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I Why normal Zeeman effect contains three components?

  1. Mar 15, 2017 #1
    Can someone explain to me why normal Zeeman effect splits spectral line into three components and not into 4, 5 or any other number?
     
  2. jcsd
  3. Mar 15, 2017 #2

    blue_leaf77

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    Do you know the selection rule for the magnetic quantum number?
     
  4. Mar 15, 2017 #3
    Yes, it can be changed by +1, -1, 0.
     
  5. Mar 15, 2017 #4

    blue_leaf77

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    In normal Zeeman effect, states with different ##m## split, accordingly the transition lines also split according to their value of ##\Delta m##.
     
  6. Mar 15, 2017 #5
    I know that normal Zeeman effect happens when total spin of electrons in an atom is S=0.
    This means that energy levels splitting in external magnetic field is done only on orbital angular momentum.

    So for example P orbital will split into three sub-levels with slightly different energies. Now, it make sense to me that if there is transition from P orbital (L=1) to S orbital (L=0) there will be splitting to three components because P is split to three sub-levels and S orbital has no splitting.
    But what about transition from say, D to P orbital? D orbital will be split into 5 sub-levels and P orbital will be split to 3 sub-levels so that gives many more combinations for transitions, even when selection rule for magnetic moment is taken into account.
    So, (assuming total spin is zero), for D orbital we have J = 2,1,0,-1,-2 and for P orbital there is J=1,0,-1.
    Possible transitions are then:
    ##D_2 \rightarrow P_1##
    ##D_1 \rightarrow P_1##
    ##D_0 \rightarrow P_1##

    ##D_1 \rightarrow P_0##
    ##D_0 \rightarrow P_0##
    ##D_{-1} \rightarrow P_0##

    ##D_0 \rightarrow P_{-1}##
    ##D_{-1} \rightarrow P_{-1}##
    ##D_{-2} \rightarrow P_{-1}##

    So there are 9 lines here, each respecting the selection rule for magnetic quantum number.
    What is wrong in this logic, where is my error in understanding?
     
  7. Mar 15, 2017 #6

    blue_leaf77

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    The formula for the transition energy under Zeeman effect is
    $$
    \Delta E = \Delta E_0 + \mu_B B \Delta m
    $$
    where ##\Delta E_0## is the energy difference without magnetic field. As you see the energy difference in the presence of magnetic field only depends on ##\Delta m##. In your example the lines ##d_2 \to p_1## and ##d_1 \to p_0## coincide.
     
  8. Mar 15, 2017 #7
    I think I am starting to get it now...

    So in my example there are 9 transitions but they are grouped such that only 3 different lines are present, right?
    If I group transitions that belong to the same line they would be like this:
    line 1:
    ##D_2 \rightarrow P_1##
    ##D_1 \rightarrow P_0##
    ##D_0 \rightarrow P_{-1}##

    line 2:
    ##D_1 \rightarrow P_1##
    ##D_0 \rightarrow P_0##
    ##D_{-1} \rightarrow P_{-1}##

    line 3:
    ##D_0 \rightarrow P_1##
    ##D_{-1} \rightarrow P_0##
    ##D_{-2} \rightarrow P_{-1}##

    Is this correct?

    Btw, This formula you gave me, it is valid only for normal Zeeman effect right? I mean g-factor is 1 in the formula which applies to singlet states with total spin equal to zero.
     
  9. Mar 15, 2017 #8

    blue_leaf77

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    Yes.

    Yes.
     
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