The Fusion Reaction
The formation of the “He-d” bond discussed in the previous chapters is the prelude to the formation of the “He-d2” molecule. In this chapter we will calculate the half-life of the fusion reactions for the two stable angular momentum states of the “He-d2” molecule. Remember that there are three states of “He-d2” originally formed in the “activated complex”:
1) [He-d2]1,1
2) [He-d2]1,2
3) [He-d2]2,2
Where the superscripts refer to the angular momentum states (l=1 or 2) of the two deuterium nuclei, coupled to the helium nuclei. We argue that only two of these states (1 & 3 above) lead to the alpha particle fusion products to maintain the chain reaction. The other state, number 2 above, has a more ambiguous fate. This mixed “odd” state molecule is shown to have a higher energy (less stable) than the even states. The overlap of the helium “1s” electrons with the nuclear wave function of the “He-d2”, provides the mechanism for very fast transitions from the “odd” states to the “even” states. Normally we think of energy transitions involving changes of spin or angular momentum as photon emitting interactions with the dipoles created by the oscillation between initial and final states. Photon transitions are orders of magnitude slower than their uncertainty principle energy difference would predict (~ħ/ΔE) due to the dipole emission time. The transitions in the “He-d2” molecule are more akin to molecular collision induced transitions, which are mediated by the electrons of the respective atoms. In this case, the “He-d2” molecule sits inside of and is essentially the nucleus of the parent helium atom. This provides for very good overlap of states toward reaction probabilities. These transitions will follow the ~ħ/ΔE uncertainty principle time constraints.
We will analyze the Schrodinger equation with both deuterium nuclei present with the helium nuclei. This will introduce a small change in the reduced mass for the wave equation, with the energy of the bond changing very little, but the normalization factor having an important energy effect. The change in the reduced mass produces a slightly larger diameter orbit for the deuterium, reducing their mutual repulsion, thus lowering total energy.
The analysis of the “He-d2” molecule is very similar to the analysis done with helium to find the energy of the two “1s” electrons. We will briefly describe the physics of putting two electrons into helium as a comparison. The first electron interacting with the helium nucleus to make [He+], has the energy expected from the exact hydrogen wave function of 54.4eV. This is exactly 4 times (z2=22=4) the energy of the electron in the hydrogen atom. The second electron going into helium has an energy of only 24.6eV (Actually once in the helium atom, both electrons will have the same energy, as they are in identical environments). The two energy values are the ionization energies. The first electron obviously comes off much easier than the second. The reason for this is the energy required to bring the two electrons into close proximity in the same “1s” state. It is the repulsive energy of the two electrons. A very simple calculation of this “repulsive” energy shows the principle physics involved. Assume the “1s” helium wave function doesn’t change with the addition of the second electron. The average diameter of this orbit is 0.53 angstroms (twice the hydrogen diameter). If we assume this separation for the electrons and calculate the energy to bring them this close we get;
E=e2/d =1.44x10-9/0.53x10-10 eV= 27.2 eV
Subtracting this energy from the bond energy of the second electron gives;
E(2nd electron)= (54.4- 27.2)eV= 27.2 eV
This quick estimate of 27.2 is not far from the actual 24.6 eV ionization energy, demonstrating the controlling physics.
This same mechanism is at work in the “He-d2” molecule, with a critical role in determining the energy levels of “He-d2”. We will show the energy involved in bringing the second deuterium nucleus into the molecule and how the spin of the deuterium nuclei affects the energy states.
It might be expected that the “He-d2” molecule having angular momentum, would emit a photon, as helium does, and drop to a lower energy and angular momentum state; ending up in the ground state with no angular momentum. If this were to happen, we would expect the standard fusion reactions, with the emission of neutrons and the production of tritium. This would not sustain a chain reaction, and there is no experimental evidence supporting this as a major reaction channel. In the next chapter we will show how the angular momentum and spin states of “He-d2” vastly minimizes the formation of tritium and forbids the reaction to He3 plus a neutron.
Figure 5-1
We will analyze the “He-d2” molecule and find that states with both deuterium nuclei in the same “orbit” having a much lower energy than if the two deuterium atoms are in different states. From the wave functions of the “He-d2” molecule we calculate the fusion half-lives of the stable “He-d2” angular momentum states. This will require a little review of some relatively simple time dependent probability calculations derived from the time dependent Schrodinger equation. Basically, we calculate the probability of finding both deuterium nuclei at the helium surface at the same time using the square of the wave functions (=probability). Once the three nuclei are all together, forming a beryllium compound nucleus, there are no angular momentum constraints to continuing the fusion reaction. The orbital angular momentum of the “He-d2” molecule is lost to the linear momentum state electrons when the nuclear force attracts the three particles together breaking the correlation of the electrons with their respective deuterium-helium bonds. The reaction sequence is depicted in figure 5-1. The beryllium compound nucleus then decomposes into two helium nuclei, giving off ~24MeV of energy; 12MeV for each alpha particle (Helium nucleus).