chapter 8 (19): vibrational and rotational spectroscopy...
Transcript of chapter 8 (19): vibrational and rotational spectroscopy...
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electromagnetic spectrumchapter 8 (19): vibrational and rotational spectroscopy of diatomics
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Einstein coefficients for a two-level system
B =2π2
3h2ϵ0
|μ1 2|2 =
|μ1 2|2
6ℏ2ϵ0
whereμ1 2 =∫Ψ1
*μ̂ Ψ2 d τ
chapter 8 (19): vibrational and rotational spectroscopy of diatomics
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Sections 3-5: infrared vibrational spectroscopy
Photon energy = change in molecular energy
Δ E= E n f− E ni
chapter 8 (19): vibrational and rotational spectroscopy of diatomics
P (n) = f n =e−(E n − E 0) /(kB T )
qvib where qvib =
1
1 − e−h ν/(kB T )
Harmonic oscillator: E n= (n+12)h ν so E n−E 0=n h ν
f n = e−n h ν/(kB T ) (1−e
−hν/(kB T ))
Example: At room temperature, kT ≈ 200 cm-1. Let ν = 1000 cm-1.e-hν/(kT) = e-1000/200 = e-5 = 0.007 ≈ 0fn ≈ (e-hν/(kT))n
fn≈0 unless n=0. 4
Sections 3-5: infrared vibrational spectroscopy
fundamental absorption frequency equals ν0≡E 1 − E 0
h(harmonic)
chapter 8 (19): vibrational and rotational spectroscopy of diatomics
~ν0 =1
2π c (kμ )1 /2
If harmonic, ~ν0=~νe
The effect of isotopic mass on ν0 comes from μ.
Example: Calculate harmonic fundamental ν0, λ and for 12C16O.
Data: k = 1902 N/m (Table 8.3)mass of 12C = 12 exactlymass of 16O = 15.9949 (inside back cover of Engel’s book)
μ = 12 × 15.9949 /(12 + 15.9949) = 6.85621 amu = 1.13850×10-26kg
ν = 1/(2π) (1902 kg×s-2 / 1.13850×10-26kg)1/2 = 6.505×1013s-1
λ = c/ν = 4.6086×10-6m = 4.6086×10-4cm
~ν0
~ν0 = 1/ λ = 2170 cm−1
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Sections 3-5: infrared vibrational spectroscopy
FTIR spectrum of CO gas.
chapter 8 (19): vibrational and rotational spectroscopy of diatomics
Source: Zach Gardner and Kelby Donnay, UMD
~ν0
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Sections 3-5: infrared vibrational spectroscopy
chapter 8 (19): vibrational and rotational spectroscopy of diatomics
Example: Compare vibrations of 12C16O and 13C16O.Data: k = 1902 N/m is independent of isotope.mass of 13C = 13.0034 (inside back cover of Engel’s book)mass of 16O = 15.9949 (inside back cover of Engel’s book)
μ = 13.0034 × 15.9949 /(13.0034 + 15.9949) = 7.1724 amu (cf 6.85621)μ = 1.1910×10-26kg = 6.360×1013s-1
= 1/λ = 2121cm-1 (cf 2170)
Larger μ lowered frequency.
~ν0
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Section 4: infrared selection rule Δn=1.Light's electric field interacts with molecule's dipole moment.
chapter 8 (19): vibrational and rotational spectroscopy of diatomics
Transition dipole moment, for n → m:
μ = μ(Re )+(∂μ
∂R )e x +⋯ where x=R−Re
μxn m
=∫ψm*μψn d x
μxn m
=∫ψm*μ(Re )ψn d x + ∫ψm
* (∂μ
∂ R )e x ψn d x
μxn m
=μ(Re)∫ψm*ψn d x + (
∂μ
∂R )e∫ψm* x ψn d x
S m n=0
μxn m
= (∂μ
∂ R )e∫ψm* x ψn d x
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Section 4: infrared selection rule Δn=1, continued.
chapter 8 (19): vibrational and rotational spectroscopy of diatomics
μxn m
= (∂μ
∂R )e∫ψm* x ψn d x
Recursion formula from chapter 7:
ψn+1 = (2α
n+1 )1/2
x ψn − (n
n+1 )1/2
ψn−1 (ψ−1≡0)
μxn m
= (∂μ
∂R )e [∫ψm* (n+1
2α )1 /2
ψn+1 d x + ∫ψm* ( n
2α )1 /2
ψn−1 d x ]μx
n m= (
∂μ
∂R )e [(n+12α )
1 /2S m , n+1 + ( n
2α )1 /2
S m , n−1]μx
n , m= 0 unless (
∂μ
∂R )e≠0 and n=m−1 or n=m+1
x ψn = (n+12α )
1/2ψn+1 + ( n
2α )1 /2
ψn−1
Selection rules:●gross: dipole moment changes during the vibration●fine: quantum number changes by 1
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Anharmonicity.Problems with the harmonic oscillator:
●The harmonic-oscillator potential V=(1/2)kx2 supports an infinite number of vibrational energy levels.
●Realistic potentials do not allow x<0.●Harmonic-oscillator energy levels are equally spaced.
En+1-En=hν, the same for all n.
chapter 8 (19): vibrational and rotational spectroscopy of diatomics
Engel, Figure 8.6 (19.6). 10
Anharmonicity, continued.
chapter 8 (19): vibrational and rotational spectroscopy of diatomics
E n = (n+ 12)h νe − (n+ 1
2)2
xe h νe + (n+ 12)3
ye h νe
Anharmonic vibrational energy levels:
xe h νe =(h ν)2
4 De in Engel's equation 19.5
Fundamental frequency
E 0 =12
h νe −14
xe h νe
E1 =32
h νe −94
xe h νe~ν0 =~νe − 2 xe
~νe
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Experimental (i.e., calculated from spectral data)
Molecule xe (%)
H2 4403.2 121.3 3
LiH 1406.1 23.2 2
BeH 2060.8 36.3 2BH 2366.7 49.3 2
anharmonicities in light hydrides
xe ν̃e (cm−1)ν̃e (cm−1
)Source: Jan M. L. Martin, Chemical Physics Letters, 283, 283-293, 1998.
chapter 8 (19): vibrational and rotational spectroscopy of diatomics
Example: Consider H2. Calculate E0, E1 and hν0=ΔE using xe.
Data: νe = 4403.2 cm-1 and xeνe = 121.3 cm-1.E0 =h( 0.5 νe - 0.25 xeνe) = 0.5× 4403.2 - 0.25× 121.3 = 2171.3 cm-1
E1 =h( 1.5 νe - 2.25 xeνe ) = 1.5× 4403.2 – 2.25×121.3 = 6331.9 cm-1
ΔE = 4161 cm-1 (< harmonic 4403.2) = hν0 (defines ν0, the fundamental vibration frequency)
GAMESS quantum calculation gives ~νe (cm−1)
IR or Raman spectrum shows ~ν0 (cm−1)12
Morse potential for HI
chapter 8 (19): vibrational and rotational spectroscopy of diatomics
De = D0 + E 0
Re = 174 pm
V ( x) = De [1 − e− α(x−xe )]
2
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E n = (n+12 ) νe − (n+1
2 )2
xe h νe
At dissociation, d E nd n = 0 = h νe−2 (n+1
2 ) xe h νe
so (n+12 )
max=
12 x e
De = E nmax=
h νe2 xe
−xe h νe
(2 xe)2
De =h νe
4 xe or De =
~νe4 xe
in cm−1
Dissociation energy and anharmonicity
chapter 8 (19): vibrational and rotational spectroscopy of diatomics
Example: Dissociation energy of HI.~ν e= 2309cm−1 xe
~νe= 51.1 cm−1
nmax =1
2 xe−
12=
23092×51.1
−12= 22
De =~νe
2
4 xe~νe
=23092
4×51.1= 26084 cm−1 14
Morse and harmonic potentials for HI
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Section 6. Rotational Spectroscopy
chapter 8 (19): vibrational and rotational spectroscopy of diatomics
Recall that ψrot= Y Jm(θ ,ϕ) and E J = J (J +1)ℏ
2
2 I , for J=0,1, 2,…
degeneracy gJ = 2 J+1
in wavenumbers ~ν = J (J +1)B
where B =h
8π2 I c in cm−1 is the rotation constant
In a pure rotational transition, J changes but vibrational n does not.Selection rules come from the transition dipole moment.Consider the z component of the transition dipole moment. (same from x,y)
μzJ , J '
= μ0∫Y J 'm ' ∗cos(θ) Y J
m sinθd θd ϕ where μ0 is the permanent dipole moment,
μ0 cosθ is the z component of the dipole The integral is zero unless m=m ' and J '=J±1Selection rules for a pure transition: The molecule must have a permanent dipole moment. The rotational quantum number J changes by 1 or -1. 16
Section 6. Rotational Spectroscopy
chapter 8 (19): vibrational and rotational spectroscopy of diatomics
Spectral line locations can be predicted from the Δ J selection rule.in wavenumbers ~ν = J ' (J ' +1)B−J (J+1)Babsorption, J '=J+1: ~ν = (J+1)(J+2)B−J (J +1)B
~ν = 2 (J +1)B
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rotational spectrum intensities depend on T
P (J ) =(2 J +1)e
−E J /(kB T )
qrot
Example: B= 2 cm-1 T=700 K qrot= 243
chapter 8 (19): vibrational and rotational spectroscopy of diatomics
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rotational spectrum Jmax
tells T
P (J ) =gJ e
−E J /(kB T )
qrot where gJ=2 J +1 and E J= J (J +1)B
d Pd J
= 0 ⇒d gJd J
− g J1
kB Td E Jd J
= 0
2 − (2 J +1)1
k B T(2 J +1)B = 0
2 k B TB
= (2 J+1)2
T =B
2kB( 2 Jmax+1)2
chapter 8 (19): vibrational and rotational spectroscopy of diatomics
Example: Estimate the temperature of the CO spectrum.
