Molecular Spectroscopy

Spectroscopy is the study of the absorption, emission and scattering of electromagnetic radiation by matter.

Weak em radiation with gas phase molecules

EM radiation consists of oscillating electric and magneticfields that carry electric and magnetic energy from asource to a detector. These fields oscillate at the samefrequency ν (cycles s-1) Hz and propagate at speed c (= 299792458 m s-1).

Distance between adjacent field oscillation crests is thewavelength λ, where λ = c/ν.

Emission spectrum from region of auroral activity in upper atmosphere of Jupiter.

Astrophys. J. 363, L37 (1990)

Wavenumber ν = 1/wavelength; invariably in cm-1 units~

Absorption spectrum of an electrical discharge through H2. JMS 191, 215 (1998).

H atom emission line weak H2 emission lines

↓

Transmittance τ = Itr(ν)/I0(ν) = exp[-ℓ c* ε(ν)]~ ~ ~

Absorption coefficient↑

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Concentration↓

Path length↑

Absorption spectrum of an electrical discharge through H2. JMS 191, 215 (1998).

H atom emission line weak H2 emission lines

↓

Transmittance τ = Itr(ν)/I0(ν) = exp[-ℓ c* ε(ν)]~ ~ ~

Absorption coefficient↑

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Concentration↓

Path length↑

H3+

1000 K

287 K

Jupiteremission

Lababsorption

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

The positions of spectral lines

▬▬▬▬▬▬▬▬▬▬

▬▬▬▬▬▬▬▬▬▬ Ei

Ef

hνif = Ef – Ei = ΔEif

νif

The Bohr frequency condition

= νif /c = Ef /hc – Ei /hc = ΔEif /hc Divide by hc: νif = 1/λif~

frequency

One task of experimental molecular spectroscopy is to assign the upper and lower levels of each transition to their position in the ladder of molecular energy levels.

λ = c/ν

Analytical use of spectroscopy based on fact that eachmolecule has a unique spectrum that characterizes it.

Molecular spectra also give temperature of the sampleand its concentration.

The spectrum of a molecule yields energy level separationsand using quantum mechanics the energies can beinterpreted to give the structure, bond strengths and other properties of the molecule involved.

Internal energy of a molecule is called therovibronic energy, and it is quantized.

Rotation-vibration-electronic

Rotational term values for some simple molecules.

Energies divided by hc and quoted in cm-1 are called TERM VALUES.

The positions of spectral lines

▬▬▬▬▬▬▬▬▬▬

▬▬▬▬▬▬▬▬▬▬ Ei

Ef

hνif = Ef – Ei = ΔEif

νif

The Bohr frequency condition

= νif /c = Ef /hc – Ei /hc = ΔEif /hc Divide by hc: νif = 1/λif~

frequency

One task of experimental molecular spectroscopy is to assign the upper and lower levels of each transition to their position in the ladder of molecular energy levels.

λ = c/ν

We also look at intensities of spectral lines

For an isolated ideal gas at thermal equilibrium the Maxwell distribution of speeds between v and v+dv isgiven by P(v)dv where P(v) = 4π(M/2πkT)3/2 v2e-Mv2/2kT

CO molecule

10 K (77 m/s)

300 K (422 m/s)1000 K (772 m/s)

Most probablespeed vMP is(2kT/M)1/2

For an isolated ideal gas at thermal equilibrium the Maxwell distribution of speeds between v and v+dv isgiven by P(v)dv where P(v) = 4π(M/2πkT)3/2 v2e-Mv2/2kT

CO molecule

10 K (77 m/s)

300 K (422 m/s)1000 K (772 m/s)

Most probablespeed vMP is(2kT/M)1/2

1520 km/hr

Molecular speeds and temperature

For CO, M ~ 4.65 x 10-26 kg and vMP = 24.4√T m/s.Thus at T = 1 μK CO has vMP = 2.4 cm/s.

At such low speeds the wave nature of the moleculepredominates over its particle nature. The de Broglie wavelength, λ = h/p = h/√(2πMkT),which for CO gives λ = 3.32/√T Å.At T = 1 μK CO has a de Broglie λ of 3320 Å.Trap such cold molecules to get BE condensation.

Back to CO gas at thermal equilibrium at ordinary Twith molecules colliding with each other to distributetheir internal energy levels according to M-B distribution.

Fraction of CO mols in each rotational state

10 K

300 K1000 K

↓

gie-Ei/kT

∑gje-Ej/kT▬▬▬▬j

F(Ei ) =

↓

↓

Boltzmannfactor

Absorption spectrum of an electrical discharge through H2. JMS 191, 215 (1998).

