(revised 5/17/07)

ZEEMAN SPLITTING

Advanced Laboratory, Physics 407

University of Wisconsin
Madison, Wisconsin 53706

Abstract
When a magnetic field is applied to atoms, some of the atomic
energy levels may be changed and some levels which had identical
energies may be split into levels with different energies. This was
discovered by Zeeman in 1896 and is called the Zeeman Effect.
In this experiment you will measure the energy level splitting and
polarizations due to the Zeeman Effect and will apply the theory for
the Zeeman effect to calculate the Bohr magneton. You will also study
and use a Fabry-Perot Etalon.
You may perform this experiment before you discuss the full theory
in your Quantum Mechanics class. For that reason, this description is
more detailed than the descriptions of most other experiments in the
Advanced Laboratory.
Theory
In 1896 Zeeman discovered a broadening effect in the yellow lines of sodium
when the light source was placed in a strong magnetic field. Soon afterward,
Lorentz predicted from classical electromagnetic theory that each line should
split into 3 components.
We now understand the splitting of lines by a magnetic field in terms
of the Dirac theory for the electron. If an atom has an eigenstate with the
angular momenta and spins of all the electrons giving vector totals:
q
• (Σ angular momenta) has magnitude L0 = `0 (`0 + 1) h̄

• (Σ angular momenta) in a given direction = m0` h̄ − ` ≤ m0` ≤ `

q
• (Σ intrinsic spins) has magnitude S 0 = s0 (s0 + 1) h̄

• (Σ intrinsic spins) in a given direction = m0s h̄

(We use a notation with primes indicating totals for all electrons and lower
case referring to quantum numbers.)
Then that state is split by the magnetic field B which causes an additional
potential energy term:
∆E = −~µ · B~
where ~µ is the effective magnetic moment of the state.
e ~ e ~
    
∆E = − Σi −g` Li + Σi −gs Si ~
·B
2me 2me
e h ~0 i
~ 0 · B.
~
= g` L + gs S
2me
~ i and S
Here L ~i are the angular momentum and spin of the ith electron. The
charge and mass of an electron are designated by e and me respectively.
The factors g` and gs from the Dirac and QED Theory are:

g` = 1 exactly and gs = 2.00232 .

Often the 2.00232 is approximated by 2 and the equation is written

µB h ~ 0 i
~0 · B
~
∆E = L + 2S ,
h̄
eh̄
where µB = 2me
called the Bohr Magneton.

2
 The total angular momentum

J~0 = L
~0 + S
~0

has a z component which is quantized into (2j 0 + 1) discrete values:

Jz0 = m0j h̄ m0j = (−j 0 ), (−j 0 + 1) . . . (j 0 − 1), (j 0 ).

However the magnetic moment

µB h ~ 0 ~0
i
~µ = L + 2S
h̄
is not parallel to J~0 and so the z component of µ depends upon both L ~ 0 and
~ 0.
S
The z component can be derived either easily by using the “vector model”
which is semi-classical or by using full quantum mechanics with perturbation
calculations. The student should read and understand the vector model
method which is given in most standard texts.
Both methods give the z component of the magnetic moment as
µB
µz = [g m0g h̄] = g m0j µB
h̄
and ∆E = g m0j µB B.
The g is called the Landé g factor. We would expect g = 1 for a simple
orbital angular momentum but the intrinsic spin causes:
j 0 (j 0 + 1) + s0 (s0 + 1) − `0 (`0 + 1)
g =1+ .
2j 0 (j 0 + 1)
The energy of the photons emitted then depends upon both the split-
ting of the initial state and the splitting of the final state. Further, there
are restrictions upon the changes which can occur in the various quantum
numbers.

Selection Rules
The most likely radiation to be emitted is that due to “electric dipole” tran-
sitions of single electrons.

