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Transcript of Slide 1 Chapter 4 Rigid-Rotor Models and Angular Momentum Eigenstates.
Slide 1
Chapter 4
Rigid-Rotor Models andAngular Momentum Eigenstates
Slide 2
Outline
• Math Preliminary: Products of Vectors
• Rotational Motion in Classical Physics
• The 3D Quantum Mechanical Rigid Rotor
• Angular Momentum in Quantum Mechanics
• Angular Momentum and the Rigid Rotor
• The 2D Quantum Mechanical Rigid Rotor
• The 3D Schrödinger Equation: Spherical Polar Coordinates
• Rotational Spectroscopy of Linear Molecules
Not Last Topic
Slide 3
Outline (Cont’d.)
• Application of QM to Molecular Structure: Pyridine
• Statistical Thermodynamics: Rotational contributions to the thermodynamic properties of gases
Slide 4
Mathematical Preliminary: Products of Vectors
x y zA A i A j A k
x y zB B i B j B k
Scalar Product (aka Dot Product)
x x y y z zA B A B A B A B
Note that the productis a scalar quantity(i.e. a number)
c o s ( )A B A B
Magnitude:
Parallel Vectors: 0c o s ( 0 )A B A B
A B
Slide 5
x y zA A i A j A k
x y zB B i B j B k
Cross Product
The cross product of two vectors is also a vector.
Its direction is perpendicular to both A and B and is given by the “right-hand rule”.
s i n ( )A x B A B
Magnitude:
Parallel Vectors: 0s i n ( 0 )A x B A B
BxA
0s i n ( 9 0 )A x B A B
Perpendicular Vectors:
0
A B
Slide 6
x y zA A i A j A k
x y zB B i B j B k
A x B
x y z
x y z
i j k
AxB A A A
B B B
y z x yx z
y z x yx z
A A A AA AA xB i j k
B B B BB B
Expansion byCofactors
( ) ( ) ( )y z z y z x x z x y y xA x B A B A B i A B A B j A B A B k
Slide 7
Outline
• Math Preliminary: Products of Vectors
• Rotational Motion in Classical Physics
• The 3D Quantum Mechanical Rigid Rotor
• Angular Momentum in Quantum Mechanics
• Angular Momentum and the Rigid Rotor
• The 2D Quantum Mechanical Rigid Rotor
• The 3D Schrödinger Equation: Spherical Polar Coordinates
• Rotational Spectroscopy of Linear Molecules
Slide 8
Rotational Motion in Classical Physics
s i n ( )L r p Magnitude:
Angular Momentum (L)
m
Circular Motion: 0s i n ( 9 0 )L r p r p
or: 2 2( )v
L rp rm v m r m rr
L I ω where 2 vI m r
r
Energy
2 2
2 2
p mvE
m
2 2 2( )
2 2
m r m r
2
2
I or:
2 2( )
2 2
I LE
I I
Momentof Inertia
AngularFrequency
L rxp
Slide 9
Comparison of Equations for Linear and Circular Motion
Linear Motion Circular Motion
Mass 2I m r Moment of inertiam
Velocityv
r Angular velocityv
Momentum L I Angular momentump=mv
Energy2
2
LE
I Energy
2
2
pE
m
or2
2
IE
Energy
2
2
mvE
Slide 10
Modification: Rotation of two masses about Center of Mass
m
m1 m1r
2I m r 2I r
where 1 2
1 2
m m
m m
Slide 11
Outline
• Math Preliminary: Products of Vectors
• Rotational Motion in Classical Physics
• The 3D Quantum Mechanical Rigid Rotor
• Angular Momentum in Quantum Mechanics
• Angular Momentum and the Rigid Rotor
• The 2D Quantum Mechanical Rigid Rotor
• The 3D Schrödinger Equation: Spherical Polar Coordinates
• Rotational Spectroscopy of Linear Molecules
Slide 12
Angular Momentum in Quantum Mechanics
L r x p
Classical Angular Momentum
( ) ( ) ( )z y x z y xL y p z p i z p x p j x p y p k
r x i y j z k
x y zp p i p j p k
x y z
i j k
x y z
p p p
x z yL y p z p
y x zL z p x p
z y xL x p y p
Slide 13
Angular Momentum in Quantum Mechanics
QM Angular Momentum Operators
x z yL y p z p
y x zL z p x p
z y xL x p y p
ˆ xp ii x x
ˆ yp ii y y
ˆ zp ii z z
Classical QM Operator
ˆxL i y z
z y
ˆyL i z x
x z
ˆzL i x y
y x
2 ˆ ˆ ˆ ˆ ˆ ˆx x y y z zL L L L L L L
^
Slide 14
Operator Commutation and Simultaneous Eigenfunctions
It can be shown that: ˆ ˆ, 0x yL L Do not commute
ˆ ˆ, 0y zL L Do not commute
ˆ ˆ, 0x zL L Do not commute
Because the operators for the individual components do not commute,one cannot determine two separate components simultaneously.
i.e. they cannot have simultaneous eigenfunctions.
In contrast, it can be shown that:2ˆ , 0zL L
^ Do commute
Because these operators commute, one can determine Lz and L2
simultaneously; i.e. they can have simultaneous eigenfunctions.
