On Jul 10, 2009, at 4:55 AM, Jones Beene wrote:
Horace,
Ø This makes no sense to me Jones. Mass Spectrometers work on
ionized species. There would thus have to exist a reduced energy
orbital for a D2+ dideuterino species. Further, even if such a
species exists, to the degree no He is present, no mass 4 He++
species will show up in the mass spectrograph.
OK – does it makes more sense to suggest the ionization potential
at 54.4 eV can be identical for both species in many cases?
Jones
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.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
