San José State University

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 Nucleonic Balance and the Binding Energies of Nuclides

Nuclides that have an excessive number of protons compared to their numbers neutrons are unstable. This shows up as a relative low level of binding energy compared to nuclides with approximately equal numbers of protons and neutrons. This is quite understandable because an excessive number of protons means that the distances between them reaches the range where the electrostatic repulsion of the protons outweighs the strong force attraction between them. Having neutrons mixed in means the strong force attraction between the protons and neutrons counterbalances the eleetrostatic repulsion.

What is not easily understandable is that an excessive number of neutrons compared to the number of protons also results in relatively low levels of binding energy and instability. The material below is an attempt to explain, at least statistically, the effect of an imbalance of protons and neutrons on the binding energies of nuclides.

Below is the graph of the binding energies of all nuclides with 24 nucleons. The data is plotted versus the numbr of protons in the nuclide. The maximum binding energy occurs for Magnesium 24 with 12 protons and 12 neutrons.

## A Simple Model

Let p and n be the numbers of protons and neutrons, respectively, in a nuclide. The number of proton-proton bonds is equal to ½p(p-1). Let this number be denoted as #pp. Likewise the number of neutron-neutron bonds is equal to #nn=½n(n-1). The number of neutron-proton bonds is denoted as #np and it is equal simply to np.

Now suppose there is a certain amount of binding energy associated with each type of bond. The total binding energy BE is then given by

#### BE = bpp#pp + bnp#np + bnn#nn

Note that if all three coefficients are equal then

#### BE = b[½p(p-1) + n*p + ½n(n-1)] or, equivalently BE = ½b[p²-p + 2n*p + n² - n] and also BE = ½b[(p+n)² − (p+n)] and thus BE = ½b[a(a-1)]

where a=p+n. Thus if the coefficients are all equal the plot such as shown in the previous graph would be a level (constant) line.

The effect associated with the minimum value of the b's is then simply a constant amount. The shape of the line is determined by the relative values of (bmax−bmin) and (bmed−bmin.

If bnp is the maximum the plot will have a peak. If not the plot will have a dip.

Now consider a regression of the form

#### BE = bpp#pp + bnp#np + bnn#nn

based upon the 2931 nuclides. The results of the regression are

#### BE = -0.68818#pp + 0.89685#np − 0.69377#nn       [-2.9]    [5.4]    [-5.8] R² = 0.92388

The numbers shown in the brackets are the t-ratios for the coefficients, the ratio of the coefficient to its standard deviation. For a coefficient to be statistically significantly different from zero at the 95 percent level of confidence its magnitude must be at least about 2. The t-ratios are all greater than 2 in magnitude.

The coefficient of determination (R²) for this equation is reasonably good. It indicates that about 92.4 percent of the variation in the binding energies of nuclides is explained by the numbers of the three types of nucleonic bonds.

The statistical performance can be improved by taking into account the formation of spin pairs of the various types. The number of spin pairs of protons is p divided by 2 rounded down to an integer. This denoted as p%2. Likewise the number of neutron spin pairs is n%2. The number of neutron-proton spin pairs is min(p,n). Thus the extended regression equation is

#### BE = bpp#pp + bnp#np + bnn#nn + cppp%2 + cnpmin(n,p) + cnnn%2

The results of the regression are:

#### BE = -0.48958#pp + 0.27804n*p − 0.19254#nn + 4.310005984(p%2) + 10.31174min(p,n) + 13.83460453(n%2) [-40.8] [35.9] [-37.3] [8.3] [34.7] [77.0] R² = 0.999882068

For this regression the t-ratios are extremely large and the coefficient of determination indicates that more than 99.988 percent of the variation in binding energies of 2931 nuclides is explained by the six variables. Put another way, the correlation between the actual binding energies of the 2931 nuclides and the regression estimate is 0.99994. Although this is impressive the standard error of the estimate, the standard deviation of the unexplained variation, is 12.88 MeV, a small value for the larger nuclides but a very large value for the smaller nuclides. This error of the estimate can be reduced to 6.1 MeV if a constant is allowed in the regression equation.

## Conclusions

The values of the regression coefficient for the proton-proton bond indicate that the proton-proton bonds, on balance, have a negative effect on binding energy. This means that at the separation distances of the protons in nuclides the electrostatic repulsion more than counterbalances the strong force attraction. The regression coefficient for the neutron-neutron bonds indicates, surprisingly, that the same effect applies on a lesser scale for the neutrons; i.e., there is a net repulsion. Although the net charge on the neutron is zero there is a radial distribution of charge which means that there can be a net repulsion at some separation distances if the charge distribution is not spherically symmetric. The regression coefficient indicates that this is the case. A nucleus is therefore held together by the net attraction of protons and neutrons for each other. Elastic scattering experiments indicate that the outer distribution of charge in a neutron is negative, with a positive charge within and then a negative charge within that positive shell. Thus to the extent that that there is any electrostatic force between a proton and a neutron it would be an attraction between the positive charge of the proton and the negative charge of the outer shell of the neutron.

The regression coefficients for the effects of spin pairs indicate the neutron-proton spin pairs have by far the greatest effect on binding energy with neutron-neutron pairs having only about two-thirds as much effect. The proton-proton spin pairs have only about one fifth as much effect on binding energy as the neutron-proton pairs. So all in all it is mainly the interactions of neutrons with the protons that holds the nuclei together.

In the 1930's Werner Heisenberg conjectured that the neutron and proton were basically the same particle, differing only in that the charge was turned on for the proton and not for the neutron. The name nucleon was given to this common particle. Quark theory indicates that the neutron and proton are not essentially the same particle. The interaction of neutrons with each other and with protons is a matter of the interaction of their constituent quarks for each other. The quarks in two particles of the same type do not attract each other; they repel. Thus there is no helium isotope consisting only of two protons. Likewise there appears to be no evidence of a nuclide consisting solely of two neutrons, although if there was only the strong force attraction involved there should be such a nuclide.

The results of the regression indicate that notion of there being simply a strong force attraction between nucleons is not adequate to explain nuclear structure.

## Finding the combination of neutrons and protons which maximizes binding energy for a fixed total of nucleons

Let binding energy be expressed as

#### BE = bpp(½p(p-1)) + bnp(n*p) + bnn(½n(n-1))

Now consider maximizing BE subject to the constraint that p+n=q. The Lagrangian multiplier method can be used to solve this optimization problem. The first order conditions are:

#### ∂BE/∂p = λ ∂BE/∂n = λ which evaluate to bpp(p−½) + bnpn = λ and bnpp + bnn(n−½) = λ

This system of two linear equations in two unknows can be solved for p and n as functions of λ. First the system is rearranged to:

#### bppp + bnpn = λ + ½bppand bnpp + bnnn = λ + ½bnn

This system may be expressed in matrix form as

#### BN = λL + ½C

where N is the column vector consisting of p and n as components, L is the column vector with components of unity and C is the column vector with components of bpp and bnn. The solution for N is

#### N = λB-1L + ½B-1C

The multiplier λ must be such that the sum of p and n is equal to q. This requirement may be expressed as

#### LTN = λLTB-1L + ½LTB-1C = q

Therefore the value of λ is given by