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.
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.
If we accept the excess power he claims, the process must be different
from the one he proposes. 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. 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.
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]>