Solid State Magnetism: Calculating N2/N1 and T for 99% Ground State Population

[Petar Mali]

Homework Statement

On crystal which containing ions V^{4+}, electronic configuration 3d^1, was applied magnetic field B_0=2,5T.

If the temperature is 1K, find the relative concentration of electron states population \frac{N_2}{N_1}.

In what temperature we should expect 99% ions in ground state?

Homework Equations

m_J=\pm J

E=\pm \mu_B B_0

\frac{N_2}{N_1}=e^{-\frac{\Delta E}{k_B T}

g=1+\frac{J(J+1)+S(S+1)-L(L+1)}{2J(J+1)}

The Attempt at a Solution

I have some solution of this problem but I don't understand it.

In solution

S=\frac{1}{2}, L=3, J=\frac{1}{2}
Why?

They get g=2

If I have configuration 3d^1

then

z=1, l=2

S=S_{max}=\frac{z}{2}=\frac{1}{2}

L=L_{max}=S_{max}(2l+1-z)=2

J=|L-S|=\frac{3}{2}

And the basic term is

^2D_{\frac{3}{2}}

How they get L=3,J=\frac{1}{2}?
\frac{N_2}{N_1}=e^{-\frac{\Delta E}{k_BT}}=e^{-\frac{2\mu_BB_0}{k_BT}}=0,035

From the text of problem - In what temperature we should expect 99% ions in ground state?

\frac{N_2}{N_1}=0,01

ln(\frac{N_2}{N_1})=-\frac{\Delta E}{k_B T}

T=-\frac{\Delta E}{k_B ln(\frac{N_2}{N_1})}=0,7K

So my fundamental problem is how they get

S=\frac{1}{2}, L=3, J=\frac{1}{2}

Thanks for your answer!

 

[nickjer]

Where did you get that solution? I don't think it is possible to have J=1/2, when S=1/2 and L=3. The only possible J's are J = 5/2 and 7/2 for that choice of S and L.

Also, you never say what the N2 and N1 states are.

 

[Petar Mali]

From some book. They write L=3 but I suppose they use L=0 but I don't know why? That they use for d orbital. They say something like L is frosen?!

 

[nickjer]

Does this book solution have more than one electron? I believe the L=2 like you said originally, not sure what solution you are reading.

 

[Petar Mali]

You have text of problem in my first post. In solution in book is mistake I think.

They write in solution

S=\frac{1}{2}

L=3 (L is frosen in crystal)

and they write then

J=\frac{1}{2}

In solution they do like this only for d orbitals.