INTERFERENCE
Topics
▪ Two source interference
▪ Double-slit interference
▪ Coherence
▪ Intensity in double slit interference
▪ Interference from thin film
▪ Michelson’s Interferometer
Text Book:
PHYSICS VOL 2 by Halliday, Resnick and Krane (5th Edition)
TWO-SOURCE INTERFERENCE
When identical waves from two sources overlap at a
point in space, the combined wave intensity at that
point can be greater or less than the intensity of either
of the two waves. This effect is called interference.
The interference is constructive when the net intensity
is greater than the individual intensities.
The interference is destructive when the net intensity is
less than individual intensities.
TWO-SOURCE INTERFERENCE
Maximal constructive
interference of two
waves occurs when their
phase difference is 0, 2,
4 , … (the waves are inphase)
Complete destructive interference of two waves occur when
their phase difference is , 3 , 5 , … (the waves are 180o out
of phase)
TWO-SOURCE INTERFERENCE
INTERFERENCE PATTERN PRODUCED BY WATER
WAVES IN A RIPPLE TANK
Maxima: where the shadows show the crests and valleys
Minima: where the shadows are less clearly visible
DOUBLE-SLIT INTERFERENCE
A train of plane light waves is incident on two narrow parallel
slits separated by distance d (<<). The interference pattern
on the screen consists of bright and dark fringes.
Phase difference (Φ) between the interfering waves:
depends on the location of the point P on the screen
DOUBLE-SLIT INTERFERENCE
▪ Consider two coherent sources S1 and
S2 separated by a distance ‘d’ and kept
at a distance ‘D’ from the screen.
▪ For D>>d, we can approximate rays r1
and r2 as being parallel.
▪ Path difference between two waves
from S1 & S2 (separated by a distance
‘d’) on reaching a point P on a screen at
a distance ‘D’ from the sources is S1b =
d sin .
DOUBLE-SLIT INTERFERENCE
For maximum at point P
S1b = m
m = 0, 1, 2, . . .
Which can be written as,
d sin  = m
m = 0, 1, 2, . . .
m = 0 is the central maximum.
For minimum at point P
S1b = (m + 21 ) 
m = 0, 1, 2, . . .
Which can be written as,
d sin  = (m + 21 )  m = 0, 1, 2, . . .
DOUBLE-SLIT INTERFERENCE
• For small value of , we can make
following approximation.
sin   
sin   tan  =
y
D
• Path difference:
d sin  = S1b =
y d
D
DOUBLE-SLIT INTERFERENCE
mth maximum is located at
ym given by
m 
d
or
ym
=
ym
D
D
= m
d
where m = 0, 1, 2, . . .
DOUBLE-SLIT INTERFERENCE
Separation between adjacent maxima
(for small ) is independent of m
y = y m+1 − y m
D
D
= (m + 1)
−m
d
d
D
y =
d
The spacing between the adjacent
minima is same the spacing between
adjacent maxima.
DOUBLE-SLIT INTERFERENCE
YOUNG’S DOUBLE SLIT EXPERIMENT
• Double slit experiment was first
performed by Thomas Young in 1801.
• So double slit experiment is known as
Young’s Experiment.
• He used sun light as source for the
experiment.
• In his experiment, he allowed sun light
to pass through narrow opening (S0)
and then through two openings (S1
and S2).
DOUBLE-SLIT INTERFERENCE
Problem: SP 41-1
The double slit arrangement is illuminated by light of
wavelength 546nm. The slits are 12mm apart and the
screen on which interference pattern appears is 55cm
away.
a) What is the angular position of (i) first minima and (ii)
tenth maxima?
b) What is the separation between two adjacent
maxima?
DOUBLE-SLIT INTERFERENCE
Problem: E 41-2
Monochromatic light illuminates two parallel slits a
distance d apart The first maximum is observed at
an angular position of 15°. By what percentage
should d be increased or decreased so that the
second maximum will instead be observed at 15° ?
DOUBLE-SLIT INTERFERENCE
Problem: E 41-5
A double-slit arrangement produces interference
fringes for sodium light (wavelength = 589 nm) that
are 0.23° apart. For what wavelength would the
angular separation be 10% greater ? Assume that the
angle  is small.
DOUBLE-SLIT INTERFERENCE
Problem: E 41-8
In an interference experiment in a large ripple tank
(see Fig 41-2) the coherent vibrating sources are
placed
120 mm
apart.
The
distance
between
maxima 2.0 m away is 180 mm. If the speed of the
ripples is 25 cm/s, calculate the frequency of the
vibrating sources.
DOUBLE-SLIT INTERFERENCE
Problem: E 41-11
Sketch the interference pattern expected from using two
pin-holes rather than narrow slits.
COHERENCE
For interference
pattern to occur, the
phase difference at
point on the screen
must not change with
time.
A SECTION OF INFINITE WAVE
A WAVE TRAIN
OF FINITE LENGTH L
This is possible only when the two sources are completely
coherent.
If the two sources are completely independent light sources,
no fringes appear on the screen (uniform illumination) . This
is because the two sources are completely incoherent.
COHERENCE
A SECTION OF INFINITE WAVE
A WAVE TRAIN
OF FINITE LENGTH L
Common sources of visible light emit light wave trains of
finite length rather than an infinite wave.
The degree of coherence decreases as the length of wave
train decreases.
COHERENCE
A SECTION OF INFINITE WAVE
Two waves are said to
be coherent when
they are of :
• same amplitude
A WAVE TRAIN
OF FINITE LENGTH L
• same frequency
• same phase or are of
a constant phase
difference
Laser light is highly coherent whereas
a laboratory monochromatic light
source (sodium vapor lamp) may be
partially coherent.
INTENSITY IN DOUBLE SLIT INTERFERENCE
▪ Electric field components at P due to S1 and S2 are,
E1= E0 sin ωt & E2= E0 sin (ωt + ) respectively.
▪ Resultant field E = E1 + E2
INTENSITY IN DOUBLE SLIT INTERFERENCE
Resultant of E1= E0 sin ωt & E2= E0 sin (ωt + )
Phasor → Rotating vector.
ADDITION OF TWO VECTORS USING PHASORS
E2
Let two vectors be, E1= E0 sin ωt &
E0
E0
E1
E2= E0 sin (ωt + )
Resultant field E = E1 + E2
ωt + 
ωt
INTENSITY IN DOUBLE SLIT INTERFERENCE
Resultant of E1= E0 sin ωt & E2= E0 sin (ωt + )
From phasor diagram,
E = E1 + E2
E0
= E sin(t + )

E2
E
= 2E0 cos  sin(t + )
But  = /2. So above eqn can be
written as,
E = 2 E0 cos(/2) sin(wt+/2)

E

E0
E1
ωt
INTENSITY IN DOUBLE SLIT INTERFERENCE
▪ E = 2 E0 cos(/2) sin(wt+/2)
▪ So intensity at an arbitrary point P on the screen due to
interference of two sources having phase difference ;
I


4 E cos  
2
2
0
2

  4  0 cos  
2
where  = E2 is intensity due to single source
0
0
2
INTENSITY IN DOUBLE SLIT INTERFERENCE
PHASE AND PATH DIFFERENCE
Phase difference Path difference
=
2

Path difference  corresponds
to phase difference of 2.
INTENSITY IN DOUBLE SLIT INTERFERENCE



4  0 cos  
2
2
where  = E 2 is intensity due to single source
0
0
Since  = 2dsin/ ,
 =
2  d
4  0 cos 


sin  



From above equation,
At maxima :  = 2 m 
or
At minima :  = ( 2 m + 1) 
where m = 0,  1,  2, . . .
d sin 
or
= m
d sin 
= (m + 1 ) 
2
Light intensity (I) versus d sin θ for a double-slit
interference pattern when the screen is far from the
two slits (D>> d).
INTENSITY IN DOUBLE SLIT INTERFERENCE
INTENSITY IN DOUBLE SLIT INTERFERENCE
Problem: SP 41-2
Find graphically the resultant E(t) of the following wave
disturbances.
E1 = E0 sin t
E2 = E0 sin (t + 15o)
E3 = E0 sin (t + 30o)
E4 = E0 sin (t + 45o)
INTENSITY IN DOUBLE SLIT INTERFERENCE
Problem: E 41-15
Source A of long-range radio waves leads source B by 90
degrees. The distance rA to a detector is greater than the
distance rB by 100m. What is the phase difference at the
detector?
Both sources have a wavelength of 400m.
INTENSITY IN DOUBLE SLIT INTERFERENCE
Problem: E 41-18
Find the sum of the following quantities (a) graphically,
using phasors; and (b) using trigonometry:
y1 = 10 sin (t)
y2 = 8.0 sin (t + 30°)
INTERFERENCE FROM THIN FILMS
▪ A film of thickness of the order of a
micron.
▪ Thickness of the film is comparable
with the wavelength.
▪ Greater thickness spoils the coherence
of the light to produce colour.
A soapy water film on a
vertical loop viewed by
reflected light
Thickness and color in a thin film
INTERFERENCE FROM THIN FILMS
The region ac looks bright or dark for an observer depending
on the path difference between the rays r1 and r2.
INTERFERENCE FROM THIN FILMS
Phase change on Reflection
It has been observed that if the medium beyond the interface
has a higher index of refraction, the reflected wave undergoes a
phase change of  (=180o).
If the medium beyond the interface has a lower index of
refraction, there is no phase change of the reflected wave.
Phase changes on reflection at a
junction between two strings of
different linear mass densities.
INTERFERENCE FROM THIN FILMS
OPTICAL PATH
•
Distance traveled by light in a medium in the time interval
of ‘t’ is d = vt
•
Refractive index n = c/v
•
Hence, ct = nd
•
nd → Optical path.
•
Optical path is the distance traveled by light in vacuum in
same time ‘t’.
•
If n is wavelength in the film of refractive index n and  is
the wavelength in vacuum then n =  / n
INTERFERENCE FROM THIN FILMS
Equations for Thin Film Interference:
Normal incidence (i = 0)
Path difference = 2 d + (½) n
Constructive interference:
2 d + (½) n = m n
m = 1, 2, 3, . . .
(maxima)
Destructive interference:
2 d + (½) n = (m+½) n
m = 0, 1, 2, . . .
(minima)
What should be the minimum thickness (in terms of wavelength used) and refractive
index of a non- reflective coating on lens made up of glass?
Light is reflected from both surfaces of the coating. In both reflections
the light is reflected from a medium of greater index than that in which it
is traveling, so the same phase change occurs in both reflections.
The thickness of the non-reflective coating can be a quarter-wavelength. (t = λ/4) .
INTERFERENCE FROM THIN FILMS
WEDGE SHAPED FILM
In wedge – shaped thin film,
constructive interference occurs in
certain part of the film [2 d + (½) n =
m n] and destructive interference in
others [2 d + (½) n = (m+½) n].
Then bands of maximum and
minimum intensity appear, called
fringes of constant thickness.
Air Wedge – the air between two sheets of flat
glass angled to form a wedge
Air Wedge – the air between two sheets of flat
glass angled to form a wedge
Minima (destructive)
2𝑡 = 𝑚 𝜆
𝑥
𝛼
Maxima (constructive)
1
2𝑡 = (𝑚 − )𝜆
2
𝑡
tan α = t /x
𝑡 = 𝑥𝛼
𝑥𝑚
𝜆
1
=
𝑚 −
2𝛼
2
Δ𝑥𝑚
𝜆
=
2𝛼
INTERFERENCE FROM THIN FILMS
Problem: SP 41-3
A soap film (n=1.33) in air is 320nm thick. If it is
illuminated with white light at normal incidence, what
color will it appear to be in reflected light?
INTERFERENCE FROM THIN FILMS
Problem: SP 41-4
Lenses are often coated with thin films of transparent
substances such as MgF2 (n=1.38) to reduce the
reflection from the glass surface. How thick a coating is
required to produce a minimum reflection at the center
of the visible spectrum? ( wavelength = 550nm)
INTERFERENCE FROM THIN FILMS
Problem: E 41-23
A disabled tanker leaks kerosene (n=1.20) into the Persian
Gulf, creating a large slick on top of water (n = 1.33).
(a)If you look straight down from aeroplane on to the region
of slick where thickness is 460nm, for which wavelengths
of visible light is the reflection is greatest?
(b)If you are scuba diving directly under this region of slick,
for which wavelengths of visible light is the transmitted
intensity is strongest?
INTERFERENCE FROM THIN FILMS
Problem: E 41-25
If the wavelength of the incident light is λ = 572 nm,
rays A and B in Fig 41-24 are out of phase by
1.50 λ. Find the thickness d of the film.
INTERFERENCE FROM THIN FILMS
Problem: E 41-29
A broad source of light (wavelength = 680nm) illuminates
normally two glass plates 120 mm long that touch at one
end and are separated by a wire 0.048mm in diameter at
the other end. How many bright fringes appear over 120
mm distance?
Newton’s Rings
• Newton's rings are formed due to interference between
the light waves reflected from the top and bottom
surfaces of the air film formed between a plane and
curved surface of large radi of curvature.
Why different colors? for any given difference in path length, the
condition ΔL = (m-1/2) n might be satisfied for some wavelength but
not for some other. A given color might or might not be present in the
visible image.
Newton’s Ring Experiment
A plano-convex lens of large radius of curvature is placed on a plane glass plate to get
an air film of circular symmetry. This set up is placed below a traveling microscope.
The air film is illuminated normally by reflecting the horizontal beam of sodium
light using an inclined glass plate.
The traveling microscope is focused and the Newton’s rings
(bright and dark circular interference fringes) are observed.
INTERFERENCE FROM THIN FILMS
Newton’s rings (sample problem 41-5):
Constructive interference
2d = (m - ½) 
(n = 1 for air film)
d
=
=
r R  1
R−
R2 − r 2
2

r

 
R − R 1 − 
 
R 



1
2
using binomial expansion
2


1 r 
d = R − R 1 − 
 + . . .
2
R







r2
2R
INTERFERENCE FROM THIN FILMS
Newton’s rings
Substituting d in
2d = (m - ½) 
we get
r =
(m − 21 )  R
m = 1, 2, . . .
(maxima)
INTERFERENCE FROM THIN FILMS
Problem: E41-33
In a Newton’s ring experiment, the radius of curvature R of
the lens is 5.0m and its diameter is 20mm.
(a) How many ring are produced?
(b) How many rings would be seen if the arrangement is
immersed in water (n = 1.33)?
(Assume wavelength = 589nm)
MICHELSON’S INTERFEROMETER
▪ Light from an extended monochromatic
source P falls on a half-silvered mirror M.
▪ The incident beam is divided into reflected
and transmitted beams of equal intensity.
▪ These
two
beams
travel
almost
in
perpendicular directions and will be
reflected normally from movable mirror
(M2) and fixed mirror (M1).
MICHELSON’S INTERFEROMETER
▪ The two beams finally proceed
towards a telescope (T) through
which interference pattern of
circular fringes will be seen.
▪ The interference occurs because
the two light beams travel
different paths between M and M1
or M2.
▪ Each beam travels its respective
path twice. When the beams
recombine, their path difference is
2 (d2 – d1)
MICHELSON’S INTERFEROMETER
The path difference can be changed
by moving mirror M2. As M2 is
moved, the circular fringes appear to
grow or shrink depending on the
direction of motion of M2. New rings
appear at the center of the
interference pattern and grow
outward or larger rings collapse
disappear at the center as they
shrink.
MICHELSON’S INTERFEROMETER
For the center of the fringe pattern to
change from bright dark and to bright
again, the path difference between two
beams must change by one
wavelength, which means that mirror
M2 moves through a distance of /2. If
N fringes cross the field of view when
mirror M2 is moved by d, then
d = N (/2)
d is measured by a micrometer
attached to M2. Thus microscopic
length measurements can be made by
this interferometer.
Note, that the light rays going to mirror M2 traverse the beam splitter
three times before reaching the observer, whereas the rays going to mirror
M1 traverses it only once.
In order to achieve exact quality of path difference through glass, the
compensator plate of exactly the same thickness , refractive index and at
same inclination as Beam Splitter is introduced between M1 and beam
splitter.
Experimental set up
MICHELSON’S INTERFEROMETER
Problem: SP 41-6
Yellow light (wavelength = 589nm) illuminates a Michelson
interferometer. How many bright fringes will be counted as
the mirror is moved through 1.0 cm?
MICHELSON’S INTERFEROMETER
An airtight chamber 5.0 cm
long with glass windows is
placed in one arm of a
Michelson’s interferometer as
indicated in Fig 41-28 . Light
of wavelength λ = 500 nm is
used. The air is slowly
evacuated from the chamber
using a vacuum pump. While
the air is being removed, 60
fringes are observed to pass
through the view. From these
data find the index of
refraction of air at atmospheric
pressure.
Problem: E41-40
QUESTIONS – INTERFERENCE
What is the necessary condition on the path length difference
(and phase difference) between two waves that interfere (A)
constructively and (B) destructively ?
Obtain an expression for the fringe-width in the case of
interference of light of wavelength λ, from a double-slit of slitseparation d.
Explain the term coherence.
Obtain an expression for the intensity of light in double-slit
interference using phasor-diagram.
QUESTIONS – INTERFERENCE
Draw a schematic plot of the intensity of light in a double-slit
interference against phase-difference (and path-difference).
Explain the term reflection phase-shift.
Obtain the equations for thin-film interference.
Explain the interference-pattern in the case of wedge-shaped
thin-films.
Obtain an expression for the radius of mth order bright ring in
the case of Newton’s rings.
Explain Michelson’s interferometer. Explain how microscopic
length measurements are made in this.
B.TECH FIRST YEAR
ACADEMIC YEAR: 2020-2021
COURSE NAME: ENGINEERING PHYSICS
COURSE CODE
: PY1001
LECTURE SERIES NO : 01 (ONE)
CREDITS
:
4
MODE OF DELIVERY : ONLINE (POWER POINT PRESENTATION)
FACULTY
:
DR. PUSHPENDRA KUMAR
EMAIL-ID
:
pushpendra.kumar@jaipur.manipal.edu
DATE OF DELIVERY: 10 October 2021
SESSION OUTCOME
“UNDERSTAND THE BASIC
PRINCIPLES OF WAVE OPTICS”
ASSIGNMENT
QUIZ
MID TERM EXAMINATION –I
END TERM EXAMINATION
ASSESSMENT CRITERIA’S
Diffraction
Topics
▪
▪
▪
▪
▪
▪
▪
▪
▪
Diffraction and wave theory of light
Single-slit diffraction
Intensity in single-slit diffraction
Diffraction at a circular aperture
Double-slit interference and diffraction
combined
Multiple slits
Diffraction gratings
Dispersion and resolving power
X-ray diffraction
Text Book:
PHYSICS VOL 2 by Halliday, Resnick and Krane
DIFFRACTION AND INTERFERENCE
Both involve superposition of coherent light waves.
Diffraction and interference are similar phenomena.
Interference is the effect of
superposition of 2 coherent waves.
Here Slit width a<<λ very small and
neglected. So, fringes are of equal
width and intensity on the screen is of
uniform distribution.
Diffraction is the superposition
of many coherent waves. Slit
width ‘a’ is finite. Intensity on the
screen is non-uniform.
Diffraction v/s Interference
▪ Bending of light around the obstacle.
▪ The interfering beam originate from
continuous distribution of sources
(Huygens’ principle).
▪ The waves emerging from different
paths of the same wave front
superimpose with each other to
produce Diffraction pattern.
▪ The width of the diffraction fringes are
not equal.
▪ Minimum intensity point will not be
perfectly dark
▪ Bright fringes in the diffraction pattern
are not of same intensity.
▪ Meeting of two waves.
▪ The interfering beam originate from
discrete number of sources.
▪ The superposition of waves coming
from two different wave front
originating from the same source,
produce Interference pattern.
▪ The width of the interference fringes
may/ may not be equal.
▪ Minimum intensity point will be
perfectly dark.
▪ Bright fringes in the interference
pattern are of uniform intensity.
DIFFRACTION AND WAVE THEORY OF LIGHT
The phenomenon of bending of light around the edges of
obstacles or slits, and hence its encroachment into the region
of geometrical shadow is known as diffraction.
For diffraction effects to be noticeable, the size of the object
causing diffraction should have dimensions comparable to the
wavelength of light falling on the object.
Diffraction pattern of razor blade viewed in
monochromatic light
DIFFRACTION AND WAVE THEORY OF LIGHT
•
•
Diffraction pattern occurs when coherent wave-fronts of light
fall on opaque barrier B, which contains an aperture of
arbitrary shape. The diffraction pattern can be seen on screen
C.
When C is very close to B a geometric shadow is observed
because the diffraction effects are negligible.
DIFFRACTION AND WAVE THEORY OF LIGHT
Ꙫ A single slit placed between a distant light source and a
screen produces a diffraction pattern.
Ꙫ It will have a broad, intense central band called the
central maximum
Ꙫ The central band will be flanked by a series of narrower,
less intense secondary bands called side maxima or
secondary maxima
Ꙫ The central band will also be bordered by a series of
dark bands called minima.
Ꙫ The diffraction pattern consists of the central maximum
and a series of secondary maxima and minima.
Ꙫ The pattern is similar to an interference pattern as
shown in figure.
Huygens’ Principle
• Every point on a propagating
wavefront serves as the source
of spherical wavelets, such that
the wavelets at sometime later
is the envelope of these
wavelets.
• If a propagating wave has a
particular frequency and speed,
the secondary wavelets have
that same frequency and
speed.
“Isotropic”
The phenomenon of diffraction is caused by the interference of
innumerable secondary wavelets that are produced by
unobstructed portions of the same wave front or from the portions
of the wave front which are allowed to pass through a aperture.
Fresnel diffraction & Fraunhofer diffraction
Diffraction patterns are usually classified into two
categories depending on the source and screen are placed.
Fresnel diffraction: When either the source or the screen is
near the aperture or obstruction, the wavefronts are
spherical and the pattern is quite complex. (near-field)
Fraunhofer diffraction: When both the source and the screen
are at a great distance from the aperture or obstruction,
the incident light is in the form of plane wave and
the pattern is simpler to analyze. (far-field)
DIFFRACTION AND WAVE THEORY OF LIGHT
The pattern formed on the screen depends on the
separation between the screen C and the aperture B. Let us
consider the following three cases.
Case 1: Very small separation: when screen C is very close to B.
Aperture
From
distant
source
Screen
Geometrical
shadow of the
aperture
B
C
The waves travel only a short distance after leaving the
aperture, and the rays diverge very little. The effects
diffraction are negligible, and the pattern on the screen is
the geometric shadow of the aperture.
Case 2: Very large separation: When screen C is far from
the aperture (Fraunhofer diffraction).
Plane wave front
Very large separation
Plane wave front
Very large separation
C
When the screen is so far from the aperture, then we can
regard the rays as parallel or wavefronts as planes.
In this case , we also assume the source to be far from the
aperture, so that the incident wavefronts are also planes.
This is one way of achieving Fraunhofer diffraction.
f
f
In the laboratory, two converging lenses are used to achieve
this condition.
The first lens converts the diverging light from the source in
to a plane wave, and the second lens focuses plane waves
leaving the aperture parallel to the point on screen.
Case 3: Intermediate separation: When screen C and
source (S) are at finite distances from the aperture (Fresnel
diffraction):
Spherical
wave front
Spherical
wave front
P
Source (S)
Finite distance
B
Finite distance
C
In Fresnel diffraction:
1. the incident and the diffracted wave fronts are
spherical.
2. The source and the screen are at finite distances from
the aperture/slit or obstacle causing diffraction.
3. No lenses/ mirrors are used.
Plane wave front
Very large separation
Plane wave front
Very large separation
C
In Fraunhofer type of diffraction:
1. Both the source and the screen are effectively at infinite
distances, from the aperture causing diffraction.
2. both the incident and emergent wavefronts are plane. That is,
both the incident and the diffracted beams are parallel.
3. can be realized in practice by using a pair of converging lenses
of suitable focal lengths (L1 and L2) and placing the source
and the screen at the foci of L1 and L2 respectively.
Fraunhofer diffraction is a special (limiting) case of
the more general Fresnel diffraction.
But, analysis of Fresnel diffraction is complicated
compared to Fraunhofer. That is, Fraunhofer
diffraction is easier to handle mathematically.
So, in our study on diffraction phenomenon, we
deal only with Fraunhofer diffraction.
SINGLE-SLIT DIFFRACTION
All the diffracted rays arriving at P0 are in-phase.
Hence they interfere constructively and produce maximum
(central maximum) of intensity I0 at P0.
SINGLE-SLIT DIFFRACTION
⁕The finite width of slits is the basis for understanding Fraunhofer diffraction.
⁕According to Huygens’s principle, each portion of the slit acts as a source of light waves.
⁕Therefore, light from one portion of the slit can interfere with light from another portion.
⁕The diffraction pattern is actually an interference pattern.
⁕The different sources of light are different portions of the single slit.
SINGLE-SLIT DIFFRACTION
At point P1,
path difference between r1
and r2 is
(a/2) sin
So the condition for first minimum,
a

sin  =
2
2
or
a sin  = 
This is satisfied for every pair of rays, one of which is from upper half
of the slit and the other is a corresponding ray from lower half of the
slit.
SINGLE-SLIT DIFFRACTION
At point P2,
path difference between
r1 and r2 is (a/4) sin
So the condition for second minimum,
a

sin  =
or
a sin  = 2
4
2
This is satisfied for every pair of rays, separated by a distance a/4.
In general, the condition for m TH minima,
a sin  = m
m =  1,  2,  3, . . .
There is a secondary maximum approximately half way between
each adjacent pair of minima.
SINGLE-SLIT DIFFRACTION
Problem: 1
A slit of width a is illuminated by white light. For what value
of a does the minimum for red light ( = 650nm) fall at  =
15o?
SINGLE-SLIT DIFFRACTION
Problem: 2
In P-1, what is the wavelength ’ of the light whose first
diffraction maximum (not counting the central maximum)
falls at 15o, thus coinciding with the first minimum of red
light?
SINGLE-SLIT DIFFRACTION
Problem: E42-5
A single slit is illuminated by light whose wavelengths are
a and b, so chosen that the first diffraction minimum of
a component coincides with the second minimum of the
b component.
(a) What is the relationship between the two
wavelengths?
(b) Do any other minima in the two patterns coincide?
INTENSITY IN SINGLE – SLIT DIFFRACTION
• Aim is to find an expression for the intensity of the entire
pattern as a function of the diffraction angle.
• The phase difference between two waves arriving at point P
from two points on the slit (with separation x) is,
2
 =
x sin 

INTENSITY IN SINGLE – SLIT DIFFRACTION
Phasor showing
a) Central maximum
b) A direction slightly shifted
from central maximum
c) First minimum
d) First maximum beyond the
central maximum
(corresponds to N = 18)
INTENSITY IN SINGLE – SLIT DIFFRACTION
From diagram,
E = 2 R sin

2
Em
Also  =
R
Combining,
Em

E =
sin

2
2
sin 
Or , E = Em
where  =

2

INTENSITY IN SINGLE – SLIT DIFFRACTION
 is the phase difference
between rays from the top
and bottom of the slit.
So we can write,
2
=
a sin 

 a
So,  = =
sin 
2

INTENSITY IN SINGLE – SLIT DIFFRACTION
The intensity    E 2
2
=
2  sin  
Em


  
2
 sin  
2
m 
 where  m  Em is the max. intensity
  
From the above eqn., for minima, sin = 0
  = m where m = 1,2,3,.....
or, a sin  = m where m = 1,2,3,.....
 =
INTENSITY IN SINGLE – SLIT DIFFRACTION
The intensity distribution in
single-slit diffraction for three
different values of the ratio a/
Position of dark fringes in single-slit diffraction
m
sin  =
a
If, like the Young’s 2-slit treatment we assume small angles, sin ≈ tan  =ymin/D, then
ymin
Dm
=
a
Positions of intensity
MINIMA of diffraction
pattern on screen,
measured from central
position.
Very similar to expression derived for 2-slit experiment:
m
ym = D
d
But remember, in this case ym are positions of MAXIMA
In interference pattern
Width of central maximum
•We can define the width of the central maximum to be the distance
between the m = +1 minimum and the m=-1 minimum:
D
D 2 D
y =
−−
=
a
a
a
the narrower the slit,
the more the diffraction
pattern “spreads out”
a sin θ = λ first minima
Intensity
distribution
image of diffraction
pattern
If we narrow the slit the
angle must get bigger more flaring
- what happens when a
= λ?
INTENSITY IN SINGLE – SLIT DIFFRACTION
Problem: SP42-3
Calculate, approximately, the relative intensities of the
maxima in the single slit Fraunhofer diffraction pattern.
INTENSITY IN SINGLE – SLIT DIFFRACTION
Problem: SP42-4
Find the width  of the central maximum in a single slit
Fraunhofer diffraction. The width can be represented as the
angle between the two points in the pattern where the
intensity is one-half that at the center of the pattern.
INTENSITY IN SINGLE – SLIT DIFFRACTION
Problem: E42-11
Monochromatic light with wavelength 538 nm falls on a slit
with width 25.2m. The distance from the slit to a screen is
3.48m. Consider a point on the screen 1.13cm from the
central maximum. Calculate (a)  (b)  (c) ratio of the
intensity at this point to the intensity at the central
maximum.
DIFFRACTION AT A CIRCULAR APERTURE
DIFFRACTION PATTERN DUE TO A CIRCULAR APERTURE
DIFFRACTION AT A CIRCULAR APERTURE
The mathematical analysis of diffraction by a circular aperture
shows that the first minimum occurs at an angle from the

central axis given by sin  = 1.22
d
where d is the diameter of aperture.
The equation for first minimum in single slit diffraction is

sin  =
a
where a is the slit width
In case of circular aperture, the factor 1.22 arises when we divide the
aperture into elementary Huygens sources and integrate over the
aperture.
DIFFRACTION AT A CIRCULAR APERTURE
Raleigh’s criterion for optical resolution: The images of two closely
spaced sources is said to be just resolved if the angular separation of
the two point sources is such that the central maximum of the
diffraction pattern of one source falls on the first minimum of the
diffraction pattern of the other.


R = sin−11.22 
d

since R is very small, it can be appoximated as
R = 1.22

d
R is the smallest angular
separation for which we
can resolve the images of
two objects.
a. Not resolved
b. Just resolved
c. Well resolved
Using a circular instrument (telescope, human eye),
when can we just resolve two distant objects?


 R = sin 1.22 
d

 R is very small, since the sources are distant and closely spaced
−1
R is the smallest angular separation
for which we can resolve the images of two objects.
DIFFRACTION AT A CIRCULAR APERTURE
Problem: SP42-5
A converging lens 32mm in diameter has a focal length f of 24
cm. (a) What angular separation must two distant point
objects have to satisfy Rayleigh’s criterion? Assume that  =
550nm. (b) How far apart are the centers of the diffraction
patterns in the focal plane of the lens?
DOUBLE-SLIT INTERFERENCE AND
DIFFRACTION
1.Young’s double-slit experiment is an idealized
situation in which slits were assumed to be very narrow
(a << λ).
2.This cannot occur with actual slits because the
condition a << λ cannot usually be met.
In case of double slit experiment:
Each of the two slit has a finite width and
the light is diffracted through it in a single slit diffraction pattern
Usually, the width of each of the slits (a) is quite a bit less than the
separation (d) between their centers
Two single slit diffraction pattern is to be superposed
to get the final intensity
The final result can be obtained by drawing a two slit interference
pattern inside the envelope of a single slit diffraction pattern
DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COMBINED
DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COMBINED
Interference
(cos β)
2
I  , INT = I m, INT
Diffraction
I , DIF =
 sin α 
 α 


 m, DIF
Interference + Diffraction
I =
m
(cos )
2
 sin α 


α


2
2
DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COMBINED
Each of the two slits is divided into N zones. Electric field at P is
found by adding the phasors. There is phase difference of  =
/N between each of the N phasors where  is the phase
difference between1st phasor and Nth phasor.
DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COMBINED
Adding all the phasors, we get the resultant E1 due to the first slit.
 is the phase difference between the light waves at the point P,
emitted from bottom edge of the first slit and top edge of the
second slit. E2 is the resultant due to the second slit. E is the
resultant of E1 and E2.
DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COMBINED
From the figure,

E  = 2E1 sin
2


where +  + +  = 
2
2
or
 =  − (  + )

 + 
+
Also sin = sin −
 = cos
 .........( A )
2
2 
2
 2 
 
and
= (d − a) sin 
2 
 a
Adding =
sin  to both sides of above eqn, we get,
2

+ 
= d sin  which is 
2

DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COMBINED
Substituting this in eqn( A ), we get,

sin = cos
2
From sin gle − slit diffraction, we have ,
the electric amplitude at P due to one slit,
 sin  
E1 = Em 

  

 E  = 2E1 sin
2
 sin  
ie, E  = (2Em )
 cos
  
 sin  
   =  m (cos)2 

  
DOUBLE-SLIT
INTERFERENCE PATTERN
2
SINGLE-SLIT DIFFRACTION
PATTERN
DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COMBINED
Problem: SP42- 6
Ina double slit experiment, the distance D of the screen
from the slits is 52cm, the wavelength is 480nm, slit
separation d is 0.12mm and the slit width a is 0.025mm.
a) What is the spacing between adjacent fringes?
b) What is the distance from the central maximum to the
first minimum of the fringe envelope?
DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COMBINED
Problem: SP42- 7
What requirements must be met for the central maximum
of the envelope of the double-slit interference pattern to
contain exactly 11 fringes?
Interference pattern from 3 slits
MULTIPLE SLITS: Width of the Central maximum
For N slits, intensity of the primary maxima is N 2 times greater than that due to a
single slit.
For N slits, intensity of the primary maxima is N 2 times greater than that due to a
single slit.
As the number of slits increases, the primary maxima increase in intensity and become
narrower, while the secondary maxima decrease in intensity
As number of slits increases, number of secondary maxima also increases. In fact,
the number of secondary maxima is always N - 2, where N is the number of slits.
MULTIPLE SLITS
Multiple slit arrangement
will be the interference
pattern multiplied by the
single slit diffraction
envelope. This assumes
that all the slits are
identical.
MULTIPLE SLITS
Condition for principal
maxima,
d sin  = m 
where d is the
separation between
adjacent slits.
Location of principal
maxima is independent
of number of slits.
MULTIPLE SLITS
Intensity pattern for
(a) Two-slit diffraction
(b) Five-slit diffraction
(diffraction effect is
neglected)
MULTIPLE SLITS
Width of the maxima: Central maximum
▪ The pattern contains central maximum with minima on
either side.
▪ At the location of central maximum, the phase difference
between the waves from the adjacent slits is zero.
▪ At minima, the phase difference is such that,
2
 =
where N is the number of slits
N
▪ Corresponding path difference is,

  
L = 
 =
N
 2 
MULTIPLE SLITS
Width of the maxima: Central maximum

  
L = 
 =
N
 2 
▪ Also we know,
L = d sin 0

= d sin 0
N

sin 0 =
Nd

0 
Nd
From the equation, for given  and
d if we increase number of slits (N),
then the angular width of principal
maximum decreases. ie the
principal
maximum
becomes
sharper.
MULTIPLE SLITS
Width of the maxima: Other principal maxima
For the mth principal
maximum at  by a
grating: d sin = m .
For the first minimum
at  +  after the mth
principal maximum
d sin(θ + θ)
λ
= mλ +
N
MINIMUM AT θ
+θ
mth PRINCIPAL
MAXIMUM AT θ
MULTIPLE SLITS
Width of the maxima: Other principal maxima
d sin(θ + θ)
=
λ
mλ +
N


d sin  cos
 + cos  sin




 
1

d
sin
 + (d cos) 

m + (d cos) 

 =
N d cos 
=
=

= m +
N
m +  N
m +  N
ANGULAR HALF WIDTH OF mTH
PRINCIPAL MAXIMUM AT 
The principal maximum become sharper as
number of slits (N) increases
MINIMUM AT θ
+θ
mth PRINCIPAL
MAXIMUM AT θ
MULTIPLE SLITS
Problem: SP43- 1
A certain grating has 104 slits with a spacing of d = 2100 nm.
It is illuminated with yellow sodium light ( = 589 nm). Find
(a) the angular position of all principal maxima observed
and (b) the angular width of the largest order maximum.
MULTIPLE SLITS
Problem: E43-5
Light of wavelength 600 nm is incident normally on a
diffraction grating. Two adjacent principal maxima occur at sin
 = 0.20 and sin  = 0.30. The fourth order is missing. (a) what
is the separation between adjacent slits? (b) what is the
smallest possible individual slit width? (c) Name all orders
actually appearing on the screen with the values derived in (a)
and (b).
Grating contains : greater
number of slits, or rulings,
as many as several 1000
per millimeter
Diffraction Gratings
Grating contains : greater number of slits,
or rulings, as many as several 1000 per
millimeter
Light passed through the grating forms
narrow interference fringes that can be
analyzed to determine the wavelength
As the number of rulings increases beyond 2
the intensity plot changes from that of a
double slit pattern to one with very narrow
maxima (called lines) surrounded by
relatively wide dark regions
DIFFRACTION GRATINGS
The diffraction grating, a useful device for analyzing light sources,
consists of a large number of equally spaced parallel slits.
▪ A transmission grating can be made by cutting parallel grooves on
a glass plate with a precision ruling machine. The spaces between
the grooves are transparent to the light and hence act as separate
slits.
▪ A reflection grating can be made
by cutting parallel grooves
on the surface of a reflective
material.
The reflection of light from
the spaces between the grooves
is specular, and the reflection
from the grooves cut into
the material is diffuse.
72
the condition for
interference maxima at a
specific angle () is
We can use this expression
to calculate the wavelength
if we know the grating
spacing d and the angle ().
If the incident radiation contains several
wavelengths, the mth-order maximum
for each wavelength occurs at
a specific angle.
All wavelengths are seen at  = 0,
corresponding to m = 0, the
zeroth-order maximum.
The first-order maximum (m = 1) is
observed at an angle that satisfies
the relationship sin  = λ/d.
The second-order maximum (m =2) is
observed at a larger angle , and so on.
DIFFRACTION GRATINGS
Grating spectrometer
m=0
m=1
m=2
m=3
Sample spectra of visible light emitted by a gaseous source
DIFFRACTION GRATINGS
Problem: SP43-2
A diffraction grating has 1.20 x 104 rulings uniformly
spaced over W= 2.50cm. It is illuminated at normal
incidence by yellow light from sodium vapor lamp which
contains two closely spaced lines of wavelengths 589.0nm
and 589.59nm. (a) At what angle will the first order
maximum occur for the first of these wavelengths? (b)
What is the angular separation between the first order
maxima of these lines? (c) How close in wavelength can
two lines be (in first order) and still be resolved by this
grating? (d) How many rulings can a grating have and just
resolve the sodium doublet lines?
DIFFRACTION GRATINGS
Problem: E43-9
Given a grating with 400 rulings/mm, how many orders of
the entire visible spectrum (400-700nm) can be produced?
A grating has 315 rulings / mm. For what wavelengths in the
visible spectrum can fifth-order diffraction be observed?
DIFFRACTION GRATINGS
Problem: E43-11
White light (400 nm <  < 700 nm) is incident on a grating .
Show that, no matter what the value of the grating spacing d,
the second- and third-order spectra overlap.
DISPERSION AND RESOLVING POWER
The ability of a grating to produce spectra that permit precise
measurement of wavelengths is determined by two intrinsic
properties of the grating,
(1) Dispersion: High dispersive powers refers to the wide separation
of the spectral lines
(2) Resolving power: Ability of the instrument to show the
close spectral lines as the separate ones
DISPERSION
A grating with high dispersive power must widely spread apart the
diffraction lines associated with nearly equal wavelengths.
Angular separation between spectral lines
Dispersion=
Difference between wavelength of spectral lines
D =
Δθ
Δλ
DISPERSION AND RESOLVING POWER
Dispersion
Δθ
D =
Δλ
d sin = m 
Differentiating the above equation,
d cos   = m 
Δθ
D=
Δλ
=
m
d cos θ
To achieve higher dispersion we must use a grating of smaller
grating spacing and work in higher order m .
RESOLVING POWER
Ability of grating to resolve two nearby spectral lines so that
two Lines can be viewed or photographed as separate lines.
To resolve lines whose wavelengths are close together, the lines
should be as narrow as possible.
For two close spectral lines of wavelength 1 and 2, just
resolved by the grating, the resolving power is defined as
1 +  2



