Particle properties of wave
Electromagnetic wave (EM)

In 1864 the British physicist James Clerk Maxwell made the remarkable suggestion that
accelerated electric charges generate linked Electric and magnetic disturbances that can travel
indefinitely through the space. If the charges oscillate periodically, the disturbances are waves
whose electric and magnetic components are perpendicular to each other and to the direction of
motion.

According to Faraday, a changing magnetic field can induce a current in a wire loop. Thus
changing magnetic field is equivalent in its effect to an electric field. Maxwell proposed the
converse: a changing electric field has a magnetic field associated with it. The electric fields
produced by EM induction are easy to demonstrate because metals offer little resistance to the
flow of charge; even a weak field can lead to a measurable current in a metal. Weak magnetic
fields are much harder to detect, however, and Maxwell’s hypothesis was based on a symmetry
argument rather than on experimental findings.
Thus, electromagnetic waves must occur in which constantly varying electric and magnetic
fields are coupled together.
 Maxwell was able to show that the speed of electromagnetic waves in free space is given by
1
c
 2.998  10 8 m / s
 0 0
with 0 the electric permittivity of free space and 0  magnetic permeability in free space.
This is the same as the speed of light waves.
 The correspondence was too great to be accidental, and Maxwell concluded that light consists
of EM waves.
 During Maxwell’s lifetime the notion of electromagnetic waves remained without direct
experimental support. Finally, in 1888, German Physicist Henrich Hertz experimentally proofs
the existence of EM wave.
 Hertz determined the wavelength and speed of the waves he generated and showed that they
have both electric and magnetic components, and found that they show the optical properties
such as reflection, refraction, diffraction, interference and polarization etc.
 A characteristic property of all waves is that they obey the principle of superposition:
“When two or more waves of the same nature travel past a point at the same time, the
instantaneous amplitude is the sum of the instantaneous amplitudes of the individual
waves at that point”.

