Solid state magnetism

  • Thread starter Petar Mali
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  • #1
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Homework Statement


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

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

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


Homework Equations


[tex]m_J=\pm J[/tex]

[tex]E=\pm \mu_B B_0[/tex]

[tex]\frac{N_2}{N_1}=e^{-\frac{\Delta E}{k_B T}[/tex]

[tex]g=1+\frac{J(J+1)+S(S+1)-L(L+1)}{2J(J+1)}[/tex]

The Attempt at a Solution



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

In solution

[tex]S=\frac{1}{2}, L=3, J=\frac{1}{2}[/tex]



Why?

They get [tex]g=2[/tex]

If I have configuration [tex]3d^1[/tex]

then

[tex]z=1[/tex], [tex]l=2[/tex]

[tex]S=S_{max}=\frac{z}{2}=\frac{1}{2}[/tex]

[tex]L=L_{max}=S_{max}(2l+1-z)=2[/tex]

[tex]J=|L-S|=\frac{3}{2}[/tex]

And the basic term is

[tex]^2D_{\frac{3}{2}}[/tex]

How they get [tex]L=3,J=\frac{1}{2}[/tex]???



[tex]\frac{N_2}{N_1}=e^{-\frac{\Delta E}{k_BT}}=e^{-\frac{2\mu_BB_0}{k_BT}}=0,035[/tex]

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

[tex]\frac{N_2}{N_1}=0,01[/tex]

[tex]ln(\frac{N_2}{N_1})=-\frac{\Delta E}{k_B T}[/tex]

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

So my fundamental problem is how they get

[tex]S=\frac{1}{2}, L=3, J=\frac{1}{2}[/tex]

Thanks for your answer!
 

Answers and Replies

  • #2
674
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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.
 
Last edited:
  • #3
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From some book. They write [tex]L=3[/tex] but I suppose they use [tex]L=0[/tex] but I don't know why? That they use for [tex]d[/tex] orbital. They say something like [tex]L[/tex] is frosen?!
 
  • #4
674
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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.
 
  • #5
290
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You have text of problem in my first post. In solution in book is mistake I think.

They write in solution

[tex]S=\frac{1}{2}[/tex]

[tex]L=3[/tex] ([tex]L[/tex] is frosen in crystal)

and they write then

[tex]J=\frac{1}{2}[/tex]

In solution they do like this only for [tex]d[/tex] orbitals.
 

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