Lithium: fine structure, internal magnetic field and Zeeman Effect PHYS2332 Modern Physics II W2018

Estimating the Internal Magnetic Field Method I Recall that the fine structure of atoms is due to the coupling between the electron’s spin with the internal magnetic field,

!Bint , of the atoms. To show how to estimate

!Bint for

specific atomic states, consider the fine structure of Lithium discussed in class: 2 2S1/2

In the diagram, a horizontal line indicates a state of lithium, with the symbol,n 2LJ . The doublet states 2 2P3/2 and 2 2P1/2 have different energies due to an internal field

!Bint arising

from the orbital angular momentum of a 2p ( L = 1) electron. Compare this to the degenerate 2 2S1/2 where there is no energy splitting (differences) since the internal field due to the 2s ( L = 0 ) is

!Bint = 0 . As discussed in class the internal field is proportional to

the orbital angular momentum !Bint ∝

!L (problem 27 of chapter 8), and all S states with

the same principal quantum number, n, are degenerate (i.e. have the same energy). The diagram above shows the transition 2 2P3/2 → 2 2S1/2 (dashed line) that emits a photon of wavelengthλ1 = 670.775nm , and of the 2 2P1/2 → 2 2S1/2 (dotted line) with wavelength λ2 = 670.790nm . To use this data to estimate Bint , consider the diagram below:

!Bint

!S - electron’s spin

Interaction energy Eint = −

!µS i!Bint =

emBintSz , m is electron’s mass

Recall that the z-component of the spin is Sz = ms!, ms = ±1 / 2

Eint = 2µBBintms , where µB =

e!2m

is the Bohr’s magneton

The SI value of µB is given in A1, but we can also use µB = 5.788 ×10−5eV /T

!µs = −

em!S ≡magnetic moment due to electron’s spin

Lithium

Let the energy of the photon be emitted by the 2 2P3/2 → 32S1/2 transition be E1 =hcλ1

,

with λ1 = 670.775nm . Let the energy of the photon be emitted by the 2 2P1/2 → 32S1/2

transition be E2 =hcλ2

, with λ1/2 = 670.790nm . Using hc = 1239.8eV / nm , the energy

difference, due to spin-orbit coupling, is

ΔE = hc 1

λ1−1λ2

⎛⎝⎜

⎞⎠⎟= 1239.8eV inm( ) 1

670.775nm−

1670.790nm

⎛⎝⎜

⎞⎠⎟= 4.21×10−5eV

The energy splitting of the 2 2P3/2 and 2 2P1/2 is due mostly to the spin-orbit coupling with

the internal magnetic field, Bint . Assume that the 2 2P3/2 state has spin up msup = 1 / 2

and the 2 2P1/2 state has spin down msdown = −1 / 2 , then the change in energy is

ΔE = Eintup − Eint

down = 2µBBint msup − ms

down( ) = 2µBBint (see example 8.3 on page 285) .

Hence we haveBint =ΔE2µB

=4.1×10−5ev

2 × 5.788 ×10−5eV /T= 0.35T .

Estimating the Internal Magnetic Field Mehod II In class, I made a semi-classical argument that the internal magnetic field can be

approximated by a ring current model to give Bint =µ0eL4πmr3

, where

µ0 = 4π ×10−7T im / A is the magnetic permeability in vacuum, and m = 9.11×10−31kg is

the mass of the electron. Using the formula Bint =µ0eL4πmr3

, with L = ℓ ℓ +1( )" = 2" (

ℓ = 1 for 2p) and from example 7.11 on page 265, replace r = 4a0 (most probable distance for 2p state). This gives

Bint =2µ0e!

4πm 4a0( )3=

2 4π ×10−7N iA−2( ) 1.6 ×10−19C( ) 1.0546 ×10−34 J is( )4π 9.1×10−31kg( ) 4 × 5.2918 ×10−11m( )3

= 0.28T

This is actually remarkably close to the answer obtained by method I. The Anomalous Zeeman Effect The left hand side (LHS) diagram below shows an atom with spin

!S and orbital angular

momentum !L in an external magnetic field,

!Bext = Bext

⌢k that points along the z axis. The

right hand side (RHS) diagram shows the correct illustration of how the atom interacts with

!Bext ; it interacts with the atoms total angular momentum

!J .