=

−

1
2
=
R=
2

The limit of resolution is determined by
the Rayleigh criterion
the minimum wavelength separation we
can resolve min  2-1 occurs when
the maximum of 2 overlaps with the
first diffraction minimum of 1.
84
DISPERSION AND RESOLVING POWER
Resolving power
We have,
Δθ
D=
Δλ
=
m
d cos θ

 =
N d cos 
Putting second equation in first equation,





 N d cos  

=
m
d cos 

R=
= Nm

Resolving power increases with increasing N
DISPERSION AND RESOLVING POWER
Resolving power
N = 5,000
d = 10 m
R = 5,000
D = 1.0 x 10-4 rad/m
N = 5,000
d = 5 m
R = 5,000
D = 2.0 x 10-4 rad/m
N = 10,000
d = 10 m
R = 10,000
D = 1.0 x 10-4 rad/m
Intensity patterns of two close
lines due to three gratings A, B, C.
DISPERSION AND RESOLVING POWER
Problem: SP43-3
A grating has 9600 lines uniformly spaced over a width
3cm and is illuminated by mercury light.
a) What is the expected dispersion in the third order, in
the vicinity of intense green line ( = 546nm)?
b) What is the resolving power of this grating in the fifth
order?
DISPERSION AND RESOLVING POWER
Problem: SP43-4
A diffraction grating has 1.20 X 104 rulings uniformly spaced
over a width W = 2.50cm. It is illuminated at normal
incidence by yellow light from a sodium vapor lamp. This
light contains two closely spaced lines of wavelengths 589.0
nm and 589.59 nm. (a) At what angle does the first
maximum occur for the first of these wavelengths? (b) What
is the angular separation between these two lines (1st
order)? (c) How close in wavelength can two lines be (in first
order) and still be resolved by this grating? (d) How many
rulings can a grating have and just resolve the sodium
doublet line?
DISPERSION AND RESOLVING POWER
Problem: E43-17
The sodium doublet in the spectrum of sodium is s pair of
lines with wavelengths 589.0 and 589.6 nm. Calculate the
minimum number of rulings in a grating needed to resolve
this doublet in the second-order spectrum.
DISPERSION AND RESOLVING POWER
Problem: E43-21
In a particular grating, the sodium doublet is viewed in
third order at 10.2 to the normal and is barely resolved.
Find (a) the ruling spacing and (b) the total width of
grating.
X-RAY DIFFRACTION
For the observation of diffraction phenomenon by grating, the
grating space should have the dimension of the wavelength of
the wave diffracted. Since the x-ray wavelength and the interplanar spacing in crystals are of the same order, a crystal can be
a suitable grating for observing the diffraction of x-rays.
x-ray diffraction
producing Laue’s
pattern
X-ray tube
X-RAY DIFFRACTION
▪ When a monoenergetic x-ray beam is
incident on a sample of a single crystal,
diffraction occurs resulting in a pattern
consisting of an array of symmetrically
arranged diffraction spots, called Laue’s
spots.
▪ The single crystal acts like a grating
with a grating constant comparable
with the wavelength of x-rays, making
the diffraction pattern distinctly visible.
▪ Since the diffraction pattern is decided
by the crystal structure, the study of
the diffraction pattern helps in the
analysis of the crystal parameters.
A Laue pattern of a
single crystal.
Each dot
represents a
point of
constructive
interference.
X-RAY DIFFRACTION
A plane through a crystal of NaCl
NaCl crystal (a0 = 0.563nm)
NaCl unit cell
X-RAY DIFFRACTION
(a) Electron density contour of an organic molecule
(b) A structural representation of same molecule
The x-rays are diffracted by the electron concentrations in the
material. By studying the directions of diffracted x-ray beam, we
can study the basic symmetry of the crystal. By studying the
intensity, we can learn how the electrons are distributed in a unit
cell.
X-RAY DIFFRACTION
Bragg’s Law
▪ In every crystal, several sets of parallel planes called the Bragg
planes can be identified.
▪ Each of these planes have an identical and a definite
arrangement of atoms.
▪ Different sets of Bragg planes are oriented at different angles
and are characterized by different inter planar distances d.
X-RAY DIFFRACTION
Bragg’s Law
→ Glancing angle. ie angle
between the incident x-ray beam
and the reflecting crystal planes.
For constructive interference of
diffracted x-rays the path
difference for the rays from the
adjacent planes, (abc in the
figure) must be an integral
number of wavelength.
ie
2d sin  = n 
X-RAY DIFFRACTION
Problem: SP43-5
At what angles must an x-ray
beam with wavelength = 0.110
nm fall on the family of planes
in figure if a diffracted beam is
to exist? Assume material to
be sodium chloride (a0 =
0.563nm)
X-RAY DIFFRACTION
Problem: E43-25
A beam of x-rays of wavelength 29.3 pm is incident on a
calcite crystal of lattice spacing 0.313 nm. Find the smallest
angle between the crystal planes and the beam that will
result in constructive reflection of the x-rays.
QUESTIONS – DIFFRACTION
Discuss the diffraction due to single-slit. Obtain the
locations of the minima and maxima qualitatively.
Obtain an expression for the intensity in single-slit
diffraction pattern, using phasor-diagram.
Calculate, approximately, the relative intensities of the first
three secondary maxima in the single-slit diffraction
pattern.
Discuss qualitatively diffraction at a circular aperture.
QUESTIONS – DIFFRACTION
Explain Rayleigh’s criterion for resolving images due to a
circular apperture.
Obtain an expression for the intensity in double-slit
diffraction pattern, using phasor-diagram.
Discuss qualitatively the diffraction due to multiple slits
(eg, 5 slits).
Obtain an expression for the width of the central
maximum in diffraction pattern due to multiple slits.
QUESTIONS – DIFFRACTION
Obtain an expression for the width of a principal
maximum at an angle in diffraction pattern due to
multiple slits.
Obtain an expression for dispersion by a diffraction
grating.
Obtain an expression for resolving power of a diffraction
grating.
Discuss Bragg’s law for X-ray diffraction.
08-11-2022
Polarisation
TOPICS
Polarization (also polarisation) is a property of certain types
of waves that describes the orientation of their oscillations.
 POLARIZATION OF
ELECTROMAGNETIC WAVES
Electromagnetic waves, such as light, exhibit polarization;
 POLARIZING SHEETS
Acoustic waves (sound waves) in a medium such as a gas or
liquid do not have polarization because the direction of
vibration and direction of propagation are the same.
 POLARIZATION BY REFLECTION
 DOUBLE REFRACTION
 CIRCULAR POLARIZATION
 Specific Rotation
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2
POLARIZATION OF ELECTROMAGNETIC WAVES
POLARIZATION OF ELECTROMAGNETIC WAVES
• By convention, we define the
direction of polarization of
the EM wave to be the
direction of the electric vector
(E).
(a)Representation of a
linearly polarized wave
with the electric field
vibrating in the
vertical direction
viewed along the
direction of
propagation.
• The plane determined by
electric vector and direction
of propagation of wave is
called plane of polarization
(xy plane in figure)
MIT-MANIPAL
3
Plane electromagnetic wave
BE-PHYSICS-POLARIZATION-2010-11
3
(b)Representation of an unpolarized wave viewed along the direction of
propagation (perpendicular to the page). The transverse electric field
can vibrate in any direction in the plane of the page with equal
probability.
(c)An equivalent representation of the unpolarized wave, as two waves
linearly polarized at right angles to one another and with a random
phase difference between them. Orientation of y and z axis: arbitary.
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08-11-2022
Polarizer- a device/sheet that allows only light with an electric
field along a singe direction to pass through
Polarized Light – light waves that vibrate in a single plane
Unpolarized light – light waves that vibrate in many
different planes
Polaroid  Polarizing
material
The polarizing direction
is established during the
manufacturing process
of the sheet.
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6
Two polarizing sheets
whose transmission
axes make an angle 
with each other. Only a
fraction of the polarized
light incident on the
analyzer is transmitted
through it.
Law of Malus
The law states that the intensity (I) of the polarised light
emerging from the analyser, is directly proportional to
the square of the cosine of the angle ()
between the polariser and the analyser
 Malus law is valid for POLARISED light
 In UNPOLARISD light: The transverse electric field can vibrate in
any direction in the plane perpendicular to the direction of the
polarisation with equal probability.
If E is the magnitude of electric
vector, only E cos  (y component)
passes through the analyser.
Transmitted intensity
I = Im cos2  [Law of
Malus]
where Im = Imax  maximum
intensity (ie  = 0 or 180 degree)
7
 For Unpolarised light : I = Im <cos2 > = Im /2
 An ideal polariser is one that transmits 50% of the
incident unpolarised light as plane-polarized one.
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08-11-2022
POLARIZING SHEETS
POLARIZING SHEETS
Problem: SP44-1
Problem: E44-8
A beam of light is linearly polarized in the vertical direction.
The beam falls at normal incidence on a polarizing sheet
with its polarizing direction at 58.8 to the vertical. The
transmitted beam falls, also at normal incidence, on a
second polarizing sheet with its polarizing direction
horizontal. The intensity of the original beam is 43.3 W/m2.
Find the intensity of the beam transmitted by the second
sheet.
Two polarizing sheets have their polarizing directions
parallel so that the intensity Im of the transmitted light is a
maximum. Through what angle must either sheet must be
turned if the intensity is to drop by one-half?
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POLARIZATION BY REFLECTION
Brewster’s Law
When an unpolarized light beam is reflected from a surface, the
polarization of the reflected light depends on the angle of incidence.
The reflected beam is completely
polarized when the angle of
incidence equals the polarizing
angle p, which satisfies the
equation n2/n1 = tan p.
(a) When unpolarized light is incident
on a reflecting surface, the reflected
and refracted beams are partially
polarized.
(b) The reflected beam is completely
polarized when the angle of incidence
equals the polarizing angle p, which
satisfies the equation n2/n1 = tan p.
At this incident angle p,
the reflected and refracted rays are
perpendicular to each other
p = polarizing angle
11
POLARIZATION BY REFLECTION
At this incident angle p,
the reflected and refracted rays are
perpendicular to each other.
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08-11-2022
POLARIZATION BY REFLECTION
Brewster’s Law
When angle of incidence is p, it
is observed that, p + r = 90
n2/n1 = tan p
From Snell’s Law,
n1 sin p = n2 sin r
Therefore,
n1 sin p = n2 sin (90- p)
n1 sin p = n2 cos (p)
This expression is called Brewster’s law, and the
polarizing angle p is called Brewster’s angle.
So n2/n1 = tan p
Because wavelength varies with n for a given
substance (n =  / n ), Brewster’s angle is also a
function of wavelength.
 1/  2 = tan p
If 1st medium is air then,
n = tan p
MIT-MANIPAL
BE-PHYSICS-POLARIZATION-2010-11
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14
POLARIZATION BY REFLECTION
POLARIZATION BY REFLECTION
Polarization of light by stack
of glass plates
Problem: E44-12
When red light in vacuum is incident at the polarizing
angle on a certain glass slab, the angle of refraction is
31.8. What are (a) the index of refraction of the glass and
(b) the polarizing angle?
MIT-MANIPAL
15
BE-PHYSICS-POLARIZATION-2010-11
Unpolarized light is incident
at the angle p. All reflected
lights are polarized
perpendicular to the plane of
figure. After passing through
the several layers, the
transmitted wave no longer
contains any appreciable
component polarized
perpendicular (normal) to
the figure (page).
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POLARIZATION BY REFLECTION
Isotropic material (= cubic)
Problem: SP44-2
We wish to use a plate of glass (n = 1.50) in air as
polarizer. Find the polarizing angle and angle of refraction.
Sphere
n is constant is every direction isotropic minerals do not change
the vibration direction of the
light - no polarisation
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POLARIZATION BY DOUBLE REFRACTION
In optically isotropic substances
(liquids,
amorphous solid such as glass,
crystalline solids having cubic symmetry)
the speed of light and the index of refraction are
independent of
the direction of propagation in the medium
the state of polarization of light.
In anisotropic crystalline materials
(such as calcite and quartz)
the speed of light is not the same in all directions.
In anisotropic crystalline materials,
the speed of light and the index of refraction depends on:
the direction of propagation & the plane of polarization of the light.
Such materials are characterized by two indices of refraction.
Hence, they are referred to as double-refracting or birefringent
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08-11-2022
If light enters a birefringent material (such as Calcite) at an angle to
the optic axis, the different indices of refraction will cause the two
polarized rays to split and travel in different directions :
a phenomenon called double refraction.
What is Bi-refringent material? Explain the phenomenon of
double refraction with diagram indicating the directions
of polarizations for two beams.
Unpolarized light incident on a birefringent material splits into an
ordinary (o) ray & extraordinary (e) ray.
These two rays [ o & e] are polarized in mutually perpendicular directions.
This phenomenon is a result of
different propagation velocities, v||
and v , associated with each electric
field component.
the propagation velocities can be
expressed as follows:
v|| = ve = c/ne
ne = 1.486
v┴ = vo =c/no no = 1.658
For Calcite no > ne
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22
DOUBLE REFRACTION:
Optic Axis
The o-ray polarized perpendicular to optic axis (v┴ )
 The o-wave travels in the crystal with the same speed vo in
all directions.
 The o-ray obeys Snell’s Law of refraction.
The characteristic direction in
the crystal in which vo = ve is
called optic axis. !!!!!!
 The crystal has a single index of refraction no for o-wave.
v|| = ve = c/ne
v┴ = vo =c/no no = 1.658
The e-ray polarized parallel to (along the) optic axis (v|| ).
 The e-wave travels in the crystal with a speed that varies
with direction from vo to ve.
A point source S inside a double-refracting crystal produces a
spherical wave front corresponding to the ordinary ray and an
elliptical wave front corresponding to the extraordinary ray.
The two waves propagate with the same velocity along the
optic axis.
24
 It does not obey the Snell’s Law.
 The index of refraction of the crystal varies with direction
from no to ne for the e-wave.
23
ne = 1.486
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08-11-2022
DOUBLE REFRACTION
Principal indices of refraction (no, ne) of some doubly
refracting crystals for sodium light
A calcite crystal produces a double image because it is a
birefringent (double-refracting) material. These two images
correspond to one formed by the ordinary ray and one formed by
the extraordinary ray.
Crystal
Formula
no
ne
ne-no
Ice
H2O
1.309
1.313
+0.004
Quartz
SiO2
1.544
1.553
+0.009
Wurzite
ZnS
2.356
2.378
+0.002
Calcite
CaCO3
1.658
1.486
-0.172
For Calcite no > ne
Birefringence = Δn = ne − no
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26
If the two images are viewed through a sheet of rotating polarizing
glass, they alternately appear and disappear because the ordinary
(o) and extraordinary (e) rays are plane-polarized along mutually
perpendicular directions.
For Calcite no > ne
The shape of the ellipsoids depends on sign of n (+ or -) as shown.
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08-11-2022
QUESTIONS – POLARIZATION
QUESTIONS – POLARIZATION
Sketch the schematic graph of a travelling electromagnetic
wave showing the electric and magnetic vectors.
Explain the phenomenon of double refraction with a
diagram indicating the directions of polarizations for the
two beams.
Explain the law of Malus with a diagram.
Sketch schematically the wave surfaces produced by a
point source in calcite explaining the reason for this.
Explain with diagram, the polarization of reflected light,
incident at Brewster’s angle.
Explain circular polarization of light and its production
with a diagram.
Explain the method of producing plane-polarized light by
refraction in a stack of glass plates.
29
Explain optical activity with a diagram.
30
18-11-2022
INTRODUCTION TO QUANTUM PHYSICS
BLACKBODY RADIATION & PLANK’S HYPOTHESIS
Definition of a Black-Body
Black-Body Radation Laws
Text Book
1- The Stefan-Boltzmann Law
1. PHYSICS for Scientists and Engineers
with Modern Physics (6th ed)
By Serway & Jewett
2- The Wien‘s Displacement Law
3- The Rayleigh-Jeans Law
Reference Book
4- The Planck Law
2. CONCEPTS of MODERN PHYSICS (6th ed)
By Arthur Beiser
Summary
2
1
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Definition of a black body
BLACKBODY RADIATION & PLANK’S HYPOTHESIS
Blackbody – a hypothetical (idealized) body
Laboratory Approximation of a black body
A black body is an ideal body which allows the whole of the
incident radiation incident on it, regardless of frequency
A small hole cut into a
cavity of a hollow container
of iron or copper painted
inside with lampblack is the
realistic example.
None of the incident
radiation escapes.
This propety is valid for radiation corresponding to all
wavelengths and to all angels of incidence.
Therefore, the black body is an ideal absorber of incident
radaition as well as an ideal radiator
This is to be expected , since a body at a constant temperature (T) is
in thermal equilibrium with its surroundings and must emit
energy at the same rate as it absorbs energy.
Any radiation incident on the hole from outside the cavity enters
the hole and is reflected a number of times on the interior walls of
the cavity; hence, the system acts as a perfect absorber.
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18-11-2022
BLACKBODY RADIATION & PLANK’S HYPOTHESIS
The electromagnetic radiation emitted by the black body
is called black-body radiation.
Any radiation is absorbed in the walls of the cavity causes a
heating of the cavity walls.
The nature of the radiation leaving the cavity
through the hole depends only on the temperature
of the cavity walls and not on the material of which
the walls are made.
The oscillators in the cavity walls vibrate and cavity walls reradiate at wavelengths corresponding to the temperature of
the cavity.
As the radiation reflects from the cavity’s walls, standing
electromagnetic waves are established within the threedimensional interior of the cavity.
The spaces between lumps of hot charcoal emit light that
is very much like blackbody radiation.
Many standing-wave modes are possible, and the
distribution of the energy in the cavity among these modes
determines the wavelength distribution I (,T) d of the
radiation leaving the cavity through the hole.
I (,T) d is the intensity or power per unit area emitted in the
wavelength interval d from a blackbody.
5
5
6
Spectral distribution of Black Body Radiation
Wien’s displacement law is consistent with the behavior of “ The spaces between
lumps of hot charcoal”
At room temperature, the object does not appear to glow because the peak is in the infrared
region of the electromagnetic spectrum.
At higher temperatures, it glows red because the peak is in the near infrared with
some radiation at the red end of the visible spectrum.
At still higher temperatures, it glows white because the peak is in the visible so that all colors
are emitted.
7
• As the temperature increases, the peak wavelength emitted by
the black body decreases.
• As temperature increases, the total energy emitted increases,
because the total area under the curve increases.
8
18-11-2022
Basic Laws of Radiation
1) All objects emit radiant energy.
2) Hotter objects emit more energy than colder objects
(per unit area). The total power of the emitted radiation
increases with temperature.
This is Stefan-Boltzmann Law.
3) The peak of the wavelength distribution shifts to
shorter wavelengths as the black body temperature
increases
This is Wien’s Law.
At about 5800 K, the peak-wavelength is in the center of the
visible wavelengths in the spectral distribution and the object
appears white.
Intensity Vs wavelength curve
Spectral distribution
The above are empirical laws: derived from experiment & observation rather than theory
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Black-Body Radiation Laws (1)
Black-Body Radiation Laws (2)
 Stefan Boltzmann Law.
I (T) = e  T 4 P = A e  T 4
 Wien’s Displacement Law.
I = I nt e nsi t y o f t h e e m it t e d r a d ia t io n (p o w e r ra d ia t e d
p e r u n it a re a )
λm T = constant = 2.898 × 10-3 m- K, or λm  T-1
P = power radiated from the surface of the object (W)
where λm - peak of the wavelength distribution in the black
body emission spectrum.
T = temperature (K)
= 5.670 x 10-8 W/m2 K4 (Stefan-Boltzmann constant)
T- equilibrium temperature of the blackbody.
A = surface area of the object (m2)
(A black-body reaches thermal equilibrium when the incident
radiation power is balanced by the power re-radiated).
e = emissivity of the surface (for an Idealized blackbody e =1).
The emissivity is the ratio of the emissive power of an object to that of
an ideal black body and is always less than 1
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18-11-2022
Black-Body Radiation Laws (3)
A successful theory for blackbody radiation must
predict
• the shape of the curves
• the temperature dependence expressed in Stefan’s
law
• the shift of the peak with temperature described by
Wien’s displacement law.
 The Rayleigh-Jeans Law.
Rayleigh-Jeans Law :
classical theory of Electromagnetism & Statistical thermodynamics
u() d is the energy density / the total energy per unit
volume in the cavity in the frequency range  and  + d /
Intensity of the emitted radiation / power per unit area
emitted in the frequency interval d from a blackbody.
Planck’s Radiation Law:


radical assumption of Quantized energy states
led to the birth of Quantum theory
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14
The ultraviolet catastrophe
BLACKBODY RADIATION & PLANK’S HYPOTHESIS
There are serious flaws in the reasoning by Rayleigh and Jeans
8 2
u ( ) d  
kTd 
c3
Black-Body Radiation Laws (4)
the result does not agree with experiment
The Raleigh Jeans law, predicts an infinite energy density as
∞!
 The Planck Law
(This discrepancy became known as the ultraviolet catastrophe)
8 π h 3
1
u()d =
h
c3
e kT - 1
h = Planck’s constant
Agreement between Raleigh
Jean’s law and experiment is
only to be found at very long
wavelengths but at short
wavelengths i.e. in the UV
range a discrepancy exists .
15
8 2
kTd 
c3
k – Boltzmann's constant
(1.381 x 10-23 J/K )
T- equilibrium blackbody
temperature
c- velocity of light.
This law tries to explain the Intensity distribution or
distribution of energy with respect to frequency from a
black body.
u ( ) d 
This law correctly explains the distribution of energy from
a black body
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18-11-2022
The law fitted the experimental data for all wavelength
regions and at all temperatures.
But for this to happen,
Plank made two bold and controversial assumptions concerning
the nature of the charged oscillators in the cavity walls.
BLACKBODY RADIATION & PLANK’S HYPOTHESIS
(2) Each discrete energy value corresponds to a different quantum
state, represented by the quantum number n .
The oscillators emit or absorb energy only when making a transition from
one quantum state to another.
Difference in energy will be integral multiples of
Planck postulated
to n = ∞
(1) the energies of oscillators could only take on discrete
values equal to multiples of a fundamental energy e = h
where  is the frequency of the oscillator.
En = n e = n h
E
Figure shows allowed energy
levels for an oscillator with
n = 0, 1, 2...
frequency  , and the allowed
Here, h is a fundamental constant, known as Planck's constant.
17
Transitions.
ENERGY
Then,
E = h 
4
3h
3
2h
2
h
1
0
0
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BLACKBODY RADIATION & PLANK’S HYPOTHESIS
Summary
The characteristics of blackbody radiation cannot
be explained using classical concepts.
Plank introduced the quantum concept and
Plank’s constant when he assumed that atomic
oscillators existing only in discrete energy states
are responsible for this radiation.
In Plank’s model, radiation is emitted in
single quantized packets whenever an oscillator
makes a transition between discrete energy states .
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n
4h
11/24/2022
B.TECH FIRST YEAR
ACADEMIC YEAR: 2020-2021
SESSION OUTCOME
COURSE NAME: ENGINEERING PHYSICS
COURSE CODE
“UNDERSTAND FOUNDATION OF
QUANTUM PHYSICS”
: PY1001
LECTURE SERIES NO : 04 (FOUR)
CREDITS
:
4
MODE OF DELIVERY : ONLINE (POWER POINT PRESENTATION)
FACULTY
:
DR. NILANJAN HALDER
EMAIL-ID
:
nilanjan.halder@jaipur.manipal.edu
DATE OF DELIVERY: 2 November 2020
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THE PHOTOELECTRIC EFFECT
•Introduction
•What is Photoelectric Effect
•Apparatus for studying Photoelectric Effect
•Experimental Observations
•Classical Predictions
QUIZ
MID TERM EXAMINATION –II
END TERM EXAMINATION
•Clash between Classical predictions
ASSESSMENT CRITERIA’S
& Observed Experimental results
• Einstein’s model of the Photoelectric Effect
• Explanation for the observed features of PE
•Application
•Conclusion
•Summary
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THE PHOTOELECTRIC EFFECT
Apparatus for studying Photoelectric Effect
What is Photoelectric Effect?
When plate E is illuminated
by light of suitable
frequency, electrons are
emitted from E and a current
is detected in A.
Apparatus for studying Photoelectric Effect
T – Evacuated
glass/ quartz tube
E – Emitter Plate/
T
At large values of V , the I reaches a
maximum value;
all the electrons emitted from E are
collected at C, and the current cannot
increase further.
Photosensitive
material /Cathode
C – Collector Plate /
Anode
V – Voltmeter
A - Ammeter
the maximum I increases as the
intensity of the incident light increases,
because more electrons are ejected by
the higher-intensity light.
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6
THE PHOTOELECTRIC EFFECT
When V is negative—that is, when the
battery in the circuit is reversed to
make plate E positive and plate C
negative—the current drops because
many of the photoelectrons emitted
from E are repelled by the now
negative plate C.
Experimental Observations
1. When plate E is illuminated by light of
suitable frequency, electrons are emitted
from E and a current is detected in A.
2. Photocurrent
produced Vs
potential difference
applied graph
shows that
maximum kinetic
energy of the
emitted electrons,
Kmax = e Vs
In this situation, only those
photoelectrons having a kinetic energy
greater than e|V| reach plate C,
where e is the magnitude of the
charge on the electron.
When V is equal to or more negative than Vs, where Vs
is the stopping potential, no photoelectrons reach C and the
current is zero.
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THE PHOTOELECTRIC EFFECT
THE PHOTOELECTRIC EFFECT
Classical Predictions
3. Maximum kinetic energy of the
photoelectron is independent of light
intensity.
1. If light is really a wave, it was thought that
if one shine light of any fixed wavelength, at
sufficient intensity on the emitter surface,
electrons should absorb energy continuously
from the em waves and electrons should be
ejected.
2. As the intensity of light is increased (made it
brighter and hence classically, a more
energetic wave), kinetic energy of the
emitted electrons should increase.
4. Electrons are emitted from the surface of the
emitter almost instantaneously.
5. No electrons are emitted if the incident light
frequency falls below a cutoff frequency.
6. Maximum kinetic energy of the
photoelectrons increases with increasing
light frequency.
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THE PHOTOELECTRIC EFFECT
THE PHOTOELECTRIC EFFECT
Einstein’s Interpretation of em radiation
4. Ejection of photoelectron should not
depend on light frequency
(A new theory of light)
Electromagnetic waves carry discrete energy
packets (light quanta called photons now).
5. Photoelectron kinetic energy should not
depend upon the frequency of the
incident light.
The energy E, per packet depends on frequency f.
E = hf.
More intense light corresponds to more photons,
not higher energy photons.
In short experimental results
contradict classical predictions.
Each photon of energy E moves in vacuum at the
speed of light c, where c = 3x 108 m/s. Each
photon carries a momentum p = E/C.
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THE PHOTOELECTRIC EFFECT
THE PHOTOELECTRIC EFFECT
Einstein’s model of the photoelectric effect
All the observed features of photoelectric effect
could be explained by Einstein’s photoelectric
equation.
A photon of the incident light gives all its
energy hf to a single electron (Absorption
of energy by the electrons is not a
continuous process as envisioned in the
wave model) and Kmax = hf - 
1. Equation shows that Kmax depends only on
frequency of the incident light.
2. Almost instantaneous emission of photoelectrons
due to one -to –one interaction between photons
and electrons.
3. Ejection of electrons depends on light frequency
since photons should have energy greater than
the work function  in order to eject an electron.
4. The cutoff frequency fc is  related to by fc =  /h.
If the incident frequency f is less than fc , no
emission of photoelectrons.
 is called the work function of the metal.
It is the minimum energy with which an
electron is bound in the metal.
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14
THE PHOTOELECTRIC EFFECT
Einstein predicted that a
graph of the maximum
kinetic energy Kmax Vs
frequency f would be a
straight line, given by the
linear relation,
Kmax = hf - 
eV0 =Kmax
And indeed such a linear
relationship was observed.
And this work won Einstein his Nobel Prize in 1921
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THE PHOTOELECTRIC EFFECT
THE PHOTOELECTRIC EFFECT
Summary
Application of photoelectric effect
Explain the device, theory, and its working
Photomultiplier tube
Einstein successfully extended Plank’s quantum
hypothesis to explain photoelectric effect.
In Einstein’s model, light is viewed as a stream of
particles, or photons, each having energy E = hf ,
where h is Plank’s constant and f is the frequency.
The maximum kinetic energy Kmax of the ejected
photoelectron is
Kmax = hf - 
Where  is the work function of the photocathode.
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18
INTRODUCTION TO QUANTUM PHYSICS
QUESTIONS
1. Explain (a) Stefan’s law (b) Wien’s displacement law
(c) Rayleigh-Jeans law.
[1 EACH]
2. Sketch schematically the graph of wavelength vs intensity of
radiation from a blackbody.
[1]
3. Explain Planck’s radiation law.
[2]
4. Write the assumptions made in Planck’s hypothesis of
blackbody radiation.
[2]
5. Explain photoelectric effect.
[1]
6. What are the observations in the experiment on photoelectric
effect?
[5]
7. What are the classical predictions about the photoelectric
effect?
[3]
8. Explain Einstein’s photoelectric equation.
[2]
19
INTRODUCTION TO QUANTUM PHYSICS
QUESTIONS
10.Which are the features of photoelectric effectexperiment explained by Einstein’s photoelectric
equation?
[2]
11.Sketch schematically the following graphs with
reference to the photoelectric effect: (a) photoelectric
current vs applied voltage (b) kinetic energy of mostenergetic electron vs frequency of incident light.
[1EACH]
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• Q. ultraviolet light of wavelength 350 nm and
intensity 1 w/m2 is directed to a potassium
surface .
(a) Find the maximum kinetic energy of the
photoelectrons.
(b) if 0.5% of the incident photons produce
photoelectrons , how many are emitted per
second if the potassium surface has an area
of 1.00 cm2 .
Work function of potassium is 2.2eV.
Q. Under favorable circumstances the human
eye on detect 10-18 J of the electromagnetic
energy. How many 600 nm photon does this
represent?
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THE PHOTOELECTRIC EFFECT
THE PHOTOELECTRIC EFFECT
SJ: Section 40.2 P-14.
Electrons are ejected
from a metallic surface with speeds up to 4.60 x 105
m/s when light with a wavelength of 625 nm is used.
(a) What is the work function of the surface? (b)
What is the cut-off frequency for this surface?
SJ: Section 40.2 P-17.
Two light sources are
used in a photoelectric experiment to determine
the work function for a metal surface. When
green light from a mercury lamp ( = 546.1 nm)
is used, a stopping potential of 0.376 V reduces
the photocurrent to zero. (a) Based on this
what is the work function of this metal? (b)
What stopping potential would be observed
when using the yellow light from a helium
discharge tube ( = 587.5 nm)?
SJ: Section 40.2 P-16.
The stopping
potential for photoelectrons released from a
metal is 1.48 V larger compared to that in
another metal. If the threshold frequency for
the first metal is 40.0 % smaller than for the
second metal, determine the work function for
each metal.
Assignment: Try to answer the questions in
page no. 1313, chapter 40 of the reference
book.
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THE COMPTON EFFECT
THE COMPTON EFFECT : General Information
A relativistic particle is a particle which moves with a relativistic
speed; that is, a speed comparable to the speed of light (v˜c)
•Introduction
•What is Compton Effect
The total energy ‘E’ relativistic particle is given by
•Schematic diagram of Compton’s apparatus
E2  p2c2  m2c4
•Experimental Observations
•Classical Predictions
•Explanation for Compton Effect
Here p and m are the momentum and mass of the
particle: c is the speed of light.
• Derivation of the Compton Shift Equation.
•Conclusion
But, photon mass m=0,
So, the relativistic energy of a photon: E2=p2c2
•Summary
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25
26
When a subatomic particle like electron travels with a
speed comparable with the speed of light (v~c):
Mechanics) is a particle which moves with a speed very
Its relativistic energy :
small compared to the speed of light (v<<c).
Ee2=pe2c2+m2c4
Where
pe=  mv
Where
(Relativistic momentum of electron)
Lorentz
NOTE: A non relativistic particle (Newtonian
factor
γ 
For non relativistic particle: =1.
So, p=mv
1
1 -
v
c
1
1-
v2
c2
2
Newtonian definition of momentum is
2
v = speed of the electron & c = speed of light in vacuum,
27
m is the mass of the electron.
27
γ
valid at low speeds !!!!!
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THE COMPTON EFFECT
SUMMARY OF PHOTON PROPERTIES
What is Compton Effect ?
Relation between particle and wave properties of light
Energy, frequency, and wavelength, E
Compton (1923)
measured intensity of
scattered X-rays from
solid target (scattering of
X-rays from electrons), as
function of wavelength
for different angles. In
such a scattering, a shift
in wavelength for the
scattered X-rays takes
place, which is known as
Compton Effect.
= hf = hc / 
Also we have relation between momentum and wavelength of a
photon as follows
For photon (light), m = 0,  E= pc . Also c =  f
p
=
E
c
=
hf
λf
=
h
λ
scattered beam
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THE COMPTON EFFECT
THE COMPTON EFFECT
Classical Predictions
Oscillating electromagnetic waves of frequency f0
incident on electrons should have two effects. (a)
oscillating electromagnetic field causes oscillations in
electrons, which re-radiate in all directions (b)
radiation pressure should cause the electrons to
accelerate in the direction of propagation of the waves.
Because different electrons will move at different
speeds after the interaction, depending on the amount
of energy absorbed from em waves, for a particular
angle of incidence of the incoming radiation, the
scattered wave frequency should show a distribution of
Schematic diagram of Compton’s apparatus
Explain the experimental
details and results
Here X- ray photons are
scattered through 90°
from a carbon target.
The wavelength is
measured with a
rotating crystal
spectrometer using
Bragg’s law. Intensity
of the scattered X-rays
are measured using the
ionization chamber.
Doppler- shifted values.
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THE COMPTON EFFECT
In the scattering
process, the total
energy and total
linear momentum of
the system must be
conserved.
Compton could explain the experimental result by
taking a “billiard ball” type collisions between particles
of light (X-ray photons) and electrons in the material.
The incident X-ray photon with energy (E0=hc/λ0), when
collides with the electron, it gives some of its energy to
(a) the electron which recoils with a velocity (v) in the direction
making an angle (φ) with the direction of the incident
photon.
and remaining energy to
(b) the photon with reduced energy (E’=hc/λ’) scattered in the
direction θ with the original direction.
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34
THE COMPTON EFFECT
Experimental Observations
Compton could explain the experimental result by
taking a “billiard ball” type collisions between
particles of light (X-ray photons) and electrons in
the material.
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THE COMPTON EFFECT
Reason for the λo peak
The graphs for three nonzero angles show two peaks,
one at 0 and one at ’ > 0 . The shifted peak at ’ is
caused by the scattering of X-rays from free electrons.
Shift in wavelength was predicted by Compton to
depend on scattering angle as
(1) λC = 2.43 pm for an electron and
less for particles with large rest mass.
The maximum wavelength change
in Compton effect is 4.86 pm.
h
λ' - λ0 
(1- cosθ)
mec
This is known as Compton shift equation, and the
factor is h
called the Compton wavelength.
h
m ec
m ec
= 0.00243
nm
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37
(2) The eqn. shows that at scattered angle the scattered wavelength shows
the initial un-scattered wavelength (λo ). This is because in deriving the equation
it was assumed that the scattering particle is able to move freely, which is reasonable
since many of the electrons in matter are only loosely bound to their parent atom.
The tightly bound electrons when struck by a photon, the entire atom recoils
instead of the single electron. So in this case the mass m in the equation is to be
replaced by the mass of the entire atom, which is 10000 times greater than the
mass of the electrons. Hence the resultant reduced Compton shift (λ’ - λo).
38
THE COMPTON EFFECT
THE COMPTON EFFECT
Derivation of the Compton Shift Equation
Applying the law of conservation of energy to the
process gives
Photon is treated as a particle having energy E = hf =
hc/ and zero rest energy. They collide elastically with
free electrons initially at rest as shown in figure.
where
+ K
e
hc/ 0 = E0 is the energy of the incident photon,
hc/ ’ = E’ is the energy of the scattered photon,
and Ke is the kinetic energy of the recoiling electron.
Substituting for Ke we get
In the scattering
process, the total
energy and total
linear momentum of
the system must be
conserved.
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39
hc
hc
=
λ0
λ'
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THE COMPTON EFFECT
THE COMPTON EFFECT
Rewriting the above equations as
Applying law of conservation of momentum to this
collision, both in x and y components of momentum
are conserved independently.
y component
x component :
h = h cos θ +  m v cos φ
e
λ
λ'
0
y component :
0 = h sin θ -  m v sin φ
e
λ'
p - p ' cos θ = p cos φ
x component :
p ' sin θ = p e sin φ
:
p 2 - 2 p p' cos θ + p' 2 = p
0
0
(40.14)
2
e
(a
)
Rewriting equation 40.12 in terms of respective
energy notations as
where
h/0 = p0
h/’ = p’
EP
(40.14)
Squaring and adding the above equations give
(40.13)
is the momentum of the incident photon
is the momentum of the scattered photon
 m e v = P e is the momentum of the scattered
electron
(40.13)
e
0
E0 = E’ + Ee - mc2
i.e., E0 - E’ + mc2 = Ee ---------(b)
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42
THE COMPTON EFFECT
THE COMPTON EFFECT
Summary
X-rays are scattered at various angles by electrons in
a target. In such a scattering, a shift in wavelength is
observed for the scattered X-rays and the
phenomenon is known as Compton Effect. Classical
physics does not predict the correct behaviour in this
effect. If x-ray is treated as a photon, conservation of
energy and linear momentum applied to the photonelectron collisions yields for the Compton shift:
Square equation (b), substitute for
E2e  p2ec 2  m2c 4
and for pe from equation (a). Write the resulting
equation in terms of respective wavelengths, we
get the Compton shift equation as
2
λ' - λ0 
h
(1- cosθ)
mec
λ' - λ0 
h
(1- cosθ)
mec
Where me is the mass of the electron, c is the speed of
light, and  is the scattering angle.
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THE COMPTON EFFECT
Q.
Compton scattering at 45°
X-rays of wavelength 0 = 0.20 nm are scattered from
a block of material. The scattered X-rays are observed
at an angle of 45° to the incident beam. Calculate
their wavelength.
Q.
Calculate the energy and momentum of a
photon of wavelength 700 nm.
45
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46
THE COMPTON EFFECT
Q.
A 0.00160 nm photon scatters from a free
electron. For what photon scattering angle does
the recoiling electron have kinetic energy equal
to the energy of the scattered photon?
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48
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Q.
A 0.880 MeV photon is scattered by a free
electron initially at rest such that the scattering angle
of the scattered electron is equal to that of the
scattered photon ( = ). (a) Determine the angles 
& . (b) Determine the energy and momentum of the
photon. © Determine the energy and momentum of
the scattered electron
Q.
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PHOTONS AND ELECTROMAGNETIC WAVES
INTRODUCTION TO QUANTUM PHYSICS
TOPICS
•
•
•
•
•
•
Photons and Electromagnetic Waves
Evidence for wave-nature of light
• Diffraction
• Interference
Photons and Electromagnetic Waves
The Quantum Particle
de Broglie hypothesis
Davisson-Germer Experiment
Quantum particle
The Uncertainty Principle
Evidence for particle-nature of light
• Photoelectric effect
• Compton effect
•This means true nature of light is not describable in
terms of any single classical picture.
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2
MATTER WAVES
PHOTONS AND ELECTROMAGNETIC WAVES
De Broglie
In 1923 Prince Louis de Broglie postulated that
an electrons or any other material particle must
exhibit wave like properties in addition to particle
properties
Waves associated with a material particle is
called matter waves
 
