In reply to Jones Beene's message of Sat, 9 May 2009 06:04:41 -0700 (PDT): Hi, [snip] >By "always there" - this would imply that it serves as a "trigger" for D+D >fusion, where the hot alpha from the Pd reacts with an adjacent 'target >deuteron' - which is kind of a like nanoscale inertial confinement type of hot >fusion. [snip] Which reaction do you envisage here?
Assuming you mean that the alpha collides with one of the D's providing it with enough energy to overcome the Coulomb barrier and fuse with another one, then there are two problems:- 1) The alpha + Be9 reaction is commonly used as a neutron source, however googling has revealed in the past that only 1 neutron is created for about every 12000 alphas. This is because most alphas lose their energy ionizing atoms in the solid, and nuclei are so very small compared to atoms. IOW the chances of hitting another nucleus are pretty slim, particularly since both carry a positive charge. If roughly the same ratio holds true for your scenario, then the energy from the DD reactions would be swamped by that from the Pd reactions, and would be unnoticeable, thus back to square one. 2) The resulting normal D-D fusion would essentially produce T + P & He3 + n, not He4. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html
