There is a claim often made about cold fusion, that it is excluded theoretically. The main theoretical argument is that electronic energies are too low to overcome the Coulomb barrier, since d-d fusion only takes place at KeV energies, while chemistry is at eV energies.
This is belied by inner shells, which in Palladium store 3 or 20 KeV of energy per ejected electron, depending on whether the first or second shell is excited. These inner shell vacancies can decay either by x-rays, or by absorbing an electron into the vacancy and simultaneously ejecting a different electron (this second process is electrostatic). The cross section for ejecting a deuteron with tens of KeV's instead of an electron should be larger, since a deuterium is heavier. So I believe deuterated metal with excited inner shells has KeV deuterons running around.
If two KeV deuterons do a fusion to an alpha in a dense environment, close to a nucleus or to an electron, I don't know why the process cannot end without a proton or neutron ejected. There are electrostatic matrix elements that allow an unstable alpha-resonance to decay by giving its energy to a charged particle nearby, instead of ejecting a constituent.
After a fusion, the resulting alpha leaves an energetic track behind, and charged particles leave behind trails of atoms with ejected inner-shell electrons. So the K-shell holes produce fast deuterons, and fusion in deuterons produces K-shell holes. I don't see why this can't make a chain reaction.
I have explained this idea before. I would like to know whether somebody knows a sound theoretical argument which rules it out. Can such a chain reaction in a Pd be excluded theoretically? I am not asking if it is likely, I am asking whether it can be firmly theoretically excluded.
Anna v. asks how this process gets started— it requires a random charged particle to pass through the deuterated material, from spontaneous environmental radioactive decay, or a cosmic ray muon. Charged particles produce K-shell holes.
To make my biases clear: I can't exclude it. Regardless of the quality of the experiments, I don't see an argument against cold fusion.
Best Answer
A seemingly problematic aspect of the proposed mechanism is that it allegedly requires two hot deuterons. (By contrast, U-235 fission requires just one neutron.)
Why is that so problematic? If $n$ is the number of 20keV particles (i.e. hot deuterons, or K-shell holes, or some superposition of them), then we expect something like:
$$dn/dt = An^2 - Bn$$
where the coefficient $A>0$ describes fusion and $B>0$ describes cooling into lower-energy modes.
This describes a very badly behaved chain reaction. This differential equation supports explosive growth of the number of hot deuterons, and it supports very easily fizzling out entirely. I don't see how it could support a reaction that goes on for 50 hours, which is the alleged observation in cold-fusion experiments. (You can object that it moves from hot-spot to hot-spot within the electrode, but even so, I find it implausible. I wouldn't expect local hot-spots; I would expect a fast-growing hot region that would fuse almost every deuteron in the entire electrode within a second! Unless I'm mistaken...)
A U-235 reaction, where there is no $n^2$ term, is relatively easy to stabilize. In theory, all you need is a negative temperature coefficient. But even a negative temperature coefficient would fail to stabilize this kind of quadratic reaction rate (if I'm not mistaken).
I won't say this disproves the mechanism, but I would say that this is something that warrants explanation and discussion.
I also wonder whether you really need the energy of two K-shell holes, not just one, to overcome the Coulomb barrier. If one hole is enough, then the problem above does not apply. Moreover, you wouldn't really need to worry about hot-deuteron-lifetime (the other major potential problem with the mechanism), because maybe the hot deuteron doesn't travel around the lattice at all. Maybe there is a K-shell hole and two deuterons all together in the lattice, and the hole's energy simply shoves one deuteron into the other. (The hole's energy becomes Coulomb potential energy, not kinetic energy.)
Again, I don't know if 20keV is enough energy. But if it were, that would make the story much more plausible in my opinion. :-D