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INTRODUCTION TO QUANTUM PHYSICS
TOPICS

Blackbody Radiation & Plank’s Hypothesis
The Photoelectric Effect
The Compton Effect
Photons and Electromagnetic Waves
The Quantum Particle
The Double-Slit Experiment
Revisited
• The Uncertainty Principle

•
•
•
•
•
•

Text Book
PHYSICS for Scientists and Engineers
with Modern Physics (6th ed)
By Serway & Jewett
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INTRODUCTION TO QUANTUM PHYSICS

INTRODUCTION
Failure of classical mechanics
Brief summary of chapter 40
of the text book

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BLACKBODY RADIATION & PLANK’S HYPOTHESIS

INTRODUCTION
Origin of thermal radiation –
the classical view point

Concept of oscillators

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BLACKBODY RADIATION & PLANK’S HYPOTHESIS

Definition of a Black-Body
Black-Body Radation Laws
1- The Stefan-Boltzmann Law
2- The Wien‘s Displacement Law
3- The Rayleigh-Jeans Law
4- The Planck Law
Application for Black Body
Conclusion
Summary
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BLACKBODY RADIATION & PLANK’S HYPOTHESIS

Definition of a black body
An object that absorbs
all incident radiation
...

⇒None of the
⇒
incident radiation
escapes
...
This causes a heating of the cavity walls
...
Some of
the energy from these standing waves can leave
through the opening
...

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BLACKBODY RADIATION & PLANK’S HYPOTHESIS

•The black body is an ideal absorber of
incident radaition
...

•Emitted radiation from a blackbody does not
depend on the material of which the walls are
made
...

At about 6000 K (not
shown in fig
...

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BLACKBODY RADIATION & PLANK’S HYPOTHESIS
Basic Laws of Radiation
1) All objects emit radiant energy
...
The total power of
the emitted radiation increases with
temperature
...

3) The peak of the wavelength distribution
shifts to shorter wavelengths as the black
body temperature increases
This is Wien’s Law
...


P = σ Ae T 4
P = power radiated from the surface of the
object(W)
T = temperature (K)
σ= 5
...

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BLACKBODY RADIATION & PLANK’S HYPOTHESIS

Black-Body Radiation Laws (2)
Wien’s Displacement Law
...
898 × 10−3 m
...

T- equilibrium temperature of the blackbody
...

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BLACKBODY RADIATION & PLANK’S HYPOTHESIS
SJ: P-SE 40
...


Find the peak wavelength of the blackbody
radiation emitted by each of the following
...
The human body when the skin temperature is
35°C
B
...
The Sun, which has a surface temperature of
about 5800 K
...


I (λ , T ) =

2 πck B T
λ4

This law tries to explain
the distribution of energy
from a black body
...

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BLACKBODY RADIATION & PLANK’S HYPOTHESIS
kB – Boltzmann's constant
T- equilibrium blackbody
temperature
c- velocity of light
...

* It predicts an energy
output that diverges towards
infinity as wavelengths grow
smaller
...

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BLACKBODY RADIATION & PLANK’S HYPOTHESIS
Black-Body Radiation Laws (4)
The Planck Law

I(λ, T)

=

2 π hc 2
λ5
e

1
hc
λk B T

- 1

This law too explains the distribution of
energy from a black body
...
Only the extra quantity
(compared to the Rayleigh-Jeans Law)
coming here is the constant known as
Plank’s constant introduced by Max Plank in
this revolutionary theory
...
In short the law fitted the
experimental data for all wavelength
regions and at all temperatures
...

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BLACKBODY RADIATION & PLANK’S HYPOTHESIS

1) The energy of an oscillator can have only
certain discrete values E n

E n = nhf
where n is a positive integer called a
quantum number, f is the frequency of
oscillation, and h is a constant called Planck’s
constant
...
Each discrete energy value
corresponds to a different quantum state,
represented by the quantum number n
...

