Download Atomic 2

Quantum
Physics
2002
Atomic Physics
Recommended Reading:
Harris Chapter 7
Energy Levels in Hydrogen
Every possible state of the hydrogen atom has a distinct wavefunction
that is specified completely by four quantum numbers (n, l, m L , mS).
In many cases the energy levels associated with the quantum numbers
mL and mS are degenerate and we can describe the states by the n and l
quantum numbers alone, e.g 1s, 2p, 3p, 3d, ...
We know that an atom can emit characteristic electromagnetic radiation
when it makes transitions to states of lower energy. An atom in the
ground state cannot emit radiation but it can absorb electromagnetic
radiation and make a transition to higher (excited) states.
We can use the hydrogen wave functions to calculate transition
probabilities for an electron to change from one state to another when it
absorbs or emits radiation.
These calculations show that transitions in which the angular
momentum quantum number l, changes by 1 are allowed to take place,
(  l =  1) these are called allowed transitions while transitions for
which  l   1 are not allowed to occur and are called forbidden
transitions.
Selection Rules
The rules on how the quantum numbers can change when a transition
takes place are called selection rules. We can summarise them as
follows:
1) There is no restriction on the change n, in the principle quantum
number
2) the orbital angular momentum quantum number l can only change
by 1 .
3) The magnetic quantum number mL, can only change by 0 or 1
4) The spin quantum number mS, can change from -1/2 to +1/2 or vice
versa but it does not have to change for a transition to occur
Δn  anything
Δl  1
ΔmL  0,  1
l=1
Δms  1
 p  s ( l = -1) and p  d ( l = +1) are allowed
 s  p ( l = +1) is allowed but s  d ( l = +2) is
forbidden
 d  p ( l = -1) and d  f ( l = +1) are allowed
Energy Level Diagram of Hydrogen
En  
13.6
n2
eV
E1  13.6 eV
E2  3.4 eV
E3  1.5 eV
E 4  0.8 eV
Atom in a Magnetic Field
In the absence of a magnetic field we can (in principle) find the energy
levels of an atom by solving the Schrodinger equation
2 2

 Ψ  UΨ  EΨ
2m
for the hydrogen atom or hydrogen like ion (He+, Li2+, Be3+…)
Ze2
U 
4 πε or
while for a multielectron atom the potential energy is much more
complicated
Ze2
e2
U  
i 4 πε ori

i, j 4 πε o ri  r j
where the first term is the interaction
between each electron and the nucleus
(with charge +Ze), while the second term
represents the Coulomb repulsion between
each pair of electrons.
r1 - r2
r1
r2
r3
In general, in the absence of a magnetic field we can write the
Schrodinger equation in the following form
Ĥ0 Ψ  E0 Ψ
where
2 2
Η0 
  Û
2m
is the total energy operator (the Hamiltonian operator) and En are the
allowed energy levels for the electrons in the atom.
If we now place the atom in a magnetic field then we will have an
additional term in the potential energy due to the interaction between the
magnetic moment of the atom and the magnetic field. This magnetic
interaction energy is given by
V  μ  B
where  is the magnetic moment of the atom due to orbital and spin
angular momentum and B is the applied magnetic field. The total energy
operator then becomes
Ĥ  Ĥ0  V̂  Ĥ0  μˆ  B
We have seen that the magnetic moment is proportional to the angular
momentum, So, for an atom with total angular momentum J, the magnetic
moment operator is
e
μˆ  g j
Ĵ
2m
where gj is a proportionality constant called the Lande g-factor (see
below).
B
If the magnetic field B, is along the z-axis then the
Interaction Energy is
V̂  μˆ  B  μˆ B cosθ   μˆ zB

and
z
e
Ĥ  Ĥ0  μˆ zB  Ĥ0  g j
ĴzB

2m
If we operate on the wavefunction with this operator
we obtain
eB
ĤΨ  Ĥ0Ψ  g j
Ĵz Ψ
2m
but
Ĵz Ψ  m j Ψ

so
ĤΨ  Ĥ0Ψ  g j
E  E0  g jm j
e
B
2m
eB
m jΨ
2m
This shows us that when an atom is placed in a magnetic field B, the
energy levels are shifted and the angular momentum degeneracy is
removed, that is , the energy levels are split.
For example, if J = 1 then mj = -1, 0, 1 and this energy level will be split
into three when a magnetic field is applied
e
E  E0  g j
B
2m
E  E0
m j  1
mj  0
E0
m j  1
No field
Applied field B
e
E  E0  g j
B
2m
E0  g j
e
B
2m
E0
E0  g j
e
B
2m
Note that the size of the splitting depends on the strength of the
magnetic field B
Lande g-factor
The magnetic moment is related to the angular momentum, by
e
μˆ  g j
Ĵ
2m
where gj is a proportionality constant called the Lande g-factor. This
factor depends on the state the atom is in and is given by
JJ  1  LL  1  SS  1
g j  1
2JJ  1
Example 1: If the atom is in a 1D2 state, then
L = 2 (D-state), J = 2, and S = 0 (since 2S+1 = 1) and the g-factor for this
state is
g j  1
22  1  22  1  00  1
1
42  1
Example 2: If the atom is in a 2P1/2 state, then
L = 1 (P-state), J = 1/2, and S = 1/2 (since 2S+1 = 2) and the g-factor for
this state is
g j  1
1 2 3 2   12   1 2 3 2  2