Slide 15
Outline
• Math Preliminary: Products of Vectors
• Rotational Motion in Classical Physics
• The 3D Quantum Mechanical Rigid Rotor
• Angular Momentum in Quantum Mechanics
• Angular Momentum and the Rigid Rotor
• The 2D Quantum Mechanical Rigid Rotor
• The 3D Schrödinger Equation: Spherical Polar Coordinates
• Rotational Spectroscopy of Linear Molecules
Slide 16
The 2D Quantum Mechanical Rigid Rotor
Assume that two masses are attached by a rigid rod (i.e. ignorevibrations) at a fixed distance, r,and are free to rotate about the Center of Mass in their x-y plane.
m1
m1
r x
y
The angle represents the angle of rotation relative to the x-axis.
The 2D Schrödinger equation for therelative motion of two masses is:
22
2V E
1 2
1 2
m m
m m
2 22
2 2x y
Two Dimensional Laplacianin Cartesian Coordinates
Slide 17
22
2V E
If one (a) converts the Laplacian to polar coordinates
(b) assumes that the potential energy is constant (arbitrarily 0)
(c) holds r fixed (i.e. neglects derivatives with respect to r)
It can be shown that the Schrödinger Equation for a 2D RigidRotor becomes: 2 2
2 22E
r
or2 2
22E
I
where 2I r is the moment of inertia
Slide 18
The Solution2 2
22E
I
2
2 2
2 IE
C o n s t a n t
22
2IEm or
2 2
2
mE
I
Note: So far, m can have any value;
i.e. there is no energy quantization
imA e
Assume
( ) imdim Ae
d
2
22
( ) imdim Ae
d
2
22
dm
d
22
2 IEm
Slide 19
2 2
0, 1, 2, 3,2
mE m
I
Application of the Boundary Conditions:Quantization of Energy
imA e
To be a physically realistic
solution, one must have: ( 2 ) ( )
Therefore: 2i m m i i mA e e A e
or 2 1m ie
c o s ( 2 ) s i n ( 2 ) 1m i m
This is valid only for: 0 , 1 , 2 , 3 ,m
Therefore, only certain values for the energy are allowed;i.e. the energy is quantized:
Slide 20
Zero Point Energy
2 2
0, 1, 2, 3,2
mE m
I0 0E
There is no minimum Zero Point Energy.
One encounters a ZPE only when the particle is bound (e.g. PIB,Harmonic Oscillator, Hydrogen Atom), but not in freely movingsystems (e.g. 2D and 3D Rigid Rotor, free particle)
Slide 21
Application of the 2D Rigid Rotor
We have solved the 2D Rigid Rotor primarily as a learning exercise,in order to demonstrate the application of angular Boundary Conditions.
However, the model has a real world application, in that it can be used to characterize the rotation of molecules adsorbed on surfaces.
Example
When an H2 molecule is chemisorbed on a crystalline surface, itsrotation can be approximated as that of a 2D rigid rotor.
The H2 bond length is 0.74 Å Calculate the frequency (in cm-1) of the lowest energy rotational transition of chemisorbed H2.
Slide 22
2H
H H
m
m m
2 2
0, 1, 2, 3,2
mE m
Ir = 0.74 Å = 0.74x10-10 m1 amu = 1.66x10-27 kgħ = 1.05x10-34 J•sh = 6.63x10-34 J•sc = 3.00x1010 cm/s
2(1 )0 .50
1 1
amuamu
amu amu
2 72 81 .6 6 1 0
8 .3 0 1 0x kg
x kgam u
22 2 8 1 08 .3 0 1 0 0 .7 4 1 0I r x k g x m 4 8 24 . 5 5 1 0x k g m
2 2 2 22 1
2 2
m mE
I I
m1 = 0
m2 = 12
2I
234
48 2
1.05 10
2 4.55 10
x J s
x kg m
2 11 . 2 1 1 0x J
E
hc
21
34 10
1.21 10
6.63 10 3.00 10 /
x J
x J s x cm s
16 0 . 9 c m
Slide 23
Outline
• Math Preliminary: Products of Vectors
• Rotational Motion in Classical Physics
• The 3D Quantum Mechanical Rigid Rotor
• Angular Momentum in Quantum Mechanics
• Angular Momentum and the Rigid Rotor
• The 2D Quantum Mechanical Rigid Rotor
• The 3D Schrödinger Equation: Spherical Polar Coordinates
• Rotational Spectroscopy of Linear Molecules
Slide 24
The Three Dimensional Schrödinger Equation
In Cartesian Coordinates, the 3D Schrödinger Equation is:
2 2 2 2 2 2
2 2 2( , , )
2 2 2V x y z E
m x m y m z
( , , )x y z
The Laplacian in Cartesian Coordinates is: 22 2 2x y z
T(x) T(y) T(z) V(x,y,z)
Therefore:2
2 ( , , )2
V x y z Em
It is sometimes not possible to solve the Schrödinger exactly inCartesian Coordinates (e.g. the Hydrogen Atom), whereas itcan be solved in another coordinate system.
The “Rigid Rotor” and the Hydrogen Atom can be solved exactlyin Spherical Polar Coordinates.
