Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Zeeemans effect

  1. May 19, 2010 #1
    1. The problem statement, all variables and given/known data
    I have problems with calculation of the Zeemans effect.
    A natrium lamp is being placed in magnetic field with B=0.4T. Calculate the Zeemans effect for the states 3s j=1/2 and 3 p j=1/2
    The only problem i get in this calculations is the Landes factor.....can get the right value

    i get for the 3s g= 1 (i know i should get 2) and for the 3p i get 1/3 (i should get 2/3) Am I missing something? can someone check it please?
    here is the solution
    http://img535.imageshack.us/img535/2053/namnlsp.jpg [Broken]
    but i still do not understand how they got this numbers....
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 19, 2010 #2
    I get the same answer as them. Please show us your work so we can see where you went wrong.
  4. May 20, 2010 #3
    for 3s 1/2 i get
    [tex] g_{j} = 1+\frac{ \frac{1}{2}\frac{3}{2}-1+\frac{1}{2}\frac{3}{2}}{\frac{3}{2}}=1+1-\frac{2}{3}=2-\frac{2}{3}=\frac{4}{3}[/tex] now i got something else.....its a bit embarrassing;)

    i think i caclulate this one correctly
    for 3p 1/2 i get
    [tex] g_{j} = 1+\frac{ \frac{1}{2}\frac{3}{2}-2+\frac{1}{2}\frac{3}{2}}{\frac{3}{2}}=1+\frac{\frac{3}{2}-2}{\frac{3}{2}}= 1-\frac{1}{3}=\frac{2}{3}[/tex]
  5. May 20, 2010 #4
    What equation are you using? Because I see quite a few mistakes. For example, your denominator only has one term. Also the 2nd term in the numerator looks incorrect as well.
  6. May 20, 2010 #5
    [tex] g_{j}=1+\frac{j(j+1)-l(l+1)+s(s+1)}{2j(j+1)}[/tex]
  7. May 20, 2010 #6
    For the s-state, you know l=0. So l*(l+1) = 0*1 = 0, but you said it is 1 in the other post.

    By the way your denominator is right, I was just confused since I just saw the one term. Didn't realize you already did the math.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook