The Quantum Theory ofAtoms and Molecules:

An overview of spectroscopy

Hilary Term 2008

Dr Grant Ritchie

Background

WHY should we study quantum mechanics?

Is it difficult?

DON’T PANIC!!!!!!

Perhaps – I am not sure exactly…..

Because quantum theory is the most important discovery of the 20th century!

Allows us to understand atoms → molecules → chemical bond → chemistry →basis of biology;

Understand solids → conduction of electricity (development transistors, computers,solar cells etc.)

Classical mechanics

Classical Mechanics Laws – Determinism

Can predict a precise trajectory for particles, with precisely specified locations andmomenta at each instant;

Example: If a particle of mass m, initially at rest at position x0 , is subject to a timevarying force of the form F = (1 − αt), derive an expression for the particle’s positionat some later time t.

Classical Mechanics allows the translation, rotation and vibrational modes of motion tobe excited to any energy simply by controlling the forces applied. i.e. Energy iscontinuous.O.K. for planets, cars, bullets, etc. but fails when applied to transfers of very smallquantities of energy and to objects of very small masses (atomic/molecular level, lightinteraction).

Quantum theory

See introductory course Michaelmas Term

Energy is not continuous. E.g. atomic line spectra

Example: On the left is the image is a helium spectral tubeexcited by means of a 5kV transformer. At the right are thespectral lines through a 600 line/mm diffraction grating. Heliumwavelengths (nm): (s = strong, m = med, w = weak)

UV 438.793 w443.755 w447.148 s471.314 m

Blue 492.193 mYellow 501.567 sYellow 504.774 wRed 587.562 sDark red 667.815 m

The classical atom

Classical model for the hydrogen atom consists of an electron with mass me moving in acircular orbit of radius r around a single proton. The electrostatic attraction between theelectron and the proton provides the centripetal force required to keep the electron in orbit

2

0

22

4

r

e

r

vmF

e

!"==

r

evmKEe

0

2

2

82

1

!"==

r

ePEKEE

tot

0

2

8!"#=+=

The kinetic energy, KE, is

and the total energy, Etot , for the orbiting electron is simply

Application of Newton’s laws of motion and Coulomb’s law of electric force: inagreement with the observation that atoms are stable; in disagreement withelectromagnetic theory which predicts that accelerated electric charges radiate energy inthe form of electromagnetic waves.

An electron on a curved path is accelerated and therefore should continuously loseenergy, spiralling into the nucleus!

The Bohr condition

hnvrmLe

==

e

pe

pem

mm

mm

)(

!

+=µ

22

22

2

82

1

rm

hnvmKE

e

e!

==

2

0

22

em

hnr

e!

"=

22220

4

8

)(

nnh

emnE

e

tot

!"="=

#

!

Bohr postulated that the electron is only permitted to be in orbits that possess anangular momentum, L, that is an integer multiple of h/2π. Thus the condition for astable orbit is

* To be strictly accurate we should not use me but the reduced mass of the electron-proton system, µ.

Hence, we can write the kinetic energy as

Comparing our two expressions for the kinetic energy we have

The above result predicts that the orbital radius should increase as n increases wheren is known as the principal quantum number. Hence the total energy, Etot (n) , is

where ℜ is known as the Rydberg constant.

Bohr modelQuantum constraint limits the electrons motion to discrete energy levels (quantumstates) with energy E(n). Radiation is only absorbed/emitted when a quantum jumptakes place:

leading to transition energies of:2 2

1 2

1 1E

n n

! "# =$ %& '

( )

This is the same as the Rydberg formula; Bohr’s theory is in agreement withexperimentally observed spectra.

!2

nhprmvr == !" n

p

nhr == 2

Furthermore, combining de Broglie’s relation with Bohr condition shows that thecircumference of the orbit of radius r must be an integer number of wavelengths, λ.

Failures of the Bohr model

Bohr’s postulate: An electron can circle the nucleus only if its orbit contains anintegral number of De Broglie wavelengths: nλ = 2πrn (n = 1, 2, 3,…);

therefore rn = n2 h2 ε0 / πme2 ≡ a0 (for n = 1) Bohr radius H-atom.

This postulate combines both the particle and the wave characters of the electron in a singlestatement, since the electron wavelength is derived from the orbital velocity required to balancethe attraction of the nucleus.

