1
Rotational Spectroscopy
Outlines
- Rotation of diatomic molecules : rigid rotor model
- Classical physics for rigid rotor rotation
- Quantum mechanics of rigid rotor rotation
- Rotational operators
- Rigid rotor wavefunctions
- Rotational energy levels
- Energy levels, Transition & Rotational constant
- Rotational Spectra
2
Introduction
Some spectral lines in rotational spectrum of CO
A molecule with
permanent electric
dipole is microwave
active.
N. W. B. Stone, Applied Optics, 1978.
- Rational spectrum contains many peaks (many transitions).
-
Spectral lines are of almost equal spacing.
-
Slight changes in their relative intensities are observed.
-
The bond length can be determined from the spectrum.
3
Rotation of diatomic molecules : Rigid rotor model
Two atoms (mass m1 and m2) are connected with a fixed distance
and spin around the center of mass (CM).
“Rigid” implies the bond length
is fixed because no vibrational
motion of the bond.
The molecule rotates with
rotational frequency of rot
CM : center of mass
r0: the bond length (fixed)
m1, m2 : mass
v1, v2 : velocity (perpendicular
to r1 and r2, respectively)
4
Classical physics for rigid rotor rotation
p
Angular Momentum (L)
v
f
m
r
w
L
  
L  r p
f = 90o (r  p)
Magnitude:
L  rp
 r p sin f
 r p  mvr
p : linear momentum
r : radius of rotation
w: angular velocity
L: angular momentum
5
Angular velocity (w) & Moment of inertia
Velocity
Momentum
Moment of inertia (I)
L  mw r 2
 m r 2w
 Iw
I  m r2
Linear
Angular
Relationship
rd
v
dt
d
ω
dt
vωr
p  mv
L  mw r 2
6
Kinetic energy of rigid rotor rotation
1
1
2
T  m1v1  m2 v 22
2
2
Rotation distance in
v  2 r0  rot
w  2  rot
1 cycle × rot
The angular frequency
v  w r0
The molecule rotates at
fixed rotational frequency
rot
1
1
2 2
T  m1w r1  m2w 2 r22
2
2
7
1
1
2 2
T  m1w r1  m2w 2 r22
2
2
1 2
T  w m1 r12  m2 r22
2


Moment of inertia, I
I  m1 r12  m2 r22
1 2
T  Iw
2
1 ( Iw )
T
2 I
L2
T
2I
2
L = angular momentum
L  Iw
The kinetic energy can be described in
terms of angular momentum and moment
of inertia.
8
Importance terms & Equations
Terms
Equations
I  mr 2   r 2
Moment of inertia:
Angular velocity:
v  w r0  2 r0  rot
Angular momentum:
L  I ω   r2
(Linear momentum, p =
mv)
Kinetic energy
2
1
1

r
KE   v 2   v 2  2
2
2
r
1 (  v r)
1L


2
2 r
2 I
2
2
Reduced Mass
m1m2

m1  m2
v
 vr
r
L2
KE 
2I
In classical physics, the
direction or magnitude of L
can be any value.
9
Quantum mechanics of rigid rotor rotation
To solve the Schrödinger equation
H  E
We need to set up the Hamiltonian:
H or Hˆ total  T  V
One need to define the kinetic and potential energy
10
ASSUMPTION
The potential energy of rigid rotor diatomic molecule is
zero:
IMPORTANCE ASSUMPTION
V 0
The Hamiltonian operator of rigid rotor diatomic
molecule corresponds to the kinetic energy operator:
Hˆ total
2
2
ˆ

L
 Tˆ  
2 
2
2I
11
Rotational operators: spherical polar coordinates
I  m1 r12  m2 r22   r02
m1m2

m1  m2
v  2 r0  rot  w r0
L  Iw
V 0
 

2
ˆ
ˆ
ˆ
ˆ
H total  T  V  
 V 
 2

2
0
where 2 is the Laplacian Operator
In Cartesian coordinates
2
2
2



2  2  2  2
x
y
z
12
Transformation to spherical polar coordinates
For rigid rotor model, we usually transform the 2 Cartesian coordinates
into spherical polar coordinates .
Cartesian
coordinates
Spherical polar
coordinates
 ( x, y , z )
 (r , , f )
  x  
  y  
  z  
0r 
0  
0  f  2
r  x2  y2  z 2
z  r cos 
z
x  r sin  cos f   cos 1
r
y  r sin  sin f
1  y 
f  tan  
x
13
Transform the 2 Cartesian coordinates into spherical polar coordinates

