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.