Ph 101-7 WAVE PARTICLES

Quantum Mechanics: (i) Particle properties of waves:
Black body radiation, photoelectric effect, x-ray
Diffraction and Compton effect, pair production, photon
and gravity.
References:
A. Beiser, Concept of Modern Physics (or Perspective of Modern
Physics), Tata-McGraw Hill, 2005
Course Outline...

ELECTROMAGNETIC SPECTRUM
Particle aspect of radiation
Classical physics: a particle : energy E and momentum p
Wave : an amplitude and wave vector k
4.3 x 1014 7.5 x 1014

Principle of superposition
A characteristic properties of all waves
Particle: e.g e-: charge, mass : laws of particle mechanics
General Consensus:
Waves: em waves: diffraction, interference, polarization etc

Young’s experiment: Light consists of waves
Black Body Radiation
Classical approach
Only the quantum theory of
light can explain its origin
An ideal “blackbody”, is a material object that absorbs all of the radiation
falling on it, and hence its appears as black under reflection when
illuminated from outside.
Black Body
When an object is heated , it radiates electromagnetic energy as a result of the
thermal agitation of the electrons in its surface
Blackbody radiation: radiation leaving the hole of a heated hollow cavity or
the radiation emitted by a blackbody when hot.

Classical: continuous energy distribution
EXPERIMENTAL
Radiation inside a cavity:
series of standing waves.
Density of standing
waves in cavity:
Classical average energy
per standing wave:
Rayleigh’s assumption: a standing wave can exchange any
amount(continum) of energy with matter,
Rayleigh & Jeans
2
2
8
)(
c
d
dG

 
kT
Rayleigh-Jeans formula: 

 d
c
kT
dGdu 2
2
8
)()( 

Classical physics failure  introduction of Q.M.
As  increases toward the ultraviolet end of the spectrum, the energy density
should increase as 2. In the limit of infinitely high , u()d should also go to .
In really, of course, the energy density falls to 0 as   . This discrepancy called
ultraviolet catastrophe.
Consequences:
Ultraviolet catastrophe
Planck’s formula
The energy exchange between radiation and matter must be
discrete and energy of radiation E = nh
1/
3
2
8
)( 
 kTh
e
d
c
h
du 


1/ 
 kTh
e
h


Average energy per standing wave
Planck’s modifications
h = 6.626 x 10-34 J.s
Max Planck (1918 Nobel prize)

SUMMARYAt low  At high 







d
c
kT
d
h
kT
c
h
du
2
3
3
2
8
8
)(








Rayleigh & Jeans
h >> kT
kTh
e /
0)(  du
h << kT


hkT
kThe kTh
/
1/1
11
1/



# of modes per
unit frequency
per unit volume
Probability of
occupying modes
Average
energy per
mode
CLASSICAL
Equal for all modes kT
QUANTUM
Quantized modes:
require h energy to
excite upper modes,
less probable
3
2
8
c

3
2
8
c

1kT
h
e
h



Particle properties of waves:
[1] Photoelectric effect
[2] Compton effect
[3] Pair production
EXPERIMENTAL CONFIRMATION

Photoelectric effect
Provides a direct confirmation for
the energy quantization of light.
The phenomenon of ejection of electron from the surface of a metal when light
of a suitable frequency strikes on It, is called photoelectric effect. The emitted
electrons are called photo-electrons.
The time between the incidence
and emission of a photoelectron is
very small,  10–9 s.
V0
Heinrich Hertz in 1887
later on, it was explained by Albert Einstein in 1905
It was during this study of electromagnetic waves that he saw that if you send ultraviolet
light onto metals, you'll get sparks coming off
Experimental
Classical: raise intensity  more electrons
independent of frequency
Observed: No electron below threshold frequency

The following experimental laws were discovered:
The kinetic energy of the ejected electrons depends on the
frequency but not on the intensity of the beam; the kinetic energy of
the ejected electron increases linearly with the incident frequency.
 If the frequency of the incident radiation is smaller than the
metal’s threshold frequency(a frequency that depends on the
properties of metal), no electron can be emitted regardless of the
radiation’s intensity.
 No matter how low the intensity of the incident radiation,
electrons will be ejected instantly the moment the frequency of the
radiation exceeds the threshold frequency.
 At a fixed frequency, the number of ejected electrons increases
with the intensity of the light but doesn’t depend on its frequency.