H atom emission line weak H2 emission lines

↓

Transmittance τ = Itr(ν)/I0(ν) = exp[-ℓ c* ε(ν)]~ ~ ~

Absorption coefficient↑

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Concentration↓

Path length↑

F(Ei ) = [ gie-Ei/kT ] / gje-Ej/kT Boltzmann factor∑j

S(f ← i) = | ∫ Φf* μA Φi dτ |2 Line strength∑A=X,Y,Z

Rstim(f→i) = 1 – exp (-hνif /kT ) Stimulated emission

I(f ← i) = ∫

8π3 Na______(4πε0)3hc2

F(Ei ) S(f ← i) Rstim(f→i)

Integrated absorption intensity for a line is:

= νif

~line

Frequency factor

ε(ν)dν~

The shapes of spectral lines

Using P(v) = 4π(M/2πkT)3/2 v2e-Mv2/2kT

Doppler effect.

Absorbs at ν0(1+v/c)

ν0(1-v/c)Absorbs at

Lineshape function is S( = S( ) exp ) ν0ν

ν- ν0Mc2___

2kT_

ν0

____[ ( ) ]2

This is a Gaussian function. FWHH = 7.15 x 10 -7 (T/M)½ ν0

in amu

Lifetime broadening

1/2πτ

+ (1/2πτ)2(ν – ν0)2

________________1π__S(ν) =

1/πτFWHH = FWHH/GHz = 318/τ

in ps

Finite lifetime from spontaneous emission and collisions

This is a Lorentzian function.

Simultaneous Doppler and lifetime broadening leadsto a Voigt profile which is a convolution of the Gaussianand Lorentzian functions.

Pressure broadening

Transmittance τ = Itr(ν)/I0(ν)

Power broadening

For weak radiation this ratio is independent of I0

and absorption is a linear function of I0.

As power is increased molecules absorbing in thecentre of the line (where absorption is the greatest)will start to absorb at a greater rate than that at whichthey can return to the lower level to maintain thermalequilibrium. Centre of line saturates and line broadens.This is called power broadening.

Pressure broadening: Higher pressures gives shiftand unsymmetrical line…function of intermol forces

~ ~

Laser induced fluorescence (LIF) spectroscopy

_________________________

_____________________________________________________________________________________________________________________________

Laser tuned

_________________________

_________________________

Measure totalfluorescencefor upper state.

Disperse thefluorescencefor lower state

Very sensitive technique.

Resonance enhanced multiphoton ionization(REMPI) Spectroscopy

________________________

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

________________________

________________________

Ionization limit

M+ + e- Detect (and weigh) M+

This shows laser tuned to 2-photon resonance

Raman scattering (10 -7)Raleigh scattering

Stokes anti-Stokes

Part of the Raman spectrum of methane. Owyoung et al., CPL 59, 156 (1978)

Fourier Transform spectroscopy

Source FT interferometer sample detector

http://www.irgas.com/ftir_spectroscopy.html

http://en.wikipedia.org/wiki/Fourier_transform_spectroscopy

http://spectroscopy.lbl.gov/FTIR-Martin/

The positions of spectral lines

▬▬▬▬▬▬▬▬▬▬

▬▬▬▬▬▬▬▬▬▬ Ei

Ef

hνif = Ef – Ei = ΔEif

νif

The Bohr frequency condition

= νif /c = Ef /hc – Ei /hc = ΔEif /hc Divide by hc: νif = 1/λif~

frequency

One task of experimental molecular spectroscopy is to assign the upper and lower levels of each transition to their position in the ladder of molecular energy levels.

λ = c/ν

Theoreticians calculate molecular energy levels andline strengths and then explain to spectroscopistswhat they have seen and why.

The molecular Hamiltonian and

Schroedinger equation

The coordinatesand wavefunctions

P. R. Bunker and Per Jensen:

Molecular Symmetry and Spectroscopy, 2nd Edition,

2nd Printing, NRC Research Press, Ottawa, 2006

(ISBN 0-660-19628-X). $49.95 for 747 pages.

paperback. Chapters 9-11

P. R. Bunker and Per Jensen:

Fundamentals of Molecular Symmetry,

Taylor and Francis, 2004

(ISBN 0-7503-0941-5). $57.95

paperback. Chapters 2-5.

Read the details in:

Per JensenProfessor, Ph.D.

