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Scale Effects in the Quantum Mechanics of a
Cluster-Rotor Model of Small Nuclei


The masses of nuclei are less than the masses of the protons and neutrons that make them up. These mass deficits expressed in terms of energy through the Einstein equation E=mc² are called their binding energies. They are expressed in units of millions of electon volts (MeV), the enegy an electron acquired upon passing through an electrical potential difference of one million volt.

The binding energies of nuclei arise from two sources. The nucleons (protons and neutrons) may form spin pairs and this leads to an increase in binding energy. The nucleon can also interact through what is usually called the nuclear strong force. The problem with this name is that this force is not so strong as the force involved in the formation of spin pairs. Consequently the binding energy associated with the interaction of nucleons through the strong force may be much smaller than that associated with the formation of spin pairs.

However spin pair formation is exclusive in the sense that a neutron can form a spin pair with one proton and with one other neutron, but no more. There is no exclusivity connected with the strong force and consequently that the binding energy associated with the interaction of large numbers of nucleons may be greater than that due to their spin pair formations. In small nuclei the binding energies associated with spin pair formation may be far larger than that due to the interaction of the nucleon through the so-called nuclear strong force.

The binding energies of the deuteron, triteron and alpha particle are, respectively, 2.225 MeV, 8.48 MeV and 28.3 Mev. Nuclei of these nuclides may be considered as two clusters rotating about their centers of mass. A deuteron is a rotor involving a proton and a neutron. Each cluster is of size 1. The triteron may have one cluster of a proton and a neutron spin pair as the other cluster. The geometric average cluster size is then √2. The triteron cannot have linear arrangement of n-p-n because that would violate the exclusivity of the spin pairs for the proton. The alpha particle may be a proton spin pair and a neutron spin pair rotating about their center of mass. The cluster size is 2.

The graph of the logarithm of binding energy versus the logarithm of average cluster size is as follows.

Visually the graph appears to be a perfectly straight line. It is not quite that. The coefficient of determination (R²) for the regression is 0.99908. The slope of the regression line is 3.67. That is to say, the binding energy is proportional to the 11/3 power of the avergage cluster size, roughly the fourth power of average cluster size.

What follows is a more detailed analysis of the scale effect of cluster size on the binding energy of rotor models of nuclei.

The Quantum Mechanics of Rotor Models

Let the centers of two clusters of nucleons be separated by a distance s. Let the masses of the two clusters be denoted as M1 and M2 with the numbers of nucleons being N1 and N2. This formulation appears to be very general, but in fact N1 and N2 can take on only the values 1 and 2.

The scale of the model is a function of N1 and N2 such that if N1 and N2 are equal then the scale is equal to their common value . One possibility for the scale function is the geometric mean of N1 and N2. Another possibility is twice the reduced number ν, defined as

1/ν = 1/N1 + 1/ N2
which is equivalent to
ν = N1N2/(N1+N2)

The clusters are attracted toward each other by what is usually called the nuclear strong force. Let the nucleonic charges of the clusters be denoted as Q1 and Q2.

The radii of the orbits of the clusters are r1 and r2, with r1+r2=s. The center of mass is such that

r1M1 = r2M2
and thus
r2 = (M1/M2)r1

For convenience let the ratio of cluster masses be denoted as ρ. Thus r2 = ρr1.


r1 + r2 = r1(1 + ρ) = s
and hence
r1 = s/(1+ρ)
r2 = s[ρ/(1+ρ)]

The Quantization of Angular Momentum

Suppose the rotor is rotating at an angular rate of ω. The angular momentum L of the rotor is

M1r1²ω + M2r2²ω
which can be expressed as
L = [M1/(1+ρ)² + M2ρ²/(1+ρ)²]s²ω
and with the factoring out of an M2
L = M2[ρ/(1+ρ)² + ρ²/(1+ρ)²]s²ω
which further reduces to
L = M2[ρ/(1+ρ)]s²ω
which is equivalent to
L = [M1/(1+ρ)]s²ω

This equation is inconveniently asymmetrical with respect to the cluster masses. The quantity M1/(1+ρ) can be expressed as

M1/(1+ρ) = M1/(1+M1/M2
= M1M2/(M1 + M2)
and dividing the numerator
and denominator by the
product of the masses gives
= 1/[1/M2 + 1/M1]

