# Zeeemans effect

1. May 19, 2010

### rayman123

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. May 19, 2010

### nickjer

I get the same answer as them. Please show us your work so we can see where you went wrong.

3. May 20, 2010

### rayman123

for 3s 1/2 i get
$$g_{j} = 1+\frac{ \frac{1}{2}\frac{3}{2}-1+\frac{1}{2}\frac{3}{2}}{\frac{3}{2}}=1+1-\frac{2}{3}=2-\frac{2}{3}=\frac{4}{3}$$ now i got something else.....its a bit embarrassing;)

i think i caclulate this one correctly
for 3p 1/2 i get
$$g_{j} = 1+\frac{ \frac{1}{2}\frac{3}{2}-2+\frac{1}{2}\frac{3}{2}}{\frac{3}{2}}=1+\frac{\frac{3}{2}-2}{\frac{3}{2}}= 1-\frac{1}{3}=\frac{2}{3}$$

4. May 20, 2010

### nickjer

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.

5. May 20, 2010

### rayman123

$$g_{j}=1+\frac{j(j+1)-l(l+1)+s(s+1)}{2j(j+1)}$$

6. May 20, 2010

### nickjer

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.