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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.
Consider a system of two clusters in which one cluster has q_{1} nucleons and a mass of M_{1} and the other q_{2} nucleons and a mass of M_{2}. 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 reduced mass μ is such that
Let r_{1} and r_{2} be the distances of the centers of the two clusters from the center of mass of the system. Then
The separation distance s for the centers of the two clusters is then
The above expressions for r_{1} and r_{2} can be rewritten as
Let ω be the rate of rotation of the system. The angular momentum of one cluster is M_{1}ωr_{1}² and that of the other is M_{2}ωr_{2}². The total angular momentum of the system, p_{θ} can then be expressed as
But both [M_{1}r_{1}] and [M_{2}r_{2}] are equal to [M_{1}M_{2}/(M_{1}+M_{2})]s which is equal to μs. Therefore
The system angular momenum is quantized so
where n is an integer, called the principal quantum number, and h is Planck's constant
divided by 2π.
Thus
The nuclear strong force F between two particles of nucleonic charges z_{1} and z_{2} at a distance s from each other is taken to be
where H and s_{0} 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 q_{1} and q_{2} with nucleonic charges Z_{1} and Z_{2} is then
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
Likewise the centrifugal force on the second cluster is
Thus for each cluster
Equating the two expressions found for ω² gives
This is the quantization condition for the separation distance of the centers of the two clusters. Note that if M_{1}=mq_{1} and M_{2}=mq_{2} then μ is equal to mq_{1}q_{2}/(q_{1}+q_{2}). This would mean that
But Z_{1}=z_{1}q_{1} and Z_{2}=z_{2}q_{2} so
In effect, this would make s*exp(−s/s_{0}) inversely proportional to the third power of the average cluster size. This is seen by replacing both q_{1} and q_{2} by their geometric mean q=(q_{1}q_{2})^{½}; i.e.,
Since s*exp(−s/s_{0}) 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 (q_{1}+q_{2})/(q_{1}²q_{2}²) for the three relevant cases
Case | (q_{1}+q_{2})/(q_{1}²q_{2}²) |
q_{1}=q_{2}=1 | 2 |
q_{1}=1, q_{2}=2 | 3/4 |
q_{1}=q_{2}=2 | 1/4 |
A quantization condition for ω may be obtained from
Since s is a function of n, the above equation quantizes ω.
The kinetic energy K of the system is given by
Replacing ω²
by n²h²/(μs^{4}) gives
This quantizes kinetic energy but the dependence of K on n and the cluster sizes q_{1} and q_{2} is obscure. A simple approximation helps reveal that dependence.
At least over some range the function s*exp(−s/s_{0}) can be approximated by γs, where γ is a constant. This follows from s*exp(−s/s_{0}) being zero at s=0. Thus
Since K = ½n²h²/s²
and thus K is inversely proportional to n².
Thus for q_{1}=1 and q_{2}=1, a deuteron, the quantity q_{1}^{4}q_{2}^{4}/((q_{1}+q_{2})² is equal to 1/4. For q_{1}=2 and q_{2}=1, a proton and a neutron pair revolving about their center of mass, the quantity q_{1}^{4}q_{2}^{4}/((q_{1}+q_{2})² is equal to 2^{4}/3^{2}=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 q_{1}=2 and q_{2}=2, a deuteron of deutrons, the quantity q_{1}^{4}q_{2}^{4}/((q_{1}+q_{2})² would have a value of 2^{4}2^{4}/4^{2}=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 z_{1}z_{1} needs to be taken into account. But z_{1} and z_{2} 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 (z_{1}z_{1})² is (−2/3)²=4/9 for all of the four small nuclides being considered.
The potential energy of the system is a function only of the separation distance of the centers of the clusters is given by
Over some range the integral ∫_{s}^{+∞}exp(−p/s_{0})/p²)dp can be approximated by α/s^{ζ}, where ζ≥1. Let q denote the average cluster size. This would mean that s*exp(−s/s_{0}) 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^{3ζ}. 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 q^{3ζ−2}.
For exp(-s/s_{0}) the value of ζ depends upon the value of s/s_{0}. For s<s_{0} the value of ζ is about 2. In that case V(s) would be proportional to q^{4}. 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 q^{4} 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.
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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