3
(a) Induced Transitions
If an atom is placed in an oscillating electric field so that the electric
potential V is a function of x and t; V (x, t) then the transition rate
from states n to state m can be shown to be
dam −i X
= an e−i(En −Em )t/h̄ Vmn ,
dt h̄ n
where Vmn is the matrix element depending upon the electric potential
V , and the wave functions of the initial and final states:
Z
∗
Vmn = ψm V (x, t)ψn .

If
(i) the perturbation is small,
~
(ii) the electric field has only one frequency f with electric field E,
(iii) the wave length of the electromagnetic field λ = c/f is large com-
pared with the diameter of the atom,
(iv) the energy difference (Em − En ) = hf ,
(v) most of the atoms are in one state with energy En , then the tran-
sition rate can be shown to be:
d ∗ 1
(am am ) = 2 µ∗mn µmn E 2 ,
dt h̄
where µmn has been derived from Vmn by assuming that V (x, t) =
E ex. Z
∗
µmn = ψm e xψn dx .
Hence an electric field with an electric vector in the x direction
will cause transitions from state n to state m only if the integral is
non zero. Remember that if m 6= n, then the states are orthogonal
and so: Z
∗
ψm ψn = 0 .
The integral we are interested in now is different because it in-
cludes the coordinate x:
Z
∗
ψm xψn =?

4
 but is also often zero. The non-zero conditions depend upon the
quantum numbers of the initial and final states and can be derived
by integrating the initial and final Associated Legendre functions
of cos θ and the Associated Laguerre functions of the radius.

The transitions occur only if all of the following are true:

(i) ∆` = ±1 for the electron making the transition. This means that
the parity of the electron wavefunction must change.
R ∗
A change in ` is necessary (otherwise ψm xψn = 0) but changes
in ` greater than ±1 require the electric field to be nonuniform
within the atom and so are very unlikely if the wavelength >
atomic diameter.
(ii) ∆S 0 = 0 for the whole atom. The intrinsic spin and magnetic mo-
ment of an electron interacts only with a magnetic field gradient
(as in the Stern Gerlach Expt). We are considering electromag-
netic radiation with long wavelengths and so the gradient over the
atom will be very small.
These unlikely transitions which change S would be called “spin-
flip” transitions.
(iii) ∆`0 = 0 or ±1 for the whole atom. Changes of ` can occur in
the separate electron wavefunctions without causing the total ` to
change. However, the total ` cannot change more than the ` of
the electron making the transition.
(iv) ∆j 0 = 0 or ±1 for the whole atom (but j 0 = 0 to j 0 = 0 is not
allowed). The total angular momentum cannot change by more
than 1 since the spin of a photon is 1 and angular momentum
must be conserved.
(v) ∆m0j = 0 or ±1 for the whole atom (but if ∆j = 0 then mj = 0
to mj = 0 is not allowed).

(b) Spontaneous Transitions

The laws of Relativistic Quantum Mechanics predict that a state will
decay to a lower state only in the presence of an external alternating
electromagnetic field. Hence Relativistic Quantum Mechanics will not
allow “spontaneous transitions.”

5
 However, the theory of Quantum Electrodynamics (QED) goes further
by allowing for quantization of the electric field and predicts that a state
may decay due to quantum fluctuations in its own electric field. These
fluctuations arise from the “zero-point energy” of the electromagnetic
field which is similar to the “zero-point energy” of a particle in the
lowest energy level of a potential well. The fluctuations can be regarded
as “virtual photons” being continuously emitted and absorbed. The
fluctuations are of all frequencies and can cause spontaneous transitions
in the same way as external fields induce transitions.
Although the transition rates can only be calculated from QED, the
selection rules for spontaneous transitions are identical to those of in-
duced transitions since the rules are derived from the same integrals of
the initial and final wave functions.

