Physical Chemistry III
(Quantum Chemistry)
Course Code: Chem 3064
Credit Hour: 3
Lec.: 2 Hrs. , Tut. : 3 Hrs.
1
1. Introduction to Quantum theory
1.1. Historical Development of Quantum Mechanics
 In the late 17th century, Isaac Newton discovered classical
mechanics (Newtonian mechanics) that describes the laws of
motion of macroscopic objects.
 In the early twentieth century, it was found that classical
mechanics does not correctly describe the behavior of very
small particles in the order of 10-12 m diameter, such as the
electrons and the nuclei of atoms and molecules.
 The behavior of such particles is described by a set of laws
called quantum mechanics.
 The development of quantum mechanics began in 1900 with
Planck’s study of the light emitted by heated solids.
2
1. Introduction to Quantum theory
 Quantum chemistry applies quantum mechanics to problems
in chemistry.
 Physical chemistry uses quantum mechanics to :
 calculate the thermodynamic properties of gases such as the
entropy and heat capacity;
 interpret molecular spectra, thereby allowing experimental
determination of molecular properties such as bond length,
bond angle, dipole moment, barrier to internal rotation and
energy difference between conformational isomers;
 calculate properties of transition states in chemical reactions,
thereby allowing estimation of rate constants;
 understand intermolecular forces; and
 deal with bonding in solids.
3
1. Introduction to Quantum theory
 Organic chemistry applies quantum mechanics to
 estimate the relative stabilities of molecules;
 calculate properties of reaction intermediates;
 investigate the mechanisms of chemical reactions; predict
aromaticity of compounds and analyze nuclear magnetic
resonance spectra.
 Inorganic chemistry uses quantum mechanics to
 ligand field theory, an approximate quantum – mechanical
method,
 predict and explain the properties of transition metal complex
ions.
 Analytical chemistry uses spectroscopic methods extensively.
 The frequencies and intensities of lines in the spectrum can
be properly understood and interpreted only through the use
of quantum mechanics.
4
1. Introduction to Quantum theory
1.3. Description of wave motion
Traveling wave: The entire waveform moves in one direction
Standing (stationary) wave: A wave which oscillates in time but
whose peak amplitude profile does not move in space.
 The peak amplitude of the wave oscillations at any point in
space is constant with time.
Wave Equations
For a wave traveling in the +x direction, the one-dimensional
differential wave equation is:
𝜕2𝑦
1 𝜕 2𝑦
= 2 2
2
𝜕𝑥
 𝜕𝑡
Y = displacement = f(x, t),  = non-negative constant
5
1. Introduction to Quantum theory
Solution
𝑥
𝑦 = 𝐴𝑒
−𝑖𝜔 𝑡−
A = maximum displacement in the y direction
 = angular frequency
y is a complex quantity that contains real and imaginary parts.
1. Standing wave equation:
The variation of displacement with time is given by:
𝑦 = 𝐴𝑠𝑖𝑛𝑡 = 𝐴𝑠𝑖𝑛2𝜋𝑡
Where  = /2 = the frequency of the wave
6
1. Introduction to Quantum theory
2. Traveling wave Equation:
Distance – time relation is considered.
2𝜋𝑥
𝑦 = 𝐴𝑠𝑖𝑛2𝜋𝑡 = 𝐴𝑠𝑖𝑛
𝑐
Where, x = ct - the distance covered in time t at speed c.
Wavelength (): Distance traveled during a complete cycle
𝑐
=

