26771284 matriculation physics waves properties of particle

7/29/2019 26771284 Matriculation Physics Waves Properties of Particle

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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)


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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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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PHYSICS CHAPTER 10

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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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PHYSICS CHAPTER 10

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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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PHYSICS CHAPTER 10

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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)


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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PHYSICS CHAPTER 10

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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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PHYSICS CHAPTER 10

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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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PHYSICS CHAPTER 10

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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)


By using the de Broglie wavelength formulae, thus

Example 5 :

KKK == pe

meV

h

2=

mK

h

2

=

and KeV =


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PHYSICS CHAPTER 10

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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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PHYSICS CHAPTER 10

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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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PHYSICS CHAPTER 10

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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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PHYSICS CHAPTER 10

21Figure 10.2Figure 10.2 Figure 10.3Figure 10.3


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PHYSICS CHAPTER 10

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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

26771284 matriculation physics waves properties of particle

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