Light is not the only example of EM wave. Although all such waves have the same fundamental
nature, many features of their interaction with matter depend upon their frequencies.
The EM spectrum spans the entire range of electromagnetic radiation from the lowest
frequency/longest wavelength to the highest frequency/shortest wavelength. Various segments of
the spectrum are useful in exploring the Universe, and enable us to "see" into the universe beyond
what we are able to see from visible light.
Electromagnetic Spectrum Bands (Overview)
Band
Radio
Microwave
Sub-millimeter
Far Infrared
Mid Infrared
Near Infrared
Optical
Ultraviolet
X-ray
Gamma-ray
"Cosmic-ray"
Wavelength
600 m - 0.187 m
187 mm - 1 mm
539 - 616 μm
40 - 350 μm
5 - 40 μm
1 - 5 μm
380 - 780 nm
10 - 400 nm
10 - 0.01 nm
0.01 - 0.000006 nm
10 - 0.000006 nm
Frequency
30 Hz - 1.6x106 Hz
1 - 300 106 Hz
487 - 556x106 Hz
300x106 Hz - 30x1012 Hz
30 - 120x1012 Hz
120 - 440x1012 Hz
400 - 790x1012 Hz
750x1012 Hz - 30x1015 Hz
30x1015 Hz - 100x1018 Hz
100x1018 Hz - 3,862x1021 Hz
30x1015 Hz - 3,862x1021 Hz
Temperature
Range
11.6 – 140 K
140 - 740 K
740 - 3,000 K
106 - 108 K
Quantum Energy
2x10-9 - 0.6x10-5 eV
0.6x10-5 - 0.1x10-2 eV
2.0x10-3 - 2.3x10-3 eV
3.1x10-2 - 0.35x10-2 eV
3.1x10-2 - 2.5x10-1 eV
2.5x10-1 - 1.2 eV
1.59 - 3.3 eV
3.1 - 124 eV
124 - 1.24x105 eV
1.24x105 - 2.07x108 eV
124 - 2.07x108 eV
Please Note: The table above lists the full range of the electromagnetic spectrum. Not all bands are
used for astronomical purposes. The boundaries between some of the spectra can vary and are not
universally agreed upon. Data listed here represents what is generally agreed upon by the sources
used. "Cosmic-rays" are high energy sub-atomic particles that exhibit properties within the X-ray
and Gamma-ray bands.
Energy – Electron Volt
An electron volt (eV), is a unit of energy used to describe the total energy carried by a particle. It is
the energy gained by an electron (or proton, same size of electric charge) moving through a voltage
difference of one volt.
 1 keV = 1 kilo-electron volt = 1,000 eV – typical of dental X-rays.
 1 MeV = 1 mega-electron volt = 1 million eV – typical of radioactive decay particles.
 1 GeV = 1 giga-electron volt = 1 billion eV – the equivalent energy of a proton (hydrogen
nucleus) at rest.
The molecules in our atmosphere have energies around 0.03 eV. The Sun's plasma and Earth's
magnetosphere contain particles that are much more energetic. Protons in the magnetosphere
typically have energies of 1 keV to 10 keV. And particles having still higher energies are quite
common throughout the Universe.
Particle properties of wave
Introduction
Classical mechanics or Newtonian mechanics deals with macroscopic systems-like big bodies,
systems of huge number of atoms or molecules that are directly observable or observable with
instruments like microscope. Classical mechanics concerns the motion of a particle under the
influence of applied forces. This includes measurements of particle’s position, mass, velocity,
acceleration etc. Classical mechanics predicts for observable magnitudes (say: v = u+at, where u, a,
t are given) agree with measured values.
However, the particles which are not observable, can we apply laws of classical mechanics for
them? No, classic concepts cannot be applied to observe the behaviour of atoms, molecules,
electrons, protons, nuclei, etc. These are called microscopic particles. At the end of the 19th century,
with the discovery of electrons, x-rays and radioactivity, the possibility arose to study the behaviour
of individual atoms or molecules. But the results obtained from the studies could not be explained
by the classical mechanics. Then a new theoretical branch of physics, called Quantum mechanics
was developed by Erwin Schrödinger, Warner Heisenberg, P. M. Dirac and others.
Inadequacy of classical physics and concept of Quantum
i) Consider the case of an electron moving round the
nucleus. Classical mechanics tells that accelerated
charged particle loses energy in the form of
electromagnetic waves. Therefore, its velocity should
decrease continuously. The electron should come
closer to the nucleus until it collapses. Actually, the
atom does not collapse. So, classical mechanics fails
to explain the stability of atom.
Question: If the electrons do not produce light when they are in their allowed stable orbits, where is
the source of the light that comes from hydrogen?
Answer: According to Bohr, electrons have more energy when they are in larger orbits. If an
electron falls from a larger orbit down to a smaller orbit, it loses energy. According to the law of
conservation of energy, the energy lost by the electron must go somewhere. Bohr explained that a
photon carries away the lost energy from the hydrogen atom; that is,
photon energy = (electron energy in larger orbit) - (electron energy in smaller orbit)
ii) Classical mechanics can not explain the existence of spectrum in black body radiation,
Photoelectric effect, Compton Effect, pair production, pair annihilation etc.
iii) Classical electrodynamics tells that excited atoms (heated solids) emit radiations in which all
wavelengths are present. But the observed fact is that the emitted radiations contain certain
wavelengths only. So, classical mechanics fails to explain the radiation of emitted atoms.
Blackbody Radiation
Any object with a temperature above absolute
zero emits light at all wavelengths. If the
object is perfectly black (so it doesn't reflect
any light), then the light that comes from it is
called blackbody radiation.
The energy of blackbody radiation is not
shared evenly by all wavelengths of light.
The spectrum of blackbody radiation (above)
shows that some wavelengths get more
energy than others. Three spectra are shown,
for three different temperatures. (One of the
curves is for the surface temperature of the
Sun, 5770 K.)
Here are some experimental facts about blackbody radiation:
a.
b. The blackbody spectrum depends only on the temperature of the object, and not on what it is
made of. An iron horseshoe, a ceramic vase, and a piece of charcoal --- all emit the same
blackbody spectrum if their temperatures are the same.
c. As the temperature of an object increases, it emits more blackbody energy at all wavelengths.
d. As the temperature of an object increases, the peak wavelength of the blackbody spectrum
becomes shorter (bluer). For example, blue stars are hotter than red stars.
e. The blackbody spectrum always becomes small at the left-hand side (the short wavelength, high
frequency side).
Concept of Photon/Quantum
At about 1900, Max Planck came up with the solution. He proposed that the classical idea that each
frequency of vibration should have the same energy must be wrong. Instead, he said that energy is
not shared equally by electrons that vibrate with different frequencies. Planck said that energy
comes in clumps. He called a clump of energy a quantum. The size of a clump of energy --- a
quantum --- depends on the frequency of vibration. Here is Planck's rule for a quantum of energy
for a vibrating electron:
Energy of a quantum = (a calibration constant) x (frequency of vibration)
or
E = hf
where h, the calibration constant, is today called Planck's constant. Its value is about 6 x 10-34, very
tiny!
So how does this explain the spectrum of blackbody radiation? Planck said that an electron
vibrating with a frequency f could only have an energy of 1 hf, 2 hf, 3 hf, 4 hf, ... ; that is,
energy of vibrating electron = (any integer) x hf