Substitution yields,
2𝜋𝑥
𝑦 = 𝐴𝑠𝑖𝑛

7
1. Introduction to Quantum theory
1.4. Description of particle motion
1. Newton’s equation:
 In classical mechanics the motion of a particle is governed by
Newton’s second law:
𝐹x = 𝑚𝑎 =
𝜕2𝑥
𝑚 2
𝜕𝑡
𝐹y = 𝑚𝑎 =
𝜕2𝑦
𝑚 2
𝜕𝑡
• Change of motion is proportional to the applied force and takes
place in the direction of the force.
• If all of the forces acting on the particle are accounted for, an exact
energy and trajectory for the particle can be determined.
 For a projectile motion
𝑥 𝑡 = 𝑥𝑜 + (o𝑐𝑜𝑠)𝑡 𝑦 𝑡 = 𝑦𝑜 + o𝑠𝑖𝑛 𝑡 − 2𝑔𝑡2
Where  = The angle from the horizontal, o initial velocity, xo, yo =
initial projectile, g = gravitational force
8
1. Introduction to Quantum theory
2. Hamiltonian function
 In this approach, the Hamiltonian, H, is obtained from the kinetic
energy, T, and the potential energy, V, of the particles.
H = T + V(x)
22
1
1
𝑚
1 2
2
𝑇 = 𝑚 =
=
𝑃
2
2 𝑚
2𝑚
Where  is the velocity and P is the momentum of the particle = m 
 The potential energy of the particles will depend on the positions of
the particles.
𝜕𝐻
𝜕𝑥 P
=−
𝑑𝑃
𝑑𝑡
𝜕𝐻
𝜕𝑃 𝑥
=
𝑑𝑥
𝑑𝑡
Simultaneous solution of these differential equations will result in the
trajectory of a particle.
𝑥 𝑡 = 𝑥o + (o𝑐𝑜𝑠)𝑡
𝑦 𝑡 = 𝑦o + o𝑠𝑖𝑛 𝑡 − 2𝑔𝑡2
9
1. Introduction to Quantum theory
Classical physics:
(1) predicts a precise trajectory for particles, with precisely specified
locations and momenta at each instant, and
(2) allows the translational, rotational, and vibrational modes of
motion to be excited to any energy simply by controlling the forces
that are applied.
If a particle of mass m is initially stationary and is subjected to a
constant force F for a time , then its kinetic energy increases from
zero to:
𝐹22
𝐾. 𝐸. =
2𝑚
 Because the applied force and the time for which it acts may be
varied at will, the solution implies that the energy of the particle
may be increased to any value.
10
2. Experimental Foundation of Quantum Theory
2.1 The Blackbody Radiation
Blackbody:- An object or system which absorbs all radiation incident
upon it and re-radiates energy which is characteristic of the
radiating system
E.g. A hallow metal with a pinhole
 The radiation emitted is called blackbody radiation.
 The emitted light does not dependent upon the type of radiation
which is incident upon the object
 When the blackbody is heated, it emits light of different
frequencies, i.e.,
 A hot object emits electromagnetic radiation.
11
Blackbody Radiation
 As the temperature of the blackbody increases,
 the body radiates more and
 the peak in the energy output shifts towards a shorter wavelength
(to higher frequency).
Fig.2.1. The distribution of the intensity of the radiation emitted by a blackbody
Vs. wavelength for various temperatures
12
Blackbody Radiation
 The intensities of various wavelengths of radiation emitted
by a blackbody depend only on its temperature.
 The higher the temperature of a blackbody, the more
energy emitted in each band of wavelengths.
Stefan – Boltzmann law
 It shows the quantitative dependence of the energy emitted
(the area under each curve) on the temperature of the
blackbody
 The total energy emitted () by a blackbody is proportional
to T4 (T in Kelvin).
𝜌 = 𝜍𝑇4
σ = the Stefan - Boltzmann constant = 5.67  10-8 J m-2K-4s-1.
13
Blackbody Radiation
Wein displacement law
 It is the second quantitative observation of the blackbody
radiation experiment
T MAX  Cons tan t
1
 C
5
 0.29 cmK
C = 1.45 cm K
max is the wavelength at which the energy emitted is
maximum.
14
Blackbody Radiation
Theoretical Explanations to explain blackbody radiation results
1) Rayleigh – Jeans law
 Basis: classical mechanics
 Assumption:
The energies of the electronic oscillators
responsible for the emission of the radiation can have any value
dE   d  
where,
8 kT