de Broglie wavelength
h
p
The electron accelerated through a potential
difference of V has a non relativistic kinetic energy
1929,
Nobel Prize
Planck’s constant
h  6.6 3  1 0  3 4 Js
K  12 m v 2  e V
p=mv
and
f = E
h
=
m = mass,
v = velocity
2meV
h
h
h
λ= h =


p mv
2mK
2meV
Energy of the particle
frequency of the particle
3
p = momentum of the particle,
p = m v for a non-relativistic particle
m = mass of the particle
V = velocity of the particle
h
λ= h=
p mv
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PHOTONS AND ELECTROMAGNETIC WAVES
PHOTONS AND ELECTROMAGNETIC WAVES
SJ: P-SE 40.5 The wavelength of an Electron
Calculate the de- Broglie wavelength for an
electron moving at 1.0 x 107 m/s.
SJ: P-SE 40.7 An Accelerated Charged Particle
A particle of charge q and mass m has been accelerated
from rest to a nonrelativistic speed through a potential
difference of V. Find an expression for its de Broglie
wavelength.
SJ: P-SE 40.6 The Wavelength of a Rock
A rock of mass 50 g is thrown with a speed of
40 m/s. What is its de Broglie wavelength?
5
SJ: Section 40.5 P-35 (a) An electron has a
kinetic energy of 3.0 eV. Find its wavelength.
(b) Also find the wavelength of a photon having
the same energy.
6
PHOTONS AND ELECTROMAGNETIC WAVES
Davisson -Germer experiment
&
Electron Diffraction pattern
Davisson-Germer experiment (1927)
Proof of existence of Matter waves
Scatters beam of electrons from a
Ni crystal.
These two experiments confirmed de- Broglie
relationship p = h /.
Davisson G.P. Thomson
1937 : Nobel prize.
Subsequently it was found that atomic beams, and beams
of neutrons, also exhibit diffraction when reflected from
regular crystals. Thus de Broglie's formula seems to apply
to any kind of matter.
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 A beam of electrons from a heated filament is accelerated through a
potential difference V.
Results
For beam of 54eV energy have sharp maximum
in the electron distribution occurred at angle
50o .
 The collimated beam of electrons strikes a single crystal of nickel.
 Electrons are scattered in all directions by the atoms in the crystal.
 The electron intensity is measured by a detector which can be
moved to any angle  relative to the incident beam.
The kink for 54 V
electrons gives the
strong existence of
electron waves, since
strong diffraction peak
is observed at ɸ =50⁰
The crystal surface acts like a diffraction grating with spacing d
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Davisson & Germer experiment
•
Bragg’s equation:
2𝑑𝑠𝑖𝑛𝜃 = 𝑛 𝜆
For Nickel crystal:
Energy of the
electron beam,
the angle at
which they reach
the target , and
position of the
detector could be
varied.
d=0.91Å; θ=65⁰; n=1
2 × 0.91 × sin 65 = 1.65Å
𝝀 = 𝟏. 𝟔𝟓Å
•
De Broglie formula:
𝜆=
=
ℎ
ℎ
=
𝑝
2𝑚𝑒𝑉
.
.
.
𝝀 = 𝟏. 𝟔𝟔Å
Since electron KE is 54 eV (small compared with its rest energy mc2=0.51 MeV );
we can write pe=  mv = mv ( as  =1)
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PHOTONS AND ELECTROMAGNETIC WAVES
PHOTONS AND ELECTROMAGNETIC WAVES
Now the dual nature of matter and radiation is
an accepted fact. And it is stated in the principle
of complementarity. This states that wave and
particle models of either matter or radiation
compliment each other.
Subsequently it was found that atomic beams,
and beams of neutrons, also exhibit diffraction
when reflected from regular crystals. Thus de
Broglie's formula seems to apply to any kind of
matter.
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PHOTONS AND ELECTROMAGNETIC WAVES
PHOTONS AND ELECTROMAGNETIC WAVES
Q. In the DavissonGermer experiment, 54.0
eV electrons were
diffracted from a nickel
lattice. If the first
maximum in the
diffraction pattern was
observed at = 50.0°,
what was the lattice
spacing a between the
vertical rows of atoms in
the figure?
Q. A particle of charge q and mass m has been
accelerated from rest to a nonrelativistic speed through
a potential difference of V. Find an expression for its de
Broglie wavelength.
Q. (a) An electron has a kinetic energy of 3.0
eV. Find its wavelength. (b) Also find the
wavelength of a photon having the same
energy.
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THE QUANTUM PARTICLE
Electron Diffraction pattern-Experiment
• What is a Quantum Particle?
Double –slit experiment with electrons (1989)
• How to represent a quantum particle?
• Wave packet
•Phase velocity
•Group velocity
•Conclusion
•Summary
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THE QUANTUM PARTICLE
THE QUANTUM PARTICLE
How to represent a quantum particle?
What is a Quantum Particle?
To represent a quantum wave, we have to
combine the essential features of both an ideal
particle and an ideal wave.
Quantum particle is a model by which
particles having dual nature are represented.
We must choose one appropriate behavior for
the quantum particle (particle or wave) in
order to understand a particular behavior.
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An essential feature of a particle is that it is
localized in space. But an ideal wave is
infinitely long (unlocalized) as shown in figure
below.
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THE QUANTUM PARTICLE
THE QUANTUM PARTICLE
Now to build a localized entity from an
infinitely long wave, waves of same
amplitude, but slightly different frequencies
are superposed. The result of superposition of
two such waves are shown below.
If we add up large number of waves in a similar
way, the small localized region of space where
constructive interference takes place is called a
wavepacket, which represents a particle.
In the figure, large
number of waves are
Combined. The result
is a wave packet,
which represents a
particle.
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THE QUANTUM PARTICLE
THE QUANTUM PARTICLE
Beat
Where k = k1 – k2 and  = 1 – 2. frequency
Mathematical Representation of a wave packet
The resulting wave oscillates with the average
frequency, and its amplitude envelope varies
according to the difference frequency.
superposition of two waves of equal amplitude, but
with slightly different frequencies, f1 & f2 and
wavelengths, traveling in the same direction are
considered. The waves are written as
(
y = A cos k 1 x - ω 1 t
1
)
The resultant wave is,
[ (
y = 2A cos
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(
and
y 2 = A cos k 2 x - ω 2 t
)
y = y1 + y2
)] cos( k +2k x - ω +2ω t)
Δk Δω
xt
2
2
1
2
1
Amplitude varies
with t and x
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THE QUANTUM PARTICLE
THE QUANTUM PARTICLE
This envelope can travel through space with a different
speed than the individual waves. This speed is called
the group speed or the speed of the wave packet (the
group of waves)
The group speed, υg =
(Δω 2)
(Δk 2)
=
Δω
Δk
For a superposition of large number of waves to form a
wave packet, this ratio is
υg =
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dω
dk
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A realistic wave is characterized by two different
speeds. The phase speed, the speed with which
single wave moves, which is given by
& the group speed, the speed with which the
envelope moves. This is given by
In general these two speeds are not the same.
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THE QUANTUM PARTICLE
THE QUANTUM PARTICLE
Relation between group speed and particle speed
Relation between group speed and phase speed
ω
=fλ
k
i.e., ω = k υphase = k υp
we have, υphase =
But υg =
ω = 2π f = 2π
d ( kvp )
dυp
dω
=
=k
+ υp
dk
dk
dk
g = p – 
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k=
2π
dE
h
2π
dp
h
2π
2π
2π p
=
=
λ
hp
h
=
dE
dp
For a classical particle moving with speed u, the
kinetic energy E is given by
E=
dυ p
dλ
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E
and
h
dω
vg =
=
dk
Substituting for k in terms of , we get
27
26
1 2
p2
mu =
2
2m
i.e., υg =
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and
dE =
2 p dp
2m
or
dE
p
=
= u
dp
m
dω
dE
speed
=
= v, the particle velocity
dk
dp
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THE DOUBLE–SLIT EXPERIMENT REVISITED
THE QUANTUM PARTICLE
we should identify the group speed with the
Electron interference
particle speed, speed with which the energy
Experimental details
moves.
And the discussion of
the results
To represent a realistic wave packet, confined to
a finite region in space, we need the
superposition of large number of harmonic waves
with a range of
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k
values.
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The slit separation d is much greater than the individual slit
widths and much less than the distance between the slit and
the detector. The electron detector is movable along the y
direction in the drawing and so can detect electrons
diffracted at different values of . The detector acts like the
“viewing screen” of Young’s double-slit experiment with
light as learned in interference of light.
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THE DOUBLE–SLIT EXPERIMENT REVISITED
THE DOUBLE–SLIT EXPERIMENT REVISITED
This experiment proves the dual nature of electrons.
The electrons are detected as particles at a localized
Photograph of a double-slit
interference pattern produced
by electrons.
spot at some instant of time, but the probability of
arrival at that spot s determined by finding the
intensity of two interfering waves.
If slit 2 is blocked half the time, keeping slit 1
open, and slit 1 blocked for remaining half the time,
keeping 2 open, the accumulated pattern of
counts/ min is shown by blue curve. That is
the min imum occurs when dsinθ  λ/2
Electron wavelengt h is given by λ  h /p.
For small angle θ, sin θ ≈ θ 
h
2pd
interference pattern is lost and the result is simply
the sum of the individual results.
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THE DOUBLE–SLIT EXPERIMENT REVISITED
THE DOUBLE–SLIT EXPERIMENT REVISITED
The observed interference pattern when both the
slits are open, suggests that each particle goes
through both slits at once. We are forced to
conclude that an electron interacts with both the
slits simultaneously shedding its localized
behaviour.
If we try to find out which slit the particle goes
through the interference pattern vanishes! Means,
Results of the two-slit electron diffraction
experiment with each slit closed half the time (blue).
The result with both slits open (interference pattern)
is shown in brown.
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the fringes. We can only say that the electron
passes through both the slits.
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• A wave packet is localized – a good representation
for a particle!
• A wave packet is a group of waves with slightly
different wavelengths interfering with one another in
a way that the amplitude of the group (envelope) is
non-zero only in the neighbourhood of the particle
• If several waves of different wavelengths
(frequencies) and phases are superposed
together, one would get a resultant which is a
localized wave packet
35
if we know which path the particle takes, we lose
• The velocities of the individual waves which
superpose to produce the wave packet representing
the particle are different - the wave packet as a whole
has a different velocity from the waves that comprise
it
• Phase velocity: The rate at which the phase of the
wave propagates in space
• Group velocity: The rate at which the envelope of the
wave packet propagates
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For a particle represented by a single
wavelength wave existing throughout
space,  is precisely known, and
according to de- Broglie hypothesis, its p
is also known accurately. But the position
(x) of the particle becomes uncertain.
The spread of wave packet in wavelength depends on
the required degree of localization in space – the
central wavelength is given by
This means  = 0, p =0; but x = 
h
 
p
The uncertainty principle is confirmed by experiment,
and is a direct consequence of the de Broglie’s hypothesis
In contrast, if a particle
whose momentum is
uncertain (combination/
a range of wavelengths
are taken to form a
wavepacket ), so that x
is small, but  is large.
If x =0, , & thereby
p = 
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THE UNCERTAINTY PRINCIPLE
Quantum theory predicts that, it is fundamentally
impossible to make simultaneous measurements
 
x
 ?
x : small |∆p large
39
of a particle’s position & momentum with
infinite accuracy. This is known as Heisenberg
uncertainty principle. The uncertainties arise from
the quantum structure of matter.
( x ) ( px) ≥ h / 4
In short
h
, we have the following
p
Also ( E ) ( t) ≥

h / 4
These uncertainties are inherent in the physical world
and have nothing to do with the skill of the observer or
with quality of the experimental equipment
x
x : large |∆p small
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THE UNCERTAINTY PRINCIPLE
THE UNCERTAINTY PRINCIPLE
Locating an electron
The speed of an electron is measured to be
5.00 x 103 m/s to an accuracy of 0.0030%.
Find the minimum uncertainty in determining
the position of this electron.
[0.386 mm]
Q.
Find the minimum kinetic energy of a
proton confined within a nucleus having a
diameter of 1.0 x 10-15 m.
[5.21 MeV]
The Line Width of Atomic Emissions
The lifetime of an excited atom is given as 1.0 x 10-8 s.
Using the uncertainty principle, compute the line width
f produced by this finite lifetime? [8.0 × 106 Hz]
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Q. A typical atomic nucleus is about 5×10-15 m. use
the uncertainty principle to place a lower limit on
the energy an electron must have if it is to be part
of nucleus. [20MeV]
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INTRODUCTION TO QUANTUM PHYSICS
QUESTIONS
1. Explain (a) Stefan’s law (b) Wien’s displacement law
(c) Rayleigh-Jeans law.
[1 EACH]
2. Sketch schematically the graph of wavelength vs intensity of
radiation from a blackbody.
[1]
3. Explain Planck’s radiation law.
[2]
4. Write the assumptions made in Planck’s hypothesis of
blackbody radiation.
[2]
5. Explain photoelectric effect.
[1]
6. What are the observations in the experiment on photoelectric
effect?
[5]
7. What are the classical predictions about the photoelectric
effect?
[3]
8. Explain Einstein’s photoelectric equation.
[2]
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INTRODUCTION TO QUANTUM PHYSICS
QUESTIONS
10.Which are the features of photoelectric effect-experiment
explained by Einstein’s photoelectric equation?
[2]
11.Sketch schematically the following graphs with reference to
the photoelectric effect: (a) photoelectric current vs applied
voltage (b) kinetic energy of most-energetic electron vs
frequency of incident light.
[1EACH]
12.Explain Compton effect.
[2]
13.Explain the experiment on Compton effect.
[5]
14.Derive the Compton shift equation.
[5]
15.Explain the wave properties of the particles.
[2]
16.Explain a wavepacket and represent it schematically.
[2]
17.Explain (a) group speed (b) phase speed, of a wavepacket.
[1+1]
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INTRODUCTION TO QUANTUM PHYSICS
QUESTIONS
20.Show that the group speed of a wavepacket is equal to the
particle speed.
[2]
21.Explain Heisenberg uncertainty principle.
[1]
22.Write the equations for uncertainty in (a) position and
momentum (b) energy and time.
[1]
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QUANTUM
MECHANICS
AN INTERPRETATION OF QUANTUM MECHANICS
Experimental evidences proved that both
matter and electromagnetic radiation exhibit
wave and particle nature depending on the
phenomenon being observed.
TOPICS
•
•
•
•
•
Interpretation of quantum mechanics
Wave function and its significance
Schrodinger equation
Particle in a box, Particle in a well of finite height
Tunnelling through a potential barrier and its
applications
• The simple harmonic oscillator (No derivation)
Making a conceptual connection between particles
and waves, for an electromagnetic radiation, we
have the probability per unit volume of finding a
photon in a given region of space at an instant of
time as
Probability
V
Probability/V ∞ N/V
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 E2
2
2
2
AN INTERPRETATION OF QUANTUM MECHANICS
•
•
For electromagnetic radiation and matter-
associated
the probability per unit volume of finding
probability
the particle is proportional to the square
the particle
associated
with
a
particle
is
generally not a measureable quantity.
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a
particle
amplitude,
or
is
called
the
wave
In general, the complete wave function  for
a system depends on the positions of all the
particles in the system and on time and can
be written as (rj,t) = (rj) e–it , where rj is
the position vector of the jTH particle in the
system.
even if the amplitude of the de Broglie
wave
with
function, and is denoted by .
of the amplitude of a wave representing
•
The amplitude of the de Broglie wave
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AN INTERPRETATION OF QUANTUM MECHANICS
For any system in which the potential energy
|| is always real and positive,
is time-independent and depends only on the
proportional
positions of particles within the system, the
volume,.
important information about the system is
If  represents a single particle,
2
to
the
probability
per
unit
contained within the space part of the wave
|| the probability density
function.
the probability per unit volume that the
2
The wave function  contains within it all the
particle will be found at any given point in
information that can be known about the
the volume.
particle.
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The probability of a particle being in
the interval a ≤ x ≤ b is the area
under the probability density curve
from a to b.
One-Dimensional Wave Functions and
Expectation Values
 = Wave function for a particle moving
along the x axis
2
P(x) dx = || dx is the probability to find
The total probability of finding the particle is 1.
Forcing this condition on the wave function is called
∞
normalization:
 2 dx  1
the particle in the infinitesimal interval dx
around the point x. The probability of finding
∫
-∞
the particle in the arbitrary interval a ≤ x ≤ b is
b
Pab 

2
dx
a
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The wave equation satisfied by  is the Schrodinger
equation and  can be computed from it. All the
measureable quantities of a particle, such as its
energy and momentum, can be derived from a
MITMANIPAL
knowledge
of . BE-PHYSICS-QUANTUM MECHANICS-2010-11
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eg, the average position at which one expects to
find the particle after many measurements is
called the expectation value of x and is defined by
the equation

The important mathematical features of a physically
acceptable wave function (x) for a system are
(i) (x) may be a complex function or a real function,
depending on the system;
x   x  dx
(ii) (x), must be finite, continuous and single valued

every where;
The expectation value of any function f(x)
associated with the particle is
(iii) The space derivatives of , must be finite,
continuous and single valued every where;
(iv)  must be normalizable.

f (x)   f (x)  dx

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AN INTERPRETATION OF QUANTUM MECHANICS
Explain the Physical Significance of ψ and |ψ|2.
The wave function ψ is described as a mathematical function whose
variations constitute matter waves. The value of the wave function
associated with a moving particle at a particular point (x,y,z) in space at
time t is related to the likely hood of finding that particle at that point.
ψ does not have any direct physical significance. It cannot be observed or
measured through an experiment. It can only be derived mathematically
but it is not an observable quantity. ψ may be a complex function or a real
function, depending upon the system.
If ψ represents a particle, |ψ|2 represents the probability density-the
probability per unit volume that the particle will be found at any given
point in the volume at a given time. This can be measured experimentally.
A large value of |ψ|2 implies a strong possibility of the presence of particle.
As long as |ψ|2 is not actually zero somewhere, there is a definite chance
of detecting the particle.
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Q. A wave Function for a particle
A particle wave function is given by the
2
equation  (x) = A e–ax
(A)What is the value of A if this wave function
is normalized?
(B) What is the expectation value of x for this
particle?
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THE SCHRÖDINGER EQUATION
AN INTERPRETATION OF QUANTUM MECHANICS
The appropriate wave equation for matter waves was developed
by Schrödinger. Schrödinger equation as it applies to a particle
of mass m confined to moving along x axis and interacting with
its environment through a potential energy function U(x) is
Q. A particle limited to the x axis has the wavefunction  (x)
= ax between x=0 and x=1;
 (x) = 0 elsewhere.
(a)Find the probability that the particle can be found
between x=0.45 and x=0.55.
 2 d2 

 U
2 m dx 2
(b) find the expectation value 𝑥 of the particle
position.
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
E
Where E is a constant equal to the total energy of the system.
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THE SCHRÖDINGER EQUATION

 2 d2 
 U
2 m dx 2

E
Schrodinger’s equation can not be derived from other basic principle
of physics; it is a basic principle in itself.
The above equation is referred to as the onedimensional, time-independent Schrödinger
equation.
Application of Schrödinger equation to the
PARTICLE IN A BOX or particle in an
infinite square well.
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PARTICLE IN A BOX
PARTICLE IN A BOX
U(x) = 0,
for
0 < x < L,
And U (x) =  , for x < 0, x > L
Since U (x) =  , for x < 0, x > L ,
(x) =0 in these regions.
Also (x =0) =0 and (x =L) =0.
Only those wave functions that satisfy these
In figure (a), a particle of mass m and velocity v,
boundary conditions are allowed.
In the region 0 < x < L, where U = 0, the
Schrödinger equation takes the form
confined to bouncing between two impenetrable
walls separated by a distance L is shown. Figure
d2
dx 2
(b) shows the potential energy function for the
system.
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
0,
18
18
18
PARTICLE IN A BOX
2
d
2mE
  k 2 , where k2  2 or k 
2
dx

PARTICLE IN A BOX
since A  0 , sin(kL) = 0 .
2mE


k
above equation is (x) = A sin(kx) + B cos(kx)
where A and B are constants determined by the
k 
boundary and normalization conditions.
( n = 1, 2, 3, ………..)
2
n


L
2mE
,


or

L  n 
 2
kL
2mE
L  n

Each value of the integer n corresponds to a
quantized energy value, En , where
Applying the first boundary condition,
i.e., at x = 0,  = 0
leads to
0 = A sin 0 + B cos 0 or
B=0 ,
and at x = L ,  = 0 ,
0 = A sin(kL) + B cos(kL) = A sin(kL) +
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n ;
kL =
The most general form of the solution to the
19
2m
E
2
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

En
0,
19
 h2  2

 n Where
2 
8mL 
n = 1, 2 , 3 …..
h2
The lowest allowed energy (n  1), E1 
8 m L2
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PARTICLE IN A BOX
PARTICLE IN A BOX
This is the ground state energy for the particle
in a box. Excited states corresponds to n = 2, 3,
4 ---- have energies given by 4E1 , 9E1 , 16E1 ---.
The corresponding
n (x)
wave functions are
To find the constant A, apply
normalization condition

Energy level diagram for
a particle confined to a
one-dimensional box of
length L. The lowest
allowed energy is
E1 = h2/8mL2.
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According to quantum mechanics, the
particle can never be at rest.
21

L
dx  1
or

nx 
A sin 

 L 
2
  n  x 
0 A sin L  dx  1
L

 2 n  x 
A 2  ½ 1  cos
 dx
L



0
solving , we get ,  n x  
2

1
2
nx 
sin 

L
 L 
The first three allowed states for a particle
confined to a one-dimensional box are shown next..
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2

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PARTICLE IN A BOX
PARTICLE IN A BOX
A Bound Electron
Fig. (a) The wave functions for n = 1, 2, and 3.
Fig. (b) The probability densities for n = 1, 2, and 3.
Q. An electron is confined between two
impenetrable walls 0.20 nm apart. Determine
the energy levels for the states n =1 ,2 , and 3.
Q. An electron is in a box 0.1 nm across, which is the
order of magnitude of atomic dimensions. Find its
permitted energies. [38n2 eV]
Q. Find the expectation value <x> of the position of
a particle trapped in a box L wide.
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A PARTICLE IN A WELL OF FINITE HEIGHT
PARTICLE IN A BOX
(PARTICLE IN A SQUARE WELL POTENTIAL)
A proton is confined to move in a one-dimensional
Potential energy
I
box of length 0.20 nm. (a) Find the lowest possible
energy of the proton. (b) What is the lowest
III
II
U
finite height U and
E
length L. A particle is
possible energy for an electron confined to the
0
same box?
diagram of a well of
trapped in the well.
L
The total energy E of
the particle-well
X
system is less than U
Explain the conditions, U(x) = 0 ,
0 < x < L,
U (x) = U , x < 0, x > L
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A PARTICLE IN A WELL OF FINITE HEIGHT
A PARTICLE IN A WELL OF FINITE HEIGHT
Particle energy = E < U ; classically the particle
The Schrödinger equation outside the finite well in
is permanently bound.
regions I and III is:
a finite probability exists that the particle can be
d2 
found outside the well even if E < U.
dx 2
the wave function is generally nonzero in regions I
and III..
where
In regions II, where U = 0, the allowed wave
conditions no longer require that the wave
(U  E )   C 2 
2
2m
C2 
(U  E)
2
 (x )  A e C x  B e C x
function must be zero at the ends of the well.
27
2m
General solution of the above equation is
functions are again sinusoidal. But the boundary
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A PARTICLE IN A WELL OF FINITE HEIGHT
A PARTICLE IN A WELL OF FINITE HEIGHT
Schrodinger equation inside the square well
potential in region II, where U = 0 is
A must be 0 in Region III and B must be zero in
Region I, otherwise, the probabilities would be
infinite in those regions. Solution should be finite.
d2
I
where C 

III
= B e–C x for x > L
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2 m (U

2m
E 
II
ћ2
+
dx2
ie., the wave functions outside the finite
potential well are
 = A e C x for x < 0
II
=
0
k2
General solution of the above equation is
_ E)
ψ II
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




 2mE 
2mE
F sin 
 x  G cos  
 
 









k

k

 
 
x 
 
 
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A PARTICLE IN A WELL OF FINITE HEIGHT
A PARTICLE IN A WELL OF FINITE HEIGHT
wave function outside the potential well decay
exponentially with distance.
I
U
Boundary conditions
so the wave function is countinuous
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E
0
L
ψ I (0)  ψ II (0)
To determine the constants A, B, F, G, & the
allowed values of energy E, apply the four
boundary conditions and the normalization
condition.
31
dψ I
dx
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
x 0
III
II
dψ II
dx
x 0
Boundary conditions
ψ II ( L )  ψ III ( L )
d ψ II
dx

x L
d ψ III
dx
x L
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A PARTICLE IN A WELL OF FINITE HEIGHT
Wave functions
differences between particle confined in a potential well and that in a finite potential well
Probability densities
The following are the differences:
In case of a finite potential well, the wave function exists outside the walls, i.e. x<0 and
x>L, while for an infinite potential well, ψ=0 for 𝒙 ≤ 𝟎 𝒂𝒏𝒅 𝒙 ≥ 𝑳.
The wavelengths that fit into the finite well are longer than those for an infinite well of the
same width. Hence the corresponding particle momenta are lower (we recall that 𝝀 =
𝒉/𝒑)
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The energy levels En for each n are lower for a finite potential well than an infinite
potential well.
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Tunneling Effect
TUNNELING THROUGH A POTENTIAL ENERGY BARRIER
Consider a particle of energy E approaching a potential barrier
of finite height U (E<U) and finite width L.
Classically:
Since E < U, the regions II and III shown in the figure are
forbidden to the particle incident from left.
Quantum mechanically,
There is a certain probability of the particle passing through the
barrier and emerging on the other side, regardless of its energy.
That is, although the particle does
not possess enough energy to go over
the top of the barrier,
there is a finite probability that it
can tunnel through it.
This is known a TUNNELING EFFECT
35
Potential energy function and wave function for a
particle incident from the left on a barrier of height
U and width L. The wave function is sinusoidal in
regions I and III but exponentially decaying in
region II.
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TUNNELING THROUGH A POTENTIAL ENERGY BARRIER
. The movement of the particle to the far side of the
barrier is called tunneling or barrier penetration.
The probability of tunneling can be described with a
transmission coefficient T and a reflection
coefficient R.
TUNNELING THROUGH A POTENTIAL ENERGY BARRIER
The transmission coefficient represents the
probability that the particle penetrates to the
other side of the barrier, and reflection
coefficient is the probability that the particle
is reflected by the barrier.
Because the particles must be either reflected or
transmitted we have, R + T = 1.
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TUNNELING THROUGH A POTENTIAL ENERGY BARRIER
The probability of TUNNELING
An approximate expression for the
transmission coefficient, when T << 1 is
T
≈
e
_
2 C L
when T << 1
where C 
2 m (U

transmission coefficient (T) : probability that
the particle penetrates to the other side of
the barrier
reflection coefficient (R): probability that the
particle is reflected by the barrier
_ E)
We have, R + T = 1.
An approximate expression for the
transmission coefficient, when T << 1
(i.e. a very wide barrier or a very high barrier
i.e. U>>E)
This violation of classical physics is allowed by the
uncertainty principle. The particle can violate
classical physics by E for a short time,
T
t ~ ħ / E.
≈
e
_
2C L
where C 
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when T << 1
2 m (U

_ E)
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Examples of tunneling mechanism
Examples of tunneling mechanism
alpha particles emission by certain
radioactive nuclei (i.e. unstable heavy
nuclei).
Inside the nucleus, an alpha particle
feels the strong, short-range attractive
nuclear force as well as the repulsive
Coulomb force between the alpha
particle and the rest of the nuclei.
The nuclear force dominates inside
the nuclear radius where the
potential is ~ a square well.
The Coulomb force dominates
outside the nuclear radius:
decreases with the distance
from the nucleus
 A semiconductor tunnel diode is
based on tunneling effect.
 Electrons pass through potential
barriers even though their
kinetic energies are smaller
than the barrier heights.
The potential wall at the nuclear
radius: 25-30 MeV
The kinetic energy of the alpha
particles : 3-5 MeV
The alpha particles manage to escape
the potential wall of the nucleus: a
Phenomenon called RADIOACTIVE
DECAY
 The tunnel diodes are high
impurity density p-n junction
devices with a narrow width of
junction barrier.
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42
TUNNELING THROUGH A POTENTIAL ENERGY BARRIER
Q.
TUNNELING THROUGH A POTENTIAL ENERGY BARRIER
Q. An electron with kinetic
A 30-eV electron is incident on a square
energy E = 5.0 eV is
barrier of height 40 eV. What is the probability
incident on a barrier with
that the electron will tunnel through the barrier if
thickness L = 0.20 nm and
its width is (A) 1.0 nm? (B) 0.10 nm?
height U = 10.0 eV as
shown in the figure.
What is the probability that the electron (a) will
tunnel through the barrier? (b) will be reflected?
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Q.
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THE SCANNING TUNNELING
MICROSCOPE
Application of
Tunnel effect.
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Quantum mechanical model of harmonic oscillator deals with
the quantum treatment of the vibrating charges
(for eg. blackbody emitting radiation)
as simple harmonic oscillators.
Particles (vibrating charges) is subject to a linear restoring force
F = -(k x),
where x is the position of the particle relative to equilibrium (x = x0)
k is force constant.
Potential energy of the system is, 𝑼 = ½ 𝒌 (𝒙 − x0)𝟐
Letting x0 = 0, 𝑼 = ½ 𝒌 𝒙𝟐 = ½ 𝒎 𝝎𝟐 𝒙𝟐
Simple Harmonic Oscillator
Simple harmonic oscillators describe many physical situations:
springs, diatomic molecules and atomic lattices.
where the angular frequency
of vibration is 𝝎 = 𝒌/𝒎
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50
THE SIMPLE HARMONIC OSCILLATOR
THE SIMPLE HARMONIC OSCILLATOR
total energy E = K + U = ½ k A2 = ½ m ω2 A2
The solution of the above equation is given by
(classically)
ψ 
Classically, the particle oscillates between the points
x = A and
x = – A,
where A is the amplitude of
In the classical model, any value of E is allowed,
where C 
including E = 0, which is the total energy when
the particle is in rest at x = 0.
The Schrödinger equation for this problem is
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mω
2
ψ  B e  (m 
 2 d2 ψ 1

m 2 x 2 ψ  E ψ
2
2m dx
2
51
C x2
Where B is a constant determined from the
normalization condition.
the motion.

B e
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and
E
/2)x
2
1
2
ω
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THE SIMPLE HARMONIC OSCILLATOR
THE SIMPLE HARMONIC OSCILLATOR
U (x)
The energy levels of a harmonic oscillator are
quantized. The energy for an arbitrary quantum
Energy level diagram
number n is given by
= E3
for a simple harmonic
oscillator,
= E2
superimposed on the
= E1
potential energy
n = 0 , ground state, E0 = (½)ħω;
= E0
function.
x
n = 1 , first excited state, E1 = (3/2) ħω, and so
0
on.
The levels are equally spaced with E = ħω.
The ground state energy is E0 = (½)ħω.
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THE SIMPLE HARMONIC OSCILLATOR
THE SIMPLE HARMONIC OSCILLATOR
The classical and quantum mechanical probabilities
ψn
2
ψn
2
-3 -2 -1
0
1
2
3
x
x
-3
-3
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-2 -1
-2 -1
0
0
1
1
2
2
3
3
The blue curves represent the classical probabilities and
2
the red ones the quantum probabilities ψn for a
simple harmonic oscillator.
x
55
MUJ
55
56
56
56
14
12/12/2022
THE SIMPLE HARMONIC OSCILLATOR
Comparison of energy of particle in a box and a
harmonic oscillator
In classical physics, probability densities are
greatest near the endpoints of its motion where
they have least kinetic energies. This is in sharp
contrast to the quantum case for small n. In the
limit of large n, the probabilities start to resemble
each other.
Quantum mechanical modification to the classical picture
MUJ
57
MUJ
57
57
58
58
Why it is impossible for the lowest-energy state of a harmonic oscillator to be zero?
Problem # 02
The lowest vibrational state (ν=0) has the zero point energy ½ hν0:
not the classical value of “0”.
• A one dimensional harmonic oscillator wave function is ψ = A x e-bx² ,
ψ satisfies the equation
This is result is in accord with the uncertainty principle.
Because: if the oscillating particle is stationary (E=0);
the uncertainty in its position would be Δx=0;
the momentum uncertainty would then have to be infinite.
[We recall: ∆𝒙 . ∆𝒑 ≥
ℏ
𝟐
Find b and total energy E. Is this a ground state or first excited state?
]

But a particle with E=0;
cannot have an infinitely uncertain momentum.
 2 d2 ψ 1

m 2 x 2 ψ  E ψ
2m dx 2
2
½ hν0 is called the Heisenberg's limit
MUJ
59
60
60
15
12/12/2022
Problem # 03
• A quantum simple harmonic oscillator consists of an electron
bound by a restoring force proportional to its position
relative to a certain equilibrium point. The proportionality
constant is 8.99 N/m. What is the longest wavelength of light
that can excite the oscillator?
MUJ
61
61
MUJ
62
62
QUESTIONS – QUANTUM MECHANICS
[MARKS]
1. What is a wave function ? What is its physical
interpretation ?
[2]
2. What are the mathematical features of a wave
function?
[2]
3. By solving the schrödinger equation, obtain the wavefunctions for a particle of mass m in a onedimensional “box” of length L.
[5]
4. Apply the schrodinger equation to a particle in a
one-dimensional “box” of length L and obtain the
energy values of the particle.
[5]
5. Sketch the lowest three energy states, wavefunctions, probability densities for the particle in a
one-dimensional “box”.
[3]
Which of the wave functions can not have
physical significance in the given interval (why
not ?).
MUJ
63
63
MUJ
64
64
16
12/12/2022
QUESTIONS – QUANTUM MECHANICS
QUESTIONS – QUANTUM MECHANICS
[MARKS]
6. The wave-function for a particle confined to moving
in a one-dimensional box is
[MARKS]
8. Sketch the potential-well diagram of finite height U
and length L, obtain the general solution of the
Schrodinger equation for a particle of mass m in
it.
[5]
9. Sketch the lowest three energy states, wavefunctions, probability densities for the particle in a
potential well of finite height.
[3]
Use the normalization condition
on  to show that
[2]
7. The wave-function of an electron is
Obtain an expression for the probability of finding
the electron between x = a and x = b.
[3]
MUJ
65
65
10. Give a brief account of tunneling of a particle
through a potential energy barrier.
[4]
11. Give a brief account of the quantum treatment of a
simple harmonic oscillator.
[5]
MUJ
66
66
END
MUJ
67
67
17
1/11/2023
THE X-RAY SPECTRUM OF ATOMS
• X-rays are high energy photos (0.01 to 10 nm)
• When fast moving e interact with an atom, x rays are produced,
which are emw which is,
– Unaffected by electric ang magnetic field
– Penetrate through opaque material
– Travel in a straight line.
• Faster the e, more penetrating is radiation
• More number of e, more intense will be the x rays.
X-ray Spectrum:-
Properties of X-ray
• X-rays are em waves of very short wavelength
(≈1 Å)
• The speed of the X-rays in vacuum is equal to
the speed of light.
• X-rays undergo reflection and refraction
according to laws of visible light.
• Continuous spectrum
(Bremsstrahlung S)
• Characteristics spectrum
1
2
Continuous Spectrum
Typical X ray Spectrum
The plot between the intensity and the wavelength of X-rays is
known as X-ray spectrum.
It is a continuous curve superimposed with several sharp
peaks.
The x-ray spectrum has two parts:
Continuous spectrum (the Curve)
Characteristic spectrum (the Peaks)
3
This type of spectrum consists of radiation of all
possible wavelengths within a range starting
with a minimum value (λMIN) called short
wavelength limit to a upper value which
depends upon the voltage across the tube.
4
1/11/2023
Origin of the continuous spectrum
 An accelerated electric charge emits electromagnetic radiation.
 The x-rays are the result of the slowing down of high-energy electrons
as they strike the target.
 It may take several interactions with the atoms of the target before the
electron loses all its kinetic energy.
 The amount of kinetic energy lost in any given interaction can vary from
zero up to the entire kinetic energy of the electron.
 Therefore, the wavelength of radiation from these interactions lies in a
continuous range from some minimum value up to infinity.
 It is this general slowing down of the electrons that provides the
continuous spectrum, which shows the cutoff of x-rays below a
minimum wavelength value (λMIN) that depends on the kinetic energy of
the incoming electrons.
 X-ray radiation with its origin in the slowing down of electrons is called
bremsstrahlung, the German word for “braking radiation.”
 Consider an electron accelerated through a
potential difference of ∆V (x-ray tube voltage) ,
hitting a target atom.
 The electron’s initial kinetic energy is K = e ∆V.
 The electron loses its kinetic energy by an amount
∆K = hf, which appears in the form of x-ray
photon energy (Bremsstrahlung).
The amount of kinetic energy lost (∆K) can vary from
zero up to the entire kinetic energy (K) of the electron.
 Therefore, the wavelength of Bremsstrahlung radiation
lies in a continuous range from some minimum value up
to infinity.
e V  hfMAX 
hc
 MIN 
e V
hc
MIN
λMIN depends only on ∆V
Duane and Hunt law
5
6
Nuclear explanation of production of Characteristic X-rays
 In Bohr model of the atom, shells of electrons surround the nucleus of
the atom containing the protons and neutrons.
 The innermost shell, called the K- shell, is surrounded by the L and M shells.
 When the energy of the electrons accelerated toward the target
becomes high enough to dislodge K- shell electrons, electrons from the
L - and M - shells move in to take the place of those dislodged.
 Each of these electronic transitions produces an X-ray with a
wavelength that depends on the exact structure of the atom being
bombarded.
 A transition from the L - shell to
the K- shell produces a Kα X-ray,
while the transition from an M - shell
to the K- shell produces a Kβ X-ray.
7
When the energy of the electrons accelerated
toward the target becomes high enough to
dislodge K- shell electrons,
electrons from the L - and M - shells move
in to take the place of those dislodged.
hc
hf 
 En  Em