Rm. L13.12Theoretische ChemieBergische Universität

Wuppertal

Tel. 0202 - 439 2468jensen@uni-wuppertal.de

The classical expression for the total energy of particlesnuclei [1 to N] and electrons [(N+1) to ].

Etot = ½ Σ mr (Xr2 + Yr

2 + Zr2)+ Σ CrCse2/(4πε0Rrs)

ℓ ..

Rrs = [(Xr –Xs)2 + (Yr – Ys)2 + (Zr – Zs)2 ]½

where

In a SPACE FIXED XYZ axis system.

We know translational energy = ½ Mv2

where v is velocity of centre of mass.So we change coords to separate that offby introducing coords of c of m (X0Y0Z0).

ℓ

r=1

ℓ

r<s=1

.

ℓ

NEW COORDSOLD COORDS

Italic fontUpright font

Complete separation of variables

IN NEW COORDINATES

Translational Sch. Eq.

Ttrans = ½ M (X02 + Y0

2 + Z02)

Htrans = -(ħ2/2M) 02

Put molecule in a cube box with side L

Htrans = -(ħ2/2M) 02 + V(cubic box side L)

.. .

E1 = 6h2/(8ML2) (nX,nY,nZ) = (1,1,2), (1,2,1) or (2,1,1)

E0 = 3h2/(8ML2) (nX,nY,nZ) = (1,1,1)

E1 – E0 = 3h2/(8ML2).

For 12C16O (M = 4.65 x 10-26 kg) in a L = 10-2 m cubic box

= 3.5 x 10-38 J. Divide by hc gives 1.8 x 10-15 cm-1

~

Etrans = (nX2 + nY

2 + nZ2)h2/(8ML2)

where nX, nY and nZ are positive integers.

Molecule-in-a-box energies

E1 – E0

Alternatively if put an electron in a cubic boxof side length 0.3 nm get an energy separation(/hc) of 1.01 x 105 cm-1

Alternatively if put an electron in a cubic boxof side length 0.3 nm get an energy separation(/hc) of 1.01 x 105 cm-1

Or a proton to move in 1 dimension constrainedto a length of 0.03 nm get an energy separation(/hc) of 5500 cm-1

Alternatively if put an electron in a cubic boxof side length 0.3 nm get an energy separation(/hc) of 1.01 x 105 cm-1

Or a proton to move in 1 dimension constrainedto a length of 0.03 nm get an energy separation(/hc) of 5500 cm-1

The first excited electronic state has an energy~20x larger than that of the first excited vibrationalstate because the electron mass is much smallerthan the nuclear mass – compensates for the factthat electronic motions are less constrained thannuclear motions. Leads to success of BO approx.

Now we look at the rovibronic energy and then convert to Hrve

^

The rovibronic Hamiltonian

int

+ Hes + Hhfs

HintΦint = EintΦint

The complete internal (em) H

Φtot = ΦtransΦint

We usually begin by solving the rovibronic problem and then deal with the effect of the spin terms afterwards.

Include spin

^ ^

The rovibronic Schroedinger Equation

The rovibronic Schroedinger Equation

Vee + VNN + VNe

THE GLUE

The rovibronic Schroedinger Equation

Obviously we can write this down for any molecule. In a worldof infinitely powerful computers we could then solve the Sch. equation numerically and that would be that (or would it?)

The Role of Approximations

We are forced to make approximations in order to solvethe Sch. Eq. but in so doing we introduce concepts thatenable us to UNDERSTAND molecules.

Concepts that come about because we make approximations:

Electronic state, molecular orbital, electronic configuration,potential energy surface, equilibrium structure, force constants,electronic angular momentum, vibrational state, vibrational angular momentum, rotational state, Coriolis coupling, centrifugal distortion,…

We could call the approach that aims to solve the Sch. Eq. numerically using a gigantic computer “The few-concepts-little-understanding big-computer” approach.

The Role of Approximations

We are forced to make approximations in order to solvethe Sch. Eq. but in so doing we introduce concepts thatenable us to UNDERSTAND molecules.

Concepts that come about because we make approximations:

Electronic state, molecular orbital, electronic configuration,potential energy surface, equilibrium structure, force constants,electronic angular momentum, vibrational state, vibrational angular momentum, rotational state, Coriolis coupling, centrifugal distortion,…

We could call the approach that aims to solve the Sch. Eq. numerically using a gigantic computer “The few-concepts-little-understanding big-computer” approach.I sometimes call it“The small-brain big-computer approach.”

The Born-Oppenheimer Approximation

First we change coordinates by shifting to a (ξηζ) axis systemthat is oriented as the XYZ, and space-fixed XYZ, axis systemsbut which has origin at the NUCLEAR centre of mass.