4
d
dE is the radiant energy density between  and  + d of an
oscillator
 = A proportionality constant between dE and d
k = Boltzmann constant = 1.381  10-23 J K-1
15
Blackbody Radiation
Fig 2.2. The Rayleigh-Jeans law predicts an infinite energy density
at short wavelengths.
16
Blackbody Radiation
 At longer wavelengths, Rayleigh-Jeans law reproduces the
experimental data
 In the longer wavelength region, as λ decreases, the energy
density increases
 At smaller wavelengths, however, the Rayleigh-Jeans law,
deviates from the experimental results:
 predicts that the radiant energy density increases without
limit at shorter wavelengths even at low temperatures.
 This absurd result, which implies that a large amount of
energy is radiated in the high-frequency region of the
electromagnetic spectrum, is called the ultraviolet
catastrophe.
 This was the first failure to explain an important naturally
occurring phenomenon based on classical physics
17
Blackbody Radiation
2) Planck’s distribution law
Assumptions:
 The radiation emitted by the blackbody was caused by the
oscillations of the electrons in the constituent particles of the
material body
 The energies of the oscillator were discrete and had to be
proportional to an integral multiple of the frequency given by the
equation:
𝐸 = 𝑛ℎ
Where
n = positive integer
 = radiation’s frequency
h = Planck’s constant = 6.626  10-34 Js
18
Blackbody Radiation
 Planck’s work marked the beginning of quantum mechanics.
Planck’s hypothesis:
 Only certain quantities of light energy can be emitted, i.e., the
emission is quantized.
 Using this quantization of energy and statistical thermodynamic
ideas, Planck derived the equation for the energy density
between λ and λ + d λ which is called Planck distribution law
dE   d  
8 hc
 5 (e
hc
 kT
1)
d
19
Blackbody Radiation
Figure 2.3. Energy density distribution as a function of wavelength using the
Planck’s distribution law.
Exercise: Show that d has a unit of energy per unit volume, J m-3.
20
Blackbody Radiation
 For very short ,


hc
1
e kT 1  0
Therefore  approaches zero as  approaches zero.
 This agrees with the experimental observation.
 For longer , it can be shown mathematically that
hc
1 and
kT
e
hc
 kT
1 hc
kT
This leads to:
d 
8 kT

4
d
 Planck’s distribution equation reduces to the Rayleigh – Jeans
law
21
Blackbody Radiation
 At maximum energy density,
𝑑𝜌
𝑑
=0
Differentiating  with respect to  leads to
ℎ𝑐
𝑇max =
= 0.29𝑐𝑚𝐾
4.965 𝐾
 The Planck’s equation reduces to the Wein displacement law
 Integration of the energy density over all the wavelength range gives,

8 5 k 4 4
   (  ) d   2 3 T   T 4
15c h
0
 The Planck’s equation reduces to the Stefan – Boltzmann law
22
Blackbody Radiation
 Planck’s quantum theory of blackbody radiation explains the
experimentally based laws of Stefan – Boltzmann and Wein.
 Therefore, the energy emitted or absorbed by a blackbody is
quantized.
23
2.2. Photoelectric Effect
 Metals eject electrons instantaneously when they are
irradiated by light of a certain minimum frequency called a
threshold frequency.
Observations :
1. Below the threshold frequency of the incident light, no
photoelectrons are ejected.
2. Above the threshold frequency, the number of electrons
ejected is directly proportional to the intensity of the light.
3. As the frequency of the incident light increases, the maximum
24
kinetic energy of the photoelectrons increases.
Photoelectric Effect
 The energy of a wave is proportional to its intensity and it is not
related to its frequency.
 So the electromagnetic wave picture of light expects the kinetic
energy of an emitted photoelectron to increase as the light
intensity increases but would not change as the light frequency
changes.
 Instead, it is observed that the kinetic energy of the emitted
electron is independent of the light’s intensity but increases as
the light’s frequency increases.
 Albert Einstein showed that these observations could be
explained by viewing light as composed of particle-like entities
called photons, with each photon having an energy given by
𝐸photon = ℎ
where Ephoton represents particle property and  represents the
wave property.
25
Photoelectric Effect

