In reply to Eric Walker's message of Sat, 24 Nov 2012 11:16:46 -0800: Hi, [snip] >I am reminded of a pinball machine, where the palladium atoms are the >devices with the rubber bouncers that flip the ball across to the other >side. Once you drop a pinball into the machine, it can go for quite a >while before falling through the opening at the bottom. Another image that >comes to mind is letting go of several marbles at the top of a board with >nails nailed into it in a cross-hatch fashion.
As mentioned previously, the fast D nuclei will lose most of their energy to valence Pd electrons, this analogy doesn't work very well. It would be more like a pinball machine that was coated with glue. The ball doesn't get very far. [snip] >One question I have has to do with the energies. At 20 keV, would an >incident D nucleus impart enough momentum to the palladium atom enough to >unseat it from the surrounding lattice? The amount of energy required to unseat any atom from it's lattice can be obtained by determining the specific energy required to vaporize the substance. If you do the calculations, I think you will find something on the order of chemical energies, i.e. a few eV at most. So I think Ron's suggestion that 20 keV won't do the job is obviously wrong. A more likely reason that they stay in place is simply that there is no moment acting on the nucleus. (If they do stay in place - how would one know if a particular Pd atom moved to a different spot in the lattice?) >If so, it does not seem like such >an interaction could last for very long before the local region was altered >significantly. Note that microphotographs of CF sites on some cathodes have revealed multiple pits where the lattice has been destroyed locally. (See a.o. the cover of Mizuno's book). Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