Problems

1. It fails to provide any understanding of why certain spectral lines are brighter thanothers. There is no mechanism for the calculation of transition probabilities.

2. The Bohr model treats the electron as if it were a miniature planet, with definiteradius and momentum. This is in direct violation of the uncertainty principle whichdictates that position and momentum cannot be simultaneously determined.

3. Results were wrong even for atoms with two electrons – He spectrum!

Molecular spectra

Energy

Photoionisation

Molecules are even more interesting – more degrees of freedom!

Overview of molecular spectra

The most commonly observed molecular spectrainvolve electronic, vibrational, or rotationaltransitions. For a diatomic molecule, the electronicstates can be represented by plots of potentialenergy as a function of internuclear distance.

Electronic transitions are vertical or almost verticallines on such a plot since the electronic transitionoccurs so rapidly that the internuclear distance can'tchange much in the process.

Vibrational transitions occur between differentvibrational levels of the same electronic state.

Rotational transitions occur mostly betweenrotational levels of the same vibrational state,although there are many examples of combinationvibration-rotation transitions for light molecules.

Some examples……..

UV/Visible/near IR spectroscopy

Example: Atmospheric absorption

IR spectroscopy

The Beer Lambert law

I I0ε = molar absorption co-efficient (molecule specific).

[c] = concentration of sample

l = sample length

[ ]

010

c lI I

!"=

ε (= ε(ν)) is greatest where the absorption is most intense and has units of L mol−1 cm−1.

Absorbance: 0log [ ]I

A c lI

!" #

= =$ %& '

Allows spectroscopy to determine absolute concentrations of species present..

(Later in the course we will see that ε ∝ µfi2)

Optical activity

A sample can also change the polarisation of the incident light.

Identifying and distinguishing enantiomers is inherently difficult, since their physicaland chemical properties are largely identical ⇒ use polarimetry.

For example, right- and left-handedenantiomers of a chiral compoundrotate the plane of polarized lightin opposite ways ⇒ opticallyactive.

Specific rotation, [α]D

[ ][ ]

D

c l

!! =

where l = cell length in dm, [c] = concentration in g/ml and Ddenotes that the measurement is conducted using the 589 nm lightfrom a sodium lamp (i.e. the sodium D lines – see atomicspectroscopy later).

Each enantiomer of a stereoisomeric pair is optically active and has an equal butopposite-in-sign specific rotation.

[α]D is an experimentally determined constantthat characterizes and identifies pureenantiomers. E.g. the carvone enantiomers havethe following specific rotations.Carvone from caraway: [α]D = +62.5º(S-(+)-carvone or d-carvone)Carvone from spearmint: [α]D = –62.5º(R-(–)-carvone or l-carvone)

A 50:50 mixture of enantiomers (racemates) has no observable optical activity.

What are we going to do?

We want to develop a set of equations akin to Newton’s laws for microscopic particles.

This will lead us to the Schrödinger equation and the concept of a wavefunction, Ψ(x, t).

We will show how the wavefunctions and their associated energies can be used to interpretspectra and obtain information about atoms and molecules.

Firstly, we need to review some of the basic properties of electrostatics and electromagneticwaves to understand how radiation interacts with molecules….

Electrostatics – Coulomb’s law

A fundamental property of electrons and protons is charge. Charged particles are eitherpositively or negatively charged, and experiments show that like charges repel whilstopposite charges attract. It is these electrostatic interactions that are responsible for theexistence of molecules, and their chemical and thermodynamic properties.

Most fundamental particles possess an electrical charge

e.g. proton, q = 1.602189 × 10-19 coulomb = eelectron, q = −1.602189 × 10-19 coulomb = -eneutron, q = 0.0 coulomb

Macroscopic particles can also have a charge which must be a multiple of theelementary charge, e.

Two charges interact Ψ force, given quantitatively by Coulomb’s law.

!!"

#$$%

&=

2

21

4

1

r

qqF

'(

Coulomb’s law continued

The magnitude of the force between the particles is dependent on the medium inwhich the charges are situated, and this is taken into account by the factor ε which isknown as the permittivity of the medium.

The permittivity of a medium is ε = εr ε0 where ε0 is a fundamental constant known asthe vacuum permittivity and εr is the relative permittivity of the medium. ε0 has thevalue 8.854×10−12 J−1 C 2 m−1. εr > 1 and so the interaction potential in the medium isreduced from that in a vacuum.

