Correction noted below. I continue to make errors willy nilly as
usual. Sorry!
On Jul 11, 2009, at 6:19 PM, Horace Heffner wrote:
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.
The above sentence has a typo and lacks clarity. It should read: "In
other words, as you push up the ionization energy, He+ will disappear
from their (m/Q) = 4 peak and migrate into their distinguishable He++
(m/Q) = 2 peak, while, if any fully ionizable dihydrinos are
present, the deuterons freed by the dihydrino ionization will
increase the size of their separate and identifiable (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/
