On Jul 9, 2009, at 4:36 PM, Jones Beene wrote:
One comment on The Letter to the Editor from John Sutherland -
Hamilton, Ontario, Canada.
It raises a point that should be clarified. He may be unaware that
we can tolerate mention of Randell Mills or Blacklight Power here.
Actually, "welcome mention" might apply, since it is so very much on
list topic.
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.
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. This would be a very evident feature of
the mass spectrograph. 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. To the
extent you don't ionize dideuterionss, then they remain neutral, and
thus not in the mass spectrograph, or show up as mass 4 singly
charged species. Any in-between condition, i.e. a mixture of
species, would still show up as mass spectrograph which would be
mysterious or anomalous without a hydrino explanation, and thus to
most CF researchers that did He mass spec. , other than possibly
Mills' staff.
It may be of use to look at US Patent 6,024,935, which discusses much
about dihydrino isolation and mass spec. , and which also makes no
sense to me for the same reasons. Maybe I'm just missing something
obvious. I didn't take much time to look at it.
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
Following is the beginning of a fairly long old thread, "Mass Spec.
question", which may be of interest.
On Dec 2, 2002, at 11:56 AM, Jones Beene wrote:
Here is a question for Ed Storms or anyone else who has done mass
spectrometry on ash from a cold fusion cell. My apology if this
question has been answered previously. I remember it having been
brought up in the past, but couldn't find a satisfactory direct
answer in searching the vortex archives or on LENR-CANR.
Question: Have experimenters absolutely eliminated the possibility
that the 4He atom that is found in CF cells, and is usually
identified through mass spectrometry - and is offered as evidence
of D+D fusion - could not instead be a tightly bound "below ground
state" deuterium pair, a.k.a. the di-deuterino molecule?
According to Mills, the first ionization energy of the dihydrino
molecule is 62.27 eV and IP2 is 65.39 eV which, of course, are far
higher than D2 and higher than Helium, but IP2 is fairly close.
The mass of a di-deuterino molecule would be somewhere between 4He=
4.0026 amu and that of the deuteron molecular mass of 4.0280 amu,
but exactly where is not clear.
AFAIK, Mills doesn't specify that a di-deuterino ionization energy
would be any different than a di-hydrino, but then again, he seldom
mentions "below ground state" deuterium at all. I suspect that lack
of mention is for reasons that relate to protecting his
intellectual property.
I know Ed has proposed that the lack of O2 buildup in his closed
cells is indicative of no deuterinos, and that is a good argument
that would have to be answered by anyone wishing to equate Mills
work with cold fusion - but this question relates *only* to mass
spectrometry results and whether or not the putative di-deuterino
has been eliminated.
Regards,
Jones Beene
On Dec 2, 2002, at 12:01 PM, Edmund Storms wrote:
Most mass spectrometers used to detect He are able to
distinguish between He and D2. In addition, the D2 is
removed chemically from the gas. We would have to assume
that the di-deuterino molecule was not chemically active
so that it remained in the gas and was mistaken for He.
In addition, we would have to assume this molecule can be
ionized to the +1 ion by 70 eV electron bombardment.
According to Mills, the first ionization energy of the
dihydrino molecule is 62.27 eV and IP2 is 65.39 eV which,
of course, are far higher than D2 and higher than Helium,
but IP2 is fairly close. The mass of a di-deuterino
molecule would be somewhere between 4He= 4.0026 amu and
that of the deuteron molecular mass of 4.0280 amu, but
exactly where is not clear.
I do not understand this argument. The energy of the
electrons is not converted to mass, so it would not show
up as a mass change between normal D2 and a di-deuterino
molecule.
AFAIK, Mills doesn't specify that a di-deuterino
ionization energy would be any different than a
di-hydrino, but then again, he seldom mentions "below
ground state" deuterium at all. I suspect that lack of
mention is for reasons that relate to protecting his
intellectual property.
I suspect he never uses D2, so this issue never comes up.
If he used D2 he might find nuclear products, which would
put him in the LENR field, something he tries to avoid.
I know Ed has proposed that the lack of O2 buildup in
his closed cells is indicative of no deuterinos, and that
is a good argument that would have to be answered by
anyone wishing to equate Mills work with cold fusion - but
this question relates *only* to mass spectrometry results
and whether or not the putative di-deuterino has been
eliminated.
I think it has been eliminated. In addition, the amount
of energy measured during He production equals about 23
MeV/He atom, in contrast to the much lower energy required
of deuterino formation.
Regards,
Ed
On Dec 2, 2002, at 1:23 PM, Jones Beene wrote:
Hi Ed,
Most mass spectrometers used to detect He are able to distinguish
between He and D2. In addition, the D2 is removed chemically from
the gas.
