On Mar 1, 2011, at 11:47 AM, mix...@bigpond.com wrote:
In reply to Horace Heffner's message of Sun, 27 Feb 2011 20:39:30
-0900:
Hi,
[snip]
I believe there was a statement by Rossi that there was no residual
radiation, so this should rule out 59Ni in the leftovers.
The radiation from 59Ni may be difficult to detect, as most of the
time it
decays via EC directly to the ground state (no gamma), with the
neutrino taking
the energy. (See http://atom.kaeri.re.kr/cgi-bin/decay?Ni-59%20EC).
The very
small amount of positron decay would be mostly stopped by the lead
shielding,
IIRC, statement was in regard to the cleanliness of the process, i.e.
the ash. The lead shielding during operation is irrelevant. There
should be no difficulty detecting 59Ni.
and if the initial reaction of 58Ni were rare anyway, then it might
be lost in
the background.
Regards,
Robin van Spaandonk
http://rvanspaa.freehostia.com/Project.html
You must be well behind the list in your reading. The above was
superceded by the following post:
On Feb 27, 2011, at 8:39 PM, Horace Heffner wrote:
58Ni creates 59Cu:
58Ni28 + p* --> 59Cu29 * + 3.419 MeV [-6.329 MeV]
59Cu positron decays in 81.5 s (normally) to 59Ni which has a
76,000 yr half-life.
I believe there was a statement by Rossi that there was no residual
radiation, so this should rule out 59Ni in the leftovers.
However, it may be a reasonable hypothesis that the initially
trapped electron (by 6.329 MeV deficit) stimulates the nucleus to
the point where 59Co is created quickly from the 59Cu. Now we can
have:
59Co27 + p* --> 60Ni28 + 9.533 MeV [1.588 MeV]
but we have now moved up the ladder sufficiently to convert
everything to copper. This reaction might produce 1.588 MeV gammas
by the way. Non-radiating alternatives might be:
59Co27 + 2 p* --> 60Ni28 + 1H1 + 9.533 MeV [-6.856 MeV]
59Co27 + 3 p* --> 58Ni28 + 4He2 + 17.441 MeV [-7.881 MeV]
The key question I think is whether the initially trapped electron
has the kinetic energy to stimulate a very short half-life of a
radioactive composite nucleus. Clearly it does! The trapped
electron *initially* has the kinetic energy it had when it was in
the very small deflated hydrogen state, a very high kinetic energy,
low potential energy. This explains why radioactive products are
not observed from heavy element LENR, at least not from the simple
single decay elements that have been experimentally involved. This
may not be true of very heavy elements like those in the actinide
decay chain, for example. This could be a very interesting line of
research concerning nuclear remediation.
I should have noted in the above that the hydrogen nucleus itself
takes on a large portion of the kinetic energy in the deflated state,
because the relativistic mass of the nucleus and electron are about
equal. Even if the electron is involved in an immediate weak
reaction, the kinetic energy remaining in the hydrogen nucleus should
be enough to fast trigger any pending single nuclear reaction with
long half-life.
Further stated in that post:
Actually, according to deflation fusion theory, there is an initial
energy deficit of around 4.84 MeV that would suppress the initial
gamma:
60Ni28 + p* --> 61Cu29 * + 4.801 MeV [-4.840 MeV]
So, I don't see that as a problem. The problem to me is the lack of
any evidence that heavy element LENR ever results in radioactive
nuclei. Maybe this would be a first.
The 61Cu29 decays to 61Ni29 in 3.333h, but the 4.84 MeV trapped
electron may stimulate that immediately. Now we are back to copper
with:
61Ni28 + p* --> 62Cu29 * + 5.866 MeV [-3.722 MeV]
This taken as a whole is an agreement with you, for very slightly
different reasons, reasons that make sense within the deflation
fusion model, that *all* the Ni isotopes can prospectively be
converted to copper (and/or Zn or Co in the process). This makes more
sense to me than 94% of the heavier isotopes being actually converted
to copper, leaving only 58Ni and no detectable 59 Cu.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
