In reply to Abd ul-Rahman Lomax's message of Sun, 08 Apr 2012 21:51:25 -0500: Hi, >At 12:31 AM 4/8/2012, [email protected] wrote: >>In reply to Abd ul-Rahman Lomax's message of Thu, 05 Apr 2012 11:34:24 -0500: >>Hi, >>[snip] >> >Widom-Larsen theory completely fails to explain the actual >> >experimental results of cold fusion experiments, particularly the PdD >> >reactions of the Pons-Fleischmann Heat Effect. >> >>Not that I'm a fan of WL :), but: >> >>D + e- => 2 n >> >>Pd106 + 2 n => Ru104 + He4 + 11.9 MeV > >What "experimental result" does this explain? > >Pd106 + 2 n would become Pd108, which is stable.
Actually Pd106 + 2 n would give Pd108 + 15.76 MeV. IOW a highly excited Pd108 nucleus which has to find some way of getting rid of that energy. One fast way of doing so, is some form of fission, such as e.g. emission of an 11.9 MeV alpha particle. Note that this type of reaction (i.e. a "fusion" reaction followed by a fission reaction is not without precedent. Two examples: N15 + p => C12 + He4 (rather than O16, by a wide margin; see stellar carbon cycle.) U235 + n => fission into a wide variety of fragments. In the case of Pd106, is may appear less likely because Pd is closer to the peak of the binding energy curve, however the addition of two neutrons concurrently provides roughly twice the excitement energy involved in the other two examples provided. > > >>Granted 11.9 MeV isn't 23.8 MeV, but it is about half, and I'm not convinced >>that the He4/heat ratio has been measured all that accurately. > >The problem, Robin: the difficulty in measuring helium release is in >capturing all the helium. The released energy is reasonably well >measured through the calorimetry. It is suspected that, in general, >about half the helium is trapped in the cathode. If the reaction is a >surface reaction, and if helium is born with some energy (it could be >below the 20 KeV Hagelstein limit), half the helium will have a >trajectory inward to the cathode. The rest will come off with the >evolving gas, and be measured. So the heat/helium numbers from >experiment, unless adjusted according to some assumption like this, >tend to be higher than the actual reaction Q, double or so. > >Not lower. Yes, but as you said the trouble is in measuring the He. Suppose that the reaction is not as much a surface reaction as you think, and even more than half is trapped in the cathode. That would imply a lower reaction energy. (Note also surfaces are rough at the atomic level and that cavities may have little or no line of sight connection with the environment. IOW nearly all energetic alphas formed there may end up in the cathode metal rather than in the environment.) As I said I'm not convinced the heat/helium ratio has been determined with sufficient accuracy to rule out reactions of the type I suggested. > >I personally find it frustrating that more work on measuring Q, and >improving accuracy, with more complete capture of the helium, hasn't >been done. As do I. >I've been suggesting that experiments be run with a >platinum wire cathode, on which would be plated palladium. (This is >done in some SPAWAR "co-deposition" experiments." I'm putting >codeposition in quotes because these are apparently not actually >codeposition, because they first plate out the palladium, then raise >the voltage to start evolving deuterium. The experiments I know of >with a platinum wire cathode were not designed to measure heat or >helium, though.) > >In any case, once the experiment is done and XP measured, then the >electrolysis would be reversed and the palladium dissolved, which >should release all the helium. Excellent idea! Alternatively, the cathode might be melted. >It needs to be a platinum base wire >for the cathode or it would break up. Just my idea. > >Storms, however, estimates 25 +/- 5 MeV/He-4, and it's reasonable >from the data. 12 MeV would not be. > > >>Furthermore, >> >>Pd104 + 2 n => Ru102 + He4 + 13.75 MeV and >> >>Pd102 + 2 n => Ru100 + He4 + 15 MeV > > >If I'm correct, there is no evidence that dineutrons are even formed, I said I was no supporter of WL. I was just pointing out a possibility (playing devil's advocate if you will ;) >but without the dineutrons, you would have two reactions necessary, >and a serious rate problem. There is no evidence that dineutrons >would be absorbed in toto, the dineutron is a transient phenomenon. Indeed (in fact I share your objections to WL). With two consecutive neutron absorptions, you would almost certainly get serious gamma radiation, because e.g. the Pd106 would first go to Pd107 (producing a gamma), then almost none of it would go to Pd108 (because the percentage of Pd107 present would be minute). This would thus also leave radioactive Pd107 in the cathode. > >W-L theory would predict a complex of transmutations, but none of >them release as much energy as the transmutation of deuterium -> >helium. These transmutations would show a predictable relationship to >the elemental mix in the close environment of the cathode surface. >Palladium would, of course, be a common activation target, and if the >targets decay by alpha emission, then we'd have hot alphas. ...which of course only have a range of microns. > >I have seen no experimental evidence that such a mix of transmuations >is actually found. http://www.lenr-canr.org/acrobat/MalloveEalchemynig.pdf ? >Transmutations are certainly reported from FPHE >experiments, but at very low levels compared with helium. The >reactions described would all produce anomalous isotopes of Ruthenium. Not necessarily. Fission reactions not involving He4 may also be possible. e.g. Pd106 + 2 n => Zr96 + C12 + 11.7 MeV Note that such reactions would also lower the amount of Helium found, thus also increasing the heat/helium ratio. C12 nuclei would also be much slower than He4 nuclei, so such radiation would be far less noticeable (Hagelstein). > >Hot alphas, i.e., energetic helium nuclei, above 20 KeV, break the >Hagelstein limit, they would be observed. I have a bit of a problem with this, in as much as it appears to assume that everyone is always trying to measure it in every experiment. >Charged particle radiation >from FPHE experiments are at quite low levels, not the high levels >that would be necessary if the helium is being produced by alpha emission. > >Pd-104 is stable, so why would Pd102 + 2 n not simply become Pd-104? See above. BTW note that fusion with a deuterino offers similar reaction possibilities, but without the objections of dineutron stability, and the need to find the 3 MeV required to convert the deuteron into the dineutron. :) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