Slide 25
Spherical Polar Coordinates
To specify a point in space requires three coordinates.In the spherical polar coordinate system, they are:
r 0 r < Distance of point from origin (OP)
0 < Angle of vector (OP) from z-axis
0 < 2 Angle of x-y projection (OQ) from x-axis
Slide 26
Relation of Cartesian to Spherical Polar Coordinates
rz
cos( )z
r
c o s ( )z r
x-axis
y-axis
O
Q
x
y
cos( )x
OQ
c o s ( )x O Q
s i n ( ) c o s ( )x r
sin( )y
OQ
s i n ( )y O Q
s i n ( ) s i n ( )y r
OQ=rsin()
OQ
Slide 27
The Volume Element in Spherical Polar Coordinates
In Cartesian Coordinates,
the volume element is: d V d x d y d z
In spherical polar coordinates,
the volume element is: d V d r r d O Q d
s i n ( )d V d r r d r d
2 s i n ( )d V r d r d d
Slide 28
The Laplacian in Spherical Polar Coordinates
22 2 2x y z
Cartesian Coordinates:
One example of a chain rule formula connecting a derivative withrespect to x, y, z to derivatives with respect to r, , is:
r
x r x x x
It may be shown that by repeated application of chain rule formulaeof this type (with 2-3 hours of tedious algebra), the Laplacian inspherical polar coordinates is given by:
22 2
2 2 2 2 2
1 1 1sin ( )
sin ( ) sin ( )r
r r r r r
Slide 29
Angular Momentum Operators in Spherical Polar Coordinates
ˆzL i x y
y x
It may beshown that
ˆzL i
i
2 ˆ ˆ ˆ ˆ ˆ ˆx x y y z zL L L L L L L
^
It may beshown that
22 2
2 2
1 1sin( )
sin( ) sin ( )L
^
Slide 30
Outline
• Math Preliminary: Products of Vectors
• Rotational Motion in Classical Physics
• The 3D Quantum Mechanical Rigid Rotor
• Angular Momentum in Quantum Mechanics
• Angular Momentum and the Rigid Rotor
• The 2D Quantum Mechanical Rigid Rotor
• The 3D Schrödinger Equation: Spherical Polar Coordinates
• Rotational Spectroscopy of Linear Molecules
Slide 31
The 3D Quantum Mechanical Rigid Rotor
3D Schrödinger Equation for a particle (Sph. Pol. Coords.)
2 22
2 2 2 2 2
1 1 1sin( )
2 sin( ) sin ( )r V E
m r r r r r
22 ( , , ) ( , , )
2V r E r
m
Modification: Two masses moving relative to their CM
m1 m1r
2 22
2 2 2 2 2
1 1 1sin( )
2 sin( ) sin ( )r V E
r r r r r
1 2
1 2
m m
m m
Slide 32
2 22
2 2 2 2 2
1 1 1sin( )
2 sin( ) sin ( )r V E
r r r r r
The Schrödinger Equation in terms of the L2 operator^
22 2
2 2
1 1sin( )
sin( ) sin ( )L
^
The L2 operator is:^
2 2 22
2 2 2 2
1 1 1sin( )
2 2 sin( ) sin ( )r V E
r r r r
22 2
2
1 1
2 2r L V E
r r r I
2I rwhere^
Radial KE Rotational KE
PE
Slide 33
The Quantum Mechanical Rigid Rotor
2 2 22
2 2 2 2
1 1 1sin( )
2 2 sin( ) sin ( )r V E
r r r r
The Rigid Rotor model is used to characterize the rotation ofdiatomic molecules (and is easily extended to linear polyatomic molecules)
It is assumed that: (1) The distance between atoms (r) does not change.
(2) The potential energy is independent of angle [i.e. V(,) = Const. = 0]
22 2
2
1 1
2 2r L V E
r r r I
^
Therefore:
2 2
2 2
1 1sin( )
2 sin( ) sin ( )E
I
21
2L E
I
^
Slide 34
2 2
2 2
1 1sin( )
2 sin( ) sin ( )E
I
( , )
This equation can be separated into two equations, one containing only and the second containing only .
Assume: ( , ) ( ) ( )
Solution of the Rigid Rotor Schrödinger Equation
Algebra + Separation of Variables
221
sin( ) sin( ) sin ( )2
E CI
2 2
2
1
2C
I
and
We will only outline the method of solution.
Slide 35
Solution of the equation is rather simple.
However, solution of the equation most definitely is NOT.
Therefore, we will just present the results for the quantum numbers,energies and wavefunctions that result when the two equations are solved and boundary conditions are applied.
221
sin( ) sin( ) sin ( )2
E CI
2 2
2
1
2C
I
and
Slide 36
The Rigid Rotor Quantum Numbers and Energies
The Quantum Numbers:
Note that because this is a two dimensional problem, thereare two quantum numbers.
0 , 1 , 2 , 3 , 0 , 1 , 2 , ,m
The Energy:2
( 1)2
EI
Note that the energy is a function of l only. However, there are2 l + 1 values of m for each value of l . Therefore, the degeneracyof the energy level is 2 l + 1
2 1g
Remember that for a classical Rigid Rotor:2
2
LE
I
Comparing the expressions, one finds for the
angular momentum, that: ( 1)L
Slide 37
An Alternate Notation
The Quantum Numbers: 0 , 1 , 2 , 3 ,J
0 , 1 , 2 , ,M J
The Energy:2
( 1)2JE J JI
2 1Jg J
When using the Rigid Rotor molecule to describe the rotationalspectra of linear molecules, it is common to denote the twoquantum numbers as J and M, rather than l and m.