When an electron in the metal absorbs a photon,
 part of the absorbed photon energy is used to overcome the
forces holding the electron in the metal, and
 the reminder appears as kinetic energy of the electron after it
has left the metal.
1
1
2
ℎ =  + 𝑚 = ℎo + 𝑚2
2
2
𝐾. 𝐸. = ℎ − ℎo = ℎ  − 𝑜
Where
 = ho - is the metal’s work function:- the minimum energy
needed by an electron to escape the metal
½ m2 = the kinetic energy of the electron after it has left the
metal and
0 = threshold frequency:- the minimum frequency needed to free
the electron from the nuclear attraction of the metal atom.
26
Photoelectric Effect
K.E.


Figure 2.4. The photoelectric effect: The kinetic energy of
photoelectrons as a function of incident light frequency
27
Photoelectric Effect
 The photoelectric effect shows that light can exhibit
particle-like behavior in addition to the wave-like behavior
that it shows in diffraction experiments.
Example 1
Given that the work function of Na metal (Na) is 1.82 eV,
calculate the threshold frequency, o, for Na.
(1 eV = 1.60210-19 J)
Solution

0 =
ℎ
1.821.60210−19𝐽
14𝑠−1
o =
=
4.410
6.62610−34𝐽𝑠
28
Photoelectric Effect
Example 2
• When lithium is irradiated with light of  = 300 nm, the kinetic
energy of the ejected electrons is 2.935  10-19 J. Calculate (a) the
work function of lithium and (b) the threshold frequency.
(c = 3108 ms-1)
Solution
(a)
=
𝑐

𝑐
 = ℎ − 𝐾. 𝐸. = ℎ − 𝐾. 𝐸.

8 𝑚𝑠−1
310
−19𝐽
= 6.62610−34𝐽𝑠
−
2.93510
30010−9𝑚
= 3.69110−19𝐽 = 2.3 𝑒𝑉
(b)o =

ℎ
=
3.69110−19𝐽
6.62610−34𝐽𝑠
= 5.571014𝑠−1=5.571014Hz
29
The Compton Effect
2.3. The Compton Effect
• It showed particle-like behavior of radiation by making use of the
scattering experiment.
3
2
X – rays, i

Scattered photons
1
i
Fig.2.5. X-rays striking a carbon surface.
30
The Compton Effect
 A.H. Compton observed the scattering of x-rays from
electrons in a carbon target and found scattered x-rays with
a longer wavelength than those incident upon the target.
 There is only one value of  shift at a given scattering angle
and the shift of the wavelength increased with scattering
angle.
 Accordingly, i < 1< 2 < 3.
 The shift in the scattering angle is given by the Compton
formula:
31
The Compton Effect
 According to the Compton scattering experiment, the radiation
(photon) is a stream of particles, a basis for a quantum wave
theory.
 If a wave acts as a particle, except the absence of rest mass, it
should have a momentum, P.
E photon  Pc
Using Planck’s relation,
Pc  h
h h
P

c 
Therefore,
 the momentum transfer takes place in a discrete manner, not in a
continuous way.
32
2.4 Line Spectra of Atoms
 Every gaseous atom, when heated to high temperatures, emits
electromagnetic radiation of characteristic wavelengths.
 Each atom has a characteristic emission spectrum.
 The light emitted is not continuous, but rather a line spectra.
 Each line represents light of a certain wavelength.
 Energy is associated with each wavelength.
Therefore,
 the energy emitted or absorbed by atoms is quantized.
33
Line Spectra of Atoms
 For H-like atoms, the wavelength of the light that can be
emitted obeys the Rydberg equation:
 1
1 
 R 2  2 

 n1 n2 
1
where n2 > n1, being positive integers and
R is the Rydberg constant = 109678 cm-1
Example
For the Balmer series of the hydrogen atomic spectrum,
calculate (a) the wavelength of the first line
(b) what happens to wavelength of the emitted light as the
value of n increases?
34
Line Spectra of Atoms
Solution
(a) For the Balmer series, n1= 2 and for the first line n2 = 3.
 1
1 
 1 1  1
 R  2  2   109678  2  2  cm  1.523 104 cm1

