# Bohr's Quantization of Angular Momentum

by msavg
Tags: angular momentum, bohr quantization, planck's constant, quantum
 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 $$\hbar$$. 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.)
 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 $$C=2 \pi r$$ 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)
P: 3
 Quote by Edgardo Welcome to physicsforums msavg, the argument goes like this: You interpret the electron as a standing wave as depicted here. 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.

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