Difference in energy will be integral multiples of hf
...

to n = ∞
E

n

Figure shows allowed energy
levels for an oscillator with
frequency f, and the allowed
Transitions
...

i
...
I (λ, Τ → 0 as λ → 0
...
626 × 10−34 J
...
e
...

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BLACKBODY RADIATION & PLANK’S HYPOTHESIS
Summary
The characteristics of blackbody radiation cannot
be explained using classical concepts
...

In Plank’s model, radiation is emitted in
single quantized packets whenever an oscillator
makes a transition between discrete energy states
...
2

The Quantized Oscillator

A 2
...
The
spring is stretched 0
...

A
...

B
...

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BLACKBODY RADIATION & PLANK’S HYPOTHESIS

SJ: P-SE 40
...


C
...
4 x 1033 state to the state
corresponding to n = 5
...
By how
much does the energy of the oscillator change n
this one-quantum change
...
1 P-1 The human eye is most
sensitive to 560 nm light
...
1 P-3 A blackbody at 7500 K
consists of an opening of diameter 0
...
Find the number of photons
per second escaping the hole and having
wavelengths between 500 nm and 501 nm
...
1 P-5 The radius of our Sun is 6
...
77 x 1026 W
...
(b) Using
the result, find λmax for the Sun
...
1 P-7 Calculate the energy in
electron volts, of a photon whose frequency is
(a) 620 THz, (b) 3
...
0 MHz
...

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BLACKBODY RADIATION & PLANK’S HYPOTHESIS

SJ: Section 40
...
An FM radio transmitter
has a power output of 150 kW and operates
at a frequency of 99
...
How many
photons per second does the transmitter
emit?

Assignment: Try to answer the questions in page
no
...


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THE PHOTOELECTRIC EFFECT

•Introduction
•What is Photoelectric Effect
•Apparatus for studying Photoelectric Effect
•Experimental Observations
•Classical Predictions
•Clash between Classical predictions
& Observed Experimental results
• Einstein’s model of the Photoelectric Effect
• Explanation for the observed features of PE
•Application
•Conclusion
•Summary
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THE PHOTOELECTRIC EFFECT

What is Photoelectric Effect?
Apparatus for studying Photoelectric Effect

T

MIT- MANIPAL

T – Evacuated
glass/ quartz tube
E – Emitter Plate/
Photosensitive
material /Cathode
C – Collector Plate /
Anode
V – Voltmeter
A - Ammeter

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THE PHOTOELECTRIC EFFECT

Experimental Observations
1
...

2
...
Maximum kinetic energy of the
photoelectron is independent of light
intensity
...
Electrons are emitted from the surface of the
emitter almost instantaneously
...
No electrons are emitted if the incident light
frequency falls below a cutoff frequency
...
Maximum kinetic energy of the
photoelectrons increases with increasing
light frequency
...
If light is really a wave, it was thought that
if one shine light of any fixed wavelength, at
sufficient intensity on the emitter surface,
electrons should absorb energy continuously
from the em waves and electrons should be
ejected
...
As the intensity of light is increased (made it
brighter and hence classically, a more
energetic wave), kinetic energy of the
emitted electrons should increase
...
Measurable/ larger time interval between
incidence of light and ejection of
photoelectrons
...
Ejection of photoelectron should not
depend on light frequency
5
...


In short experimental results
contradict classical predictions
...

The energy E, per packet depends on frequency f
...

More intense light corresponds to more photons,
not higher energy photons
...
Each
photon carries a momentum p = E/C
...

It is the minimum energy with which an
electron is bound in the metal
...

1
...

2
...

3
...

4
...

If the incident frequency f is less than fc , no
emission of photoelectrons
...

And this work won Einstein his Nobel Prize in 1921
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THE PHOTOELECTRIC EFFECT
Application of photoelectric effect
Explain the device, theory, and its working
Photomultiplier tube

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THE PHOTOELECTRIC EFFECT
Summary
Einstein successfully extended Plank’s quantum
hypothesis to explain photoelectric effect
...

