On Jan 21, 2011, at 7:06 PM, Man on Bridges wrote:
Horace,
Based upon natural presence and absence of radiation I would
probably go for this on:
62Ni28 + p* --> 63Cu29 + 6.122 MeV [-1.984 MeV] (B_Ni:28)
62Ni28 : pres. 3.634 %
1H1 : pres. 99.985 %
63Cu29 : pres. 69.17 %
Kind regards,
MoB
If I recall correctly, Rossi stated that deuterium kills the
reaction, and that he uses pure protium.
Looking again at the mean atomic weight 63.6 and mean reaction energy
7.2, I calculated as shown below, you can see that these are weighted
averages, and are roughly 4 to 1 in favor of 62Ni vs 64Ni,
corresponding roughly to their abundances, 3.634% vs 0.926%. They
should be roughly equal in lattice half life (as I defined it in my
paper), possibly with a small edge for the larger nucleus.
Everything posted by me on this is seat of the pants accuracy. I
think for casual discussion like this slide rule accuracy is
sufficient to demonstrate the principles. I guess my age is showing! 8^)
On Jan 21, 2011, at 4:23 PM, Horace Heffner wrote:
62Ni28 + p* --> 63Cu29 + 6.122 MeV [-1.984 MeV] (B_Ni:28)
64Ni28 + p* --> 65Cu29 + 7.453 MeV [-0.569 MeV] (B_Ni:60)
64Ni28 + 3 p* --> 63Cu29 + 4He2 + 17.922 MeV [-7.605 MeV] (B_Ni:83)
Looking at the first two reactions as likely candidates, with mean
atomic weight near 63.6, and mean reaction energy about 7.2, we
have an estimated energy density of
E = (1/(63.6 gm/mol))*Na*7.2 MeV = 1.09x10^10 J/gm
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
