On Oct 25, 2008, at 3:05 PM, Mike Carrell wrote:
Ed wrote:
Subject: Re: [Vo]:Banking on BLP?
Robin, my main point is that an electron leaving an atom cannot go
to infinity under the conditions Mills has in his reactor. At
most, it will go into some other energy level, such as the
conduction band if one exists in the material. This fact is not
based on speculation, assumptions, or theory. This is a simple fact
of nature that is well understood.
MC: Which values and which electrons, Ed? In eq. 23, two electrons a
'liberated' to facilitate catalyzing H[1/3]. The physical situation
in the cell is NaH resident within the R-Ni mesh, which has an
enormous surface area. On the scale of a molecule, why can't the
electrons wander away? There are He atoms at 760 Torr hanging around
too. The electron bound to the catalyzed H doesn't go anywhere, it
just gets closer to its proton. Now I don't yet understand where the
energy to ionize the Na comes from, but the DSC plot shows
*something* happens. *That* requires eplanation.
The mechanism that Mills proposes requires the catalyst electrons to
change their energy by a required and known amount. This being the
case, a method is required to calculate how much energy a proposed
reaction is able to absorb. The ability to match a calculated energy
to a reaction or element is the unique strength of the Mills
approach. When Mills bases this match on the ionization energy of an
ion, his method breaks down in a solid. While an electron might wonder
away, he must know exactly how much energy this wandering absorbs to
apply his model. I suggest it is impossible to calculate how much
energy is absorbed under the conditions involving NaH. Therefore, he
cannot apply his model except by making some unlikely assumptions.
This is not to say that nothing happens or that hydrinos are not
involved. I'm only suggesting that the details of his mechanism in
this case makes no sense.
The values Mills uses to evaluate the process are all based on the
electron going to infinity. Therefore, these values simply cannot
apply to the real process. Instead, Mills assumes an unrealistic
process to make his numbers fit his expectation.
MC: Are you also including the ionic catalysts in the gas phase cells?
If we accept the excess power he claims, the process must be
different from the one he proposes.
MC: Why so? These solid fuel cells are a continuum with years of
work in the electrolytic and gas phases. There are dozens of reports
and papers supporting lthe reactions. Good calorimetry has been done
iwth microwace excitation by Jonatan Phillips at the University of
New Mexico. He was in town during ICCF-14 and slipped in to put up a
poster on his calorimetric studies. In an early version of his
reports there is a statement that the heat measured implied
substantial conversion to H[1/4]. Philipps is currentlyas
Distinguished Professor at the Farris center, supported by Los
Alamos. He has a long association with Mills. I very strongly
suggest that you contact him; he may be very helpful.
The gas studies clearly support his ideas and are consistent with his
calculations using the ionization energy. The problem comes when he
tries to apply this idea to solids. While the mechanism may work in
solids, his proposed path and the calculated values make no sense.
This is important to me, because I'm trying
to identify the Mills catalyst that is making hydrinos in the CF
process, which has similar restrictions.
MC: H and D atoms can autocatalyze in a three-body reaction because
2H+ provide the 27.2 energy for catalysis. Because it is a three-
body reaction, the reaction density is low but favored by H and D
rich environments such as the LENR environments. A reactions density
too low for optical observation may yet be very intense on the
particle-counting scene.
The infrequent success in CF can be explained if the required and rare
catalyst is absent in most studies. This being the case, we need to
search for this catalyst. An autocatalyze three-body reaction can ot
be the mechanism because H and D are always present, yet the required
nuclear product is rare.
An assumption on his part
that is unrealistic and impossible does me no good in trying to
use his method in this search. Therefore, I'm trying to
understand what is actually happening in his cell because the
hydrino process appears to be real under these conditions. Only
his explanation makes no sense.
MC: Granted, there are problems, as with LENR phenomena which don't
make sense either. Nature is trying to tell us something.
Yes, and I'm trying to listen.
Ed
Mike Carrell
Regards,
Ed
On Oct 24, 2008, at 9:47 PM, Robin van Spaandonk wrote:
In reply to Edmund Storms's message of Fri, 24 Oct 2008 16:05:50
-0600:
Hi,
[snip]
I think you are close to describing the process, Robin. Simply
decomposing NaH cannot result in hydrinos because the expected
ion is
not formed.
Absence of evidence is not evidence of absence, unless someone
explicitly looked
for it under the right conditions, and didn't find it.
On the other hand, as you suggest, if the decomposition
occurs on the Ni surface, the Na will have a complex ion state
because
it now is an absorbed atom, not a free, isolated atom. In
addition,
the electron that is promoted to a higher level has a place to go,
i.e. into the conduction band of the Ni. The only problem is
achieving a match between the energy change of the promoted
electron
and the energy shrinkage of the hydrino electron.
I suspect you are needlessly multiplying entities. ;)
IOW Mills provides a catalyst that has the necessary property,
and gets the
expected result. Why is it so hard to accept that he might be right?
Granted spectroscopic results indicating presence of Na++ would go
a long way to
proving him right.
Now for a question. Why must the electron that is promoted always
come from a level that is observed to form an ion during normal
ionization?
Personally, I don't think it does, and have previously suggested
that Li, which
has an x-ray absorption energy of 54.75 eV, may be an example of
this. However
Na doesn't appear to fit the bill.
For example, removal of a 2p electron from Na++ would
occur during "normal" ionization, but is this happening here?
No, but then Na++ is not the catalyst either. The whole molecule
is the
catalyst. BTW the third ionization energy of Na is 71.641 eV, and
none of the
immediate reactions have enough energy to do this. Only a further
reaction of
H[1/3] to a lower level would provide such energy. (3->4 yields
95 eV).
In
other words, why can't a 1s electron be removed from a neutral Na
without the 2p electron being affected. After the 1s electron is
removed, a 2p electron would take its place and release a small
amount of energy as X-rays. This energy would be a byproduct of
the
process just like the hydrino energy.
Do you know how much energy is required to remove a 1s electron
from
nearly neutral Na?
1073 eV. (K shell x-ray absorption energy).
The process gets more unknown because the electron
would be promoted into the conduction band, which has a lower
energy
than vacuum. In other words, perhaps Mills has the right process
but
is using the wrong electron promotion process to describe it simply
because the wrong promotion gives the expected energy.
If so, then I think you need to come up with an alternative (and
the numbers to
back it up). The work function of the metal might be a good place
to start,
however in this case we're looking at an alloy/compound, which
complicates
matters.
[snip]
Regards,
Robin van Spaandonk <[EMAIL PROTECTED]>
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