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Systems of Nucleonic Clusters |
One of the major enigmas of nuclear physics is the relatively high value of the binding energy of the Helium 4 nuclide, the alpha particle, compared to the smaller nuclides such as the Hydrogen 2 nuclide (deuteron) and the Hydrogen 3 (triteron). The binding energy of the deuteron is about 2.225 million electron volts (MeV), that of the triteron 8.48 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 has three and the alpha particle has six. The number of bonds 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 few percent by treating the alpha particle as two deuterons spinning about their center of mass. A search for a model that would give a more accurate explanation of the binding energy was not successful so this is a further investigation and generalization of nuclides as systems of clusters.
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 this investigation is that the energy levels of such systems are highly nonlinearly dependent upon the size of the clusters;i.e., the number of particles in the clusters.
Consider first the simplest case, the one in which two identical clusters revolve about their center of mass. 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. What is to be established here is the highly nonlinear character of the dependence upon cluster size. An accurate explanation of the binding energy of the alpha particle would have to take into account the moderation of the strong force attraction by the electrostatic repulsion of its two protons.
Let q be the number of nucleons is a cluster; in effect, the strong force charge of the cluster. Let m be the mass of a single nucleon. The mass of a cluster is then mq.
Let r be the radius of the orbits of the clusters and s the separation distance of their centers. Thus r=s/2.
Let ω be the rate of rotation of the system. The angular momentum of the system is 2(mq)(ωr)r, which is equal to mqωs²/2. This angular momentum is quantized so
where h is Planck's constant divided by 2π. Thus
Without any loss of generality the nuclear force between clusters of size q_{1} and q_{2} may be expressed as
where H is a constant and f(s) is a function such that f(0)=1. Later it will be assumed that f(s)=exp(-s/s_{0}), but for now the form of the force formula will be kept general.
The attractive nuclear force on a cluster must balance the centrifugal force, which is equal to (mq)ω²r=(mq)ω²s/2. Thus
Equating the two expressions derived for ω² gives
where σ=2h²/(mH).
This is the quantization condition for the separation distance s.
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 2(½(mq)(ωr)²). This reduces to mqω²s²/4. Replacing ω²
by 4n²h²/(m²q²s^{4}) gives
This quantizes kinetic energy but the dependence of K on n and q is obscure. A simple approximation helps reveal that dependence.
At least over some range the function sf(s) can be approximated by γs, where γ is a constant. This follows from sf(s) being zero at s=0. Thus
where σ was previously defined as 2h²/(mH).
So, since K = n²h²/(mqs²)
Thus a deuteron of deutrons would have kinetic energy equal to 32 times the kinetic energy of a deuteron for the same principal quantum number.
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}^{+∞}(f(p)/p²)dp can be approximated by α/s^{ζ}, where ζ≥1. Given the inverse dependence of s on q³ this means that this integral proportional to q^{3ζ}. From the above expression for V(s) this means that the potential energy and the hence the binding energy of cluster deuteron has potential energy is proportional to q^{3ζ−2}. For f(s)=1, ζ is equal to 1 and hence V(s) is proportional to q. Thus a deuteron of deuterons when the nuclear strong force is simply an inverse distance squared force would have two times the potential energy of a deuteron and its binding energy would be only twice as large.
For f(s)=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 deuterons would be about 16 times that of a deuteron.
With a fourth power dependence of binding energy on the average cluster size and the binding energy of the deuteron of 2.225 MeV this would make the binding energy of a deuteron of deuterons 35 MeV. Thus 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. 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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