!Bext

!µL

!Bext

!µJ

!S

!µs

!L

!J

Using quantum mechanics (QM) the magnetic moment of the electron’s orbital angular

momentum is

!µL = −

e2m!L and the electron’s spin is

!µS = −

em!S . It is then tempting to

assume that the actual magnetic moment of the atom is found by simply adding these two contributions, but this would be wrong, since the spin and orbital part of the angular momentum interact. Instead an atom behavior in a magnetic field is determine by its total angular momentum

!J , with a magnetic moment

!µJ = −

e"2m

g!J"= −µBg

!J"

with µB = 5.788 ×10−5eV /T being the Bohr magneton, and

g = 1+J J +1( ) + S S +1( ) − L L +1( )

2J J +1( ) (equation 8.23) is the Lande g factor. Hence when

an atom is in an external magnetic field that points along the z-axis !Bext = Bext

⌢k the

Zeeman interaction energy is Vzeeman = −

!µJ i!Bext = µBgBext

Jz"

. Using the quantization

rule Jz = mJ! , with mJ = ±J,± J −1( ),... gives Vzeeman = µBgBextmJ , as in equation 8.23. The selection rules for transitions between atomic states in an external field combine equations 8.14 and 8.24. Study the examples in section 8.3 on page 292 to 294. Below are some examples on the Zeeman effect on Lithium, Example of Anomalous Zeeman effect in Lithium Suppose a lithium atom in the state is placed in an applied magnetic field of Bext = 0.5T . Calculate the energy different between adjacent lines due to the Zeeman effect of, A) the adjacent levels of 2 2P3/2 , and B) the adjacent levels of 2 2P1/2 . C) If there were no spin how many levels would the P state ( L = 1) have? D) Briefly explain how the results of part A, B, and C illustrate the anomalous Zeeman effect.

A) For the 2P3/2 state of Lithium J = 32

and mJ =32, 12,− 12,− 32

. The Lithium Zeeman

splitting is similar to Figure 8.16 (on sodium) of example 8.10 (PP. 293-294), where we can clearly see that the 2P3/2 is split into four states. The Zeeman energy of these levels are given by equation 8.22 (page 293) Vz = µBBextgmJ . The energy separation between adjacent levels (for example mj = 3 / 2 and mj = 1 / 2 , or mj = 0 and mj = −1 / 2 ) is simply: ΔVz = µBBextgΔmj = µBBextg , where Δmj = 1 for adjacent levels. Note that for 2P3/2 state

of sodium J = 32

, L = 1, and S = 1/2, the Lande factor is calculated on p. 294, g = 1.33.

For Bext = 0.50T , we use µB = 5.788 ×10−5eV /T and g = 1.33 to obtain for the energy

difference between adjacent Zeeman 2P3/2 levels, ΔVz = µBgBext = 5.788 ×10

−5eV /T 0.5T( ) 1.33( ) = 3.84 ×10−5eV

B) For the 2P1/2 state of Lithium J = 12

and mJ =12,− 12

. The Lithium Zeeman splitting is

similar to Figure 8.16 (on sodium) of example 8.10 (PP. 293-294), where we can clearly see that the 2P3/2 is split into four states. The Zeeman energy of these levels are given by equation 8.22 (page 293) Vz = µBBextgmJ . The energy separation between adjacent levels (for example mj = 3 / 2 and mj = 1 / 2 , or mj = 0 and mj = −1 / 2 ) is simply:

ΔVz = µBBextgΔmj = µBBextg , where Δmj = 1 for adjacent levels. Note that for 2P1/2 state

of sodium J = 12

, L = 1, and S = 1/2, the Lande factor is calculated on p. 294, g = 0.67.

For Bext = 0.50T , we use µB = 5.788 ×10−5eV /T and g = 0.67 to obtain for the energy

difference between adjacent Zeeman 2P3/2 levels, ΔVz = µBgBext = 5.788 ×10

−5eV /T 0.5T( ) 0.67( ) = 1.94 ×10−5eV C) If there were no spin the Lithium 2P levels will be ℓ = 1will have three levels mℓ = ±1,0 . The Zeeman energy level is Vz = µBBextmℓ , since g = 1 when there’s no spin. D) Without spin there are three Zeeman levels. With spins there are 6 Zeeman levels. The transitions between these Zeeman levels can be observed in spectroscopic experiments. The discrepancy between the number of observed lines in experiments and in the number predicted by theory that assumes no spin, is called the anomalous Zeeman effects.