On Jul 11, 2009, at 3:42 PM, [email protected] wrote:
In reply to Horace Heffner's message of Fri, 10 Jul 2009 04:10:13
-0800:
Hi,
[snip]
If you fully ionize a dideuterino, if that is
readily feasible, since it ostensibly takes around 70 eV , then you
still get mass 2 deuterium nuclei, not a mass 4 He nucleus.
It takes 70 eV to remove the second electron from the negative
Hydrino ion. To
remove both electrons from a maximally shrunken Deuterino molecule
resulting in
two Deuterium nuclei costs several hundred keV.
First, I would point out that if removing the first electron from a
hydrino molecule (which for convenience I take here as a class which
includes one or two deuterino atoms) takes excessive voltage then it
will not ionize in the mass spectrograph and thus not show up in a
mass spectrogram at all.
Beyond that, and I expect this is my weakness that you are
addressing, I can't see how it is within reason to expect the
formation of hydrinos below N=1/2 from typical cold fusion
experiments. That is the context within which the discussion lies
(see original comments from Jones Beene below), i.e. that it is not
He that is being observed in CF experiments, but hydrino molecules. I
don't really see how it is within reason to expect all CF experiments
(at least all those in which He has been measured) to create
hydrinos, much less hydrino molecules which are readily singly
ionized in a mass spectrograph. However, given my limited vision in
this matter, I still stand by the following:
In approximate terms, suppose we say the m/Q ratio for He+ is (4/1) =
4. The m/q ratio for He++ is then (4/2) = 2. The m/Q ratio for a
singly charged dideuterino is (4/1) = 4, thus it masquerades as a He
+. If the ionization potential is pushed far enough, then the
dideuterino breaks down and becomes ordinary deuterium D+ (or at
least one of the deuterons does, since there is only one electron)
with a mass/charge ratio of (2/1) = 2, thus it masquerades (not very
well in a precision mass spec.) as He++.
Suppose the singly charged dideuterino breaks down at a very high
voltage, much higher than where He+ loses its last electron. Suppose
very little helium is present in a sample, but a lot of
dideuterinos. This would be readily detected by comparing the mass
spectrographs for (average) ionization energies just above He+ and
then just above He++, both in a high precision mass spec. The
helium will migrate from the m/Q = 4 peak down into the m/Q =2
peak. If the singly charged dihydrinos require a large ionization
energy, then they will all remain in the m/Q = 4 peak. This lack of
any migration would be recognizable as anomalous.
The above assumption of a differing second ionization energy is not
needed to make the determination of the presence of dihydrinos
though. Suppose you then push the ionization energy well beyond the
dihydrino's full ionization energy. This will result in an increase
in deuterons in the m/Q = 2 peak, but these, necessarily being
*ordinary* D+ deuterons, will be readily distinguished from any
small amount of He+ that would remain. In other words, as you push
up the ionization energy, He+ will disappear from their (m/Q) = 2
peak, while, if any dihydrinos are present, they will *increase* the
size of the m/Q = 2 deuterium peak. Further, if the ionization
energies for He+ and singly charged dihydrinos differ, then the m/Q =
2 migrations will occur at definitively different ionization
voltages, which would provide even further confirmation of an
anomalous (hydrino based) process.
Even if the first ionization potential of the CF experiment created
hydrino molecules was in range of the spectrograph ionization
chamber, and even if the second hydrino molecule electron continues
to bind both atoms and not stick with just one, and even if the
second ionization potential of the hydrino molecule is hundreds of
keV, the hydrino molecule presence will still be disclosed by the
lack of migration from the m/Q=4 peak to the m/Q=2 peak when the
ionization potential is sufficient to double ionize most all He that
is in the sample. That proves the m/Q=4 peak is not from singly
ionized helium. This is in stark disagreement with the statement
attributed to Mills (below).
On Jul 9, 2009, at 4:36 PM, Jones Beene wrote:
Which is to say that when 4He is measured as the ash from LENR, and
this has been assumed to be real helium, it could instead consist
of one molecule of ”two fractional hydrogen isotopes” - better
known as the Mills hydrino, or more specifically the Mills’ “di-
deuterino.”
Back circa 1994, if memory serves, Mills mentioned this possibility
in Fusion Technology.
The ionization potential for the “di-deuterino” would be extremely
high according to Mills, in the case of deep redundancy – and
essentially there is little way they could ever be distinguished
from helium except for the small mass difference which we have
talked about here before - and which has actually shown up in very
sophisticated Mass Spec charts before, as that small blip.
I recall posting the reference to that chart, years ago, to vortex,
and if memory serves it was done at Frascatti but I cannot find the
reference now.
Jones
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