32.6

CH3OH

78.524.32.32.2εr

H2OC2H5OHC6H6CCl4Molecule

A consequence of the reduced ionic interaction is the wide variation in the rates ofionic reactions in solution with different solvents.

Electric field

The influence of electrostatic forces is described in terms of an electric field. The electricfield strength, E, at a particular point in space is defined as the force exerted per unitcharge on a positive test charge located at that point. The force experienced by a testcharge q is F = qE, and so the electric field E is a vector that is parallel to the force F .

From Coulomb's law, the electric field at a position r away from an isolated charge q1located at the origin is

rr

q rE

2

0

1

4!"=

For several charges, the fields at a point areadditive.

E = E1 + E2 + E3 + etc

Total force F = QE.

Electrostatic energy

Intrinsically linked to the electrostatic force is the electrostatic potential V. Theelectrostatic potential at a point in an electric field is defined as the work done in bringinga unit positive charge from infinity to that point. Since

VF

r

!" #= $% &

!' (⇒

r

qV

r

0

1

4d

!"=#= $

%

• rF

and so the interaction potential energy, U, of a charge q2 with charge q1 is

r

qqVqU

0

21

24

!"

==

U is a scalar quantity ⇒ the net electrostatic potential due to an array of charges issimply the algebraic sum of their individual contributions.

NB: The Coulomb interaction is long range and continues to exert its influence atdistances that are far beyond those of any other intermolecular force. e.g. r −6 for Van derWaals’ interaction between atoms.

Applications

(a) Can be used to describe the potential energy of any assembly of static charges

(b) Important for describing the energy of ionic compounds, e.g. ionic crystals.

(c) In atoms/molecules - can be used to describe the interactions between nuclei andelectrons

(d) Useful in quantum mechanics when describing the energies of atoms and molecules.

2

1 2

0 12

2

1

0 1

2

0

........( )4

- ........( )4

........( )4

i

ij

Z Z eV nuclear nuclear

R

Z eelectron nuclear

R

eelectron electron

R

!"

!"

!"

= + #

+ #

+ + #

0

4

i j

i j ij

QQV

R!"#

= $

The electric dipole

In molecules, the charge distribution is usually not “uniform”; the centre (of gravity)of positive charge does not coincide with the centre of negative charge.

This charge asymmetry can be represented by an electric dipole.

The simplest electric dipole consists of two equal and opposite charges (∀Q) separatedby a distance d.

The value of the electric dipole moment is given by µ = Qd.(The correct SI units are C m. However the dipole moment is frequently quoted inDebye units. 1 Debye (D) = 3.336 x 10-30 C m).

0.1121.1091.8260.00µ/Debye

COHClHFH2Molecule

N.B. the dipole moment is a vector quantity because it has magnitude and direction (inthe direction from negative to positive charge).

Electric dipoles continued

Within larger molecules, to a good approximation, we can consider the overall dipolemoment as the vector sum of the dipole moments associated with the individual bondswithin the molecule.

For example, the dipole moment of benzene is zero (by symmetry). However that ofchlorobenzene takes the value of 1.45 Debye (this corresponds to the difference of the C-Cl and C-H moments). We can then estimate the dipole moments of the dichlorobenzenesby straight vector addition of bond moments. Although not quantitative, the results arereasonable.

0.00 Debye1.38 Debye2.27 Debyeµobs

2 µPhClcos90o

= 0.00Debye

2 µPhClcos60o

= 1.45Debye

2 µPhClcos30o

= 2.51 Debye

µcalc

ParaMetaOrtho

Electric dipole in an electric field

Consider a dipole in a uniform electric field. Each endof the dipole experiences equal and opposite force ⇒net force = 0.

Dipole does experience a torque, Γ, which willrotate dipole so that it is aligned with E:

E!= µ"

The work done, W, in rotating the dipole from angle θ 1 to θ 2 is:

)cos(cos d sin 12

2

1

!!µ!!µ

!

!

"="= # EEW

The work done is the change in electrical PE of the dipole in the field (i.e. W = ΔU = U 2 − U 1), and so the PE of a dipole in an electric field is given by

E•!=!= µ"cosìEU

State of lowest PE has θ = 0 and the dipole lies along the electric field direction.