21 2 3 2 
3
Lande g-factors for one electron atoms
L
Term
gj
gj mj
0
2
S1/2
2
1
1
2
P1/2
2/3
1/3
2
P3/2
4/3
2/3, 6/3
2
D1/2
4/5
2/5, 6/5
2
D3/2
6/5
3/5, 9/5, 15/5
2
3
2
F1/2
6/7
3/7, 9/7, 15/7
2
F3/2
8/7
4/7, 12/7, 20/7, 28/7
Zeeman Effect
We have seen that when an atom is placed in a magnetic field the energy
levels are split, i.e.
e
E  E0  g jm j
B
2m
Suppose that in the absence of a magnetic field, we study the transition
between two energy levels of an atom
E'0'
hν 0  E'0' - E'0
hν 0
In the presence of a magnetic field, both of
these levels may split, i.e.
'
E
e

0
E''  E'0'  g'j'm'j'
B
2m
and
e
'
'
'
'
E  E0  g jm j
B
2m
So when the atom is placed in a magnetic field the energy of the emitted
photon is
e
hν  E''  E'  E'0'  E'0  g'j'm'j'  g'jm'j
B


2m
or, in terms of the frequency
4πemB

ν  ν 0   g'j'm'j'  g'jm'j  ν L
ν  ν 0  g'j'm'j'  g'jm'j
or
where
νL 
e
B
4 πm
is the Larmor frequency
In a magnetic field the single emission line due to transitions
between the two states will now split into a series of lines. This
splitting is called Zeeman Splitting and the effect is known as the
Zeeman Effect
There are two cases we have to consider:
CASE 1: Suppose that the atom has an even number of electrons in its
outer shell so that S = 0, that is, we are looking at transitions between two
singlet states. Then, since J = L + S  J = L for singlet states and the
Lande g-factor is
JJ  1  LL  1  0
for all singlet
g j  1
1
states
2JJ  1
then, the expression for the transition frequencies becomes


ν  ν 0  m'j'  m'j ν L  ν 0  Δm j ν L
where mj = mj’’ - mj’
From the selection rules mj = 0,  1, so we are left with the following
frequencies.
ν0  νL

ν  ν0
ν  ν
 0
L
the single emission line is split into three lines when the atom is placed
into a magnetic field.
This is best illustrated by an example. The cadmium atom has an
emission line at a wavelength of 643.8 nm in the absence of a
magnetic field. This emission line is due to a transition between the
61D2 state and the 51P1 state of the atom.
61D2 state  n = 6, L = 2, J = 2, S = 0 and mj = -2, -1, 0, 1, 2
51P1 state  n = 5, L = 1, J = 1, S = 0 and mj = -1, 0, 1
Energy Level Diagram for 1D2 1P1 transition in Cd
No B field
With magnetic field B
6 1D 2
mj
2
1
0
-1
-2
1
0
-1
5 1P1
mj = -1
ν0
Polarisation 
ν0  νL

mj = 0
ν0

mj = +1
ν0  νL

Polarization of Emitted Lines
If we observe the polarisation of the emitted lines when the atom is
placed in a magnetic field, then it is found that the lines are polarised.
For lines which have:
m = 0, the emitted light is polarised parallel to the direction of the
magnetic field. These lines are referred to as  -lines or  -polarised
lines
m =  1, the emitted light is polarised perpendicular to the direction
of the magnetic field. These lines are referred to as -lines or polarised lines
B
Polariser
 -polarised
B
Polariser
-polarised
No B field
ν0
With magnetic field B
ν0  νL
ν0  νL
ν0
Δν  ν L 
The splitting between the lines is  = L , i.e.
e
B
4 πm
Note that the splitting is directly proportional to the strength of the
magnetic field, if we increase B then the splitting increases.
If we place the atom in a known magnetic field, then we can obtain the
ratio of e/m from the splitting of the lines e 4 πΔν
m

B
Very accurate method to determine e/m ratio. Vary B, and measure
splittings. plot graph of  versus B get e/m from slope
When a spectral line splits into three lines in the presence of a
magnetic field, the effect is known as the Normal Zeeman Effect
Case 2: Anomalous Zeeman Effect
We now consider the more general case where S  0, and the transition
frequencies are given by
ν  ν 0  g'j'm'j'  g'jm'j ν L


Here the upper and lower levels can have different g-factors and the line
can split into more than three components. For example consider the
following transition in sodium 2P3/2  2S1/2 (this is the D2-line that gives a
sodium lamp its characteristic yellow colour)
Lande g-factors:
The 2P3/2 line has J = 3/2, L = 1, S = 1/2 and m j = -3/2, -1/2, 1/2, 3/2
 g  1
3 2 5 2   12   1 2 3 2  4

23 2 5 2 
3
The 2S1/2 line has J = 1/2, L = 0, S = 1/2 and mj = -1/2, +1/2
1 2 3 2   01  1 2 3 2 
 g  1
2
21 2 3 2 
The 2P3/2 state splits into four levels while the 2S1/2 state splits into two
levels when a magnetic field is applied
No B field
2P
3/2
With magnetic field B
g=4/3, L = 1,
S = 1/2, J = 3/2
mj
gj m j
3/2
6/3
1/2
2/3
-1/2 -2/3
-3/2 -6/3
2S
1/2
g=2, L = 0,
S = 1/2, J = 1/2
ν0
mj = -1
mj = 0
mj = +1
1
2
3
4
5
6
Polarisation  





 
1/2
1
-1/2
-1
Frequencies of lines in 2P3/2 2s1/2 transition in Na
Referring to diagram on previous page we have, from left to right
1)
2)
3)
4)
5)
6)
5
 2

ν  ν0     1  νL  ν0  νL
3
 3

 6

ν  ν 0      1  ν L  ν 0  ν L
 3

1
 2

ν  ν0     1  νL  ν0  νL
3
 3

1
 2

ν  ν 0      1  ν L  ν 0  ν L
3
 3

 6

ν  ν0     1  νL  ν0  νL
 3

5
 2

ν  ν 0      1  ν L  ν 0  ν L
3
 3