n2 
2 3 
 n1
1
= 6.57  10-5 cm = 657 nm
(b) Let n2 goes to infinity, .
 1
1 
1  1
1
 R  2  2   109678  2  2  cm  2.742 104 cm1

n2 
2  
 n1
1
  3.65 105 cm1  365nm
Result: The wavelength of the emitted light decreases as the value
of n increases.
35
Wave Properties of Particles
2.5. The wave properties of particles
De Broglie hypothesis:
 All moving particles have wave properties besides their
particle properties.
 An electron of mass m and speed  would have a
wavelength,
h h


m P
Where, P is the linear momentum.
 De Broglie arrived at the above equation by reasoning in
analogy with photons.
36
Wave Properties of Particles
The energy of a photon (a particle) can be expressed by
E photon  mc 2
Where c is the speed of light and m is the particle’s relativistic mass.
Using,
E photon  h
and equating the above two equations gives
mc 2  h  h
Or
And
Or
mc 
c

h

m  P 
h

P
h

37
Wave Properties of Particles

The de Broglie hypothesis was confirmed by Davisson and
Germer, by reflecting electrons from metals and observing
diffraction effects.
 The “electron waves” were observed to have wavelengths
related to electron momentum in just the manner proposed by
de Broglie.
 A higher momentum corresponds to a shorter wavelength.
1
m2 2 P 2
2
kinetic energy  T  m 

2
2m
2m
P  2mT
E = T + V where E is the total energy and V is the potential energy,.
We can write the de Broglie wavelength as

h
2m  E  V 
38
Wave Properties of Particles
 The equation is useful for understanding the way in which 
will change for a particle moving with constant total energy
in a varying potential
 If the particle enters a region where its potential energy
increases (e.g., an electron approaching a negatively charged
plate),
 E – V decreases and  increases, i.e.,
 the particle slows down,
 so its momentum decreases.
39
Wave Properties of Particles
 Electrons behave in some respects like particles and in other
respects like waves.
 This is a particle – wave duality of matter (of light as well).
 How can an electron be both a particle, which is a localized
entity, and a wave, which is a non-localized entity?
 The answer is that an electron is neither a particle nor a
wave.
 Both photons and electrons show an apparent duality,
 But, they are not the same kinds of entities.
 Photons always travel at a speed c and have zero rest mass
 Electrons always have  < c and have a non-zero rest mass.
40
Wave Properties of Particles
Example
Calculate the de Broglie wavelength of an electron traveling at
2.998  106 m s-1.
Solution
me = 9.109  10-31 kg
The momentum of the electron is given by



P  m  9.109 10-31 kg 2.998 106 m s-1  2.73 1024 kg m s 1
The de Broglie wavelength of the electron is
h
6.626 1034 J s
10
 

2.43

10
m
24
1
P 2.73 10 kg m s
 The wavelength corresponds to the wavelength of X-rays.
 Thus, the equation predicts that electrons can be observed to
act like X-rays
41
2.6. Heisenberg Uncertainty Principle
 The wave – particle duality of microscopic particles imposes a
limit on our ability to measure simultaneously the position
and the momentum of such particles.
 The more precisely we determine the position, the less
accurate is our determination of momentum and the vice
versa.
 This limitation is the Heisenberg uncertainty principle.
 The principle is demonstrated by variables such as
 linear position - linear momentum,
 energy - time and
 angular momentum - angular position
where the product of the variable pairs have a dimension of h.
42
Heisenberg Uncertainty Principle
X PX  h
where X and PX are respectively a measure of the
uncertainty in linear position and linear momentum.
Example
The life-time of an excited state of a molecule is 210-9s. What
is the uncertainty in its energy in J? In cm-1?
Solution
The Heisenberg uncertainty principle gives, for the minimum
uncertainty,
E t  h
6.626 1034 Js
25
2
1
E 

3.3

10
J

1.66

10
cm
2 109 s
43