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The Effects of Changing the Type of
One or Two Nucleons on Binding Energy

Previous work on the binding energies of nuclides relied upon the concept of the incremental binding energies of nucleons. If p and n denoted the number of protons and neutrons in a nuclide then the incremental binding energies are defined as

IBEn = BE(n, p) − BE(n-1, p)
and
IBEp = BE(n, p) − BE(n, p-1)

Another construction that provides some insights is the effect of changing the type of one nucleon; i.e.,

BE(n, p) versus BE(n+1, p-1)
and
BE(n, p) versus BE(n-1, p+1)

The first is the effect of changing one proton into a neutron and the second that of changing one neutron into a proton. Here is an example. The binding energy of Helium 9, with two protons and seven neutrons, is 30.26 million electron volts (MeV) while that of Lithium 9, with three protons and six neutrons, is 45.3409 MeV. The ratio is 0.6674. This is the effect of changing one of seven neutrons in Helium 9 into a proton. The increase in binding energy can be attributed to having a better balance between neutrons and protons in Lithium 9 than in Helium 9.

The potential energy V involved in the interaction of two particles of charges q1 and q2 separated by a distance of s1,2 is of the form

V = q1q2W(s1,2)

In the following the binding energies due to the formation of spin pairs is ignored. Later these will be considere.

The binding energy due to the strong force for a collection of nucleons is of the form

BE = ΣΣi≠jqiqjW(si,j)

This can be partitioned into that which involves a particular particle, say number 1, in the arrangement and that which does not; i.e.,

BE = Σj≠1q1qjW(si,j) + ΣΣi≠1, j≠iqiqjW(si,j)
and, equivalently
BE = q1Σj≠1qjW(s1,j) + ΣΣi≠1, j≠iqiqjW(si,j)

Now we can consider changing the charge of the number 1 particle from q1 to q'1 which would result in

BE' = q'1Σj≠1qjW(s1,j) + ΣΣi≠1, j≠iqiqjW(si,j)

The second terms on the right of the above equations are just the binding energies of the nuclides without the number 1 nuclide.

Thus if the nuclide being considered has n neutrons and p protons and we considering changing one neutron to a proton

BE(n, p) = qnΣj≠1qjW(s1,j) + BE(n-1, p)
and
BE(n-1, p+1) = qpΣj≠1qjW(s1,j) + BE(n-1, p)

These equations lead to the equation

qn/qp = [BE(n, p) − BE(n-1, p)]/[BE(n-1, p+1) − B(n-1, p)]

Previous studies indicate that qn/qp is equal to −2/3 or possibly −3/4. Therefore one of the two terms in the ratio must be negative.

The matter of spin pair formation may be taken into account by considering the four cases of the odd-eveness of the neutron and proton numbers separately. However a simpler approach is to consider changes of two nucleons. A change of two neutrons into two protons results in a disappearance of one neutron-neutron pair and the formation of one proton-proton pair. Since there is evidence that these two pairs involve the same effect on binding energy there would be no net effect from the change in that matter. However, there could be some effect effect concerning the formation of neutron-proton pairs. This would depend upon the relative number of neutrons and protons. For example, in a nuclide with 10 protons and 12 neutrons the extra two neutrons are unpaired. When those two neutrons are changed into protons the two protons would also be unpaired. Thus when n=p+2 the following equation would prevail

qn/qp [BE(n, p) − BE(n-2, p)]/[BE(n-2, p+2) − B(n-2, p)]

However if there is some phenomenon that contributes to binding energy that is not accounted for, its effect will result in the the above ratio not being equal to qn/qp. Suppose the equation are

qn/qp = [BE(n, p) − BE(n-1, p)]/[BE(n-1, p+1) − B(n-1, p)] + Q
and
BE(n-1, p+1) = qpΣj≠1qjW(s1,j) + BE(n-1, p) + Q

where Q is the effect of the unaccounted factor on binding energy.

The the ratio of the differences would be

[BE(n, p) − BE(n-1, p) + Q]/[BE(n-1, p+1) − B(n-1, p) + Q]

which would not be equal to qn/qp unless Q=0.

One such unaccounted for effect is the binding energy that would result from the formation of alpha particles. The effect of alpha particle formation can be eliminated by computing the binding energies of nuclides which is in excess of that due which can be accounted for due to the formation of alpha particles.

XSBE(n, p) = BE(n, p) − 28.295674*α

where α is equal to the number of alpha particles which can be formed from n neutrons and p protons.

