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. > >> 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. >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. >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). >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. [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. Regards, Robin van Spaandonk http://users.bigpond.net.au/rvanspaa/ Competition provides the motivation, Cooperation provides the means.