The Quantum Theory ofAtoms and Molecules

The breakdown of classical physics:

Quantisation of Energy

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 of transistors, computers, solar cells etc.)

Classical mechanics

Classical Mechanics Laws – Determinism

1. Can predict a precise trajectory for particles, with precisely specified locations and momenta at each instant;

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

2. Classical Mechanics allows the translation, rotation and vibrational modes of motion to be excited to any energy simply by controlling the forces applied.i.e. Energy is continuous.

O.K. for planets, cars, bullets, etc. but fails when applied to transfers of very small quantities of energy and to objects of very small masses (atomic/molecular level, light interaction).

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 tube excited by means of a 5kV transformer. At the right are the spectral lines through a 600 line/mm diffraction grating. Helium wavelengths (nm): (s = strong, m = med, w = weak)

UV 438.793 w

443.755 w

447.148 s

471.314 m

Blue 492.193 m

Yellow 501.567 s

Yellow 504.774 w

Red 587.562 s

Dark red 667.815 m

The classical atom

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

20

22

4

r

e

r

vmF e

r

evmKE e

0

22

82

1

r

ePEKEEtot

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: in agreement with the observation that atoms are stable; in disagreement with electromagnetic theory which predicts that accelerated electric charges radiate energy in the form of electromagnetic waves.

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

r

The Bohr condition

nvrmL e

epe

pe mmm

mm

)(

22

222

82

1

rm

hnvmKE

ee

2

022

em

hnr

e

22220

4

8

)(

nnh

emnE e

tot

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

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

Hence, the kinetic energy is

Comparing our two expressions for the kinetic energy we have:

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

where is known as the Rydberg constant.

NB. L is quantised!

(see atomic orbitals later in course!)

Bohr model

Quantum constraint limits the electrons motion to discrete energy levels (quantum states) with energy E(n). Radiation is only absorbed/emitted when a quantum jump takes place:

Transition energies are:

2 21 2

1 1E

n n

Same as the Rydberg formula Bohr’s theory is in agreement with experimentally observed spectra.

2

nhprmvr

np

nhr 2

Combining de Broglie’s relation with Bohr condition shows that the circumference of the orbit of radius r must be an integer number of wavelengths, .

Failures of the Bohr model

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

therefore rn = n2 h2 0 / me2 ≡ a0 (for n = 1) Bohr radius H-atom.

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

Problems

1. It fails to provide any understanding of why certain spectral lines are brighter than others. 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 definite radius and momentum. This is in direct violation of the uncertainty principle which dictates 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 spectra involve electronic, vibrational, or rotational transitions. For a diatomic molecule, the electronic states can be represented by plots of potential energy as a function of internuclear distance.

Electronic transitions are vertical or almost vertical lines on such a plot since the electronic transition occurs so rapidly that the internuclear distance can't change much in the process.

Vibrational transitions occur between different vibrational levels of the same electronic state.

Rotational transitions occur mostly between rotational levels of the same vibrational state, although there are many examples of combination vibration-rotation transitions for light molecules.

Some examples……..

UV/Visible/near IR spectroscopy

Example: Atmospheric absorption

Atmospheric Profiling – the ozone hole is real!

IR spectroscopy

Chemical identification – different molecules/groups have different vibrational frequencies – WHY?

The Photoelectric effect

APlate

h f

Detector

V

Planck’s photon picture: E = h.

The photon supplies the energy available, = hc (ionisation energy / work function) For < c , not enough energy to ionise.For > c , hc used in ionisation, the rest is carried off by the electron as kinetic energy: KEmax = h .

Light shining onto matter causes the emission of photoelectrons.

Note:1. Photoelectrons are emitted instantly, whatever the intensity of the light.2. There is a critical frequency below which no photoelectrons are emitted.3. Maximum kinetic energy of photoelectrons increases linearly with frequency.

Frequency f

Cur

rent

I

f0

Heat capacities of monatomic solids(C

al /

K m

ol)

* I cal = 4.18 J

Some typical data…

Classical calculation - Equipartition

Treat atoms as a classical harmonic oscillator:2

21

2 2xp

E KE PE kxm

In 1 dimension

Equipartition: Every quadratic energy term contributes 1/2kT to the average energy, <E>.2

21 1 1

2 2 2 2x

B B B

pE kx k T k T k T

m

Atoms vibrate 3d:3

3 BdE k T

The average energy, <E> is just the internal energy of the system , U, and so:

3 and ( ) 3B V BV

UU k T C T k

T

i.e. Cv is independent of temperature and has a value of 3R (25 J K 1 mol1) for all monatomic solids. This is known as the Dulong-Petit law.

This classical calculation works well at “high temperatures” but utterly fails at low T.

Quantisation is the answer….

Bk T Bk T Bk T

Temperature T

H

eat c

apac

ity (

/2 J

Km

olR

-1

-1)

0

2

4

6

8Dulong & Petit

Debye

Einstein

1.What matters is size of compared to kT

2. Different solids have different sizes of Cv(T) “cuts-on” at different T but has same shape for all.

Take home message……

Evidence: Atomic + molecular spectra

Photoelectric effect

T-dependence of heat capacities

+ others…..

Energy is quantised

Quantisation is observable in the macroscopic thermodynamic properties of matter.

e.g. CV (T) H(T), S(T) etc……

cf) Kirchoff’s Law