Feynman
New member
artrius said:thanks bro, its all clear as the fission process of U-235 yeilding Xe-133 & Kr-88 + a couple of neutrons, neutrinos, antineutrinos, probably some electrons, and maybe some positrons too...![]()
Are you working on the Los Alamos project too?
And U235->Xe133+Kr88+2n gives an atomic weight of 223, and when subtracted from the original mass of 235+1n, gives a mass loss of 13 amu. Converting to pure energy gives E=(13.0)c^2=12 GeV, when it is actually closer to 200 MeV or more if including energy released from gamma rays, beta particles and neutrons.
How do you account for the loss mass?
...oh, and to answer the original question, Taurus!