!

E

Electric field and potential due to a dipole

-q +q2xA

r(a)The total potential at a point A is given by thesum of the potentials due to each charge in thedipole:

!"

#$%

&

''

+=

drdr

qV

11

40

()

Assuming that 2d « r then2

0

20 4

2

4 rr

dqV

!"

µ

!"#=$

%

&'(

) #*

The interaction PE, U, for an ion of charge Q with a dipole is then2

04

r

QU

!"

µ#=

Very similar to the Coulomb potential between two ions except that it declines morerapidly with r ⇒ as ion moves further away from the dipole the two charges of thedipole ‘merge’ (from the point of view of the ion) and produce a neutral entity.

An atom in an electric field - polarisability

An electric field E can induce a dipole moment, µind, in an atom or a non-polar moleculeby interacting with and distorting the electron distribution in the atom or molecule.

The induced dipole moment is directly proportional to E and the proportionality constantis called the polarisability α :

Eì !=ind

α related to how strongly the nuclear charges interact with the electron distribution as thisgoverns the degree to which the electron distribution can be distorted by an applied field⇒ light atoms / molecules with few electrons have low polarisabilities whereas largesystems with many electrons, where the electron distribution is more diffuse, show largepolarisabilities.

1.972

N2 (║)

1.2764.021.6420.206α‘= (α/4πε0)/10-30 m3

N2 (⊥)XeArHeAtom /Molecule

E++

+

-

-

-

Higher multipole moments

Many molecules with no dipole moments have charge distributions that are notspherically symmetric and as a consequence they create a non-zero electric field e.g.H2, N2 , O2, CO2 and C2H2.

Consider CO2: Each C=O bond possesses an electric dipole moment, whose valuescancel in the overall molecule. However since these dipole are displaced relative to eachother, the fields due to these dipole do not cancel ⇒ charge distribution can be describedby a quadrupole moment, Θ, and can be represented as two displaced dipoles.

Field due to a quadrupole fall as R −4 (compared with R−2 for a charge and R−3 for adipole). More symmetric molecules like methane have both a zero dipole moment and azero quadrupole moment. The first non-zero multipole moment is the octupole moment(and its field drops off as R −5) – see States of Matter course in Trinity Term.

OO

O

O C

C

OO C

OO COO C

- + -

+- +

+

-

-

+

-

+- -+

+

Molecular charge distribution + spectroscopy

The total charge, dipole moment, quadrupole moment, and the polarisability oftenprovide the best descriptions of the charge distribution* (charges are not static) in amolecule. *(charges are not static points and we must speak of a charge distribution)

For example, in the case of a linear molecule the electric potential φ (r) due to thecharge distribution is:

constants are CBAr

C

r

B

r

Aqr ,,......)(

32+

!++=

µ"

It is the interaction between the charge distribution and the electric field associatedwith electromagnetic radiation that is the physical basis of many spectroscopies.

For example: In Uv/Vis and IR spectroscopies the electric field of the radiationinteracts with a changing dipole moment in the molecule and causes the electronic orvibrational state (or both) of the molecule to be changes from some initial states to afinal state.

Waves and Vibrations

One of the most common forms of mechanical behaviour is that of periodic motion;that is to say, an action that is repeated at regular intervals. This includes:

i. swinging of a pendulum,

ii. ripples on a pond,

iii. the vibrational modes of molecules

iv. the wave-like properties of electromagnetic radiation.

The best illustration of oscillatory characteristics is given by simple harmonic motion(SHM), which is often used as an idealised model to analyse real-life situations……

Simple harmonic motion (revision)

An object in stable equilibrium moves back towards its ‘restingposition’ if it is nudged away from it. If restoring force, F, isproportional to the displacement from equilibrium, x, then theresultant motion is said to be simple harmonic. i.e.

xkF !=

where k is a (positive) constant having units Nm−1.

Newton’s 2nd law:x

m

k

t

x

d

d

2

2

!=

which has the solution: ) sin( !" += tAx

where ω 2 = k/m; A and φ are constantsthat are determined by two boundaryconditions, such as the values of x anddx/dt at t = 0. !A

A

x

-"/#

2$ #

t

x

x = 0

k x

m

SHM continued

- The periodicity of SHM is enshrined in the sine term, which repeats itself every 2π /ω seconds (with ω given in rad s−1).

