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Physical Chemistry (II)
Lecture 18
CHEM 3172-80
Lecturer: Hanning Chen, Ph.D.04/03/2017
Molecular Vibrational Spectra
Quiz 17
5 minutes
Please stop writing when the timer stops !
Three Types of Molecular Motion
Translation Vibration Rotation
O
O
O
O
: center of massChange? Yes
d d
d : inter-particle distance
Change? No
d
: center of massChange? No
d : inter-particle distance
Change? Yes
: center of massChange? No
d : inter-particle distance
Change? Noone-dimensional box quantum tunneling rigid rotorone-dimensional
harmonic oscillator(hard to study due to frequent
molecular collisions)
the Vibrations of Diatomic Molecules
equilibrium state: R = R0 : equilibrium bond length
vibrational state: R ' = R0 + ΔRlinear spring
Hooke’s law: F = −kΔRThe spring always tends to move the particle towards the equilibrium position.
k ↑ : stiff spring k ↓ : floppy spring
A B
vibrational energy: Evib R( ) = Evib (R0 )+∂Evib
∂R⎛⎝⎜
⎞⎠⎟ R=R0
R − R0( ) + 12
∂2Evib
∂R2⎛⎝⎜
⎞⎠⎟ R=R0
R − R0( )2 + ...
: bond distortion
0Evib R( ) ≈ 1
2∂2Evib
∂R2⎛⎝⎜
⎞⎠⎟ R=R0
ΔR2 = 12
− ∂F∂R
⎛⎝⎜
⎞⎠⎟ R=R0
ΔR2 = 12kΔR2harmonic approximation:
(Taylor expansion)
Quantum Harmonic Oscillator
system’s Hamiltonian:
A B
H = T + V = − !
2
2mA
d 2
dRA2 −!2
2mB
d 2
dRB2 +
12k(RA − RB )
2
effective mass: meff =mAmB
mA +mBif mA ≫ mB meff ≈ mB
seems like only B is moving
bond distortion: ΔR = RA − RB
rewritten Hamiltonian:
Heff = Teff + Veff = − !
2
2meff
d 2
dΔR2+ 12kΔR2
two-particle system:
effective one-particle harmonic oscillator
vibrational energy: Ev = nv +
12
⎛⎝⎜
⎞⎠⎟ !ω
nv : vibrational quantum number non − negative integerquantized
ω = kmeff
: characteristic frequency
Ladder of Energy Levels
nv = 0
nv = 1
nv = 2
nv = 3
nv = 4
equally spaced energy levels !
E nv = 0( ) = 1
2!ω zero point
energy
a quantum oscillator can NOT be entirely frozen !
Heinsberg uncertainty principle
Δx→ 0Δp→ 0×
ΔpΔx ≥ !
2
Ev = nv +
12
⎛⎝⎜
⎞⎠⎟ !ω
ΔE = !ω
ΔE = !ω
ΔE = !ω
ΔE = !ωnv = 0→ nv = 1 ΔE = !ω observednv = 0→ nv = 2 ΔE = 2!ω nevernv = 0→ nv = 3 ΔE = 3!ω nevernv = 1→ nv = 2 ΔE = !ω observednv = 1→ nv = 3 ΔE = 2!ω nevernv = 2→ nv = 3 ΔE = !ω
×
observed
Selection Rules for Vibrational TransitionsFermi’s Golden Rule:
Ti→ f =
2π!
