Yes Eric, exactly but that orbital for the hydrogen where it's orbital radius is reduced is the s orbital. My thinking on this is that the s orbital of the embedded proton under immense pressures from Ni lattice electrons and general free electrons of the Ni d state - conduction electron cloud. That should collapse the s orbital to a much smaller radius than H would have in a gas state. I think it might be so strong that it forms the "virtual neutron" that could then enable this LENR process. Great discussion in the Ni62 + (VN) -> Cu63 thread.
You know Eric, it is a matter of what the wave mechanics work out to be, It's not a hydrino, I don't think there is a sub s orbital, but the core proton is screened like one. It would not surprise me that the solid state quantum mechanics of metals is distorted in a hydrated state. It also would not surprise me to see the wave mechanics of a hydrogen atom altered by the solid state environment. If the conduction band has an over potential of electrons, it seems like it might influence the proton's wave function radial diameter to compensate. The only thing that it really need to do is shield the proton until it's in the sphere of the strong force's influence. On Tue, Apr 30, 2013 at 1:53 AM, Eric Walker <[email protected]> wrote: > On Mon, Apr 29, 2013 at 10:39 PM, Chuck Sites <[email protected]> wrote: > > What I'm suggesting is that in an electric field like the background of >> charge of the electrons in a metal will reduce the orbital radii of an H in >> that metal. That effect is seen in Rydberg atoms in an electric field. I >> see no difference here. >> > > Rydberg states are typically higher energy than normal, and although they > can be deformed under a field in various ways, in general they extend well > beyond the nucleus because of the additional energy. Are you thinking of > one of the Rydberg orbitals where the electron passes close to the nucleus > on one side? > > Eric > >
