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The Explanation of the Binding Energy
of the Triteron, the Hydrogen 3 Nuclide

One of the major enigmas of nuclear physics is the relatively values of the binding energies of the small nuclides, the deuteron, the triteron, the Helium 3 nuclide and the alpha particle. The conventional estimate of the binding energy of the deuteron is about 2.225 million electron volts (MeV), that of the triteron 8.48 MeV, that of the He 3 nuclide 7.76 MeV and the alpha particle an enormous 28.29 MeV. To a degree the differences could be due to the number of nucleon-nucleon interactions (bonds) of the different nuclides. The deuteron has only one bond; the triteron and the He 3 nuclide each have three and the alpha particle has six. The number of bonds however, is at best only a partial explanation of the differences.

A previous study accounted for the binding energy of the alpha particle to within a fraction of one percent by treating the alpha particle as a neutron pair and a proton pair spinning about their center of mass. This is an attempt to use the model of a deuteron of nucleonic clusters to explain the binding energy of the triteron.

This approach treats the clusters as particles and analyzes the tractible two-body problem when in reality such a system is an intractible n-body problem. This is based up the adage

It is better to be approximately right
than precisely undecided.

What came out of a previous investigation in which the clusters were identical and each contained q nucleons is that the potential and kinetic energy levels, respectively, of such systems are roughly proportional to the fourth and fifth powers of q. This is the investigation of the more general case in which the number of nucleons in the clusters can be different.

The Model

Consider a system of two clusters in which one cluster has q1 nucleons and a mass of M1 and the other q2 nucleons and a mass of M2. This is a deuteron of clusters. The alpha particle might be a deuteron of deuterons. On the other hand it might not be. It could be a neutron pair and a proton pair revolving about their center of mass. Or, it could be something entirely different.

Here are a few of the possible structures of the triteron, with the red circles representing protons and the black circles representing neutrons:

In the above the force on a neutron is less than what occurs in the deuteron because of the repulsion between the neutrons. It would have less binding energy than the deuteron.

An equilateral triangle would be impossible to maintain because of the differing forces between protons and neutrons compared to that between neutrons and neutrons.

This deuteron of a deuteron and a neutron would involve dual rotations; one to maintain the separation of the proton and the neutron within the deuteron and another to keep the deuteron and the other neutron separated.

This is a deuteron of a neutron spin pair and a proton. In this one the distance between the neutrons would be maintained by the combination of binding energy of the spin pairing of the neutrons and their mutual repulsion. This is shown below.

The separation distance of the neutrons would be where the potential energy is a local minimum; i.e., at the limit of the spin pairing of the neutrons.

The Analysis of the Case of a Deuteron
Made Up of a Proton and a Neutron Spin Pair

The Reduced Mass of the System

The reduced mass μ is such that

1/μ = 1/M1 + 1/M2
which is equivalent to
μ = M1M2/(M1+M2)

The Orbit Radii

Let r1 and r2 be the distances of the centers of the two clusters from the center of mass of the system. Then

M1r1 = M2r2
and thus
r2/r1 = M1/M2

The separation distance s for the centers of the two clusters is then

s = r1 + r2 = (1 + r2/r1)r1
which is then equal to
s = (1 + M1/M2)r1
and hence
r1 = s/(1 + M1/M2)
and likewise
r2 = s/(1 + M2/M1)

The above expressions for r1 and r2 can be rewritten as

r1 = [M2/(M1+M2)]s
r2 = [M1/(M1+M2)]s

Angular Momentum

Let ω be the rate of rotation of the system. The angular momentum of one cluster is M1ωr1² and that of the other is M2ωr2². The total angular momentum of the system, pθ can then be expressed as

pθ = [M1ωr1]r1 + [M2ωr2]r2

But both [M1r1] and [M2r2] are equal to [M1M2/(M1+M2)]s which is equal to μs. Therefore

pθ = μωs(r1+r2) = μωs²

The system angular momenum is quantized so

pθ = μωs² = nh

where n is an integer, called the principal quantum number, and h is Planck's constant divided by 2π. Thus

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

Dynamic Balance

The nuclear strong force F between two particles of nucleonic charges z1 and z2 at a distance s from each other is taken to be

F = Hz1z2exp(−s/s0)/s²

where H and s0 are positive constants. Elsewhere the case is made that if the nucleonic charge of a proton is taken to be +1 then the nucleonic charge of the neutron is −2/3. Thus if the nucleons are of the same type then the force between them is a repulsion. If they are of different types the force between them is an attraction.

The force between clusters of size q1 and q2 with nucleonic charges Z1 and Z2 is then

F = HZ1Z2exp(−s/s0)/s²

Dynamic Balance

The attractive nuclear force on each cluster must balance the centrifugal force on that cluster. The centrifugal force on the first cluster is equal to

M1ω²r1 = [M1M2/(M1+M2)]ω²s.
which reduces to
μω²s

Likewise the centrifugal force on the second cluster is

M2ω²r2 = [M2M1/(M1+M2)]ω²s.
which again reduces to
μω²s

Thus for each cluster

μω²s = HZ1Z2exp(−s/s0)/s²
and therefore
ω² = HZ1Z2exp(−s/s0)/(μs³)

Equating the two expressions found for ω² gives

HZ1Z2exp(−s/s0)/(μs³) = n²h²/(μ²s4)
which reduces to
s*exp(−s/s0) = n²h²/(μHZ1Z2)

This is the quantization condition for the separation distance of the centers of the two clusters. Note that if M1=mq1 and M2=mq2 then μ is equal to mq1q2/(q1+q2). This would mean that

s*exp(−s/s0) = n²h²(q1+q2)/(mHq1q2Z1Z2)

But Z1=z1q1 and Z2=z2q2 so

s*exp(−s/s0) = n²h²(q1+q2)/(mHz1z2q1²q2²)

In effect, this would make s*exp(−s/s0) inversely proportional to the third power of the average cluster size. This is seen by replacing both q1 and q2 by their geometric mean q=(q1q2)½; i.e.,

s*exp(−s/s0) = n²h²(q+q)/(mHz1z2q4)
which reduces to
s*exp(−s/s0) = n²h²2/(mHz1z2q3)

Since s*exp(−s/s0) can be approximated as γs this makes the separation distance s inversely proportional to the third power of the average cluster size.

