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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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Why was this downvoted? –  Awesome Apr 15 at 8:54
@Awesome I downvoted your question because it is not of expert-level. –  user31782 Apr 26 at 11:27
@user31782 One day I will become an expert... And then we'll see... –  Awesome Apr 26 at 11:28

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up vote 2 down vote accepted

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.

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This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. –  Danu Apr 15 at 8:00
@Danu It does. That was exactly what I was asking. Thanks! –  Awesome Apr 15 at 8:54
@Awesome I understand that it does. This message was automatically generated by the site when I flagged the answer. In doing so, I meant to convey the message that this answer should really have been a comment rather than being posted as an answer, because it's just a single line. In particular, it doesn't offer an explanation and is therefore not useful to other users. –  Danu Apr 15 at 9:05
@Danu I don't think any suitable explanation would be provided without quantum mechanics which is beyond my scope. We just have Bohr's postulates in our syllabus. –  Awesome Apr 15 at 9:31
@Awesome The answer is not just meant to enlighten you, although that's definitely part of the objective. –  Danu Apr 15 at 10:24

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