0
r
2
1


1


1





2
2  2
r
 2
 sin 
 2 2
r r  r  r sin   
  r sin  f 2
1 1  
 
1 2 

 2 
 sin 
 2
2 
r0  sin   
  sin  f 
Because of rigid bond
(fixed bond length)
r  r0
2
2
2




1


1



2
ˆ
ˆ


H T  
 
 sin 
 2
2 
2 
2
2 r0  sin   
  sin  f 
2  1  
 
1 2 

  
 sin 
 2
2 
2 I  sin   
  sin  f 
I   r02
14
After transformation, we get:
2
2



1


1



ˆ

T   
 sin 
 2
2 
2 I  sin   
  sin  f 
ˆ2
L
Because
Tˆ 
2I
The square of angular momentum operator becomes
2


1


1



ˆ

L   
 sin 
 2
2 
  sin  f 
 sin   
2
2
L2 is important operator in the solution for energy levels and
wavefunctions of rigid rotor in the Schrödinger equation.
Another importance of the angular momentum is Lz (the z-component)

ˆ
Lz  i
f
15
Thus, the Hamiltonian operator in the Schrödinger equation
L2
H
2I
Another implication of the rigid rotor (fixed bond length) is the
wave function depends on  and f :
 (r , , f )   ( , f )
H ( , f )  E ( , f )
16
Rigid rotor wavefunctions
H ( , f )  E ( , f )
0     : 0  f  2
2
2



1


1



ˆ


H  
 sin 
 2
2 
2 I  sin   
  sin  f 
The full wavefunctions
Reduce to
 (r , , f )  R(r )Y ( , f )
 ( , f )  Y ( , f )
Consider only the
angular part of the
wavefunctions
2  1  
 
1 2 
Y ( , f )  EY ( , f )
 
 sin 
 2
2 
2 I  sin   
  sin  f 
Set the Schrödinger Equation equal to zero and

  2 I sin 2  /  2

 
Y ( , f )   2Y ( , f ) 2 I
2
sin 

sin
EY ( , f )  0
 sin 

2
2
 
 
f

17
From the previous slide

sin 

Y ( , f )   2Y ( , f ) 2 I

2
sin



sin
EY ( , f )  0


2
2
 
f


Rearranging the differential equation separating the θ-dependent terms
from the -dependent terms:
2


 
  2I

Y ( , f )
2
sin    sin      2 sin E Y ( , f )   f 2




Only 
Only f
the Schrödinger equation can be solved using separation of variables.
Y ( , f )  ( ) (f )
18
From the previous slide

 
 
 2 ( ) (f )
2 
sin    sin      sin   ( ) (f )  
2

f




2 IE
 2

LHS operator does not operate on Φ(f) and RHS operator does not
operate on Θ(θ).

 (f ) sin 

( ) 
 2  (f )

2
 sin 
   sin  ( ) (f )  ( )
 
f 2

Devide by Φ(f) Θ(θ)
2
sin  d 
d( ) 
1

 (f )
2
sin



sin





( ) d 
d 
 (f ) f 2
Only f
Only 
Each side of the equation must be equal. Let give it to m2
19
From the previous slide
2
sin  d 
d( ) 
1