Classical physics fails: for dependence of the
effect on the threshold frequency
K = h - W = h - h0
0 = threshold frequency
The stopping potential : at which all of
the electrons will be turned back before
reaching the collector
e
W
e
hc
e
W
e
h
VS 


Work function: W = h0
Larger W needs more energy needed for an
electron to leave
Photoelectric effect:
metal
E = h0
KE = 0
KEmax
= h - h0
E = h
Einstein interpretation(1905)
Light comes in packets of energy (photons)
E = h
An electron absorbs a single photon to leave the material
0
K


When two UV beams of wavelengths 1 = 280nm and 2 = 490nm fall on a lead
surface, they produce photo electrons with maximum kinetic energies 8.7eV and
6.67eV, respectively. (a) calculate the value of planck constant. (b) Calculate the
work function and the cutoff frequency of lead.
ANS.(a)
K1= hc/ 1 – W and K2= hc/ 2 – W
Js
c
KK
h
34
12
2121
1062.6 






Use 1eV = 1.6 x 10-19 J
(b) W = hc/ 1 –K1 = - 4.14eV
The cutoff frequency of the metal is
0 = W/h = 1015 Hz
Exercise(estimation of the Planck constant)

EM radiation with 0.01
to 10 nm  x-rays
Photons transfer energy to electron: photoelectric effect
Can reverse be true?
X-rays
V
16
min
1024.1 

 V.m
X-ray production:

X-ray diffraction
d

2
d sin

2d sin = Path
difference
Incident
wave
Bragg’s Law:
2d sin = n
Crystal structure
determination

Compton effect
Compton scattering is another one of those really important events that
happened at the beginning of the 20th century that indicated that photons were
real. They really, like really does behave like a particle.
Large wave length small wave length collision


c
h 


cos
c
h 




sin
c
h 
cosp
sinp
p
chp
hE
/



0
2
0


p
cmE
chp
hE
/



Incident photon
Target
electron
pp
cpcmE

 2242
0
Scattered
electron
-
-
Further confirmation of
photon model
1927 Nobel
Scattering of x-rays from electrons in a carbon target and found scattered x-rays with
a longer wavelength than those incident upon the target.

In the original photon direction:
Initial momentum = final momentum


coscos0 p
c
h
c
h



In the perpendicular direction: 

sinsin0 p
c
h



Collisions: conserve energy and conserve momentum
Elastic process
We've got to know what the energy of the photon is and what its momentum is.
Photon energy: h Photon momentum: h/= E/c


sinsin
coscos


hpc
hhpc
2222
)(cos))((2)(   hhhhcp
Form the total energy expression we have:
)(2)())((2)( 2
0
2222
  hhcmhhhhcp
)cos1)()((2)(2 2
0   hhhhcm

Compton wavelength:
cm
h
C
0
 gives the scale of the wave length
change of the incident photon 


cos/
sin
tan


Angle of recoil electron:
Compton effect or shift:
In order to conserve energy and momentum, I've got to have a shift in the wavelength. that
means, the outgoing photon has a different wavelength than the incoming one. this is
impossible classically. You cannot have a wave do that. But in quantum, you can.
this shift in wavelength is associated with h/mc. This is dimensionless
h/mc is really really small compared to the wavelength of light, then the shift in wavelength
is relatively really small.
If h/mc is very very large compared to the wavelength of the radiation that you're sending in
to hit the electron, then this shift in wavelength is going to be huge.
So the Compton scattering is determined by how big the wavelength of the light that you're sending in
is, in comparison to the Compton wavelength.
Small effect if  >> c
Large effect if  << c
)cos1(
0
 
cm
h
Compton wavelength is really really small. 2.4 pm. So a pm is a trillionth of a meter. So this actually is
about 100 times smaller than the atom. So you got to send in light with a really tiny wavelength, 100
times smaller than an atom in order to get an appreciable Compton scattering effect. But we've done it
and this is the way it comes out.

