The Anomalous Zeeman Effect - Semantic Scholar

The Anomalous Zeeman Effect

Physics 273 – Anomalous Zeeman Effect

Chemical properties are (mostly) determined by the outermost electrons.Outer electrons do not feel the full nuclear charge, because inner electrons partially shield the nucleus. The is called shielding or screening.Radii of complex atoms are within a factor of 2 of the hydrogen! (inner electrons get pulled in closer to the nucleus)

Ionization energy of hydrogen: -13.6 eV. Other atoms: -5 to -25 eV.Hydrogen exhibits 𝓁-degeneracy: energy depends on n, but not on 𝓁 (up to fine-structure). i.e. 3s, 3p, 3d all have the same energy.

Complex atoms: for a given n, the energy increases with 𝓁. This is because of screening. 2


Electrons in lower angular momentum configurations spend more time closer to the nucleus than higher angular momentum configurations (at least semi-classically).For multi-electron atoms, the inner electrons shield the nuclear charge from the outer electrons.This means that smaller n electrons have a stronger binding energy than larger n electrons. (but you knew that already).It also means that smaller 𝓁 electrons have a stronger binding energy than larger 𝓁 electrons.

3


The Sodium spectrum (alkalai metal, so only one valence electrons with a filled inner shell).Notice that the energies increase with with 𝓁 (i.e. they are less tightly bound at higher 𝓁).

4


Electrons fill subshells in order of increasing energy:1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, ...note that the 𝓁-dependence of energy becomes more and more important as Z increases...

Hund’s rules:First priority is to maximize the spin angular momentum (parallel spins are preferred).Second priority is to maximize the orbital angular momentum.

5

Physics 273 – Anomalous Zeeman Effect 6

PRS Question: A certain subshell can have a maximum of 14 electrons. Which of the following could that

subshell be?A) 3pB) 3dC) 3fD) None of the above


14 electrons means that 2(2𝓁+1)=14 → 𝓁=3, i.e. the “f” sub-shell.But we need to have n>𝓁, so 3f is impossible...

7

PRS Question: A certain subshell can have a maximum of 14 electrons. Which of the following could that

subshell be?A) 3pB) 3dC) 3fD) None of the above


At this point we have calculated two different magnetic moments of the electron.One is due to its rotation about the nucleus:

Because of this magnetic moment, we deduced that by putting an atom in a magnetic field we would get splittings according to the quantum number m𝓁:

hence the name: m𝓁 is the “magnetic” quantum number. This is called the “normal” Zeeman effect.

8

~µL = �g e2m

~L with g = 1

�E = geB

2me(m`~) = m`gµBB where µB ⌘ e~

2me⇡ 5.8⇥ 10�5 eV

T

3s 0 3p 0+1

-13d 0

+1

-1

+2

-2


Another magnetic moment is due to the instrinsic spin of the electron:

The factor of 2 comes from considerations of the Dirac equation (the “relativistic Schrodinger equation”, if you like).Here we find that the interaction between the spin magnetic moment of the electron and the “magnetic field” of the nucleus creates fine- structure splitting.The interaction between the electron and the nucleus is called a spin-orbit coupling.

9

~µs = �g e2m

~S with g = 2 (!)


For a hydrogen atom (but not a generic atom), the energy splitting followed the total angular momenta (J) of the spin (S) plus the orbital angular momentum (L):

where EBohr is the usual Bohr energy levels. The splitting is approximately 10-5 eV.

Remember that j can be either 𝓁+1/2 or 𝓁-1/2 (unless 𝓁=0) 10

Efine

= EBohr

1 +

↵2

n

✓1

j + 1/2� 3

4n

◆�

3s1/2 3p1/2

3p3/2 3d3/2

3d5/2

n(2s+1)Lj


Now let’s put these two together: spin-orbit effects in a magnetic field.In order to do this, we need to calculate μJ (not just μL), that is, the magnetic moment from the total angular momentum.

Because g=2, μS+μL≠μJ:

11

~µL = �g e2m

~L with g = 1

~µs = �g e2m

~S with g = 2

L

μLμS

S

JμJ

See how S is anti-parallel to μS, and L in anti-

parallel to μL, but J is not anti-parallel to μJ?


