chapter 8 (19): vibrational and rotational spectroscopy...

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1 electromagnetic spectrum chapter 8 (19): vibrational and rotational spectroscopy of diatomics 2 Einstein coefficients for a two-level system B = 2 π 2 3 h 2 ϵ 0 | μ 12 | 2 = | μ 12 | 2 6 2 ϵ 0 where μ 12 = Ψ 1 * ^ μΨ 2 d τ chapter 8 (19): vibrational and rotational spectroscopy of diatomics 3 Sections 3-5 : infrared vibrational spectroscopy Photon energy = change in molecular energy Δ E = E n f E n i chapter 8 (19): vibrational and rotational spectroscopy of diatomics P ( n)= f n = e −( E n E 0 )/( k B T ) q vib where q vib = 1 1 e h ν/( k B T ) Harmonic oscillator: E n =( n + 1 2 ) h ν so E n E 0 =nh ν f n = e nh ν/( k B T ) ( 1e h ν/( k B T ) ) Example: At room temperature, kT ≈ 200 cm -1 . Let ν = 1000 cm -1 . e -hν/(kT) = e -1000/200 = e -5 = 0.007 ≈ 0 f n ≈ (e -hν/(kT) ) n f n ≈0 unless n=0. 4 Sections 3-5 : infrared vibrational spectroscopy fundamental absorption frequency equals ν 0 E 1 E 0 h ( harmonic ) chapter 8 (19): vibrational and rotational spectroscopy of diatomics ~ ν 0 = 1 2 π c ( k μ ) 1/ 2 If harmonic, ~ ν 0 = ~ ν e The effect of isotopic mass on ν 0 comes from μ. Example : Calculate harmonic fundamental ν 0 , λ and for 12 C 16 O. Data: k = 1902 N/m (Table 8.3) mass of 12 C = 12 exactly mass of 16 O = 15.9949 (inside back cover of Engel’s book) μ = 12 × 15.9949 /(12 + 15.9949) = 6.85621 amu = 1.13850×10 -26 kg ν = 1/(2π) (1902 kg×s -2 / 1.13850×10 -26 kg) 1/2 = 6.505×10 13 s -1 λ = c/ν = 4.6086×10 -6 m = 4.6086×10 -4 cm ~ ν 0 ~ ν 0 = 1 /λ= 2170 cm 1 5 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 ~ ν 0 6 Sections 3-5 : infrared vibrational spectroscopy chapter 8 (19): vibrational and rotational spectroscopy of diatomics Example : Compare vibrations of 12 C 16 O and 13 C 16 O. Data: k = 1902 N/m is independent of isotope. mass of 13 C = 13.0034 (inside back cover of Engel’s book) mass of 16 O = 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 -26 kg = 6.360×10 13 s -1 = 1/λ = 2121cm -1 (cf 2170) Larger μ lowered frequency. ~ ν 0

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

13

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.