With this notation, one has:
Slide 38
The Wavefunctions
When both the and differential equations have been solved,the resulting wavefunctions are of the form:
,( , ) ( ) ( ) ( )mimmN e P
The are known as the associated Legendre polynomials.( )mP
)(mP
The first few of these functions are given by:
00 1P
01 c o s ( )P
11 s in ( )P
0 22
13 cos ( ) 1
2P
12
1sin( ) cos( )
2P
2 22 s in ( )P
We will defer any visualization of these wavefunctions untilwe get to Chapter 6: The Hydrogen Atom
Slide 39
Spherical Harmonics
( , ) ( ) ( ) ( )mimlm mY N e P
The product functions of and are called “Spherical Harmonics”,Ylm(, ):
They are the angular solutions to the Schrödinger Equation for anyspherically symmetric potential; i.e. one in which V(r) is independentof the angles and .Some examples are:
0 01 0 1 0 1 1 0( , ) ( ) c o s ( )iY N e P N
11 1 1 1 1 1 1( , ) ( ) s i n ( )i iY N e P N e
0 0 22 0 2 0 2 2 0( , ) ( 0 ) 3 c o s ( ) 1iY N e P N
12 1 2 1 2 2 1( , ) ( ) s i n ( ) c o s ( )i iY N e P N e
2 2 2 22 2 2 2 2 2 2( , ) ( ) s i n ( )i iY N e P N e
11 1 1 1 1 1 1( , ) ( ) s i n ( )i iY N e P N e
12 1 2 1 2 2 1( , ) ( ) s i n ( ) c o s ( )i iY N e P N e
2 2 2 22 2 2 2 2 2 2( , ) ( ) s i n ( )i iY N e P N e
Slide 40
( , ) sin( )iY Ne One of the Spherical Harmonics is:
Show that this function is an eigenfunction of the Rigid RotorHamiltonian and determine the eigenvalue (i.e. the energy).
cos( )iYNe
sin( ) sin( ) cos( )iYNe
2 2s in ( ) co s ( )iN e
21 2 s in ( )iN e
1sin( ) 2 sin( )
sin( ) sin( )
iiY Ne
Ne
2 2
2 2
1 1sin( )
2 sin( ) sin ( )
Y YEY
I
21
2HY L Y EY
I
^or
Slide 41
( , ) s i n ( )iY N e
2 2
2 2
1 1sin( )
2 sin( ) sin ( )
Y YEY
I
1sin( ) 2 sin( )
sin( ) sin( )
iiY Ne
Ne
sin( )iYiNe
2
2sin( )iY
Ne
2
2 2
1
sin ( ) sin( )
iY Ne
2 2
2 2
1 1sin( )
2 sin( ) sin ( )
Y YHY
I
2
2 sin( )2 sin( ) sin( )
i iiNe Ne
NeI
2
2 sin( )2
iNeI
2
22
YI
Slide 42
2
22
HY YI
Therefore:2
22
EI
Note: Comparing to:2
( 1)2
EI
1we see that:
Slide 43
Outline
• Math Preliminary: Products of Vectors
• Rotational Motion in Classical Physics
• The 3D Quantum Mechanical Rigid Rotor
• Angular Momentum in Quantum Mechanics
• Angular Momentum and the Rigid Rotor
• The 2D Quantum Mechanical Rigid Rotor
• The 3D Schrödinger Equation: Spherical Polar Coordinates
• Rotational Spectroscopy of Linear Molecules
Slide 44
Angular Momentum and the Rigid Rotor
The Spherical Harmonics, Ylm(, ), are eigenfunctions of theangular momentum operators:
22 2
2 2
1 1sin( )
sin( ) sin ( )L
^
ˆzL i
Note: It is straightforward to show that L2 and Lz commute;
i.e. [L2,Lz] = 0.
Because of this, it is possible to find simultaneouseigenfunctions of the two operators which are, as shown above,the Spherical Harmonics.
^ ^
^^
2 2( , ) ( 1 ) ( , )l m lmL Y Y ^
The eigenvalues are given by the equations:
ˆ ( , ) ( , )z lm lmL Y m Y
Slide 45
As discussed earlier, the restrictions on the quantum numbersare given by:
0 , 1 , 2 , 3 , 0 , 1 , 2 , ,m
Therefore, both the magnitude, |L|, and the z-component, Lz, of theangular momentum are quantized to the values:
( 1 0 , 1, 2 , 3 ,L
0 , 1 , 2 , ,zL m m z-axis
Lz |L|
If a magnetic field is applied, its directiondefines the z-axis.
If there is no magnetic field, the z-directionis arbitrary.
Slide 46
( , ) s i n ( )iY N e One of the Spherical Harmonics is:
22 2
2 2
1 1sin( )
sin( ) sin ( )L
^ ˆ
zL i
Show that this function is an eigenfunction of L2 and Lz and determinethe eigenvalues.
^^
We’ve actually done basically the first part a short while ago.
Remember: 21
2HY L Y EY
I
^ 2
22
Y
I
Therefore: 2 22L Y Y ^ 2( 1 ) Y 1
z
YL Y
i
sin( )ii N ei
Y m Y
1m
Slide 47
Preview: The Hydrogen Atom Schrödinger Equation
2 22
2 2 2 2 2
1 1 1sin( )
2 sin( ) sin ( )r V E
m r r r r r
3D Schrödinger Equation in Spherical Polar Coordinates
The Hydrogen atom is an example of a “centrosymmetric” system,which is one in which the potential energy is a function of only r, V(r).
In this case, the Schrödinger equation can be rearranged to:
2 2 22
2 2 2 2 2
1 1 1( ) sin( )
2 2 sin( ) sin ( )r V r E
m r r r m r r
Radial Part Angular Part
Note that the Angular part of the Hydrogen atom Schrödinger equationis the same as Rigid Rotor equation, for which the radial part vanishes.
Therefore, the angular parts of the Hydrogen atom wavefunctions arethe same as those of the Rigid Rotor
Slide 48
Outline
• Math Preliminary: Products of Vectors
• Rotational Motion in Classical Physics
• The 3D Quantum Mechanical Rigid Rotor
• Angular Momentum in Quantum Mechanics
• Angular Momentum and the Rigid Rotor
• The 2D Quantum Mechanical Rigid Rotor
• The 3D Schrödinger Equation: Spherical Polar Coordinates
• Rotational Spectroscopy of Linear Molecules
Slide 49
Rotational Spectroscopy of Linear Molecules
Energy Levels2
( 1)2JE J JI
2 1Jg J
Equivalent Form:2
2( 1)
8J
hE J J
I
2( 1)
8J
hE hc J J
Ic
( 1 )JE h c B J J
RotationalConstant (cm-1):
28
hB
Ic
Note: You must use c in cm/s, even when using MKS units.
0 0 g0=1
1 g1=3B~
2
E /
hc
[cm
-1]
J EJ gJ
2 g2=5B~
6
3 g3=7B~
1 2
4 g4=9B~
2 0
Slide 50
Diatomic versus Linear Polyatomic Molecules
28
hB
Ic
In general, for linear molecules, the moment of inertia is given by:
2
1
N
i ii
I m r
N is the number of atomsmi is the mass of the atom iri is the distance of atom i from the Center of Mass.
If N=2 (diatomic molecule) the moment of inertia reduces to:
2I r1 2
1 2
m m
m m
where r is the interatomic distance
Slide 51
Selection Rules
Absorption (Microwave) Spectroscopy
For a rotating molecule to absorb light, it must have a permanentdipole moment, which changes direction with respect to the electric vector of the light as the molecule rotates.
J = 1 (J = +1 for absorption)
e.g. HCl, OH (radical) and O=C=S will absorb microwave radiation.
O=C=O and H-CC-H will not absorb microwave radiation.
Rotational Raman SpectroscopyFor a rotating molecule to have a Rotation Raman spectrum, thepolarizability with respect to the electric field direction must change as the molecule rotates. All linear molecules have Rotational Raman spectra.
J = 2
J = +2: Excitation (Stokes line)
J = -2: Deexcitation (Anti-Stokes line)
Slide 52
Intensity of Rotational Transitions
The intensity of a transition in the absorption (microwave) orRotational Raman spectrum is proportional to the number of moleculesin the initial state (J’’); i.e. Int. NJ’’
Boltzmann Distribution:''
'' ''
JE
kTJ JN g e
''( '' 1)
'' (2 '' 1)hcBJ J
kTJN J e
Slide 53
Rotational Spectra
0 0 g0=1
1 g1=3B~
2
E /
hc
[cm
-1]
J EJ gJ
2 g2=5B~
6
3 g3=7B~
1 2
4 g4=9B~
2 0Absorption (Microwave) Spectra
' ''J JE E E
J’’ J’
'( ' 1 ) ''( '' 1 )E h c B J J h c B J J
J’ = J’’+1
( '' 1 ) ( '' 2 ) ''( '' 1 )E h c B J J h c B J J
( 2 '' 2 ) '' 0 , 1 , 2 , 3 ,E h c B J J
(2 '' 2 ) '' 0 , 1, 2 , 3,E
B J Jhc
B~
2 B~
4 B~
6 B~
8~0
Slide 54
0 0 g0=1
1 g1=3B~
2
E /
hc
[cm
-1]
J EJ gJ
2 g2=5B~
6
3 g3=7B~
1 2
4 g4=9B~
2 0Rotational Raman Spectra
' ''J JE E E
J’’ J’
'( ' 1 ) ''( '' 1 )E h c B J J h c B J J
J’ = J’’+2
( '' 2 ) ( '' 3 ) ''( '' 1 )E h c B J J h c B J J
( 4 '' 6 ) '' 0 , 1 , 2 , 3 ,E h c B J J
(4 '' 6 ) '' 0 , 1, 2 , 3,E
B J Jhc
B~
1 4B~
6 B~
1 0~
0
Slide 55
The HCl bond length is 0.127 nm.
Calculate the spacing between lines in the rotational microwave
absorption spectrum of H-35Cl, in cm-1.h = 6.63x10-34 J•sc = 3.00x108 m/sc = 3.00x1010 cm/sNA = 6.02x1023 mol-1
k = 1.38x10-23 J/K1 amu = 1.66x10-27
kg
H Cl
H Cl
m m
m m
1 35
0.9721 35
amu amuamu
amu amu
2I r 22 7 91 .6 1 1 0 0 .1 2 7 1 0x k g x m 4 7 22 . 6 0 1 0x k g m
28
hB
Ic
34
2 47 2 10
6.63 10
8 3.14 2.60 10 3.00 10 /
x J s
x kg m x cm s
As discussed above, microwave absorption lines occur at 2B, 4B, 6B, ...