But the electron has to have at least one quantum of energy if it is going to vibrate. If it doesn't
have at least an energy of 1hf, it will not vibrate at all and can't produce any light. "Aha!" said
Planck: at high frequencies the amount of energy in a quantum, hf, is so large that the highfrequency vibrations can never get going! This is why the blackbody spectrum always becomes
small at the left-hand (high frequency) side.
WAVES OR PARTICLES? BOTH!
When light passes through a double-slit, an interference pattern consisting of bright bands and dark
bands is seen on a screen. This is produced when the wave from one slit combines with the wave
from the other slit. If two wave crests meet at the screen, the waves add and you get a bright band.
If a wave crest from one slit meets a wave trough from the other slit, the waves cancel and you get
a dark band. This proves that light is a wave.
On the other hand, the photoelectric effect proves that light consists of massless particles
called photons.
So which is it? Is light a wave or a stream of particles? The answer is "Yes!"
Light acts like a wave if you want to know how it propagates, how it travels from one place to
another. To describe how light travels from the double slit to the screen, you have to use the wave
characteristics of light.
Light acts like particles (photons) if you want to know how light interacts with matter. To describer
how light interacts with the electrons in a metal and how it ejects them from the metal's surface,
you have to use the particle characteristics of light.
We say that light exhibits a wave-particle duality. It can behave like either waves or particles (but
not both at the same time), depending on the situation.
Thinking about the photoelectric effect again, how can a photon (which has no mass) knock an
electron about? Einstein used his theory of relativity to show that even massless photons have
momentum. Newton defined momentum = (mass) x (velocity) for a particle with mass, but Einstein
was able to show that the momentum of a massless photon depends on its wavelength:
The proof that electrons propagate like a wave came when electrons were passed through a double
slit and counted as they hit a screen. If the electrons traveled like a stream of particles, they would
have simply piled up at two locations behind the two slits. But they didn't. They showed a doubleslit interference pattern, bright bands and dark bands just like the ones produced by light waves.
Without a doubt, electrons exhibit the wave-particle duality of nature. In fact, every massive object
exhibits the wave-particle duality of nature. It just isn't noticeable on the large scale of our
everyday world.
Wave interference in Thomas Young's double-slit experiment
Photoelectric Effect
When radiations with frequency over a thresold value falls on a metal surface, electrons are
ejected from the surface. This phenomenon, known as the photoelectric effect and the emitted
electrons are called photoelectrons.
It was discovered by Heinrich Hertz in 1887 in the process of his research into electromagnetic
radiation.
When light shines on the surface of a metallic substance, electrons in the metal absorb the energy
of the light and they can escape from the metal's surface. This is called the photoelectric effect, and
it is used to produce the electric current that runs many solar-powered devices. Using the idea that
light is a wave with the energy distributed evenly throughout the wave, classical physicists expected
that when using very dim light, it would take some time for enough light energy to build up to eject
an electron from a metallic surface. WRONG!! Experiments show that if light of a certain
frequency can eject electrons from a metal, it makes no difference how dim the light is. There is
never a time delay.
Experimental study of Photo-electric effect:
Photoelectric effect can be studied with the help of the
following simple experiment. The apparatus consists of two
photosensitive surfaces E and C enclosed inside a
evacuated glass tube. E and C are connected to variable
voltage source Vext through an ammeter. Vext has positive
polarity at E and negative polarity on C. When E is exposed
to light with Vext = 0, a current is found to exist in the
circuit A. This current is due to the emission of
photoelectrons from E due to illumination by radiation. As
Vext is increased, the current in the circuit decreases and
becomes zero as Vext reaches certain value. This reverse
potential is called stopping potential Vs. Here Vs is a
measure of maximum kinetic energy of the photoelectrons
as eVs is the work done against the most energetic
photoelectron to stop it from reaching the collector.
Fig: 1 Apparatus for observing
photoelectric effect.
Hence the maximum kinetic energy of the photoelectrons is
Kmax = eVs
Another distinguishing factor is that this value of the stopping potential and hence the maximum
kinetic energy of the photoelectron does not depend on the intensity of light.
Work function:
The surface of the metal is illuminated by an electromagnetic wave of intensity I. An electron
absorbs energy from the wave until the binding energy of the electron to the metal is exceeded at
which point the electron is released. The minimum quantity of energy needed to remove an electron
is called the work function  of the material. Table 1 lists some values of the work function of
different materials.
Table 1: Some photoelectric work-functions
Material
Na
Al
Co
Cu
Zn
Ag
Pt
Pb
Workfunction (eV)
2.28
4.08
3.90
4.70
4.31
4.73
6.35
4.14
Current I
Experimental characteristics of photoelectric effect:
Photo electrons are emitted from metal surfaces when exposed to
light of suitable frequency. Experiment reveals the following
characteristics of the photoelectric effect.
No photoelectrons are emitted for a particular surface if the Vs
frequency of light is below a certain value. It does not matter
how much the intensity is. The lowest frequency at which
photoelectric effect emitted is a characteristic of the material
Fig 2: the photo current I as a
and is called the cutoff frequency. The corresponding longest function of potential difference V
wavelength is called cutoff wavelength or threshold for two different values of intensity
wavelength. If the frequency of light is greater than the cutoff
of light
value photoelectrons are emitted.
2.
The photoelectron current i.e. the number of photoelectrons emitted from a surface is
directly proportional to the intensity of the light. Increasing the intensity of light, increase
the photo current only; the stopping potential remains the same.
3.
The stopping potential and hence the maximum kinetic
energy of the photoelectrons varies linearly with the
frequency of the light.
4.
The first photoelectrons are emitted virtually
instantaneously [within 10-9s] after the light source is
turned on.
These three experimental results all suggest the complete failure of
the wave theory to account for the photoelectric effect.
Kmax
1.
c
Fig 3: Variation of Kmax
with frequency 
Einstein’s theory of the photoelectric effect:
A successful theory of the photoelectric effect was developed by Einstein in 1905.
Based on Planck’s ideas, Einstein postulated that a beam of light consists of small packages of
energy called photons or quanta. The energy of a photon associated with an electromagnetic wave
of frequency  is
E  h
where h is Planck’s constant.
(1)
The photon frequency is given by =c/ where  is the wavelength of the photon. Thus
E
hc

