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PHYSICS CHAPTER 10
is a phenomenon where under certainphenomenon where under certain
circumstances a particle exhibits wavecircumstances a particle exhibits wave
properties and under other conditions aproperties and under other conditions a
wave exhibits properties of a particle.wave exhibits properties of a particle.
CHAPTER 10:CHAPTER 10:
Wave properties of particleWave properties of particle
(2 Hours)(2 Hours)
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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 offormulae for wave-particle duality of
de Broglie,de Broglie,
Learning Outcome:
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p
h=
10.1 de Broglie wavelength (1 hour)
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PHYSICS CHAPTER 10
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From the Plancks quantum theory, the energy of a photon is
given by
From the Einsteins 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
hcE= (10.1)(10.1)
2mcE=
(10.2)(10.2)
and pmc =pcE=
pchc =
hp =particle aspectparticle aspect
wave aspectwave aspect(10.3)(10.3)
where momentum:p
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PHYSICS CHAPTER 10
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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 likeothers likeparticle (photon)particle (photon) and this phenomenon is called wave particlewave 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 mightelectron and proton might
also have a dual naturealso have a dual nature.
Wave Particle
Youngs double slitexperiment
Photoelectric effect
Diffraction experiment Compton effect
Table 10.1Table 10.1
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PHYSICS CHAPTER 10
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He proposed that for any particle of momentumfor any particle of momentumpp shouldshould
have a wavelengthhave 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)
hwavelengtBrogliede:
particleaofmass:mparticleaofvelocity:v
constantsPlanck':h
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PHYSICS CHAPTER 10
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In a photoelectric effect experiment, a light source of
wavelength 550 nm is incident on a sodium surface. Determine themomentum and the energy of a photon used.
(Given the speed of light in the vacuum, c =3.00 108 m s1and
Plancks constant, h =6.63 1034 J s)
Solution :Solution :By using the de Broglie relation, thus
and the energy of the photon is given by
Example 1 :
m10550
9
=
p
h=
p
349 1063.610550
=
127 smkg1021.1 =p
hcE=
( )( )9
834
10550
1000.31063.6
=E
J1062.319
=E
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PHYSICS CHAPTER 10
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Calculate the de Broglie wavelength for
a. a jogger of mass 77 kg runs with at speed of 4.1 m s1.
b. an electron of mass 9.11 1031kg moving at 3.25 105 m s1.
(Given the Plancks constant, h =6.63 1034 J s)
Solution :Solution :
a. GivenThe de Broglie wavelength for the jogger is
b. Given
The de Broglie wavelength for the electron is
Example 2 :
1
sm1.4kg;77
== vm
mv
h=
( )( )1.4771063.6 34
=
m101.236
= 1531 sm1025.3kg;1011.9 == vm
( )( )531
34
1025.31011.9
1063.6
=
m1024.2 9=
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PHYSICS CHAPTER 10
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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 =3.00 108 m s1, h =6.63 1034 J s, me=9.11 1031 kg,
mp=1.67 1027 kg and e=1.60 1019 C)
Solution :Solution :
a. From de Broglie relation,
the de Broglie wavelength is inversely proportional to thewavelength is inversely proportional to themassmass of the particle. Since the electron lighter than the masselectron lighter than the mass
of the protonof the proton therefore the electron has the longer de Broglieelectron has the longer de Broglie
wavelengthwavelength.
Example 3 :
vvv == pe
mv
h=
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PHYSICS CHAPTER 10
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Solution :Solution :
Therefore the ratio of their de Broglie wavelengths is
e
p
m
m=
31
27
1011.91067.1
=
1833p
e =
vvv == pe
=
vm
h
vm
h
p
e
p
e
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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:
DescribeDescribe Davisson-Germer experiment by using aDavisson-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 electronthe wave behaviour of electron in an electronmicroscope and its advantages compared to opticalmicroscope and its advantages compared to optical
microscope.microscope.
Learning Outcome:
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10.2 Electron diffraction (1 hour)
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PHYSICS CHAPTER 10
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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 graphitewhich 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 film
anode
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 tobecome narrowernarrowershowing that the wavelength of electronswavelength 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 Vacross anode and cathode i.e.
mvh=
where electronanofmass:m
(10.5)(10.5)
electronanofvelocity: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
=m
eVm
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 massivemassive
particlesparticles because their momentatheir momenta are generally much highermuch higherand sothe wavelengthwavelength is correspondingly shortershorter.
Diffraction of the particles are observed when the wavelength is ofwavelength 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
differenceof 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 =3.00 108 m s1, h =6.63 1034 J s, me=9.11 1031 kg and
e=1.60 1019 C)
Solution :Solution :
a. Given
The de Broglie wavelength for the electron is
Example 4 :
V2000=V
meV
h
2=
( )( )20001060.11011.921063.6
1931
34
=
m1075.2 11=
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Solution :Solution :
b. Given
For an electron,
Its momentum is
and its energy is
m1021.0 9pe==
e
hp
=
9
34
1021.0
1063.6
=p124 smkg1016.3 =p
2e2
1vmK=
( )( )31224
1011.921016.3
=
19
18
1060.1
1048.5
=
andem
pv =
e
2
2m
p=
eV3.34=K
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Solution :Solution :
b. Given
For a photon,
Its momentum is
and its energy is
m1021.0 9pe==
124 smkg1016.3 =p
p
hcE=
( )( )9
834
1021.0
1000.31063.6
=
19
16
1060.1
1047.9
=
eV5919=E
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Compare the de Broglie wavelength of an electron and a proton if
they have the same kinetic energy.(Given c =3.00 108 m s1, h =6.63 1034 J s, me=9.11 10
31 kg,
mp=1.67 1027 kg and e=1.60 1019 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
=
8.42
p
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 powerof the electronelectron
microscopemicroscope is much highermuch higherthan that of an opticalopticalmicroscopemicroscope.
This is becausebecause the electrons can be accelerated to a very high
kinetic energy giving them a very short wavelengthvery short wavelength typically
100 times shorterthanthan those ofvisible lightvisible light. Therefore the
diffraction effectdiffraction effect ofelectronselectrons as a wave is much lessmuch less thanthat oflightlight.
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 orscattering of the electrons.
The electron beam is controlled by electrostatic or magneticelectrostatic 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-dimensionalthree-dimensional
qualityquality.
Figures 10.2 and 10.3 are diagram of the transmission electronmicroscope and the scanning electron microscope.
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21Figure 10.2Figure 10.2 Figure 10.3Figure 10.3
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Exercise 10.1 :Given c =3.00 108 m s1, h =6.63 1034 J s, me=9.11 10
31 kg and
e=1.60 1019 C
1. a. An electron and a photon have the same wavelengths andthe total energy of the electron is 1.0 MeV. Calculate theenergy 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 thisparticle and the electron is 1.813 104, calculate the mass
of the particle.
ANS. :ANS. : 1.621.62 1010 1313 J; 1.67J; 1.67 1010 2727 kgkg
2. a. An electron that is accelerated from rest through a
potential difference V0 has a de Broglie wavelength 0. If
the electrons wavelength is doubled, determine the
potential difference requires in terms ofV0.
b. Why can an electron microscope resolve smaller objectsthan 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 ChapterCHAPTER 11 :
Bohrs model of hydrogen atom