The maximum kinetic energy Kmax of the ejected
photoelectron is

Kmax = hf - φ
Where φ is the work function of the photocathode
...
3 The Photoelectric Effect for Sodium
A sodium surface is illuminated with light having
a wavelength of 300 nm
...
46 eV
...
The maximum kinetic energy of the ejected
photoelectrons and
B
...

SJ: Section 40
...

Molybdenum has a
work function of 4
...
(a) Find the cut off
wavelength and cut off frequency for the
photoelectric effect
...
2 P-14
...
60 x 105
m/s when light with a wavelength of 625 nm is used
...
2 P-16
...
48 V larger compared to that in
another metal
...
0 % smaller than for the
second metal, determine the work function for
each metal
...
2 P-17
...
When
green light from a mercury lamp (λ = 546
...
376 V reduces
the photocurrent to zero
...
5 nm)?
λ
Assignment: Try to answer the questions in
page no
...

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THE COMPTON EFFECT

•Introduction
•What is Compton Effect
•Schematic diagram of Compton’s apparatus
•Experimental Observations
•Classical Predictions
•Explanation for Compton Effect
• Derivation of the Compton Shift Equation
...

Expressions for relativistic momentum and relativistic
kinetic energy of a particle
Relativistic expression for
the momentum of a particle P = γ m v
where m = mass of the particle, v = speed of the
particle & c = speed of light in vacuum
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THE COMPTON EFFECT
1
and
γ =
v 2
1 c 2
And finally, the relativistic kinetic energy of a particle is
K = (γ -1) m c2
...
Also
MIT- MANIPAL

c= λf

E
hf
h
p=
=
=
c
λf
λ
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THE COMPTON EFFECT
What is Compton Effect ?
Compton (1923)
measured intensity of
scattered X-rays from
solid target (scattering of
X-rays from electrons), as
function of wavelength
for different angles
...

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scattered beam

45

THE COMPTON EFFECT

Classical Predictions
Oscillating electromagnetic waves of frequency f0
incident on electrons should have two effects
...

Because different electrons will move at different
speeds after the interaction, depending on the amount
of energy absorbed from em waves, for a particular
angle of incidence of the incoming radiation, the
scattered wave frequency should show a distribution of
Doppler- shifted values
...

The wavelength is
measured with a
rotating crystal
spectrometer using
Bragg’s law
...


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THE COMPTON EFFECT
Experimental Observations
Contrary to the classical predictions where X-rays are
treated as waves, in Compton experiment, at a given
angle, only one frequency for scattered radiation is
seen
...

Compton could explain the experimental result by
taking a “billiard ball” type collisions between particles
of light (X-ray photons) and electrons in the material
...
The shifted peak at λ’ is
caused by the scattering of X-rays from free electrons
...

h
m ec

m ec

= 0
...

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THE COMPTON EFFECT

Derivation of the Compton Shift Equation
Photon is treated as a particle having energy E = hf =
hc/λ and zero rest energy
...

In the scattering
process, the total
energy and total
linear momentum of
the system must be
conserved
...

Substituting for Ke we get

hc
=
λ0
MIT- MANIPAL

hc
+ ( γ - 1 ) m ec
λ'

2

BE-PHYSICS-INTRODUCTION TO QUANTUM PHYSICS-2011-12

(40
...


x component :

y component :

where
h/λ0 = p0
λ
h/λ’ = p’
λ

h = h c
λ
λ
0
o
'
s s
0 = h i

'

λ

n

c

θ + γ m ev o
s

φ

s

θ -γ m ev i

(40
...
14)

n

is the momentum of the incident photon
is the momentum of the scattered photon
γ m e v = P e is the momentum of the scattered
electron
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THE COMPTON EFFECT
Rewriting the above equations as

c

p 0 - p' o

x component :

e

p

φ

o

(40
...
14)
θ= pe i
φ
n
n
Squaring and adding the above equations give
2

2

2

(

e
- 2 p0 p c θ +
=
' o
p
p
s
'
Rewriting equation 40
...
e
...
Write the resulting
equation in terms of respective wavelengths, we
get the Compton shift equation as

h
λ' - λ0 =
(1- cosθ)
mec

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THE COMPTON EFFECT
Summary
X-rays are scattered at various angles by electrons in
a target
...
Classical
physics does not predict the correct behaviour in this
effect
...