(2)
Since the photons travel with the electromagnetic wave at the speed of light, they must obey the
E
p
relativistic relationship
(3)
c
Combining Eq. (2) and (3), we get
p
h

(4)
Like other particles, photons carry linear momentum as well as energy.
In Einstein’s interpretation, a photoelectron is released as a result of an encounter with a single
photon. The entire energy of the photon is delivered instantaneously to a single photoelectron. If the
photon energy h is greater than the work function  of the material, the photoelectron will be
released. If the photon energy is smaller than the work function, the photoelectric effect will not
occur. This explanation thus accounts for two of the failures of the wave theory: the existence of
the cutoff frequency and the lack of any measurable time delay.
The photo-electric work-function is given by
  hc
(5)
where  c is the threshold frequency.
If the photon energy exceeds the work function, the excess energy appears as the kinetic energy of
the electron:
or,
K max  h  
(6)
K max  h  h c
(7)
The intensity of light source does not appear in this expression. Doubling the intensity of the light
source means that twice as many photoelectrons are released, but they all have precisely the same
maximum kinetic energy.
If v is the speed of the ejected photo-electron, then the kinetic energy of the photo-electron is
K max 
1 2
mv  h  h c
2
(8)
The maximum kinetic energy corresponds to the release of the least tightly electron. Some electrons
may lose energy through interactions with other electrons in the material and emerge with smaller
kinetic energy.
For a particular metal
K max  h(   c )  h(

c


c
c
)  hc(
1


1
c
)  6.625  10
 34
1
8 1
 3  10 ( 
) joules,  and  in meter
c
 
c
34
1 1
1
6.625  10 
 3  10 8 1
(  ) eV,  and  in Angstrom   12400 (  ) eV,  and  in Angstrom
c
c
 c
10  10  1.602  10  19  c
(9)
Photo-electric work function
  h c 


hc
c
 6.625  10
 34
 3  10
8 1
34
6.625  10 
 3  10 8 1
eV,
10  10  1.602  10  19 c
12400
c
c
joules,  in meter
c
 in Angstrom
c
eV,  in Angstrom
c
(10)
Problems of Photo-electric effect
Example 1: (a) what are the energy and momentum of a photon of red light of wavelength 650 nm?
(b) What is the wavelength of a photon of energy 2.40 eV?
Solution:
hc [6.63x10 34 J .s][3.0 x10 8 m / s]
3.06 x10 19 J
19