Yes. But this wouldn't be relevant to deuterinos, which
presumably have very different properties
We would have to assume that the di-deuterino molecule was not
chemically active so that it remained in the gas and was mistaken
for He.
That is exactly the argument that might exist in principle. Mills
believes that the hydrino molecule is chemically inert.
In addition, we would have to assume this molecule can be ionized
to the +1 ion by 70 eV electron bombardment.
This is getting to the crux of the issue...
According to Mills, the first ionization energy of the dihydrino
molecule is 62.27 eV and therefore to completely ionize the
molecule so that your spectrometer reading was showing something
closer to mass ~2 rather than mass ~4, it would require
considerably more than a 70 eV electron bombardment - more like 135
eV.
Yet, If I understand your response, you are saying that you
believe that despite this extremely high ionization potential of
the deuterino molecule (giving Mills the benefit of the doubt), you
believe that most mass spectrometers would still be able to
accurately pick up the minute difference between 4.0026 and 4.0280
amu, is that correct?
Regards,
Jones
The thread then digresses into discussion of electron energy levels,
and I show the final post in the thread being:
On Dec 14, 2002, at 10:51 PM, Horace Heffner wrote:
At 9:06 AM 12/15/2, Robin van Spaandonk wrote:
This doesn't appear to be a problem for the electrons in the inner
orbitals of the higher elements.
I think none of the conventional inner orbitals violates
Heisenberg. If
they did, Heisenberg would have been struck down long ago. The
energy
levels are or at least can be very high. The 29th ionization level of
copper, for example, is 11.567 keV. The 28th ionization level of
Ni is
10.775 keV. But that is neither here or there, as both models are
in synch
with regard to Heisenberg when n=1. (After all, if n=1, then no
fractional
state is involved.)
Looking at the situation again from a simple (hopefully not too
simple?)
perspective, the kinetic energy K of an electron in Bohr orbit
radius r is
given by:
K = q^2/((8Pi)(e0)(r)) = (1/2)(m)(v^2)
So speed v is:
v = (q^2/((4pi)(e0)(r)(m))^0.5
and momentum is thus ~ 1/r:
p = mv = ((m)(q^2)/(4(pi)e0(r))^0.5
Given that the radius is quantized to:
r = (n^2) ((h^2)(e0))/((pi)(q^2)(m)), for n = 1,2,3, ...
(or in Mills' case: n = 1/2,
1/3, ...)
we have:
v = [q^2/(2(e0)(h))] (1/n) = [q^2/(2(e0)(h)(n))]
instead of the Mills valocity:
v = [q^2/(2(e0)(h))] 1/(n^0.5), for n = 1/2,1/3,1/4, ...
Uncertainty of momentum (delta mv) for a particle (electron)
constrained by
distance delta x is given by Heisenberg as:
delta mv = h/(2 Pi delta x)
but 2r acts as our delta x because the electron is contained within
the
orbitsphere, so we have (substituting 2r for delta x in the above):
delta mv = h/(2 Pi [ 2 (n^2) ((h^2)(e0))/((pi)(q^2)(m)) ] )
delta mv = [h (q^2)(m)] / [4 (n^2)(h^2)(e0)]
delta mv = (q^2)(m) / [4 (n^2)(h)(e0)]
and an uncertainty in velocity delta v:
delta v = (q^2) / (4 (n^2)(h)(e0))
So now the question is how does delta v compare to v? That is to
say is
the uncertainty on v small in comparison to v? To see, let's look
at the
ratio:
(delta v)/v = [ (q^2) / (4 (n^2)(h)(e0)) ] / [q^2/(2(e0)(h)(n))]
(delta v)/v = [(2)(e0)(h)(n)] / [(4)(n^2)(h)(e0)]
(delta v)/v = [(2)(n)] / [4 (n^2)]
(delta v)/v = 1 / (2n)
However, with normal (non-Mills) orbitals, n is a whole number, so
deltav
remains small with respect to v. There is not the large imbalance I
pointed out in the earlier post of mine which is the subject of
discussion
here, which occurs because (1) n is a fraction, and (2) the
exponents in
the Mills equations differ such that as n goes to increasingly lower
values, i.e. n = 1/x as x gets large, we have
delta v = (1/2) (x^0.5) v
for Mill's, and the resulting energy gets way out of whack in
states other
than n=1.
The above relations for K, v, p remain valid for inner electons in
the
Bohr or Mills model (ignoring relativisitc effects)1, With n = 1
in these
inner states, it appears r is valid for either model, thus either
model
works fine for the innner electrons. Neither then violates
Heisenberg. It
is only the hypothesized (by Mills) fractional quantum states that
violate
Heisenberg. At least that's the way it appears to me.
Regards,
Horace Heffner
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
