# Landé Factor

1. Dec 14, 2004

### kuengb

Hello everyone

I did an experiment about magnetic susceptibility of $Dy_2O_3$ and $Er_2O_3$. For data evaluation I have to calculate the theoretical Landé factor for the paramagnetic atoms in these salts which is defined as
$$g_j=1+\frac{j(j+1)+s(s+1)-l(l+1)}{2j(j+1)}$$
where l and s are the net quantum numbers for orbital momentum and electron spin respectively, j=l+s total angular momentum. The factor gives the relation between angular momentum and magnetic momentum.

The problem is: I don't completely understand how to find the correct values for l and s. That's how I understood it:

1)l is the sum of the l values for all electrons in non-filled orbitals, and s is the sum of the spin numbers (i.e. 1/2) of all unpaired electrons. For Erbium with configuration $(Xe)4f^{12} 6s^2$ this would then be:12 electrons in the 4f orbital, hence l=12*3, and two unpaired electrons in that orbital, i.e. s=2*1/2. Does this make sense?

2)In the crystal structure of $Er_2O_3$, Erbium gives away three elecrons to Oxygen. Is it then reasonable to take the electron configuration of Terbium, which is three numbers below, for the above calculation? If not, how do I have to deal with this ionic bindings?

Thanks
Bruno

2. Dec 14, 2004

### Pieter Kuiper

Er3+ has 11 electrons in its 4f shell (3 holes). These are the only ones you need to consider.

3. Dec 15, 2004

### kuengb

Thank you. So, what I've written in 1) is correct, i.e. in this case l=11*3 and s=3/2 ?

Is there an easy way to understand why Er3+ has this particular configuration and not 4f9-6s2?

Regards
Bruno

4. Dec 15, 2004

### Pieter Kuiper

No, that requires computer calculations. But it is what the rare earths do (similar to the transition metals).

Your result for the total l is wrong, because one needs to add orbital moments vectorially. It is easiest to add the l_z components of the 4f-holes, and then one can see that l = maximum l_z = 3 + 2 + 1.