Note Physics 2 SP025

- End of Chapter 9 -

35

SP025 CHAPTER 10 Physics Unit KMJ
Session
Wave Properties of Particle
(1 HOURS) 2020/2021

SUBTOPICS :
10.1 De Broglie wavelength (0.5 hour)
10.2 Electron diffraction (0.5 hour)

MODE Face to face Non Face to face
SLT SLT
Lecture
Tutorial 1 1

2 2

1

SP025 CHAPTER 10 Physics Unit KMJ
Wave Properties of particle Session 2020/2021

Learning Outcome:

At the end of this chapter, students should be able to:

10.1 De Broglie Wavelength

a) State wave-particle duality

(Lecture : C2,PLO1, MQF LOD1)

b) Use de Broglie wavelength,

(Tutorial : C3, CLO3,PLO4, CTPS3, MQF LOD6)
2

OVERVIEW

Wave properties of particle

Wave-particle Electron
duality diffraction

de Broglie Davisson-Germer Electron
principle experiment microscope

  h
p

3

Wave particle duality

properties that are mutually exclusive.

- has mass, m
- when in motion it has

momentum
Waves - have wavelength

- interfere and diffracted.

4

Wave-particle duality is ..

At certain time, a particle
exhibits wave

properties, and at other
times, it exhibits particle

properties. But we
cannot observe both
behavior simultaneously.

5

SP025 10.1 De Broglie Wavelength

The De - Broglie Relation
Particle Properties of a wave

From the Planck’s quantum theory, energy of a
photon is

where E  hc ( Equation 1 )



h : Planck constant (6.63×10–34 J s)

c : speed of light in vacuum ( 3×108 m s–1 )

5

By Einstein’s Theory of special relativity, the
energy equivalent of a mass m is given by:

E  mc2 ( Equation 1 )

Equating (1) & (2):

Mass × speed
=Momentum, p

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Momentum,p of a photon of wavelength λ is :

Particle wave
aspect aspect

Thus, light (wave) has momentum and exhibits particle

properties.

This show light is behaving in some situations like waves
and in others like particles (photons).

Evidences to show wave particle duality of light.

Wave Particle (Photon)

Young’s double slit experiment Photoelectric effect

and diffraction experiment 8

Louis de Broglie suggested that matter (electron
and proton) might also have a dual nature.

He proposed :

Any particle of momentum p should
have a wavelength λ (the de Broglie
wavelength) given by:

wavelength, λ represents
property of wave

Mproompeernttyuomf,pparrteipclreesents9

Example 10.1 :
In a photoelectric effect experiment, a light source of wavelength 550 nm is
incident on a sodium surface. Determine the momentum and the energy of the
photon used.
(Given the speed of light in the vacuum,

)

Solution:

By using the de Broglie relation, thus,

And the energy of the photon is given by,

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Example 10.2: .
Calculate the de Broglie wavelength for: )

a. A jogger of mass 77 kg runs at speed of
b. An electron of mass

(Given the Planck’s constant,

Solution:
a. Given

The de Broglie wavelength for the jogger is:

b. Given
The de Broglie wavelength for the electron is:

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SP025 CHAPTER 10 Physics Unit KMJ
Wave Properties of particle Session 2020/2021

Learning Outcome:

At the end of this chapter, students should be able to:

10.2 Electron diffraction

a) Describe the observations of electron
diffraction in Davisson-Germer experiment.

b) Explain the wave behavior of electron in an
electron microscope.

c) State the advantages of electron microscope
compared to optical microscope.

(Lecture : C1 & C2,PLO1, MQF LOD1)

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10.2 Electron Diffraction

Electron can
behaves like wave !

How? Show
me the proof!!

Davisson 13
Germer
Experiment

Davisson and Germer Experiment

A beam of accelerated electrons strikes on a layer
of graphite which is extremely thin and a diffraction
pattern (ring shape) is seen on the tube screen. 14

Davisson and Germer Experiment
- Proves the de Broglie relation was right.
- The wavelength of the electron is ;

Electrons appear to act as
waves whose wavelength is
inversely proportional to
their velocity

• If the velocity of electrons is increased, the rings
are seen to become narrower, showing that the
wavelength of electrons decreases.

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How they control the velocity of
electron?

Varies the applied voltage V

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p  2meV (1)

Knowing that :   h (2)
(1) into (2) : p

Wavelength of
electron decreases
with increasing
voltage V

17

Note: Important

18

Electron Microscope

A device that relies on the wave
properties of electron.

Transmission Scanning
Electron Microscope Electron
Microscope

19

Scanning
electron
microscope

Transmission
electron
microscope

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Transmission Electron Microscope- TEM

TEM image contrast is due to A TEM image of the polio virus.
absorption of electrons in the The polio virus is 30 nm in size.
material, due to the thickness and
composition of the material. 21

Scanning electron microscope

Final image projected
on screen

22

Wave behavior of electron in
an Electron Microscope (EM)

1. Electron is accelerated through a high potential
difference, so they have very high speed.

Then, from

2. As velocity is very high, electron gain very short

wavelength of order (shorter de Broglie

wavelength), resulting in high resolving power.

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Electron Microscope compared to Compound Microscope

- Both are to see small objects, with some differences

ELECTRON OPTICAL / COMPOUND
MICROSCOPE MICROSCOPE

Using wave properties Using visible light
of electron
λvisible light ≈ 10–7 m. So, can
λelectron ≈ <10–9 m. So, can only distinguish clearly in
distinguish clearly 2 points a distance of micrometer.

separated by a distance
of nanometer.

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ADVANTAGES OF ELECTRON MICROSCOPE
COMPARED TO COMPOUND MICROSCOPE

ELECTRON OPTICAL /
MICROSCOPE COMPOUND
MICROSCOPE
Has high resolving power
because of the short The visible light used
have a certain range
wavelength of electron (Can (below 0.25 to 0.30 )

resolve down to 0.001 )

Resolving power can be Resolving power can be
adjusted with the voltage adjusted by changing
applied between cathode
lens magnification
and anode
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Example 10.3
a. An electron is accelerated from rest through a potential difference of 2000 V.

Determine its de Broglie wavelength.
b. An electron and a photon has the same wavelength of 0.21 nm. Calculate the

momentum and energy (in eV) of the electron and the photon.

(Given

)

Solution: The de Broglie wavelength for the electron is:
a. Given

  

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b. Given .× = 3.16 .
For an electron, .×

Its momentum is and

And its kinetic

Given and its energy is
For an electron,

Its momentum is

.×

. × 27

Example 10.4:
Compare the de Broglie wavelength of an electron and a proton if they have the
same kinetic energy.
(Given

)
Solution:
By using the de Broglie wavelength formulae, thus

 

 

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Therefore the ratio of their de Broglie wavelength is

 
 
  

29

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