 (f )
2
2

m
 sin 
   sin   
( ) d 
d 
 (f ) f 2
sin  d 
d( ) 
2
2
sin


   sin   m
( ) d 
d 
1  2  (f )
2


m
 (f ) f 2
(I)
(II)
20
Let solve Eq.(II)
1  2  (f )
2


m
 (f ) f 2
 2  (f )
1
2
 m
2
f
 (f )
 (f )  Am eimf or A m e  imf
Consider periodicity conditions
 (f  2 )   (f )
The solution for Φ(f) is :
1 imf
 (f ) 
e ; m  0,  1,  2,..
2
Normalization constant
21
The Legendre Equation
For solving Eq.(I),
sin  d 
d( ) 
2
2
sin


   sin   m
( ) d 
d 
Change variables:
Since:
0  
x  cos  , ( )  P( x),
dx
 d
 sin 
 1  x  1
sin 2   1  cos 2   1  x 2
The eq(I) is transformed into:
2
2


d
d
m
2
(1  x ) 2 P( x)  2 P( x)    
P( x)  0
2
dx
dx
1 x 

The Legendre Equation
22
Associated Legendre polynomials
The solutions of the Legendre Eq. are :
( )  Am PJ|m| (cos  )
PJ|m| (cos  ) The Associated Legendre polynomials
Normalization constant
 2 J  1   J  | m | !
Am  

 2   J  | m | !
 2 J  1   J  | m |! |m|
( )  
PJ (cos  )

 2   J  | m | !
23
The Associated Legendre polynomials
PJ|m| (cos  )
; J  0, 1, 2,... and m  0,  1,  2,..
P00 (cos  )  1
P10 (cos  )  cos 
P11 (cos  )  sin 
1
P (cos  )  (3 cos 2   1)
2
P21 (cos  )  3 sin  cos 
0
2
P2 2 (cos  )  3 sin 2 
24
Spherical harmonic wavefunctions
Y ( , f )  YJm ( , f )  ( ) (f )
1 imf
 2 J  1   J  | m |! |m|
 
PJ (cos  )
e

2
 2   J  | m | !
( )
 (f )
 2 J  1   J  | m |! |m|
Y ( , f )  
PJ (cos  ) eimf

 4   J  | m |!
m
J
; J  0, 1, 2,... and m  0,  1,  2,..
25
The wavefunctions for J =0, 1, 2
Y00  , f  
1
4 1/ 2
1/ 2
 3 
0
Y1  , f   

 4 
cos 
1/ 2
 3 
Y  , f    
 8 
sin  e if
1
1
1/ 2
 5 
0
Y2  , f   

 16 
3 cos
2
  1
1/ 2
 5 
Y  , f   

 16 
sin  cos  e if
1
2
1/ 2
 15 
2
Y2  , f   

 32 
sin 2  e  2if
26
Rotational energy levels of rigid rotor
The Hamiltonian (Laplacian) operator:
2
2



1


1



ˆ

H  T   
 sin 
 2
2 
2 I  sin   
  sin  f 
The spherical harmonic wavefunctions with Legendre functions :
2  1  
 
1 2 
Y ( , f )  EY ( , f )
 
 sin 
 2
2 
2 I  sin   
  sin  f 
 2 J  1   J  | m |! |m|
Y ( , f )  
PJ (cos  ) eimf

 4   J  | m |!
m
J
Because YJm ( , f ) are Eigenfunctions of Ĥ
solving for this Schrödinger gives the quantized J(J+1) result :
2

Hˆ YJm ( , f ) 
J ( J  1) YJm ( , f )
2I
2
E
J ( J  1)
2I
, J  0,1,2 ,...
J is Rotational
quantum number
27
Energy levels, Transition & Rotational constant
2
Energy levels:
EJ 
J  J  1  BJ ( J  1) , J  0,1,2 ,...
2I
2
h2
B
 2
in Joule
2 I 8 I
Rotational constant :
B
B
in cm-1
hc
J = 0 there is no rotational energy at
ground state.
Energy levels
J
EJ
0
0
1
2B
2
6B
3
12B
4
20B
Energy levels are not equally spaced.
28
States of the system (Degeneracy)
Because each J value has mJ = 2J+1 values
State of the system:
-mJ , –mJ+1 ,…, 0, 1, 2, ...+ mJ-1+ mJ
J
3
12B
mJ
-3
-2
-1
0
1
2
2
6B
mJ
1
0
3
mJ
mJ
-2
-1
-1
0
0
0
1
1
2
2B
0
29
YJm ( , f ) are also Eigenfunctions of L̂2 and L z
For L2 :
L̂2YJm ( , f )   2 J ( J  1) YJm ( , f )
2