Polarization of Radiation
The polarization of the emitted light depends upon the change ∆mj . Assume
the magnetic field B is along the z axis. Here are three explanations of the
same phenomena:

Explanation 1
The same theory which gave the ∆m restrictions also makes predictions
for the electric field.
(a) If ∆m0j = 0, then the alternating E ~ is parallel to the magnetic field.
This light is then linearly polarized and mostly emitted perpendicularly
to the magnetic field. An observer in the x direction would see E ~ k ~z,
an observer in the z axis would see no radiation.
~ is perpendicular to the magnetic
(b) If ∆m0j = ±1, then the alternating E
field (z axis).

• An observer on the x axis will see E ~ k ~y

• An observer on the z axis will see E~ in both ~y and ~x directions A
careful analysis of the integrals shows that the observer on the z
axis will see:

6
 right hand circular polarization if ∆m0j = +1 and
left hand circular polarization if ∆m0j = −1 .

Explanation 2

This can be partly understood by remembering that photons have a spin

angular momentum of one unit of h̄ and that angular momentum must be
conserved.

(a) If ∆m0j = 0, then no angular momentum about the z axis can be carried
off by a photon and so an observer on the z axis will see no light.
An observer on the x axis will see photons due to ∆m0j = 0 and ∆m0j =
±1. The linearly polarized light is equivalent to a superposition of
equal amplitudes of oppositely circularly polarized light.

(b) If ∆m0j = +1, then the angular momentum about the z axis must be
conserved by the photon taking off with an s = −1. Hence an observer
on the z axis will see 100% right hand circularly polarized light.
Similarly with ∆m0j = −1, the photons will have an s = +1 and an
observer on the z axis will see 100% left hand circularly polarized light.

7
Explanation 3

From G.R. Fowles “Introduction to Modern Optics”:

Coherent States Consider a system that is in the process of changing

from one eigenstate Ψ1 to another Ψ2 . During the transition the state func-
tion is given by a linear combination of the two state functions involved,
namely
Ψ = c1 ψ1 e−iE1 t/π + c2 ψ2 e−iE2 t/π (1.1)
Here c1 and c2 are parameters whose variation with time is slow in comparison
with that of the exponential factors. A state of the above type is known as
a coherent state. One essential difference between a coherent state and a
stationary state is that the energy of a coherent state is not well defined,
whereas that of a stationary state is.
The probability distribution of the coherent state represented by Eq. 1.1
is given by the following expression:

Ψ∗ Ψ = c1 ∗ c1 ψ1 ∗ ψ1 + c2 ∗ c2 ψ2 ∗ ψ2 + c1 ∗ c2 ψ1 ∗ ψ2 eiωt + c2 ∗ c1 ψ2 ∗ ψ1 e−iωt (1.2)

where
E1 − E2
ω= (1.3)
h̄
or, equivalently
E1 − E2
f=
h
The above result shows that the probability density of a coherent state
undergoes a sinusoidal oscillation with time. The frequency of this oscillation
is precisely that given by the Bohr frequency condition.
The quantum-mechanical description of a radiating atom may be stated
as follows. During the change from one quantum state to another, the proba-
bility distribution of the electron becomes coherent and oscillates sinusoidally.
This sinusoidal oscillation is accompanied by an oscillating electromagnetic
field that constitutes the radiation.

8
Fig. 1.1 Probability density for the first excited state (n = 2) of the hydrogen
atom.

Radiative Transitions and Selection Rules

When an atom is in the process of changing from one eigenstate to another,

the probability density of the electronic charge becomes coherent and oscil-
lates sinusoidally with a frequency given by the Bohr frequency condition.
The way in which the charge cloud oscillates depends on the particular eigen-
states involved. In the case of a so-called dipole transition, the centroid of the
negative charge of the electron cloud oscillates about the positively charged
nucleus. The atom thereby becomes an oscillating electric dipole.

Fig. 1.2 Charge distribution in the coherent state 1s + 2p0 as a function of

time. The atom is an oscillating dipole.