Let the reduced mass μ be defined as

1/μ = 1/M2 + 1/M1

Therefore the angular momentum be expressed as

L = μs²ω

The angular momentum must be equal to an integral multiple n of h, Planck's constant divided by 2π. This means that L=nh where n is a positive integer and h is Planck's constant divided by 2π. Therefore

ω = nh/(μs²)
and, for later use
ω² = n²h²/(μ²s4)

Dynamic Balance

Dynamic equilibrium requires that the attractive force between the clusters be balance by the centrifugal forces on the clusters. The attractive force between the clusters can be expressed, without loss of generality, as


where H is a constant and f(s) is positive-valued function. For a strictly inverse-distance-squared force f(s)=1.

The centrifugal forces on the clusters are of the form Mjrjω². But the center of mass is such that

M1r1 = M2r2 = μs

Thus the centrifugal force is μsω². Dynamic balance then requires that

HQ1Q2f(s)/s² = μsω²
and therefore
ω² = HQ1Q2f(s)/(μs³)

When the two expressions for ω² are equated and reduced the result is

HQ1Q2f(s)/(μs³) = n²h²/[μ²s4]
and hence
HQ1Q2sf(s) =n²h²/μ

This last expression may be expressed as

sf(s) = n²h²/[HQ1Q2μ]
or, equivalently
sf(s) = Γn²
Γ = h²/[HQ1Q2μ]

This is a quantization condition for the separation distance s. The variable Γ is a function only of the parameters.

For a strictly inverse-distance-squared force f(s)=1 and s can be arbitrarily large. For f(s)=exp(−s/s0) the maximum value of sf(s) is limited.

Let z equal s/s0.

The quantization condition for z is

z·exp(−z) = Γn²/s0

The maximum value of z·exp(−z) is 1/e=0.36788 and it occurs at z=1. If Γn²/s0 is less than 0.36788 there are two values of z and hence two values of s. If Γn²/s0 is greater than 0.36788 there are no solutions for z. Thus n must be such that

Γn²/s0 ≤ 0.36788
and therefore
n ≤ (s0/Γ)½

Thus for this case there is a maximum energy for the cluster-rotor and the input of energy to it above that level would result in its dissolution.

It is appropriate here to find the quantization conditions for the other variables of the model.

Rotation Rate

One equation for ω² may be expressed as

ω² = HQ1Q2sf(s)/(μs4)

Replacing sf(s) with Γn² gives

ω² = HQ1Q2Γn²/(μs4)
which reduces to
ω² = h²n²/(μ²s4)
and hence to
ω = ±hn/[μs2]

This is a quantization condition for ω. Note that trivially there are two solutions. The rate of rotation depends upon the principal quantum number n in two ways. It depends directly on n and indirectly through the dependence of s on n.

Kinetic Energy

The kinetic energy K of the rotor is

K = ½M1(ωr1)² + ½M2(ωr2

Since M1r1=M2r2=μs, kinetic energy can be expressed as

K = ½μs[r1+r2]ω²
and hence
K= ½μs²ω²

When ω² is replaced in the above equation by its quantization condition the result is a quantization condition for kinetic energy; i.e.,

K= ½μ s²h²n²/(μ²s4)
and hence
K= ½h²n²/(μs2)

Again we have the quantization of a variable depending upon the principal quantum number n in two ways; directly proportional to n² and indirectly through its dependence upon s.

Since s is inversely proportional to β³ and μ is proportional to β, the dependence of kinetic energy on scale is β7.

Potential Energy

The potential energy V of the system as a function of the separation distance of the centers of the clusters is given by

V(s) = ∫s+∞[−HQ1Q1f(p)/p²]dp
which reduces to
V(s) = −HQ1Q1s+∞(f(p)/p²)dp

Over some range the integral ∫s+∞(f(p)/p²)dp can be approximated by α/sζ, where ζ≥1.

Now consider the dependence of s on the size of the clusters. The sizes of clusters are manifested in the magnitudes of the nucleonic charges Q1 and Q2 and also in the masses M1 and M2. These are proportional to the numbers N1 and N2 of nucleons in the clusters. If the masses and charges are denoted as mj and qj then

sf(s) = h²n²/(HQ1Q2μ)
sf(s) = h²n²/(Hq1q2N1N2μ)

The reduced mass can be expressed as

μ = M1M2/(M1+M2)
which means that
μ = m1m2N1N2/(m1N1+m2N2)

Over some range of s, sf(s) can be approximated by a constant γ times s, γs, since sf(s) goes to zero as s goes to zero.

