Fusion Reaction Rates- He-d2 => [Be] => 2 He
With the two deuterium nuclei stuck in their respective even angular momentum states, it is only a matter of time before the two deuterium atoms end up at the helium nucleus simultaneously. We will use a technique first developed in the 1930’s to explain alpha emissions due to tunneling away from radioactive nuclei. Instead of the alpha particle tunneling out to escape a nuclei, we have the simultaneous “tunneling” of the deuterium in, to the surface of the helium nucleus. Both problems involve the penetration of the coulomb barrier, with the alpha particle trying to get out, in the case of alpha emissions, and the deuterium nuclei trying to get in, with Fleischmann Pons fusion. The presence of the stable wave function for the deuterium nuclei around the helium nucleus allows this calculation to be made.
First a brief derivation of the probability flow using the Schrodinger equation.
(taken almost verbatim from “An Introduction to Quantum Physics”).
Figure 5-4
We are looking for an expression that will give the time variation in probability for a particle tunneling out of a stationary state. The alpha particle emission problem is shown in figure 5-4. The free particle wave function is shown as Aeikx. This is the standard free particle wave function. The coefficient “A” is the value of the “stationary state” wave function at the point that it “leaves” the potential that the particle is tunneling through, becoming a “free particle”. The wave function in the nucleus is Boeikx+ Be-ikx. If we put the “free particle” wave function into equation 5-16 we get (changing the variable from “x” to ”r” for radius):
For the Fleishmann-Pons fusion reaction the only information required to calculate the “He-d2” half-life is the ratio A2/B2 (more accurately A12A22/B12B22). We will assume that the reaction is so fast that when the compound beryllium nucleus forms, we can assign the free particle wave function to the two alpha particles leaving the reaction. This is reasonable since the half-life of the compound nucleus will be on the order of 10-22 seconds, about the time it takes the alpha particle to traverse the nucleus.
Figure 5-5
Figure 5-6Once we calculate the wave functions for the He-d pairs in “He-d2”, it is straight forward to get A2/B2. All that needs to be done in the wave function calculation is to force a constant value for the wave function in the nucleus. This is basically telling the Schrodinger equation that the particle is a free particle in the nucleus. This is shown schematically in figure 5-5. The square of the n=2,L=1 wave function used for the calculation is shown in figure 5-6. The zone near the nucleus is expanded to show how the wave function is held constant in the nucleus, depicting a free particle. These calculations assumed a nuclear radius of 2 Fermi. When the wave function is normalized, this value of the wave function in the nucleus automatically becomes “A”. Since at t=0; B2=1 => A2/B2 =A2. We have neglected the change in the function from a linear one- dimensional problem to the three-dimensional problem. This just requires that we use U=rΨ instead of Ψ. This assures that the probability flow has the units of (probability/area). Since there are two particles, and thus two wave functions involved, we must manipulate the equation to give the form that we need. This just requires taking the differential of the product of the two wave functions as below.
Now, we must decide what mass, or reduced mass, to use in “k” above. The mass, m, in equation 5-17 is the reduced mass between the helium and deuterium nuclei. The mass leaving the helium nuclei will be 2*deuterium nucleus = helium nucleus. The full center of mass velocity is appropriate for this mass as it represents the departing of the two deuterium nuclei as helium from the compound nucleus. Thus we get equation 5-18.
These half-life calculations are sensitive to the assumed radius of nuclear interaction; 2 Fermi in this case. Experimental data will be needed to calculate accurate half-lives. The half-life for the total L=2 “He-d2” state is only ~0.00024 seconds. We will see later that this short half-life explains the bursts of heat in micron sized areas in the Navy work with co-deposition of deuterium and palladium. The ~54 minute half-life, from the total L=4 state, explains the time lag in getting bulk palladium reactions started; as it takes on the order of 100-300 half-lives to achieve a reasonable amount of steady state heat. This can be as much as 5 to 15 days depending on the efficiency of the helium channeling. This long half-life is also responsible for the infamous Fleischmann-Pons “melt down”. This will also be elaborated on in a later chapter.