9
 Figure 1.2 is a diagram showing the time variation of the charge distribu-
tion for the hydrogen atom when it is in the coherent state represented by the
combination 1s + 2p(m = 0). It is seen that the centroid of the charge moves
back and forth along the z axis. The associated electromagnetic field has a
directional distribution that is the same as that of a simple dipole antenna
lying along the z axis. Thus, the radiation is maximum in the xy plane and
zero along the z axis. The radiation field, in this case, is linearly polarized
with its plane of polarization parallel to the dipole axis.

Fig. 1.3 Charge distribution in the coherent state 1s + 2p1 as a function of

time. The atom is a rotating dipole.

A different case is shown in Figure 1.3. Here the coherent state is the
combination 1s + 2p(m = +1). The centroid of the electronic charge now
moves in the circular path around the z axis. The angular frequency of the
motion is also that given by the Bohr frequency formula ω = ∆E/h̄.
Instead of an oscillating dipole, the atom is now a rotating dipole. The
associated radiation field is such that the polarization is circular for radiation
traveling the direction of the z axis and linear for radiation traveling in
a direction perpendicular to the z axis. For intermediate directions, the
polarization is elliptical. The cases are illustrated in Figure 1.4. The coherent
state 1s + 2p(m = −1) is just the same as the state 1s + 2p(m = +1)
except that the direction of rotation of the electronic charge is reversed.
Consequently, the sense of rotation of the associated circularly polarized
radiation is also reversed.

10
Fig. 1.4 Polarization of the associated electromagnetic radiation (E vector)
for (a) an oscillating dipole and (b) a rotating dipole.

The Spectra
We will study the 546.1 nm green line of mercury. Each atom has 80 electrons
and in the ground state these have the distribution:
n = 1 shell filled 2
n = 2 shell filled 8
n = 3 shell filled 18
n = 4 shell filled 32
n = 5 shell ` = 0 subshell 2
` = 1 subshell 6
` = 2 subshell 10
` = 3 subshell empty
` = 4 subshell empty
n = 6 shell ` = 0 subshell 2

11
 The excited states have one electron lifted from the n=6, `=0 subshell
and are shown in the following diagram from Herzberg.

Fig. 2.1 Energy Level Diagram of Hg I [Grotrian (8)]. The wavelengths of

the more intense Hg Lines are given (cf. Fig. 6, p. 6). The symbols 6p, etc.,
written near each level, indicate the true principal quantum number and the
` value of the emission electron.

12
 The levels are labelled by the s0 , `0 , and j 0 of the total atom. The full
shells and subshells have zero contributions and so the net s0 , `0 and j 0 are due
to one electron left in the n = 6, ` = 0 subshell and to the excited electron.
Remember the traditional spectroscopic notation:

• (superscript = 2s0 + 1)

• (S, P , D, F etc. for L = 0, 1, 2, 3, etc.)

• (Subscript = j 0 )

The notation sometimes includes the n and ` of the excited electron. For
example:

• 7s 3 S1 has the excited electron with n = 7, ` = 0 with the totals s0 = 1,

`0 = 0, j 0 = 1.

Notice that most transitions have no change in s0 . The levels with s0 = 0

are called singlets (no spin-orbit splitting) and the levels with s0 = 1 are
called triplets (spin-orbit interactions give 3 close levels).
We will study the bright 546.1 nm green line which is easily distinguished
from the fainter 579.0 and 577.0 nm yellow and the bright 435.8 nm blue.
The 546.1 nm is between two states with different Landé g factors:

• 7s 3 S1 s0 = 1 `0 = 0 j 0 = 1 g = 2 3 values of m0j

• 6p 3 P2 s0 = 1 `0 = 1 j 0 = 2 g = 3/2 5 values of m0j

The restriction that ∆m0j = 0, ±1 gives nine transitions which are split
by the Zeeman Effect. Let δ = (g m0j )initial − (g m0j )final .
The two unequal g factors cause the 9 possible transitions to have 9
different wave numbers ν ≡ λ1 = ∆E/hc:
µB B
ν = ν0 + δ . (2.1)
hc

