# Proof using mainly classical mechanics

1. Mar 3, 2012

### physicsjock

Hey,

http://img822.imageshack.us/img822/407/25944209.jpg [Broken]
\begin{align} & \frac{m{{v}^{2}}}{r}=\frac{Z{{e}^{2}}}{4\pi {{\varepsilon }_{0}}{{r}^{2}}} \\ & L=\frac{Z{{e}^{2}}}{4\pi {{\varepsilon }_{0}}v} \\ \end{align}

and I know by using the v derived using Bohr's equations it will give the answer but that v is derived using L=nh so it's not that simple.

I can't figure out how to quantize L without breeching the conditions of the question. Would anyone have any ideas?

Last edited by a moderator: May 5, 2017
2. Mar 3, 2012

### genericusrnme

You want to quantise L without using any premises of quantum mechanics?

3. Mar 3, 2012

### physicsjock

Haha, that's what's messing this up for me, that's how I've been thinking, I don't see how you can get a result like that using means that don't agree with the result.

There's a few results I just found which I can use,

$E=-\frac{hR}{{{n}^{2}}};\frac{{{\left| E \right|}^{3}}}{{{\omega }^{2}}}=\frac{{{Z}^{2}}m{{\alpha }^{4}}}{8}=\frac{R{{h}^{3}}}{16{{\pi }^{2}}}=\operatorname{Constant},\alpha =\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}}$

At first glance the first equation looks like the most useful but h and n and inversely related, and I can't shake the squared on the n.

4. Mar 3, 2012

### genericusrnme

If you're using only classical ideas, where did the n and the h come from in E?

5. Mar 3, 2012

### physicsjock

It was derived using Ritz principal and Rydbergs formula

${{E}_{n}}-{{E}_{m}}=h\,{{v}_{nm}}$

λ-1=$R\left( \frac{1}{{{m}^{2}}}-\frac{1}{{{n}^{2}}} \right)$

So I'm trying to find a way to quantize L using classical mechanics in conjunction with the results I posted before

6. Mar 4, 2012

### physicsjock

I'm having trouble deriving

$E=-\frac{hR}{{{n}^{2}}}$

as well,

This is what I've been doing,
\begin{align} & \frac{1}{\lambda }=R\left( \frac{1}{{{m}^{2}}}-\frac{1}{{{n}^{2}}} \right) \\ & \frac{ch}{\lambda }=vh=E=chR\left( \frac{1}{{{m}^{2}}}-\frac{1}{{{n}^{2}}} \right) \\ & E=-chR\left( \frac{1}{{{n}^{2}}} \right) \\ \end{align}

It's supposed to be the ionization energy, so the 1/m disappears because you take the limit as m -> infinity

Not sure how to get rid of the c