Spin-orbit coupling in a mercury atom

  • Thread starter juzbe
  • Start date
  • #1
2
0

Homework Statement



The problem is to determine which has a more dominant effect on the energy of a given state in mercury, spin-orbit interaction or the Zeeman effect, when the applied magnetic field B is about 2T.

Homework Equations



As long as the spin-orbit interaction is the dominant effect, I can calculate the Zeeman energy from
[tex]E_{Z}=\mu_{B}\g_{J}Bm_{J}[/tex],
but I'm at a loss trying to figure out whether this is a lot or very little compared to the spin-orbit interaction.

The Attempt at a Solution



The first-order spin-orbit correction to the energy of a hydrogen level is given by

[tex]E^{1}_{so}=\frac{E_{n}^{2}}{2mc^{2}}\left(\frac{n[j(j+1)-l(l+1)-s(s+1)]}{l(l+1/2)(l+1)}\right)[/tex]

How does this generalize to atoms with more than 1 electron? Can I just substitute for j, l, and s the atomic quantum numbers J, L, and S, and get a rough estimate? and if so, what do I substitute for [tex]E_{n)[/tex]? (I understand that for the exact answer I'd have to take into account the two electron's interaction with each other's orbitals plus some exchage terms as well, but all I really need to know is if the two figures are in the same ballpark...)
 

Answers and Replies

Related Threads on Spin-orbit coupling in a mercury atom

  • Last Post
Replies
0
Views
5K
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
6
Views
4K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
5
Views
788
  • Last Post
Replies
5
Views
448
  • Last Post
Replies
1
Views
692
Replies
0
Views
323
Top