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Bohr's Quantization of Angular Momentum

  1. Aug 27, 2009 #1
    Bohr's second postulate says that it is only possible for an electron to move in an orbit for which its orbital angular momentum L is an integral multiple of [tex]\hbar[/tex].

    Can somebody please derive and explain L= n[tex]\hbar[/tex] for me?

    I feel like a total dummy for not understanding this, but this is what I have so far:

    L= mrv

    L=pr, p= hf/c, f= w/2pi, where w is the angular frequency and w= v/r

    L= [tex]\hbar[/tex]wr/c = [tex]\hbar[/tex]v/c ??

    Yeah... I'm obviously missing something...


    (Thank you in advance.)
    Last edited: Aug 27, 2009
  2. jcsd
  3. Aug 27, 2009 #2
    Welcome to physicsforums msavg,

    the argument goes like this:
    You interpret the electron as a standing wave as depicted http://www.personal.psu.edu/faculty/g/x/gxm21/A/Mayer-RingofFire_files/image003L.jpg [Broken]. A circle has circumference [tex]C=2 \pi r[/tex] and the condition for a standing wave is [tex]C=n \lambda[/tex]. From these two equations we get [tex]n \lambda = 2 \pi r[/tex].

    De Broglie says [tex]\lambda = h / p[/tex]. Can you proceed?

    (Edit: I changed the letter for circumference from L to C since it collides with the notation for the angular momentum)
    Last edited by a moderator: May 4, 2017
  4. Aug 27, 2009 #3

    Thank you.

    I knew I was missing something. This makes a whole lot more sense in context of standing waves.
    Last edited by a moderator: May 4, 2017
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