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the Helium 4 Nuclide, Between That Due to the Strong Force and That Due to Nucleon Spin Pair Formation |
The Helium 4 nuclide contains four nucleons, two neutrons and two protons. The conventionally accepted figure for the binding energy of the Helium 4 nuclide is 28.295674 MeV. Because this is computed using a mass for the neutron that is too low by 0.98638 MeV the binding energy of the Helium 4 nuclide is 28.295674+2(0.98638)= 30.268434 MeV.
The nuclear strong force between two nucleons is presumed to be given by
where Z_{1} and Z_{2} are the nuclear charges of the particles, s is the separation distance of their centers and H and s_{0} are parameters. The potential energy V(s) at a separation distance of s is give by
By a change of the variable of integration the expression for potential energy can be put into the form
where y=x/s_{0}.
The formula for potential energy may be expressed as
According to the article, "Precise Radii of Light Nuclei from Electron Scattering," by I. Sick published in Precision Physics of Simple Atoms and Molecules (edited by Savely G. Karshenboin), the root-mean-square (rms) charge radius of a Helium 4 nuclide is 1.68 fermi; and that of a proton is 0.895 fermi. Thus the distance from the centroid of the tetrahedron to the center of a proton is 0.785 fermi.
For a triangle with two sides equal to 1 and an internal angle of 120°, as shown above the distance between adjacent vertices is √3. In this case the sides are equal to 0.785 fermi so the distance between the centers of the nucleons is √3*(0.785)=1.36 fermi. Relative to the scale parameter of 1.522 fermi for nuclei this is 0.89. The value of W(0.89) is 0.493693.
The nucleonic (strong force) charge of a neutron is −2/3 if that of a proton is defined as +1. This means that between a neutron and proton there is an attraction and a negative potential energy while between a neutron and a neutron there is a repulsion and a positive potential energy based upon separation distance. Likewise between two protons there is strong force repulsion. The factor of Z_{1}Z_{2} in the formula for potential energy for a neutron-proton interaction is then −2/3. For a neutron-neutron interaction it is +4/9 and for a proton-proton interaction it is +1.
The Helium 4 nuclide is also subject to the electrostatic repulsion between the two protons. The magnitude of the potential energy due to this repulsion can be measured by comparing the binding energies of the triteron and the Helium 3 nuclide. The difference in those binding energies is [(8.481821+0.98638)−7.718058]=1.750143 MeV.
Finally there could be an increment in binding energy due solely to the formation of a quadruple particle, analogous to the formation of a pair. Let Q_{α} denote this enhancement of binding energy due to the formation of a quadruple.
There is evidence from scattering experiments that the spatial arrangement of the nucleons in an alpha particle is that of a tetrahedron. The equation to be sastified for the Helium 4 nuclide is, if its spatial arrangement is that of a tetrahedron,
This figure is surprisingly high but within the realm of plausibility. The discrepancy could be due to a number of sources. One, of course, is that the model may be in error. Another could be that the estimate of the parameter H is in error. If the value of H were 82.6 percent higher the value of Q_{α} would be reduced to zero. Another is that the value used for distances between nucleons could be too high. If W(z) were 82.6 percent higher the value of Q_{α} would be reduced to zero. In order for W(z) to be 82.6 percent higher the value of z would have to be something on the order 0.72, which is only 19 percent smaller than the value used in the above computation. This means the average distances between nucleons instead of being 1.36 fermi would have to be 1.1 fermi.
There is a simple relationship between the potential energy due to the electrostatic force and the separation of charged particles. If V is the potential energy in MeV then the separation distance s in fermi is given by
Thus if the potential energy due to the repulsion of two protons is 1.750143 MeV then their separation distance is 0.82279 fermi. Thus it would be very easy to find modifications of the tetrahedral arrangement of nucleons which would result in an average separation distance below the value of 1.36 fermi used in the above computation.
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