& Tornado Alley
of the Tritium Nucleus, 1H2
The tritium nucleus, 1H2, is the third simplest nucleus, consisting of three nucleons (one proton and two neutrons). The other three-nucleon nucleus, 2He1, is a bit more complicated because of the electrostatic repulsion between the two protons. The binding energy of a tritium nucleus 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.
The model used here is developed elsewhere. It presumes the nucleons are attracted to each other with a force given by the following formula.
It is additionally hypothesized for this analysis that a proton-neutron pair exists as a deuteron and that the extra neutron revolves around the center of mass of the deuteron-neutron combination. This system is a two-nucleon mass particle combined with a unit mass particle. Since the binding energy of tritium is 8.482 Mev and the deuteron has a binding energy of 2.225 Mev the formation of tritium involves a gain in binding energy of 6.257 Mev.
Let Particle 1 be the single neutron and Particle 2 the deuteron. Therefore n2/n1=2. Let θ be the angle between the angle between the line between either of the nucleons in the deuteron and the axis of rotation of the deuteron.
The value of the separation distance and the angle θ must be found by iteration. There are more than one equilibrium positionings of the singleton neutron with respect to the deuteron. The value so found for cos(θ) is 0.2764 which corresponds to an angle of 74°. The value found for the potential energy is 4.72 Mev. This is substantially below the empirical value 6.257 Mev, but the method used does not take into account the effect of the single neutron on the separation distance of the deuteron nucleons. The component of the attractive force of the single neutron is equivalent to that of a nucleon situated at the the center of rotation of the deuteron but multiplied by sin³(θ), which in this case is equal to (0.961)³=0.8876. This generates a higher potential energy for the deuteron. The effective H would be (H*/4 + H*(0.8876))=1.211H* instead of 0.25H*. This would induce a smaller orbital radius for the nucleons in the deuteron, thereby changing the value of θ. The final values would have to be found by iteration.
(To be continued.)
HOME PAGE OF Thayer Watkins