13
 δ −2 −1.5 −1 −0.5 0 0.5 1 1.5 2
initial m0j −1 0 +1 −1 0 +1 −1 0 +1
final m0j 0 +1 +2 −1 0 +1 −2 −1 0
∆m0j +1 +1 +1 0 0 0 −1 −1 −1
Not seen on axis π π π
Seen on axis as
σ σ σ σ σ σ
circularly polarized

mj'

+1

3
S1 0
g=2
–1

+2

+1
3P 0
2
g=3/2
–1

–2

14
 Modified Section from Melissinos:

Reduction of the Data Obtained from a Fabry-Perot

Constructive interference occurs for a Fabry-Perot etalon of spacing t when:

nλ = 2t cos θ . (3.1)

When parallel rays of angle θ are brought to a focus by the use of a lens of
focal length f and Eq. 3.1 is satisfied, bright rings will appear in the focal
plane. Their radius given by:

r = f tan θ ' f θ . (3.2)

Thus the interference rings formed in the focal plane have radii:

rn = f θ n (3.3)

where µ is the index of refraction and the angle θn is given by:

!
2µt θn
n= cos θn = n0 cos θn = n0 1 − 2 sin2 . (3.4)
λ 2

Since θ is always small, we obtain:

s
θ2
!
2(n0 − n)
n = n0 1− n or θn = . (3.5)
2 n0

Now if θn is to correspond to a bright fringe, n must be an integer; however,

n0 , which gives the interference at the center (cos θ = 1 or θ = 0 in Eq. 1.1),
is in general not an integer:
2µt
n0 = . (3.6)
λ
(There is no bright spot in the center of the pattern, in general.) If n1 is the
interference order of the first ring, clearly n1 < n0 since n1 = n0 cos θ1 . We
then let n1 = n0 − , with 0 <  < 1 where n1 is the closest integer to n0
(smaller than n0 ). Thus, we have in general for the pth ring of the pattern,
as measured from the center out,

np = (n0 − ) − (p − 1) . (3.7)

15
Combining Eq. 3.7 with Eqs. 3.5 and 3.3, we obtain for the radii of the rings
s
2f 2 q
rp = (p − 1) +  . (3.8)
n0
We note (a) that the difference between the squares of the radii of adjacent
rings is a constant
2 2 2f 2
rp+1 − rp = (3.9)
n0
and (b) that the fraction of an order  can be found by extrapolating to
rp2 = 0 (according to the slope 2f 2 /n0 )∗
Now,† if there are two components of a spectral line with wavelengths
λ1 and λ2 , very close to one another, they will have fractional orders at the
center 1 and 2 :
2t
1 = − n1 (1) = 2tν̄1 − n1 (1)
λ1
2t
2 = − n1 (2) = 2tν̄2 − n1 (2)
λ2
where n1 (1), n1 (2) is the order of the first ring. Hence, if the rings do not
overlap by a whole order (n1 (1) = n1 (2)), the difference in wave numbers
between the two components is simply
1 − 2
ν̄1 − ν̄2 = . (3.10)
2t
If the orders are overlapped x times,
x + 1 − 2
ν̄1 − ν̄2 = . (3.11)
2t
From Eq. 3.10 we see that we do not need to know t much more accurately
than 1 −2 . The fractional order 1 , 2 , . . ., can hardly be measured to 1/1000;
therefore knowledge of t to this accuracy of 1/1000 is amply adequate; this
can be easily achieved with a micrometer or a microscope.
The resolution of the Fabry-Perot etalon can be obtained from Eq. 3.1:
1 n
ν̄ = =
λ 2t cos θ
∗
See figure on page 18.
†
From here on we set µ = 1.