Therefore, the spacing is 2B
~ ~ ~
~
0 . 9 7 2 a m u 27
271.66 101.61 10
1
x kgx kg
amu
1 11 0 . 7 8 1 0 . 8c m c m
1S p a c i n g 2 2 1 0 . 8 2 1 . 6B x c m
Slide 56
h = 6.63x10-34 J•sc = 3.00x1010 cm/sk = 1.38x10-23 J/K
B = 10.8 cm-1~
Calculate the ratio of intensities (at 250C): 3 4
1 2
I
I
0 0 g0=11 g1=3B
~2
E /
hc
[cm
-1]
J EJ gJ
2 g2=5B~
6
3 g3=7B~
1 2
4 g4=9B~
2 0
3 4 3
1 2 1
I N
I N
3
1
/3
/1
E kT
E kT
g e
g e
12 /
2 /
7
3
hcB kT
hcB kT
e
e
10 /7
3hcB kTe
34 10 1
23
10 6.63 10 3.00 10 / 10.810
(1.38 10 / ) 298
x J s x cm s cmhcB
kT x J K K
= 0.52
0.523 4
1 2
7
3
Ie
I
1 .4
Note: This is equivalent to asking for the ratio of intensites offourth line to the first line in the rotational microwave spectrum.
Slide 57
The first 3 Stokes lines in the rotational Raman spectrum of 12C16O2 are found at 2.34 cm-1, 3.90 cm-1 and 5.46 cm-1.
Calculate the C=O bond length in CO2, in nm.
h = 6.63x10-34 J•sc = 3.00x1010 cm/sk = 1.38x10-23 J/K1 amu = 1.66x10-27 kg
0 0 g0=11 g1=3B
~2
E /
hc
[cm
-1]
J EJ gJ
2 g2=5B~
6
3 g3=7B~
12
4 g4=9B~
2012 . 3 4 6 0 6c m B B
13 . 9 0 1 2 2 1 0c m B B B
15 . 4 6 2 0 6 1 4c m B B B
10 .3 9B c m
10 .3 9B c m
10 .3 9B c m
28
hB
Ic
34 2
2 1 10
6.63 10 /
8 3.14 0.39 3.00 10 /
x kg m sI
cm x cm s
28
hI
Bc
4 6 27 . 1 8 1 0x k g m
Slide 58
CM
rCO rCO
The first 3 Stokes lines in the rotational Raman spectrum of 12C16O2 are found at 2.34 cm-1, 3.90 cm-1 and 5.46 cm-1.
Calculate the C=O bond length in CO2, in nm.
h = 6.63x10-34 J•sc = 3.00x1010 cm/sk = 1.38x10-23 J/K1 amu = 1.66x10-27 kg
4 6 27 . 1 8 1 0I x k g m
2i i
i
I m r C OO 22 20O C O C O C Om r m m r 22 O C Om r
2COO
Ir
m 27
26
16 1.66 10 /
2.66 10
Om amu x kg amu
x kg
46 2
26
7.18 10
2 2.66 10CO
x kg mr
x kg
1 01 .1 6 1 0x m 0 . 1 1 6 1 . 1 6n m A n g s t r o m s
Slide 59
h = 6.63x10-34 J•sc = 3.00x1010 cm/sk = 1.38x10-23 J/K
Calculate the initial state (i.e. J’’) corresponding to the mostintense line in the rotational Raman spectrum of 12C16O2 at 25oC.
Hint: Rather than calculating the intensity of individualtransitions, assume that the intensity is a continuousfunction of J’’ and use basic calculus.
B = 0.39 cm-1~
'' '' 2 ''J J JI N '' '' 1
2 '' 1J J hcB
kTJ e
2'' ''2 '' 1
J J hcBJ e
kT
NJ'' is at a maximum for dNJ''/dJ''=0.
2 2'' '' '' '''' 0 2 '' 1 2 '' 1
'' '' ''
J J J JJdN d dJ e e J
dJ dJ dJ
2 2'' '' '' ''0 2 '' 1 2 '' 1 2
J J J JJ e J e
2'' '' 2
0 2 '' 1 2J J
e J
Slide 60
Therefore: 22 '' 1 2 0J
22 '' 1J
2'' '' 2
0 2 '' 1 2J J
e J
hcB
kT
h = 6.63x10-34 J•sc = 3.00x1010 cm/sk = 1.38x10-23 J/K
B = 0.39 cm-1~
23
34 10 1
2 1.38 10 / 2982 '' 1
6.63 10 3.00 10 / 0.39
x J K KJ
x J s x cm s cm
1 0 6 0 3 2 .6
32.6 1''
2J
1 5 . 8 1 6
Slide 61
Consider the linear molecule, H-CC-Cl.
There are two major isotopes of chlorine, 35Cl (~75%) and37Cl (~25%). Therefore, one will observe two series of linesin the rotational spectrum, resulting from transitions ofH-CC-35Cl and H-CC-37Cl.
Can the structure of H-CC-Cl be determined from these two series?
No. There are 3 bond distances to be determined, but only 2 moments of inertia.
What additional information could be used to determine all threebond distances?
The spectrum of D-CC35Cl and D-CC37Cl
Slide 62
Iz
Ix
Iy
Non-Linear Molecules
Non-linear molecules will generally have up to3 independent moments of inertia, Ix, Iy, Iz.
The Hamiltonian will depend upon the angularmomentum about each of the 3 axes.
22 2
2 2 2yx z
x y z
LL LH
I I I
The Schrödinger Equation for non-linear rotors is more difficult to solve,but can be done using somewhat more advanced methods, and therotational spectra can be analyzed to determine the structure(sometimes requiring isotopic species).