8
6
1/11/2023
Characteristic X-ray spectrum
If you wish to produce 10.0 nm X rays in the
laboratory, what is the minimum voltage you
must use in accelerating the electrons?
 The peaks in the x-ray spectrum have wavelengths
characteristic of the target element in the x-ray tube
and hence they form the characteristic x-ray spectrum.
 These characteristic X-rays have a much higher intensity
than those produced by the continuous spectra, with Kα Xrays having higher intensity than Kβ X-rays.
 The important point here is that the wavelength of these
characteristic X-rays is different for each target atom in
the periodic table (of course only those elements with
higher atomic number have L- and M - shell electrons that
can undergo transitions to produce X-rays)..
What is the minimum voltage applied to an X
ray tube to produce X-ray of wavelength 1
Angstrom?
9
10
In X-ray production, electrons are accelerated
through a high voltage ΔV and then decelerated
by striking a target. Show that the shortest
wavelength of an x-ray that can be produced is
1240 𝑛𝑚. 𝑉
𝜆
=
Δ𝑉
11
12
1/11/2023
Moseley’s relation
Bohr theory and the Moseley plot:
Moseley’s observation on the characteristic K x-rays
shows a relation between the frequency (f) of the K
x-rays and the atomic number (Z) of the target element
in the x-ray tube:
According to Bohr’s formula:
The frequency of radiation corresponding to a transition
in a one-electron atom between any two atomic levels
differing in energy by ΔE
f  C Z  1
C is a constant.
f 
Importance of Moseley’s Law:
According to this law, it is the
atomic number and not atomic
weight of an element which
determines its characteristic
properties, both physical and
chemical. Therefore atoms must be
E
m Z 2e 4  1
1 
 2  2 

2 3 
h
8  oh  n f
ni 
In a many-electron atom, for a K transition, the effective
nuclear charge felt by an L-electron can be thought of as
equal to +(Z–b)e instead of +Ze,
where b is the screening constant due to the
screening effect of the K-electron.
arranged in the periodic table
according to their atomic numbers
and not according to their atomic
weights.
13
14
THE X-RAY SPECTRUM OF ATOMS
Bohr theory and the Moseley plot:
A K x-ray results due to the transition of the electron
from L-shell to K-shell. A K x-ray results due to the
transition of the electron from M-shell to K-shell.
When the vacancy arises in the L-shell, an L-series (L,
L, L) of x-rays results.
 Frequency of the K x-ray is
m Z  b e 4  1
1 
 2  2 
f 
2 3
8  oh
 1 1 2 
2
and
or
 3 m e4  2
 Z  b 
f  
2 3 
 32  oh 
f  C Z  1
Q: Calculate the cutoff wavelength for the continuous
spectrum of x-rays emitted when 35-keV electrons fall
on a molybdenum target.
sin ce b  1
16
15
15
16
1/11/2023
THE X-RAY SPECTRUM OF ATOMS
HRK-Exercise 48.1: Show that the short-wavelength
cutoff in the continuous x-ray spectrum is given by
 MIN 
1240 pm
V
where ΔV is the applied potential difference in
kilovolts.
HRK-Exercise 48.5: Electrons bombard a
molybdenum target, producing both continuous and
characteristic x-rays. If the accelerating potential
applied to the x-ray tube is 50.0 kV, what values of
(a) λMIN (b) λKβ (c) λK result ? The energies of the
K-shell and L-shell in the molybdenum atom are –20.0
keV and –2.6 keV, respectively.
17
17
18
THE X-RAY SPECTRUM OF ATOMS
HRK-Exercise 48.12: The binding energies of K-shell
and L-shell electrons in copper are 8.979 keV and
0.951 keV, respectively. If a K x-ray from copper is
incident on a sodium chloride crystal and gives a firstorder Bragg reflection at 15.9 when reflected from
the alternating planes of the sodium atoms, what is
the spacing between these planes ?
HRK-Exercise 48.9: X-rays are produced in an x-ray
tube by a target potential of 50.0 keV. If an electron
makes three collisions in the target before coming to
rest and loses one-half of its remaining kinetic energy
on each of the first two collisions, determine the
wavelengths of the resulting photons. Neglect the
recoil of the heavy target atoms.
19
20
1/11/2023
X-RAYS AND THE NUMBERING OF THE ELEMENTS
Q. Which element has a Kα x- ray line whose
wavelength is 0.18 nm? (R=1.097*10^7m-1)
HRK-Sample Problem 48-3: A cobalt target is
bombarded with electrons, and the wavelengths of its
characteristic x-ray spectrum are measured. A second,
fainter characteristic spectrum is also found, due to an
impurity in the target. The wavelengths of the K
lines are 178.9 pm (cobalt) and 143.5 pm (impurity).
What is the impurity ?
22
21
22
1. Explain the continuous x-ray spectrum with a schematic plot of the
spectrum.
[2]
2. Obtain an expression for the cutoff wavelength in the continuous x-ray
spectrum.
[4]
3. Explain the characteristic x-ray spectrum with a schematic plot of the
spectrum.
[2]
4. Explain the origin of characteristic x-ray spectrum with a sketch of x-ray
energy level diagram.
[3]
5. Write Moseley’s relation for the frequency of characteristic x-rays. Sketch
schematically the Moseley’s plot of characteristic x-rays. What is the
importance of Moseley’s law
6. Obtain Moseley’s relation for characteristic x-ray frequency from Bohr
theory.
[4]
23
11-01-2023
Laser Fundamentals

The light emitted from a laser is monochromatic, that is,
it is of one color/wavelength. In contrast, ordinary white
light is a combination of many colors (or wavelengths)
of light.

Lasers emit light that is highly directional, that is, laser
light is emitted as a relatively narrow beam in a specific
direction. Ordinary light, such as from a light bulb, is
emitted in many directions away from the source.

The light from a laser is said to be coherent, which
means that the wavelengths of the laser light are in
phase in space and time. Ordinary light can be a mixture
of many wavelengths.
LASERS
These three properties of laser light are what can make it
more hazardous than ordinary light. Laser light can
deposit a lot of energy within a small area.
2
1
2
Incandescent vs. Laser Light
1. Many wavelengths
1. Monochromatic
2. Multidirectional
2. Directional
3. Incoherent
3. Coherent
Interaction of radiation with matter
3
3
4
1
11-01-2023
Absorption: Absorption of a photon of frequency f takes
place when the energy difference E2 – E1 of the allowed
energy states of the atomic system equals the energy hf of
the photon. Then the photon disappears and the atomic
system moves to upper energy state E2 (see figure).
5
Spontaneous Emission: The average life time of the atomic
system in the excited state is of the order of 10–8 s. After
the life time of the atomic system in the excited state, it
comes back to the state of lower energy on its own accord
by emitting a photon of energy hf = E2– E1 (see figure).
In an ordinary light source the radiation of light from
different atoms is not coherent. The radiations are emitted
in different directions in random manner. Such type of
emission of radiation is called spontaneous emission.
6
Stimulated Emission: When a photon (stimulating photon) of
suitable frequency interacts with an excited atomic system, it
comes down to ground state before its life time. Such an
emission of radiation is called stimulated emission. In
stimulated emission, both the stimulating photon and the
stimulated photon are of same frequency, same phase and
are in same state of polarization, they are emitted in the
same direction. In other words, these two photons are
coherent.
Thus we get
amplified radiation
by stimulated
emission
(see figure).
7
population inversion
An incident photon can cause atomic energy transitions either
upward (stimulated absorption) or downward (stimulated
emission).
The two processes are equally probable.
When light is incident on a collection of atoms, a net absorption
of energy usually occurs because when the system is in thermal
equilibrium, many more atoms are in the ground state than in
excited states.
If the situation can be inverted so that more atoms are in an
excited state than in the ground state, however, a net emission
of photons can result. Such a condition is called population
inversion.
This is a non equilibrium condition and is facilitated by the
presence of “metastable states”.
8
2
11-01-2023
population inversion
if
E 2 > E1
Figure a : This is the normal thermal equilibrium condition in
which the population of the atoms in upper energy state is
less than that in lower energy state i.e. n(E2) < n(E1)
Figure b : This condition is called population inversion [n(E2) > n(E1)]
which is a non equilibrium condition
9
Metastable state
 A metastable state is an excited energy state of an atomic
system from which spontaneous transitions to lower
states is forbidden (not allowed by quantum mechanical
selection rules).
 The average life time of the atomic system in the
metastable state is of the order of 10–3 s which is much
longer than that in an ordinary excited state.
 Stimulated transitions from the metastable state are
allowed.
 An excited atomic system goes to metastable state
(usually a lower energy state) due to transfer of its extra
energy by collision with another atomic system (non
radiative transition).
 Thus it is possible to have “population inversion” of
atomic systems in a metastable state relative to a lower
energy state.
10
The conditions of stimulated emission for buildup of photons in a
system for laser light emission.
Condition#1: The system must be in a state of population
inversion: there must be more atoms in an excited state
than in the ground state. That must be true because the
number of photons emitted must be greater than the
number absorbed.
11
12
3
11-01-2023
Common Components of all Lasers
Condition#2: The excited state of the system must be a
metastable state, meaning that its lifetime must be long
(10-3 s )compared with the usually short lifetimes of excited
states, which are typically 10-8 s. In this case, the population
inversion can be established and stimulated emission is
likely to occur before spontaneous emission.
1. Active Medium
The active medium may be (a) solid crystals such as ruby or Nd:YAG, (b)
liquid dyes, (c) gases like CO2 or Helium/Neon, (d) semiconductors such
as GaAs.
Active mediums contain atoms whose electrons may be excited to a
metastable energy level by an energy source.
The atomic systems in the active medium may have energy levels
including a ground state (E1), an excited state (E3) and a metastable
state (E2).
Condition#3: The emitted photons must be confined in the
system long enough to enable them to stimulate further
emission from other excited atoms. That is achieved by
using reflecting mirrors at the ends of the system (Resonant
Cavity). One end is made totally reflecting, and the other is
partially reflecting. A fraction of the light intensity passes
through the partially reflecting end, forming the beam of
laser light
13
14
14
Common Components of all Lasers
Does the intensity of light from a laser fall off as 1/r2?
2. Excitation Mechanism
Excitation mechanisms pump energy into the active medium by one or
more of three basic methods; (i) optical (ii) electrical (iii) chemical.
3. High Reflectance Mirror
A mirror which reflects essentially 100% of the laser light.
4. Partially Transmissive Mirror
A mirror which reflects less than 100% of the laser light and transmits the
remainder.
15
15
16
4
11-01-2023
Why is stimulated emission is so important in the operation of laser?
• Stimulated emission causes atoms to emit photons along a specific axis,
rather than in the random directions of spontaneously emitted photons.
• The photons that are emitted through stimulation can be made to
accumulate over time in the resonant cavity.
• The fraction allowed to escape constitutes the intense, collimated, and
coherent laser beam.
• If this process relied solely on spontaneous emission, the emitted
photons would not exit the laser tube in the same direction. Neither
would they be coherent with one another.
Ruby Laser
In ruby laser the lasing medium is a ruby rod. Ruby is Al2O3 doped with Cr2O3.
Cr3+ ions are the active centres, which have approximately similar energy level
structure shown above.
The resonant cavity is a pair of parallel mirrors to reflect the radiation back into
the lasing medium. Pumping is a process of exciting more number of atoms in
the ground state to higher energy states, which is required for attaining the
population inversion. In Ruby laser the pumping is done by xenon flash lamp.
The atoms in the state E3 may come down to state E1 by spontaneous emission
or they may come down to metastable state (E2) by collision. The atoms in
the state E2 come down to state E1 by stimulated emission.
18
17
18
LASERS AND LASER LIGHT
Ruby LASER
He-Ne Laser has a
glass discharge tube
filled with He (80%)
These radiations may be reflected due to mirror action of
the end faces (see figure). When population inversion takes
place at E2, a stray photon of right energy stimulates
chain reaction, accumulates more photons, all coherent. The
reflecting ends turn the coherent beam back into active
region so that the regenerative process continues and part
of the light beam comes out from the partial mirror as a
laser pulse. The out put is an intense beam of coherent
light. The ruby laser gives red light
and Ne (20%) at low pressure.
He-gas is the “pumping”
medium and Ne-gas is the
“lasing” medium. The simplified
energy level diagram (see
figure) shows 4 levels: Eo, E1,
E2 and E3. Electrons and ions
in the electrical gas discharge
occasionally collide with Heatoms, raising them to level E3
(a metastable state).
19
19
20
20
5
11-01-2023
LASERS AND LASER LIGHT
During collisions
between He- and Neatoms, the
excitation energy of He-atom is transferred to Ne-atom (level
E2). Thus, population inversion occurs between levels E2 and
E1. This population inversion is maintained because (1) the
metastability of level E3 ensures a ready supply of Ne-atoms
in level E2 and (2) level E1 decays rapidly to Eo. Stimulated
emission from level E2 to level E1 predominates, and red laser
light is generated. The mirror M1 is fully reflective and the
mirror M2 is partially reflective to allow the laser beam to
come out. The Brewster’s windows W & W are at polarizing
angles to the mirrors, to make the laser light linearly
polarized.
21
21
22
23
24
6
11-01-2023
Mention the characteristics of a laser beam.
25
[2]
Explain the following terms with reference to lasers:
(a) spontaneous emission
(b) stimulated emission
(c) metastable state
(d) population inversion
(e) pumping
(f) active medium
(g) resonant cavity.
[2]
[2]
[2]
[2]
[1]
[2]
[1]
Explain the principle of a laser.
[5]
Give a brief account of a He-Ne laser.
Give a brief account of a Ruby laser.
[3]
[3]
26
7
19-01-2023
Solid State Physics
• Band theory of solids
Text Book for reference:
• Modern Physics, Mc Graw Hill, 6th
edition., 2009
by Beiser & Mahajan
Syllabus
• Energy electrical conduction in metals,
insulators and semiconductors
• Superconductivity
• Type-I and Type-II superconductors
• Meisner effect, BCS theory
• Applications of superconductivity.
• PHYSICS for Scientists and
Engineers with Modern Physics
(6th ed)
by Serway & Jewett
1
1
2
BAND THEORY OF SOLIDS
• Applying Quantum Mechanics to solids
Band Theory
of Solids
• Able to explain many physical properties of
solids, such as electrical resistivity and
optical absorption
There are two different wave functions  S (r) and  S (r ) (for an atom with single selectron outside of a closed shell) for which the probability density is the same.
The wave functions of two atoms combine to form a composite wave function for
the two-atom system when the atoms are close together. In Figure (a), two atoms
with wave functions  S (r) combine. In Figure (b), two atoms with wave functions
 S (r ) and  S (r ) combine.
• Foundation of the understanding of all solidstate devices (transistors, solar cells, etc.)
3
These two possible combinations of wave functions
represent two possible states of the two-atom system
which have slightly different energies. Thus, each energy
level of an atom splits into two close energy levels when
the wave functions of the two atoms overlap.
4
19-01-2023
BAND THEORY OF SOLIDS
Isolated Sodium (Na) atom
Number of electrons: 11: How this electrons will be filled??
• Occupancy of levels: Pauli’s Exclusion Principle
Splitting of 1s and 2s
levels
when
two
sodium atoms are
brought together
Splitting of 1s and 2s
levels when five sodium
atoms are brought
together
• For crystalline solid there are a large number of allowed
energy bands.
For n (Avogadro number)
sodium atoms the levels
as so close they will
appear as band
• Some bands may be wide enough in energy so that there is
an overlap between the adjacent bands.
• The 1s, 2s, and 2p bands of solid sodium are filled
completely with electrons.
• Since there are 2(2l+1) energy states in a subshell each
energy band has 2(2l+1)N energy states, where N = number
of atoms in the crystal.
• The 3s band (2N states) of solid sodium has only N electrons
and is partially full; The 3p band and the bands above this
are completely empty.
5
3p
3s
E
2p
2s
1s
6
EMPTY
CB
Eg = 0
PARTIALLY FILLED
VB & CB
CONDUCTOR
7
ENERGY BANDS OF SODIUM
CRYSTAL
• Forbidden energy gaps occur between the allowed bands.
E=EF
E=0
•
The 3s level has capacity to accommodate 2N electrons, and for
Na only N electrons are available the band is partially filled
•
Since the band is partially filed the electron can move freely and
hence give rise to good electric conductivity of metals in general.
•
In other words the conduction and valance bands are overlapping
EF
Eg
Band Theory of
Solids/
Conductors
FILLED
VB
INSULATOR
• The outermost energy bands are filled valence band and
empty conduction band with a large energy gap (Eg>>kT,
kT = thermal energy).
• The Fermi-level lies in the energy gap.
• Thermal energy at room temperature is not sufficient to
excite the electrons from valence-band to conduction
band.
• Since the free-electron density is nearly zero, these
materials are bad conductors of electricity.
8
Band Theory of
Solids/Insulators
19-01-2023
CONDUCTION BAND
EF
Eg
Intrinsic Semiconductor
VALENCE BAND
SEMICONDUCTOR
These have the band structure similar to an insulator but the energy gap
is much smaller ( 1 eV).
At zero K, all electrons in semiconductors are in the valence band, and
no energy is available to excite them across the energy gap. Thus,
semiconductors are poor conductors at very low temperatures.
Band Theory of
Solids/
Semiconductors
• The charge carriers in a semiconductors are electrons and holes.
• When an electron moves from the valence band into the
conduction band, it leaves behind a vacant site, called a hole
(particle with a positive charge +e).
• In an intrinsic semiconductor (pure semiconductor) there are
equal number of conduction electrons and holes.
At room temperature a small fraction of valence electrons are thermally
excited to conduction band.
• In the presence of an external electric field, the holes move in
the direction of field and the conduction electrons move opposite
to the direction of the field. Both these motions correspond to
the current in the same direction.
Because the thermal excitation of electrons across the narrow gap is
more probable at higher temperatures, the conductivity of semiconductors
increases rapidly with temperature. [The conductivity of a metal
decreases slowly with increasing temperature.]
9
• Doping is the process of adding impurities to a semiconductor. Doped
semiconductor is also called extrinsic semiconductor. By doping both the
band structure of the semiconductor and its resistivity are modified.
• If a tetravalent semiconductor (Si or Ge) is doped with a pentavalent
impurity atom (donor atom), four of the electrons form covalent bonds
with atoms of the semiconductor and one is left over.
• Since the energy Ed between the donor levels and the bottom of the
conduction band is small, at room temperature, the extra electron is
thermally excited to the conduction band. This type of semiconductors
are called n-type semiconductors because the majority of charge
carriers are electrons (negatively charged).
• At zero K, this extra electron resides in the donor-levels, that lie in
the energy gap, just below the conduction band.
• At zero K, this hole resides in the acceptor levels that lie in the energy
gap just above the valence band.
• Since the energy Ea between the acceptor levels and the top of the
valence band is small, at room temperature, an electron from the valence
band is thermally excited to the acceptor levels leaving behind a hole in
the valence band.
• This type of semiconductors are called p-type semiconductors because the
majority of charge carriers are holes (positively charged).
12
CONDUCTION
BAND
VALENCE BAND
APPLIED E-FIELD
Extrinsic Semiconductor (p-type)
• If a tetravalent semiconductor is doped with a trivalent impurity atom
(acceptor atom), the three electrons form covalent bonds with neighboring
semiconductor atoms, leaving an electron deficiency (a hole) at the site of
fourth bond.
ELECTRONS
HOLES
ENERGY GAP
10
Extrinsic Semiconductor (n-type)
11
CONDUCTION
ELECTRONS
19-01-2023
SJ-PROBLEM 43.37: Light from a hydrogen discharge tube is incident on a CdS
crystal (Eg= 2.42 eV). Which spectral line from the Balmer series are absorbed and
which are transmitted ?
SJ-PROBLEM 43.39: Most solar radiation has a wavelength of 1 μm or
less. What energy gap should the material in solar cell have in order
to absorb this radiation ? Is silicon (Eg= 1.14 eV) appropriate ?
Answer: All Balmer lines absorbed except the red line (656 nm) which is
transmitted.
Answer: 1.24 eV or less; yes
SOLUTION TO SJ-PROBLEM 43.39
SOLUTION TO SJ-PROBLEM 43.37
13
14
Meissner Effect
Superconductivity
When you place a superconductor in a
magnetic field, the field is expelled below T=TC.
Discovered by H. K. Onnes (1911)
Therefore,
ρ
• a superconductor is more than a perfect
conductor (resistivity ρ = 0);
The resistivity of mercury drops to zero at 4.2 K.
• It is also a perfect diamagnet (B= 0).
The temperature at which normal metal transform to
superconducting state is known as critical temperature
(𝑇𝑐)
Materials with high critical temperature (𝑇𝑐) are known as
high temperature superconductor.
Not all metals transforms to superconducting state.
0
A superconductor in the form
of a long cylinder in the
presence of an external magnetic
field.
T
TC
15
15
16
• If the magnitude of the applied magnetic
field exceeds a critical value
Bc, the material’s superconducting
properties gets destroyed i.e. the field
again penetrates the sample.
19-01-2023
 The superconductors are perfect conductor (𝜌 → 0)
type-I superconductors.
 The superconductors are also Perfectly diamagnetic
𝑆𝑢𝑠𝑐𝑒𝑝𝑡𝑖𝑏𝑖𝑙𝑖𝑡𝑦, χ =
= −1 .
The transition from a superconducting state to a normal state due to the external magnetic field is sharp and abrupt for
type-I superconductors.
Meissner effect is the exclusion of magnetic flux from the interior of
superconductors when their temperature is decreased below the
critical temperature.
type-II superconductors.
The transition from a superconducting state to a normal state due to the external magnetic field is gradually but not so
abrupt. At lower critical magnetic field (BC1), type-II superconductor starts losing its superconductivity. At upper
critical magnetic field (BC2), type-II superconductor completely loses its superconductivity. The state between lower
critical magnetic field and upper magnetic field is known as an intermediate state or mixed state.
A superconductor expels magnetic flux (Meissner effect) by forming
surface currents.
These surface currents induced in the
superconductor produce a magnetic field that exactly cancels the
externally
applied magnetic field inside the superconductor.
 Normal metal: Magnetic field permeates the sample
 Superconductor: Magnetic flux is expelled by sample
If a magnetic field B is applied on the superconductor and its value is
increased, the superconductivity disappears when B > BC = critical
magnetic field.
17
18
BCS theory of superconductivity in metals
Type-1 superconductor
Type 2 superconductor
•
Isotope effect: The critical temperature of two different isotopes of same
element is given by 𝑇 ∝ 𝑀 ⁄ , where M is atomic mass. If thermal
motion of ions is assumed to be SHM the frequency of vibrations is also
proportional to 𝑀 ⁄
• Two electrons can interact via distortions in the array of
lattice ions so that there is a net attractive force between the
electrons.
19
•
As a result, the two electrons, are bound into an entity
called a Cooper pair. The Cooper pair behaves like a boson (=
particle with integral spin that do not obey the Pauli exclusion
principle).
•
At very low temperature, it is possible for all bosons in a
collection of such particles to be in the lowest quantum state.
20
Bardeen
Cooper
Schrieffer
19-01-2023
BCS theory of superconductivity in metals
QUESTIONS – BAND THEORY OF SOLIDS
• As a result, the entire collection of Cooper pairs in the metal
is described by a single wave function.
1. Explain briefly the energy band theory of solids.
Bardeen
Cooper
Schrieffer
• Above the energy level associated with this wave function is
an energy gap equal to the binding energy of a Cooper pair.
• Under the action of an applied electric field, the Cooper pairs
experience an electric force and move through the metal.
• A random scattering event of a Cooper pair from a lattice ion would represent resistance to
the electric current. Such a collision would change the energy of the Cooper pair because
some energy would be transferred to the lattice ion.
But there are no available energy levels below that of the Cooper pair (it is already in the
lowest state) and none available above, because of the energy gap. As a result, collisions do
not occur and there is no resistance to the movement of Cooper pairs.
•
21
2. Explain the classification of solids regarding the electrical properties,
based on their energy band diagram.
3. Indicate the position of (a) Fermi-level (b) donor levels (c) acceptor
levels, in the energy band diagram of a semiconductor.
4. Explain the terms: Conductor, insulator, semiconductor, intrinsic
semiconductor, extrinsic semiconductor, n-type semiconductor, p-type
semiconductor, valence band, conduction band, donor levels, acceptor
levels.
22
QUESTIONS – SUPERCONDUCTIVITY
[MARKS]
5. Sketch schematically the plot of resistance of a superconducting
material vs temperature, near the critical temperature.
6. Explain Meissner effect.
7. Give a brief account of superconductivity.
Thank You
8. Explain briefly the BCS theory of superconductivity in metals.
23
[MARKS]
24
ENGINEERING
PHYSICS
[SUBJECT CODE: PHY1051]
COMMON COURSE MATERIAL FOR FIRST YEAR BTech STUDENTS
DEPARTMENT OF PHYSICS
MANIPAL INSTITUTE OF TECHNOLOGY
MIT | Physics | 2020
Table of Contents
1
INTERFERENCE OF LIGHT WAVES................................................... 1
1.1
Young’s Double-Slit Experiment .................................................................... 1
1.2
Analysis Model: Waves in Interference ......................................................... 3
1.3
Intensity Distribution of the Double-Slit Interference Pattern .................... 4
1.4
Change of Phase Due to Reflection ................................................................6
1.5
Interference in Thin Films .............................................................................. 7
1.6
Newton’s Rings ................................................................................................8
1.7
Michelson Interferometer............................................................................. 10
1.8
Questions........................................................................................................ 11
1.9
Problems ......................................................................................................... 11
2 DIFFRACTION PATTERNS AND POLARIZATION ......................... 15
2.1
Introduction to Diffraction Patterns ............................................................. 15
2.2
Diffraction Patterns from Narrow Slits ........................................................ 16
2.3
Intensity of Single-Slit Diffraction Patterns ................................................ 18
2.4
Intensity of Two-Slit Diffraction Patterns ................................................... 18
2.5
Resolution of Single-Slit and Circular Apertures ........................................ 19
2.6
Diffraction Grating .........................................................................................21
2.7
Diffraction of X-Rays by Crystals ................................................................. 24
2.8
Polarization of Light Waves ......................................................................... 25
2.9
Polarization by Selective Absorption ........................................................... 26
2.10
Polarization by Reflection ............................................................................ 27
2.11
Polarization by Double Refraction ............................................................... 29
2.12
Polarization by Scattering ............................................................................ 30
2.13
Optical Activity .............................................................................................. 31
2.14
Questions........................................................................................................ 31
2.15
Problems ........................................................................................................ 32
I
3 QUANTUM PHYSICS .........................................................................34
3.1
Blackbody Radiation and Planck’s Hypothesis ............................................ 34
3.2
Photoelectric Effect ....................................................................................... 38
3.3
Compton Effect .............................................................................................40
3.4
Photons and Electromagnetic Waves [Dual Nature of Light] ....................44
3.5
de Broglie Hypothesis - Wave Properties of Particles .................................44
3.6
The Quantum Particle .................................................................................. 45
3.7
Double–Slit Experiment Revisited ............................................................... 47
3.8
Uncertainty Principle ....................................................................................48
3.9
Questions.......................................................................................................49
3.10
Problems ........................................................................................................ 50
4 QUANTUM MECHANICS ................................................................. 54
4.1
An Interpretation of Quantum Mechanics .................................................. 54
4.2
The Schrödinger Equation ............................................................................ 56
4.3
Particle in an Infinite Potential Well (Particle in a “Box”) .......................... 57
4.4
A Particle in a Potential Well of Finite Height ............................................ 59
4.5
Tunneling Through a Potential Energy Barrier ........................................... 61
4.6
The Simple Harmonic Oscillator ................................................................. 62
4.7
Questions....................................................................................................... 63
4.8
Problems ........................................................................................................64
5 ATOMIC PHYSICS ............................................................................. 66
5.1
The Quantum Model of the Hydrogen Atom ..............................................66
5.2
Wave functions for hydrogen .......................................................................68
5.3
More on Atomic Spectra: Visible and X-Ray ............................................... 70
5.4
X-Ray Spectra .................................................................................................71
5.5
Spontaneous and Stimulated transitions ..................................................... 73
5.6
LASER (Light Amplification by Stimulated Emission of Radiation) .......... 76
5.7
Applications of laser...................................................................................... 78
II
5.8
Questions....................................................................................................... 78
5.9
Problems ........................................................................................................ 79
6 MOLECULES AND SOLIDS .............................................................. 82
6.1
Molecular bonds............................................................................................ 82
6.2
Energy States and Spectra of Molecules ......................................................86
6.3
Rotational Motion of Molecules ...................................................................86
6.4
Vibrational Motion of Molecules ................................................................. 87
6.5
Molecular Spectra .........................................................................................89
6.6
Bonding in Solids .......................................................................................... 91
6.7
Free-Electron Theory of Metals ....................................................................94
6.8
Band Theory of Solids .................................................................................. 99
6.9
Electrical Conduction in Metals, Insulators and Semiconductors ............ 101
6.10
Superconductivity-Properties and Applications ....................................... 104
6.11
Questions..................................................................................................... 106
6.12
Problems ...................................................................................................... 108
III
Reference book:
Physics for Scientists and Engineers with Modern
Physics by Raymond Serway and John Jewett (Cengage Learning,
Seventh Edition 2012)
1 INTERFERENCE OF LIGHT WAVES
OBJECTIVES
•
•
•
To understand the principles of interference.
To explain the intensity distribution in interference under various
conditions.
To explain the interference from thin films.
Wave optics (Physical Optics): It is the study of interference, diffraction, and
polarization of light. These phenomena cannot be adequately explained with the ray
optics.
1.1 YOUNG’S DOUBLE-SLIT EXPERIMENT
Light waves also interfere with one another like mechanical waves. Fundamentally,
all interference associated with light waves arises when the electromagnetic fields
that constitute the individual waves combine.
Figure 1.1 (a) Schematic diagram of Young’s double-slit experiment. Slits S1 and S2 behave as coherent
sources of light waves that produce an interference pattern on the viewing screen (drawing not to
scale). (b) An enlargement of the center of a fringe pattern formed on the viewing screen.
Interference in light waves from two sources was first demonstrated by Thomas
Young in 1801. A schematic diagram of the apparatus Young used is shown Fig. 1.1a.
Plane light waves arrive at a barrier that contains two slits S1 and S2. The light from
S1 and S2 produces on a viewing screen a visible pattern of bright and dark parallel
bands called fringes (Fig. 1.1b). When the light from S1 and that from S2 both arrive
at a point on the screen such that constructive interference occurs at that location,
1
a bright fringe appears. When the light from the two slits combines destructively at
any location on the screen, a dark fringe results.
Figure 1.2 Waves leave the slits and combine at various points on the viewing screen.
Fig. 1.2 shows different ways in which two waves can combine at the screen. In Fig.
1.2a, the two waves, which leave the two slits in phase, strike the screen at the central
point O. Because both waves travel the same distance, they arrive at O in phase. As
a result, constructive interference occurs at this location and a bright fringe is
observed. In Fig. 1.2b, the two waves leave the slits in phase, but the wave leaving
from S2 has to travel longer distance compare to wave from S1. However, the
difference in the path is exactly one wavelength and they arrive in phase at P and a
second bright fringe appears at this location. At point R in Fig. 1.2c, wave from S2 has
fallen half a wavelength behind the wave from S1 and a crest of the upper wave
overlaps a trough of the lower wave, giving rise to destructive interference at point
R.
If two lightbulbs are placed side by side so that light from both bulbs combines, no
interference effects are observed because the light waves from one bulb are emitted
independently of those from the other bulb. The emissions from the two lightbulbs
do not maintain a constant phase relationship with each other over time. Therefore,
the conditions for constructive interference, destructive interference, or some
intermediate state are maintained only for short time intervals. Since the eye cannot
follow such rapid changes, no interference effects are observed. Such light sources
are said to be incoherent.
To observe interference of waves from two sources, the following conditions must
be met:
• The sources must be coherent; that is, they must maintain a constant phase
with respect to each other.
• The sources should be monochromatic; that is, they should be of a single
wavelength.
A common method for producing two coherent light sources is to use a
monochromatic source to illuminate a barrier containing two small openings,
2
usually in the shape of slits, as in the case of Young’s experiment illustrated in Fig.
1. The light emerging from the two slits is coherent because a single source produces
the original light beam and the two slits serve only to separate the original beam
into two parts. Any random change in the light emitted by the source occurs in both
beams at the same time. As a result, interference effects can be observed when the
light from the two slits arrives at a viewing screen.
1.2 ANALYSIS MODEL: WAVES IN INTERFERENCE
Figure 1.3 (a) Geometric construction for describing Young’s double-slit experiment (not to scale). (b)
The slits are represented as sources, and the outgoing light rays are assumed to be parallel as they
travel to P.
The viewing screen is located a perpendicular distance L from the barrier containing
two slits, S1 and S2 (Fig. 1.3a). These slits are separated by a distance d, and the source
is monochromatic. To reach any arbitrary point P in the upper half of the screen, a
wave from the lower slit must travel farther than a wave from the upper slit by a
distance d sin  (Fig. 1.3b). This distance is called the path difference . If we
assume the rays labeled r1 and r2 are parallel, which is approximately true if L is
much greater than d, then  is given by
  r2  r1  d sin 
(1.1)
The value of  determines whether the two waves are in phase when they arrive at
point P.
Angular positions of bright and dark fringes: If  is either zero or some integer
multiple of the wavelength, the two waves are in phase at point P and constructive
interference results. Therefore, the condition for bright fringes, or constructive
interference, at point P is,
d sin  bright  m ;
 m  0,  1,  2, ...
(1.2)
The number m is called the order number.
3
When d is an odd multiple of  / 2 , the two waves arriving at point P are 180° out of
phase and give rise to destructive interference. Therefore, the condition for dark
fringes, or destructive interference, at point P is,
1

d sin  dark   m    ;
2

 m  0,  1,  2, ...
(1.3)
Linear positions of bright and dark fringes: From the triangle OPQ in Fig. 1.3a,
y
(1.4)
tan  
L
Using this result, the linear positions of bright and dark fringes are given by
ybright  L tan  bright
ybright  L
m
d
(1.5)
(small angle approximation)
ydark  L tan  dark
ydark
(1.6)
(1.7)
1

m  
2
 L
d
(small angle approximation)
(1.8)
1.3 INTENSITY DISTRIBUTION OF THE DOUBLE-SLIT
INTERFERENCE PATTERN
Consider two coherent sources of sinusoidal waves such that they have same angular
frequency  and phase difference . The total magnitude of the electric field at
point P on the screen in Fig. 1.3a is the superposition of the two waves. Assuming
that the two waves have same amplitude E0 , we can write the magnitude of electric
field at point P due to each source as
E1  E0 sin  t
and
E2  E0 sin  t   
(1.9)
Although the waves are in phase at the slits, their phase difference  at P depends
on the path difference   r2  r1  d sin  . A path difference of  (for constructive
interference) corresponds to a phase difference of 2 radians.

2


2

d sin 
(1.10)
Using the superposition principle and Equation (1.9), we obtain the following
expression for the magnitude of the resultant electric field at point P:
4
EP  E1  E2  E0 sin  t  sin  t   
(1.11)

 

EP  2 E0 cos   sin   t  
2
2

(1.12)
This result indicates that the electric field at point P has the same frequency  as
the light at the slits but that the amplitude of the field is multiplied by the factor
2 cos( / 2) . If   0, 2 , 4 ,......the magnitude of the electric field at point P is 2E0 ,
corresponding to the condition for maximum constructive interference. Similarly, if
   , 3 , 5 ,...... the magnitude of the electric field at point P is zero.
Intensity of a wave is proportional to the square of the resultant electric field
magnitude at that point. Using Equation 1.12, we can express the light intensity at
point P as
I

 

 EP2  4 E02 cos 2   sin 2   t  
2
2

(1.13)
Most light-detecting instruments measure time-averaged light intensity, and the


time averaged value of sin 2   t   over one cycle is ½. Therefore, we can write the
2

average light intensity at point P as,
 
I  I max cos2  
2
(1.14)
where I max is the maximum intensity on the screen. Substituting the value for  from
Equation 1.10;
 d sin  
I  I max cos2 

 

Alternatively, since sin  
d 
I  I max cos 2 
y
 L 
(1.15)
y
for small values of  , we can write;
L
(1.16)
A plot of light intensity versus d sin  is given in Fig. 1.4. The interference pattern
consists of equally spaced fringes of equal intensity.
5
Figure 1.4 Light intensity versus d sin for a double-slit interference pattern when the screen is far
from the two slits (L >> d).
1.4 CHANGE OF PHASE DUE TO REFLECTION
Young’s method for producing two coherent light sources involves illuminating a
pair of slits with a single source. Another simple arrangement for producing an
interference pattern with a single light source is known as Lloyd’s mirror (Fig. 1.5).
Figure 1.5 Lloyd’s mirror. The reflected ray undergoes a phase change of 180°.
A point light source S is placed close to a mirror, and a viewing screen is positioned
some distance away and perpendicular to the mirror. Light waves can reach point P
on the screen either directly from S to P or by the path involving reflection from the
mirror. The reflected ray can be treated as a ray originating from a virtual source S.
As a result, we can think of this arrangement as a double slit source where the
6
distance d between sources S and S in Fig. 1.5 is analogous to length d in Fig. 1.3a.
Hence, at observation points far from the source (L >> d), waves from S and S form
an interference pattern exactly like the one formed by two real coherent sources.
But, the positions of the dark and bright fringes, however, are reversed relative to
the pattern created by two real coherent sources (Young’s experiment). Such a
reversal can only occur if the coherent sources S and S differ in phase by 180°.
In general, an electromagnetic wave undergoes a phase change of 180° upon
reflection from a medium that has a higher index of refraction than the one in which
the wave is traveling. Analogy between reflected light waves and the reflections of a
transverse pulse on a stretched string is shown in Fig. 1.6.
Figure 1.6 Comparisons of reflections of light waves and waves on strings.
The reflected pulse on a string undergoes a phase change of 180° when reflected from
the boundary of a denser string or a rigid support, but no phase change occurs when
the pulse is reflected from the boundary of a less dense string or a freely-supported
end. Similarly, an electromagnetic wave undergoes a 180° phase change when
reflected from a boundary leading to an optically denser medium, but no phase
change occurs when the wave is reflected from a boundary leading to a less dense
medium.
1.5 INTERFERENCE IN THIN FILMS
Interference effects are commonly observed in thin films, such as thin layers of oil
on water or the thin surface of a soap bubble. The varied colors observed when white
light is incident on such films result from the interference of waves reflected from
the two surfaces of the film. Consider a film of uniform thickness t and index of
refraction n. Assume light rays traveling in air are nearly normal to the two surfaces
of the film as shown in Fig. 1.7. If  is the wavelength of the light in free space and n
is the index of refraction of the film material, then the wavelength of light n in the
film is n 

n
.
7
Figure 1.7 Light paths through a thin film.
Reflected ray 1, which is reflected from the upper surface (A) in Fig. 1.7, undergoes a
phase change of 180° with respect to the incident wave. Reflected ray 2, which is
reflected from the lower film surface (B), undergoes no phase change because it is
reflected from a medium (air) that has a lower index of refraction. Therefore, ray 1
is 180° out of phase with ray 2, which is equivalent to a path difference of n/2. We
must also consider that ray 2 travels an extra distance 2t before the waves recombine
in the air above surface A. (Remember that we are considering light rays that are
close to normal to the surface. If the rays are not close to normal, the path difference
is larger than 2t). If 2t = n/2, rays 1 and 2 recombine in phase and the result is
constructive interference. In general, the condition for constructive interference in
thin film is,
1

2t   m   n
2

(m  0, 1, 2, ...)
(1.17)
(m  0, 1, 2, ...)
(1.18)
Or,
1

2nt   m   
2

If the extra distance 2t traveled by ray 2 corresponds to a multiple of n the two
waves combine out of phase and the result is destructive interference. The general
equation for destructive interference in thin films is
2nt  m
(m  0, 1, 2, ...)
(1.19)
1.6 NEWTON’S RINGS
When a plano-convex lens is placed on top of a flat glass surface as shown in Fig.
1.8a, interference fringes are formed, and these fringes can be seen under the
8
traveling microscope. With this arrangement, the air film between the glass surfaces
varies in thickness from zero at the point of contact to some value t at point P. If the
radius of curvature R of the lens is much greater than the distance r and the system
is viewed from above, a pattern of light and dark rings is observed as shown in Fig.
1.8b. These circular fringes, discovered by Newton, are called Newton’s rings.
Figure 1.8 (a) The combination of rays reflected from the flat plate and the curved lens surface gives
rise to an interference pattern known as Newton’s rings. (b) Photograph of Newton’s rings.
Expressions for radii of the bright and dark rings:
Using the geometry shown in Fig. 1.8a, we can obtain expressions for the radii of the
bright and dark rings in terms of the radius of curvature R and wavelength . For
the thin air film trapped between the two glass surfaces as shown in the figure above,
the conditions for constructive (bright rings) and destructive (dark rings)
interference are given by equations (1.18) and (1.19).
Consider the dark rings (destructive interference)
2nt

m ,
For air film, n  1 ,
m  0, 1, 2, 3...

