Robin van Spaandonk wrote:
In reply to Edmund Storms's message of Fri, 18 Nov 2005 14:49:56
-0700:
Hi,
[snip]
Yea, I changed my mind based on the way you described how the Hy is
thought to behave.
Note that most of the behavioral aspects are my interpretation,
not necessarily Mills' opinion.
That's ok, we might even arrive at a better understanding than he has.
Just to be clear, both electrons are in fractional quantum states
according to Mills. (Otherwise the binding energy of the second
electron wouldn't increase with shrinkage level).
Yes, that is what I initially assumed. However, for a compound to form,
the normal quantum levels must be involved.
Why? In a "perfect" ionic compound, solidity results from the
binding energy of positive and negative ions. IOW the attractive
force between ions of opposite charge pulls the ensemble together.
There is no real need for electrons to be interchanged at a local
level as would be the case in a covalent bond. Granted, with
normal substances there is more often a "polar" bond than a pure
ionic bond. In short, the hyh "bond" with a positive ion would be
the most extreme ionic bond imaginable. You may calculate the
degree of electron sharing if you wish, but given an ionization
potential of around 70 eV for hyh[n=1/16], I think you will find
that it is so negligible as to be immeasurable.
A "normal" ionic compound results from electrons being moved from one
atom to the other. For example, in making NaCl, the electron moves from
the Na atom to reside for most of the time at the Cl atom. This is
different from the situation with Hy, which I'm trying to understand, so
be patient. When Hy is involved, the situation involves a preionized
atom, so to speak, which as a negative charge that can not be removed by
chemical interaction. Consequently for it to form a bond, the other
atom must also be preionized to form a positive ion. Of course, this is
easily done. You would propose that if Hy were bubbled through a
solution of Na+ Cl- in H2O, a compound should form having the formula
NaHy. In a similar fashion, Hy bubbled through an acid should result in
HHy. In addition to ionic bonding, both compounds have the potential for
some covalent bonding as electrons from the "normal" atom briefly occupy
"normal" energy levels in the Hy structure. I suggest this addition of
pure ionic and covalent components makes the bond exceptionally strong,
not just the ionic part. I suggest the ionic part can not be any
stronger than the stability of the positive ion respect to regaining its
electron from other sources in the environment. Does this fit with your
understanding?
The Hy levels might be
filled by one or more electrons, but these only give the assembly a
negative charge, rather like a really big electron.
...or a very small negative ion.
Bonds are formed by
electrons interacting between similar quantum levels.
Bonds are formed by two forces. Electrostatic, and magnetic. Pure
covalent bonds are pure magnetic bonds. Pure Ionic bonds are pure
electrostatic bonds. Polar bonds are a mixture of the two. In the
case of hyh, because the second electron is so tightly bound, the
bond is the purest electrostatic bond of all compounds. IMO.
No, I don't agree. The positive ion can get an electron from many
sources other than the Hy. Consequently, the bond is no more stable
than any other pure ionic bond. Thermodynamic stability is based on the
components of the compound being returned to their elemental state.
Returning NaHy to its elements would be equivalent to adding an electron
to the Na+ and doing nothing with the Hy, because its state can not be
changed. Hy is essentially a pure element when viewed from a chemical
viewpoint.
This is something
the Hy electrons can not do. However these Hy electrons would modify
the energetics of normal quantum levels and cause such compounds to have
unusual properties without the Hy electrons being directly involved.
Yes, this is also possible, particularly if the hyh can replace a
deeper electron from the normal shell. Such an atom may simply
appear to the outside world as an the original atom with a proton
converted to a neutron, and a neutron added. E.g. if one started
out with K39 and hyh replaced an inner electron, then the result
may look like Ar40, both chemically and for SIMS.
If D were used iso H, then it would look like Ar41.
Because of this possibility of "fooling" SIMS, it's imperative
that NAA also be used to identify new atoms in CF experiments.
(Tightly bound hydrinos can't fool NAA).
This is an interesting possibility. The question is, can a highly
reduced Hy actually act like a neutron that is stuck to an atom outside
of the nucleus? On the other hand, I would expect such a structure to
be so close to being neutral that interaction with the electron quantum
states would not be possible. This seems to be an idea worth exploring.
Would this explain the Fisher-Oriani super heavy carbon?
Because the charge is stable, the Hy should act like a really heavy
electron when focused by electric and magnetic fields. In fact, if they
were caused to bombard a metal plate, they could be used to build up
very high static potentials. Unlike electron, they could not leak away
by conduction. This might produce some unusual effects.
I doubt it, because, while the hyh may not be able to leak away,
the electron it replaces can leak away. This would still leave a
neutral charge over all.
Yes, over all. But immediately at the surface of the metal, the
positive changes left behind would generate a voltage gradient. Would
this gradient be large enough to do something unusual?
[snip]
Since the second electron has a "binding energy curve" (for want
of a better term), it would be helpful here if you elucidate your
remarks with level numbers.
("lowest" and "highest" are terms I try to avoid, because they
depend on one's point of view. I.e. is 1 the highest or lowest
level?)
I'm assuming Hy1 is a level closest to the Bohr zero level and Hy22 is a
level that releases the most energy when it is occupied, in which the
electron occupies an orbit close to the proton.
<g> . Yes, but it wasn't the definition of 1 or 22 that was
ambiguous, but rather the definition of high and low.
Yes, but I'm trying to show that my high is equivalent 22 and my low is
equivalent to 1.
Regards,
Ed
Regards,
Robin van Spaandonk
http://users.bigpond.net.au/rvanspaa/
Competition provides the motivation,
Cooperation provides the means.
