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 else.
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 carbon. 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”. http://www.nanalysis.com/?gclid=CjwKCAjwlrnNBRBMEiwApKU4PEbIHuJZGUNC8CFU4ibwPuplnUU4h2204xt0yRJB1iL9B89RWh5Z8BoC49kQAvD_BwE 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. From: [email protected] 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 state.