Consider the quantity (q1+q2)/(q1²q2²) for the three relevant cases

Case(q1+q2)/(q1²q2²)
q1=q2=12
q1=1,
q2=2
3/4
q1=q2=21/4

Quantization of the Kinetic Energy

A quantization condition for ω may be obtained from

ω = nh/(μs²)

Since s is a function of n, the above equation quantizes ω.

The kinetic energy K of the system is given by

K = ½M1ω²r1² + ½M2ω²r2²
which reduces to
K = ½ω²[M1r1² + M2r2²]
and further to
K = ½ω²μs²

Replacing ω² by n²h²/(μs4) gives

K = ½n²h²/s²

This quantizes kinetic energy but the dependence of K on n and the cluster sizes q1 and q2 is obscure. A simple approximation helps reveal that dependence.

At least over some range the function s*exp(−s/s0) can be approximated by γs, where γ is a constant. This follows from s*exp(−s/s0) being zero at s=0. Thus

γs = n²h²(q1+q2)/(mHz1z2q1²q2²)
and hence
s = n²h²(q1+q2)/(γmHz1z2q1²q2²)
and therefore
s² = n4h4(q1+q2)²/(γ²m²H²z1²z2²q14q24)

Since K = ½n²h²/s²

K = [½γ²m²H²(z1z2)²q14q24/((q1+q2h²)]/n²

and thus K is inversely proportional to n².

Thus for q1=1 and q2=1, a deuteron, the quantity q14q24/((q1+q2)² is equal to 1/4. For q1=2 and q2=1, a proton and a neutron pair revolving about their center of mass, the quantity q14q24/((q1+q2)² is equal to 24/32=16/9 and thus and the kinetic energy of such a system would have (16/9)/(1/4)=64/9=7.111 times the kinetic energy of a deuteron

And for q1=2 and q2=2, a deuteron of deutrons, the quantity q14q24/((q1+q2)² would have a value of 2424/42=16 and thus the kinetic energy of a deuteron of deuterons would have 16/(1/4)=64 times the kinetic energy of a deuteron for the same principal quantum number.

It might seem that the factor z1z1 needs to be taken into account. But z1 and z2 are the average nucleonic charges in the clusters and these are +1 for a proton cluster and −2/3 for a neutron cluster. The quantity (z1z1)² is (−2/3)²=4/9 for all of the four small nuclides being considered.

The Quantization of Potential Energy

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

V(s) = ∫s+∞[HZ1Z2exp(−p/s0)/p²]dp
which reduces to
V(s) = HZ1Z2s+∞(exp(−p/s0)/p²)dp
and hence
V(s) = Hz1z2q1q2s+∞(exp(−p/s0)/p²)dp

Over some range the integral ∫s+∞exp(−p/s0)/p²)dp can be approximated by α/sζ, where ζ≥1. Let q denote the average cluster size. This would mean that s*exp(−s/s0) and hence s would be inversely proportional to the third power of this average cluster size q. Then given the inverse dependence of s on q³ this means that the above integral and hence potential energy is proportional to q. From the above expression for V(s) this means that the potential energy and the hence the binding energy of a cluster deuteron has potential energy is proportional to q3ζ−2.

Some Interesting Computations

For exp(-s/s0) the value of ζ depends upon the value of s/s0. For s<s0 the value of ζ is about 2. In that case V(s) would be proportional to q4. Thus the potential energy and hence the binding energy of a deuteron of nucleonic pairs would be about 16 times that of a deuteron. With the binding energy of the deuteron of 2.225 MeV this would make the binding energy of a deuteron of nucleonic pairs equal to 35 MeV.

Consider a triteron as a deuteron of a proton and a neutron pair revolving about their center of mass. The geometric mean q of the cluster sizes of 1 and 2 is √2. With the potential energy and binding energy proportions to q4 this would make the binding energy of the triteron equal to (√2)4=4 times that of the deuteron. With binding energy of the deuteron being 2.225 MeV this would make the binding energy of the triteron equal to 8.9 MeV whereas the conventional estimate is 8.5 MeV.

An accurate explanation of the binding energy of the triteron would have to take into account the moderation of the strong force attraction and the formation of spin pairs by the neutrons. Furthermore the conventional estimates of the binding energies of nuclides may be in error because the estimate of the mass of the neutron is based upon the assumption that the binding energy of the deuteron is equal to the energy of the gamma photon associated with its formation or disassociation. That assumption does not take into account the change in kinetic energy of the particles making up the deuteron.

Conclusions

It is not surprising that an alpha particle has a binding energy of 28.29 MeV compared to a binding energy of 2.225 MeV for a deuteron, or that the binding energy of the triteron is 8.48 MeV. It is 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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