MC: remember to look at the DSC scan in Fig. 7. NaH goes strongly exothermic
all by itself in an He atmosphere.
Regards,
Mike Carrell
----- Original Message -----
From: "Robin van Spaandonk" <[EMAIL PROTECTED]>
To: <[email protected]>
Sent: Friday, October 24, 2008 11:47 PM
Subject: Re: [Vo]:Banking on BLP?
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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