In reply to Eric Walker's message of Mon, 25 Feb 2013 19:15:06 -0800: Hi, [snip] >On Mon, Feb 25, 2013 at 6:59 PM, <[email protected]> wrote: > >If the energy is released in the form of a fast particle, then it does not >> have >> to depend on diffusion. A fast particle will rip through a lattice at high >> speed, leaving a trail of ionized atoms in it's wake. >> > >That provides a nice segue to a recent set of responses from Ron Maimon to >questions that have been raised here about his Auger deuteron proposal. > The questions he is responding to have been posted here [1], as is his >reply in the comments section. Copied below is the text of the reply >with brief notes referring to the questions. > >Eric > > >"1+2 [concerning the probability of deuterons meeting and ROI in general]. >The dynamics of charged particles in materials is not as "frictional" as >you imagine. Charged particles don't slow down through bulk dynamics like a >baseball in water, they have to transfer their energy through ionization of >atoms, and in the case of 20KeV deuterons, they need to ionize other inner >shells. They will eventually slow down, but it isn't clear how quickly, >because it depends the precise density of states. If you have banded >excitations, there are sometimes band-gaps where you have no states, so you >can sometimes have extremely long-lived high energy states just because >there aren't any available states to allow a simple de-excitation process. >So if there are no available states at 17 KeV to transition to when you >kick out 3KeV in exciting the n=1 p-wave electrons, these things could live >a long time. I am not saying that this is necessarily so, I have no idea >how long lived a 20 KeV deuteron is, but it can have a lifetime from a >fraction of a second to minutes. But in any case, there is certainly not a >barrier to having a 20KeV particle go microns or a millimeter through a >metal before slowing down to 5KeV. The penetration depth of 20 KeV charged >particles is not 10 atoms.
A fast particle doesn't need a matching energy band to lose energy. It loses a portion of what it has simply by ionizing other atoms. IOW the charge on the deuteron attracts electrons from other atoms as the deuteron passes through, and tears them loose from their original atom. This slows the deuteron slightly, as it costs energy. IOW it doesn't particularly need to interact with inner electrons. Any electron will do, as long as the ionization energy for that electron is less than the kinetic energy of the deuteron. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
