Chapter 8: Quantum Theory

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Chapter 8 Quantum Theory

• Failures of classical physics
• Black-body radiation
• Heat capacities
• Photoelectric effects
• Diffraction of electrons
• Atomic spectra
• Quantum mechanics
• Schroedinger equation
• Born interpretation
• Uncertainty principle

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• Applications of quantum mechanics
• Translation motion a particle in a box
• Rotation motion a particle on a ring
• Vibration motion the harmonic oscillator

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• Failures of classical physics

• Around the turn of 19th century, 3 principles of
  classical physics were challenged by new
  findings.
• A particle travels in a trajectory
• -location and velocity (momentum) of a particle
  are known at all time (Newtons law)
• A particle can possess any arbitary energy
• -There exists continuous excited states
• Waves and particles are distinct concepts

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• Black-body radiation

Black-body is a body that can emit and absorb
all frequencies of electromagnetic
radiations. Fig 12.3 p 287, Atkins
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All substances contain black-bodies. It was
found that radiation emitted or absorbed by
black-body changes with temperature Iron turns
from red to white (redblue) when heated. Fig
12.3 p 287, Atkins
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Not all frequencies are emitted or absorbed with
the same energy density. There exists a maximum
corresponding to a particular wavelenght, lmax,
in the curve. This lmax changes with
temperature As temperature is raised, lmax moves
toward smaller wavelenght (from red to blue)
called blue shift Wiens displacement
law c2, second radiation constant, 1.44 cm K
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Another features of black-body radiation given by
Josef Stefan called Stephan-Boltzmann law is
E aT4
E ? total energy density or M ?T4
M emittance i.e. brightness of emission ?
5.67 x 10-8 wm-2k-4
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Rayleigh-Jeans proposed that black-body consists
of oscillators which can emit or absorb light at
any frequency Using classical physics and
equipartition principle, they arrived at
Rayleigh-Jeans Law
dE ? d? where ? 8pkT/ ?4
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Rayleigh-Jeans law suggested there can be
emission of very short wavelength even at room
temperature. In other word, there should in fact
be no darkness. (objects should glow in the
dark.) Failure of Reyleigh-Jeans law are called
UV-catastrophe Planck took Rayleigh-Jeanss
idea but instead of allowing oscillators to be
able to emit or absorb light at any frequency
each oscillator possesses only discrete values of
energy
E nhn
h Plancks constant 6.626 x 10-34
Js Plancks hypothesis also implies that energy
is quantized and is regarded as Quantum theory
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By using statistical interpretation Planck
proposed that
dE ? d?
where
For very large ?
Rayleigh-Jeans Law
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By integrating ? from 0 to ?, one obtains
2p5k4
-Stefan-Boltzman law
By differentiating
-Wiens displacement law
5k
c2 1.439 cm K
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Example 8.1 Find the surface temperature of sun
where the maximum emission occurs at 490 nm.
Wiens displacement law
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Example 8.2 Calculate the number of photons
emitted by a 100 W yellow lamp in 10.0 s. Take
the wavelength of yellow light as 560 nm and
assume 100 percent efficiency.
Energy emitted by lamp Eemitted
(Power)(time) (100 W) (10.0 s) 1,000 J
Photon energy Ephoton hn h
c/l (6.626x10-34 J s)(2.998x108
ms-1) (560x10-9 m) 3.547x10-9 J
Nphoton Eemitted/Ephoton 1,000/3.547x10-9
J 2.82x1021
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• Heat capacity

• In 1819, Dulong Petit proposed that for all
  monoatomic solids, heat capacities are about 25
  J/Kmol (3R) at all range of temperature
• Dulong Petit arrived at this conclusion by
  assuming the concept of classical physics
• heat capacities of solid is the energy used to
  oscillate or vibrate atom at the lattice position
• Each atom can vibrate in 3 orthogonal directions
• For N atoms solid, their will be 3N motions
• From equipartition theory, each motions uses
  energy of kT

