|San José State University|
& Tornado Alley
of the Triteron (H3) and He3 Nuclides
After the deuteron the next simplest nuclide is the triteron, the triple weight hydrogen nuclide. The conventional estimate of the binding energy of the triteron (Hydrogen 3, H3) is about 8.48 million electron volts (MeV) whereas that of the Helium 3 (He3) nuclide is about 7.72 MeV. The difference might be due simply to the He3 having two protons whereas the triteron has only one. The two protons would experience the electrostatic repulsion in addition to the attraction due to the nuclear strong force. If this were the only difference between the triteron and the He3 nuclide then the separation distance of the protons would be 1.89 fermi (1.89×10-15 meters). Since the separation distance of the nucleons in the deuteron (H2) is about 2.25 fermi this means that the difference in binding energies is due to something more than the electrostatic repulsion of the protons in He3.
In working with binding energies it is important to keep in mind that the conventional values may be in error due to an underestimate of the mass of the neutron. This underestimate is of an order of magnitude of 2 MeV. Such an error would result in the estimated separation distance of the protons in He3 nuclide being even smaller than the improbable 1.89 fermi mentioned above.
The most plausible difference is a difference in structure. The nucleons in the triteron may be arranged in the form of an equilateral triangle whereas those in the He3 nuclide could have a linear arrangement with the proton in the middle.
Consider first an equilateral triangular arrangement for the triteron.
The net force on a nucleon in an equilateral triangle would be directed toward the center of mass of the three nucleons. The magnitude of this force would be equal to 2cos(30°) times the magnitude of the force between any two nucleons; i.e., √3 times internucleonic force. This would lead to a tighter arrangement and correspondingly higher potential energy for the internucleonic bonds.
Consider the simplified version of H3 in which three equal mass revolve around their center of mass at a rotation rate of ω Let s be the separation distance between the nucleons. The orbit radius r is then
The angular momentum pθ of the system is 3mωr², which reduces to mωs². This
is quantized to n
h, where n is an interger and h is Planck's constant divided
by 2π. Such an integer value is usually called the principal quantum number of the system. Thus
The kinetic energy K of the system is 3(½mω²r²), which reduces to ½mω²s². Therefore
The potential energy is three times the potential energy of two nucleons separated by a distance s. Let V be the potential energy of the system and W(s) the potential energy function for two nucleons separated by a distance s. Thus
The net radial force on a nucleon in an equilateral triangular arrangement is equal to F(s)/√3, where F(s) is the force between two nucleons at a distance s. This must be enough to generate the centripetal acceleration of a particle traveling in a circle; i.e., the net radial force must balance the centrifugal force.
It was previously found that ω is equal to n
h/(ms²) and thus
ω² is equal to n² h²/(m²s4).
Equating the two expressions for ω² gives
This is the quantization condition for the separation distance s.
The formula for the nuclear force is uncertain but it no doubt has an inverse distance squared dependence. Let it be expressed as F(s)=Hf(s)/s², where H is a constant. The term f(s) is such that f(0)=1. For the electrostatic and gravitational forces the function f(s) is the constant 1.
The quantization condition for the separation distance then takes the form
For the nuclear force the function f(s) is likely to be an exponential function due to the decay of the π mesons which carry the force. This means that
where s0 is a scale parameter equal to 1.522 fermi (from the Yukawa condtion). Thus
Let s/s0 be denoted as z and (
h²/(ms0))/H as σ.
The quantization condition is then
The left-hand side (LHS) of the above equation is a function that is zero at z=0, rises to a maximum of e-1 at z=1 and decline asymptotically toward zero as z increases without bound.
This means that there is a maximum allowable n. If n takes a value above that maximum the system disintergrates.
For allowable values of n there would appear to be two values of z satisfying the quantization condition. The total energy H=T+V would differ and so the solution with the minimum total energy would be the only relevant one. Intutition or instinct would lead one to expect the branch of the function below z=1 would be relevant but that implies separation distances of less than 1.522 fermi. As noted above the separation distance for the nucleons in a deuteron is about 2.25 fermi.
The root-mean-square (rms) charge radius of the triteron is estimated to be 1.755 fermi. From this must be deducted the radius of the proton, which is 0.895 fermi. This means the radius of the orbits of the nucleon centers is 1.755-0.895=0.860 fermi.
From the geometry of an equilateral triangle and the Law of Cosines it follows that
The separation distance of the centers of the nucleons is then about √3(0.860)=1.489563695 fermi. Thus the empirical data indicates that the branch of the sf(s) function that is relevant is the one to the left, but the most striking aspect is that the empirical value is very close to the maximum of the sf(s) function.
The parameter σ is defined as (
h²/(ms0))/H. Using the mass of the
proton for m, the term h²/(ms0) evaluates to
An estimate of H based upon the dimensions of the deuteron is 2.591310×10-26 kg*m≥/sec≤. This makes the value of σ equal to 0.1505. The solution for z for this value is z=2.99. This would make the separation distances of the nucleons in the triteron equal to 4.55 fermi.
Since the separation distances of the nucleons in the triteron based upon the empirically measured rms charge radius of the triteron are about 1.489563695 fermi this means the value of z is should be about 0.978688367. For z=0.978688367 the value of ze-z is equal to 0.367794702.
Thus if the principal quantum number n is equal to 1 the force constant H has to be equal to 3.9×10-27 divided by 0.367794702, which is 9.4306×10-27 which is 36.78 percent of the value found from the dimension of the deuteron.
While the model using the estimate of the parameter H based on the dimensions of the deuteron does not give an accurate estimate of the separation distance of the centers of the nucleons in a triteron, they do give an estimate of the right order of magnitude.
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