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Magnetic resonancy, Zeeman effect

  • Thread starter fluidistic
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  • #1
fluidistic
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Homework Statement


A magnetic resonancy experiment is realized using hydrogen atoms in their ground state. A constant magnetic field [itex]B_0[/itex] duplicate the magnetic energy levels in the atoms and an oscillating magnetic field [itex]B_ \omega[/itex] is synchronized to the frequency that corresponds to the transition between these levels. Calculate the value of the frequency of resonance for a field [itex]B_0 =2000G[/itex].


Homework Equations


Somes.


The Attempt at a Solution


I think I know how to solve the problem if the atoms weren't in their ground state.
What makes me doubt about my whole understanding of the quantum numbers and the hydrogen atom is...
If n=1, the quantum number l must be worth 0.
Since [itex]m_l[/itex] goes from [itex]-l[/itex] to [itex]l[/itex], it must also be worth 0.
So how can there be any duplication of lines?

Edit: It isn't stated but I guess I must assume that the atoms absorb photons to reach the shell n=2. Is this right?
 

Answers and Replies

  • #2
ehild
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The electrons have magnetic moment due to their spin, too.
There is no lines mentioned; you need to calculate the splitting of the level.

ehild
 
  • #3
fluidistic
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Ah I see. I totally confused 2 effects.
Is this formula right (I found it in hyperphysics with some substitutions) [itex]\mu _S=-\frac{2 \mu _B S}{\hbar}[/itex]? If so, this is worth [itex]-\sqrt 3 \mu _B[/itex].
I don't really know how to calculate the difference of energy of the electrons due to the external magnetic field.
 
  • #4
ehild
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The interaction energy of dipole and field is equal to the product of the field multiplied by the parallel component of the momentum. The magnetic momentum of the electron can align only parallel and antiparallel to the field.

ehild
 
  • #5
fluidistic
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The interaction energy of dipole and field is equal to the product of the field multiplied by the parallel component of the momentum. The magnetic momentum of the electron can align only parallel and antiparallel to the field.

ehild
Thank you very much, I understand now.
So you mean [itex]\Delta E = \pm \mu _B B[/itex].
Also, [itex]E= \hbar \omega[/itex].
This gives me [itex]\omega = \frac{\mu _B B_0}{\hbar}[/itex]. [itex]B_0[/itex] is worth 2 teslas.
I reach [itex]\omega \approx 1.75 \times 10 ^{11} Hz[/itex]. Does this looks good?
 

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