Fred wrote: > Michel Jullian (The new broom of Vo) wrote: :-) >> >> >> Fred wrote: >> >> >A single electron orbiting"Hydrogenic" atom has a Potential V = 13.6 * > Z^2 volts >> >> Wrong for H in any case, it's 27.2V Fred (did you get my 2nd post of > yesterday in thread "Electronium (Bound Ps-) Orbits vs Fractional Electron > Orbits"?). >> > The last I heard the proton H+ has 13.6 eV "potential energy" at the ground > state Bohr Radius > 5.29E-11 meters. V = k*q/2r = 1.44E-9/(2*5.29E-11) = 13.6 eV
Dear Fred, you should be ashamed of tweaking formulae to get good-looking results, you're not a schoolboy anymore ;) Potential energy (J or eV) is not the same as potential (V), furthermore P.E. in this case is equal to -27.2 eV (twice what you said, and opposite sign), it's the total energy (K.E. + P.E.) which is equal to -13.6 eV (again sign matters). In order not to duplicate my brooming efforts could you please review my 2nd post of yesterday in which those things are derived? (beware there is stuff towards the bottom of that post too) > For Oxygen 8+, (hydrogenic) 870 eV, Scandium 21+, 5998 eV etc, > with a bunch of shell groups of lower energy electrons betwixt and between. >> >> > >> > Thus for hydrogen Z = 1, V = 13.6 and for oxygen Z = 8, 13.6 * 64 = > 870 volts >> > Argon 13.6 Z^2 = 13.6 * 18^2 = 4,406 volts, Potassium 4910 volts etc. >> > >> > Since the potential V at a distance r from a charge = k * q/2r volts >> >> V=k * q/r (cf same post, come on you had it right two days ago) >> > That is total energy, Potential plus Kinetic.A 0.5 kilogram raindrop > (hailstones like those > that trashed my roof in October 2004) formed at 1 Km or more has a > potential energy of mgh, > but while it is falling it's gaining kinetic energy too. No? >> >> > and the >> > Electrostatic attractive or repulsive force Fes = k * Z1 * Z2 * > q^2/r^2, how >> > close can a bare proton (H+) approach a hydrogenic (one electron) atom? >> > >> > Or, how far into an oxygen atom of an H2O molecule does a hydrogen atom >> > need to go in order to capture one of the high energy inner shell > electrons of the >> > oxygen atom which can then be taken up to orbit the proton with the > same energy >> > it had in it's oxygen orbit, allowing that one of the outer (low > energy) electrons >> > of the H2O molecule will replace the proton-captured oxygen electron > with >> > a commensurate energy release? >> >> well it's exactly... >> >> > Prompt answer not required. :-) >> >> ... oh I won't tell you then ;) >> > Please do. I want to see if high energy "fractional orbit" hydrinos are > formed in the > water molecules in an ICE after "Preconditioning" in the Joe Cell. > Do they evaporate out of the "highly conductive-syrupy liquid polywater" of > the Cell > as Dimers (H2OH2O) ? Later, we must sort out the basics first. Michel_the_broom ;-)