Let β denote the geometric mean of N1 and N2. It represents a scale parameter for the clusters. The scale of μ is then roughly proportional to β. This means that

γs = h²n²/Gβ³

where G is a constant. Hence the separation distance s is inversely proportional to the cube of the scale. If the scale is twice as large the separation distance is one eighth.

Given the inverse dependence of s on β³ this means that this integral ∫s+∞(f(p)/p²)dp which is approximated by α/sζ is proportional to β. In the above expression for V(s) it is seen that V(s) is proportional to Q1Q1 and hence to β². This means that the potential energy of a cluster-rotor is directly proportional to β3ζ+2.

For f(s)=1, ζ is equal to 1 and hence V(s) is inversely proportional to β to the fifth power. Thus a cluster-rotor model of the alpha particle involving clusters of double size when the nuclear strong force is simply an inverse distance squared force would have thirty two times the potential energy of a deuteron.

For f(s)=exp(-s/s0) the value of ζ depends upon the value of s/s0. For s<s0 the value of ζ is about 1.9 In that case V(s) would be proportional to q7.7. Thus the potential energy due to the nuclear strong force of a triteron (having a scale of √2) would be 14.42 times that of a deuteron. For an alpha particle it would be about 207.9 times that of a deuteron.

Spin Pair Bonds Among the Three Nuclides

The deuteron has one neutron-proton spin pair bond. The triteron has a neutron-neutron spin pair bond and a neutron-proton bond. The alpha particle has four such bonds; a neutron-neutron one, a proton-proton one and two neutron-proton ones.

The binding energies for the different types of spin pair bonds are not necessarily equal but they are of the same order of magnitude, about 2 to 3 MeV. For the purpose of analysis they are assumed to be equal. That common value is denoted as δ. Let ε denote the binding energy in the deuteron due to the nuclear strong force. The scales of the clusters in the deuteron, triteron and alpha particle are 1, √2 and 2, respectively. The conditions to be satisfied then are

ε + δ = 2.225
ε(√2)7.7 + 2δ = 8.482
ε27.7 + 4δ = 28.29567

This set of equations is overdetermined. If the first and third are used to solve for ε and δ the result is:

ε(207.9 − 4) = 28.29567 − 4(2.225)
and hence
ε = 19.3957/203.9 = 0.0951 MeV
and therefore
δ = 2.13 MeV.

When these values are substituted into the LHS of the second equation the result is 5.631 MeV.

This is substantially different for the RHS of 8.482 MeV but it is of the same order of magnitude.

Binding Energies, Potential Energies and Kinetic Energies

In an atom the potential and kinetic energy of an electron are quantized. When an electron moves rom a higher energy state to a lower one the decrease in potential energy is greater than the increase in kinetic energy. The discrepancy is manifested in the emission of a photon whose energy matches the discrepancy. Because the force involved has a strictly inverse-distance-squared law the decrease in potential energy is equally divided between the increase in kinetic energy and the energy of the emitted photon.

When a proton and a neutron come together to form a deuteron a gamma ray is emitted with an energy of 2.225 MeV. There is a disappearance of mass which is assumed to be equal to 2.225 MeV. (This assumption is used to deduce the mass of the neutron and that mass is used to compute the mass deficits and binding energies of other nuclides.)

When a deuteron is hit with a gamma ray of at least 2.225 MeV in energy it comes apart. The mass that disappeared in the formation of the deuteron comes back in its dissociation. The mass deficit has something to do with the proximity of nucleons just as the loss of potential energy does.


With a high power dependencies of potential energy and kinetic energy on the average cluster size and the binding energy of the deuteron of 2.225 MeV this could make the binding energy of an alpha particle, a rotor of two spin pairs, equal to 28.3 MeV. Thus it is not surprising that an alpha particle has a binding energy so large compared to a binding energy of 2 a deuteron. It may be just a matter of scale; i.e., cluster size, given the nature of the dependence of the nuclear force on distance.

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