Experimental Demonstration
A.H. Compton, Phys. Rev. 22 409
(1923)

Ex-1
High energy photos (-rays) are scattered from electrons initially at rest. Assume
the photons are backscattered and their energies are much larger than the
electron’s rest mass energy, E >> mec2.
(a) calculate the wavelength shift,
(b) show that the energy of the scattered photons is half the rest mass energy of the
electron, regardless of the energy of the incident photons,
(c) calculate the electrons recoil kinetic energy if the energy of the incident photons
is 150 MeV.
ANS:
(a) Here  = , wave length shift or Compton shift:
m
cm
c
cm
h
cm
h
ee
12
2
0
108.4
42
)cos1( 



(b) Energy of scattered photon E :
2/2)/()/(2 2
2
2
2








Ecm
cm
hccm
cm
cmh
hchc
E
e
e
e
e
e 
MeV
cm
E
cmcm
E
cmcm
E eeeee
25.0
2
)(
2
2
1
2
2222122








(c) Kinetic energy of recoil electron: Ke = E - E  150 MeV-0.25 MeV = 149.75 MeV
If E >> mec2 we can approximate by

Pair production
+
-
Nucleus
Photon
Electron
Positron
p
p
cosp
cosp

ch /
Electromagnetic energy is converted into matter
Energy and linear momentum could not both be conserved if pair
production were to occur in empty space, so it does not occur there
Pair production requires a photon
energy of at least 1.02 MeV.
e- or e+
m0c2 = 0.51 MeV (rest mass energy),
 additional photon energy becomes K.E of e- and e+.
Pair annihilation: e- + e+   + 
The pair production : direct consequences of the Einstein mass-energy relation; E = mc2.

Ex-1
Calculate the minimum energy of a photon so that it converts into
an electron-positron pair. Find the photon‘s frequency and
wavelength
The Emin of a photon to produce e- and e+ pair is equal to sum of
rest mass energy of both. So Emin = 2mec2 = 1.02 MeV.
The photon’s frequency and wavelength can be obtained at once
from Emin = h = 2mec2 and =c/:
  = 2.47 x 1020 Hz and  = 1.2 x 10-12 m
Ans:

Photon and gravity
H


  hgH
c
h
hE 2
hE 
KE = mgH
KE = 0 Gravitational behaviour of light: follows from
the observation that, although a photon has
no rest mass, it nevertheless interacts with
electrons as though it has the inertial mass
Photon “mass” 2
c
h
v
p
m


Photon energy after falling through height H






 2
1
c
gH
hh 
Although they lack rest mass, photons behave as though they have gravitational
mass

Assignment-IV
PH101: Physics-I Due Date: 28/10/2013 S. N. Dash
1. A 45 kW broadcasting antenna emits radio waves at a frequency of 4 Mhz. (a) How many photons
are emitted per second? (b) Is the quantum nature of EM radiation important in analyzing the
radiation emitted from this antenna? [2+2]
2. When a light of a given 1 is incident on a metallic surface, the stopping potential for the
photoelectrons is 3.2 V. If a second light source whose wavelength is double that of the first is
used, the stopping potential drops to 0.8 V. From this data, calculate (a) the  of the first radiation
(b) the work function and cutoff frequency of the metal. [4+2]
3. Consider a photon that scatters from an electron at rest. If the Compton’s wavelength shift is
observed to be triple the wavelength of the incident photon and if the photon scatters at 60o ,
calculate (a) the wavelength of the incident photon, (b) the energy of the recoiling electron, (c)
the angle at which the electron scatters. [2+2+2]
4. What voltage must be applied to an x-ray tube for it to emit x-rays with a minimum wave length
of 30 pm (10-12 m) ? [2]
5. The distance between adjacent atomic planes in calcite is 0.3 nm. Find the smallest angle of
Bragg scattering for 0.03 nm x-rays ? [2]
6. A positron collides head on with an electron and both are annihilated. Each particle had a kinetic
energy of 1 MeV. Find the wavelength of the resulting photons. [2]

Ph 101-7 WAVE PARTICLES

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Ph 101-7 WAVE PARTICLES