1


1



2
2
ˆ

L   
 sin 
 2
2 
  sin  f 
 sin   
L2   2 J ( J  1)
L   J ( J  1)
For Lz :
L̂ zYJm ( , f )  mJ ( J  1) YJm ( , f )

ˆ
Lz  i
f
Lz  m
Lz  m
L
L   J ( J  1)
30
• In QM, rotational motion is quantized not continuous.
Only certain states of motion are allowed that are
determined by quantum numbers J and m.
• J determines the length of the angular momentum
vector
L   J ( J  1)
• m indicates the orientation of the angular
momentum with respect to z-axis
Lz  m
31
J
m
EJ
E J  BJ ( J  1)
32
3
E
2
2
1
0
10
-1 -2
-3
12B
L
L   J ( J  1)
12
Lz m
3
2

2
2
0
2
10
-1 -2
1 0
-1
0
6B
2B
0
6
2

2

0
0
0
Lz  m
32
Rotational Transitions
Change in quantum number:
Selection rule:
J  J final  J initial
J  1
Absorption:
J  1
E  E J 1  E J
E J  BJ ( J  1) , J  0,1,2 ,...
 B( J  1) J  2   BJ  J  1
 2 B( J  1)
E  hvobs  2 B( J  1)
In unit of Hz
In cm-1
wave number
vobs  2 B( J  1) / h
vobs
 w  2 B( J  1) / hc
c
w  2 B ( J  1)
2
h2
B
 2
2 I 8 I
B
h
B
 2
hc 8 Ic
33
(Pure) Rotational Spectra of Diatomic Molecule
Energy levels:
E J  BJ ( J  1) , J  0,1,2 ,...
Transition energy :
E J   J 1  2 B( J  1)
J  1
J
EJJ+1
01
2B
12
4B
23
6B
34
8B
45
10B
2 B( J  1)
Energy space between each transition levels: 2B
34
Energy levels for a rigid rotor.
The spectrum observed
through absorption of
microwave Radiation.
35
Why do we observe many
absorption peaks in
rotational microwave
spectrum? How many peaks
do we observe in vibrational
IR spectrum? Why ?
1)
2)
36
Ex. From the rotational microwave spectrum of 1H35Cl,
we find that the rotational constant
B  10.5934 cm -1
Determine the bond length of the 1H35Cl molecule.
Giving the masses of 1H and 35Cl are 1.008 and 34.969 amu
B
r0 

h
8 2 Ic

h
8 2 r02 c
h
8 2 B c
8 2 (1.64 10-27
 1.27 10 10 m
m1m2

 1.64 10-27 kg
m1  m2
6.626 10 34 J s
kg )(10.5934 cm -1 )(2.998 1010 cm s -1 )
37
Ex. Spectral line spacing of rotational microwave spectrum
of a diatomic molecule is 3.8626 cm-1 . Determine the
bond length of this molecule.
Spectral line spacing (E) = 2B
Spectrum (cm-1)
3.8626 cm-1
vobs  3.8626 (cm -1 )
แก้เป็ น
CO
Wave number
vobs  3.8626 (cm -1 )  c  1.16 1011 Hz
B  hvobs / 2  3.84 10-23 J
hvobs  2 B
B
h2
8 I
2
I r
2
0
I
h2
8 2 B
 1.45 10-46
m1m2

 1.15 10-26 kg
m1  m2
r0  I /   1.12 10 10
m
Example: Using the following total energy eigenfunctions for the
three-dimensional rigid rotor, show that the J=0 → J=1
transition is allowed, and that the J=0 → J=2 transition is
forbidden:
M
Y j jis used for the preceding functions.
Providing the notation
Y00  , f  
1/ 2
 3 
Y10  , f   

4



1
4 
1/ 2
1/ 2
 5 
0
Y2  , f   

 16 
3 cos
2
cos 
  1
Assuming the electromagnetic field to lie along the z-axis,
the transition dipole moment takes the form
2

0
0
 zJ 0    df  YJ0  , f cos  Y00  , f sin d
 z   cos
For the J=0 → J=1 transition,
Y  , f  
0
0
2

