Hi Robin, I think you are right but even a few tens of eV may be a welcome complement to the hypothesized main screening process, soon to be tested experimentally in a reputed lab BTW.
In a nutshell it assumes that the excess surface electrons can get within tunneling distance, and often enough, two low energy simultaneous deloading and incident deuterons headed to the same ~0.02 angstrom thick "electron pool" at the center of each unit cell of the 3 angstrom square surface Pd grid. The rationale is that only intervening negative charge can make up for the lack of kinetic energy, and we happen to have a large mine of such potential screening charge at the cathode surface (the negative side of the electrical double layer), as well as the required aligned counterfluxes in some occasions, as rare as the CF phenomenon itself but fortunately they can be enhanced e.g. by judiciously conducted double sided electrolysis. The exact QM mechanism for this hypothetic process is far from elucidated but in a classical approximation, when the respective arrival times are sufficiently close, I picture the screening electron as hesitating between the two deuterons and getting trapped half-way in between, oscillating locally in something similar to the L1 stable point in gravitational 3-body interactions, at least for the short time required for the deuterons to tunnel to each other. The rest (most) of the time, each deuteron would of course pick up an electron of its own, associate with another D and bubble up. Does this make any sense to you? Michel ----- Original Message ----- From: "Robin van Spaandonk" <[EMAIL PROTECTED]> To: <[email protected]> Sent: Monday, October 01, 2007 10:49 AM Subject: Re: [Vo]:Re: #CF hypothesis (was Re: surface electron layer catalyzed fusion hypothesis) In reply to R.C.Macaulay's message of Sun, 30 Sep 2007 19:37:55 -0500: Hi, [snip] > >Robin wrote.. >>It can't be any more than the chemical binding energy of hydrogen to the >>lattice. A few eV at most. When the lattice springs back, it's not going to yield any more energy than it took to expand it in the first place. In fact you also need to break the Pd-H bond, though that will largely be compensated for by the formation of H-H bonds. These energies are not large. To rip Pd atoms apart altogether, you need to vaporize it, which happens at 2940 deg. C. That implies an average energy of 3/2 x k x 3213 K = 0.415 eV /atom. You can expect much less from contraction. Even if a few atoms all contract at once, you need to consider that the amount of contraction will get less as you get farther from the emission point. My guess would be that by the time you are three lattice cells removed, the contribution would be negligible. Even 3^3 = 27 * .415 eV is only about 11 eV. That's not even going to make a dent in the Coulomb barrier. [snip] Regards, Robin van Spaandonk The shrub is a plant.
