Adding an Interstitial Atom
Now, let us look at the wave function of the lowest energy level, n=1, if another atom is added to the crystal so that it is between two of the other atoms, say between atom 4 and 5 from the left. Figure 2-12 shows the potential, and the wave function for this situation, which uses the same atomic potential as the other atoms in the crystal. Note how it is possible to localize a free electron to these three closely spaced atoms, at least in this one dimensional model.
Figure 2-12
This does not happen in a similar situation in a real 3-dimensional crystal. This of course would not be an equilibrium separation, as this small decreases in the energy level of the electron, is not enough to overcome the repulsive potential of the positively charged nuclei. This could be a short-lived state (activation state), however, as a high energy particle passes through the crystal, using kinetic energy to overcome the repulsive force and approach the nuclei. This would be analogous to how high temperatures aid standard chemical reactions to achieve activated complexes. The wave function of an electron can easily adjust to a passing particle, if the particle’s speed is much less than the electron speed. Free electrons in metals have energies on the order of a few electron volts, say 3 eV. This would mean that an alpha particle with much less than ~22,000 eV as shown below would interact chemically with the free electrons.
The electrons around an alpha particle with 100 to 500 eV could easily adjust their wave function as if the alpha particle were stationary in the crystal.
An alpha particle, with ~300 eV, approaching a deuteron would have a center of mass energy of about 200 eV, leaving 100 eV to overcome the repulsive potential energy of the nuclei. This would allow the alpha particle to come within about 0.2 angstroms of the deuteron. So let us see what happens to the wave function in the crystal, at this separation.
Figure 2-13
Figure 2-13 shows the potential, and the wave function for this situation. Note that again the wave function is localized to the two closely spaced atoms. The energy of this electron is ~19.9 eV, which is higher than the other free electrons, but far short of the over 70 eV required to overcome the electrostatic repulsion. This is where the lack of core electrons of hydrogen and helium, coupled with the solid state, linear momentum state electrons, can create extremely strong bonds. We will not discuss the full wave function quite yet, but show a part of the mechanism.
Under most circumstances, hydrogen will behave as the two closely spaced atoms above. Although there are no core electrons, the electron density around the “metallic” hydrogen is filled by all of the free electrons. To picture this look back at figure 2-9, which shows the sum of all the linier momentum states, giving the total electron density. Around each nucleus, there is some probability of finding every free electron. The presence of some probability from all the electrons builds up around the proton giving an electron probability density that is not too different from a lone hydrogen atom charge density. The difference is that the charge is made up of states from every free electron in the crystal, with no single electron with much of a presence.
figure 2-14
Now, look at figure 2-14. This is the ground state (n=1) wave function for the same situation shown in figure 2-13, except the “core” electron potential is removed from the two nuclei that are close together. Figure 2-14 shows the potentials as well as the ground state, n=1, wave function. The energy of the ground state electron, from these crude calculations, is 1,544 eV. This is more than enough to overcome the ~72eV repulsive energy of the protons at the separation 0f 0.2 angstroms. Figure 2-15 shows an expanded length scale for the same n=1, and also the n=2 wave functions. The linear momentum state has allowed for a very high energy for the electron, which lies well down in the potential well of the bare nuclei. This is the first hint of the extraordinary bond energy obtainable via the localized linear momentum state. Recall that the three dimensional filling of space is made possible by the partially overlapping nuclear wave functions.
Figur2-15
Now we replace the interstitial proton with a helium nucleus with a +2 charge. Keeping the separation at 0.2 angstrom, the n=1 energy state is 4993 eV, as shown in an expanded length scale in figure 2-16. The n=2 state (figure 2-16) is much lower at 1400 eV. The n=3 energy level is 24.8 eV and is still localized around the helium as shown in figure 2-17.
It is this state that is critical to the “activation complex”. Note that the electron wave function average radius ~0.3 angstroms, is about the same as helium, 0.26 angstroms. This is part of the overlap of the molecular state with the linear momentum states. Note that with the proton, the n=3 state is not a localized state. The state is still spread over the entire sample as a free electron. Hydrogen would need to engage the n=2 state to reach an “activation complex”. This state is on the order of 1000 eV.
Figure 2-6
This is probably responsible for some of the tritium, from a d+d reaction, found in glow discharge experiments with deuterium and palladium. Since channeling at 1000 eV is not very probable, this will be a low cross section reaction, but not nearly as low as the d+d fusion reaction at 1000eV
Figure 2-17
