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) . :-)
Ah, we have another controversy Fred. V is positive and so am I ;-) Consider my previous comment right below, plus this: Electric field created by the proton is minus the derivative of the potential wrt radius right? (e field is V/m) E = -dV/dr => dV/dr = -E = -k q/r^2 => V(r) = k q/r (+ constant which me make zero so V(infinity)=0) So the sign of V is given by the sign of _the other_ particle, the one which acts on the particle under study, positive in this case :) >> 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 Could hardly be more wrong (sign, value _and_ unit!), V = + k*q/r = 27.2 volts, I guess you haven't had your coffee yet :) >> >> 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. OK I will, but let's solve the above discrepancies first :) Michel
