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 
of 12C.
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 

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