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The total vibrational energy U 3NkT For 1
mole of atoms
3NA kT
25 J/Kmol
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Later on with the advance in measurement, it was
found that at low temperature all substances have
heat capacities lower than 25 J/Kmol. Diamond
has Cm 6.1 J/K mol at 25C
Dulong-Petit
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Albert Einstein borrowed Plancks idea. He
proposed that each atom can possess only a
discrete amount of energy which is an integral
multiple of hn
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Debye corrected Einsteins formula by allowing
atom to oscillate at more than one frequency value
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• Photoelectric effects

In late 1800s, it was demonstrated that electrons
can be ejected from surface of certain metals
when exposed to light of at least a certain
minimum frequency.
This phenomena is called photoelectric effect
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In 1905, Albert Einstein using Plancks quantum
concept was able to explain the photoelectric
effect
Einsteins proposal 1) Light is composed of
light particle called photon. 2)
Each photon has discrete value of energy E
hn h c/l 3) Let F, work function of
metal, be the minimum energy that required to
knock electrons out of metal (binding energy,
each metal has different F).
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If photon energy is equal to F, electrons will be
ejected from metal when exposed to light.
If photon energy is greater than F, electrons
will be ejected from metal and carry kinetic
energy EK when exposed to light.
EK hn - F
EK could be measured, F could then be determined
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Example 8.3 The work function for metallic
caesium is 2.14 eV. Calculate the kinetic energy
and the speed of the electron ejected by light of
wavelength 250 nm.
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• The diffraction of electrons

• In 1925, Davisson and Germer observed the
  diffraction of electrons by a crystal
• Davisson-Germers experiment has been repeated
  using other particles such as H2
• particles have wave-like properties
• photoelectric effect suggests that
• waves have particle-like properties
• This phenomena is called
• wave-particle duality

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In 1924, de Broglie suggested that any particle
traveling with a linear momentum, p , should have
a wavelength ? according to ? h/p faster
object will have smaller wavelength heavier
object will have smaller wavelength
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Example 8.4 Estimate the wavelength of electrons
that have been accelerated from rest through a
potential difference of 1.00 kV.
Example 8.4 Estimate the wavelength of electrons
that have been accelerated from rest through a
potential difference of 1.00 kV
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• Atomic spectra

Radiation is emitted (or absorbed) at a series of
discrete frequencies
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This phenomena can be explained if we assume
energy of atomic state is quantized and light
omits or absorbs as electron transfer from lower
to higher state or vice versa and releases photon
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Quantum Mechanics
Failures of classical physics made scientists
realized that new kind Of mechanics was needed.
This new mechanics should not give accurate
trajectories of particles. This new mechanics was
later called Quantum mechanics
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Uncertainty principle
In 1927, Heisenberg pointed out that for very
small particles (quantum particles) like
electrons it is not possible to measure
accurately their positions (x) and linear
momentum (px) at the same time. This statement is
now known as the uncertainty principle.
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For us to find the position of electron, light
(h?) must hit electron and reflect back to the
lens of the microscope.
The accuracy of measurement can be determined
from
where ? is the wavelength of light and 2sin? is
the dimension of the lens
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From de Broglies relation,
p
p
As photon hits electron, momentum is
transferred. The leaving electron carries
momentum of mv while leaving photon has momentum
of h/? (or energy h?)
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From law of conservation
incoming momentum outgoing momentum
Along x-axis
(1)
Along y-axis
(2)
From (1)
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Generally ? gt ? but for simplicity assuming ? ?
? then
px
90 - ? lt ? lt- 90 ?
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Thus
lt px lt
lt px lt
The accuracy of measurement of momentum is then
Dpx
The accuracy of measurement of position is
DxDp
DxDp
uncertainty principle
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-Schrödinger equation
Wave concept was introduced to explain quantum
particle. Schrödinger adapted Maxwells equation
for electromagnetic radiation for describing
behavior of quantum particle.
Maxwells equation
Wamplitude of wave
where ? is de Broglie wavelength of the particle
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velocity of the particle can be given by
Replacing waves amplitude by wave function ?
(psi) which gives the information about
positions of electron
?(x,y,z,t) is not trajectory, it gives
information about position but not exactly like
the amplitude of the wave
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For localized or standing wave
Thus
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Kinetic energy Tp2/2m and total kinetic
potential (ETV)
p2
-2m4p2
-8p2m
8p2m
? Laplacian operator
operator
Schrodinger equation or SE
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SE is differential equation SE is an eigen-value
equation E is an eigen-value which represents
energy of particle Solution of SE is wave
function ?
SE can be solved using methods of differential
equation
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Born Interpretation
In wave theory, square amplitude of wave is
interpreted as intensity. Using the same
analogy, Born gave interpretation of wave
function as
The probability of finding a particle in a small
region of space dV is proportional to l?l2dV,
where ? is the value of wave function in the
region. In other word, l?l2 is the probability
density
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Applications of quantum mechanics

