Bohr's Quantization of Angular Momentum


by msavg
Tags: angular momentum, bohr quantization, planck's constant, quantum
msavg
msavg is offline
#1
Aug27-09, 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.)
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Edgardo
Edgardo is offline
#2
Aug27-09, 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)
msavg
msavg is offline
#3
Aug27-09, 04:14 AM
P: 3
Quote Quote by Edgardo View Post
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)

Thank you.
:)

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


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