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General quantization of circular motion/spectrum in potential

  1. Mar 28, 2013 #1
    1. The problem statement, all variables and given/known data

    The general quantization of motion in circular orbits is obtained by combining the equation of motion ## \frac{mv^2}{r} = |\frac{dU(r)}{dr}| ## with the angular momentum quantization condition ## mvr=n\hbar ## Use this procedure to calculate the spectrum for circular motion in the potential ## U = (F_0)r ##

    2. Relevant equations
    I think you need to use one of the series to find the spectrum but I'm quite lost on how to get there.


    3. The attempt at a solution

    I assume you make a substitution from ## mvr=n\hbar ## to quantize the equation of motion. I don't know if you use the given potential at first, and use its derivative ## |F_0| ##?

    The form of the answer highly suggests using a series, however there are terms in it that I don't have in the initial conditions so I don't think I know where to go from here.
     
  2. jcsd
  3. Mar 28, 2013 #2

    TSny

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    You can find dU/dr from your expression for U. That leaves you with two equations with two unknowns (r and v). After solving for r and v you can set up the total energy expression.
     
  4. Mar 29, 2013 #3
    Thanks for your reply.
    So the total energy expression is ## E_n = \frac{1}{2}mv_n^2 + F_0r_n ##??

    I am still lost after solving for r and v, and substituting them into the E expression. r and v should both be quantized, containing an n term correct? I know how to calculate emission spectra for transitions with the Bohr model but I'm unsure of how exactly to get the spectrum from here. Is it still going to end up as something like ## \frac{hc}{λ} = ΔE ##?
     
    Last edited: Mar 29, 2013
  5. Mar 29, 2013 #4

    TSny

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    Yes. So, you will get an expression for E in terms of n. These are the quantized energy levels. You should be able to simplify E by combining the kinetic energy and potential energy into a single term.

    That's right. Good.
     
  6. Mar 29, 2013 #5
    Okay, so substituting ## v_n^2 = \frac{F_0r}{m} = \frac{F_0r^2v}{n\hbar} ## and ## r_n = \frac{n\hbarω}{F_0} ## into ## E_n ##

    which yields ## E_n = \frac{1}{2n\hbar}(mF_0r_n^2v+2n^2\hbar^2ω) ##

    This does not seem convincing with two integral n's. Did I make an arithmetic error? Also it is still not only in terms of n...


    The answer given is ## λ = \frac{4c}{3}\sqrt[3]{\frac{m\hbar}{u_0^2}}(\sqrt[2/3]{n}-\sqrt[2/3]{m})^-1 ##

    u_0 is reduced mass correct? if m under the cubed root is electron mass then m in the series term is just a variable for an energy level?
     
    Last edited: Mar 29, 2013
  7. Mar 29, 2013 #6

    TSny

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    You need to find expressions for ##r## and ##v## in terms of just ##n, m, \hbar,## and ##F_0##.

    For example, to find the expression for ##r##, you can first solve ##mvr = n\hbar## for ##v## in terms of ##r## and then substitute that expression for ##v## in the other equation ##mv^2/r = |dU/dr|##.
     
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