For small to moderate sized molecules (I would guess 10-15 atoms),rotational microwave spectroscopy is the most accurate method fordetermining molecular structure.
Slide 63
Outline (Cont’d.)
• Application of QM to Molecular Structure: Pyridine
• Statistical Thermodynamics: Rotational contributions to the thermodynamic properties of gases
Slide 64
N 1.35 Å1.34
1.40 Å1.38
1.39 Å1.38
117o
117
124o
124
119o
119
118o
118
The Structure of Pyridine
Calculated: MP2/6-31G(d) – 4 minutes
Experimental: Crystal Structure
Slide 65
#MP2/6-31G(d) opt freq Pyridine 0 1 C -1.236603 1.240189 0.000458 C -1.236603 -0.179794 0.000458 C -0.006866 -0.889786 0.000458 C 1.222870 -0.179794 0.000458 C 1.053696 1.280197 0.000458 N -0.104187 1.989731 0.000458 H 1.980804 1.872116 -0.008194 H 2.205566 -0.673935 -0.009628 H -0.006866 -1.989731 -0.009064 H -2.189194 -0.729767 -0.009064 H -2.205551 1.760818 0.009628
The Command File for the Structure of Pyridine
Slide 66
The Frontier Orbitals of Pyridine
HOMO LUMO
Calculated: HF/6-31G(d) – <3 minutes
Experimental: Huh???
Slide 67
N 1.35 Å1.40
1.40 Å1.33
1.39 Å1.44
117o
116
124o
120
119o
119
118o
120
The Structure of Excited State Pyridine
S0: Calculated: MP2/6-31G(d) – 4 minutes
T1: Calculated: MP2/6-31G(d) – 11 minutes
Experimental: None
Slide 68
Outline (Cont’d.)
• Application of QM to Molecular Structure: Pyridine
• Statistical Thermodynamics: Rotational contributions to the thermodynamic properties of gases
Slide 69
Statistical Thermodynamics: Rotational Contributions to Thermodynamic Properties of Gases
A Blast from the Past
2
,
ln
V N
QU kT
T
2
, ,
ln ln
lnV N T N
Q QH kT kT
T V
,V
V N
UC
T
,P
P N
HC
T
lnU
S k QT
l n ( )A U T S k T Q
,
lnln
ln T N
QG H TS kT Q kT
V
Slide 70
The Rotational Partition Function: Linear Molecules
The Energy:2
( 1)2J J JI
2 1Jg J
0
Jrot kT
JJ
q g e
The Partition Function: 2( 1) / 2
0
2 1J J I
kT
J
J e
( 1)
0
2 1RJ Jrot T
J
q J e
2
2R Ik
It can be shown that for most molecules at medium to hightemperatures:
1J J k T
Thus, the exponent (and hence successive terms in thesummation) change very slowly.
Therefore, the summation in qrot can be replaced by an integral.
Slide 71
( 1)
0
2 1RJ Jrot T
J
q J e
2
28R
h
Ik
( 1)
0
2 1RJ Jrot Tq J e dJ
This integral can be solved analytically by a simple substitution.
( 1) Ru J JT
(2 1) Rdu J dJ
T
Therefore:( 1)
0
(2 1)RJ Jrot RT
R
Tq e J dJ
T
0
u
R
Te du
0
1rot u
R
Tq e
1 0 1
R
T
rot
R
Tq
Slide 72
rot
R
Tq
A Correction
For homonuclear diatomic molecules, one must account for the factthat rotation by 180o interchanges two equivalent nuclei.
Since the new orientation is indistinguishable from the original one, onemust divide by 2 so that indistinguishable orientations are counted once.
For heteronuclear diatomic molecules, rotation by 180o produces adistinguishable orientation. No correction is necessary.
rot
R
Tq
is the "symmetry number"
Homonuclear Diatomic Molecule: = 2
Heteronuclear Diatomic Molecule: = 1
Slide 73
How good is the approximate formula?
( 1)
0
12 1
RJ Jrot T
J
q J e
Exact:rot
R
Tq
Approx:
H2: I = 4.61x10-48 kg•m2 R = 87.5 K
O2: I = 1.92x10-46 kg•m2 R = 2.08 K
Compd. T qrot(ex) qrot(app) Error
O2 298 K 71.8 71.7 0.2%
H2 298 1.88 1.70 10%
H2 100 0.77 0.57 26%
H2 50 0.55 0.29 47%
For molecules like H2, HCl, …, with small moments of inertia, the integral approximation gives poor results, particularly at lowertemperatures.
Slide 74
The Total Partition Function for N Molecules (Qrot)
rot
R
Tq
and Nro t ro tQ q
N
rot
R
TQ
where
2
28R
h
Ik
Slide 75
Internal Energy
,
lnln ln(
rot
R
V N
Q dN T
T dT
N
rot
R
TQ
2
,
ln rotrot
V N
QU kT
T
N
T
Therefore: 2rot NU kT
T N k T An N k T
r o tU n R T or r o tU R T
This illustrates Equipartition of Rotational Internal Energy.
[(1/2)RT per rotation].
ln lnN
rot
R
TQ
ln
R
TN
ln ln ( )RN T
Slide 76
2
, ,
ln ln
ln
rot rotrot
V N T N
Q QH kT kT
T V
Enthalpy
,
ln
ln
rotrot
T N
QU kT
V
Qrot independent of V
r o t r o tH U R T
Heat Capacities
,
rot rotrot VV
V N
C UC R
n T
,
rot rotrot PP
P N
C HC R
n T
Remember that these results are for Diatomic and LinearPolyatomic Molecules.
