----- Original Message ----- From: "Horace Heffner" <[EMAIL PROTECTED]> To: <vortex-l@eskimo.com> Sent: Tuesday, January 08, 2008 11:50 PM Subject: Re: [Vo]:Re: Cold Fusion-Treated Palladium-Lithium-Boron Laser Fusion Target Factory
> Let me try that one more time! As usual I made a slight error. > ' > From the electric potential energy Pe for separating an electron and > proton we have: > ' > Pe = k (-q)(q)(1/r) = -(2.88x10^-9 eV m) (1/r) > ' > which we can rearrange to obtain r for a given potential energy, > ' > r = (1.439965x10^-9 eV m) (1/Pe) Mmm, all you did is multiply both sides by r/Pe it seems, how come your constant got halved and changed sign from one line to the next? Michel > ' > and we have for 0.78 MeV: > ' > r = (1.44x10-9 eV m) (1/(0.78x10^6 eV)) > ' > r = 1.846x10-15 m <=== exponent -15, not -16 > ' > > [snip] > > Here's another issue I think is not commonly recognized or > considered, but which I think is valid. The size of the nucleus is > dependent on its de Broglie wavelength in the frame of observation. > However, the reference frame that is important to the nuclear > reactions discussed is that of the electron, not the laboratory. In > the electron's reference frame, the nucleus is very small, and can be > orders of magnitude smaller than in the lab frame, when the electron > is at near c velocity. This explains why a nucleus radius of > 1.846x10-15 m can be no problem for the calculation made above, and > further why neutron formation is an unlikely event. > > Horace Heffner > http://www.mtaonline.net/~hheffner/ > > >
