Neutrons binding subject to the nuclear strong force should form free pairs in the same manner as neutrons and protons forming deuterons. If this occurs there should be a gamma ray emitted. In the formation of a deuteron a 2.22457 MeV gamma is emitted.

Free neutrons decay into protons and electrons with a reaction half-time of about 15 minutes. From this decay a weak gamma ray with energy of about 0.76 MeV is emitted.

Since empirical measurements of gamma rays associated with the formation of neutrons pairs are apparently not available an alternate procedure will be use. Binding energies are available for almost three thousand nuclides. A data set for the binding energies of nuclides that could contain an integral number of alpha particles plus two neutrons. The binding energy in excess of that which is accounted for by the formation of alpha particles is displayed below.

The significant relationship is the second of the ones above. The procedure to be used is to extrapolate a first linear section of the relationship down to the number of alpha particles being zero. This is done through regression analysis. However, clearly from the above graph the data for the first four alphas do not fit the pattern. The regression is carried out for the number of alphas being from 5 to 14.

The resulting regression equation is C

The magnitude of this intercept, which indicates that a neutron pair has a binding energy of about 14 MeV is unexpected because the binding energy of a neutron-proton pair is only 2.2 MeV. But this result is not impossible.

For four neutrons the results are as follows:

ΔBE = 10.11524 + 0.82892#α
[0.897]
R² = 0.8924

Again the procedure produces an egregious result. The value for the binding energy of a neutron quartet, 10.1 MeV, is not only smaller than for a neutron pair, it is much less than the value for an alpha particle, two neutron-proton pairs.

The procedure can be tested by applying it to the corresponding data for neutron-proton pairs (deuterons). The binding energy of the deuteron is believed to be 2.22457 MeV.

ΔBE = 12.38912 + 0.18697#α
[0.269]
R² = 0.3832

The results indicate that the procedure is not a suitable way to estimate the binding energy of a free-standing deuteron.

Interestingly enough if regression is applied to the first three data points the result is

ΔBE = 2.94589 + 2.23985#α
[0.803]
R² = 0.617

The intercept in this case is a reasonable approximation of the binding energy of the deuteron.

The procedure cannot also be tested by treating an alpha particle in the same manner as was done for the neutron-proton and neutron-neutron pairs. The results are

ΔBE = 36.14064 − 0.05181#α
[-0.023]
R² = 0.0088

In this case the procedure gives a reasonable estimate of the binding energy of an alpha particle. d

(To be continued.)