Yes, in my view, the DDL state diatomic hydrogen (shrunken hydrogen) reacts with Ni58. Should both atoms be in shrunken state? Is the DDL atoms small enough to go in the lattice?
We can consider as well with pD or DD DDL state if natural hydrogen is used.
_____________________________________________
From: Jones Beene [mailto:jone...@pacbell.net]
Sent: lundi 22 septembre 2014 18:28
To: vortex-l@eskimo.com
Subject: RE: [Vo]:Mizuno, Rossi & copper transmutation
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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