-2

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)?

Qmechanic
  • 201,751
evil999man
  • 1,812

2 Answers2

3

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.

  • 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 '14 at 08:00
  • 2
    @Danu It does. That was exactly what I was asking. Thanks! – evil999man Apr 15 '14 at 08:54
  • 3
    @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 '14 at 09:05
  • @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. – evil999man Apr 15 '14 at 09:31
  • 1
    @Awesome The answer is not just meant to enlighten you, although that's definitely part of the objective. – Danu Apr 15 '14 at 10:24
  • @Danu this case seems more like an appropriate use of a downvote, rather than a flag. – David Z Apr 15 '14 at 17:42
  • @DavidZ Okay. I'll do that next time then. – Danu Apr 15 '14 at 22:43
  • Sorry, this is my first answer to a question ever. –  Apr 17 '14 at 04:16
0

As per the logic of conservation of angular momentum, the net external torque should be zero. As we know, electron revolves around the nucleus, which also consist of proton inside it. even if the path is not circular however if the net external torque is zero, conservation of angular momentum is valid. Therefore it should revolve around nucleus, as COM may shift due to external distortion.

Amit_choubey
  • 1