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
I don't see what this has to do with a mass spectrogram. Mass
spectrography determines mass to charge, m/Q, ratios. Ionization is
typically a kinetic process, be it via chemical, plasma, or beam
methods. Kinetic processes are random processes, so the ionization
method used in a mass spectrometer may have some effect on the
percentage of a specific m/Q species measured, or may not, depending
on the method used. However, the ionization potential has nothing
to do with the actual measurement of m/Q, which uses an electrostatic
field that is separate from the ionization field/method to accelerate
the ions. It admittedly does take high precision mass spectrometry
to determine the difference between D and He peaks though, but if I
recall this was addressed in the CF literature and thoroughly debated.
As Ed Storms pointed out, the reaction energies are dramatically
different between fusion and hydrino creation. So, in the absence of
a continuous catalytic process of some kind, one which provides
energy from some source other than dropping to the fractional state
(e.g. from ZPE expansion of the orbital), the two processes should be
readily sorted out by energy production alone, without the use of
mass spectrometry. This does admittedly require the assumption that
the full measure of 23.9 MeV per He atom created is obtained. This
measure of energy production is not guaranteed at all under various
theories, including deflation fusion, which predicts the energy
obtained in a given reaction to be sampled from a random distribution
with 23.9 MeV to be a maximal and therefore improbable amount, the
mean being much lower.
BTW, there is an apparently good method to determine whether you have
helium or deuterium, or the ratio of the two, without precision mass
spectrometry. See:
http://www3.interscience.wiley.com/journal/118680814/abstract
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
