Reaction Times
The reaction time is controlled by the wave function overlap and the energy difference between the two states as discussed above. For intuitive reasons we will first estimate the time available for a reaction to take place. To do this we start at the point of closest approach and estimate how much time is required to move a significant distance; so as to move the overlapping wave functions far enough to slow the reaction significantly (i.e. reduce the overlap probability). We will guess that 0.01 angstroms of movement will not affect the nuclear wave functions that are on the order of 0.2 angstroms across. We will do the estimation for the shortest of the two reaction separations as a worst case; since the acceleration will be much higher at the 50ev of the 0.19 angstrom, compared to the 12ev at 3.3 angstroms.
The acceleration of the particles away from each other is estimated from the energy of closest approach = 50eV, from Figure 4-7.
We will assume that when the particles travel the 0.01 angstroms, the stabilizing electron energy will not change much, thus the force will be due to the 1/r2 nuclear repulsion.
To find the transition time from the initial standard collision state to the final linear momentum state, we will use the ~half life from above; t1/2 =2π ħ /ΔE. This gives a transition time of ~0.8x10-16seconds. If the overlap of the two states is say 50% then the times are on the same order. Thus, there is plenty of time for half the probability to make the transition (a half life).
A crude calculation of the particle separation versus time can be made using a semi-classical argument coupled with the uncertainty principle. If we assume that the reaction time is determined by the electron "communication" between the two nuclei, and all the kinetic energy going to the nuclei goes into the velocity towards each other, we get the separation versus time graph of figure 4-10d.
figure 4-10d
Consequence of the Three Angular Momentum States of “He-d2”
We will see in the next chapter that the three angular momentum states of “He-d2” will give three different half-lives (remember the two angular momentum states of “He-d” can combine in “He-d2” 3 different ways). The difference in the half-lives can be understood intuitively by recognizing that higher angular momentum leads to higher centrifugal force, and hence the particles spending more time further apart. The concentration of“He-d2” in each of the three angular momentum states can be calculated from the helium channeling probability, coupled to the reaction rates of each reaction. We will outline the general problem here and solve it in detail in the chapter on macroscopic behavior.
figure 4-15aThere will be 6 different reactions, 2 giving the “He-d” “intermediate”,and 4 giving final products, “He-d2” (see Figure 4-15a). The [alpha] is the unchanneled helium atom and [He] the channeled helium. First, a note on nomenclature. The reaction products will be either “He-d” or “He-d2”. These products and the angular quantum number, l=1 or 2, will be written with the angular quantum number as superscripts, separated with a coma for the “He-d2”(example: [He-d2]1,2 ). Each reaction will have its own rate designated as k1, k2 ……etc. The rates are calculated using a probability of transition as determined by the time for reaction coupled with the time the wave function “overlaps” at the critical separation. This will be developed quantitatively in the macroscopic chapter. As discussed above, too high of an energy will reduce the transition probability for the linear momentum bonding, and too low an energy will not allow the particle to reach the critical separations determined by angular momentum. The helium will be assumed to have constant energy loss in this energy range. This will be elaborated on in the macroscopic behavior discussion. The [d] represents the concentration of deuterium in the tetrahedral sites. The other reactants and products in brackets “[ ]” represent the probability “concentration” of each constituent. The unchanneled "alpha" starts out at “one”. As it moves through the crystal there is a probability of channeling, f(KE). the probability of each reaction below can be described by a differential equation (Figure 4-15).
figure 4-15 When coupled with the helium capture probability as a function of energy, the solutions to the differential equations will give the probability of the helium reacting to each species. Intuitively, it would be expected that when the helium is captured at lower energies, the l=2 state would be preferred, as it takes less energy to get to the larger reaction separation of 0.33 compared to 0.19 angstroms. The 0.33 reaction separation is estimated to have only a 12 eV activation energy. The center of mass energy comes into play here, which significantly lowers the available “activation energy”. This too, will be elaborated on later.