Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
1  
and how are the K shell electrons kicked out of the k shell? –  anna v Aug 4 '12 at 10:54
1  
@annav: By the fast charged particles produced in the fusion process. To start the process, you need a cosmic ray, or random spontaneous charged particle produced by radioactivity in the environment, to produce an inner shell hole, then its self-sustaining. I said this in the question. –  Ron Maimon Aug 4 '12 at 15:55
    
"and simultaneously ejecting a different electron (this second process is electrostatic). The cross section for ejecting a deuterium with tens of KeV's instead of an electron should be larger, since a deuterium is heavier".Deuterium is neutral, deuteron (the ion) is positively charged. Why would a neutral deuterium get involved in an electrostatic ejection in tandem with the absorption of the electron in the k-shell?. From what I know the deuteriums fill dislocations in the crystals, are interstitial. –  anna v Aug 4 '12 at 16:56
    
@annav: I meant deuteron, I fixed it just now. The deuterons fill the interstitials, the electrons are just shared. The electrostatic process is always particle by particle, so it ejects electrons/deuterons at KeV energies, but the phase space for deuterons is bigger by a factor of 50 (although there are more electrons around than deuterons, also by a factor of about 50). –  Ron Maimon Aug 4 '12 at 16:59
1  
In that case one has to calculate the density of deuterons and the probability of meeting another deuteron to start the chain and sustain it.As fusion is a strong interaction and deuterons are interstitial, i.e. at electromagnetic distances, I would guess that there are orders of magnitude missing. –  anna v Aug 4 '12 at 17:12
show 8 more comments

2 Answers 2

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

share|improve this answer
    
Ron emailed me. Summary: One K-shell hole is enough energy, but the deuterons would be too far from the Pd nucleus for it to participate. So an electron would have to be the 3rd body instead. This is possible but there are some reasons to think it is less likely. About kinetics and stability: He says there is a series of micro-explosions. Maybe each explosion is centered on a region of unusually high D-density (drawn by a low electric potential), and the reaction can't engulf the whole electrode because D density is lower elsewhere, below $B/A$. There is more discussion I'm leaving out. :-P –  Steve B Mar 19 at 2:56
add comment

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

An inner shell electron can be ejected from an atom by high energy X-rays (or other high energy processes). This is not particular to palladium, but can occur for any element. If fusion occurred due to exposing palladium deuteride to a high energy source such as X-rays to eject inner shell electrons, this would not be “cold” fusion, but the real question is whether or not fusion can take place by the mechanism described in the question.

It is misleading to state “palladium stores 3 to 20 KeV of energy per ejected electron”. The lifetime of the inner shell hole left by an ejected electron is only about 1 femtosecond (source: Core Level Spectroscopy of Solids, page 11).

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).

As atomic number increases, the decay via X-rays is increasingly dominant. For palladium, decay by emission of an X-ray and no electron is 5 times as likely as by ejecting an electron. The process of ejecting a different electron is not exactly electrostatic. Instead an X-ray formed by the transition of an electron from a higher energy level to the lower energy level of the hole (such as 2p to 1s) sometimes excites another electron causing it to be ejected (such as another 2p electron) (sources: Core Level Spectroscopy of Solids, page 13; Burhop and Asaad, Advances in Atomic and Molecular Physics, vol. 8, page 165).

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.

It seems reasonable that the energy of an X-ray photon could be transferred to a deuteron. See Proton ejection from molecular hydride clusters exposed to strong X-ray pulses . However, the X-ray formed by the hole being filled would be less energetic than the X-ray that formed the hole in the first place. There would be a better chance of forming an energetic deuteron from the original X-ray that created the hole. Also, the deuterons in the palladium lattice are obviously outside the inner shells of the palladium atoms. (See Nelin "A Neutron Diffration Study of Palladium Hydride" phys. stat. sol. (b) 45, 527 (1971) for the structure). The X-ray from the hole-filling transition would have to pass through all the inner shell electrons to reach a deuteron.

In conclusion, if enough X-ray energy could be focused upon a deuterium-containing material, fusion is possible (like the Nation Ignition Facility does with laser focusing to indirectly yield X-rays). However, placing deuterium in a palladium lattice does not offer any benefit. Deuterium could be made more concentrated in other forms such as solid deuterium. Invoking indirect energy transfer via X-ray photons from the filling of inner shell holes offers no benefit over direct transfer of energy from the high energy X-rays needed to create the holes in the first place.

share|improve this answer
    
This is a clear argument. 1)One "objection", it is based on probabilities of scattering , i.e. no calculations or specific data. 2)it does not take into account the lattice structure which can have collective enhancement effects on probabilities. For example, irrelevant to topic, muons pass crystals without scattering when in the direction of the crystal axis. –  anna v Mar 16 at 4:35
    
A deuteron cannot just absorb an x-ray and wind up with extra kinetic energy. Such a process cannot satisfy energy-momentum conservation. (Proof: Think about it in the center-of-mass frame.) That's why your description of ordinary Auger emission -- first an x-ray is literally emitted, then another electron absorbs it -- cannot be correct, even if you found a book that says it. The book is wrong, it's electrostatic. If you look at theoretical papers calculating Auger rates and shifts you'll find that specialists agree it's electrostatic. –  Steve B Mar 16 at 14:00
    
In the mechanism under discussion, the (alleged) inner-shell holes are created not by x-rays but by fast charged particles. A 20MeV alpha could theoretically create a hundred 20keV deuterons by slamming into them all almost dead-on, but this is extraordinarily unlikely. OTOH, if the alpha is traveling through Pd, it can quite frequently create a hundred inner-shell holes. ALSO, this is allegedly a three-body fusion process where the palladium atom is one of the bodies. So in summary I don't know how you could say that the palladium lattice is not helping ... it's critical! –  Steve B Mar 16 at 14:14
    
@SteveB "The book is wrong". This is the standard theory, no just one book. In addition to Core Level Spectroscopy of Solids, see Burhop and Asaad, "The Auger Effect", in Advances in Atomic and Molecular Physics, vol. 8, page 165: "an X-ray quantum produced in an ordinary inner shell radiative transition being scattered by an electron in an outer shell of the same atom, the electrom being ejected in the process". –  DavePhD Mar 16 at 17:46
    
@SteveB "In the mechanism under discussion, the (alleged) inner-shell holes are created not by x-rays but by fast charged particles" High energy X-rays are just an example of how holes can be created in the inner shell. Historically, this is how the Auger Effect was discovered. Pierre Auger applied high energy X-rays and observed the Auger Effect electrons. Certainly holes can be created by other high energy processes such as through a high energy electron beam. –  DavePhD Mar 16 at 18:03
show 6 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.