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THE COMPTON EFFECT
SJ: P-SE 40
...
20 nm are scattered from
a block of material
...
Calculate
their wavelength
...
3 P-21
Calculate the energy
and momentum of a photon of wavelength 700
nm
...
3 P-23
...
00160 nm
photon scatters from a free electron
...
3 P-25
...
880 MeV photon is
scattered by a free electron initially at rest such that
the scattering angle of the scattered electron is
equal to that of the scattered photon (θ = φ)
...
(b) Determine the
energy and momentum of the scattered electron and
photon
...
1313, chapter 40 of the reference
book
...

• Photoelectric effect and Compton effect can only
be explained taking light as photons/ particle
• This means true nature of light is not describable in
terms of any single classical picture
...


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PHOTONS AND ELECTROMAGNETIC WAVES

De Broglie

The Wave Properties of Particles
We have seen that light comes in discrete units
(photons) with particle properties (energy and
momentum) that are related to the wave-like
properties of frequency and wavelength
...
63 × 10−34 Js

and

E
f=
h

Energy of the particle

frequency of the particle
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PHOTONS AND ELECTROMAGNETIC WAVES

h= h
λ=
p mv

p = momentum of the particle,
p = m v for a non-relativistic particle
m = mass of the particle
v = velocity of the particle

The electron accelerated through a potential
difference of V has a non relativistic kinetic energy
1
2

2

mv =eV

m = mass,
p=mv
MIT- MANIPAL

=

v = velocity

2 m eV

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PHOTONS AND ELECTROMAGNETIC WAVES

Davisson -Germer experiment
&

Electron Diffraction pattern
(Go through the details of the experiments)
These two experiments confirmed de- Broglie
relationship p = h /λ
...
Thus de
Broglie's formula seems to apply to any kind of
matter
...
And it is stated in the principle
of complementarity
...

SJ: P-SE 40
...
0 x 107 m/s
...
6 The Wavelength of a Rock
A rock of mass 50 g is thrown with a speed of
40 m/s
...
7 An Accelerated Charged Particle
A particle of charge q and mass m has been accelerated
from rest to a nonrelativistic speed through a potential
difference of ∆V
...


SJ: Section 40
...
0 eV
...

(b) Also find the wavelength of a photon having
the same wavelength
...
5 P-38
In the Davisson-Germer
experiment, 54
...
If
the first maximum in the
diffraction pattern was
observed at φ= 50
...

We must choose one appropriate behavior for
the quantum particle (particle or wave) in
order to understand a particular behavior
...

An essential feature of a particle is that it is
localized in space
...


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THE QUANTUM PARTICLE

Now to build a localized entity from an
infinitely long wave, waves of same
amplitude, but slightly different frequencies
are superposed
...


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THE QUANTUM PARTICLE

If we add up large number of waves in a similar
way, the small localized region of space where
constructive interference takes place is called a
wavepacket, which represents a particle
...
The result
is a wave packet,
which represents a
particle
...
The waves are written as

(

y = A cos k 1 x - ω 1 t
1

)

and

The resultant wave is,

Δk Δω
y = 2A cos x t
2
2

[ (

(

y 2 = A cos k 2 x - ω 2 t

)

y = y1 + y2

k1 +k2
ω1 + ω2
cos
xt
2
2

)] (

)

Amplitude varies with t and x
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THE QUANTUM PARTICLE
Where ∆k = k1 – k2 and ∆ω = ω1 – ω2
...


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THE QUANTUM PARTICLE
This envelope can travel through space with a different
speed than the individual waves
...
The phase
speed, the speed with which wave crest moves,
which is given by

vp = f λ

& the group speed, the speed with which the
envelope (energy) moves
...

