Interesting comments here, thanks.
I do believe that NMR is the way to go for easiest and most reliable detection,
especially with carbon. Pure carbon should have only one peak – 13C. Nothing
Basically there would be two kinds of carbon to test – old and young. Old
carbon as defined herein comes from mineral graphite and is at least 100
million years in the making, while young carbon comes from activated carbon
(made from charcoal biomass, thus comparatively young). Both must be pure
The premise would be that young carbon has one peak only - since most of the
13C in young carbon is the isotope, just as expected, BUT in old carbon a
significant fraction is 12C+UDH and this gives two peaks.
In short, the actual NMR signature of young would have only one peak, and that
of old would have two peaks and it does not get much simpler than that.
The two peaks could look something like methanol, actually. There is a simple
route for testing called “benchtop NMR”. This one is on my “wish list”.
Nice. They have a database page with spectra which can be used as
identification. There is a page for carbon 13C compounds – check out ethanol.
This is what I would expect to see from graphite with significant UDH – two
peaks instead of one. The UDH should look like a proton in NMR which is the
second peak but it could be at a different frequency.
The magnetic resonance states of the fC13 would be easy to test for and should
say much about the nature of the entity, if it exists. An unpaired electron
in a 1p H(0) would certainly have a unique signature IMHO. The key would be to
get enough to test.
From: Bob Higgins
As I understand it, there are two hydrino-like transitions that could occur,
perhaps on a 12C atom. Suppose that the 12C is subject to catalytic hydrino
formation wherein one of its electron enters a (1/p) state. Such an electron
would enter an orbital around the nucleus that is smaller than the s orbital
and would screen one of the protons from the remainder of the electrons. This
would cause it chemical and spectral properties to appear as 12B instead of
12C. This would be a very unusual find because real 12B decays with a
half-life of 20ms and should not be seen in the experiment. Finding a stable
signature of 12B would be a likely indicator of formation of the hydrino state
Now consider that a hydrino hydride ion, described by Mills as H-(1/p) could
enter a hydrogen nucleus and bind so tightly as to become an innermost orbital
below the s orbital. A similar thing would happen in that this tightly bound
negative charge would screen a proton as far as the remainder of the 12C
electrons are concerned - it would have a mass of 13, but would chemically and
spectrally appear as 13B, not 13C. 13B has the same uniqueness in discovery as
the 12B - because real 13B has a half-life of only 17ms and hence should not be
found in the experiment. It would only be determined to be 13C accidentally if
there were no spectra taken - I.E. in a high resolution mass spectrometer test
only. This aspect is certainly not out of the question, as 13B would not be
anticipated to be found because real 13B would quickly decay most of the time
to 13C anyway. If they were to test for the x-ray spectra of B, perhaps the
hydrino hydride of 12C could be detected.
Note, however, that 13C is stable and is about 1% of natural C. It is not used
for dating. Interestingly, the natural variation of 13C is nearly +/-1%.
Could the hydrino hydride of 12C cause a measurement uncertainty in the
isotopic ratio of 13C/12C?
I estimate that hydrino states would be as stable in atoms with multiple
electrons as they are with hydrogen having a single electron. The reason is
that the additional electrons of, say a 12C, provide a possible means of
evanescent coupling to the innermost (hydrino) electron and provides some
opportunity to transfer energy without photon transfer and relieve the hydrino