. The conventional estimate of the mass of a neutron is based on the assumption that the binding energy of a deuteron is 2.224573 MeV. This value expressed as a mass defit, the measured masses of the deuteron and the proton give a value for the mass of a neutron.

The problem with this procedure is that in the analogous situation for electrons, if an electron moves from a higher orbit state to a lower one a photon is emitted but the decrease in potential energy is exactly twice the energy of the photon because an equal amount of potential energy is converted into kinetic energy for the electron in the lower potential energy state. The equality of the increase in kinetic energy with the energy of the emitted photon is because the electrons are subject to the inverse distance squared law of electrostatic fprce.

The nuclear strong force drops off faster than inverse distance squared so the above equality does not prevail, but the change in potential energy should be on the order of magnitude of 2.224573 MeV.

That there is an error in the mass of the neutron is obvious because there is one nuclide for which the binding energy is negative. That nuclide is the isotope with four protons and one neutron, Be 5, whose binding energy is computed with the conventional mass of the neutron to be −0.768

Therefore the error in the binding energy of any nuclide is proportional to the number of neutrons it contains. This matter is taken into account by including the number of neutrons as one of the explanatory variables in the regression equation. The resulting regression equation has a coefficient of determination (R²) of 0.9999544. The standard error of the estimate is 3.42 MeV, which with an average binding energy of 1072 MeV corresponds to a coefficient of variation of 0.32 of 1 percent. These are only small improvements over the previous case but there is room only for small improvements.

The other item of interest is the coefficient for the number of neutrons. It is 6.244 MeV. Its magnitude is too large and it is of the wrong sign, but its t-ratio is 18.3. The wrong sign is crucial; it is inconsistent with the model.

An alternate ploy may be useful. Since it is known that the error in the neutron mass is of the same order of magnitude as the energy of the gamma ray created upon the formation of a deuteron let it be assumed that the error in the neutron mass is equal to that value and so adjust the binding energies,

BE' = BE + 2.22457*n

The regression of BE' on the variables of the model gives an equation with a coefficient of deterimination of 0.9999604 and a coefficient of variation of the errors in the regression equation of 0.30 of 1 percent. This looks good, but an estimate of the neutron mass error of ten times this looks even better. The coefficient of determination for that regression is 0.99998 and the coefficient of variation is 0.224 of 1 percent. What is apparently happening is when a multiple of the number of neutrons is added the LHS the result looks a bit more like the sum of the neutron numbers on the RHS of the equation and thus there is a better correlation.

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