In these coordinates we have (see p 185 in BJ1)

Hrve = Te + TN + Vee + VNN + VNe

The KE is completely separable into electronic and nuclear parts.

BUT the V is not separable because of VNe. VNe is the “glue”that holds the molecule together so we cannot neglect it.

HOWEVER, based on the understanding that the first excited electronic state is at an energy likely to be more than 10X higher than that of the first excited vibrational state, we can make an APPROXIMATE separation of nuclear and electronic degrees of freedom. This is the Born-Oppenheimer approximation. We will see that this approximation is really only valid for the ground electronic states of molecules that do not have any low lying excited electronic states.

The Born-Oppenheimer Approximation

Hrve = Te + TN + Vee + VNN + VNe

We initially focus on the electron dynamics andneglect TN and VNN.So we solve the electronic Sch. Eq.

[Te + Vee + VNe]Φelec,n = Velec,nΦelec,n

where a particular nuclear geometry is chosenin VNe, and n labels successive electronic states.This is a calculation at a fixed nuclear geometry.

[VNN + Velec,n ] provides the potential energy surface for the nuclei in the electronic state n.So within the electronic state n the nuclearmotion (or rotation-vibration) Sch. Eq. is

[TN + VNN + Velec,n]Φrv,nj = Erve,njΦrv,nj

Where j labels successive rotation-vibration states.Introduce Eelec,n as min value of (VNN+Velec,n).

0

Now we look at the nuclear dynamics

[TN + VN,n]Φrv,nj = Erv,njΦrv,nj

VNN+Velec,n-Eelec,n Erve,nj–Eelec,n

0

0

[Te + Vee + VNe]Φelec,n = Velec,nΦelec,n

where a particular nuclear geometry is chosenin VNe, and n labels successive electronic states.This is a calculation at a fixed nuclear geometry.

The electronic Sch. Eq.

Elec. energy Eelec,n is min value of (VNN+Velec,n).

[TN + VN,n]Φrv,nj = Erv,njΦrv,nj

VNN+Velec,n-Eelec,n Erve,nj–Eelec,n

The rotation-vibration Sch. Eq.

0

In the BOA we solve two equations:

Φrve,nj = Φelec,nΦrv,nj

0

and HrveΦrve,nj ≠ Erve,njΦrve,nj0 0

because off-diagonal M.E. is not zero:

∫Φelec,nΦrv,njTNΦelec,mΦrv,mk dτ* *

Extent of BO wavefunction mixing depends on:

H(nj,mk)ΔE(nj,mk) ~

ΔEvib

ΔEelec

The Born-Oppenheimer approx

H(nj,mk) =

Erve,nj = Eelec,n + Erv,nj0

We now see that the Born-Oppenheimer approximation is really only valid for the ground electronic states of molecules that do not have any low lying excited electronic states.

[TN + VN ]Φrv,j = Erv,jΦrv,j

VNN + Velec,0 - Eelec,0 Erve,0j - Eelec,00

For an isolated ground electronic state the rotation-vibration Sch. Eq. is:

In coords ξiηiζi and conjugate momenta

Solving the rotation-vibration Sch. Eq.We change coordinates (again!). Based onintuition that dynamics involves two types ofmotion - “rotation” and “vibration”

ξ1η1ζ1…ξNηNζN θ,φ,χ,(xi-xie),(yi-yi

e),(zi-zie)

xyz axes aremolecule-fixed axes

The node line ON

ξηζ axes arespace-fixed axes

Δxi Δyi Δzi

ξ

ζ

η

Solving the rotation-vibration Sch. Eq.We change coordinates (again!). Based onintuition that dynamics involves two types ofmotion - “rotation” and “vibration”

ξ1η1ζ1…ξNηNζN θ,φ,χ,(xi-xie),(yi-yi

e),(zi-zie)

xyz axes aremolecule-fixed axes

Δxi Δyi ΔziGiven these

We want to determine these

The equi coords are determined using principal axes so wereally want to determine and the xi,yi,ziθ,φ,χ

Solving the rotation-vibration Sch. Eq.We change coordinates (again!). Based onintuition that dynamics involves two types ofmotion - “rotation” and “vibration”

ξ1η1ζ1…ξNηNζN θ,φ,χ,(xi-xie),(yi-yi

e),(zi-zie)

xyz axes aremolecule-fixed axes

Δxi Δyi ΔziGiven these

We want to determine these

and the xi,yi,ziθ,φ,χ