2
$\begingroup$

I wish to understand how to solve this atomic structure question:

For the ground state, the electron in the H-atom has an angular momentum = $\hslash$ , according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true, because:

  1. Bohr model gives incorrect values of angular momentum.
  2. only one of these would have a minimum energy.
  3. angular momentum must be in the direction of spin of electron.
  4. electrons go around only in horizontal orbits.

Actually, we have been taught Bohr orbits a lot in class, with formulae for everything (atomic radius/electron orbital speed/various energy levels/etc.) and also basics of the quantum mechanical model. However, we didn't go into such a great depth of Bohr orbits as asked in the above question.

Specifically, I don't know of any reasonable arguments against points 1,2,3. Point 4 is evidently wrong based on the quantum mech model.

How do I develop any claim for/against the points 1,2,3? What are the facts involved here?

$\endgroup$
2
$\begingroup$

The only correct answer is #1.

In actuality, the ground state of the hydrogen atom has zero angular momentum.

Regarding #2 and #4, there is no preferred plane in which one point should orbit another point. By symmetry, they are all equivalent.

Regarding #3, there is no angular momentum in the ground state, and in a state whether there is angular momentum, it doesn't need to be in the same direction as the electron spin angular momentum.

$\endgroup$
  • $\begingroup$ Thank you! Can you please add some detail as to why it has zero angular momentum? And why are points 2 and 3 wrong? $\endgroup$ – Gaurang Tandon Jan 16 '18 at 15:23
  • $\begingroup$ Oh, I just noticed your elaboration (next time please comment if you update your answer :)) I understand your logic for point 2, 3 and 4. For why there is zero angular momentum, I came across an old answer by you (coincidence ;))? Does that answer also apply in this question? $\endgroup$ – Gaurang Tandon Jan 24 '18 at 1:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.