Michel wrote: > > Fred wrote: > >> ... > >> >> > The potential V of a `particle with charge - q at a distance r > >> >> > from a particle with charge + q equals V = k*q/r independent > >> >> > of the mass of either particle. k = 1/4(pi)eo > >> ... > > The "q" in the V formula is that of the particle with charge + q, agreed? > Not really, The rule is that the sign of the Potential V is given as the sign of the particle, so for an electron (- q) Potential in this case is - V (negative) . :-)
> > The voltage at the particle we are concerned with (the -q one, the electron) does not depend on it's own charge, > but only on external charges, and the only one around is the proton +q. > The proton creates a voltage at distance r from itself equal to V = k*q/r, independently > of there being a charge there, or of it's value. More below. > > >> >> > The velocity v = [2 V*q/r * (1/m)]^1/2 = [2 V*q/r (1/2m)]^1/2 at > > that > >> >> > point is also the same (c * alpha or c/137 at a distance > >> >> > r = 5.29E-11 meters, the Bohr radius). > >> >> > >> >> Where does this come from? > >> >> > >> > The velocity in the classical Bohr ground state orbit. > >> ... > >> > >> OK one step at a time so Bohr proposed in 1913 (cf this article > > http://en.wikipedia.org/wiki/Bohr_model ) an ad hoc not-too-bad > > semi-classical model of the H atom where the electron's angular momentum > > can only take some discrete values: > >> > >> L=n*(h/2pi) > >> Where n = 1,2,3,. is called the principal quantum number, and h is > > Planck's constant. > >> > >> Angular momentum L is r*m*v isn't it, so for ground state n=1 we have: > >> > > Yes. typically written mvr = h/2(pi) or as the de Broglie wavelength > > lambda = 2(pi)r = h/mv. > > > >> r*m*v=1*(h/2pi) > >> => v=1/r * 1/m * h/2pi > >> > > That is okay for algebraic acrobatics. > >> > >> How does one get from this to your v formula above? > >> Wait a minute, your v formula simply results from equating centripetal > > coulombic force k*q^2/r^2 = V*q/r to > >> centrifugal force m*v^2/r doesn't it? > >> > > Yes. > > k* q^2/r^2 is the electrostatic force between two particles each with > > identical unit charge +/- q > > For the picky it should be k* +/- q1* +/- q2/r^2 newtons > > Or if you are into Fusion Coulomb Barriers: Z1 * Z2 * k*q^2/r^2 which is > > > > A handy constant at r = 1 meter is 2.306E-28 newtons. I keep it and alpha > > (0.00729729) > > along with E = hc/lambda = 1.9878E-25 in my Hp 11C storage registers. > >> > >> But then there is a mistake, the "2" factor in front of V*q/r shouldn't > > be there, > >> which is confirmed by your second expression for v where the "2" factor > > cancels out. > >> > >> Or maybe your second expression was for your electronium (same charge as > > electron, twice the mass, right?) in which case it's wrong too! > >> > >> Please let me know if you agree with the above and we'll proceed from > > there. > >> > > Velocity v = [2*V*q/m]^1/2 derived from K.E. = 1/2 mv^2 was the intent, > > This amounts to saying that K.E. 1/2 m*v^2 is equal to V*q, which can't be right either since "coulombic=centripetal" yields: > > m*v^2/r = k*q^2/r^2 > => m*v^2 = k*q^2/r = V*q > => 1/2 m*v^2 = 1/2 V*q (one half of what you wrote) > > Maybe you were trying to write the law of conservation of energy (Energy = Kinetic Energy + Potential Energy = constant). In this case you have to be picky about signs: > I wasn't trying, I used to applying cook book equations off the top of my head without going through their derivation. Hazardous, no? > > K.E. = 1/2 V*q (from "coulombic=centripetal" as we just saw) > P.E. = -V*q (potential energy of -q charge at potential V) > > => E = K.E.+ P.E. = -1/2 V*q > > Now if you do the computation for Bohr's ground state radius r=0.53 x 10^-10 m you find V = k*q/r = 27V, so: > > E= -1/2 27*q = = -1/2 27*e J = -13.5*e J = -13.5 eV > K.E. = +13.5 eV > P.E. = -27 eV > That's okay. Potential V = - k*q/2r = - 13.6 eV at the 5.29E-11 meter Bohr Radius > > The Wikipedia Bohr model page (link above) says En= -13.6eV/n^2 so the above must be right (n=1). > > Hope this helps, it helped me in any case, a good exercise before trying to understand fractional orbits and corresponding energies. Note I have assumed r was known, this is cheating, could a good soul do the derivation of r as a function of n based on the results in this page? > If you want to wing it, Michel. "The orbit of an electron in an atom must have a circumference equal to an integral number of wavelengths" And you will find that the orbital velocity at the Bohr radius ( n = 1) is c/137. Bon Voyage. Fred. > > Michel
