# Angular momentum Bohr's model

I have been trying to derive speed, radius etc. in hydrogen atom using Bohr's postulates and not neglecting the coulombic attraction on proton.

I know that they will be revolving around their centre of mass with same angular speed. But, I have this one doubt. Do we write $$L=\frac{nh}{2\pi}$$

of electron with respect to their centre of mass or wrt proton(nucleus)?

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Related: physics.stackexchange.com/q/91895/2451 , physics.stackexchange.com/q/78664/2451 , and links therein. – Qmechanic Apr 15 '14 at 8:09
@Awesome I downvoted your question because it is not of expert-level. – user Apr 26 '14 at 11:27
@user31782 One day I will become an expert... And then we'll see... – evil999man Apr 26 '14 at 11:28

The expression you have there looks like that of the electron relative to the proton. The equation $$L=\frac{nh}{2\pi}$$ can be derived from the de Broglie relation $p = h/\lambda$.
Consider electron "orbiting" (classically speaking) about a proton (we take to be the origin). Its orbital angular momentum will be given by $$L=rp$$ $r$ and $p$ of course, being the radius and angular momentum respectively. By demanding that an integer number of wavelengths fit into the radius, $$\lambda = \frac{2\pi r}{n}$$, then $$L = rp = r\left(\frac{nh}{2\pi r}\right) = \frac{nh}{2\pi} = n\hbar$$ as required.