ψ i"r ψ f
2ρ f density of final states
transition dipole moment !r : polarization direction of incident light
harmonic oscillator: ϕv = NvHv y( )e− y2 /2 y = ΔR
α,α = !2
meff k⎛
⎝⎜⎞
⎠⎟
14
ϕv ' = Nv 'Hv ' y( )e− y2 /2Hv : Hermite polynomials
Decomposition of total wavefunction: ψ =ϕvϕe ϕe ΔR( ) : electronic wavefunction
ϕe ΔR( ) =ϕe(Re )+dϕe
dR (R=Re )ΔR + ...variation of electronic wavefunction:
ψ i!r ψ f = ϕe(Re )
!r ϕe(Re ) ϕv ϕv ' +∂ ϕe(R)
!r ϕe(R)∂R
ϕv!r ϕv ' = ∂µ
∂R⎛⎝⎜
⎞⎠⎟ ϕv
!r ϕv ' ≠ 00Hermitian Hamiltonian
molecular dipole for allowed
transitions
(Born-Oppenheimer approximation)
δ vv '
Allowed Vibrational Transitions
µif = ψ i
!r ψ f = ∂µ∂R
⎛⎝⎜
⎞⎠⎟ ϕv
!r ϕv ' ≠ 0transition dipole moment:
1. ϕv!r ϕv ' ≠ 0
NvNv ' Hv y( ) !rHv ' y( )∫ e− y
2
dr ≠ 0
only satisfied when Δv = v − v ' = ±1A vibrational transition can ONLY occur between two neighboring vibrational states !
2. ∂µ∂R
⎛⎝⎜
⎞⎠⎟ ≠ 0
A vibrational transition is ONLY possible when changing molecule dipole moment !
!E
initial final initial
µi = 0
A
A A
A
finalµ f = 0 ΔH = 0
NO photon is absorbed in order to conserve the energyH =
!E i!µ
Anharmonicity of Molecular Vibration
H2
potential energy profile
experimental data
harmonic approximation
OK
terrible Evib R( ) = 12
∂2Evib
∂R2⎛⎝⎜
⎞⎠⎟ R=R0
ΔR2 + 16
∂3Evib
∂R3⎛⎝⎜
⎞⎠⎟ R=R0
ΔR3 + 124
∂4Evib
∂R4⎛⎝⎜
⎞⎠⎟ R=R0
ΔR4 + ...
higher-order terms are neededwhen ΔR is large
F ≠ −kΔR the restoring force is no longer linear !
Morse potential:
VMorse = De(1− e−αΔR )2 α =
meffω2
De
Morse parameter
for a perfect harmonic oscillator, α → 0,De →∞
E nv( ) = nv +
12
⎛⎝⎜
⎞⎠⎟ !ω − nv +
12
⎛⎝⎜
⎞⎠⎟2
xe!ω
corrected energy levels:
xe : anharmonicity constant
Vibrational OvertonesWith the help of anharmonicity:A vibrational transition is NO LONGER limited between two neighboring vibrational states !
ψ v!r ψ v ' ≠ 0 even when Δv = v − v ' ≠ ±1
ΔEnv→nv+2= 2!ω − (4nv − 6)xe!ω
Vibrational overtones:
ΔEnv→nv+3= 3!ω − (6nv −12)xe!ω ΔEnv→nv+4
= 4!ω − (8nv − 20)xe!ω
overtone intensity drops substantially with increasing Δv
ΔEnv→nv+1= !ω − (2nv − 2)xe!ω
why ΔE1→2 < ΔE0→1 ? redshift ?
ΔE0→1 = !ω + 2xe!ω ΔE1→2 = !ω
difference maker
greater redshifts for vibrational overtones
the Birge-Sponer Plot
nv = 0
nv = 1
nv = 2
nv = 3
nv = 4
ΔE = !ω + 2xe!ω
ΔE = !ω
ΔE = !ω − 2xe!ω
ΔE = !ω − 4xe!ω ΔE = !ω − 8xe!ω
nv = 5 E nv( ) = nv +
12
⎛⎝⎜
⎞⎠⎟ !ω − nv +
12
⎛⎝⎜
⎞⎠⎟2
xe!ω
D0 = E0→1 + E1→2 + E2→3 + ...= Ei→i+1i=1
∞
∑Dissociation energy of a chemical bond:
assuming all vibrational transitions are detectable:
In most experiments, only the first several transitions can be observed
linear assumption:
ΔEnv→nv+1= !ω + 2xe!ω( )− 2!ω xenv
y axis interception: !ω + 2xe!ω
x axis interception: 1+ 12xe
Sshade =
(2xe +1)2
4xe!ω ≈ D0
Vibration-Rotation SpectraCoupling between molecular vibration and molecular rotation
~ 10 TetraHertz (1013 / s) ~ 100 GigaHertz (1011 / s)In general, molecular vibration is ~100 times faster than molecular rotation
ΔEvib >> ΔErot
molecular vibrational energy gap is much much greater than that of molecular rotation !