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THE QUANTUM PARTICLE

Relation between group speed and phase speed

ω
we have, υphase = = f λ
k
i
...
, ω = k υphase = k υp
d ( kvp )
dυp
dω
=
=k
+ υp
But υg =
dk
dk
dk
Substituting for k in terms of λ, we get

dυ p
υg = υp – λ
dλ
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THE QUANTUM PARTICLE

Relation between group speed and particle speed

E
and
ω = 2π f = 2π
h

2π
2π
2π p
k=
=
=
h
λ
hp

2π
dE
h
2π
dp
h

dω
=
vg =
dk

dE
=
dp

For a classical particle moving with speed u, the
kinetic energy E is given by
2
1
p2
E = mu =
2
2m

and

2 p dp
dE =
2m

or

dE
p
=
= u
dp
m

dω
dE
speed
i
...
, υg =
=
= v, the particle velocity
dp
dk
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THE QUANTUM PARTICLE

ie
...


To represent a realistic wave packet, confined to
a finite region in space, we need the
superposition of large number of harmonic waves
with a range of

MIT- MANIPAL

k

values
...
6 P-43

Consider a freely moving

quantum particle with mass m and speed u
...
Determine the phase
speed of the quantum wave representing the
particle and show that it is different from the
speed at which the particle transports mass and
energy
...
The electron detector is movable along the y
direction in the drawing and so can detect electrons
diffracted at different values of θ
...

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THE DOUBLE–SLIT EXPERIMENT REVISITED

Photograph of a double-slit
interference pattern produced
by electrons
...

The electrons are detected as particles at a localized
spot at some instant of time, but the probability of
arrival at that spot s determined by finding the
intensity of two interfering waves
...
That is
interference pattern is lost and the result is simply
the sum of the individual results
...

The result with both slits open (interference pattern)
is shown in brown
...
We are forced to
conclude that an electron interacts with both the
slits simultaneously shedding its localized
behaviour
...
We can only say that the electron
passes through both the slits
...
7 P-46 Electrons are incident on a
pair of narrow slits 0
...
The ‘bright
bands’ in the interference pattern are separated
by 0
...
0 cm from the slits
...


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THE UNCERTAINTY PRINCIPLE

Quantum theory predicts that, it is fundamentally
impossible to make simultaneous measurements
of a particle’s position & momentum with
infinite accuracy
...
The uncertainties arise from
the quantum structure of matter
...
But the position of the
particle in this case becomes uncertain
...
If ∆x is made zero, ∆λ &
∆λ,
thereby ∆p will become ∞
...
8
Locating an electron
The speed of an electron is measured to be
5
...
0030%
...


SJ: P-SE 40
...
0 x 10-8 s
...
8 P-51

Use the uncertainty

principle to show that if an electron were
confined inside an atomic nucleus of diameter
2x 10-15 m, it would have to be moving
relativistically, while a proton confined to the
same nucleus can be moving
nonrelativistically
...
8 P-52

Find the minimum

kinetic energy of a proton confined within a
nucleus having a diameter of 1
...

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INTRODUCTION TO QUANTUM PHYSICS
QUESTIONS
1
...

[1 EACH]
2
...

[1]
3
...

[2]
4
...

[2]
5
...

[1]
6
...
What are the classical predictions about the photoelectric
effect?
[3]
8
...

[2]

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INTRODUCTION TO QUANTUM PHYSICS
QUESTIONS
10
...
Sketch schematically the following graphs with reference to
the photoelectric effect: (a) photoelectric current vs applied
voltage (b) kinetic energy of most-energetic electron vs
frequency of incident light
...
Explain compton effect
...
Explain the experiment on compton effect
...
Derive the compton shift equation
...
Explain the wave properties of the particles
...
Explain a wavepacket and represent it schematically
...
Explain (a) group speed (b) phase speed, of a wavepacket
...
Show that the group speed of a wavepacket is equal to the
particle speed
...
Explain Heisenberg uncertainty principle
...
Write the equations for uncertainty in (a) position and
momentum (b) energy and time

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