2t  m
From the above figure, t  R  R 2  r 2
  r 2 
t  R  R 1    
  R  
12
Binomial theorem is, 1  y   1  ny 
n
n(n  1) 2
y  .......
2!
If r / R  1, using binomial theorem and neglecting higher order terms,
9
 1  r 2
 r2
t  R  R 1     ........ 
 2  R 
 2 R
(1.20)
Substituting the value of t from equation (1.20) into equation (1.19), we get
rdark  mR
(m  0, 1, 2, ...)
(1.21)
In general, for any thin film of refractive index n film , the expression for the radii of
the dark rings is given by rdark 
mR
n film
(m  0, 1, 2, ...)
(1.22)
Similarly, the expression for the radii of the bright rings is given by,
1

 m   R
2
(1.23)
rbright  
(m  0, 1, 2, ...)
n film
That is, the diameters of Newton’s dark rings are proportional to square root of the
natural numbers and the diameters of Newton’s bright rings are proportional to
square root of natural odd numbers.
1.7 MICHELSON INTERFEROMETER
The interferometer, invented by American physicist A. A. Michelson (1852–1931),
splits a light beam into two parts and then recombines the parts to form an
interference pattern. A schematic diagram of the interferometer is shown in Fig. 1.9.
A ray of light from a monochromatic source is split into two rays by mirror M0, which
is inclined at 45° to the incident light beam. Mirror M 0, called a beam splitter,
transmits half the light incident on it and reflects the rest. One ray is reflected from
M0 to the right toward mirror M1, and the second ray is transmitted vertically
through M0 toward mirror M2. Hence, the two rays travel separate paths L1 and L2.
After reflecting from M1 and M2, the two rays eventually recombine at M0 to produce
an interference pattern, which can be viewed through a telescope.
The interference condition for the two rays is determined by the difference in their
path length. When the two mirrors are exactly perpendicular to each other, the
interference pattern is a target pattern of bright and dark circular fringes. As M1 is
moved, the fringe pattern collapses or expands, depending on the direction in which
M1 is moved. For example, if a dark circle appears at the center of the target pattern
(corresponding to destructive interference) and M1 is then moved a distance /4
toward M0, the path difference changes by /2. This replaces dark circle at center by
bright circle. Therefore, the fringe pattern shifts by one-half fringe each time M1 is
moved a distance /4. The wavelength of light is then measured by counting the
number of fringe shifts for a given displacement of M1. So it can also be used to
10
detect small change in path length as in the laser interferometer gravitational-wave
observatory.
Figure 1.9 Schematic diagram of Michelson Interferometer
1.8 QUESTIONS
1. What is interference of light waves?
2. What is coherence? Mention its importance.
3. Write the necessary condition for the constructive and destructive
interference of two light waves in terms of path/phase difference.
4. Obtain an expression for intensity of light in double-slit interference.
5. Write the conditions for constructive and destructive interference of
reflected light from a thin soap film in air, assuming normal incidence.
6. Explain the formation of fringes in Michelson interferometer.
1.9 PROBLEMS
1. A viewing screen is separated from a double slit by 4.80 m. The distance
between the two slits is 0.0300 mm. Monochromatic light is directed toward
the double slit and forms an interference pattern on the screen. The first dark
fringe is 4.50 cm from the center line on the screen. (A) Determine the
11
2.
3.
4.
5.
6.
wavelength of the light. (B) Calculate the distance between adjacent bright
fringes. Ans: 562 nm and 9 cm
A light source emits visible light of two wavelengths:  = 430 nm and / = 510
nm. The source is used in a double-slit interference experiment in which L =
1.50 m and d = 0.025 0 mm. Find the separation distance between the thirdorder bright fringes for the two wavelengths. Ans: 1.44 cm
A laser beam ( = 632.8 nm) is incident on two slits 0.200 mm apart. How far
apart are the bright interference fringes on a screen 5.00 m away from the
double slits? Ans: 15.8 mm
A Young’s interference experiment is performed with monochromatic light.
The separation between the slits is 0.500 mm, and the interference pattern
on a screen 3.30 m away shows the first side maximum 3.40 mm from the
center of the pattern. What is the wavelength? Ans: 515 nm
Young’s double-slit experiment is performed with 589-nm light and a
distance of 2.00 m between the slits and the screen. The tenth interference
minimum is observed 7.26 mm from the central maximum. Determine the
spacing of the slits. Ans: 1.54 mm
Two radio antennas separated by d = 300 m as shown in figure simultaneously
broadcast identical signals at the same wavelength. A car travels due north
along a straight line at position x = 1000 m from the center point between
the antennas, and its radio receives the signals. (a) If the car is at the position
of the second maximum after that at point O when it has traveled a distance
y = 400 m northward, what is the wavelength of the signals? (b) How much
farther must the car travel from this position to encounter the next minimum
in reception? Note: Do not use the small-angle approximation in this
problem. Ans: 55.7 m and 124 m
7. Two narrow parallel slits separated by 0.250 mm are illuminated by green
light ( = 546.1 nm). The interference pattern is observed on a screen 1.20 m
away from the plane of the slits. Calculate the distance (a) from the central
maximum to the first bright region on either side of the central maximum
and (b) between the first and second dark bands. Ans: 2.62 mm and 2.62 mm
12
8. In a Young’s interference experiment, the two slits are separated by 0.150 mm
and the incident light includes two wavelengths: 1 = 540 nm (green) and
2 = 450 nm (blue). The overlapping interference patterns are observed on a
screen 1.40 m from the slits. Calculate the minimum distance from the center
of the screen to a point where a bright fringe of the green light coincides with
a bright fringe of the blue light. Ans: 2.52 cm
9. In a double slit experiment, let L = 120 cm and d = 0.250 cm. The slits are
illuminated with coherent 600-nm light. Calculate the distance y above the
central maximum for which the average intensity on the screen is 75.0% of
the maximum. Ans: 48 micrometer
10. Show that the two waves with wave functions E1  6.00 sin(100t ) and
E2  8.00 sin(100t   / 2) add to give a wave with the wave function
E R sin(100t   ) . Find the required values for ER and  . Ans: 10 and 53.1
11. Calculate the minimum thickness of a soap-bubble film that results in
constructive interference in the reflected light if the film is illuminated with
light whose wavelength in free space is  = 600 nm. The index of refraction
of the soap film is 1.33. (b) What if the film is twice as thick? Does this
situation produce constructive interference? Ans: 113 nm and No
12. Solar cells—devices that generate electricity when exposed to sunlight—are
often coated with a transparent, thin film of silicon monoxide (SiO, n = 1.45)
to minimize reflective losses from the surface. Suppose a silicon solar cell (n
= 3.5) is coated with a thin film of silicon monoxide for this purpose.
Determine the minimum film thickness that produces the least reflection at
a wavelength of 550 nm, near the center of the visible spectrum. Ans: 95 nm
13. A thin film of oil (n = 1.25) is located on smooth, wet pavement. When viewed
perpendicular to the pavement, the film reflects most strongly red light at
640 nm and reflects no green light at 512 nm. How thick is the oil film? Ans:
512 nm.
14. An oil film (n = 1.45) floating on water is illuminated by white light at normal
incidence. The film is 280 nm thick. Find (a) the wavelength and color of the
light in the visible spectrum most strongly reflected and (b) the wavelength
and color of the light in the spectrum most strongly transmitted. Explain your
reasoning. Ans: 541 nm and 406 nm
15. An air wedge is formed between two glass plates separated at one edge by a
very fine wire of circular cross section as shown in Figure 12. When the wedge
is illuminated from above by 600-nm light and viewed from above, 30 dark
fringes are observed. Calculate the diameter d of the wire. Ans: 8.7
micrometer
13
16. When a liquid is introduced into the air space between the lens and the plate
in a Newton’s-rings apparatus, the diameter of the tenth ring changes from
1.50 to 1.31 cm. Find the index of refraction of the liquid. Ans: 1.31
17. A certain grade of crude oil has an index of refraction of 1.25. A ship
accidentally spills 1.00 m3 of this oil into the ocean, and the oil spreads into a
thin, uniform slick. If the film produces a first-order maximum of light of
wavelength 500 nm normally incident on it, how much surface area of the
ocean does the oil slick cover? Assume the index of refraction of the ocean
water is 1.34. Ans: 5km2
18. In a Newton’s-rings experiment, a plano-convex glass (n = 1.52) lens having
radius r = 5.00 cm is placed on a flat plate as shown in Figure 1.8a. When light
of wavelength 650 nm is incident normally, 55 bright rings are observed, with
the last one precisely on the edge of the lens. (a) What is the radius R of
curvature of the convex surface of the lens? (b) What is the focal length of
the lens? Ans: 70.5 m and 138 m
19. Monochromatic light is beamed into a Michelson interferometer. The
movable mirror is displaced 0.382 mm, causing the interferometer pattern to
reproduce itself 1700 times. Determine the wavelength of the light. What
color is it? Ans: 449 nm, Blue
20. Mirror M1 in Figure 1.9 is moved through a displacement L. During this
displacement, 250 fringe reversals (formation of successive dark or bright
bands) are counted. The light being used has a wavelength of 632.8 nm.
Calculate the displacement L. Ans: 39.6 micrometer
21. One leg of a Michelson interferometer contains an evacuated cylinder of
length L, having glass plates on each end. A gas is slowly leaked into the
cylinder until a pressure of 1 atm is reached. If N bright fringes pass on the
screen during this process when light of wavelength  is used, what is the
index of refraction of the gas? Ans: n = 1 + (Nλ)/(2L)
14
2 DIFFRACTION PATTERNS AND
POLARIZATION
OBJECTIVES
•
•
•
•
To understand the principles of diffraction.
To explain the intensity distribution in diffraction under various
conditions.
To explain the diffraction of light waves at single, multiple slits and
circular apertures.
To understand polarization phenomena and various techniques used to
produce polarized light.
2.1 INTRODUCTION TO DIFFRACTION PATTERNS
Light of wavelength comparable to or larger than the width of a slit spreads out in
all forward directions upon passing through the slit. This phenomenon is called
diffraction. When light passes through a narrow slit, it spreads beyond the narrow
path defined by the slit into regions that would be in shadow if light traveled in
straight lines. Other waves, such as sound waves and water waves, also have this
property of spreading when passing through apertures or by sharp edges.
A diffraction pattern consisting of light and dark areas is observed when a narrow
slit is placed between a distant light source (or a laser beam) and a screen, the light
produces a diffraction pattern like that shown in Figure 2.1 (a). The pattern consists
of a broad, intense central band (called the central maximum) flanked by a series
of narrower, less intense additional bands (called side maxima or secondary
maxima) and a series of intervening dark bands (or minima).
Figure 2.1 (a) The diffraction pattern that appears on a screen when light passes through a narrow vertical
slit. (b) Diffraction pattern created by the illumination of a penny, with the penny positioned midway
between the screen and light source.
15
Figure 2.1 (b) shows a diffraction pattern associated with the shadow of a penny. A
bright spot occurs at the center, and circular fringes extend outward from the
shadow’s edge. From the viewpoint of ray optics (in which light is viewed as rays
traveling in straight lines), we expect the center of the shadow to be dark because
that part of the viewing screen is completely shielded by the penny. We can explain
the central bright spot by using the wave theory of light, which predicts constructive
interference at this point.
2.2 DIFFRACTION PATTERNS FROM NARROW SLITS
Let’s consider light passing through a narrow opening modeled as a slit and
projected onto a screen. To simplify our analysis, we assume the observing screen is
far from the slit and the rays reaching the screen are approximately parallel. In
laboratory, this situation can also be achieved experimentally by using a converging
lens to focus the parallel rays on a nearby screen. In this model, the pattern on the
screen is called a Fraunhofer diffraction pattern.
Until now, we have assumed slits are point sources of light. In this section, we
abandon that assumption and see how the finite width of slits is the basis for
understanding Fraunhofer diffraction. We can explain some important features of
this phenomenon by examining waves coming from various portions of the slit as
shown in Figure 2.2.
According to Huygens’s principle, each portion of the slit acts as a source of light
waves. Hence, light from one portion of the slit can interfere with light from another
portion, and the resultant light intensity on a viewing screen depends on the
direction . Based on this analysis, we recognize that a diffraction pattern is an
interference pattern in which the different sources of light are different portions of
the single slit.
Figure 2.2(a) Geometry for analyzing the Fraunhofer diffraction pattern of a single slit. (b) Photograph of a
single-slit Fraunhofer diffraction pattern.
16
Figure 2.3 Paths of light rays that encounter a narrow slit of width a and diffract toward a screen in the
direction described by angle .
To analyze the diffraction pattern, let’s divide the slit into two halves as shown in
Figure 2.3. Keeping in mind that all the waves are in phase as they leave the slit,
consider rays 1 and 3. As these two rays travel toward a viewing screen far to the
right of the figure, ray 1 travels farther than ray 3 by an amount equal to the path
difference (a/2) sin , where a is the width of the slit. Similarly, the path difference
between rays 2 and 4 is also (a/2) sin , as is that between rays 3 and 5. If this path
difference is exactly half a wavelength (corresponding to a phase difference of 180°),
the pairs of waves cancel each other and destructive interference results.
This cancellation occurs for any two rays that originate at points separated by half
the slit width because the phase difference between two such points is 180°.
Therefore, waves from the upper half of the slit interfere destructively with waves
from the lower half when
𝑎

sin  = ±
2
2
Dividing the slit into four equal parts and using similar reasoning, we find that the
viewing screen is also dark when
sin  = ± 2

𝑎
Likewise, dividing the slit into six equal parts shows that darkness occurs on the
screen when
sin  = ± 3

𝑎
Therefore, the general condition for destructive interference is
17
sin dark  m

a
m  1,  2,  3, ...
(2.1)
2.3 INTENSITY OF SINGLE-SLIT DIFFRACTION PATTERNS
Analysis of the intensity variation in a diffraction pattern from a single slit of width
‘a’ shows that the intensity is given by
I  I max
 sin  a sin  /   


  a sin  / 

2
(2.2)
where Imax is the intensity at  = 0 (the central maximum) and  is the wavelength
of light used to illuminate the slit. Intensity variation plot and photograph of the
pattern are shown below.
Figure 2.4 A plot of light intensity I versus (/)a sin  for the single-slit Fraunhofer diffraction
pattern. (b) Photograph of a single slit Fraunhofer diffraction pattern.
2.4 INTENSITY OF TWO-SLIT DIFFRACTION PATTERNS
When more than one slit is present, we must consider not only diffraction patterns
due to the individual slits but also the interference patterns due to the waves coming
from different slits. Intensity due to combined effect is given by
  d sin 
I  I max cos 


2
  sin  a sin  /   


   a sin  / 

2
(2.3)
18
Above equation represents the single-slit diffraction pattern (the factor in square
brackets) acting as an “envelope” for a two slit interference pattern (the cosinesquared factor).
We have seen that angular position of interference maxima is given by d sin  = m,
where d is the distance between the two slits. Also, the first diffraction minimum
occurs when a sin  = , where a is the slit width. Dividing interference equation by
diffraction equation,
𝑑
=𝑚
𝑎
In this case, mth interference maximum coincides with first diffraction minimum.
Figure 2.5 The combined effects of two-slit and single-slit interference.
2.5 RESOLUTION OF SINGLE-SLIT AND CIRCULAR
APERTURES
The ability of optical systems to distinguish between closely spaced objects is limited
because of the wave nature of light. To understand this limitation, consider Figure
2.6, which shows two light sources far from a narrow slit of width a. The sources can
be two noncoherent point sources S1 and S2; for example, they could be two distant
stars. If no interference occurred between light passing through different parts of
the slit, two distinct bright spots (or images) would be observed on the viewing
screen. Because of such interference, however, each source is imaged as a bright
central region flanked by weaker bright and dark fringes, a diffraction pattern. What
is observed on the screen is the sum of two diffraction patterns: one from S1 and the
other from S2.
19
Figure 2.6 Two-point sources far from a narrow slit each produce a diffraction pattern. (a) The
sources are separated by a large angle. (b) The sources are separated by a small angle.
When the central maximum of one image falls on the first minimum of another
image, the images are said to be just resolved. This limiting condition of resolution
is known as Rayleigh’s criterion.
From Rayleigh’s criterion, we can determine the minimum angular separation min
subtended by the sources at the slit in Figure 2.6 for which the images are just
resolved. Equation 2.1 indicates that the first minimum (m = 1) in a single-slit
diffraction pattern occurs at the angle for which

sin  = 𝑎
(2.4)
where a is the width of the slit. According to Rayleigh’s criterion, this expression
gives the smallest angular separation for which the two images are resolved. Because
 << a in most situations, sin  is small and we can use the approximation sin   .
Therefore, the limiting angle of resolution for a slit of width a is
𝑚𝑖𝑛 =

𝑎
(2.5)
where min is expressed in radians. Hence, the angle subtended by the two sources
at the slit must be greater than /a if the images are to be resolved.
Many optical systems use circular apertures rather than slits. The diffraction pattern
of a circular aperture as shown in the photographs of Figure 2.7 consists of a central
circular bright disk surrounded by progressively fainter bright and dark rings. Figure
2.7 shows diffraction patterns for three situations in which light from two point
sources passes through a circular aperture. When the sources are far apart, their
images are well resolved (Fig. 2.7a). When the angular separation of the sources
satisfies Rayleigh’s criterion, the images are just resolved (Fig. 2.7b). Finally, when
the sources are close together, the images are said to be unresolved (Fig. 2.7c) and
the pattern looks like that of a single source. Analysis shows that the limiting angle
of resolution of the circular aperture is
20
𝑚𝑖𝑛 = 1.22

𝐷
(2.6)
where D is the diameter of the aperture. This expression is similar to Equation 2.4
except for the factor 1.22, which arises from a mathematical analysis of diffraction
from the circular aperture.
Figure 2.7 Individual diffraction patterns of two-point sources (solid curves) and the resultant
patterns (dashed curves) for various angular separations of the sources as the light passes through a
circular aperture. In each case, the dashed curve is the sum of the two solid curves.
2.6 DIFFRACTION GRATING
The diffraction grating, a useful device for analyzing light sources, consists of
many equally spaced parallel slits. A transmission grating can be made by cutting
parallel grooves on a glass plate with a precision ruling machine. The spaces between
the grooves are transparent to the light and hence act as separate slits. A reflection
grating can be made by cutting parallel grooves on the surface of a reflective
material. The reflection of light from the spaces between the grooves is specular,
and the reflection from the grooves cut into the material is diffuse. Therefore, the
spaces between the grooves act as parallel sources of reflected light like the slits in
a transmission grating.
21
Figure 2.8 Side view of a diffraction grating. The slit separation is d, and the path difference between
adjacent slits is d sin.
A plane wave is incident from the left, normal to the plane of the grating. The
pattern observed on the screen far to the right of the grating is the result of the
combined effects of interference and diffraction. Each slit produces diffraction, and
the diffracted beams interfere with one another to produce the final pattern. The
waves from all slits are in phase as they leave the slits. For an arbitrary direction 
measured from the horizontal, however, the waves must travel different path
lengths before reaching the screen. Notice in Figure 2.8 that the path difference 
between rays from any two adjacent slits is equal to d sin . If this path difference
equals one wavelength or any integral multiple of a wavelength, waves from all slits
are in phase at the screen and a bright fringe is observed. Therefore, the condition
for maxima in the interference pattern at the angle bright is
d sin bright  m
m  0,  1,  2,  3, ...
(2.7)
22
Figure 2.9 Intensity versus sin  for a diffraction grating. The zeroth-, first-, and
second-order maxima are shown.
The intensity distribution for a diffraction grating obtained with the use of a
monochromatic source is shown in Figure 2.9. Notice the sharpness of the principal
maxima and the broadness of the dark areas compared with the broad bright fringes
characteristic of the two-slit interference pattern.
Figure 2.10 Diagram of a diffraction grating spectrometer.
A schematic drawing of a simple apparatus used to measure angles in a diffraction
pattern is shown in Figure 2.10. This apparatus is a diffraction grating spectrometer.
The light to be analyzed passes through a slit, and a collimated beam of light is
incident on the grating. The diffracted light leaves the grating at angles that satisfy
Equation 2.7, and a telescope is used to view the image of the slit. The wavelength
can be determined by measuring the precise angles at which the images of the slit
appear for the various orders. The spectrometer is a useful tool in atomic
spectroscopy, in which the light from an atom is analyzed to find the wavelength
components. These wavelength components can be used to identify the atom.
23
2.7 DIFFRACTION OF X-RAYS BY CRYSTALS
In principle, the wavelength of any electromagnetic wave can be determined if a
grating of the proper spacing (on the order of ) is available. X-rays, discovered by
Wilhelm Roentgen (1845–1923) in 1895, are electromagnetic waves of very short
wavelength (on the order of 0.1 nm). It would be impossible to construct a grating
having such a small spacing by the cutting process. The atomic spacing in a solid is
known to be about 0.1 nm, however. In 1913, Max von Laue (1879–1960) suggested
that the regular array of atoms in a crystal could act as a three-dimensional
diffraction grating for x-rays. Subsequent experiments confirmed this prediction.
The diffraction patterns from crystals are complex because of the three-dimensional
nature of the crystal structure. Nevertheless, x-ray diffraction has proved to be an
invaluable technique for elucidating these structures and for understanding the
structure of matter.
Figure 2.11 Crystalline structure of sodium chloride (NaCl).
The arrangement of atoms in a crystal of sodium chloride (NaCl) is shown in Figure
2.11. Each unit cell (the geometric solid that repeats throughout the crystal) is a cube
having an edge length a. A careful examination of the NaCl structure shows that the
ions lie in discrete planes (the shaded areas in Fig. 2.11). Now suppose an incident xray beam makes an angle  with one of the planes as in Figure 2.12. The beam can
be reflected from both the upper plane and the lower one, but the beam reflected
from the lower plane travels farther than the beam reflected from the upper plane.
The effective path difference is 2dsin. The two beams reinforce each other
(constructive interference) when this path difference equals some integer multiple
of . The same is true for reflection from the entire family of parallel planes. Hence,
the condition for constructive interference (maxima in the reflected beam) is
2d sin   m
m  1, 2, 3, ...
(2.8)
This condition is known as Bragg’s law, after W. L. Bragg, who first derived the
relationship. If the wavelength and diffraction angle are measured, Equation 2.8 can
be used to calculate the spacing between atomic planes.
24
Figure 2.12 A two-dimensional description of the reflection of an x-ray beam from two parallel
crystalline planes separated by a distance d.
2.8 POLARIZATION OF LIGHT WAVES
An ordinary beam of light consists of many waves emitted by the atoms of the light
source. Each atom produces a wave having some orientation of the electric field
vector ⃗𝑬, corresponding to the direction of atomic vibration. The direction of
polarization of each individual wave is defined to be the direction in which the
electric field is vibrating. In Figure 2.13, this direction happens to lie along the y axis.
⃗⃗ vector
All individual electromagnetic waves traveling in the x direction have an 𝑬
parallel to the yz plane, but this vector could be at any possible angle with respect
to the y axis. Because all directions of vibration from a wave source are possible, the
resultant electromagnetic wave is a superposition of waves vibrating in many
different directions. The result is an unpolarized light beam, represented in Figure
2.14a. The direction of wave propagation in this figure is perpendicular to the page.
The arrows show a few possible directions of the electric field vectors for the
individual waves making up the resultant beam. At any given point and at some
instant of time, all these individual electric field vectors add to give one resultant
electric field vector.
⃗ vibrates in
A wave is said to be linearly polarized if the resultant electric field ⃗𝑬
the same direction at all times at a particular point as shown in Figure 2.14b.
(Sometimes, such a wave is described as plane-polarized, or simply polarized.) The
plane formed by ⃗𝑬 and the direction of propagation is called the plane of polarization
of the wave. If the wave in Figure 2.14b represents the resultant of all individual
waves, the plane of polarization is the xy plane. A linearly polarized beam can be
obtained from an unpolarized beam by removing all waves from the beam except
those whose electric field vectors oscillate in a single plane.
25
Figure 2.13 Schematic diagram of an electromagnetic wave propagating at velocity c in the x direction.
The electric field vibrates in the xy plane, and the magnetic field vibrates in the xz plane.
Figure 2.14 (a) A representation of an unpolarized light beam viewed along the direction of
propagation. The transverse electric field can vibrate in any direction in the plane of the page with
equal probability. (b) A linearly polarized light beam with the electric field vibrating in the vertical
direction.
2.9 POLARIZATION BY SELECTIVE ABSORPTION
The most common technique for producing polarized light is to use a material that
transmits waves whose electric fields vibrate in a plane parallel to a certain direction
and that absorbs waves whose electric fields vibrate in all other directions. Polaroid,
that polarizes light through selective absorption. This material is fabricated in thin
sheets of long-chain hydrocarbons. The sheets are stretched during manufacture so
that the long-chain molecules align. After a sheet is dipped into a solution
containing iodine, the molecules become good electrical conductors. Conduction
takes place primarily along the hydrocarbon chains because electrons can move
easily only along the chains.
If light whose electric field vector is parallel to the chains is incident on the material,
the electric field accelerates electrons along the chains and energy is absorbed from
26
the radiation. Therefore, the light does not pass through the material. Light whose
electric field vector is perpendicular to the chains passes through the material
because electrons cannot move from one molecule to the next. As a result, when
unpolarized light is incident on the material, the exiting light is polarized
perpendicular to the molecular chains. It is common to refer to the direction
perpendicular to the molecular chains as the transmission axis. In an ideal polarizer,
⃗ parallel to the transmission axis is transmitted and all light with 𝑬
⃗
all light with 𝑬
perpendicular to the transmission axis is absorbed.
Figure 2.15 Two polarizing sheets whose transmission axes make an angle  with each other. Only a
fraction of the polarized light incident on the analyzer is transmitted through it.
Figure 2.15 represents an unpolarized light beam incident on a first polarizing sheet,
called the polarizer. Because the transmission axis is oriented vertically in the figure,
the light transmitted through this sheet is polarized vertically. A second polarizing
sheet, called the analyzer, intercepts the beam. In figure, the analyzer transmission
axis is set at an angle  to the polarizer axis. We call the electric field vector of the
⃗ 𝟎 . The component of 𝑬
⃗⃗ 𝟎 perpendicular to the analyzer axis
first transmitted beam 𝑬
is completely absorbed. The component of ⃗𝑬𝟎 parallel to the analyzer axis, which is
transmitted through the analyzer, is E0 cos . Because the intensity of the
transmitted beam varies as the square of its magnitude, we conclude that the
intensity I of the (polarized) beam transmitted through the analyzer varies as
I  I max cos2 
(2.9)
where Imax is the intensity of the polarized beam incident on the analyzer. This
expression, known as Malus’s law.
2.10 POLARIZATION BY REFLECTION
When an unpolarized light beam is reflected from a surface, the polarization of the
reflected light depends on the angle of incidence. If the angle of incidence is 0°, the
27
reflected beam is unpolarized. For other angles of incidence, the reflected light is
polarized to some extent, and for a particular angle of incidence, the reflected light
is completely polarized.
Figure 2.16 (a) When unpolarized light is incident on a reflecting surface, the reflected and refracted
beams are partially polarized. (b) The reflected beam is completely polarized when the angle of
incidence equals the polarizing angle p, which satisfies the equation n2/n1 = tan p. At this incident
angle, the reflected and refracted rays are perpendicular to each other.
Now suppose the angle of incidence 1 is varied until the angle between the reflected
and refracted beams is 90° as in Figure 2.16b. At this angle of incidence, the reflected
beam is completely polarized (with its electric field vector parallel to the surface)
and the refracted beam is still only partially polarized. The angle of incidence at
which this polarization occurs is called the polarizing angle p. Using Snell’s law
of refraction
𝑛2
𝑛1
=
sin 𝑝
sin 2
2.10
But, 2 = 90 - p. So, we can write,
tan 𝑝 =
𝑛2
𝑛1
2.11
This expression is called Brewster’s law, and the polarizing angle p is sometimes
called Brewster’s angle, after its discoverer, David Brewster. Because n varies with
wavelength for a given substance, Brewster’s angle is also a function of wavelength.
28
2.11 POLARIZATION BY DOUBLE REFRACTION
In certain class of crystals like calcite and quartz, the speed of light depends on the
direction of propagation and on the plane of polarization of the light. Such materials
are characterized by two indices of refraction. Hence, they are often referred to as
double-refracting or birefringent materials. When unpolarized light enters a
birefringent material, it may split into an ordinary (O) ray and an extraordinary
(E) ray. These two rays have mutually perpendicular polarizations and travel at
different speeds through the material. There is one direction, called the optic axis,
along which the ordinary and extraordinary rays have the same speed.
Figure 2.17 Unpolarized light incident at an angle to the optic axis in a calcite crystal splits into an
ordinary (O) ray and an extraordinary (E) ray
Figure 2.18 Point source S inside a double-refracting crystal (calcite) produces a spherical wave front
corresponding to the ordinary (O) ray and an elliptical wave front corresponding to the extraordinary
(E) ray.
Some materials such as glass and plastic become birefringent when stressed.
Suppose an unstressed piece of plastic is placed between a polarizer and an analyzer
so that light passes from polarizer to plastic to analyzer. When the plastic is
unstressed, and the analyzer axis is perpendicular to the polarizer axis, none of the
polarized light passes through the analyzer. In other words, the unstressed plastic
has no effect on the light passing through it. If the plastic is stressed, however,
29
regions of greatest stress become birefringent and the polarization of the light
passing through the plastic changes. Hence, a series of bright and dark bands is
observed in the transmitted light, with the bright bands corresponding to regions of
greatest stress. Engineers often use this technique, called optical stress analysis, in
designing structures ranging from bridges to small tools. They build a plastic model
and analyze it under different load conditions to determine regions of potential
weakness and failure under stress.
Figure 2.19 The pattern is produced when the plastic model is viewed between a polarizer and analyzer
oriented perpendicular to each other. Such patterns are useful in the optimal design of architectural
components
2.12 POLARIZATION BY SCATTERING
Figure 2.20 The scattering of unpolarized sunlight by air molecules.
30
When light is incident on any material, the electrons in the material can absorb and
reradiate part of the light. Such absorption and reradiation of light by electrons in
the gas molecules that make up air is what causes sunlight reaching an observer on
the Earth to be partially polarized. An unpolarized beam of sunlight traveling in the
horizontal direction (parallel to the ground) strikes a molecule of one of the gases
that make up air, setting the electrons of the molecule into vibration. These
vibrating charges act like the vibrating charges in an antenna. The horizontal
component of the electric field vector in the incident wave results in a horizontal
component of the vibration of the charges, and the vertical component of the vector
results in a vertical component of vibration. If the observer in Figure 2.20 is looking
straight up (perpendicular to the original direction of propagation of the light), the
vertical oscillations of the charges send no radiation toward the observer. Therefore,
the observer sees light that is completely polarized in the horizontal direction as
indicated by the orange arrows. If the observer looks in other directions, the light is
partially polarized in the horizontal direction.
2.13 OPTICAL ACTIVITY
Many important applications of polarized light involve materials that display optical
activity. A material is said to be optically active if it rotates the plane of polarization
of any light transmitted through the material. The angle through which the light is
rotated by a specific material depends on the length of the path through the material
and on concentration if the material is in solution. One optically active material is a
solution of the common sugar dextrose. A standard method for determining the
concentration of sugar solutions is to measure the rotation produced by a fixed
length of the solution.
2.14 QUESTIONS
1. Explain the term diffraction of light.
2. Discuss qualitatively, the Fraunhofer diffraction at a single-slit.
3. Draw a schematic plot of the intensity of light in single slit diffraction against
phase difference.
4. Explain briefly diffraction at a circular aperture.
5. State and explain Rayleigh’s criterion for optical resolution.
6. Effect of diffraction is ignored in the case of Young’s double slit interference.
Give reason.
7. Discuss qualitatively, the diffraction due to multiple slits.
8. What is diffraction grating? Write the grating equation.
9. Briefly explain x-ray diffraction and Bragg’s law.
10. Distinguish between unpolarized and linearly polarized light.
11. Explain Malus’s law.
12. How to produce linearly polarized light by (a) selective absorption, (b)
reflection, (c) double refraction, (d) scattering ? Explain.
31
2.15 PROBLEMS
1. Light of wavelength 540 nm passes through a slit of width 0.200 mm. (a) The
width of the central maximum on a screen is 8.10 mm. How far is the screen
from the slit? (b) Determine the width of the first bright fringe to the side of
the central maximum. Ans: (a) 1.5 m (b) 4.05 mm
2. Helium–neon laser light ( = 632.8 nm) is sent through a 0.300-mm-wide
single slit. What is the width of the central maximum on a screen 1.00 m from
the slit? Ans: 4.22 mm
3. A screen is placed 50.0 cm from a single slit, which is illuminated with light
of wavelength 690 nm. If the distance between the first and third minima in
the diffraction pattern is 3.00 mm, what is the width of the slit?
Ans: 2.3x10-4 m
4. A beam of monochromatic light is incident on a single slit of width 0.600
mm. A diffraction pattern forms on a wall 1.30 m beyond the slit. The distance
between the positions of zero intensity on both sides of the central maximum
is 2.00 mm. Calculate the wavelength of the light. Ans: 462 nm
5. A diffraction pattern is formed on a screen 120 cm away from a 0.400-mmwide slit. Monochromatic 546.1-nm light is used. Calculate the fractional
intensity I/Imax at a point on the screen 4.10 mm from the center of the
principal maximum. Ans: 0.0162
6. Yellow light of wavelength 589 nm is used to view an object under a
microscope. The objective lens diameter is 9.00 mm. (a) What is the limiting
angle of resolution? (b) Suppose it is possible to use visible light of any
wavelength. What color should you choose to give the smallest possible angle
of resolution, and what is this angle? (c) Suppose water fills the space
between the object and the objective. What effect does this change have on
the resolving power when 589-nm light is used? Ans: (a) 79.8 x 10-6 rad (b)
400nm, 54.2 x 10-6 rad (c) Resolving power will improve with minimum resolvable
angle 60 x 10-6 rad
7. The angular resolution of a radio telescope is to be 0.100° when the incident
waves have a wavelength of 3.00 mm. What minimum diameter is required
for the telescope’s receiving dish? Ans: 2.1 mm
8. White light is spread out into its spectral components by a diffraction grating.
If the grating has 2000 grooves per centimeter, at what angle does red light
of wavelength 640 nm appear in first order? Ans: θ = 7.35o
9. Light of wavelength 500 nm is incident normally on a diffraction grating. If
the third-order maximum of the diffraction pattern is observed at 32.0°, (a)
what is the number of rulings per centimeter for the grating? (b) Determine
the total number of primary maxima that can be observed in this situation.
Ans: 3530 rulings/cm (b) 11
32
10. If the spacing between planes of atoms in a NaCl crystal is 0.281 nm, what is
the predicted angle at which 0.140-nm x-rays are diffracted in a first-order
maximum? Ans: θ = 14.4o
11. The first-order diffraction maximum is observed at 12.6° for a crystal having
a spacing between planes of atoms of 0.250 nm. (a) What wavelength x-ray is
used to observe this first-order pattern? (b) How many orders can be
observed for this crystal at this wavelength? Ans: (a) 0.109 nm (b) 4
12. Plane-polarized light is incident on a single polarizing disk with the direction
of E parallel to the direction of the transmission axis. Through what angle
should the disk be rotated so that the intensity in the transmitted beam is
reduced by a factor of (a) 3.00, (b) 5.00, and (c) 10.0? Ans: (a) 54.70 (b) 63.40
(c) 71.60
13. Unpolarized light passes through two ideal Polaroid sheets. The axis of the
first is vertical, and the axis of the second is at 30.0° to the vertical. What
fraction of the incident light is transmitted? Ans: 0.375
14. The angle of incidence of a light beam onto a reflecting surface is continuously
variable. The reflected ray in air is completely polarized when the angle of
incidence is 48.0°. What is the index of refraction of the reflecting material?
Ans: 1.1
15. The critical angle for total internal reflection for sapphire surrounded by air is
34.4°. Calculate the polarizing angle for sapphire. Ans: 60.5o
33
3 QUANTUM PHYSICS
OBJECTIVES:
 To learn certain experimental results that can be understood only by
particle theory of electromagnetic waves.
 To learn the particle properties of waves and the wave properties of the
particles.
 To understand the uncertainty principle.
3.1 BLACKBODY RADIATION AND PLANCK’S
HYPOTHESIS
A black body is an object that absorbs all incident radiation. A small hole cut into a
cavity is the most popular and realistic example. None of the incident radiation
escapes. The radiation is absorbed in the walls of the cavity. This causes a heating of
the cavity walls. The oscillators in the cavity walls vibrate and re-radiate at
wavelengths corresponding to the temperature of the cavity, thereby producing
standing waves. Some of the energy from these standing waves can leave through
the opening. The electromagnetic radiation emitted by the black body is called
black-body radiation.
Figure 3.1 A physical model of a blackbody
•
•
•
The black body is an ideal absorber of incident radiation.
A black-body reaches thermal equilibrium with the surroundings when the
incident radiation power is balanced by the power re-radiated.
The emitted "thermal" radiation from a black body characterizes the equilibrium
temperature of the black-body.
34
•
The nature of radiation from a blackbody does not depend on the material of
which the walls are made.
Basic laws of radiation
(1) All objects emit radiant energy.
(2) Hotter objects emit more energy (per unit area) than colder objects. The total
power of the emitted radiation is proportional to the fourth power of temperature.
This is called Stefan’s Law and is given by
P =  A e T4
(3.1)
where P is power radiated from the surface of the object (W), T is equilibrium
surface temperature (K), σ is Stefan-Boltzmann constant (= 5.670 x 10−8 W/m2K4 ),
A is surface area of
the object (m2) and e is emissivity of the surface (e =1 for a perfect blackbody).
(3) The peak of the wavelength distribution shifts to shorter wavelengths as the
black body temperature increases. This is Wien’s Displacement Law and is given
by
λm T = constant = 2.898 × 10−3 m.K , or λm  T−1
(3.2)
where λm is the wavelength corresponding to peak intensity and T is equilibrium
temperature of the blackbody.
Figure 3.2 Intensity of blackbody radiation versus wavelength at two temperatures
(4) Rayleigh-Jeans Law: This law tries to explain the distribution of energy from
a
black body. The intensity or power per unit area I (,T)d, emitted in the
wavelength interval  to +d from a blackbody is given by
2  c kB T
(3.3)
I(  ,T ) 
4