Jmax=11 and B=2.0 cm-1
T =2 cm−1
2×0.695 K⋅cm−1232
= 760 K
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~νe=2170 cm−1
xe~νe=13 cm−1
Re=112.8 pm
too-large B=116 cm−1
Data:
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R
inte
nsit
y
wavenumbers
spectrum:
P
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R
inte
nsit
y
wavenumbers
spectrum:
P
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R
inte
nsit
y
wavenumbers
spectrum:
P
R branchP branch
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rotation - vibration spectrum of CO
Engel, Figure 8.18 (or 19.18)
Quantum Chemistry & Spectroscopy
P branch
R branch
chapter 8 (19): vibrational and rotational spectroscopy of diatomics
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positions of rotation - vibration spectral lines
R branch: J → J + 1
Q branch: J does not change (forbidden for diatomic molecules)
P branch: J → J - 1
chapter 8 (19): vibrational and rotational spectroscopy of diatomics
~ν =~ν0 +~ν rot
~ν =~ν0 + 2 B (J+1) ; J=0,1,2,…
~νrot =(J−1)(J )B − J (J+1)B =−2 J B~ν =~ν0 − 2 B J ; J=1,2, 3,…
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rotation-vibration spectrum of HCl
Jessica Woodward, U Mn Duluth
chapter 8 (19): vibrational and rotational spectroscopy of diatomics
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Fourier-transform infrared spectroscopy
http://www.hazmatid.com/Applications/App043.pdf
sample
chapter 8 (19): vibrational and rotational spectroscopy of diatomics
computer → Fourier transform
beamsplitter isGe-coated KBr
He-Ne laser measures mirror Δx, "optical lag" or "optical delay"
source is a hot glowing ceramic or wire
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Fourier-transform infrared
spectroscopy
chapter 8 (19): vibrational and rotational spectroscopy
center burst
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Fourier-transform infrared spectroscopy
http://www.midac.com/files/Tn-100.PDF
chapter 8 (19): vibrational and rotational spectroscopy
FT
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Section 8. Raman Spectroscopy
Raman relies on scattering rather than absorption of light.●Intense monochromatic laser light hits the sample. ●Scattered light is measured to a side or back toward the source.●Some scattered photons have ν shifted by vibration frequency.●Intensity of ν-shifted light is plotted versus ν.
chapter 8 (19): vibrational and rotational spectroscopy of diatomics
Rayleigh: νfinal = νsource filtered out
Stokes: νfinal = νsource - νvib intense
anti-Stokes: νfinal = νsource + νvib weak30
Raman Spectra
N2 spectrum taken by Keith Carron, DeltaNU, Inc.
chapter 8 (19): vibrational and rotational spectroscopy
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Section 8. Anti-Stokes lines are weaker than Stokes lines
Raman relies on scattering rather than absorption of light.● Stokes intensity depends upon P(n=0)● anti-Stokes depends upon P(n=1)
chapter 8 (19): vibrational and rotational spectroscopy
The ratio has been used to measure temperature.E.g., T of CO2 at high pressure in a diamond anvil cell.
Our Raman instrument measures Stokes lines only.
I anti−StokesI Stokes
=P (1)P (0)
= e−
h νkT
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Section 8. Classical theory of Raman spectroscopy
Raman relies on scattering rather than absorption of light.● Stokes intensity depends upon P(n=0)● anti-Stokes depends upon P(n=1)
chapter 8 (19): vibrational and rotational spectroscopy
electric field of light : E = E 0 cos(2π νsource t )E 0 induces a dipole moment in the molecule
μinduced = α E 0 cos(2π νsource t ) where α is polarizabilityExpand α in a vibrational coordinate x
α = α0 + (d α
d x )0
x +⋯
x varies periodically
x = xmax cos (2π νvib t ) so α = α0 + (d α
d x )0
xmax cos(2π νvib t )
μinduced = E0α0 cos(2π νsource t ) + E 0 (d αd x )0 x max cos(2π νsource t ) cos(2π νvib t )
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Section 8. Classical theory of Raman spectroscopy, continued
chapter 8 (19): vibrational and rotational spectroscopy
From the previous slide
μinduced = E0α0 cos(2π νsource t ) + E 0 (dαd x )0 x max cos(2π νsource t ) cos(2π νvib t )
Recall that cos(a)cos(b ) = 12[cos (a+b) + cos(a−b )]
μinduced =E 0α0 cos(2π νsource t ) Rayleigh
+ 12
E 0(d α
d x )0 x max [ cos(2π(νsource+νvib) t ) anti-Stokes
+ cos(2π(νsource−νvib) t ) ] Stokes
Gross selection rule: polarizability α must change as the molecule vibrates.