From: Arnaud Kodeck
Jones,
Why not consider also the Ni58 + 2p -> Zn60 -> Cu60 -> Ni60? Zn60
has a spin 0.
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the suggested reaction is Ni58 + D -> Cu60 -> Ni60
Arnaud, This would be a three body reaction, no?
You may be suggesting this reaction - in the event that Rossi does not use
deuterium. That is wise to consider - since he professes not to, despite a
tank of it being seen in his lab, early on.
There is an even better possibility when two protons densified as a DDL
molecule, and would act like the two needed neutrons, to make this reaction
work.
If my understanding is correct, nickel-58 is active ONLY because it is
neutron-deficient, and the two protons do not help the immediate situation,
at least not on the surface - even if both protons decay to neutrons,
eventually. However, all bets are off with the DDL, since it allows the
protons to look like "virtual neutrons".
There is nothing out there, which fits all of the parameters seamlessly, so
in the end - we need reliable data. But it looks like we are framing a
workable situation with enough variable to accommodate either D, H or H+D as
the active gases.
In short, your suggestion may work well - an especially if Rossi uses
hydrogen only, and even more so - if the signature of the DDL formation
(soft x-ray) is documented.
Jones
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