35
kB is Boltzmann’s constant, c is speed of light in vacuum, T is equilibrium blackbody
temperature. It agrees with experimental measurements only for long wavelengths.
It predicts an energy output that diverges towards infinity as wavelengths become
smaller and is known as the ultraviolet catastrophe.
Figure 3.3 Comparison of experimental results and the curve predicted by the Rayleigh–Jeans law for
the distribution of blackbody radiation
(5) Planck‘s Law:
Max Planck developed a theory of blackbody radiation that leads to an equation for
I (,T) that is in complete agreement with experimental results. To derive the law,
Planck made two assumptions concerning the nature of the oscillators in the cavity
walls:
(i) The energy of an oscillator is quantized hence it can have only certain discrete
values:
En = n h f
(3.4)
where n is a positive integer called a quantum number, f is the frequency of cavity
oscillators, and h is a constant called Planck’s constant. Each discrete energy
value corresponds to a different quantum state, represented by the quantum
number n.
(ii) The oscillators emit or absorb energy only when making a transition from one
quantum state to another. Difference in energy will be integral multiples of hf.
36
Figure 3.4 Allowed energy levels for an oscillator with frequency f
Planck’s law explains the distribution of energy from a black body which is given by,
I(  ,T ) 
2 h c 2

1
5
e
hc
λkB T
(3.5)
1
where I (,T) d is the intensity or power per unit area emitted in the wavelength
interval d from a blackbody, h is Planck’s constant, kB is Boltzmann's constant,
c is speed of light in vacuum and T is equilibrium temperature of blackbody .
The Planck‘s Law gives a distribution that peaks at a certain wavelength, the peak
shifts to shorter wavelengths for higher temperatures, and the area under the curve
grows rapidly with increasing temperature. This law is in agreement with the
experimental data.
The results of Planck's law:
 The denominator [exp(hc/λkT)] tends to infinity faster than the numerator (λ–5),
thus resolving the ultraviolet catastrophe and hence arriving at experimental
observation:
I (λ, T)  0 as λ  0.
hc
exp( hckT )  1   k T  I(  ,T )  2  c 4 k T
 For very large λ,
i.e. I (λ, T)  0 as λ  .
From a fit between Planck's law and experimental data, Planck’s constant was
derived to be h = 6.626 × 10–34 J-s.
37
3.2 PHOTOELECTRIC EFFECT
Ejection of electrons from the surface of certain metals when it is irradiated by an
electromagnetic radiation of suitable frequency is known as photoelectric effect.
A
E
V
C
Figure 3.5(a) Apparatus (b) circuit for studying Photoelectric Effect (T – Evacuated
glass/ quartz tube, E – Emitter Plate / Photosensitive material / Cathode, C –
Collector Plate / Anode, V – Voltmeter, A - Ammeter)
Experimental Observations:
Figure 3.6 Photoelectric current versus applied potential difference for two light intensities
1.
When plate E is illuminated by light of suitable frequency, electrons are emitted
from E and a current is detected in A (Figure 3.5).
38
2.
Photocurrent produced vs potential difference graph shows that kinetic energy
of the most energetic photoelectrons is,
Kmax = e Vs
3.
4.
5.
6.
(3.6)
where Vs is stopping potential
Kinetic energy of the most energetic photoelectrons is independent of light
intensity.
Electrons are emitted from the surface of the emitter almost instantaneously
No electrons are emitted if the incident light frequency falls below a cutoff
frequency.
Kinetic energy of the most energetic photoelectrons increases with increasing
light frequency.
Classical Predictions:
1. If light is really a wave, it was thought that if one shine of light of any fixed
wavelength, at sufficient intensity on the emitter surface, electrons should
absorb energy continuously from the em waves and electrons should be ejected.
2. As the intensity of light is increased (made it brighter and hence classically, a
more energetic wave), kinetic energy of the emitted electrons should increase.
3. Measurable / larger time interval between incidence of light and ejection of
photoelectrons.
4. Ejection of photoelectron should not depend on light frequency
5. In short experimental results contradict classical predictions.
6. Photoelectron kinetic energy should not depend upon the frequency of the
incident light.
Einstein’s Interpretation of electromagnetic radiation:
1. Electromagnetic waves carry discrete energy packets (light quanta called
photons now).
2. The energy E, per packet depends on frequency f: E = hf.
3. More intense light corresponds to more photons, not higher energy photons.
4. Each photon of energy E moves in vacuum at the speed of light: c = 3 x 108 m/s
and each photon carries a momentum, p = E/c.
Einstein’s theory of photoelectric effect:
A photon of the incident light gives all its energy hf to a single electron (absorption
of energy by the electrons is not a continuous process as envisioned in the wave
model) and the kinetic energy of the most energetic photoelectron
Kmax = hf −  (Einstein’s photoelectric equation)
(3.7)
39
 is called the work function of the metal. It is the minimum energy with which an
electron is bound in the metal.
All the observed features of photoelectric effect could be explained by Einstein’s
photoelectric equation:
1. Equation shows that Kmax depends only on frequency of the incident light.
2. Almost instantaneous emission of photoelectrons due to one -to –one
interaction between photons and electrons.
3. Ejection of electrons depends on light frequency since photons should have
energy greater than the work function  in order to eject an electron.
4. The cutoff frequency fc is related to  by fc =  /h. If the incident frequency
f is less than fc , there is no emission of photoelectrons.
The graph of kinetic energy of the most energetic photoelectron Kmax vs frequency
f is a straight line, according to Einstein’s equation.
Figure 3.7 A representative plot of Kmax versus frequency of incident light for three different metals
3.3 COMPTON EFFECT
When X-rays are scattered by free/nearly free electrons, they suffer a change in their
wavelength which depends on the scattering angle. This scattering phenomenon is
known as Compton Effect.
Classical Predictions: Oscillating electromagnetic waves (classically, X-rays are
em waves) incident on electrons should have two effects: i) oscillating
electromagnetic field causes oscillations in electrons. Each electron first absorbs
radiation as a moving particle and then re-radiates in all directions as a moving
40
particle and thereby exhibiting two Doppler shifts in the frequency of radiation. ii)
radiation pressure should cause the electrons to accelerate in the direction of
propagation of the waves. Because different electrons will move at different speeds
after the interaction, depending on the amount of energy absorbed from
electromagnetic waves, the scattered waves at a given angle will have all frequencies
(Doppler- shifted values).
Compton’s experiment and observation: Compton measured the intensity of
scattered X-rays from a solid target (graphite) as a function of wavelength for different
angles. The experimental setup is shown in Figure 3.8. Contrary to the classical
prediction, only one frequency for scattered radiation was seen at a given angle. This
is shown in the Figure 3.9.
The graphs for three nonzero angles show two peaks, one at o and the other at ’
>o . The shifted peak at ’ is caused by the scattering of X-rays from free electrons.
Shift in wavelength was predicted by Compton to depend on scattering angle as
λ′ − λ =
h
(1−cos θ)
mc
(3.8)
where m is the mass of the electron, c is velocity of light, h is Planck’s constant.
This is known as Compton shift equation, and the factor
Compton wavelength and 𝑚ℎ𝑐 = 2.43 pm.
ℎ
𝑚𝑐
is called the
Figure 3.8 Schematic diagram of Compton’s apparatus. The wavelength is measured with a rotating
crystal spectrometer for various scattering angles θ.
41
Figure 3.9 Scattered x-ray intensity versus wavelength for Compton scattering at  = 0°, 45°, 90°,
and 135° showing single frequency at a given angle
Derivation of the Compton shift equation:
Compton could explain the experimental result by treating the X-rays not as waves
but rather as point like particles (photons) having energy E = hfo = hc/o ,
momentum p = hf/c = h/ and zero rest energy. Photons collide elastically with free
electrons initially at rest and moving relativistically after collision.
Let o , po = h/o and Eo = hc/o be the wavelength, momentum and energy of
the incident photon respectively. ’, p’ = h/’ and E’ = hc/’ be the corresponding
quantities for the scattered photon.
We know that, for the electron, the total relativistic energy 𝐸 = √𝑝2 𝑐 2 + 𝑚2 𝑐 4
Kinetic energy K = E − m c2
1
And momentum p =  mv.
where  
2
1  vc 2
v and m are the speed and mass of the electron respectively.
Figure 3.10 Quantum model for X-ray scattering from an electron
In the scattering process, the total energy and total linear momentum of the
system must be conserved.
For conservation of energy we must have, Eo = E’ + K
42
Eo = E’ + (E − m c2)
ie,
Eo − E’ + m c2 = 𝐸 = √𝑝2 𝑐 2 + 𝑚2 𝑐 4
Or
Squaring both the sides, (𝐸𝑜 − 𝐸′)2 + 2(𝐸𝑜 − 𝐸′) 𝑚𝑐 2 + 𝑚2 𝑐 4 = 𝑝2 𝑐 2 + 𝑚2 𝑐 4
For conservation of momentum, x-component: 𝑝𝑜 = 𝑝′ 𝑐𝑜𝑠 𝜃 + 𝑝 𝑐𝑜𝑠 𝜙
y-component: 0 = 𝑝′ 𝑠𝑖𝑛 𝜃 − 𝑝 𝑠𝑖𝑛 𝜙
Rewriting these two equations
𝑝𝑜 − 𝑝′ 𝑐𝑜𝑠 𝜃 = 𝑝 𝑐𝑜𝑠 𝜙
𝑝′ 𝑠𝑖𝑛 𝜃 = 𝑝 𝑠𝑖𝑛 𝜙
Squaring both the sides and adding,
𝑝𝑜2 − 2𝑝𝑜 𝑝′ 𝑐𝑜𝑠 𝜃 + 𝑝′2 = 𝑝2
Substituting this 𝑝2 in the equation :
(𝐸𝑜 − 𝐸′)2 + 2(𝐸𝑜 − 𝐸′) 𝑚𝑐 2 = 𝑝2 𝑐 2 , one gets
(𝐸𝑜 − 𝐸′)2 + 2(𝐸𝑜 − 𝐸′) 𝑚𝑐 2 = (𝑝𝑜2 − 2𝑝𝑜 𝑝′ 𝑐𝑜𝑠 𝜃 + 𝑝′2 )𝑐 2
Substituting photon energies and photon momenta one gets
(
ℎ𝑐
𝜆𝑜
−
2
ℎ𝑐
𝜆
) + 2(
′
ℎ𝑐
𝜆𝑜
−
ℎ𝑐
ℎ𝑐
𝜆
𝜆𝑜
2
) − 2(
ℎ𝑐
) 𝑚𝑐 2 = (
′
ℎ𝑐
) 𝑚𝑐 2 = (
′
ℎ𝑐
𝜆𝑜
ℎ𝑐
2
) ( 𝜆′ ) 𝑐𝑜𝑠 𝜃 + ( ′ )
𝜆
Simplifying one gets
2
ℎ𝑐
ℎ𝑐
ℎ𝑐
ℎ𝑐
𝜆𝑜
𝜆
𝜆
2
( ) − 2 ( ) ( ′ ) + ( ′ ) + 2 ℎ𝑐 (
𝜆𝑜
i.e.,
−
ℎ𝑐
𝜆𝑜 𝜆′
OR,
+ ( 𝜆1 −
𝑜
′
1
𝜆′
) 𝑚𝑐 2 = −
(𝜆𝜆−𝜆𝜆𝑜′ ) 𝑚𝑐 2 =
𝑜
ℎ𝑐
𝜆𝑜 𝜆′
1
𝜆𝑜
−
ℎ𝑐
𝜆𝑜 𝜆′
1
𝜆
𝜆𝑜
2
) − 2(
ℎ𝑐
𝜆𝑜
ℎ𝑐
ℎ𝑐
2
) ( 𝜆′ ) 𝑐𝑜𝑠 𝜃 + ( ′ )
𝜆
𝑐𝑜𝑠 𝜃
(1 − 𝑐𝑜𝑠 𝜃)
Compton shift:
𝝀′ − 𝝀𝒐 =
𝒉
𝒎𝒄
(𝟏 − 𝒄𝒐𝒔 𝜽)
43
3.4 PHOTONS AND ELECTROMAGNETIC WAVES [DUAL
NATURE OF LIGHT]
•
•
•
Light exhibits diffraction and interference phenomena that are only explicable
in terms of wave properties.
Photoelectric effect and Compton Effect can only be explained taking light as
photons / particle.
This means true nature of light is not describable in terms of any single picture,
instead both wave and particle nature have to be considered. In short, the
particle model and the wave model of light complement each other.
3.5 de BROGLIE HYPOTHESIS - WAVE PROPERTIES OF
PARTICLES
We have seen that light comes in discrete units (photons) with particle properties
(energy E and momentum p) that are related to the wave-like properties of
frequency and wavelength. Louis de Broglie postulated that because photons have
both wave and particle characteristics, perhaps all forms of matter have wave-like
properties, with the wavelength λ related to momentum p in the same way as for
light.
de Broglie wavelength: 𝜆 =
ℎ
𝑝
=
ℎ
(3.9)
𝑚𝑣
where h is Planck’s constant and p is momentum of the quantum particle, m is mass
of the particle, and v is speed of the particle. The electron accelerated through a
potential difference of V, has a non-relativistic kinetic energy
1
𝑚 𝑣 2 = 𝑒 ∆𝑉
where e is electron charge.
2
Hence, the momentum (p) of an electron accelerated through a potential difference
of V is
𝑝 = 𝑚 𝑣 = √2 𝑚 𝑒 ∆𝑉
Frequency of the matter wave associated with the particle is
(3.10)
𝐸
ℎ
, where E is total
relativistic energy of the particle
Davisson-Germer experiment and G P Thomson’s electron diffraction experiment
confirmed de Broglie relationship p = h /. Subsequently it was found that atomic
beams, and beams of neutrons, also exhibit diffraction when reflected from regular
crystals. Thus de Broglie's formula seems to apply to any kind of matter. Now the
dual nature of matter and radiation is an accepted fact and it is stated in the
44
principle of complementarity. This states that wave and particle models of either
matter or radiation complement each other.
3.6 THE QUANTUM PARTICLE
Quantum particle is a model by which particles having dual nature are represented.
We must choose one appropriate behavior for the quantum particle (particle or
wave) in order to understand a particular behavior.
To represent a quantum wave, we have to combine the essential features of both an
ideal particle and an ideal wave. An essential feature of a particle is that it is localized
in space. But an ideal wave is infinitely long (non-localized) as shown in Figure 3.11.
Figure 3.11 Section of an ideal wave of single frequency
Now to build a localized entity from an infinitely long wave, waves of same
amplitude, but slightly different frequencies are superposed (Figure 3.12).
Figure 3.12 Superposition of two waves Wave 1 and Wave 2
If we add up large number of waves in a similar way, the small localized region of
space where constructive interference takes place is called a wave packet, which
represents a quantum particle (Figure 3.13).
45
Figure 3.13 Wave packet
Mathematical representation of a wave packet:
Superposition of two waves of equal amplitude, but with slightly different
frequencies, f1 and f2, traveling in the same direction are considered. The waves are
written as
𝑦1 = 𝐴 𝑐𝑜𝑠(𝑘1 𝑥 − 𝜔1 𝑡)
and 𝑦2 = 𝐴 𝑐𝑜𝑠(𝑘2 𝑥 − 𝜔2 𝑡)
where 𝑘 = 2𝜋/𝜆 ,
The resultant wave
𝜔 = 2𝜋𝑓
y = y 1 + y2
𝛥𝑘
𝑦 = 2𝐴 [𝑐𝑜𝑠 ( 2 𝑥 −
𝛥𝜔
𝑡)
2
𝑘1 +𝑘2
𝑥
2
𝑐𝑜𝑠 (
−
𝜔1 +𝜔2
𝑡)]
2
where k = k1 – k2 and  = 1 – 2.
Figure 3.14 Beat pattern due to superposition of wave trains y1 and y2
The resulting wave oscillates with the average frequency, and its amplitude envelope
(in square brackets, shown by the blue dotted curve in Figure 3.14) varies according
to the difference frequency. A realistic wave (one of finite extent in space) is
characterized by two different speeds. The phase speed, the speed with which wave
crest of individual wave moves, is given by
𝑣𝑝 = 𝑓 𝜆
or
𝑣𝑝 =
𝜔
𝑘
(3.11)
The envelope of group of waves can travel through space with a different speed than
the individual waves. This speed is called the group speed or the speed of the wave
packet which is given by
46
(𝛥𝜔
)
2
𝑣𝑔 =
(𝛥𝑘
)
2
𝛥𝜔
=
(3.12)
𝛥𝑘
For a superposition of large number of waves to form a wave packet, this ratio is
𝑣𝑔 =
𝑑𝜔
𝑑𝑘
In general these two speeds are not the same.
Relation between group speed (vg) and phase speed (vp):
𝜔
𝑣𝑃 =
But
𝑑𝜔
𝑣𝑔 =

= 𝑓𝜆
𝑘
𝑑𝑘
𝑑(𝑘𝑣𝑃 )
=
𝑑𝑘
= 𝑘
𝑑𝑣𝑃
𝑑𝑘
𝜔 = 𝑘 𝑣𝑃
+ 𝑣𝑃
Substituting for k in terms of λ, we get
𝑣𝑔 = 𝑣𝑃 − 𝜆 (
𝑑𝑣𝑃
𝑑𝜆
)
(3.13)
Relation between group speed (vg) and particle speed (u):
𝜔 = 2𝜋𝑓 = 2𝜋
𝑣𝑔 =
𝑑𝜔
𝑑𝑘
𝐸
2𝜋
ℎ
2𝜋
ℎ
=
𝑘 =
and
ℎ
𝑑𝐸
𝜆
=
2𝜋
ℎ⁄𝑝
2𝜋𝑝
=
ℎ
𝑑𝐸
=
𝑑𝑝
2𝜋
𝑑𝑝
For a classical particle moving with speed u, the kinetic energy E is given by
𝐸 =
1
2
𝑚 𝑢2 =
𝑣𝑔 =
𝑑𝜔
𝑑𝑘
𝑝2
2𝑚
=
𝑑𝐸 =
and
𝑑𝐸
𝑑𝑝
=
2 𝑝 𝑑𝑝
2𝑚
or
𝑑𝐸
𝑑𝑝
=
𝑢
𝑝
𝑚
= 𝑢
(3.14)
i.e., we should identify the group speed with the particle speed, speed with which
the energy moves. To represent a realistic wave packet, confined to a finite region
in space, we need the superposition of large number of harmonic waves with a range
of k-values.
3.7 DOUBLE–SLIT EXPERIMENT REVISITED
One way to confirm our ideas about the electron’s wave–particle duality is through
an experiment in which electrons are fired at a double slit. Consider a parallel beam
of mono-energetic electrons incident on a double slit as in Figure 3.15. Let’s assume
the slit widths are small compared with the electron wavelength so that diffraction
effects are negligible. An electron detector screen (acts like the “viewing screen” of
Young’s double-slit experiment) is positioned far from the slits at a distance much
greater than d, the separation distance of the slits. If the detector screen collects
47
electrons for a long enough time, we find a typical wave interference pattern for the
counts per minute, or probability of arrival of electrons. Such an interference pattern
would not be expected if the electrons behaved as classical particles, giving clear
evidence that electrons are interfering, a distinct wave-like behavior.
Figure 3.15 (a) Schematic of electron beam interference experiment, (b) Photograph of a double-slit
interference pattern produced by electrons
If we measure the angle θ at which the maximum intensity of the electrons arrives
at the detector screen, we find they are described by exactly the same equation as
that for light: 𝑑 𝑠𝑖𝑛 𝜃 = 𝑚 𝜆 , where m is the order number and λ is the electron
wavelength. Therefore, the dual nature of the electron is clearly shown in this
experiment: the electrons are detected as particles at a localized spot on the
detector screen at some instant of time, but the probability of arrival at the
spot is determined by finding the intensity of two interfering waves.
3.8 UNCERTAINTY PRINCIPLE
It is fundamentally impossible to make simultaneous measurements of a particle’s
position and momentum with infinite accuracy. This is known as Heisenberg
uncertainty principle. The uncertainties arise from the quantum structure of
matter.
For a particle represented by a single wavelength wave existing throughout space, 
is precisely known, and according to de Broglie hypothesis, its p is also known
accurately. But the position of the particle in this case becomes completely
uncertain.
This means  = 0, p =0; but x = 
In contrast, if a particle whose momentum is uncertain (combination of waves / a
range of wavelengths are taken to form a wave packet), so that x is small, but  is
large. If x is made zero,  and thereby p will become .
48
In short
( x ) ( px) ≥ h / 4
(3.15)
where x is uncertainty in the measurement of position x of the particle and px is
uncertainty in the measurement of momentum px of the particle.
One more relation expressing uncertainty principle is related to energy and time
which is given by
( E ) ( t ) ≥
h / 4
(3.16)
where E is uncertainty in the measurement of energy E of the system when the
measurement is done over the time interval t.
3.9 QUESTIONS
1. Explain (a) Stefan’s law (b) Wien’s displacement law (c) Rayleigh-Jeans law.
2. Sketch schematically the graph of wavelength vs intensity of radiation from a
blackbody.
3. Explain Planck’s radiation law.
4. Write the assumptions made in Planck’s hypothesis of blackbody radiation.
5. Explain photoelectric effect.
6. What are the observations in the experiment on photoelectric effect?
7. What are the classical predictions about the photoelectric effect?
8. Explain Einstein’s photoelectric equation.
9. Which are the features of photoelectric effect-experiment explained by
Einstein’s photoelectric equation?
10. Sketch schematically the following graphs with reference to the photoelectric
effect: (a) photoelectric current vs applied voltage (b) kinetic energy of mostenergetic electron vs frequency of incident light.
11. Explain Compton effect.
12. Explain the experiment on Compton effect.
13. Derive the Compton shift equation.
14. Explain the wave properties of the particles.
49
15. Explain a wave packet and represent it schematically.
16. Explain (a) group speed (b) phase speed, of a wave packet.
17. Show that the group speed of a wave packet is equal to the particle speed.
18. (a) Name any two phenomena which confirm the particle nature of light.
(b) Name any two phenomena which confirm the wave nature of light.
19. Explain Heisenberg uncertainty principle.
20. Write the equations for uncertainty in (a) position and momentum (b) energy
and time.
21. Mention two situations which can be well explained by the uncertainty
relation.
3.10 PROBLEMS
1 Find the peak wavelength of the blackbody radiation emitted by each of the
following.
A. The human body when the skin temperature is 35°C
B. The tungsten filament of a light bulb, which operates at 2000 K
C. The Sun, which has a surface temperature of about 5800 K.
Ans: 9.4 μm, 1.4 μm, 0.50 μm
2 A 2.0- kg block is attached to a spring that has a force constant of k = 25 N/m.
The spring is stretched 0.40 m from its equilibrium position and released.
A. Find the total energy of the system and the frequency of oscillation
according to classical calculations.
B. Assuming that the energy is quantized, find the quantum number n for the
system oscillating with this amplitude.
C. Suppose the oscillator makes a transition from the n = 5.4 x 1033 state to the
state corresponding to n = 5.4 x 1033 – 1. By how much does the energy of
the oscillator change in this one-quantum change.
Ans: 2.0 J, 0.56 Hz, 5.4 x 1033, 3.7 x 10–34 J
3 The human eye is most sensitive to 560 nm light. What is the temperature of a
black body that would radiate most intensely at this wavelength?
Ans: 5180 K
4 A blackbody at 7500 K consists of an opening of diameter 0.050 mm, looking into
an oven. Find the number of photons per second escaping the hole and having
wavelengths between 500 nm and 501 nm.
50
5
6
7
8
9
10
11
12
13
Ans: 1.30 x 1015/s
The radius of our Sun is 6.96 x 108 m, and its total power output is 3.77 x 1026 W.
(a) Assuming that the Sun’s surface emits as a black body, calculate its surface
temperature. (b) Using the result, find max for the Sun.
Ans: 5750 K, 504 nm
Calculate the energy in electron volts, of a photon whose frequency is (a) 620 THz,
(b) 3.10 GHz, (c) 46.0 MHz. (d) Determine the corresponding wavelengths for
these photons and state the classification of each on the electromagnetic
spectrum.
Ans: 2.57 eV, 1.28 x 10–5 eV, 1.91 x 10–7 eV, 484 nm, 9.68 cm,
6.52 m
An FM radio transmitter has a power output of 150 kW and operates at a
frequency of 99.7 MHz. How many photons per second does the transmitter emit?
Ans: 2.27 x 1030 photons/s
A sodium surface is illuminated with light having a wavelength of 300 nm. The
work function for sodium metal is 2.46 eV. Find
A. The maximum kinetic energy of the ejected photoelectrons and
B. The cutoff wavelength for sodium.
Ans: 1.67 eV, 504 nm
Molybdenum has a work function of 4.2eV. (a) Find the cut off wavelength and
cut off frequency for the photoelectric effect. (b) What is the stopping potential if
the incident light has wavelength of 180 nm?
Ans: 296 nm, 1.01 x 1015 Hz, 2.71 V
Electrons are ejected from a metallic surface with speeds up to 4.60 x 105 m/s
when light with a wavelength of 625 nm is used. (a) What is the work function of
the surface? (b) What is the cut-off frequency for this surface?
Ans: 1.38 eV, 3.34 x 1014 Hz
The stopping potential for photoelectrons released from a metal is 1.48 V larger
compared to that in another metal. If the threshold frequency for the first metal is
40.0 % smaller than for the second metal, determine the work function for each
metal.
Ans: 3.70 eV, 2.22 eV
Two light sources are used in a photoelectric experiment to determine the work
function for a metal surface. When green light from a mercury lamp ( = 546.1
nm) is used, a stopping potential of 0.376 V reduces the photocurrent to zero. (a)
Based on this what is the work function of this metal? (b) What stopping potential
would be observed when using the yellow light from a helium discharge tube ( =
587.5 nm)?
Ans: 1.90 eV, 0.215 V
X-rays of wavelength o = 0.20 nm are scattered from a block of material. The
scattered X-rays are observed at an angle of 45° to the incident beam. Calculate
their wavelength.
51
14
15
16
17
18
19
What if we move the detector so that scattered X-rays are detected at an angle
larger than 45°? Does the wavelength of the scattered X-rays increase or decrease
as the angle  increase?
Ans: 0.200710 nm, INCREASES
Calculate the energy and momentum of a photon of wavelength 700 nm.
Ans: 1.78 eV, 9.47 x 10–28kg.m/s
A 0. 00160 nm photon scatters from a free electron. For what photon scattering
angle does the recoiling electron have kinetic energy equal to the energy of the
scattered photon?
Ans: 70°
A 0.880 MeV photon is scattered by a free electron initially at rest such that the
scattering angle of the scattered electron is equal to that of the scattered photon
( = ).
(a) Determine the angles  & . (b) Determine the energy and
momentum of the scattered electron and photon.
Ans: 43°, 43°, 0.602 MeV, 3.21 x 10–22 kg.m/s, 0.278 MeV, 3.21 x 10–22 kg.m/s
Calculate the de- Broglie wavelength for an electron moving at 1.0 x 107 m/s.
Ans: 7.28 x 10–11 m
A rock of mass 50 g is thrown with a speed of 40 m/s. What is its de Broglie
wavelength?
Ans: 3.3 x 10–34 m
A particle of charge q and mass m has been accelerated from rest to a
nonrelativistic speed through a potential difference of V. Find an expression for
its de Broglie wavelength.
Ans: λ =
h
√2 m q Δv
20 (a) An electron has a kinetic energy of 3.0 eV. Find its wavelength. (b) Also find
the wavelength of a photon having the same energy.
Ans: 7.09 x 10–10 m, 4.14 x 10–7 m
21 In the Davisson-Germer experiment, 54.0 eV electrons were diffracted from a
nickel lattice. If the first maximum in the diffraction pattern was observed at =
50.0°, what was the lattice spacing a between the vertical rows of atoms in the
figure?
Ans: 2.18 x 10–10 m
22 Consider a freely moving quantum particle with mass m and speed u. Its energy is
E= K= mu2/2. Determine the phase speed of the quantum wave representing the
particle and show that it is different from the speed at which the particle
transports mass and energy.
Ans: vGROUP = u ≠ vPHASE
52
23 Electrons are incident on a pair of narrow slits 0.060 m apart. The ‘bright bands’
in the interference pattern are separated by 0.40 mm on a ‘screen’ 20.0 cm from
the slits. Determine the potential difference through which the electrons were
accelerated to give this pattern.
Ans: 105 V
24 The speed of an electron is measured to be 5.00 x 103 m/s to an accuracy of
0.0030%. Find the minimum uncertainty in determining the position of this
electron.
Ans: 0.383 mm
25 The lifetime of an excited atom is given as 1.0 x 10-8 s. Using the uncertainty
principle, compute the line width f produced by this finite lifetime?
Ans: 8.0 x 106 Hz
26 Use the uncertainty principle to show that if an electron were confined inside an
atomic nucleus of diameter 2 x 10–15 m, it would have to be moving relativistically,
while a proton confined to the same nucleus can be moving nonrelativistically.
Ans: vELECTRON  0.99996 c, vPROTON  1.8 x 107 m/s
27 Find the minimum kinetic energy of a proton confined within a nucleus having a
diameter of 1.0 x 10–15 m.
Ans: 5.2 MeV
53
4 QUANTUM MECHANICS
OBJECTIVES:
 To learn the application of Schrödinger equation to a bound particle and
to learn the quantized nature of the bound particle, its expectation values
and physical significance.
 To understand the tunneling behavior of a particle incident on a potential
barrier.
 To understand the behavior of quantum oscillator.
4.1 AN INTERPRETATION OF QUANTUM MECHANICS
Experimental evidences proved that both matter and electromagnetic radiation
exhibit wave and particle nature depending on the phenomenon being observed.
Making a conceptual connection between particles and waves, for an
electromagnetic radiation of amplitude E, the probability per unit volume of finding
a photon in a given region of space at an instant of time as
PROBABILITY
𝑉
∝ 𝐸2
Figure 4.1 Wave packet
Taking the analogy between electromagnetic radiation and matter-the probability
per unit volume of finding the particle is proportional to the square of the amplitude
of a wave representing the particle, even if the amplitude of the de Broglie wave
associated with a particle is generally not a measurable quantity. The amplitude of
the de Broglie wave associated with a particle is called probability amplitude, or the
wave function, and is denoted by .
In general, the complete wave function  for a system depends on the positions of
all the particles in the system and on time. This can be written as
(r1,r2,…rj,…,t) = (rj) e–it
54
where rj is the position vector of the jth particle in the system.
For any system in which the potential energy is time-independent and depends only
on the position of particles within the system, the important information about the
system is contained within the space part of the wave function. The wave function
 contains within it all the information that can be known about the particle. | |2
is always real and positive, and is proportional to the probability per unit volume, of
finding the particle at a given point at some instant. If  represents a single particle,
then ||2 - called the probability density - is the relative probability per unit
volume that the particle will be found at any given point in the volume.
One-dimensional wave functions and expectation values: Let  be the wave
function for a particle moving along the x axis. Then P(x) dx = ||2dx is the
probability to find the particle in the infinitesimal interval dx around the point x.
The probability of finding the particle in the arbitrary interval a ≤ x ≤ b is
𝑏
𝑃𝑎𝑏 = ∫𝑎 |𝜓|2 𝑑𝑥
(4.1)
The probability of a particle being in the interval a ≤ x ≤ b is the area under the
probability density curve from a to b. The total probability of finding the particle is
one. Forcing this condition on the wave function is called normalization.
+∞
∫−∞ |𝜓|2 𝑑𝑥 = 1
(4.2)
Figure 4.2 An arbitrary probability density curve for a particle
All the measurable quantities of a particle, such as its position, momentum and
energy can be derived from the knowledge of . e.g., the average position at which
one expects to find the particle after many measurements is called the expectation
value of x and is defined by the equation
+∞
⟨𝑥⟩ ≡ ∫−∞ 𝜓 ∗ 𝑥 𝜓 𝑑𝑥
(4.3)
55
The important mathematical features of a physically reasonable wave function
(x) for a system are
 (x) may be a complex function or a real function, depending on the
system.
 (x) must be finite, continuous and single valued everywhere.
 The space derivatives of, must be finite, continuous and single valued
everywhere.
  must be normalizable.
4.2 THE SCHRÖDINGER EQUATION
The appropriate wave equation for matter waves was developed by Schrödinger.
Schrödinger equation as it applies to a particle of mass m confined to move along x
axis and interacting with its environment through a potential energy function U(x)
is
−
ℏ2
𝑑2 𝜓
2 𝑚 𝑑𝑥 2
+𝑈𝜓 = 𝐸𝜓
(4.4)
where E is a constant equal to the total energy of the system (the particle and its
environment) and ħ = h/2.This equation is referred to as the one dimensional, timeindependent Schrödinger equation.
Application of Schrödinger equation:
1.
2.
3.
4.
Particle in an infinite potential well (particle in a box)
Particle in a finite potential well
Tunneling
Quantum oscillator
56
4.3 PARTICLE IN AN INFINITE POTENTIAL WELL
(PARTICLE IN A “BOX”)
Figure 4.3 (a) Particle in a potential well of infinite height, (b) Sketch of potential well
Consider a particle of mass m and velocity v, confined to bounce between two
impenetrable walls separated by a distance L as shown in Figure 4.3(a). Figure
4.3(b) shows the potential energy function for the system.
U(x) = 0,
for
0 <x<L,
U (x) =  ,
for x≤ 0, x≥L
Since U (x)=  , for x< 0, x>L , (x) = 0 in these regions. Also (0) =0 and (L) =0.
Only those wave functions that satisfy these boundary conditions are allowed. In
the region 0 <x<L, where U = 0, the Schrödinger equation takes the form
𝑑2 𝜓
2𝑚
+
𝐸 𝜓 = 0
𝑑𝑥 2
ℏ2
Or
𝑑2 𝜓
𝑑𝑥 2
= − 𝑘2 𝜓 ,
where 𝑘 2 =
2𝑚𝐸
ℏ2
or
𝑘 =
√2𝑚𝐸
ℏ
The most general form of the solution to the above equation is
(x) = Asin(kx) + B cos(kx)
where A and B are constants determined by the boundary and normalization
conditions.
Applying the first boundary condition,
i.e.,
at x = 0,  = 0
leads to
0 = A sin 0 + B cos 0
or B = 0 ,
And at x = L ,  = 0 ,
57
0 = A sin(kL) + B cos(kL) = A sin(kL) + 0 ,
Since A  0 ,
k L = n π ; ( n = 1, 2, 3, ……….. )
sin(kL) = 0 .
𝜓𝑛 (𝑥) = 𝐴 𝑠𝑖𝑛 (
Now the wave function reduces to
𝑛𝜋𝑥
𝐿
)
To find the constant A, apply normalization condition
+∞
∫−∞ |ψ|2 dx = 1
𝐿1
𝐴2 ∫0
2
[1 − 𝑐𝑜𝑠(
2𝑛𝜋𝑥
)] 𝑑𝑥
𝐿
2
We get,
√2𝑚𝐸
ℏ
∴
𝑛𝜋𝑥
𝐿
2
)] 𝑑𝑥 = 1 .
= 1
2
Thus 𝜓𝑛 (𝑥) = √𝐿 𝑠𝑖𝑛 (
𝑘 =
𝐿
∫0 𝐴2 [𝑠𝑖𝑛 (
𝐴 = √𝐿
Solving we get
Since
or
𝑛𝜋𝑥
𝐿
)
√2𝑚𝐸
and
ℏ
𝐿 =
is the wave function for particle in a box.
kL = nπ
𝑛𝜋.
ℎ2
𝐸𝑛 = ( 8 𝑚 𝐿2) 𝑛2 ,
n = 1, 2, 3,
. . . . .
(4.5)
Each value of the integer n corresponds to a quantized energy value, En .
The lowest allowed energy (n = 1),
𝐸1 =
ℎ2
8 𝑚 𝐿2
.
This is the ground state energy for the particle in a box. Excited states correspond
to n = 2, 3, 4,…which have energies given by 4E1 , 9E1 , 16E1…. respectively. Energy level
diagram, wave function and probability density sketches are shown in Figure 4.4
and 4.5 respectively. Since ground state energy E1 ≠0, the particle can never be at
rest.
Figure 4.4 Energy level diagram for a particle in potential well of infinite height
58
Figure 4.5 Sketch of (a) wave function, (b) Probability density for a particle in potential well of infinite
height
4.4 A PARTICLE IN A POTENTIAL WELL OF FINITE
HEIGHT
Figure 4.6 Potential well of finite height U and length L
Consider a particle with the total energy E, trapped in a finite potential well of height
U such that
U(x) = 0 ,
0 <x<L,
U(x) = U ,
x≤ 0, x≥L
Classically, for energy E<U, the particle is permanently bound in the potential well.
However, according to quantum mechanics, a finite probability exists that the
particle can be found outside the well even if E<U. That is, the wave function is
generally nonzero in the regions I and III. In region II, where U = 0, the allowed wave
functions are again sinusoidal. But the boundary conditions no longer require that
the wave function must be zero at the ends of the well.
59
Schrödinger equation outside the finite well in regions I & III
𝑑2 𝜓
𝑑𝑥 2
=
2𝑚
ℏ2
(𝑈 − 𝐸) 𝜓
𝑑2 𝜓
or
𝑑𝑥 2
= 𝐶 2 𝜓 where
𝐶2 =
2𝑚
ℏ2
(𝑈 − 𝐸)
General solution of the above equation is
(x) = AeCx + B e−Cx
where A and B are constants.
A must be zero in Region III and B must be zero in Region I, otherwise, the
probabilities would be infinite in those regions. For solution to be finite,
I = AeCx
for x≤ 0
III = Be-Cx
for x≥L
This shows that the wave function outside the potential well decay exponentially
with distance.
Schrodinger equation inside the square well potential in region II, where U = 0
𝑑2 𝜓𝐼𝐼
𝑑𝑥 2
+ (
2𝑚
ℏ2
2𝑚𝐸
𝐸) 𝜓𝐼𝐼 = 0 ,
ℏ2
= 𝑘2
General solution of the above equation
𝜓𝐼𝐼 = 𝐹 𝑠𝑖𝑛[𝑘𝑥] + 𝐺 𝑐𝑜𝑠[𝑘𝑥]
To determine the constants A, B, F, G and the allowed values of energy E, apply the
four boundary conditions and the normalization condition:
𝑑𝜓
At x = 0 , I(0) = II(0) and [ 𝑑𝑥𝐼 ]
𝑥=0
At x = L , II(L) = III(L)
=
[
[
and
𝑑𝜓𝐼𝐼
𝑑𝑥
𝑑𝜓𝐼𝐼
𝑑𝑥
]
]
𝑥=0
𝑥=𝐿
=
[
𝑑𝜓𝐼𝐼𝐼
𝑑𝑥
]
𝑥=𝐿
+∞
∫
|𝜓|2 𝑑𝑥 = 1
−∞
Figure 4.7 shows the plots of wave functions and their respective probability
densities.
60
Figure 4.7 Sketch of (a) wave function, (b) Probability density for a particle in potential well of finite
height
It is seen that wavelengths of the wave functions are longer than those of wave
functions of infinite potential well of same length and hence the quantized energies
of the particle in a finite well are lower than those for a particle in an infinite well.
4.5 TUNNELING THROUGH A POTENTIAL ENERGY
BARRIER
Consider a particle of energy E approaching a potential barrier of height U, (E<U).
Potential energy has a constant value of U in the region of width L and is zero in all
other regions. This is called a square barrier and U is called the barrier height. Since
E<U, classically the regions II and III shown in the figure are forbidden to the particle
incident from left. But according to quantum mechanics, all regions are accessible
to the particle, regardless of its energy.
Figure 4.8 Tunneling through a potential barrier of finite height
61
By applying the boundary conditions, i.e. and its first derivative must be
continuous at boundaries (at x = 0 and x = L), full solution to the Schrödinger
equation can be found which is shown in figure. The wave function is sinusoidal in
regions I and III but exponentially decaying in region II. The probability of locating
the particle beyond the barrier in region III is nonzero. The movement of the particle
to the far side of the barrier is called tunneling or barrier penetration. The
probability of tunneling can be described with a transmission coefficient T and a
reflection coefficient R.
The transmission coefficient represents the probability that the particle penetrates
to the other side of the barrier, and reflection coefficient is the probability that the
particle is reflected by the barrier. Because the particles must be either reflected or
transmitted we have, R + T = 1.
An approximate expression for the transmission coefficient, when T<< 1 is
T ≈ e−2CL , where 𝐶 =
√ 2 𝑚 (𝑈−𝐸)
ℏ
.
(4.6)
4.6 THE SIMPLE HARMONIC OSCILLATOR
Consider a particle that is subject to a linear restoring force 𝐹 = −𝑘𝑥, where k is a
constant and x is the position of the particle relative to equilibrium (at equilibrium
position x=0).
Classically, the potential energy of the system is,
𝑈=
1 2 1
𝑘𝑥 = 𝑚𝜔2 𝑥 2
2
2
where the angular frequency of vibration is 𝜔 = √𝑘/𝑚.
The total energy E of the system is,
1
1
𝐸 = 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 + 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 = 𝐾 + 𝑈 = 𝑘𝐴2 = 𝑚𝜔2 𝐴2
2
2
where A is the amplitude of motion. In the classical model, any value of E is allowed,
including E= 0, which is the total energy when the particle is at rest at x=0.
A quantum mechanical model for simple harmonic oscillator can be obtained by
1
substituting 𝑈 = 2 𝑚𝜔2 𝑥 2 in Schrödinger equation:
ℏ2 𝑑 2  1
−
+ 𝑚𝜔2 𝑥 2  = 𝐸
2𝑚 𝑑𝑥 2 2
The solution for the above equation is
 = 𝐵𝑒 −𝐶𝑥
2
62
1
where 𝐶 = 𝑚𝜔/2ℏ and 𝐸 = 2 ℏ𝜔. The constant B can be determined from
normalization condition.
In quantum model, the energy levels of a harmonic oscillator are quantized. The
energy of a state having an arbitrary quantum number n is given by
1
𝐸𝑛 = (𝑛 + 2) ℏ𝜔;
𝑛 = 0, 1, 2 ….
(4.7)
1
The state n = 0 corresponds to the ground state, whose energy is 𝐸0 = 2 ℏ𝜔 the state
3
n = 1 corresponds to the first excited state, whose energy is 𝐸1 = 2 ℏ𝜔 and so on. The
energy-level diagram for this system is shown in Figure 4.9. The separations between
adjacent levels are equal and given by ∆𝐸 = ℏ𝜔.
Figure 4.9 Energy-level diagram for a simple harmonic oscillator, superimposed on the potential
energy function.
4.7 QUESTIONS
1 What is a wave function ? What is its physical interpretation ?
2 What are the mathematical features of a wave function?
3 By solving the Schrödinger equation, obtain the wave-functions
for a particle of mass m in a one-dimensional “box” of length
L.
4 Apply the Schrödinger equation to a particle in a onedimensional “box” of length L and obtain the energy values of
the particle.
5 Sketch the lowest three energy states, wave-functions, probability
densities for the particle in a one-dimensional “box”.
6 The wave-function for a particle confined to moving in a onedimensional box is
63
ψ(x) = A sin(nπx
) . Use the normalization condition on  to
L
show that 𝐴 = √2𝐿 .
7 The wave-function of an electron is ψ(x) = A sin(nπx
) . Obtain
L
an expression for the probability of finding the electron between
x = a and x = b.
8 Sketch the potential-well diagram of finite height U and length
L, obtain the general solution of the Schrödinger equation for
a particle of mass m in it.
9 Sketch the wave-functions and the probability densities for the
lowest three energy states of a particle in a potential well of
finite height.
10 Give a brief account of tunneling of a particle through a
potential energy barrier.
11 Give a brief account of the quantum mechanical treatment of a
simple harmonic oscillator.
4.8 PROBLEMS
1 A particle wave function is given by the equation  (x) = A 𝑒 −𝑎𝑥 2
(A) What is the value of A if this wave function is normalized?
(B) What is the expectation value of x for this particle?
Ans: A = (2a/π)¼ ,
x = 0
2 A free electron has a wave function ψ(x) = 𝐴 exp[𝑖(5.0 × 1010 )𝑥]
where x is in meters. Find (a) its de Broglie wavelength, (b) its momentum, and (c)
its kinetic energy in electron volts.
Ans: 1.26 x 10–10m, 5.27 x 10–24kg.m/s, 95.5 eV
3 An electron is confined between two impenetrable walls 0.20 nm apart. Determine
the energy levels for the states n =1 ,2 , and 3.
Ans: 9.2 eV, 37.7 eV, 84.8 eV
4 A 0.50 kg baseball is confined between two rigid walls of a stadium that can be
modeled as a “box” of length 100 m. Calculate the minimum speed of the baseball.
If the baseball is moving with a speed of 150 m/s, what is the quantum number of
the state in which the baseball will be?
Ans: 6.63 x 10–36 m/s, 2.26 x 1037
5 A proton is confined to move in a one-dimensional “box” of length 0.20 nm. (a)
Find the lowest possible energy of the proton. (b) What is the lowest possible
64
6
7
8
9
10
energy for an electron confined to the same box? (c) Account for the great
difference in results for (a) and (b).
Ans: 5.13 x 10–3 eV, 9.41 eV
(A) Using the simple model of a particle in a box to represent an atom, estimate
the energy (in eV) required to raise an atom from the state n =1 to the state n
=2. Assume the atom has a radius of 0.10 nm and that the moving electron
carries the energy that has been added to the atom.
(B) Atoms may be excited to higher energy states by absorbing photon energy.
Calculate the wavelength of the photon that would cause the transition from
the state n =1 to the state n =2.
Ans: 28.3 eV, 43.8 nm
A 30-eV electron is incident on a square barrier of height 40 eV. What is the
probability that the electron will tunnel through the barrier if its width is (A) 1.0
nm? (B) 0.10 nm?
Ans: 8.5 x 10–15, 0.039
An electron with kinetic energy E = 5.0 eV is incident
on a barrier with thickness L = 0.20 nm and height U =
10.0 eV as shown in the figure. What is the probability
that the electron (a) will tunnel through the barrier?
(b) will be reflected?
Ans: 0.0103, 0.990
A quantum simple harmonic oscillator consists of an electron bound by a restoring
force proportional to its position relative to a certain equilibrium point. The
proportionality constant is 8.99 N/m. What is the longest wavelength of light that
can excite the oscillator?
Ans: 600nm
A quantum simple harmonic oscillator consists of a particle of mass m bound by a
restoring force proportional to its position relative to a certain equilibrium point.
The proportionality constant is k. What is the longest wavelength of light that can
𝑚
excite the oscillator? Ans: 2𝜋𝑐√ 𝑘
65
5 ATOMIC PHYSICS
OBJECTIVES:






