In reply to Eric Walker's message of Sat, 11 May 2013 18:04:19 -0700: Hi Eric,
I think your summation is quite good. >Very interesting discussion. If can summarize the main points in my own >words, it would be something like this -- for a hypothetical heavy water >electrolytic system in which watts of prompt alphas are being produced in a >palladium cathode by way of a hypothetical d+d+Pd?4He+Pd + Q (22.9 MeV >kinetic energy) reaction, you can expect the following: > > - There will be plenty of spallation neutrons due to collisions between > alphas and heavy water molecules, where the deuterium nuclei are broken > apart. These neutrons will either escape the system or activate the > surrounding material, resulting in easily detectable gammas. > - (To add Hagelstein's point, if I have understood it: in addition, > prompt alphas can be expected to collide with deuterium nuclei in the > cathode, and deuterium nuclei in the cathode scattered by alphas can be > expected to collide with one another, providing an additional source of > neutrons.) > - The number of spallation neutrons can be expected to be large, since > it takes only a fraction of the energy of an alpha with 22.9 MeV to break > apart a deuterium nucleus. > - There would also be x-rays, with a peak in the 3 keV range. These > x-rays will be stopped before leaving the system. > - The doubling of the radiation level above background described in an > earlier post would be easy to detect, although there would be some > subtleties relating to the specific detector that is used. > >For the mean free paths of 3 keV x-rays, I get small numbers: 46 microns in >water, 2 microns in palladium and 6.2 cm in air. So it seems pretty clear >that the x-rays are unlikely to make it to a detector after traversing the >outer layer of palladium, the heavy water and the glass. > >On the basis of all of this, I'm wondering if it is safe to conclude the >following: any radiation exiting a system of this kind involving fast >alphas would be attributable solely to neutrons (spallation and fusion) and >the activation gammas they lead to, and to inelastic collisions between >alphas and other nuclei and the resulting gammas. Any other radiation can >be expected to be quenched and to be undetectable apart from a general >increase in temperature of the system. Is this conclusion too broad? > >Just to call out two important assumptions here: > > - Prompt alphas will escape from the cathode and make it into the heavy > water at still-high energies. > - Prompt alphas travelling within a palladium lattice will scatter a > significant number of deuterium nuclei, and a significant number of fast > deuterium nuclei will scatter with one another; i.e., there is no mechanism > that somehow segregates the locations of the two types of nuclei into > separate channels. > >Eric > > >On Sat, May 11, 2013 at 3:37 PM, <[email protected]> wrote: > >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 >> >> Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
