Bohr's Quantization of Angular Momentum

[msavg]

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 $$\hbar$$.

Can somebody please derive and explain L= n$$\hbar$$ 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= $$\hbar$$wr/c = $$\hbar$$v/c ??

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

Help?(Thank you in advance.)

 

[Edgardo]

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 . A circle has circumference $$C=2 \pi r$$ and the condition for a standing wave is $$C=n \lambda$$. From these two equations we get $$n \lambda = 2 \pi r$$.

De Broglie says $$\lambda = h / p$$. 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]

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 . A circle has circumference $$C=2 \pi r$$ and the condition for a standing wave is $$C=n \lambda$$. From these two equations we get $$n \lambda = 2 \pi r$$.

De Broglie says $$\lambda = h / p$$. 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.