3
.
06
x
10
J

 1.91 eV
(a) E  h 

650 x10 9 m
1.60  10 19 J / eV
hc 1240 eV .nm

 517 nm
(b)  
E
2.40 eV
Example 2: Calculate the threshold frequency and the corresponding wavelength of radiation
incident on a certain metal whose work function is 3.3110-19 joules. Given, Planck’s constant =
6.6210-34 Joul-sec.
Solution:
(a) Work function,   h c   c 
(b) c 
c
c


h
 c 
3.31  10 19 J
 5  1014 Hertz
34
6.625  10 J .s
3  10 m / sec
 6  10 7 m  6000 Angstrom
5  1014 / sec
8
Example 3: The work-function for tungsten metal is 4.52 eV. (a) What is the cut-off wavelength 
c for tungsten? (b) What is the maximum K. E of the electron when radiation of wavelength 198 nm
is used? (c) What is the stopping potential in this case?
Solution:
hc
(a) We know that the stopping potential can be written as   h c 
c
Therefore, we can write the cutoff wavelength c as
c 
hc 1240 eV . nm

 274 nm

4.52 eV
(b) Maximum kinetic energy of the photoelectron can be written as
1240 eV . nm
hc
K .E max  h  hvc 

 4.52 eV  1.74 eV

198 nm
(c) The stopping potential Vs is given by Vs 
K .Emax 1.74 eV

 1.74 V
e
e
Example 4: Light of wavelength 4300 Angstrom is incident on (a) nickel surface of work fuinction
5 eV and (b) potassium surface of work function 2.3 eV. Find out, if electrons will be emitted, and
if so, the maximum velocity of the emitted electrons in each cases.
Solution:
Here
Light wavelength,  = 4300 Angstrom
Work function for Nickel,  Ni  5 eV
Work function for Potassium,  P  2.3 eV
(i) For the nickel surface

hc 12400
12400 12400 

A  2480 A

eV   c 
 Ni
5
c
c
Since  c  , no electrons will be emitted from the Nickel surface.
We know that work function,  Ni  h c 
(ii) For the Potassium surface

12400
12400 12400 
eV   c 

A  5391 A
Work function,  P 
c
P
2.3
Since  c  , electrons will be emitted from the Potassium surface
Maximum kinetic energy of photo-electrons is
1
1
K max  12400( 
) eV,  and  in Angstrom
c
 c
 12400(
1
1
 5391  4300 

) eV  12400
 eV  0.583 eV
4300 5391
 4300  5391 
1
Or, mv 2  0.583 eV  0.583  1.602  10 19 J  0.9349  10 19 J
max
2
 v max 
2  0.9349  10 19 J
1.8698  10 19 J
12
5

 0.2054  10  4.6  10 m/s
31
m
9.1  10 kg
Example 5: The photoelectric threshold of copper is 3200 Angstrom. If ultra-violet light of
wavelength 2500 Angstrom falls on it, find (a) the maximum kinetic energy of the photo-electrons
ejected, (b) maximum velocity of the photo-electrons and (c) the value of the work function.
Solution:
(a) Maximum kinetic energy of photo-electrons is
1
1
K max  12400( 
) eV,  and  in Angstrom
c
 c
1
1
 12400(

) eV  1.087 eV
2500 3200
(b) Maximum velocity of the photo-electrons is given by
1
K max  mv 2  1.087 eV  1.087  1.602  10 19 J
max
2
 v max
2  1.087  1.602  10 19 J
3.482748  10 19 J
5