16
and by differentiation
" #
∆n 1 n sin θ
∆ν̄ = − . (3.12)
2t cos θ cos2 θ
Since θ is always small
1
∆ν̄ '∆n (3.13)
2t
where ∆n is the fraction of order by which one ring pattern is shifted with
respect to another. (Note that when this fraction of order is measured at the
center, Eq. 3.13 becomes exact, and since ∆n(θ = 0) ≡ 1 − 2 we get back
Eq. 3.10)
The fraction of order ∆n that can be measured experimentally depends
on the quality of the plates, on the proper alignment and focusing of the
optical system, and on the width and relative intensity of the components
that are being measured. Values of ∆n ≈ 1/100 are common, and with some
care this value can be exceeded.
Using ∆n = 1/100, we then find for the resolving power of the Fabry-
Perot at a wavelength of λ = 5000 Å and a spacing t = 5.0 mm
!
∆ν̄ λ
= ∆n = 10−2 × 5 × 10−5 = 5 × 10−7 (3.14)
ν̄ 2t
which is quite satisfactory.
In reducing the data our aim is to obtain the orders of fractional inter-
ference 1 , 2 , . . ., for all the components of the line, and also to know if any
of the comonents overlap in order, and in that case by how many orders.
Let Rp be the radius of the pth ring as measured on the photographic
plate. Note that it is possible to measure Rp only if the center of the pattern
is included on the plate. From Eq. 3.8 it follows that:

2 2 2f 2
Rp+1 = (Rp+1 − Rp2 ) × (p + ) = × (p + ) (3.15)
n0
and taking into account the magnification m of the camera we have:
2f 2
Rp2 = [p − (1 − )]m2 (3.16)
n0
and !
no
p= Rp2 + (1 − ) . (3.17)
2f 2 m2

17
 We note that since n0 /2f 2 m2 is a constant, any adjacent pair of rings
can yield a value for . However, since the squares of the radii of successive
rings are linearly related (they form an arithmetic progression), in order to
utilize all available information a least squares fit to Eq. 3.17 will considerably
improve the accuracy in the determination of .
Alternately, a plot of the integer p against Rp2 gives an intercept of (1 − ).

p 5

1-

Rp2

18
Apparatus
The equipment is similar to that described in Melissinos, page 320.

• Discharge tube. The tube contains a low pressure of the mercury iso-
tope Hg198 and 3 mTorr of Argon buffer gas. The spectral lines are
sharp because:

(a) the nuclear spin I of Hg198 is zero and so there is no hyperfine

structure.
(b) All of the atoms have the same mass (unlike those of natural
mercury) and so all the atoms will give the same spectra.
(c) The discharge is maintained by RF and the tube is cool so the
Doppler broadening is less than in an arc discharge.

The tube is expensive and shockmounted. Do not attempt to move the

tube.

• Tesla Coil. The Tesla Coil should be used only for ionizing the Hg so
that the RF can maintain the discharge. This is done by moving the
tesla coil near the glass Hg tube and then turning the tesla coil on for
a moment. The tesla coil is a high voltage transformer and the tip is
at high voltage so do not place the tesla coil near the Hall probe or the
Hall probe circuitry may be damaged. Do not touch the tip.

• Tuned RF Oscillator. The oscillator has a frequency of about 100 MHz.

It is used to drive the Hg discharge tube. If the discharge cannot be
maintained or occupies less than 2/3 of the tube, ask the instructor to
check the oscillator tuning.

• Magnet. The magnet supply is stabilized. The control must be turned

to zero (anti-clockwise) before the supply is turned on or off. The
magnet is mounted on a ball bearing turntable so that it may be rotated
to view the discharge tube via a hole in one pole face.

19
• Hall Probe. The Hall probe is fragile and should be handled with care.
The probe is used to measure the magnetic field. The RF oscillator
should be turned off during these measurements.

• Lens. The lens focuses the light onto the slit of the constant deviation
spectrometer. A linear polarizing or circularly polarizing filter may be
placed before or after the lens.