• Solution to SE

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For 1-dimensional system
2
gt Free electron V 0
2
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y
2
-k2
i2-1
-k2
-k2
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gt 1D Square potential well
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Area 1 prob.
Probability
III
II
I
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We can substitute
At x 0
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yII yIII
If D 0, trivial solution is obtained. Then D
must not be equal to 0 and sinkL must be equal to
0. sinkL 0 only when kL np
k np/L
npx
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Using Borns interpretation
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From cos (AB) cosAcosB - sinAsinB cos2x
cosxcosx - sinxsinx cos2x
- sin2x (1-sin2x) - sin2x
cos2x 1- 2sin2x sin2x
(1 cos2x)/2
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The lowest energy is achieved when n 1, and
equal to h2/8mL2. The lowest energy is non-zero
called zero-point energy. Excitation or
transition to adjacent wall could be calculated as
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gtExpanding to 3D particle in a
box
c
b
a
Particle in a box is a model representing
translation motion
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Particle on a ring
Particle is held on the ring, hence travels in 2
dimension with V0 Thus
To solve this equation, one must transform from
cartesian (x,y) to polar coordinate (r,?)
r is fixed
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I ?moment of inertia mr2
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From boundary condition
Thus, 2m must be positive or negative even
integer
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From classical interpretation
Angular momentum j pr
Linear momentum
p2
from Schrödinger equation
From deBroglies relation ph/?
Particle on a ring represents rotation motion
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Harmonic oscillator
V 1/2kx2
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(1)
(2)
(a - b2x2)
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(3)
(4)
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which
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The first 5 polynomial are
H0(x) 1
H1(x) 2x
H2(x) 4x2-2
H3(x) 8x3-12x
H4(x) 16x4-48x212
If we set 2n, then (4) is the
Hermites equation.
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Equation (4) has Hermite polynomials for
solution. The wave function for the harmonic
oscillator is then given by,
v

N0,N1 are normalizing factor
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Take definition of a and ß
( )
From classical physics (vibration frequency)
Thus
v 0,1,2,
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Harmonic oscillator represents mode of
vibration
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Example 8.5 Show whether (a) or (b)
is eigenfunction of the operator d/dx and
find the corresponding eigenvalue
f(x) would be an eigenfunction of an operator
only when
where is an eigenvalue
(a)
( )
(b)
( )
is an eigenfunction of d/dx with
eigenvalue a is not an eigenfunction of
d/dx
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Example 8.6 Proof that sinx and sin2x
which are eigenfunction of d2/dx2 are mutually
orthogonal.
Orthogonal means
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(2x x)-
eigenfunctions of the same operator must be
orthogonal
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Example 8.7 Find the normalizing constant for
ground state and first excited state of harmonic
oscillator
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even n0
0
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y1(x) y1(x) dx
even n1
0