Proof using mainly classical mechanics

AI Thread Summary
The discussion focuses on the challenge of quantizing angular momentum (L) using classical mechanics without invoking quantum mechanics principles. The user references equations related to classical mechanics and attempts to derive energy levels, specifically E = -hR/n², while grappling with the implications of constants like h and n that arise from quantum theory. There is confusion regarding how to reconcile these classical equations with quantum results, particularly in the context of ionization energy. The user seeks guidance on how to eliminate certain variables and derive results consistent with classical mechanics. The thread highlights the complexities of bridging classical and quantum frameworks in physics.
physicsjock
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Hey,

http://img822.imageshack.us/img822/407/25944209.jpg
\begin{align}<br /> &amp; \frac{m{{v}^{2}}}{r}=\frac{Z{{e}^{2}}}{4\pi {{\varepsilon }_{0}}{{r}^{2}}} \\ <br /> &amp; L=\frac{Z{{e}^{2}}}{4\pi {{\varepsilon }_{0}}v} \\ <br /> \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?

Thanks in advanced
 
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You want to quantise L without using any premises of quantum mechanics?
 
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.
 
If you're using only classical ideas, where did the n and the h come from in E?
 
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
 
I'm having trouble deriving

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

as well,

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