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, <mix...@bigpond.com> 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, <mix...@bigpond.com> 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 > >
