Fate of “He-d2” Angular Momentum
Bringing angular momentum into the He+d+d nuclear reaction would necessitate the emission of a gamma ray to shed the two l=1 units of orbital angular momentum. Obviously, this does not happen, as no gamma rays are found experimentally. So what happens to the angular momentum? To answer this we must look back at the Schrodinger equation in chapter 3 (eq 3-6 and 3-7) for the “He-d-electron” system of particles, as reproduced below in equations 5-20 and 5-21. (recall “B” is the electron and “A” the reduced mass nuclear wave functions: particle 1 & 2 are the deuterium and helium nuclei, particle 3 is the electron);
These equations are still valid, but the potential V1,2 changes from the repulsive force
V1,2= 2e2/r ,
to an overwhelming attractive nuclear force. The attractive force will break the correlation of the linear momentum state electron with the helium and deuterium nuclei, making the E* in equation 5-21 vanish. The electrostatic potential “felt” by the electron does not change, but the ability to “do work” on the two nuclei has gone away. Without the ability to do work on the nuclei, the electron can no longer remain in the low energy state. This would be like taking a “1s” electron in a hydrogen atom and putting a photon (the work done to drop to a lower state) back into the electron that was emitted when the electron dropped to the “1s” state (i.e. it puts the electron back to an excited state).
Remember that the original wave function, Ψ, is composed of the multiplication of the two wave functions “A” and ”B” (A=He-d and B=electron wave functions) ;
Ψ = AB= A(ro,θ,φ )B(x) = R(ro) Θ(θ) Φ(φ)B(x)) = R Θ Φ B
In this equation the “R” is the radial component of the reduced mass He-d wave function, Θ and Φ are the angular components of the He-d wave function and “B” the radial component of the electron in a linear momentum state, which is correlated to the position of He-d. This makes R, Θ and Φ components of the “B” electron wave function. Equation 5-20 is still correct after the nuclear interaction, but the interpretation of the del squared operator must change. The two nuclei are arguably still there, but the electron can no longer do work on the nuclei repelling each other because the attractive nuclear force is now controlling. This is what breaks the correlation. The del squared operator has changed from a linear momentum state operator, to a space filling operator. In essence, the difference is in the normalization of the electron wave function. When the electron was correlated with the helium and deuterium nuclei, the normalization was done to the linear momentum state. This is a one-dimensional state, much like any radial component of a central force wave function, given volume by the angular distribution of the nuclei wave function. With the correlation broken, the standard central force, radial space filling del squared operator must be used, as in equation 5-22. The solution to this equation, for the electron, is the standard hydrogen solution modified for the presence of more than one electron and a different charge. The potentials from chapter 3 between the electron and the helium and deuterium nuclei, V1,3 and V2,3 , are still the proper electron potentials for the problem, but they are held in close proximity. The potential is simply the nucleus of beryllium, with a plus 4 charge. Note that equation 5-22 has the standard angular components, and will thus have the solution with angular momentum given by L=l(l+1)1/2ħ, and shown in equation 5-23. When the helium and two deuterium nuclei collide simultaneously, to create the beryllium compound nucleus and break the correlation with the linear momentum state electrons, the electrons simply stay in the same combined wave function Ψ (= AB). The solution to the angular part of the wave function remains exactly the same. The potential that the electron “sees” is exactly the same; the difference in the radial dependence of the electron wave function is due to the radial space filling operator. This is manifest in the rearranged Schrodinger equation as the radius times the wave function as seen in equation 5-23, instead of just the lone wave function. This simple change to a space-filling operator with rΨ instead of just Ψ, is the difference between an electron bond energy with 10’s of electron volts and ~150,000eV.
The electron is then “excited” back up to near the energy that it started with as it entered the “activated complex” discussed in chapter 4. Since the electron kinetic energy in the He-d bond was due to the work done against the He-d repulsion, when that repulsion disappeared, due to the much stronger nuclear attraction, so did the wave function that provided the ~150,000 eV of stationary state kinetic energy. The loss of the very high kinetic energy can be thought of as going into pushing the electron out of a very deep potential well. This does not mean that the electron has time to move anywhere(it doesn’t); it just means that it entered the new state as part of breaking the correlation of the linear momentum state. The overlap of the two states facilitate this. The nuclear interaction and the breaking of the correlation are simultaneous occurrences, with a half-life determined by the 23.8 MeV fusion energy.
A quick calculation shows where the electron energy has gone. The electron had a kinetic energy of ~150,000eV in the He-d bond, with an average linear momentum state radius of ~25x10-15meters (25 Fermis). With a total charge on the deuterium plus helium of three, this gives a potential of;
E=3*1.44x10-9/25x10-15~170,000eV
The 170,000eV is on the same order as the He-d bond of 150,000 eV. This brings up the question of where the kinetic energy of the nuclei has gone. The nuclear kinetic energy was used to tunnel through the helium coulomb barrier, thus converting the nuclei kinetic energy into potential energy. The coulomb potential energy is stored in the beryllium compound nucleus, which breaks into two alpha particles within ~10-22 seconds.
The two linear momentum state electrons are forced into the “2p” state of beryllium as part of the nuclear attraction; each electron with a unit of angular momentum (3d state in the case of the longer half-life “He-d2”). Generally, electron transitions can not happen this fast, but since the electron is intimately linked to the nuclear wave function, the overall half-life of the reaction is determined by the nuclear force. There is no photon emission involved to slow the transition of the electron. Photon emission half-lives are on the order of 10-16 to 10-7 seconds, much too slow to affect the d+d+He fusion reaction with a half-life of ~10-22 seconds. This is why we can use the alpha decay model, since the rate-determining step is completely controlled by the simultaneous tunneling of the deuterons through the helium potential barrier, which simultaneously breaks the electron-“He-d” correlation. The electron does not have time to move; there is simply a change in the electron wave equation operator. The alpha particles leave the compound nucleus with 12MeV each and of course, totally disrupt the electron configuration leaving behind free electrons (this has no bearing on the fusion reaction).
We have shown how the overall fusion chain reaction takes place, giving two intermediates with vastly different half-lives, and only 12 MeV alpha particles as by-products. These alpha particles have a range in palladium of less than 20 microns, so few escape the palladium for detection. There are no radioactive by-products or gamma radiation from this reaction except for the low level x-rays from the alpha particle collisions with electrons that are common to any alpha emitter. The different intermediate half-lives will lead to interesting macroscopic behavior to be discussed later. We have not yet justified the suppression of the expected “d+d” fusion reaction to [tritium + proton] and [helium3 + neutron]. We will elaborate on this in the next chapter, which will set the stage to describe the overall “macroscopic” mechanism as promised in chapter 4.