Consider the thorium isotope Th220 which has 90 protons and 130 neutrons. Its excess binding energy is 417.31867 MeV. whereas that of U218 is 403.47267 MeV and their difference is 13.846 MeV. On the other hand the excess binding energy of the nuclide with two more protons and two less neutrons, the uranium isotope U220 is 379.098996 MeV. When the excess binding energy of the nuclide with 90 protons and 128 neutron is subtracted from 379.098996 MeV the result is −24.373674 MeV. The ratio of 13.846 MeV and −24.373674 MeV is −0.56807.

The same computation starting out with the radium isotope Ra220 gives a ratio of −0.57594. Starting with radon isotope Rn210 the resulting ratio is −0.67677. Starting with the polonium isotope Po200 the ratio −0.69984 results.

For smaller nuclides the ratio is not close to −2/3. For example, starting with the tin isotope Sn110 the ratio which results is −0.85188. Starting with the zirconium isotope Zr90 the ratio is −1.2395. For the zinc isotope Zn70 the ratio is −1.19269. For smaller nuclides the ratio deviates even more drastically, such as −1.76747 for the calcium isotope Ca50. For a smaller nuclide such as the neon isotope Ne30 the ratio changes sign to +0.80846. It does not switch signs for all of the lighter nuclides. For example, for the helium isotope He8 the ratio is negative.

So the phenomenon is more complicated than the model allows for. However the fits the data for the heavier nuclides and all but the lightest nuclides the ratio is negative.

Examination of the Detailed Effects

The Increments in Excess Binding Energy for Changes of
Two Nucleons for the Isotopes of Radon (p=86) and Radium (p=88)
Number of
Neutrons
XSBE(n,86)-XSBE(n-2,86) XSBE(n-2,88)-XSBE(n-2,86) Ratio
117 18.19 -26.455674 -0.68756517
118 17.87 -26.065674 -0.68557598
119 17.7 -25.685674 -0.68910008
120 17.27 -25.285674 -0.682995438
121 17.02 -24.995674 -0.680917826
122 16.631 -24.545674 -0.677553201
123 16.47 -24.165674 -0.681545236
124 16.098 -23.786674 -0.676765486
125 15.946 -23.515674 -0.678100913
126 15.202 -23.129674 -0.657250941
127 13.085 -22.811674 -0.57360981
128 11.805 -22.475674 -0.525234527
129 11.615 -21.948674 -0.529189144
130 11.567 -21.330674 -0.542270722
131 11.312 -20.775674 -0.544482937
132 11.179 -20.113674 -0.55579105
133 10.964 -19.449674 -0.563711248
134 10.742 -18.773674 -0.572184219
135 10.568 -17.847674 -0.592121976
136 10.381 -17.422674 -0.595832764
137 10.22 -16.547674 -0.61760946
138 10.115 -16.169674 -0.625553737
139 10 -15.382674 -0.650082034
140 9.8 -14.985674 -0.653957907
141 9.6 -14.424674 -0.665526306
142 9.4 -13.916674 -0.675448746

The graphs of the data, shown below, indicate a regular pattern. Note however that [XSBE(n-2,88)-XSBE(n-2,86)] is plotted versus n so that for it what appears to be occurring at n is occurring at n-2.

The abrupt drop at 126 neutrons is due to the filling of a neutron shell and the subsequent neutrons going into a higher shell. There is a constancy of the slopes within the shells.

There is a near constancy of the ratio for the neutron shell up to and including 126. The average value of the ratios is -0.67241.

The case for the isotopes of tin (p=50) and tellurium (p=52) replicates the constancy of the slopes of the relationships of the increments except at the point where the shell is filled at 82 neutrons.

The ratios however involve no constancy whatsoever.

The case for the isotopes of krypton (p=36) and strontium (p=38) also has near constancy of the slopes of the relationships except at the point where the neutron shell is filled at 50 neutrons. What is different here is the extreme jump at the point where the number of neutrons equals the number of protons. This is where the addition of a neutron after that point no longer involves the formation of a neutron-proton spin pair.

The case for the isotopes of chromium and iron confirms the pattern of the previous example. There is an extreme jump in the level where the number of neutrons equals the number of protons, 24. The drop at the point where the neutron shell is filled, n=28, is not as great as the previous cases but it is there.

Although the graph below for the isotopes of magnesium and silicon looks more complicated it fits the pattern of the previous cases. The only thing that is different is the extreme jump comes for the red line at a value one beyond the value where the number of neutrons equals the number of protons.

Conclusions

The proposition that the ratio of particular increments in the excess binding energies of nuclides is constant for all nuclides is not borne out. In particular it is not always equal to −2/3 or even that it is always negative, although generally it is. What is found for the cases considered is that the relationships between the increments in excess binding energies for two nucleon changes and the number of neutrons in the nuclide are piecewise linear. They are linear for ranges of neutrons within a shell. This means that the second differences and the cross differences with respect to the number of neutrons are constant with the neutrons shells.


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