- The maximum displacement from equilibrium is x = ± A , with A referred to asthe amplitude of the oscillation.

- Any temporal offset with respect to x = Asin(ω t), which has x increasing as itpasses through the origin (x = 0, t = 0), is encapsulated in the phase constant φ by x = Asin(ω t+φ ).

[ ] mAtktmA

kxdt

dxm

222222

22

2

1) (sin) (cos

22

1

2

1!"!"!! =+++=+#

$

%&'

(

since ω 2 = k/m and sin2θ + cos2θ = 1. While the energy continually interchangesbetween kinetic and potential their sum is independent of time.

- The total energy (KE+PE) is conserved:

Damped oscillations

The oscillations represented by x (t) = A sinω t are executed indefinitely once they areinitiated because there is no mechanism for losing energy in the system.

The amplitude of the oscillations in a real (e.g. pendulum-like) system diminishes ⇒modelled by introducing a dissipative term proportional to the speed, dx/dt, intoequation on motion:

t

xqx

t

x

d

d 2

d

d 2

2

2

!!= "

(factor of 2 is included to simplify the subsequent algebra, and q > 0)

With trial solution x = α e pt, the auxiliary equation becomes p2+2qp+ω 2 = 0 and hasthe complex roots p = −q ± (q2−ω 2)1/2. Hence, the general solution is:

!"

#$%

&+=

'''+'2222

ee e(( )* qtqtqt

x

Damped oscillations - solutions

!"

#$%

&+=

'''+'2222

ee e(( )* qtqtqt

x

1. When 0 < q < ω , we obtain damped SHM:

x = [α eiω0 t +β e−iω0 t] e−qt = Ae−qt sin(ω0 t +φ )where ω0

2 = ω 2 − q2 ⇒ Oscillation has an angularfrequency of (ω 2 − q2)1/2 but an amplitude that decaysexponentially with time.

2. If q > ω, there is only a decay and no oscillations:

x = α e−(q+q0)t +βe−(q-q0)t

where q0 = (q2 − ω2)1/2 < q.

3. If q = ω we have critical damping: x = (α +β ) e−qt, sothat there is a fast decay to x = 0.

2 2

2

q

!

" #

over-damped

x

t

critically damped

Forced oscillationsTo overcome the loss of amplitude due to friction and air-resistance when, for example,pushing a swing, we often tend to give it periodic jolts ⇒ driven SHM.

Consider pulling down a mass attached to the ceiling by a spring and letting it go whilethe fixture is itself made to vibrate at a certain frequency ω . How does the system react?

If ω0 is the natural frequency of the spring-and-mass system (ω 0 = (k/m)1/2) that issubject to a damping force defined by q, then the equation of motion is:

22

02

d d sin ( )

d d

x xq x t

t t! !+ + =

Try Z = Beiω t as the solution of d2Z/dt2+ q dZ/dt + ω02Z = eiω t; this yields

1220 )( !

+!= """ iqB

ResonanceThe magnitude of the resultant forced oscillations of frequency ω is then

( ) 21

222220

* )( !

+!== qBBB """

where B* is the complex conjugate of B.

B has the largest value when ω = ω0, of1/(qω0), and diminishes as the differenceincreases.

This enhancement of response as thedriving frequency matches the naturalone is called resonance ⇒ the physicalbasis of absorption spectroscopy. Driving frequency !

1

q!0

!00

A classical picture for IR spectroscopy

Consider a heteronuclear diatomicmolecule ⇒ has a dipole moment.

Electric field of IR radiation (see later)oscillates in time and exerts aninstantaneous force on a dipole thatchanges bond length.

Case 1: If ω » ω 0 then field reverses itselfmany times during a single vibrational periodand motion is only slightly perturbed.

Classical model for IR spectroscopy continued.

Case 2: If ω « ω 0 and field canonly slightly effect a change in thevalue of the charge separation andonly for some cycles ⇒ no uptakeof energy from the field.

Case 3: If ω = ω0 then field stretchesbond when it is expanding andcompresses it when it is contracting ⇒maximises energy exchange betweenmolecule and the field.

Such an exchange of energy is called an electric dipole transition.

WARNING!!!!! DANGER!!!!!!