Michel Jullian 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 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,
but correcting my errors keeps me busy.
Sorry about that.
Fred
>
> Michel
>