198 Constant
Hg Deviation
Discharge Spectrometer
between
poles slit
field
lens
filter Fabry-Perot
Etalon

Telescope Camera

• Constant Deviation Spectrometer. The spectrometer contains a lens

system to make a virtual image of the slit at infinity and a Pellin-
Broca prism. The prism provides dispersion to separate the spectral
lines and deflects the beam through exactly 90◦ . The wavelength is
chosen by rotating the prism.

• Fabry-Perot Etalon. The parallel light from the spectrometer is sent

through the Fabry-Perot etalon. The gap width is 4.600 ± 0.005 mm.
The Etalon should be covered to provide some protection from air cur-
rents and to keep the refractive index of its air gap constant. The etalon
looks small and cheap - it is delicate and expensive.

20
 • Viewing Telescope. The telescope is used for visually inspecting the
spectra. It can be dismounted by unscrewing the clamp screws.

• Camera. The camera is a Santa Barbara Instrument Group Model ST-

7 digital camera with 765 × 510 pixels of size 9 × 9 µ forming a 6.9 × 4.6
mm array. The camera lens has a 130 mm focal length and an aperture
variable to f/2.8. The camera is linked to a PC running the CCDOPS
software which controls the camera.

• Mercury lamp with green filter and diffuser. This lamp is used to check
the Fabry-Perot pattern.

• Polarizers. Linear polarizers and a quarter wave plate are used to check
polarizations.

Procedure
1. Understand the theory and check the g factors for 3 S1 and 3 P2 levels.

2. Read the section on apparatus and note the precautions. Read the
section on the Fabry-Perot etalon.

3. Start the Mercury Light Source. Be sure to wear the eye protection
goggles. The tube emits a considerable amount of UV.

(a) Turn on the radio frequency (RF) coil power supply.

(b) Bring the tesla coil near the discharge tube. Turn the tesla coil
on for a few seconds and then off.
(c) The tube of low pressure Hg198 should glow a whitish blue. If the
glow is weak, ask the instructor to adjust the tuning of the RF
system.

4. Align the source and Spectrometer.

21
 (a) The magnet has two viewing conditions: perpendicular and par-
allel to the magnetic field. The configuration parallel to the mag-
netic field requires a rotation of the magnet so that light from the
source passes through the axial hole in the pole piece. Be careful
with the oscillator connecting wires.
(b) Adjust the lens position so that the source is imaged onto the
collimator slit.
(c) Gently rotate the magnet and check that light emitted from the
source continues to be focussed on the slit.
(d) Adjust the collimator slit width such that the entire source image
just passes through the slit.
(e) Mount the viewing telescope. At this point, the Faby-Perot etalon
is not yet in the optical path.
(f) After focussing the telescope, rotate the constant deviation dial to
center the green spectral line.(This is the Mercury 546.1 nm green
line. In the vicinity of this line, a yellow line should be observed.)

5. Check of the Fabry-Perot Etalon.

(a) Install the Fabry-Perot Etalon on the spectrometer platform. (Con-

sult with the instructor for doing this.)
(b) With the magnet off, look through the telescope for the ring pat-
tern. The pattern should be sharp. There are two adjustments
on the Fabry-Perot platform: a screw underneath the platform to
make vertical adjustments, and a horizontal screw on the side to
adjust the left-right positioning of the etalon. You will have to
adjust these screws to center the Fabry-Perot image.

6. Setting up the magnet

(a) Turn on the magnet and increase the current until the split lines
in different orders no longer overlap.
(b) Monitor and record the magnet voltage with a multimeter since
the ammeter is not accurate enough to reproduce settings.

7. Test the polarization of the spectra

22
 (a) Inspect the light emitted perpendicularly to the magnetic field.
Increase the field until the splitting of the rings almost causes
them to overlap. You should see each original line split into 9
lines.
(b) Insert a linear polarizing filter. Watch the intensity as the filter
is rotated. The outer six lines have ∆m0j = ±1 and so have E
vertical when seen perpendicularly to the field, while the inner
3 lines have ∆m0j = 0 and so have E horizontal (parallel to the
field).
(c) Now insert a the quarter wave plate and check for circular polar-
ization. The spectra should show no signs of circular polarization.
(d) Now gently rotate the magnet and look down the hole in the axis
of the magnet. The inner 3 lines (∆m0j = 0) should be missing.
The outer 6 lines (∆m0j = ±1) should be tested with polarizers
to verify that 3 have left-handed circular polarization and 3 have
right (see page 14).

