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PHYSICS CHAPTER 10

is a phenomenon where under certain phenomenon where under certain circumstances a particle exhibits wave circumstances a particle exhibits wave properties and under other conditions a properties and under other conditions a wave exhibits properties of a particle.wave exhibits properties of a particle.

CHAPTER 10: CHAPTER 10: Wave properties of Wave properties of particleparticle(2 Hours)(2 Hours)

PHYSICS CHAPTER 10

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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: State and useState and use formulae for wave-particle duality of formulae for wave-particle duality of

de Broglie,de Broglie,

Learning Outcome:

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10.1 de Broglie wavelength (1 hour)

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

From the Einstein’s special theory of relativity, the energy of a photon is given by

By equating eqs. (10.1) and (10.2), hence

10.1 de Broglie wavelength

hc

E (10.1)(10.1)

2mcE (10.2)(10.2)

and pmc pcE

pchc

h

p particle aspectparticle aspectwave aspectwave aspect

(10.3)(10.3)

where momentum: p

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From the eq. (10.3), thus light has momentum and exhibits particle property. This also show light is dualistic in naturedualistic in nature, behaving is some situations like wavesome situations like wave and in others like others like particle (photon)particle (photon) and this phenomenon is called wave particle wave particle duality of lightduality of light.

Table 10.1 shows the experiment evidences to show wave particle duality of light.

Based on the wave particle duality of light, Louis de Broglie suggested that matter such as electron and proton might electron and proton might also have a dual naturealso have a dual nature.

Wave Particle

Young’s double slit experiment

Photoelectric effect

Diffraction experiment Compton effect

Table 10.1Table 10.1

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He proposed that for any particle of momentum for any particle of momentum pp should should

have a wavelength have a wavelength given bygiven by

Eq. (10.4) is known as de Broglie relation (principle)de Broglie relation (principle). This wave properties of matter is called de Broglie wavesde Broglie waves or

matter wavesmatter waves. The de Broglie relation was confirmed in 1927 when Davisson

and Germer succeeded in diffracting electrondiffracting electron which shows that electrons have wave propertieselectrons have wave properties.

mv

h

p

h

where

(10.4)(10.4)

h wavelengtBroglie de:

particle a of mass: mparticle a ofvelocity : v

constant sPlanck': h

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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 a photon used.

(Given the speed of light in the vacuum, c =3.00108 m s1 and

Planck’s constant, h =6.631034 J s)

Solution :Solution :

By using the de Broglie relation, thus

and the energy of the photon is given by

Example 1 :

m 10550 9

p

h

p

349 1063.6

10550

127 s m kg 1021.1 p

hc

E

9

834

10550

1000.31063.6

E

J 1062.3 19E

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Calculate the de Broglie wavelength fora. a jogger of mass 77 kg runs with at speed of 4.1 m s1.b. an electron of mass 9.111031 kg moving at 3.25105 m s1.

(Given the Planck’s constant, h =6.631034 J s)

Solution :Solution :

a. Given

The de Broglie wavelength for the jogger is

b. Given

The de Broglie wavelength for the electron is

Example 2 :

1s m 1.4kg; 77 vm

mv

h 1.477

1063.6 34

m 101.2 361531 s m 1025.3kg; 1011.9 vm

531

34

1025.31011.9

1063.6

m 1024.2 9

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An electron and a proton have the same speed.a. Which has the longer de Broglie wavelength? Explain.

b. Calculate the ratio of e/ p.

(Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg,

mp=1.671027 kg and e=1.601019 C)

Solution :Solution :

a. From de Broglie relation,

the de Broglie wavelength is inversely proportional to the wavelength is inversely proportional to the

mass mass of the particle. Since the electron lighter than the mass electron lighter than the mass

of the protonof the proton therefore the electron has the longer de Broglie electron has the longer de Broglie

wavelengthwavelength.

Example 3 :

vvv pe

mv

h

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Solution :Solution :

Therefore the ratio of their de Broglie wavelengths is

e

p

m

m

31

27

1011.9

1067.1

1833p

e

vvv pe

vm

h

vm

h

p

e

p

e

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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: DescribeDescribe Davisson-Germer experiment by using a Davisson-Germer experiment by using a

schematic diagram to show electron diffraction.schematic diagram to show electron diffraction. ExplainExplain the wave behaviour of electron in an electron the wave behaviour of electron in an electron

microscope and its advantages compared to optical microscope and its advantages compared to optical microscope.microscope.

Learning Outcome:

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10.2 Electron diffraction (1 hour)

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ee

+4000 V+4000 V

10.2.1 Davisson-Germer experiment Figure 10.1 shows a tube for demonstrating electron diffraction

by Davisson and Germer.

A beam of accelerated electrons strikes on a layer of graphite which is extremely thin and a diffraction pattern consisting of rings is seen on the tube face.