In order to deal with this, we will define μJ as the total magnetic moment (adding orbital and spin components together) along the J direction

where “g” is the Lande g factor:

This g-factor is an effective constant that compensates for the fact that the magnetic moment is not anti-parallel to the J direction.

12

~µJ = �ge

2me

~J

g = 1 +j(j + 1) + s(s+ 1)� `(`+ 1)

2j(j + 1)


Just as with the “normal” Zeeman effect, if you place the atom in a magnetic field, you get an energy shift

but this time, it’s proportional to j (not 𝓁), and the g-factor depends on j, s, and 𝓁.

13

�E = geB

2me(mj~) = mjgµBB


Example: for B=0 the Hydrogen lines 3S1/2 and 3P1/2 are degenerate (same n and j). What happens when B>0?

Since j=1/2, they split into (2j+1) states: mj=1/2 and mj=-1/2.But g is different for the 3S and 3P states because the 𝓁 is different (s versus p).

For the 3S1/2 state, j=1/2, s=1/2, and 𝓁=0, so g=2:

For the 3P1/2 state, j=1/2, s=1/2, and 𝓁=1, so g=2/3:

14

�E = geB

2me(mj~) = mjgµBB

g = 1 +12 (

12 + 1) + 1

2 (12 + 1)� 0(0 + 1)

2 12 (

12 + 1)

= 2

g = 1 +12 (

12 + 1) + 1

2 (12 + 1)� 1(1 + 1)

2 12 (

12 + 1)

=2

3


Diagramatically, this looks like:

This is the anomalous Zeeman effect. Notice that one can apply this effect as a correction to the fine-structure.When does the “normal” Zeeman effect apply? When the net spin of the electrons is 0 (and so j=𝓁).

15

3s1/2

mj=+1/2

mj=-1/2

3p1/2

mj=+1/2

mj=-1/2

ΔE=μB.B ΔE=1/3 μB.B

Physics 273 – Anomalous Zeeman Effect 16

PRS Question: Suppose a hydrogen atom in a 3d5/2 state is placed in an external magnetic field. Into how

many substates will it split?A) 3B) 4C) 5D) 6E) None of the above


j is 5/2, so the number of states is (2j+1)=6.Specificially, they are -5/2, -3/2, -1/2, 1/2, 3/2, and 5/2.

17

PRS Question: Suppose a hydrogen atom in a 3d5/2 state is placed in an external magnetic field. Into how

many substates will it split?A) 3B) 4C) 5D) 6E) None of the above

Bonus


What if we want to consider the anomalous Zeeman effect for more complicated atoms (i.e. not Hydrogen)?

Two ways to calculate the total angular momentum:LS coupling:

JJ coupling:

The JJ coupling is only useful for large Z atoms, so we’ll skip that.

19

~L =X

i

~Li, ~S =X

i

~Si, ~J = ~L+ ~S

~Ji = ~Li + ~Si, ~J =X

i

~Ji


Consider Carbon: the electron configuration is:1s2 2s2 2p2. The 1s and 2s state are completely filled, so there are two active electrons in the 2p state.Each active electron has 𝓁=1 and s=1/2. Let’s use the LS coupling scheme to determine S, L, and J:S=0 or 1. L=0, 1, or 2.

20

S L J Term Symbol Exists?0 0 0 21S0 yes0 1 1 21P1 no0 2 2 21D2 yes1 0 1 23S1 no1 1 0,1,2 23P0, 23P1, 23P2 yes1 2 1,2,3 23D1, 23D2, 23D3 no

n2s+1Lj


The term symbol of the atom determines the Zeeman splitting:

Example: Suppose an atom in the 53F2 state is placed in a magnetic field B. What is the energy shift between adjacent magnetic substates?We have the quantum numbers: S=1, L=3, J=2and mj=-2, -1, 0, 1, 2

21

�E = geB

2me(mj~) = mjgµBB

g = 1 +2(2 + 1) + 1(1 + 1)� 3(3 + 1)

2 · 2(2 + 1)=

2

3

�E = gµBB =2

3µBB

Documents

The Anomalous Zeeman Effect - Semantic Scholar