8. Photograph the spectra

(a) Remove the telescope from its mount.

(b) Set up and align the camera close to the Fabry-Perot etalon.
(There is no way to do this except by eye.)
(c) Since the image is at infinity, set the camera focus to infinity.
Using the Focus command, center the image and determine the
optimum focus and exposure time. The Focus command takes
images and displays them one after another. Subsequently, the
Grab command is used to take a single image.
(d) Align the camera and adjust the Fabry-Perot so that you see the
green ring pattern with an equal number of orders on each side
of the center. The ring pattern should be centered in the field of
view. (The room lights need to be turned off.)
(e) Adjust the constant deviation prism slightly to make the ring pat-
tern vertical when one looks through the camera.
(f) Take at least 1 exposure with no field and a series of exposures
with transverse field, no filters. Below is a sample image taken

23
 with 3.0 A in the magnet, 0.12 sec exposure and aperture f/11.
The focus is set to infinity.

(g) Make sure to measure the magnetic field with the Hall probe before
the end of the session. The instructor will help you remove the
discharge tube.
9. Measurements
(a) Measure the radii of each of the split rings using the Cross Hairs
option of the software. You can also Zoom the image. It is
of course best to measure the diameter directly as opposed to
estimating the location of the center and measuring the radius.
Any scale can be used. (Why?)
(b) You should now have a set of radii for each of the 9 values of δ
(−2, −1.5, . . . + 1.5, +2) (see page 14). In each set, label the radii
with p = 1, 2, 3 . . .
(c) Tabulate the squares of the radii for all 9 sets.
10. Analyze the data
(a) Plot the ring number p(y)(as measured from the center out) against
Rp2 (x) for the set of rings with δ = 0 (no field) as on page 18 and
fit to a straight line.
(b) Measure the intercept (1 − ) on the p axis. The quantity  is the
fractional order for that set of rings and δ = 0.
(c) Repeat a) and b) for the other 8 sets with δ nonzero (field on).
(d) Plot the fractional order (y) obtained for each of the 9 sets against
the value of δ(x). Because the g values are 3/2 and 2, the splitting
and  are equally spaced.

24
(e) Fit a straight line through the various fractional orders . The
slope of this line is proportional to the product of the electron
magnetic moment times the magnetic field. For example, the ex-
pression for the wave numbers for the two outer rings of the nine
is: (See Eq. 2.1)
µB B
ν = ν0 − 2 ∆m0j = +1, σ transition
hc
µB B
ν = ν0 + 2 ∆m0j = −1, σ transition
hc
(f) Derive an expression for µB B in terms of the measured slope and
the etalon spacing t.
(g) Now use the results from above to calculate the ratio (e/me ).
e 2µB
=
me h̄
(h) Also calculate the Bohr Magneton.
eh̄
µB =
2me
The best value for µB is
eh̄
µB = = (9.274078 ± 0.000036) × 10−24 Joule/Tesla
2me
= (5.788378 ± 0.000009) × 10−9 eV/Gauss.
(i) Be prepared to answer questions on all aspects of the experiment.

25
References
[1] R. Eisberg and R. Resnick, Quantum Physics of Atoms, Molecules,
Solids, Nuclei, and Particles, 2nd Ed., Wiley, 1985.

[2] R. Fowles, Introduction to Modern Optics, 2nd Ed., Dover, 1975.

[3] A.C. Melissinos, Experiments in Modern Physics, Academic Press, 1966.

[4] A.C. Melissinos and J. Napolitano, Experiments in Modern Physics, 2nd

Ed., Academic Press, 2003.

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