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the H3 (Tritium) and He3 Nuclides |
The triteron, the tritium nucleus, _{1}H_{2}, is the third simplest nucleus, consisting of three nucleons (one proton and two neutrons). The other three-nucleon nucleus, _{2}He_{1}, is a bit more complicated because of the electrostatic repulsion between the two protons. The binding energy of a triteron is 8.482 million electron volts (Mev) but that of the Helium 3 nucleus is only 7.718 Mev. The 0.764 Mev difference is due to the mutual repulsion of the two protons in the He 3 nucleus.
This last bit of information is valuable because it can tell us how far apart are the protons in the _{2}He_{1} nucleus. The formula for the relation between the potential energy in Mev and the distance in fermi is
(For the derivation see Electrostatic Potential.)
Thus, according to this formula, the separation distance of the centers of the protons in He 3 is d=1.884 fermi. This value however is sensitive to any error in the estimate of the mass of a neutron. The value of 1.884 fermi corresponds in order of magnitude with the estimate of the distance between the centers of nucleons in a deuteron; i.e., 2.252 fermi. Although the 1.884 fermi distance is the right order of magnitude it is notably small. It might be expected that the distance between two repelling protons in He3 would be larger than the distance between a proton and a neutron in a deuteron which are only subject to the attractive strong nuclear force. However if the nucleons in H3 and He3 are arranged in a triangle each nucleon is subject to a greater net attraction than those in a deuteron.
According to Hideki Yukawa the strong force is carried by π mesons. Any force carried by particles will have an inverse distance squared dependence. However the π mesons decay with time and there the number surviving at any distance is a negative exponential function of distance. This means that the strong force between two nucleons has the following formula.
where s is the separation distance of two particles, H is a constant and s_{0} is a scale factor that depends upon the mass of the force-carrying particle. From Yukawa's relation and the mass of the π meson the parameter s_{0} has the value 1.522 fermi.
The potential function V(s) for the force formula above is
By a change in the variable of integration from z to ζ=z/s_{0} the potential function can be converted into
where σ=s/s_{0}. Thus the crucial aspect of the separation distance is its value relative to the scale parameter s_{0}.
A group of physicists under the editorship of Savely G. Karshenboin published in 2008 a book devoted to the compilation of the best estimates of physical properties of particles, simple atoms and simple molecules (Precision Physics of Simple Atoms and Molecules, Springer-Verlag).
The best estimate of the root-mean-square-charge diameter of a deuteron from page 70 of the above mentioned work is 4.260 fermi with a margin of error of ±0.02 fermi. The recommended estimate of the rms-charge radius of the proton, given on page 49 of the above work, is 0.895 fermi. Precision Physics of Simple Atoms and Molecules does not give an estimate for the radius of the neutron. Another source gives the rms-radius of the neutron as 1.113 fermi.
Thus the separation distance of the centers of the nucleons is
The separation distance s_{1} in the physical deuteron is thus about 2.252 fermi (2.252×10^{-15} m), Thus the ratio s_{1}/s_{0} is equal to 1.4796, a pure number.
The root-mean-square (rms) charge radius of the triteron is 1.755 fermi; that of a proton is 0.895 fermi. Thus the distance from the centroid of the triangle to the center of the proton is 0.860 fermi.
For an equilateral triangle the distance between adjacent vertices is √3*(0.86)=1.49 fermi. Relative to the scale parameter of 1.522 fermi for nuclei this is 0.98. For the deuteron this figure is 1.4796. Thus the potential energy involved in each of the nucleon bonds in the triteron should be larger than that in the deuteron. The conventional estimate of the binding energy of the deuteron is 2.22457 MeV whereas for the triteron it is 8.482/3=2.827 MeV.
Previous studies found that the strong force interactions of two neutrons or two protons are repulsions, whereas between a neutron and proton it is attraction. However, despite the strong force repulsion two neutrons or two protons can for a spin pair and there is a loss of potential energy associated with such pair formations. The general shape of the relationship between pottential energy and the separation distance of the nucleons is shown in the two diagrams below. The first diagram shows the separate components and the second the combined result.
The implication of the above diagrams is that the separation distance for neutron-neutron bonds and for proton-proton bonds will be at the maximum distance at which the spin pairs can form. What was found above for the separation distance of neutrons in an equilateral triangular arrangement was the figure of 1.49 fermi. This figure is 0.98 of the scale parameter. This suggest that the maximum distance for the formation of neutron pairs is 1.0 times the scale parameter.
Nuclear stability is represented in terms of binding energy. If the binding energy of all nuclides having the same number of nucleons is plotted versus the number of protons the result is a curve that rises to a peak where the numbers of protons and neutrons are approximately equal. This is shown below for 24 nucleons.
The maximum binding energy occurs for Magnesium 24 with 12 protons and 12 neutrons. It is very significant that the binding energy drops off in both directions.
This suggests that the binding energy (BE) of nuclides might be explained in terms of the number of interaction pairs of nucleons. If the numbers of protons and neutrons are denoted by #p and #n, respectively, the number of proton-proton interaction pairs (#pp) is equal to ½#p(#p-1) and likewise the number of neutron-neutron interaction pairs (#nn) is equal to ½#n(#n-1). On the other hand the number of proton-neutron interaction pairs is #p*#n.
There is another type of pairing of nucleons, spin pairing. Spin pairing, in contrast to the interaction pairing, is exclusive. That is to say, if a proton forms a spin pair with another proton then it cannot form a spin pair with any other proton. The number of proton spin pairs is (#p%2), the number of protons divided by 2 and the result rounded down to an integer and likewise for the number of spin pairs of neutrons is (#n%2). The number of possible spin pairs of protons and neutrons is the minimum of #p and #n, min(#p, #n).
The regression of the binding energies of the 2931 nuclides gives:
The numbers shown in the brackets are the t-ratios for the coefficients, the ratio of the coefficient to its standard deviation. For a coefficient to be statistically significantly different from zero at the 95 percent level of confidence its magnitude must be at least about 2. The t-ratios are all very much greater than 2 in magnitude.
What the results mean is that there is a mutual repulsion between protons and between neutrons, although to a lesser extent. All three types of spin pairing have a positive effect on binding energy but that for proton-neutron pairing is much stronger than for proton-proton spin pairing.
The rms-charge radius of the He3 nuclide, as given in Karshenboin, is 1.959 fermi. This makes the distance from the centroid of the triangle 1.109 fermi. This, in turn, makes the separation distance of the nucleons √3*(1.109)=1.920 fermi, only 2 percent larger than the figure of 1.884 fermi found from the difference in binding energies of H3 and He3. The average of 1.884 and 1.920 is 1.902 fermi and the value of this figure relative to the scale parameter of 1.522 fermi is 1.25. This strongly suggests that the maximum distance for pair formation of protons is about 1.9 fermi, which is very close to (5/4) of the value of the scale parameter 1.522 fermi.
(To be continued.)
The Coulomb Law for the force of repulsion between two charges separated by a distance r is
Thus the potential energy V(r) is given by
The value of k_{e} is 9×10^{9} newton-meters²/coulomb². For the charges being those of protons
Thus
One newton-meter is equal to 6.2415×10^{12} Million electron-volts (MeV).
Thus the potential energy of two protons separated by a distance r (in meters) is
If r is measured in fermi (10^{-15} meters) the formula becomes
A computation using more precise values of the figures gives
Therefore,
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