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Proof using mainly classical mechanics

  1. Mar 3, 2012 #1
    Hey,

    http://img822.imageshack.us/img822/407/25944209.jpg [Broken]
    [itex]\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}[/itex]

    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
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Mar 3, 2012 #2
    You want to quantise L without using any premises of quantum mechanics?
     
  4. Mar 3, 2012 #3
    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,


    [itex]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}}}[/itex]

    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.
     
  5. Mar 3, 2012 #4
    If you're using only classical ideas, where did the n and the h come from in E?
     
  6. Mar 3, 2012 #5
    It was derived using Ritz principal and Rydbergs formula

    [itex]{{E}_{n}}-{{E}_{m}}=h\,{{v}_{nm}}[/itex]

    λ-1=[itex]R\left( \frac{1}{{{m}^{2}}}-\frac{1}{{{n}^{2}}} \right)[/itex]

    So I'm trying to find a way to quantize L using classical mechanics in conjunction with the results I posted before
     
  7. Mar 4, 2012 #6
    I'm having trouble deriving

    [itex]E=-\frac{hR}{{{n}^{2}}}[/itex]

    as well,

    This is what I've been doing,
    [itex]\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}[/itex]

    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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