To know about the quantum model of H-atom and its wave functions.
To understand more about Visible and X ray spectra
To explain basic interactions of radiation with matter.
To understand the basic principles and requirements for working of laser.
To recognize the various applications of laser.
To apply and evaluate the above concepts by solving numerical problems
5.1 THE QUANTUM MODEL OF THE HYDROGEN ATOM
The formal procedure for solving the problem of the hydrogen atom is to substitute
the appropriate potential energy function into the Schrödinger equation, find
solutions to the equation, and apply boundary conditions as we did for the particle
in a box.
The potential energy function for the H-atom is
𝑈(𝑟) = −
𝑘𝑒 𝑒 2
(5.1)
𝑟
where ke = 1/40= 8.99 x 109 N.m2/C2 Coulomb constant and r is radial distance
of electron from H-nucleus. The mathematics for the hydrogen atom is more
complicated than that for the particle in a box because the atom is threedimensional, and U depends on the radial coordinate r.
The time-independent Schrödinger equation in 3-dimensional space is
−
ℏ2
2𝑚
𝜕2 𝜓
( 𝜕𝑥 2 +
𝜕2 𝜓
𝜕𝑦 2
+
𝜕2 𝜓
𝜕𝑧 2
) +𝑈𝜓 = 𝐸𝜓
(5.2)
Since U has spherical symmetry, it is easier to solve the Schrödinger equation
in spherical polar coordinates (r, , ) where 𝑟 = √𝑥 2 + 𝑦 2 + 𝑧 2 ,
⃗ .  is the angle between the x-axis and
 is the angle between z-axis and 𝒓
⃗⃗ onto the xy-plane.
the projection of 𝒓
66
Figure 5.1 Spherical polar coordinate system
It is possible to separate the variables r, θ,  as follows:
(r, , ) = R(r) f() g()
By solving the three separate ordinary differential equations for R(r), f(), g(),
with conditions that the normalized  and its first derivative are continuous
and finite everywhere, one gets three different quantum numbers for each
allowed state of the H-atom. The quantum numbers are integers and
correspond to the three independent degrees of freedom.
The radial function R(r) of  is associated with the principal quantum number
n. Solving R(r), we get an expression for energy as,
𝑘 𝑒2
𝐸𝑛 = − ( 2𝑒𝑎 )
𝑜
1
𝑛2
= −
13.606 𝑒𝑉
𝑛2
,
n = 1, 2, 3,
. . .
(5.3)
which is in agreement with Bohr theory.
The polar function f() is associated with the orbital quantum number . The
azimuthal function g() is associated with the orbital magnetic quantum
number m .
The application of boundary conditions on the three parts of 
leads to important relationships among the three quantum numbers:
n can range from 1 to ,
 can range from 0 to n–1 ; [n allowed values].
m can range from –to + ; [(2+1) allowed values].
All states having the same principal quantum number are said to form a shell.
All states having the same values of n and  are said to form a subshell:
n=1
 K shell
=0  s
subshell
67
n = 2  L shell
=1  p
subshell
n = 3  M shell
=2  d
subshell
n = 4  N shell
=3  f
subshell
n = 5 
O shell
=4  g
subshell
n = 6  P shell
=5  h
subshell
..
..
..
..
..
..
..
..
5.2 WAVE FUNCTIONS FOR HYDROGEN
The potential energy for H-atom depends only on the radial distance r between
nucleus and electron. Therefore some of the allowed states for the H-atom can
be represented by wave functions that depend only on r (spherically symmetric
function). The simplest wave function for H-atom is the 1s-state (ground
state) wave function (n = 1,  = 0):
𝜓1𝑠 (𝑟) =
1
√𝜋 𝑎𝑜3
𝑒𝑥𝑝 (−𝑎𝑟𝑜 ) where ao is Bohr radius (0.0529 nm).
(5.4)
|1s|2 is the probability density for H-atom in 1s-state:
|𝜓1𝑠 |2 =
1
𝜋 𝑎𝑜3
𝑒𝑥𝑝 (− 2𝑎𝑜𝑟 )
(5.5)
The radial probability density P(r) is the probability per unit radial length
of finding the electron in a spherical shell of radius r and thickness dr.
P(r)dr is the probability of finding the electron in this shell.
P(r) dr = ||2 dv = ||2 4r2 dr

P(r) = 4r2 ||2
Figure 5.2 A spherical shell of radius r and thickness dr has a volume equal to 4 r2dr
68
Radial probability density for H-atom in its ground state:
𝑃1𝑠 = (
4 𝑟2
𝑎𝑜3
) 𝑒𝑥𝑝 (− 2𝑎𝑜𝑟 )
Figure 5.3 (a) The probability of finding the electron as a function of distance from the nucleus for the
hydrogen atom in the 1s (ground) state. (b) The cross section in the xy plane of the spherical
electronic charge distribution for the hydrogen atom in its 1s state
The next simplest wave function for the H-atom is the 2s-state wave function
(n = 2,  = 0):
𝜓2𝑆 (𝑟) =
1
√32𝜋𝑎𝑜3
(2 − 𝑎𝑟𝑜 ) 𝑒𝑥𝑝 (− 𝑎𝑟𝑜 )
(5.6)
2s is spherically symmetric (depends only on r). Energy corresponding to n = 2
(first excited state) is E2= E1/4 = –3.401 eV.
Figure 5.4 Plot of radial probability density versus r/a0 (normalized radius) for 1s and 2s states of
hydrogen atom
69
5.3 MORE ON ATOMIC SPECTRA: VISIBLE AND X-RAY
These spectral lines have their origin in transitions between quantized atomic states.
A modified energy-level diagram for hydrogen is shown in the Figure 5.5.
Figure 5.5 Some allowed electronic transitions for hydrogen, represented by the colored lines
In this diagram, the allowed values of , for each shell are separated horizontally.
Figure shows only those states up to  = 2, the shells from n = 4 , upward would
have more sets of states to the right, which are not shown. Transitions for which 
does not change are very unlikely to occur and are called forbidden transitions.(Such
transitions actually can occur, but their probability is very low relative to the
probability of “allowed” transitions.) The various diagonal lines represent allowed
transitions between stationary states. Whenever an atom makes a transition from a
higher energy state to a lower one, a photon of light is emitted.
The frequency of this photon is f = E/h, where E is the energy difference between
the two states and h is Planck’s constant. The selection rules for the allowed
transitions are
  1 and m  0,  1
The allowed energies for one-electron atoms and ions, such as hydrogen (H) and
helium ion (He+), are
70

ke e 2  Z 2 
13.6 eV Z 2


En  

2a0  n 2 
n2
(5.7)
This equation was developed from the Bohr theory, but it serves as a good first
approximation in quantum theory as well. For multi-electron atoms, the positive
nuclear charge Ze is largely shielded by the negative charge of the inner-shell
electrons. Therefore, the outer electrons interact with a net charge that is smaller
than the nuclear charge.
Hence, we can write
En  
13.6 eV  Z eff2
n2
(5.8)
where Zeff depends on n and 
5.4 X-RAY SPECTRA
X-rays are emitted when high-energy electrons or any other charged particles
bombard a metal target. The x-ray spectrum typically consists of a broad continuous
band containing a series of sharp lines as shown in Figure 5.6. So, the x-ray
spectrum has two parts: continuous spectrum and characteristic spectrum.
Figure 5.6 The x-ray spectrum of a metal target. The data shown were obtained when 37-keV electrons
bombarded a molybdenum target.
An accelerated electric charge emits electromagnetic radiation. The x-rays in figure
are the result of the slowing down of high-energy electrons as they strike the target.
It may take several interactions with the atoms of the target before the electron loses
all its kinetic energy. The amount of kinetic energy lost in any given interaction can
vary from zero up to the entire kinetic energy of the electron. Therefore, the
wavelength of radiation from these interactions lies in a continuous range from
some minimum value up to infinity. It is this general slowing down of the electrons
71
that provides the continuous curve, which shows the cutoff of x-rays below a
minimum wavelength value that depends on the kinetic energy of the incoming
electrons. X-ray radiation with its origin in the slowing down of electrons is called
bremsstrahlung, the German word for “braking radiation”.
Thus the emitted x-rays can have any value for the wavelength above λMIN in
the continuous x-ray spectrum. Thus
e V  hf MAX 
MIN 
hc
MIN
hc
e V
(5.9)
λMIN depends only on ∆V
The peaks in the x-ray spectrum is the characteristic of the target element
in the x-ray tube and hence they form the characteristic x-ray spectrum. When
a high energy (K = e ∆V, ∆V = x-ray tube voltage) electron strikes a target atom
and knocks out one of its electrons from the inner shells with energy
Enf (|Enf | ≤ K, nf = integer), the vacancy in the inner shell is filled up by an
electron from the outer shell (energy = Eni, ni = integer). The characteristic xray photon emitted has the energy:
hf 
hc

 Eni  Enf
Figure 5.7 Transitions between higher and lower atomic energy levels that give rise to x-ray photons
from heavy atoms when they are bombarded with high-energy electrons.
72
A K x-ray results due to the transition of the electron from L-shell to Kshell. A K x-ray results due to the transition of the electron from M-shell to
K-shell. When the vacancy arises in the L-shell, an L-series (L, L, L) of xrays results. Similarly, the origin of M-series of x-rays can be explained.
Moseley’s observation on the characteristic K x-rays shows a relation between
the frequency (f) of the K x-rays and the atomic number (Z) of the target
element in the x-ray tube:
f  C Z  1
(5.10)
where C is a constant.
Note: Based on this observation, the elements are arranged according to their
atomic numbers in the periodic table
Figure 5.8 Moseley plot
5.5 SPONTANEOUS AND STIMULATED TRANSITIONS
There are three possible processes that involve interaction between matter and
radiation.
Stimulated Absorption: Absorption of a photon of frequency f takes place
when the energy difference E2 – E1 of the allowed energy states of the
atomic/molecular system equals the energy hf of the photon. Then the photon
disappears and the atomic system moves to upper energy state E2.
73
Figure 5.9 Stimulated absorption of a photon
Spontaneous Emission: The average life time of the atomic system in the
excited state is of the order of 10–8 s. After the life time of the atomic system
in the excited state, it comes back to the state of lower energy on its own
accord by emitting a photon of energy hf = E2– E1 .
This is the case with ordinary light sources. The radiations are emitted in different
directions in random manner. Such type of emission of radiation is called
spontaneous emission and the emitted light is not coherent.
Figure 5.10 Spontaneous Emission of a photon
Stimulated Emission: When a photon (called stimulating photon) of suitable
frequency interacts with an excited atomic system, the latter comes down to
ground state by emitting a photon of same energy. Such an emission of radiation
is called stimulated emission. In stimulated emission, both the stimulating photon
and the emitted photon (due to stimulation) are of same frequency, same phase,
same state of polarization and in the same direction. In other words, these two
photons are coherent.
74
Figure 5.11 Stimulated Emission
All the three processes are taking place simultaneously to varying degrees, in the
matter when it is irradiated by radiation of suitable frequency.
Population inversion: From Boltzmann statistics, the ratio of population of
atoms in two energy states E1 and E2 at equilibrium temperature T is,
 E  E1 
nE 2 

 exp  2
nE 1 
k T 

(5.11)
where k is Boltzmann constant, n(E1) is the number density of atoms with energy
E1 , n(E2) is the number density of atoms with energy E2 . Under normal condition,
where populations are determined only by the action of thermal agitation,
population of the atoms in upper energy state is less than that in lower energy
state (i.e. n(E2)<n(E1), Figure 5.12a).
Figure 5.12 (a) Normal thermal equilibrium distribution of atomic systems (b) An inverted population,
obtained using special techniques
We have described how an incident photon can cause atomic energy transitions
either upward (stimulated absorption) or downward (stimulated emission). The two
processes are equally probable. When light is incident on a collection of atoms, a
net absorption of energy usually occurs because when the system is in thermal
equilibrium, many more atoms are in the ground state than in excited states. If the
situation can be inverted so that more atoms are in an excited state than in the
75
ground state, however, a net emission of photons can result. Such a condition is
called population inversion.
5.6 LASER (LIGHT AMPLIFICATION BY STIMULATED
EMISSION OF RADIATION)
Laser light is highly monochromatic, intense, coherent, directional and can be
sharply focused. Each of these characteristics that are not normally found in
ordinary light makes laser a unique and the most powerful tool. Population
inversion is, in fact, the fundamental principle involved in the operation of a laser.
The full name indicates one of the requirements for laser light: to achieve laser
action, the process of stimulated emission must occur.
For the stimulated emission rate to exceed the absorption rate it is necessary
to have higher population of upper energy state than that of lower energy
state. This condition is called population inversion [n(E2)>n(E1)]. This is a nonequilibrium condition and is facilitated by the presence of energy states called
‘metastable states’ where the average life time of the atom is 10-3 s which is much
longer than that of the ordinary excited state ( 10-8s).
Suppose an atom is in the excited state E2 as in the below figure and a photon with
energy hf = E2 - E1 is incident on it. The incoming photon can stimulate the excited
atom to return to the ground state and thereby emit a second photon having the
same energy hf and traveling in the same direction. The incident photon is not
absorbed, so after the stimulated emission, there are two identical photons: the
incident photon and the emitted photon. The emitted photon is in phase with the
incident photon. These photons can stimulate other atoms to emit photons in a
chain of similar processes. The many photons produced in this fashion are the
source of the intense, coherent light in a laser.
For the stimulated emission to result in laser light, there must be a buildup of
photons in the system. The following three conditions must be satisfied to achieve
this buildup:
• The system must be in a state of population inversion: there must be more atoms
in an excited state than in the ground state. That must be true because the number
of photons emitted must be greater than the number absorbed.
• The excited state of the system must be a metastable state, meaning that its
lifetime must be long compared with the usually short lifetimes of excited states,
which are typically 10-8 s. In this case, the population inversion can be established
and stimulated emission is likely to occur before spontaneous emission.
• The emitted photons must be confined in the system long enough to enable
them to stimulate further emission from other excited atoms. That is achieved by
using reflecting mirrors at the ends of the system. One end is made totally reflecting,
76
and the other is partially reflecting. A fraction of the light intensity passes through
the partially reflecting end, forming the beam of laser light.
Figure 5.13 Schematic diagram of a laser design.
Lasing medium (active medium), resonant cavity and pumping system are the
essential parts of any lasing system. Lasing medium has atomic systems (active
centers), with special energy levels which are suitable for laser action. This
medium may be a gas, or a liquid, or a crystal or a semiconductor. The
atomic systems in this may have energy levels including a ground state (E1),
an excited state (E3) and a metastable state (E2). The resonant cavity is a pair
of parallel mirrors to reflect the radiation back into the lasing medium.
Pumping is a process of exciting more number of atoms in the ground state
to higher energy states, which is required for attaining the population
inversion.
Figure 5.14 Energy-level diagram for a neon atom in a helium–neon laser.
77
In He-Ne laser, the mixture of helium and neon is confined to a glass tube that is
sealed at the ends by mirrors. A voltage applied across the tube causes electrons to
sweep through the tube, colliding with the atoms of the gases and raising them into
excited states. Neon atoms are excited to state E3* through this process (the asterisk
indicates a metastable state) and also as a result of collisions with excited helium
atoms. Stimulated emission occurs, causing neon atoms to make transitions to state
E2. Neighboring excited atoms are also stimulated. The result is the production of
coherent light at a wavelength of 632.8 nm.
5.7 APPLICATIONS OF LASER
Laser is used in various scientific, engineering and medical applications. It is used
in investigating the basic laws of interaction of atoms and molecules with
electromagnetic wave of high intensity. Laser is widely used in engineering
applications like optical communication, micro-welding and sealing etc. In medical
field, laser is used in bloodless and painless surgery especially in treating the retinal
detachment. Also used as a tool in treating dental decay, tooth extraction, cosmetic
surgery.
5.8 QUESTIONS
1 Give a brief account of quantum model of H-atom.
Explain the origin of (i) orbital quantum number (ii) magnetic orbital
quantum number and write the relation between them
2 The wave function for H-atom in ground state is ψ1S (r) =
1
√πa3o
exp (− aro) . Obtain an expression for the radial probability density
of H-atom in ground state.
Sketch schematically the plot of this vs. radial distance.
3 The wave
function for
H-atom
in 2s
state is ψ2S (r) =
1
√32πa3o
(2 − 𝑎𝑟𝑜 ) exp (− aro) .
Write
the
expression
for
the
radial
probability density of H-atom in 2s state. Sketch schematically the plot
of this vs. radial distance.
4 Sketch schematically the plot of the radial probability density vs. radial
distance for H-atom in 1s-state and 2s-state.
5 Explain the continuous x-ray spectrum with a schematic plot of the
spectrum.
6 Explain the origin of characteristic x-ray spectrum with a sketch of xray energy level diagram.
78
7 Write Moseley’s relation for the frequency of characteristic x-rays.
sketch schematically the Moseley’s plot of characteristic x-rays.
8 Explain three types of transitions between two energy levels, when radiation
interacts with matter
9 Explain the characteristics of a laser beam
10 Explain metastable state
11 What is population inversion? explain
12 Describe the principle of a laser using necessary schematic design and
energy level diagram
13 Mention any four applications of laser.
14 Describe the three important conditions need to be satisfied to achieve
laser action
5.9 PROBLEMS
1 For a H-atom, determine the number of allowed states corresponding
to the principal quantum number n = 2, and calculate the energies of
these states.
Ans: 4 states (one 2s-state + three 2p-states), –3.401 eV
2 A general expression for the energy levels of one-electron atoms and
ions is 𝐸𝑛 = −
𝜇 𝑘𝑒2 𝑞12 𝑞22
2 ℏ2 𝑛 2
,
where ke is the Coulomb constant, q1 and q2
are the charges of the electron and the nucleus, and μ is the reduced
m m
mass, given by μ = m 1+m2 . The wavelength for n = 3 to n = 2 transition
1
2
of the hydrogen atom is 656.3 nm (visible red light). What are the
wavelengths for this same transition (a) positronium, which consists of
an electron and a positron, and (b) singly ionized helium ?
Ans: 1310 nm, 164 nm
3 Calculate the most probable value of r (= distance from nucleus) for an
electron in the ground state of the H-atom. Also calculate the average
value r for the electron in the ground state.
Ans: ao , 3 ao/2
4 Calculate the probability that the electron in the ground state of Hatom will be found outside the Bohr radius.
Ans: 0.677
5 For a spherically symmetric state of a H-atom the Schrodinger equation
in spherical coordinates is−
ℏ2
2𝑚
𝜕2 𝜓
( 𝜕𝑟 2 +
2 𝜕𝜓
𝑟
) −
𝜕𝑟
𝑘𝑒 𝑒 2
𝑟
𝜓 = 𝐸 𝜓 . Show
79
that the 1s wave function for an electron in H-atom
1
√𝜋𝑎𝑜3
𝜓1𝑆 (𝑟) =
𝑒𝑥𝑝 (− 𝑎𝑟𝑜) satisfies the Schrodinger equation.
6 The ground-state wave function for the electron in a hydrogen atom is
1
𝑟
𝜓1𝑆 (𝑟) =
𝑒𝑥𝑝 (− 𝑎 )
𝑜
√𝜋𝑎𝑜3
where r is the radial coordinate of the electron and a0 is the Bohr radius.
(a) Show that the wave function as given is normalized. (b) Find the
probability of locating the electron between r1 = a0/2 and r2 = 3a0/2.
7 What minimum accelerating voltage would be required to produce an x-ray
with a wavelength of 70.0 pm? Ans: 17.7 kV
8 A tungsten target is struck by electrons that have been accelerated from rest
through a 40.0-keV potential difference. Find the shortest wavelength of the
radiation emitted. Ans: 0.031 nm
9 A bismuth target is struck by electrons, and x-rays are emitted. Estimate (a)
the M- to L-shell transitional energy for bismuth and (b) the wavelength of
the x-ray emitted when an electron falls from the M shell to the L shell.
Ans: (a) 14 keV (b) 0.885 Å
10 The 3p level of sodium has an energy of -3.0 eV, and the 3d level has an energy
of -1.5 eV. (a) Determine Zeff for each of these states. (b) Explain the
difference. Ans: (a) 1.4 and 1.0
11 The K series of the discrete x-ray spectrum of tungsten contains wavelengths
of 0.0185 nm, 0.020 9 nm, and 0.021 5 nm. The K-shell ionization energy is
69.5 keV.
(a) Determine the ionization energies of the L, M, and N shells.
(b) Draw a diagram of the transitions.
Ans: (a) L shell = 11.8 keV ; M shell = 10.2 keV ; N shell = 2.47 keV
80
12 When an electron drops from the M shell (n = 3) to a vacancy in the K shell
(n = 1), the measured wavelength of the emitted x-ray is found to be 0.101 nm.
Identify the element. Ans: Gallium (Z=31)
13 A ruby laser delivers a 10.0-ns pulse of 1.00-MW average power. If the
photons have a wavelength of 694.3 nm, how many are contained in the
pulse? Ans: 3.49x1016 photons
14 A pulsed laser emits light of wavelength . For a pulse of duration t having
energy TER, find (a) the physical length of the pulse as it travels through space
and (b) the number of photons in it. (c) The beam has a circular cross section
having diameter d. Find the number of photons per unit volume. (a) cΔt (b)
𝜆𝑇𝐸𝑅
ℎ𝑐
4𝜆𝑇
(c) 𝑛 = 𝜋ℎ𝑐 2 𝑑𝐸𝑅
2 Δ𝑡
81
6 MOLECULES AND SOLIDS
OBJECTIVES:





To understand the bonding mechanism, energy states and spectra of
molecules
To understand the cohesion of solid metals using bonding in solids
To comprehend the electrical properties of metals, semiconductors and
insulators
To understand the effect of doping on electrical properties of
semiconductors
To understand superconductivity and its engineering applications
6.1 MOLECULAR BONDS
The bonding mechanisms in a molecule are fundamentally due to electric forces
between atoms (or ions). The forces between atoms in the system of a molecule are
related to a potential energy function. A stable molecule is expected at a
configuration for which the potential energy function for the molecule has its
minimum value. A potential energy function that can be used to model a molecule
should account for two known features of molecular bonding:
1. The force between atoms is repulsive at very small separation distances. This
repulsion is partly electrostatic in origin and partly the result of the exclusion
principle.
2. At relatively at larger separations, the force between atoms is attractive.
Considering these two features, the potential energy for a system of two atoms can
be represented by an expression of the form (Lennard–Jones potential)
U r   
A B

rn rm
(6.1)
where r is the internuclear separation distance between the two atoms and n and m
are small integers. The parameter A is associated with the attractive force and B with
the repulsive force. Potential energy versus internuclear separation distance for a
two-atom system is graphed in Figure 6.1.
82
Figure 6.1 Total potential energy as a function of internuclear separation distance for a system of two
atoms.
Ionic Bonding: When two atoms combine in such a way that one or more outer
electrons are transferred from one atom to the other, the bond formed is called an
ionic bond. Ionic bonds are fundamentally caused by the Coulomb attraction
between oppositely charged ions.
Figure 6.2 Total energy versus internuclear separation distance for Na + and Cl- ions.
A familiar example of an ionically bonded solid is sodium chloride, NaCl, which is
common table salt. Sodium, which has the electronic configuration 1s22s22p63s1, is
ionized relatively easily, giving up its 3s electron to form a Na+ ion. The energy
required to ionize the atom to form Na+ is 5.1 eV. Chlorine, which has the electronic
configuration 1s22s22p5 is one electron short of the filled-shell structure of argon. The
amount of energy released when an electro joins Cl atom to make the Cl ̶ ion, called
the electron affinity of the atom, is 3.6 eV. Therefore, the energy required to form
Na+ and Cl ̶ from isolated atoms is 5.1 - 3.6 = 1.5 eV. The total energy of the NaCl
molecule versus internuclear separation distance is graphed in Figure 6.2. At very
large separation distances, the energy of the system of ions is 1.5 eV as calculated
above. The total energy has a minimum value of - 4.2 eV at the equilibrium
83
separation distance, which is approximately 0.24 nm. Hence, the energy required to
break the Na+ ̶ Cl ̶ bond and form neutral sodium and chlorine atoms, called the
dissociation energy, is 4.2 eV. The energy of the molecule is lower than that of the
system of two neutral atoms. Consequently, it is energetically favorable for the
molecule to form.
Covalent Bonding: A covalent bond between two atoms is one in which electrons
supplied by either one or both atoms are shared by the two atoms. Many diatomic
molecules such as H2, F2, and CO—owe their stability to covalent bonds. The bond
between two hydrogen atoms can be described by using atomic wave functions for
two atoms. There is very little overlap of the wave functions ψ1(r) for atom 1, located
at r = 0, and ψ2(r) for atom 2, located some distance away (Figure 6.3a). Suppose now
the two atoms are brought close together, their wave functions overlap and form the
compound wave function ψ1(r) + ψ2(r) shown in Figure 6.3b. Notice that the
probability amplitude is larger between the atoms than it is on either side of the
combination of atoms. As a result, the probability is higher that the electrons
associated with the atoms will be located between the atoms than on the outer
regions of the system. Consequently, the average position of negative charge in the
system is halfway between the atoms.
Figure 6.3 Ground-state wave functions ψ1(r) and ψ2(r) for two atoms making a
covalent bond. (a) The atoms are far apart, and their wave functions overlap
minimally. (b) The atoms are close together, forming a composite wave function
ψ1(r) + ψ2(r) for the system.
Van der Waals Bonding: Ionic and covalent bonds occur between atoms to form
molecules or ionic solids, so they can be described as bonds within molecules. The
84
van der Waals force results from the following situation. While being electrically
neutral, a molecule has a charge distribution with positive and negative centers at
different positions in the molecule. As a result, the molecule may act as an electric
dipole. Because of the dipole electric fields, two molecules can interact such that
there is an attractive force between them. There are three types of van der Waals
forces.
The first type, called the dipole– dipole force, is an interaction between two
molecules each having a permanent electric dipole moment. For example, polar
molecules such as HCl have permanent electric dipole moments and attract other
polar molecules.
The second type, the dipole–induced dipole force, results when a polar molecule
having a permanent electric dipole moment induces a dipole moment in a nonpolar
molecule. In this case, the electric field of the polar molecule creates the dipole
moment in the nonpolar molecule, which then results in an attractive force between
the molecules.
The third type is called the dispersion force, an attractive force that occurs between
two nonpolar molecules. Two nonpolar molecules near each other tend to have
dipole moments that are correlated in time so as to produce an attractive van der
Waals force.
Hydrogen Bonding: Because hydrogen has only one electron, it is expected to form
a covalent bond with only one other atom within a molecule. A hydrogen atom in a
given molecule can also form a second type of bond between molecules called a
hydrogen bond. Let’s use the water molecule H2O as an example. In the two
covalent bonds in this molecule, the electrons from the hydrogen atoms are more
likely to be found near the oxygen atom than near the hydrogen atoms, leaving
essentially bare protons at the positions of the hydrogen atoms. This unshielded
positive charge can be attracted to the negative end of another polar molecule.
Because the proton is unshielded by electrons, the negative end of the other
molecule can come very close to the proton to form a bond strong enough to form
a solid crystalline structure, such as that of ordinary ice. The bonds within a water
molecule are covalent, but the bonds between water molecules in ice are hydrogen
bonds. The hydrogen bond is relatively weak compared with other chemical bonds
and can be broken with an input energy of approximately 0.1 eV. Because of this
weakness, ice melts at the low temperature of 0°C.
85
6.2 ENERGY STATES AND SPECTRA OF MOLECULES
Consider an individual molecule in the gaseous phase of a substance. The energy E
of the molecule can be divided into four categories: (1) electronic energy, due to the
interactions between the molecule’s electrons and nuclei; (2) translational energy,
due to the motion of the molecule’s center of mass through space; (3) rotational
energy, due to the rotation of the molecule about its center of mass; and (4)
vibrational energy, due to the vibration of the molecule’s constituent atoms:
E = Eel + Etrans + Erot + Evib
Because the translational energy is unrelated to internal structure, this molecular
energy is unimportant in interpreting molecular spectra. Although the electronic
energies can be studied, significant information about a molecule can be determined
by analyzing its quantized rotational and vibrational energy states. Transitions
between these states give spectral lines in the microwave and infrared regions of the
electromagnetic spectrum, respectively.
6.3 ROTATIONAL MOTION OF MOLECULES
Let’s consider the rotation of a diatomic molecule around its center of mass (Figure
6.4a). A diatomic molecule aligned along a y axis has only two rotational degrees of
freedom, corresponding to rotations about the x and z axes passing through the
molecule’s center of mass. If  is the angular frequency of rotation about one of
these axes, the rotational kinetic energy of the molecule about that axis can be
expressed as
1
Erot  I 2
2
(6.2)
In this equation, I is the moment of inertia of the molecule about its center of mass,
given by
 mm 
I   1 2  r2   r2
 m1  m2 
(6.3)
where m1 and m2 are the masses of the atoms that form the molecule, r is the atomic
separation, and  is the reduced mass of the molecule

m1m2
m1  m2
(6.4)
86
Figure 6.4 Rotation of a diatomic molecule around its center of mass. (a) A diatomic molecule
oriented along the y axis. (b) Allowed rotational energies of a diatomic molecule expressed as
multiples of E1 = ℏ2/I.
The magnitude of the molecule’s angular momentum about its center of mass is
L=Iω, which can attain quantized values given by,
L  J  J  1
J  0, 1, 2, ...
(6.5)
where J is an integer called the rotational quantum number. Combining
Equations 6.5 and 6.2, we obtain an expression for the allowed values of the
rotational kinetic energy of the molecule:
Erot  EJ 
2
2I
J  J  1
J  0,1, 2, ...
(6.6)
The allowed rotational energies of a diatomic molecule are plotted in Figure 6.4b.
The allowed rotational transitions of linear molecules are regulated by the selection
rule ΔJ =±1. From Equation 6.6, the energies of the absorbed photons are given by,
Ephoton 
2
I
J
h2
4 2 I
J
J  1, 2, 3, ...
(6.7)
where J is the rotational quantum number of the higher energy state.
6.4 VIBRATIONAL MOTION OF MOLECULES
If we consider a molecule to be a flexible structure in which the atoms are bonded
together by “effective springs” as shown in Figure 6.5a, we can model the molecule
as a simple harmonic oscillator as long as the atoms in the molecule are not too far
from their equilibrium positions. Figure 6.5b shows a plot of potential energy versus
87
atomic separation for a diatomic molecule, where r0 is the equilibrium atomic
separation. For separations close to r0, the shape of the potential energy curve
closely resembles a parabola. According to classical mechanics, the frequency of
vibration for the system is given by
f 
1
2
k
(6.8)

where k is the effective spring constant and  is the reduced mass given by Equation
6.4. The vibrational motion and quantized vibrational energy can be altered if the
molecule acquires energy of the proper value to cause a transition between
quantized vibrational states. The allowed vibrational energies are
1

Evib   v   hf
2

v  0, 1, 2, ...
(6.9)
where v is an integer called the vibrational quantum number.
Figure 6.5 (a) Effective-spring model of a diatomic molecule. (b) Plot of the potential energy of a
diatomic molecule versus atomic separation distance.
88
Figure 6.6 Allowed vibrational energies of a diatomic molecule, where f is the frequency of vibration of
the molecule
Substituting Equation 6.8 into Equation 6.9 gives the following expression for the
allowed vibrational energies:
1 h

Evib   v  
2  2

k
v  0, 1, 2, ...

