Bohr's Quantization of Angular Momentumby msavg Tags: angular momentum, bohr quantization, planck's constant, quantum 

#1
Aug2709, 03:26 AM

P: 3

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... :\ Help? (Thank you in advance.) 



#2
Aug2709, 03:55 AM

P: 685

Welcome to physicsforums msavg,
the argument goes like this: You interpret the electron as a standing wave as depicted here. 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) 



#3
Aug2709, 04:14 AM

P: 3

Thank you. :) I knew I was missing something. This makes a whole lot more sense in context of standing waves. 


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