0
0
2

1/ 2
 3 
Y  , f   

 4 
1
cos 
0
1
4 
1/ 2
0
0



 10


d
f
Y

,
f
cos

Y
z
J  0  , f  sin d
  J 1
1/ 2
 3 
   df  

4 
0
0


3
2
4
0
0

(2 )  cos  sin d
2
0

1
4 
1/ 2
sin d

2
d
f
cos
   sin d
4
3
cos  cos  
3
2

cos
 sin d

2 0
2
 df  f 0  2  0
0
2
For the J=0 → J=1 transition,

Now consider
2
cos
  sin d
0
Use reduction or substitution method
x   , z  cos x,
dz
1
  sin x, dx  
dz
dx
sin x
1 

2
2
2
2
cos
x
sin
xdx

z
sin
xdx

z
sin
x

dz


z
dz






 sin x 
1
1
  z 3   cos 3 x
3
3
Replace the result into the original integration


1
2
 1
2
3 
3
3
0 cos  sin d   3 cos   0   3 (1  1 )  3
For the J=0 → J=1 transition,
From the previous derivation:

3
2

cos
 sin d

2 0
 10
z

10
z
3 2
3

 
2 3
3
Thus:
3
 
0
3
10
z
The J=0 → J=1 transition is allowed.

2
cos
  sin d 
0
2
3
For the J=0 → J=2 transition,
Y00  , f  
2

0
0
2

1/ 2
 5 
Y20  , f   

 16 
1
4 
1/ 2
3 cos   1
 z20    df  YJ0 2  , f cos  YJ00  , f sin d
1/ 2
 5 
   df  

16 
0
0

5
8
2

0
0
3 cos
2
  1cos  
1
4 
1/ 2
sin d
3
 sin   cos  sin  )d
cos
3
(
f
d
 
2

5
3
 sin   cos  sin  )d
cos
3
(


4 0
 df  f 0  2  0
0
Let consider by dividing into two separate terms:


0
0
3
3
cos
 sin d

 cos sin d
2
Consider


0
0
For the J=0 → J=2 transition,
 cos sin d
3
3
cos
 sin d

Use the substitution method (similar to the previous one)
dz
1
x   , z  cos x,
  sin x, dx  
dz
dx
sin x
1 
3 4
3

3
3
3
4
3
cos
x
sin
xdx

3
z
sin
xdx

3
z
sin
x

dz


z


cos
x





4
4
 sin x 
Replace x with  and integrate from 0 to , we get:


3
4
3
cos

sin

d



cos

0
4
3


0
3
  ((1) 4  14 )  0
4

Do the same for
 cos sin d
0
1 
1 2
1

2
cos
x
sin
xdx

z
sin
x

dz


z


cos
x0




2
2
 sin x 
For the J=0 → J=2 transition,
Thus:


0
0
3
3
cos
 sin d  0

 cos sin d  0
From the previous derivation:
 z20

5
3

(
3
cos
 sin   cos  sin  )d  0

4 0
Therefore:
 z20  0
Thus, the J=0 → J=2 transition is forbidden.
45
Home Work 2
1. Spectral line spacing of rotational microwave spectrum
of OH radical is 37.8 cm-1 . Determine the OH bond
length (in pm unit) and moment of inertia (in kg m2)
mO = 15.994 amu
Spectrum (cm-1)
mH = 1.008 amu
37.8 cm-1
2. Use the bond length of diatomic molecules in Table to
predict line spacing (in cm-1 unit) of rotational microwave
spectrum.
molecule bond length (pm)
HF
91.7
HI
161
HCl
128
HBr
141
46
3. Determine the bond length of these diatomic gases in
Table and arrange them in order of increasing the bond
length.
Bond length (pm)
B (cm -1 )
OH
37.80
ICl
0.11
ClF
1.03
AlH
12.60
4. Using the information in the Table to calculate the ratio
between the transition energy of rotation from J=0 to J =1
and vibration from n=0 to n=1 for H2
Atomic mass
1.008 amu
Bond length of H2
74.14 pm
Erotation , J 0 J 1
Force constant of H2
575 N/m
Evibration,n 0n 1
?
47
Why do we observe many
absorption peaks in
rotational microwave
spectrum? How many peaks
do we observe in vibrational
IR spectrum? Why ?
1)
2)
48
Reason 1) Vibration spectrum has only one peak because
Based on harmonic oscillator model of diatomic molecule,
energy-level spacing between adjacent levels is the same for
all values of the vibrational quantum number. With the
selection rule (n=1) , all transitions have the same
frequency. So we expect only one intense peak for
vibrational spectrum.
Vibrational energy level:
1
En  (n  )hv
2
n
Evib
(harmonic)
En-En-1
0
1/2 h
-
1
3/2 h
h
2
5/2 h
h
3
7/2 h
h
4
3
2
h
1
0
E
49
Rotational microwave spectrum has many peaks
because the rotational energy levels are not equally
spaced in energy. Different transitions give rise to
separate peaks.
Rotational energy level
2
EJ 
J  J  1  BJ ( J  1) , J  0,1,2 ,...
2I
Energy levels
J
EJ
0
0
1
2B
2
6B
3
12B
4
20B
Energy levels are not equally spaced.
50
Reason 2)
Nearly all the molecules in a macroscopic sample are in
their ground vibrational state (n=0) at room temperature
(even at 1000K). Only the n = 0  n = 1 transition is
observed in vibrational spectroscopy.
N1 g1  hv / k BT