 6.186  10 m / sec
31
m
9.1  10 kg
(c) Work function   h c 
hc
c

12400
c
eV, c in A.U. 
12400
eV  3.88 eV
3200
Example 6: Ultraviolet light of wavelength 350 nm and intensity 1.00 w/m2 is directed at a
potassium surface (   2.2 eV ). (a) Find the maximum KE of photoelectrons. (b) If 0.5 percent of
the incident photons produce photoelectrons, how many are emitted per second if the potassium
surface has an area of 1 cm2?
Solution:
COMPTON SCATTERING
When a beam of intermediate-energy photon/x-rays of well defined wavelength  is colliding with
a target, then the loosely bound electrons are released from the outer shell of the atom or molecule
of the target and some of the radiation/photon/x-rays is scattered in some angle  with a component
of well defined wavelength  , which is longer than  . The change in wavelength depends on the
angle  through which the radiation is scattered. This phenomenon was first observed by A.H.
Compton and is known as Compton effect.
h
[1  cos ]
The shift of wavelength is then     
m0 c
h
where, m0 is the electron rest mass. The quantity
has the unit of length and its value is:
m0 c
h
6.626 x10 34 J .s

 2.426 x10 12 m
m0 c [9.109 x10 31 kg][ 2.998 x10 8 m s]
Derivation of expression for the change in wavelength of a photon
undergoing a Compton scattering
Fig. 1: A Compton effect experiment

Fig. 2: (a) An incident photon with wavelength , energy E and momentum p approaching an
electron e at rest. The incident photon is scattered through an angle  and the struck electron
recoils.
(b) Vector diagram for the conservation of momentum.
The collision process in Compton scattering is schematically shown in the figure.
E h

collide with an electron which is
c
c
initially at rest. Thus the photon energy can be written as E  pc
 A photon with an energy E  h and momentum p 
The electron in rest has momentum p  0 .energy E  m0 c 2
 After collision, the scattered photon deflected at an angle  with the initial direction and carries
h 
different energy E  h and momentum p 
.
c
 While the electron is moving at an angle  carrying energy and momentum
E  mo c 4  p 2c 2 and momentum p  p .
 The Compton scattering is simply the elastic scattering of a photon by an electron, in which
both energy and momentum is conserved.
2
i) Conservation of energy
According to the conservation of energy
Loss in photon energy  gain in electron energy
(1)
h  h  K
ii) Conservation of momentum
According to the conservation law, Initial momentum  Final momentum
The horizontal component:
h
h 
0
cos   p cos 
c
c
 h  h  cos   pc cos 
 pc cos   h  h  cos 
The vertical component:
h 
0
sin   p sin 
c
h 
 p sin  
sin 
c
 pc sin   h  sin 
(2)
(3)
Squaring and adding eq [2] and eq [3], we get
p 2 c 2 (cos 2   sin 2  )  (h ) 2  2(h )(( h  cos  )  (h  cos  ) 2  (h  sin  ) 2
 p 2 c 2  (h) 2  2(h)(( h cos )  (h) 2
iii) Total energy expression
The total energy of a particle can be expressed as
E  K  mo c 2
E  mo c 4  p 2c 2
2
(4)
(5)
(6)
Thus from eq [5] and [6]
K  mo c 2  mo c 4  p 2 c 2
2
 ( K  mo c 2 ) 2  mo c 4  p 2 c 2
2
 p 2 c 2  K 2  m02 c 4  2 Kmo c 2  mo c 4
2
 p 2 c 2  K 2  2mo c 2 K
Using h  h  K from eq. [1] in [7], we get
p 2 c 2  ( h  h ) 2  2mo c 2 ( h  h )
(7)
(8)
From equation [4] and [8], we get
(h  h ) 2  2mo c 2 (h  h )  (h ) 2  2(h )( h  cos  )  (h ) 2
 (h ) 2  (h ) 2  2(h )( h )  2mo c 2 (h  h )  (h ) 2  2(h )( h  cos  )  (h ) 2
 2 (h  )( h )  2 mo c 2 (h   h )  2 (h  )( h  cos  )
 h   mo c 2 (   )  h  cos 
mo c 2

(   )   (1  cos  )
h
h
 (   )    
(1  cos  )
mo c 2
  h
   
   c   
(1  cos  )
c c mo
c c 
c
1
c
1


Now,  
 
and    



c


c

Eq. (9) becomes
h
1 1  1 1
(1  cos  )
 
  
        mo c
(9)
1
h
    


(1  cos  )

      mo c
h
    
(1  cos  )
(10)
mo c
This equation gives the Compton shift of the wavelength of the photon and strongly depend on the
angle through which it is deflected. It is independent of the wavelength  of the incident photon.
This quantity
h
C 
(11)
mo c
Is called the Compton wavelength of the scattering particle.
h
6.625  10 34 J .s
The value is C 

