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The Distribution of the Enhancements to
Binding Energy Due to the Formation of
Neutron-Neutron Spin Pairs and
Neutron-Proton Spin Pairs

The purpose of this webpage is to demonstrate that whenever possible the neutrons in a nucleus form spin pairs. That is to say, when the number of neutrons in a nucleus is even then they are all involved in spin pairs. It is also true that the same proposition applies to the protons in a nucleus. The propositions for neutrons and for protons can be demonstrated separately. However it is also true that whenever posslible the neutrons and protons in a nucleus form neutron-proton spin pairs. If the number of neutrons is less than the number of protons then the addition of another neutron will form a neutron-proton spin pair and if the number of neutrons is greater than the number of protons a neutron-proton spin pair will not be formed. Thus if the numbers of neutrons and protons are equal and equal to an odd number then the incremental binding energy of another neutron could possibly be negative. This would not negate the proposition that neutrons whenever possible form spin pairs.

BACKGROUND

The binding energies of 2931 nuclides are available as a function of the number of neutrons and protons they contain, BE(n, p). The binding energy of a nuclide is its mass deficit expressed in energy units via the Einstein equation, E=mc². The masses of nuclei are generally less than the masses of their constituent nucleons.

The incremental binding energy of a neutron (IBEN) is defined as

IBEN(n, p) = BE(n, p) − BE(n−1, p)

The IBEN of a nuclide represents the effect of an additional neutron on the energy and structure of a nucleus. If the number of protons is less than the number of neutrons then the addition of another proton will form a neutron-proton spin pair and if the number of prot ons is greater than the number of neutrons then no neutron-proton spin pair will form.

It is be emphasized that the term the binding energy associated with the formation of a spin pair is used to indicate that there are two components involved. One is the binding energy resulting from the formation of the spin pair itself. The other is the change in binding energy that results from any rearrangement of the nucleons as a result of the formation of the spin pair. To get a notion of the relative magnitudes of these two components, consider the case of the deuteron which is simply a neutron-proton spin pair. Its binding energy. is about 2.2 MeV and there is no other nucleons to experience rearrangement. Thus the binding energy due the formation of a neutron-proton is about 2.2 MeV. A review of the binding energies associated with the formation of a neutron-proton is on the order of 10 MeV in other situations so the energy resulting from the rearrangements is about 8 MeV.

There are a few cases in which the incremental binding energy of a neutron is negative, notably for the exotic isotopes of hydrogen.

The Binding Energies of the Isotopes of Hydrogen
Isotope Number
of Neutrons
Binding
Energy
IBEN
1H 0 0 0
2H 1 2.224573 2.224573
3H 2 8.481821 6.257248
4H 3 5.58 -2.901821
5H 4 2.7 -2.88
6H 5 5.8 3.1

There is also the special case of Beryllium 6 in which the binding energy is a negative value. This is most likely due to an error in the mass of the neutron which is used in the calculation of the mass deficit of nuclides. This makes the IBEN of Be6 negative and hence suggests that there is no enhancement of BE due to the formation of a neutron-neutron spin pair in Be6. This is not necessarily true in as much that there are extraneous factors which can account for the negative value.

(To be continued.)

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