Slide 77
5
2tranPC R
Chap. 3
Experimental Heat Capacities at 298.15 K
Compd. CP (exp)
H2 29.10 J/mol-K
O2 29.36
I2 36.88
r o tPC R
7 78 .3 1 4 / 2 9 .1 0 /
2 2tra n ro tP PC C R J m o l K J m o l K
We can see that vibrational (and/or electronic) contributionsto CP become more important in the heavier diatomic molecules.
We'll discuss this further in Chap. 5
Slide 78
Entropy
lnrot
rot rot US k Q
T
N
rot
R
TQ
lnN rot
rot
R
T US k
T
ln
R
T RTNk
T
ln
R
TnR R
O2: r = 1.202 Å = 1.202x10-10 m (from QM - QCISD/6-311G*)
8.002
O O O
O O
m m mamu
m m
2 7 2 61 . 6 6 1 0 / 1 . 3 2 8 1 0x x k g a m u x k g
2 2 6 1 0 2 4 6 2(1 . 3 2 8 1 0 ) (1 . 2 0 2 1 0 ) 1 . 9 2 1 0I r x k g x m x k g m
= 2 (Homonuclear Diatomic)
2342
2 2 46 2 23
6.63 102.08
8 8 (3.1416) 1.92 10 1.38 10 /R
x J shK
Ik x kg m x J K
Slide 79
O2 (Cont'd)
lnrot
R
TS nR R
2 .0 8R K
For one mole of O2 at 298 K:
Stran = 151.9 J/mol-K (from Chap. 3)
Stran + Srot = 195.7 J/mol-K
O2: Smol(exp) = 205.1 J/mol-K at 298.15 K
Thus, there is a small, but finite, vibrational (and/or) electronic contribution to the entropy at room temperature.
2981(8.31 / ) ln 8.31 /
(2)(2.08 )
43.8 /
rot KS J mol K J mol K
K
J mol K
Slide 80
E (Thermal) CV S KCAL/MOL CAL/MOL-K CAL/MOL-K TOTAL 3.750 5.023 48.972 ELECTRONIC 0.000 0.000 2.183 TRANSLATIONAL 0.889 2.981 36.321 ROTATIONAL 0.592 1.987 10.459 VIBRATIONAL 2.269 0.055 0.008 Q LOG10(Q) LN(Q) TOTAL BOT 0.330741D+08 7.519488 17.314260 TOTAL V=0 0.151654D+10 9.180853 21.139696 VIB (BOT) 0.218193D-01 -1.661159 -3.824960 VIB (V=0) 0.100048D+01 0.000207 0.000476 ELECTRONIC 0.300000D+01 0.477121 1.098612 TRANSLATIONAL 0.711178D+07 6.851978 15.777263 ROTATIONAL 0.710472D+02 1.851547 4.263345
Output from G-98 geom. opt. and frequency calculation on O2 (at 298 K)
QCISD/6-311G(d)
)2 9 8(/4 8.2/5 9 2.0 KR Tm o lk Jm o lk c a lUU r o tt h e r m
r o t
RKm o lJKm o lc a lC r o tV /3 1 4.8/9 8 7.1
Km o lJKm o lc a lS r o t /8.4 3/4 5 9.1 0 (same as our result)
Slide 81
Translational + Rotational Contributions to O2 Entropy
Note that the other (vibration and/or electronic) contributionsto S are even greater at higher temperature.
0 1000 2000 3000 4000 5000100
150
200
250
300
Expt T TR
S
[J/m
ol-K
]
Temperature [K]
Slide 82
Translational + Rotational Contributions to O2 Enthalpy
There are also significant additional contributionsto the Enthalpy.
0 1000 2000 3000 4000 5000
0
50
100
150
200
Expt T TR
H
[kJ/
mol
]
Temperature [K]
Slide 83
Translational + Rotational Contributions to O2 Heat Capacity
0 1000 2000 3000 4000 500015
20
25
30
35
40
45
Expt T TR
CP
[J/m
ol-K
]
Temperature [K]
Note that the additional (vibration and/or electronic) contributionsto CP are not important at room temperature, but very significantat elevated temperatures.
Slide 84
Helmholtz and Gibbs Energies
l nr o t r o tA k T Q ,
lnln
ln
rotrot rot
T N
QG kT Q kT
V
Qrot independent of V
r o tA
lnN
rot rot
R
TA G kT
ln
R
TNkT
ln
R
TnRT
For O2 at 298 K: R = 2.08 K
298ln (8.314 / ) ln 35.5 /
2 2.08rot rot
R
T KA G RT J mol K J mol K
K
Slide 85
Non-Linear Polyatomic Molecules
It can be shown that:
1/ 21/ 2 3rot
a b c
Tq
2
2
2
2
2
2
8
8
8
aa
bb
cc
hI k
hI k
hI k
The symmetry number is defined as the"number of pure rotational elements(including the identity) in the molecule's point group:NO2: = 2NH3: = 3CH4: = 12
I will always give you the value of for non-linear polyatomic molecules.
One can simply use the expression for qrot above in the sameway as for linear molecules to determine the rotational contributionsto the thermodynamic properties of non-linear polyatomic molecules(as illustrated in one of the homework problems).