 2.426  10 12 m  2.426 pm (1 pm  10 12 m)
mo c 9.1  10 31 kg  3  10 8 m / s
With C, the Eq. (10) becomes
       C (1  cos  )
(12)
(i)
When  = 1800,        C (1  cos180 0 )  2C
(ii)
When  = 900,        C (1  cos 90 0 )  C
Summery:
a. Compton shift        
h
(1  cos  ) is independent of the wavelength  of the incident
m0 c
x-rays.


h
 cons tan t
c. Compton wavelength =  c 
m0 c
E  E  h  h  c   c      





d. Fractional energy loss =
E
h
c

  
b. Fractional wavelength shift =
Example 1: Compare Compton scattering for x-rays (  20 pm) and visible light (  500 nm) at a
particular angle of scattering. Which has greater [a] Compton shift, [b] fractional wavelength shift,
[c] fractional photon energy change and [d] energy imparted to the electron?
h
[1  cos  ] is independent of wavelength and depends only on
m0 c
angle . So both these waves have same shift in wavelength
[a] Compton shift,       
For  = 900,  
6.626 10 34
h
[1  cos 90 0 ] 
 2.427  10 -12 m
m0 c
9.1 10 31  3  10 8

[b] Fractional wavelength shift =
For x-rays with  = 20 pm



=
2.427  10 -12 m
 0.12135
20  10 12 m

For visible light with  = 500 nm
[c] Fractional energy loss =
For x-rays with  = 20 pm
For x-rays with  = 500 nm

=
2.427  10 -12 m
 4.854  10 -6
9
500  10 m
E  E


E
  
2.427  10 -12 m
E  E

 0.108
E
20  10 12 m  2.427  10 -12 m
2.427 10 -12 m
E  E

 4.85 10 6
E
500 10 9 m  2.427 10 -12 m
Energy imparted to electron E  E  
For x-rays with  = 20 pm E =
hc


E
  

1240 eV nm
 62000 eV  62 keV
20 10 3 nm
Hence energy imparted to electron E  E   0.108  62 keV  6.696 keV
For x-rays with  = 500 nm E =
hc


1240 eV nm
 2.48 eV
500 nm
Hence energy imparted to electron E  E   4.85 10 6  2.48 eV  1.95 10 6 eV
DE BROGLIE WAVES/De Broglie’s hypothesis:
We know that Photoelectric effect Compton effect could be explained if electromagnetic wave was
supposed to consist of packets of energy called quanta or photon which behaved like a particle of
rest mass 0 and moved through space with momentum p = mc, i.e., the photon moves with the
velocity of light.
On the other hand, light exhibits the phenomena like interference, diffraction and polarization,
which can only be explained by the wave nature of light. Thus, electromagnetic wave was regarded
as exhibiting the behaviours of wave-particle duality. The physicist made use of either the wave or
particle nature of light to explain the experimental result as suitable. This fact led Louis de Broglie
in 1924 to make a fascinating suggestion that if light under certain circumstances behaves as waves
and other circumstances behaves as particles, then particle of matter also should show similar
behaviour as that exhibited by the light.
However, Louis de Broglie gave the reasoning of wave properties of particle based on the
assumption that nature always loves symmetry, and the two physical entities, wave and particle,
must by symmetrical.
L. D. Broglie wavelength
We know that the energy of a photon of radiation having frequency  is given by
E = h
…..
…..
…..
[1]
If the photon is considered to be a particle of mass m then its energy E = mc 2 = mc.c = p.c, where p
is the momentum of photon.
Thus, we can write h = pc
c


h
h
 
p
p
…..
…..
[2]
de Broglie assumed that equation (2) should be equally applicable to the electromagnetic waves and
the material particles. He considered that if a particle of mass m would move with a velocity of v,
the wavelength of the wave associated with the particle should be given by