10.2 Electron diffraction

screen diffraction pattern

electron diffraction

graphite filmanode

cathode

Figure 10.1: electron diffraction tubeFigure 10.1: electron diffraction tube

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This experiment proves that the de Broglie relation was right and the wavelength of the electron is given by

If the velocity of electrons is increasedvelocity of electrons is increased, the ringsrings are seen to become narrowernarrower showing that the wavelength of electrons wavelength of electrons decreasesdecreases with increasing velocityincreasing velocity as predicted by de broglie (eq. 10.5).

The velocity of electrons are controlled by the applied voltage V across anode and cathode i.e.

mv

h

where electronan of mass: m

(10.5)(10.5)

electronan ofvelocity : v

KU 2

2

1mveV

m

eVv

2 (10.6)(10.6)

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By substituting the eq. (10.6) into eq. (10.5), thus

meV

m

h

2

meV

h

2 (10.7)(10.7)

Note:Note:

Electrons are not the only particles which behave as waves. The diffraction effects are lessdiffraction effects are less noticeable with more massive massive

particlesparticles because their momentatheir momenta are generally much highermuch higher and so the wavelengthwavelength is correspondingly shortershorter.

Diffraction of the particles are observed when the wavelength is of wavelength is of the same order as the spacing between plane of the atomthe same order as the spacing between plane of the atom.

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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 c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and

e=1.601019 C)

Solution :Solution :

a. Given

The de Broglie wavelength for the electron is

Example 4 :

V 2000V

meV

h

2

20001060.11011.92

1063.61931

34

m 1075.2 11

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Solution :Solution :

b. Given

For an electron,

Its momentum is

and its energy is

m 1021.0 9pe

e

hp

9

34

1021.0

1063.6

p

124 s m kg 1016.3 p2

e2

1vmK

31

224

1011.92

1016.3

19

18

1060.1

1048.5

andem

pv

e

2

2m

p

eV 3.34K

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Solution :Solution :

b. Given

For a photon,

Its momentum is

and its energy is

m 1021.0 9pe

124 s m kg 1016.3 p

phc

E

9

834

1021.0

1000.31063.6

19

16

1060.1

1047.9

eV 5919E

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Compare the de Broglie wavelength of an electron and a proton if they have the same kinetic energy.

(Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg,

mp=1.671027 kg and e=1.601019 C)

Solution :Solution :

By using the de Broglie wavelength formulae, thus

Example 5 :

KKK pe

meV

h

2

mK

h

2

and KeV

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Solution :Solution :

Therefore the ratio of their de Broglie wavelengths is

KKK pe

Km

h

Km

h

e

p

p

e

2

2

e

p

m

m

31

27

1011.9

1067.1

8.42p

e

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A practical device that relies on the wave properties of electrons is electron microscope.

It is similar to optical compound microscope in many aspects. The advantageadvantage of the electron microscope over the optical

microscope is the resolving powerresolving power of the electron electron microscopemicroscope is much highermuch higher than that of an optical optical microscopemicroscope.

This is becausebecause the electrons can be accelerated to a very high

kinetic energy giving them a very short wavelengthvery short wavelength λ typically 100 times shorter thanthan those of visible lightvisible light. Therefore the diffraction effectdiffraction effect of electronselectrons as a wave is much lessmuch less than that of lightlight.

As a result, electron microscopes are able to distinguish details about 100 times smaller.

10.2.2 Electron microscope

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In operation, a beam of electrons falls on a thin slice of sample. The sample (specimen) to be examined must be very thin (a few

micrometres) to minimize the effects such as absorption or scattering of the electrons.

The electron beam is controlled by electrostatic or magnetic electrostatic or magnetic lenseslenses to focusfocus the beambeam to an image.

The image is formed on a fluorescent screen. There are two types of electron microscopes:

TransmissionTransmission – produces a two-dimensional imagetwo-dimensional image. ScanningScanning – produces images with a three-dimensional three-dimensional

qualityquality. Figures 10.2 and 10.3 are diagram of the transmission electron

microscope and the scanning electron microscope.

PHYSICS CHAPTER 10

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Exercise 10.1 :Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and

e=1.601019 C

1. a. An electron and a photon have the same wavelengths and the total energy of the electron is 1.0 MeV. Calculate the energy of the photon.b. A particle moves with a speed that is three times that of an electron. If the ratio of the de Broglie wavelength of this particle and the electron is 1.813104, calculate the mass of the particle.

ANS. :ANS. : 1.621.6210101313 J; 1.67 J; 1.6710102727 kg kg2. a. An electron that is accelerated from rest through a

potential difference V0 has a de Broglie wavelength

0. If the electron’s wavelength is doubled, determine the

potential difference requires in terms of V0.

b. Why can an electron microscope resolve smaller objects than a light microscope?

(Physics, 3(Physics, 3rdrd edition, James S. Walker, Q12 & Q11, p.1029) edition, James S. Walker, Q12 & Q11, p.1029)

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PHYSICS CHAPTER 10

Next Chapter…CHAPTER 11 :

Bohr’s model of hydrogen atom