San José State University

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 The Interactive Binding Energies of Neutrons in Nuclides

## The Incremental Binding Energiesof Neutrons and Protons

The binding energies (BE) of 2930 nuclides have been measured. The incremental binding energies (ΔBE) can be computed according to the following definitions

#### Δneutron(n, p) = BE(n, p) − BE(n-1, p) Δproton(n, p) = BE(n, p) − BE(n, p-1)

The incremental binding energies for each nucleon represents the interactive binding energy of the last nucleon with all of the rest of the nucleons in the nuclide. The second differences, the increments in the incremental binding energies are even more interesting. The difference in the incremental binding energies of neutrons is an approximation of the interactive binding energy of the last neutron with the previous neutron, and likewise for the difference of the incremental binding energies of protons. These can be called the second differences in binding energy. For analytical justifications of these propositions see Neutrons and Protons. For their empirical verification see Neutrons and Protons.

On the other hand, the cross differences give an exact measure of the interactive binding energy between the last neutron and the last proton in the nuclide. This is positive, reflecting the fact that the force between a neutron and a proton is an attraction.

Consider the following graphs of the incremental binding energies of neutrons. (The data for the incremental binding energies of protons will be dealt with elsewhere.)  The odd-even sawtooth pattern has to do with the binding energies involved with the formation of neutron-neutron pairs.

The fact that the curves for the different elements nest together so neatly has certain important implications.

• The slopes of the relationships, which has to do with the interactions of the last two neutrons, are independent of the number of the other nucleon in the nuclide.
• The fact that, abstracting away the effect of pair formation, the relationships are piecewise linear means that the interaction energies are constant within a shell or subshell.
• The fact that the slopes are negative is verification that the force between two neutrons is a repulsion.

## A Test of the Degree of Parallelism of the Relationships

Although visually the above displays appear to be parallel an empirical verification requires looking a the differences between IBEn's for the elements. That information is displayed below.

 Number ofNeutrons ΔnBE(Se)−ΔnBE(As)(MeV) ΔnBE(Br)−ΔnBE(Se)(MeV) ΔnBE(Kr)−ΔnBE(Br)(MeV) 32 1.2 33 0.06 1.33 34 2.84 -0.23 1.2 35 -0.09 3.08 -0.1 36 1.44 -0.61 3 37 -0.11 1.37 0.23 38 1.252 -0.41 0.89 39 -0.016 1.449 0.304 40 1.2695 -0.1635 0.905 41 0.0527 1.1935 0.043 42 0.9099 -0.1381 1.043 43 0.0904 0.8712 0.064 44 0.7999 0.1892 0.8349 45 -0.009 0.9296 -0.0202 46 1.0228 0.2429 0.8093 47 0.148 0.892 -0.128 48 0.7897 0.3093 0.935 49 -0.0399 1.0199 0.283 50 1.051 0.225 0.951 51 0.229 0.551 0.4154 52 0.685 0.104 0.7646 53 0.215 0.835 0.156 54 0.36 0.55 0.397 55 0.35 0.36 0.313 56 1.2 0.07 0.575 57 0.54 0.165 58 0.14 0.67 59 0.2

For two curves to be parallel their difference would have to be constant. In the case of the incremental binding energies of neutrons in the isotopes of Arsenic and Selenium the differences for the neutrons in the fifth shell (29 through 50 neutrons) the differences are roughly constant. There is greater differences for the even numbers of neutrons because that difference includes the difference in the binding energy associated with the formation of neutron pairs whereas the differences for the odd numbers of neutrons do not. When the successive differences in the incremental binding energies for the isotopes of Arsenic, Selenium, Bromine and Krypton are plotted together the result seems to be chaos. However if the values for the differences for Bromine and Selenium are shifted back one unit and those of Krypton and Bromine are shifted back two units the results are very orderly. This appears to mean that the differences are a function of the number of neutrons minus the number of protons. Thus if n is the number of neutrons and p the number of protons then