h
h

p mv
…..
…..
[3]
This wavelength of a particle is called de Broglie wavelength.
He derived the wavelength by picturing a material particle as a standing wave in the vicinity of the
particle.
[Note: L. D. Broglie and M. D. Broglie were two brothers. The former was a professor of history
and the latter was a professor of Physics. L. D. Broglie always had a great zeal on the
invention/discovery of the physical laws/theories at that time. He talked about the development of
physics with brother.]
de Broglie’s suggestion about the nature of particle was verified experimentally by Davisson and
Garmer in 1927 and GP Thomson in 1928 and they showed that the electrons are diffracted when a
beam of electrons fall on metal surface. Thus the truth of de Broglie’s suggestion was established
and he got the noble prize in 1929.
It should be mentioned that as in the case of electromagnetic waves, the wave and particle aspects
of moving bodies can never be observed at the same time.
Example 1:
Compute the de Broglie wavelength of the following:
[1] A 1000-kg automobile traveling at 25 m/s.
[2] A 10 gm bullet traveling at 500 m/s.
[3] A smoke particle of mass 10-9 gm moving at 1cm/s.
[4] An electron with a K. E of 1keV.
[5] An electron with a K.E of 100 MeV.
Solution:
[1] λ 
h
h
6.6x10 34 J.s


 2.6504  10 -38 m
p mv [1000kg][2 5 m/s]
[2]  
h
h
6.6 x10 34 J .s


 1.3x10 34 m
2
p mv [10 kg][500m / s]
[3]  
h
h
6.6 x10 34 J .s


 6.6 x10  20 m
12
2
p mv [10 kg][10 m / s]
[4]  
1240 eV .nm
h hc
hc
hc



 0.0387 nm
2
2
p pc
E 2  m0 c 2
( KE  m0 c 2 ) 2  m0 c 2 31984 eV
[5]  
1240 eV .nm
hc
hc


 0.12338 x 10 15 m  0.12338 fm
8
2
2
2
2
pc
1.00498 x10 eV
( KE  m0 c )  m0 c
Note that the wavelengths computed in parts [1] to [3] are far too small to be observed in the
laboratory. Only in the last two cases, in which the wavelength is of the same order as atomic or
nuclear sizes, we have the chance of observing the wavelength.
Because of the smallness of h, only for particles of atomic or nuclear size will the wave behavior be
observable.
Example 2:
An electron has a de Broglie wavelength of 2.00 pm=2.00x10-12 m. Find its K.E, phase velocity and
group velocity of its de Broglie waves.
Solution:
hc [4.136x10 15 eV.s][3.00 x10 8 m s]
pc 

 6.20x10 5 eV  620 keV
12
λ
2.00x10 m
Rest energy of the electron =E0 = 511 keV
Therefore,
K.E  E  E 0  E 02  (pc) 2  E 0  (511 keV) 2  (600 keV) 2  511 keV
= 803 keV – 511 keV = 292 keV
The electron velocity can be found from
E0
E
1 v2 c2
or,
Therefore,
  c 1  E02 E 2  c 1  [ 511 803 ] 2  0.77c
c2
c2

 1.30c
v 0.77c
g  v
p 
Fig: 4
Millikan’sExample 3:
result forFind the de Broglie wavelength of (a) 46-g golf ball with a velocity of 30 m/s, and (b) an electron
the with a velocity of 107 m/s.
photoelectrSolution:
ic effect in
h
6.63  10 34 J .s
sodium. (a) Since v<<c, we can let m = m0. Hence  

 4.8  10 34 m
mv ( 0.046 kg )( 30m / s )
The wavelength of the golf ball is so small compared with its dimensions that we would not expect
to find any wave aspects in its behaviour.
(b) Again v<<c, so m = m0 = 9.1×10-31 kg for the electron.
h
6.63  10 34 J .s
So,  

 7.3  10 11 m
31
7
m0 v ( 9.1  10 kg )( 10 m / s )
The dimension of the atoms are comparable with this figure-the radius of the hydrogen atom, for
instance, is 5.3×10-11 m. Thus, wave nature of electron is the key to understand the atomic
structure.
Example 4:
(a) A photon has an energy of 1.00 eV, and an electron has a kinetic energy of that same amount.
What are their wavelengths? (b) Repeat for an energy of 1.00 GeV.
Example 5:
(a) A photon and an electron both have a wavelength of 1.00 nm. Give the energy of the photon and
the kinetic energy of the electron. (b) Repeat for a wavelength of 1.00 fm.
Example 6:
Singly charged sodium ions are accelerated through a potential difference of 300 V. (a) What is the
momentum acquired by such an ion? (b) What is the de Broglie wavelength?