e
Boltzmann population
N0 g0
N1/N0 is very low.
51
HCl, 1<J<5; 10-22- 10-21 J
In case of rotational transition, Erotation < kBT, many
rotational energy levels can be populated.
4.14X10-21 J
Boltzmann distribution
N J 1 g J 1  E / k BT

e
NJ
gJ
52
Questions
1) In the microwave spectroscopy, is the photon energy
sufficient to excite vibrational transitions?
2) In the IR spectroscopy, is the photon energy sufficient to
excite rotational transitions?
53
Vibrational transitions usually also involve rotational
transitions.
54
Ex. Consider (Cl2) adsorbed on a platinum surface. The ClCl bond length is 2.00 Å. Assuming Cl-Cl can be modeld by
a rigid rotor. Calculate the wave number of rotational
transition of Cl2 from ml= 3 to ml= 6, how many rotational
spectral lines do we observe?
m1m2

m1  m2
B
h2
8 I
B
2
B
hc

= 2.94X10-26 kg
h2
8 r
2
2
0
= 4.72X10-24 J
= 0.238 cm-1
E J  B J ( J  1)
ml= 3  J = 3 and ml= 6  6
Selection rule:
55
J  1
E  2 B ( J  1)
J
34
45
56
E (cm-1)
1.90
2.38
2.85
For the rotational transition from ml= 3 to ml= 6, we
expect to observe three rotational spectral lines.
Calculate energy gap or spacing between spectral lines in
cm-1 unit?
energy spacing between spectral lines
2 B  2  0.238  0.476 cm -1
56
57
58
Centrifugal Distortion
59
Rigid rotor wavefunctions
H ( , f )  E ( , f )
0     : 0  f  2
2
2



1


1



ˆ


H  
 sin 
 2
2 
2 I  sin   
  sin  f 
Consider only the angular part of the wavefunctions
 ( , f )  Y ( , f )
Let
 ( , f )  ( )(f )
2
EJ 
J J
2I
 (r , , f )  R(r )Y ( , f )
Set the Schrödinger Equation equal to zero:
sin  d 
d  2 IE 2
 sin 
  2 sin 
 d 
d  
the Schrödinger equation can be solved using separation of variables.
Solution Assuming the electromagnetic field to lie along
the zaxis, z   cos,  and the transition dipole moment
takes the form
2
2
0
0
 zJ 0    df  YJ0  , f cos   , f Y00 cos  sin d
For the J=0 → J=1 transition,
Y  , f  
0
0
 
10
z
3
4
2
1
4 
1/ 2
2
 df  cos
0
0
1/ 2
2
 3 
Y  , f   

 4 
cos 
0
1
 sin d 
 3  cos  

2 
3
3
 
 
 0
 3
3
0
61
The Hamiltonian Operator can now be written:
2

2 r02
 1  
Y ( , f ) 
1  2Y ( , f ) 
 sin 
 2

  EY ( , f )
2
  sin  f
 sin   

62
63
64
65