In reply to Eric Walker's message of Fri, 10 May 2013 17:05:05 -0700: Hi, [snip]
In the situation you describe, there are going to be lots of spallation neutrons, because you only need 2.2 MeV to break a deuterium nucleus into a proton and a neutron, and heavy water is all deuterium. It's going to be a prime source of neutrons, as Jones is fond of pointing out. ;) So far from creating a situation where there will be less neutrons, you have in fact created one where there will be more. In fact this a good reason for suggesting that CF in the D/Pd experiments is not mediated by (very) fast alphas. Neutrons combining with the glass, or the cathode will produce gammas. (The neutron capture cross section of heavy water itself is quite low, which is why it's used as a moderator/coolant in some reactors.) >On Fri, May 10, 2013 at 3:21 PM, <[email protected]> wrote: > >In short, very roughly, a 1 W unshielded power source would double the >> background rate. >> > >Thank you for the numbers. Twice background doesn't sound like all that >much; presumably this is near the threshold of detection, and a signal >would be easy to swamp out with noise? > >The alphas would be Ron's alphas, at 22.9 MeV. I'm trying a thought >experiment where the secondary spallation neutrons are somehow minimized -- >I don't have an explanation for why this would be the case at this point, >but I'm curious anyway. The setup I'm thinking of is something like this: > > | air | glass | heavy water | cathode surface | active >region | cathode interior | > >Here the cathode surface is assumed to be very thin. In the scenario I'm >trying to better understand, where the spallation neutrons are somehow >avoided, I'm wondering what the activity would like like from the vantage >point at the far left, at "air". The alphas could potentially travel for >quite a while through the cathode before encountering a lattice site, I >think I remember reading, during which time they will dissipate energy by >way of low-level EMF. I assume that EMF will be stopped by the cathode >surface, the heavy water and the glass, before reaching the air -- is this >a mistaken assumption? No, in fact all of it will be stopped by the free electrons in the cathode(surface). Note however that we are specifically talking about UV and lower energy levels here. High energy X-rays &/or gamma rays will escape easily. > Like you say, there will no doubt be inelastic >collisions, metastable nuclei and gammas. But assuming little neutron >activation, do you have a sense of what the activity would be like outside >of this kind of "shielding"? Aside from the effects caused by spallation neutrons, I doubt you would see much from a purely alpha reaction such as Ron's. I say this because apart from previously mentioned things, the only other form of energy that is likely to escape the cell is high energy X-rays, and to create these, you need high energy electrons. The highest energy electron you can create in a head on collision with a 23 MeV alpha is M_e/M_alpha x 23 MeV = 3152 eV. Even if all of the kinetic energy of such an electron is converted into a maximum energy X-ray through the bremsstrahlung mechanism (and it rarely is), you only get a 3 keV X-ray, which is not very penetrating. I'll leave it up to you to figure out the mean free path and transmission fraction in the various materials of 3 keV X-rays. > >The question I'm trying to get at is whether we can say for sure that the >number of energetic particles (in this case alphas) in the cold fusion >experiments is not commensurate with heat. I think most (>99%) of the energy from the alphas would convert to heat in the cell. Note however that a doubling of the background rate is easily detected, especially if it turns on and off with the cell, and BTW so is a neutron production rate of that magnitude, if you are using neutron detectors. I said in a previous post a few weeks back that Geiger counters were not particularly good at detecting neutrons, however I forgot about the prompt gammas which would be internally generated in the Geiger counter, so I would now expect them to be more sensitive than I first thought, but the sensitivity would depend on the precise elements, quantities and geometry used in their manufacture, and therefore would vary from one brand/type to the next. [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