Combined energy term:
Etotal = Evib + Erot = v + 1
2⎛⎝⎜
⎞⎠⎟ !ω + BJ(J +1)
hω >> Bv : vibrational quantum number J : rotational quantum number
Spectral Branchesselection rules for rotational spectra:
ΔJ = 0,±1ΔJ = −1: P branch
ΔJ = +1: R branch
ΔJ = 0 : Q branch
EP = ΔEvib + ΔErot = !ω − 2BJ
ER = ΔEvib + ΔErot = !ω + 2B(J +1)
EQ = ΔEvib + ΔErot = !ωSeparation of the three branches yields
B : rotational constant
Combinational Differences
vibrational excitation
nv
BI 'A BI ''↑
A
the rotational “constant” B is actually a function of vibrational quantum number,
I ' ≠ I ''
EP = !ω − (B '+ B '')J + (B ''− B)J 2Corrected spectral differences: B '(B '') for the initial(final) vibrational state
ER = !ω + (B '+ B '') J +1( ) + (B ''− B) J +1( )2
EQ = !ω + B ''− B '( )J(J +1)Combinational differences:
ER J( )− EP J( ) = 4B'' (J + 12)
starting from the same rotational state
ER J −1( )− EP J +1( ) = 4B '(J + 12)
ending at the same rotational state
Energy Diagram of Combinational Differences
B '
B ''E R
J ()
E PJ ()
E PJ+1
()
E RJ−1
()
ER J( )− EP J( ) = 4B'' (J + 12)
ER J −1( )− EP J +1( ) = 4B '(J + 12)
J
JJ −1
J +1
J −1
J +1
For example, in HCl
Bv=0' : 10.440 cm−1 Bv=1
' : 10.136 cm−1
apparently due to the bond elongation
the reduction of B was also observed in DCl
Bv=0' : 5.392 cm−1 Bv=1
' : 5.280 cm−1
~ 3% reduction
~ 2% reduction
the bond distortion is less prominent in the heavier DCl
Review of Homework 6Review of Homework 1712.7 The microwave spectrum of 16 O12 CS gave absorption lines (in GHz) as follows:
J 1 2 3 4
32S 24.32592 36.48882 48.65164 60.81408
34S 23.73233 47.46240
Use the expressions for moments of inertia in Table 12.1 and assume that the bond lengths are unchanges by substitution; calculate the CO and CS bond lengths in OCS.
RCO = R1 and RCS = R2
I = mOmS
mOCS
R1 + R2( )2 + mC mOR12 +mSR2
2( )mOCS
!v = 2B J +1( ) = "2πcI
J +1( )
I = !2πc!v
J +1( )
Review of Homework 6Review of Homework 17
I(16O12C32S)=1.38 ×10−45kgm2 I(16O12C34S)=1.42 ×10−45kgm2
I = mOmS
mOCS
R1 + R2( )2 + mC mOR12 +mSR2
2( )mOCS
R1 = RCO = 1.16Å R2 = RCS = 1.56Å
Homework 18
Reading assignment: Chapters 12.8, 12.9, 12.10, 12.11 and 12.12
Homework assignment: Exercises 12.20 Problems 12.9
Homework assignments must be turned in by 5:00 PM, April 4th, Tuesday
to my mailbox in the Department Main Office located at Room 4000, Science and Engineering Hall