(6.10)
The selection rule for the allowed vibrational transitions is Δv = ±1. The photon
energy for transition is given by,
Ephoton  Evib 
h
2
k

(6.11)
The vibrational energies of a diatomic molecule are plotted in Figure 6.6. At ordinary
temperatures, most molecules have vibrational energies corresponding to the v = 0
state because the spacing between vibrational states is much greater than kBT, where
kB is Boltzmann’s constant and T is the temperature.
6.5 MOLECULAR SPECTRA
In general, a molecule vibrates and rotates simultaneously. To a first approximation,
these motions are independent of each other, so the total energy of the molecule is
the sum of Equations 6.6 and 6.9:
2
1

E   v   hf  J  J  1
2
2I

(6.12)
The energy levels of any molecule can be calculated from this expression, and each
level is indexed by the two quantum numbers v and J. From these calculations, an
energy-level diagram like the one shown in Figure 6.7a can be constructed. For each
allowed value of the vibrational quantum number v, there is a complete set of
rotational levels corresponding to J = 0, 1, 2, . . . . The energy separation between
successive rotational levels is much smaller than the separation between successive
vibrational levels. The molecular absorption spectrum in Figure 6.7b consists of two
groups of lines: one group to the right of center and satisfying the selection rules ΔJ
= +1 and Δv = +1, and the other group to the left of center and satisfying the selection
rules ΔJ = -1 and Δv = +1. The energies of the absorbed photons can be calculated
from Equation 6.12:
89
Ephoton  E  hf 
2
I
Ephoton  E  hf 
 J  1
2
I
J
J  0,1, 2, ...
J  1, 2, 3, ...
 J  1
 J  1
(6.13)
(6.14)
where J is the rotational quantum number of the initial state.
Figure 6.7 (a) Absorptive transitions between the v = 0 and v = 1 vibrational states of a diatomic
molecule. (b) Expected lines in the absorption spectrum of a molecule.
The experimental absorption spectrum of the HCl molecule shown in Figure 6.8.
One peculiarity is apparent, however: each line is split into a doublet. This doubling
occurs because two chlorine isotopes were present in the sample used to obtain this
spectrum. Because the isotopes have different masses, the two HCl molecules have
different values of I.
90
Figure 6.8 Experimental absorption spectrum of the HCl molecule
The second function determining the envelope of the intensity of the spectral lines
is the Boltzmann factor. The number of molecules in an excited rotational state is
given by
𝑛 = 𝑛0
−ℏ2 𝐽(𝐽+1)
𝑒 2𝐼𝑘𝐵 𝑇
where n0 is the number of molecules in the J = 0 state.
Multiplying these factors together indicates that the intensity of spectral lines
should be described by a function of J as follows:
𝐼𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 ∝ (2𝐽 + 1)𝑒
−ℏ2 𝐽(𝐽+1)
2𝐼𝑘𝐵 𝑇
(6.15)
6.6 BONDING IN SOLIDS
A crystalline solid consists of a large number of atoms arranged in a regular array,
forming a periodic structure.
Ionic Solids: Many crystals are formed by ionic bonding, in which the dominant
interaction between ions is the Coulomb force. Consider a portion of the NaCl
crystal shown in Figure 6.9a. The red spheres are sodium ions, and the blue spheres
are chlorine ions. As shown in Figure 6.9b, each Na+ ion has six nearest-neighbor Clions. Similarly, in Figure 6.9c, each Cl- ion has six nearest-neighbor Na+ ions. Each
Na+ ion is attracted to its six Cl- neighbors. The corresponding potential energy is 6kee2/r, where ke is the Coulomb constant and r is the separation distance between
each Na+ and Cl-. In addition, there are 12 next-nearest-neighbor Na+ ions at a
distance of √2r from the Na+ ion, and these 12 positive ions exert weaker repulsive
forces on the central Na+. Furthermore, beyond these 12 Na+ ions are more Cl2 ions
that exert an attractive force, and so on. The net effect of all these interactions is a
resultant negative electric potential energy
91
U attractive   ke
e2
r
(6.16)
where α is a dimensionless number known as the Madelung constant. The value
of α depends only on the particular crystalline structure of the solid (α = 1.747 for
the NaCl structure). When the constituent ions of a crystal are brought close
together, a repulsive force exists because of electrostatic forces and the exclusion
principle. The potential energy term B/rm in Equation 6.1 accounts for this repulsive
force. The repulsive forces occur only for ions that are very close together.
Therefore, we can express the total potential energy of the crystal as
U total   ke
e2 B

r rm
(6.17)
where m in this expression is some small integer.
Figure 6.9 (a) Crystalline structure of NaCl. (b) Each positive sodium ion is surrounded by six
negative chlorine ions. (c) Each chlorine ion is surrounded by six sodium ions
Figure 6.10 Total potential energy versus ion separation distance for an ionic solid, where U 0 is the
ionic cohesive energy and r0 is the equilibrium separation distance between ions
92
A plot of total potential energy versus ion separation distance is shown in Figure
6.10. The potential energy has its minimum value U0 at the equilibrium separation,
when r = r0.
U 0   ke
e2  1 
1  
r0  m 
(6.18)
This minimum energy U0 is called the ionic cohesive energy of the solid, and its
absolute value represents the energy required to separate the solid into a collection
of isolated positive and negative ions.
Covalent Solids: Solid carbon, in the form of diamond, is a crystal whose atoms are
covalently bonded. In the diamond structure, each carbon atom is covalently
bonded to four other carbon atoms located at four corners of a cube as shown in
Figure 6.11a. The crystalline structure of diamond is shown in Figure 6.11b. The basic
structure of diamond is called tetrahedral (each carbon atom is at the center of a
regular tetrahedron), and the angle between the bonds is 109.5°. Other crystals such
as silicon and germanium have the same structure. Covalently bonded solids are
usually very hard, have high bond energies and high melting points, and are good
electrical insulators.
Figure 6.11 (a) Each carbon atom in a diamond crystal is covalently bonded to four other carbon
atoms so that a tetrahedral structure is formed. (b) The crystal structure of diamond, showing the
tetrahedral bond arrangement
Metallic Solids: Metallic bonds are generally weaker than ionic or covalent bonds.
The outer electrons in the atoms of a metal are relatively free to move throughout
the material, and the number of such mobile electrons in a metal is large. The
metallic structure can be viewed as a “sea” or a “gas” of nearly free electrons
surrounding a lattice of positive ions (Fig. 6.14). The bonding mechanism in a metal
93
is the attractive force between the entire collection of positive ions and the electron
gas.
Light interacts strongly with the free electrons in metals. Hence, visible light is
absorbed and re-emitted quite close to the surface of a metal, which accounts for
the shiny nature of metal surfaces. Because the bonding in metals is between all the
electrons and all the positive ions, metals tend to bend when stressed.
Figure 6.12 Highly schematic diagram of a metal
6.7 FREE-ELECTRON THEORY OF METALS
Quantum based free electron theory of metals is centered on wave nature of
electrons. In this model, one imagines that the outer-shell electrons are free to
move through the metal but are trapped within a three-dimensional box
formed by the metal surfaces. Therefore, each electron is represented as a
particle in a box and is restricted to quantized energy levels. Each energy state
can be occupied by only two electrons (one with spin up & the other with spin
down) as a consequence of exclusion principle. In quantum statistics, it is shown
that the probability of a particular energy state E being occupied by an electrons is
given by
f E 
1
 E  EF 
exp 
  1
 kT 
(6.19)
where f(E) is called the Fermi-Dirac distribution function and EF is called the
Fermi energy. Plot of f(E) versus E is shown in figure 6.13.
94
Figure. 6.13 Plot of Fermi-Dirac distribution function f(E) Vs energy E at (a) T = 0K and (b) T > 0K
At zero kelvin (0 K), all states having energies less than the Fermi energy are
occupied, and all states having energies greater than the Fermi energy are vacant.
i.e. Fermi energy is the highest energy possessed by an electron at 0 K (Figure 6.13a).
As temperature increases (T > 0K), the distribution rounds off slightly due to
thermal excitation and probability of Fermi level being occupied by an electron
becomes half (Figure 6.13b). In other words, Fermi energy is that energy state at
which probability of electron occupation is half. The Fermi energy EF also depends
on temperature, but the dependence is weak in metals.
Density of states: From particle in a box problem, for a particle of mass m is
confined to move in a one-dimensional box of length L, the allowed states have
quantized energy levels given by,
En 
2 2
h2

2
n

n2
2
2
8mL
2mL
n = 1, 2, 3 . . .
(6.20)
An electron moving freely in a metal cube of side L, can be modeled as particle in a
three-dimensional box. It can be shown that the energy for such an electron is
2
E
n 2  n y2  nz2 
2  x
2mL
2
(6.21)
where m is mass of the electron and nx, ny, nz are quantum numbers(positive
integers). Each allowed energy value is characterized by a set of three quantum
numbers (nx, ny, nz - one for each degree of freedom). Imagine a threedimensional quantum number space whose axes represent nx, ny, nz. The
allowed energy states in this space can be represented as dots located at positive
integral values of the three quantum numbers as shown in the Figure 6.14.
95
Figure 6.14 Representation of the allowed energy states in a quantum number space (dots represent
the allowed states)
Eq. 6.21 can be written as
nx2  ny2  nz2 
where Eo 
2
2
2
2mL
E
 n2
Eo
and n 
(6.22)
E
Eo
Eq. 6.22 represents a sphere of radius n. Thus, the number of allowed energy
states having energies between E and E+dE is equal to the number of points
in a spherical shell of radius n and thickness dn. In this quantum number space
each point is at the corners of a unit cube and each corner point is shared by eight
unit cubes and as such the contribution of each point to the cube is 1/8 th. Because
a cube has eight corners, the effective point per unit cube and hence unit volume is
one. In other words, number of points is equal to the volume of the shell. The
“volume” of this shell, denoted by G(E)dE.
1
G(E) dE =    4 n 2 dn 
8

1
2
   n dn
2
96
 E  1 2 


E
G ( E ) dE  12  
 d 
 
E
E

 o   o  
 E   12
 Eo
 Eo 
G ( E ) dE 
1
2

G( E ) dE 
1
4
 2 2 

2 
 2mL 
3
1
2
E
2
E
1
1
2
using the relation n 
dE 
2
1
4
3
 Eo
2
E
1
2
E
Eo
dE
dE
3
2 m 2 L3 12
G ( E ) dE 
E dE ,
2 2 3
L3  V
Number of states per unit volume per unit energy range, called density of
states, g(E) is given by
g(E) = G(E)/V
G( E )
2 m
g ( E ) dE 
dE 
V
2 2
4 2 m
g ( E ) dE 
h3
3
2
E
1
2
3
2
3
E
1
2
dE

dE
h
2
Finally, we multiply by 2 for the two possible spin states of each particle.
8 2 m
g ( E ) dE 
h3
3
2
E
1
2
dE
(6.23)
g(E) is called the density-of-states function.
Electron density: For a metal in thermal equilibrium, the number of electrons
N(E) dE, per unit volume, that have energy between E and E+dE is equal to
the product of the density of states and the probability that a state is occupied.
that is,
N(E)dE = [ g(E)dE ] f(E)
97
8 2 m
N ( E ) dE 
h3
3
2
1
E 2 dE
 E  EF 
exp 
  1
 kT 
(6.24)
Plots of N(E) versus E for two temperatures are given in figure 6.15.
Figure 6.15 Plots of N(E) versus E for (a) T = 0K (b) T = 300K
If ne is the total number of electrons per unit volume, we require that

8 2 m
ne   N ( E ) dE 
h3
0
3
2
1

E 2 dE
 E  EF 
exp 
  1
 kT 

0
(6.25)
At T = 0K, the Fermi-Dirac distribution function f(E) = 1 for E <EF and f(E) = 0 for E
>EF. Therefore, at T = 0K, Equation 5.7 becomes
8 2 m
ne 
h3
3
2
EF

0
E
1
2
8 2 m
dE 
h3
3
2
  EF
2
3
3
2
16 2  m

3 h3
3
2
3
EF 2
(6.26)
Solving for Fermi energy at 0K, we obtain
h 2  3 ne 
EF  0  


2 m  8 
2
3
(6.27)
The average energy of a free electron in a metal at 0K is Eav = (3/5)EF.
98
6.8 BAND THEORY OF SOLIDS
When a quantum system is represented by wave function, probability density ||2
for that system is physically significant while the probability amplitude  not.
Consider an atom such as sodium that has a single s electron outside of a closed



shell. Both the wave functions  S ( r ) and  S ( r ) are valid for such an atom [ S ( r )

and  S ( r ) are symmetric and anti symmetric wave functions]. As the two sodium
atoms are brought closer together, their wave functions begin to overlap. Figure 6.16
represents two possible combinations : i) symmetric - symmetric and ii) symmetric
– antisymmetric . These two possible combinations of wave functions
represent two possible states of the two-atom system. Thus, the states are
split into two energy levels. The energy difference between these states is relatively
small, so the two states are close together on an energy scale.
When two atoms are brought together, each energy level will split into 2 energy
levels. (In general, when N atoms are brought together N split levels will occur which
can hold 2N electrons). The split levels are so close that they may be regarded as a
continuous band of energy levels. Figure 6.17 shows the splitting of 1s and 2s
levels of sodium atom when : (a) two sodium atoms are brought together (b)five
sodium atoms are brought together (c) a large number of sodium atoms are
assembled to form a solid. The close energy levels forming a band are seen clearly
in (c).
Figure. 6.16 The wave functions of two atoms combine to form a composite wave function : a)
symmetric-symmetric b) symmetric-antisymmetric
99
Figure 6.17 Splitting of 1s and 2s levels of sodium atoms due to interaction between them
Some bands may be wide enough in energy so that there is an overlap
between the adjacent bands. Some other bands are narrow so that a gap may occur
between the allowed bands, and is known as forbidden energy gap. The 1s, 2s,
and 2p bands of solid sodium are filled completely with electrons. The 3s
band (2N states) of solid sodium has only N electrons and is partially full;
The 3p band, which is the higher region of the overlapping bands, is completely
empty as shown in Figure 6.18
Figure 6.18 Energy bands of a sodium crystal
100
6.9 ELECTRICAL CONDUCTION IN METALS, INSULATORS
AND SEMICONDUCTORS
Good electrical conductors contain high density of free charge carriers, and
the density of free charge carriers in insulators is nearly zero. In
semiconductors free-charge-carrier densities are intermediate between those of
insulators and those of conductors.
Metals: Metal has a partially filled energy band (Figure 6.19a). At 0K Fermi level
is the highest electron-occupied energy level. If a potential difference is applied
to the metal, electrons having energies near the Fermi energy require only a
small amount of additional energy to reach nearby empty energy states above
the Fermi-level. Therefore, electrons in a metal experiencing a small force (from a
weak applied electric field) are free to move because many empty levels are available
close to the occupied energy levels. The model of metals based on band theory
demonstrates that metals are excellent electrical conductors.
Insulators: Consider the two outermost energy bands of a material in which the
lower band is filled with electrons and the higher band is empty at 0 K (Figure 6.19b).
The lower, filled band is called the valence band, and the upper, empty band is the
conduction band. The energy separation between the valence and conduction band,
called energy gap Eg, is large for insulating materials. The Fermi level lies
somewhere in the energy gap. Due to larger energy gap compare to thermal energy
kT (26meV) at room temperature, excitation of electrons from valence band to
conduction band is hardly possible. Since the free-electron density is nearly zero,
these materials are bad conductors of electricity.
Semiconductors: Semiconductors have the same type of band structure as an
insulator, but the energy gap is much smaller, of the order of 1 eV. The band
structure of a semiconductor is shown in Figure 6.19c. Because the Fermi level is
located near the middle of the gap for a semiconductor and Eg is small, appreciable
numbers of electrons are thermally excited from the valence band to the conduction
band. Because of the many empty levels above the thermally filled levels in the
conduction band, a small applied potential difference can easily raise the energy of
the electrons in the conduction band, resulting in a moderate conduction. At T = 0
K, all electrons in these materials are in the valence band and no energy is available
to excite them across the energy gap. Therefore, semiconductors are poor
conductors at very low temperatures. Because the thermal excitation of electrons
across the narrow gap is more probable at higher temperatures, the conductivity of
101
semiconductors increases rapidly with temperature. This is in sharp contrast with
the conductivity of metals, where it decreases with increasing temperature. Charge
carriers in a semiconductor can be negative, positive, or both. When an electron
moves from the valence band into the conduction band, it leaves behind a vacant
site, called a hole, in the otherwise filled valence band.
Figure 6.19 Band structure of (a) Metals (b) Insulators (c) Semiconductors
In an intrinsic semiconductor (pure semiconductor) there are equal number
of conduction electrons and holes. In the presence of an external electric
field, the holes move in the direction of field and the conduction electrons
move opposite to the direction of the field. Both these motions correspond
to the current in the same direction (Figure 6.20).
Figure 6.20 Movement of electrons and holes in an intrinsic semiconductor
102
Doped Semiconductors (Extrinsic semiconductors): Semiconductors in their
pure form are called intrinsic semiconductors while doped semiconductors are
called extrinsic semiconductors. Doping is the process of adding impurities to a
semiconductor. By doping both the band structure of the semiconductor and
its resistivity are modified. If a tetravalent semiconductor (Si or Ge) is doped
with a pentavalent impurity atom (donor atom), four of the electrons form
covalent bonds with atoms of the semiconductor and one is left over (Figure
6.21). At zero K, this extra electron resides in the donor-levels, that lie in the
energy gap, just below the conduction band. Since the energy Ed between the
donor levels and the bottom of the conduction band is small, at room
temperature, the extra electron is thermally excited to the conduction band.
This type of semiconductors are called n-type semiconductors because the
majority of charge carriers are electrons (negatively charged).
If a tetravalent semiconductor is doped with a trivalent impurity atom
(acceptor atom), the three electrons form covalent bonds with neighboring
semiconductor atoms, leaving an electron deficiency (a hole) at the site of
fourth bond (Figure 6.22). At zero K, this hole resides in the acceptor levels
that lie in the energy gap just above the valence band. Since the energy E a
between the acceptor levels and the top of the valence band is small, at room
temperature, an electron from the valence band is thermally excited to the
acceptor levels leaving behind a hole in the valence band. This type of
semiconductors are called p-type semiconductors because the majority of
charge carriers are holes (positively charged).
103
Figure 6.21 n-type semiconductor – two dimensional representation and band structure
When conduction in a semiconductor is the result of acceptor or donor impurities,
the material is called an extrinsic semiconductor. The typical range of doping
densities for extrinsic semiconductors is 1013 to 1019 cm-3, whereas the electron
density in a typical semiconductor is roughly 1021 cm-3.
Figure 6.22 p-type semiconductor: two-dimensional representation and band structure
6.10 SUPERCONDUCTIVITY-PROPERTIES AND
APPLICATIONS
Superconductor is a class of metals and compounds whose electrical resistance
decreases to virtually zero below a certain temperature Tc called the critical
temperature. The critical temperature is different for different superconductors.
Mercury loses its resistance completely and turns into a superconductor at 4.2K.
Critical temperatures for some important elements/compounds are listed below.
Element/Compound
La
NbNi
Nb3Ga
Tc (K)
6.0
10.0
23.8
104
Figure 6.23 Plot of Resistance Vs Temperature for normal metal and a superconductor
Meissner Effect: In the presence of magnetic field, as the temperature of
superconducting material is lowered below Tc, the field lines are spontaneously
expelled from the interior of the superconductor(B = 0, Figure 6.24). Therefore, a
superconductor is more than a perfect conductor; it is also a perfect diamagnet. The
property that B = 0 in the interior of a superconductor is as fundamental as the
property of zero resistance. If the magnitude of the applied magnetic field exceeds
a critical value Bc, defined as the value of B that destroys a material’s
superconducting properties, the field again penetrates the sample. Meissner effect
can be explained in the following way.
Figure 6.24 A superconductor in the form of a long cylinder in the presence of an external magnetic
field.
A good conductor expels static electric fields by moving charges to its surface. In
effect, the surface charges produce an electric field that exactly cancels the
externally applied field inside the conductor. In a similar manner, a superconductor
expels magnetic fields by forming surface currents. Consider the superconductor
105
shown in Figure 6.24. Let’s assume the sample is initially at a temperature T>Tc so
that the magnetic field penetrates the cylinder. As the cylinder is cooled to a
temperature T<Tc, the field is expelled. Surface currents induced on the
superconductor’s surface produce a magnetic field that exactly cancels the
externally applied field inside the superconductor. As expected, the surface currents
disappear when the external magnetic field is removed.
BCS Theory: In 1957. Bardeen, Cooper and Schrieffer gave a successful theory to
explain the phenomenon of superconductivity, which is known as BCS theory.
According to this theory, two electrons can interact via distortions in the array of
lattice ions so that there is a net attractive force between the electrons. As a result,
the two electrons are bound into an entity called a Cooper pair, which behaves like
a single particle with integral spin. Particles with integral spin are called bosons. An
important feature of bosons is that they do not obey the Pauli exclusion principle.
Consequently, at very low temperatures, it is possible for all bosons in a collection
of such particles to be in the lowest quantum state and as such the entire collection
of Cooper pairs in the metal is described by a single wave function. There is an
energy gap equal to the binding energy of a Cooper pair between this lowest state
and the next higher state.. Under the action of an applied electric field, the Cooper
pairs experience an electric force and move through the metal. A random scattering
event of a Cooper pair from a lattice ion would represent resistance to the electric
current. Such a collision would change the energy of the Cooper pair because some
energy would be transferred to the lattice ion. There are no available energy levels
below that of the Cooper pair (it is already in the lowest state), however, and none
available above because of the energy gap. As a result, collisions do not occur and
there is no resistance to the movement of Cooper pairs.
Applications: Most important and basic application of superconductors is in high
field solenoids which can be used to produce intense magnetic field.
Superconducting magnets are used in magnetic resonance imaging (MRI)
technique. Magnetic levitation, based on Meissner effect, is another important
application of superconductors. This principle is used in maglev vehicles. Detection
of a weak magnetic field and lossless power transmission are some other important
applications of superconductors.
6.11 QUESTIONS
1.
Sketch schematically the plot of potential energy and its components as a
function of inter-nuclear separation distance for a system of two atoms.
106
2.
Explain briefly (a) ionic bonding, (b) covalent bonding, (c) van der Walls
bonding, (d) hydrogen bonding.
3. Obtain an expression for rotational energy of a diatomic molecule. Sketch
schematically these rotational energy levels.
4. Obtain expressions for rotational transition photon energies and frequencies.
5. Obtain an expression for vibrational energy of a diatomic molecule. Sketch
schematically these vibrational energy levels. Obtain expression for vibrational
transition photon energies.
6. Write expression for total energy (vibrational and rotational) of a molecule.
Sketch schematically these energy levels of a diatomic molecule for the
lowest two vibrational energy values, indicating the possible transitions.
Write the expressions for the energy of the photon in the molecular
energy transitions. Write the expression for the frequency separation of
adjacent spectral lines.
7. Explain the expression for the total potential energy of a crystal. Sketch
schematically the plot of the same.
8. Define (a) ionic cohesive energy (b) atomic cohesive energy, of a solid.
9. Write the expression for Fermi-Dirac distribution function.
Sketch
schematically the plots of this function for zero kelvin and for temperature
above zero kelvin.
10. Derive an expression for density-of-states.
11. Assuming the Fermi-Dirac distribution function , obtain an expression for
the density of free-electrons in a metal with Fermi energy EF, at zero K
and, hence obtain expression for Fermi energy EF in a metal at zero K. [
3
12.
13.
14.
15.
16.
17.
8 2  m 2 12
E dE ]
Given: density-of-states function g( E ) dE 
h3
Explain the formation of energy bands in solids with necessary diagrams.
Distinguish between conductors, insulators and semiconductors on the basis of
band theory
Indicate the position of (a) donor levels (b) acceptor levels, in the energy
band diagram of a semiconductor.
What is the difference between p-type and n-type semiconductors? Explain
with band diagram.
With necessary diagrams, explain doping in semiconductors.
Why the electrical conductivity of an intrinsic semiconductor increases with
increasing temperature?
107
18. What are superconductors? Draw a representative graph of Resistance Vs
Temperature for a superconductor.
19. Explain Meissner effect.
20. Explain briefly the BCS theory of superconductivity in metals.
6.12 PROBLEMS
1. A K+ ion and a Cl– ion are separated by a distance of 5.00 x 10–10 m.
Assuming the two ions act like point charges, determine (a) the force
each ion exerts on the other and (b) the potential energy of the two-ion
system in electron volts.
Ans: a) 921 pN toward the other ion b) - 2.88 eV
2. The potential energy (U) of a diatomic molecule is
potential:
A
B
U 
 6
r 12
r
where A and B are constants. Find, in terms of A and
which the energy is minimum and (b) the energy E
diatomic molecule. (c) Evaluate ro in metres and E in
molecule.
Take A = 0.124 x 10–120 eV.m12 and B = 1.488 x 10–60 eV.m6.
Ans: a) (2A/B)1/6
b) B2/4A
c) 74.2 pm, 4.46 eV
given by Lennard-Jones
B, (a) the value of r o at
required to break up a
electron-volts for the H2
3. An HCl molecule is excited to its first rotational energy level,
corresponding to J = 1. If the distance between its nuclei is ro = 0.1275 nm,
what is the angular speed of the molecule about its center of mass ?
Ans: 5.69x1012 rad/s
4. The rotational spectrum of the HCl molecule contains lines with wavelengths of
0.0604, 0.0690, 0.0804, 0.0964, and 0.1204 mm. What is the moment of inertia of
the molecule?
Ans: 2.72x10-47 kg.m2
5. The frequency of photon that causes v = 0 to v = 1 transition in the CO molecule
is 6.42 x 1013 Hz. Ignore any changes in the rotational energy. (A) Calculate the force
constant k for this molecule. (B) What is the maximum classical amplitude of
vibration for this molecule in the v = 0 vibrational state ?
Ans: A) 1.85x103 N/m
B) 0.00479nm
6. Consider a one-dimensional chain of alternating positive and negative ions.
Show that the potential energy associated with one of the ions and its
interactions with the rest of this hypothetical crystal is
108
Ur    k e 
e2
r
where the Madelung constant is  = 2 ln 2 and r is the inter-ionic
spacing. Hint: Use the series expansion
ln 1  x   x 
x2
x3
x4


 ...
2
3
4
7. Each atom of gold (Au) contributes one free-electron to the metal.
The
28
3
concentration of free-electron in gold is 5.90 x 10 /m . Compute the Fermi Energy
of gold.
Ans: 5.53 eV
8. Sodium is a monovalent metal having a density of 971 kg/m3 and a molar mass
of 0.023 kg/mol. Use this information to calculate (a) the density of charge
carriers and (b) the Fermi energy.
Ans: 2.54 x 1028/m3, 3.15 eV
9. Calculate the energy of a conduction electron in silver at 800 K, assuming the
probability of finding an electron in that state is 0.950. The Fermi energy is
5.48 eV at this temperature.
Ans: 5.28 eV
10. Show that the average kinetic energy of a conduction electron in a metal at
zero K is (3/5) EF
Suggestion: In general, the average kinetic energy is
1
E N( E ) dE
ne 
E AV 
where the density of particles

ne   N( E ) dE
0
109
3
8 2  m2
N( E ) dE 
h3
1
E 2 dE
 E  EF
exp 
 kT

  1

11. (a) Consider a system of electrons confined to a three-dimensional box.
Calculate the ratio of the number of allowed energy levels at 8.50 eV to the
number at 7.00 eV. (b) Copper has a Fermi energy of 7.0 eV at 300 K.
Calculate the ratio of the number of occupied levels at an energy of 8.50 eV
to the number at Fermi energy. Compare your answer with that obtained in
part (a).
Ans: (a) 1.10 (b) 1.46x10-25
12. Most solar radiation has a wavelength of 1 μm or less. What energy gap should
the material in solar cell have in order to absorb this radiation ? Is silicon
(Eg= 1.14 eV) appropriate?
Ans: 1.24 eV or less; yes
13. The energy gap for silicon at 300 K is 1.14 eV. (a) Find the lowest-frequency photon
that can promote an electron from the valence band to the conduction band. (b)
What is the wavelength of this photon?
Ans: 2.7x1014 Hz, 1090 nm
14. A light-emitting diode (LED) made of the semiconductor GaAsP emits red light ( λ=
650nm). Determine the energy-band gap Eg in the semiconductor.
Ans: 1.91 eV
110
ENGINEERING
PHYSICS
[SUBJECT CODE: PHY1051]
COMMON COURSE MATERIAL FOR FIRST YEAR BTech STUDENTS
DEPARTMENT OF PHYSICS
MANIPAL INSTITUTE OF TECHNOLOGY
MIT | Physics | 2018
Table of Contents
1
INTERFERENCE OF LIGHT WAVES................................................... 1
1.1
Young’s Double-Slit Experiment .................................................................... 1
1.2
Analysis Model: Waves in Interference ......................................................... 3
1.3
Intensity Distribution of the Double-Slit Interference Pattern .................... 4
1.4
Change of Phase Due to Reflection ................................................................6
1.5
Interference in Thin Films .............................................................................. 7
1.6
Newton’s Rings ................................................................................................8
1.7
Michelson Interferometer............................................................................. 10
1.8
Questions........................................................................................................ 11
1.9
Problems ......................................................................................................... 11
2 DIFFRACTION PATTERNS AND POLARIZATION ......................... 15
2.1
Introduction to Diffraction Patterns ............................................................. 15
2.2
Diffraction Patterns from Narrow Slits ........................................................ 16
2.3
Intensity of Single-Slit Diffraction Patterns ................................................ 18
2.4
Intensity of Two-Slit Diffraction Patterns ................................................... 18
2.5
Resolution of Single-Slit and Circular Apertures ........................................ 19
2.6
Diffraction Grating .........................................................................................21
2.7
Diffraction of X-Rays by Crystals ................................................................. 24
2.8
Polarization of Light Waves ......................................................................... 25
2.9
Polarization by Selective Absorption ........................................................... 26
2.10
Polarization by Reflection ............................................................................ 27
2.11
Polarization by Double Refraction ............................................................... 29
2.12
Polarization by Scattering ............................................................................ 30
2.13
Optical Activity .............................................................................................. 31
2.14
Questions........................................................................................................ 31
2.15
Problems ........................................................................................................ 32
I
3 QUANTUM PHYSICS .........................................................................34
3.1
Blackbody Radiation and Planck’s Hypothesis ............................................ 34
3.2
Photoelectric Effect ....................................................................................... 38
3.3
Compton Effect .............................................................................................40
3.4
Photons and Electromagnetic Waves [Dual Nature of Light] ....................44
3.5
de Broglie Hypothesis - Wave Properties of Particles .................................44
3.6
The Quantum Particle .................................................................................. 45
3.7
Double–Slit Experiment Revisited ............................................................... 47
3.8
Uncertainty Principle ....................................................................................48
3.9
Questions.......................................................................................................49
3.10
Problems ........................................................................................................ 50
4 QUANTUM MECHANICS ................................................................. 54
4.1
An Interpretation of Quantum Mechanics .................................................. 54
4.2
The Schrödinger Equation ............................................................................ 56
4.3
Particle in an Infinite Potential Well (Particle in a “Box”) .......................... 57
4.4
A Particle in a Potential Well of Finite Height ............................................ 59
4.5
Tunneling Through a Potential Energy Barrier ........................................... 61
4.6
The Simple Harmonic Oscillator ................................................................. 62
4.7
Questions....................................................................................................... 63
4.8
Problems ........................................................................................................64
5 ATOMIC PHYSICS ............................................................................. 66
5.1
The Quantum Model of the Hydrogen Atom ..............................................66
5.2
Wave functions for hydrogen .......................................................................68
5.3
More on Atomic Spectra: Visible and X-Ray ............................................... 70
5.4
X-Ray Spectra .................................................................................................71
5.5
Spontaneous and Stimulated transitions ..................................................... 73
5.6
LASER (Light Amplification by Stimulated Emission of Radiation) .......... 76
5.7
Applications of laser...................................................................................... 78
II
5.8
Questions....................................................................................................... 78
5.9
Problems ........................................................................................................ 79
6 MOLECULES AND SOLIDS .............................................................. 82
6.1
Molecular bonds............................................................................................ 82
6.2
Energy States and Spectra of Molecules ......................................................86
6.3
Rotational Motion of Molecules ...................................................................86
6.4
Vibrational Motion of Molecules ................................................................. 87
6.5
Molecular Spectra .........................................................................................89
6.6
Bonding in Solids .......................................................................................... 91
6.7
Free-Electron Theory of Metals ....................................................................94
6.8
Band Theory of Solids .................................................................................. 99
6.9
Electrical Conduction in Metals, Insulators and Semiconductors ............ 101
6.10
Superconductivity-Properties and Applications ....................................... 104
6.11
Questions..................................................................................................... 106
6.12
Problems ...................................................................................................... 108
III
Reference book:
Physics for Scientists and Engineers with Modern
Physics by Raymond Serway and John Jewett (Cengage Learning,
Seventh Edition 2012)
1 INTERFERENCE OF LIGHT WAVES
OBJECTIVES
•
•
•
To understand the principles of interference.
To explain the intensity distribution in interference under various
conditions.
To explain the interference from thin films.
Wave optics (Physical Optics): It is the study of interference, diffraction, and
polarization of light. These phenomena cannot be adequately explained with the ray
optics.
1.1 YOUNG’S DOUBLE-SLIT EXPERIMENT
Light waves also interfere with one another like mechanical waves. Fundamentally,
all interference associated with light waves arises when the electromagnetic fields
that constitute the individual waves combine.
Figure 1.1 (a) Schematic diagram of Young’s double-slit experiment. Slits S1 and S2 behave as coherent
sources of light waves that produce an interference pattern on the viewing screen (drawing not to
scale). (b) An enlargement of the center of a fringe pattern formed on the viewing screen.
Interference in light waves from two sources was first demonstrated by Thomas
Young in 1801. A schematic diagram of the apparatus Young used is shown Fig. 1.1a.
Plane light waves arrive at a barrier that contains two slits S1 and S2. The light from
S1 and S2 produces on a viewing screen a visible pattern of bright and dark parallel
bands called fringes (Fig. 1.1b). When the light from S1 and that from S2 both arrive
at a point on the screen such that constructive interference occurs at that location,
1
a bright fringe appears. When the light from the two slits combines destructively at
any location on the screen, a dark fringe results.
Figure 1.2 Waves leave the slits and combine at various points on the viewing screen.
Fig. 1.2 shows different ways in which two waves can combine at the screen. In Fig.
1.2a, the two waves, which leave the two slits in phase, strike the screen at the central
point O. Because both waves travel the same distance, they arrive at O in phase. As
a result, constructive interference occurs at this location and a bright fringe is
observed. In Fig. 1.2b, the two waves leave the slits in phase, but the wave leaving
from S2 has to travel longer distance compare to wave from S1. However, the
difference in the path is exactly one wavelength and they arrive in phase at P and a
second bright fringe appears at this location. At point R in Fig. 1.2c, wave from S2 has
fallen half a wavelength behind the wave from S1 and a crest of the upper wave
overlaps a trough of the lower wave, giving rise to destructive interference at point
R.
If two lightbulbs are placed side by side so that light from both bulbs combines, no
interference effects are observed because the light waves from one bulb are emitted
independently of those from the other bulb. The emissions from the two lightbulbs
do not maintain a constant phase relationship with each other over time. Therefore,
the conditions for constructive interference, destructive interference, or some
intermediate state are maintained only for short time intervals. Since the eye cannot
follow such rapid changes, no interference effects are observed. Such light sources
are said to be incoherent.
To observe interference of waves from two sources, the following conditions must
be met:
• The sources must be coherent; that is, they must maintain a constant phase
with respect to each other.
• The sources should be monochromatic; that is, they should be of a single
wavelength.
A common method for producing two coherent light sources is to use a
monochromatic source to illuminate a barrier containing two small openings,
2
usually in the shape of slits, as in the case of Young’s experiment illustrated in Fig.
1. The light emerging from the two slits is coherent because a single source produces
the original light beam and the two slits serve only to separate the original beam
into two parts. Any random change in the light emitted by the source occurs in both
beams at the same time. As a result, interference effects can be observed when the
light from the two slits arrives at a viewing screen.
1.2 ANALYSIS MODEL: WAVES IN INTERFERENCE
Figure 1.3 (a) Geometric construction for describing Young’s double-slit experiment (not to scale). (b)
The slits are represented as sources, and the outgoing light rays are assumed to be parallel as they
travel to P.
The viewing screen is located a perpendicular distance L from the barrier containing
two slits, S1 and S2 (Fig. 1.3a). These slits are separated by a distance d, and the source
is monochromatic. To reach any arbitrary point P in the upper half of the screen, a
wave from the lower slit must travel farther than a wave from the upper slit by a
distance d sin  (Fig. 1.3b). This distance is called the path difference . If we
assume the rays labeled r1 and r2 are parallel, which is approximately true if L is
much greater than d, then  is given by
  r2  r1  d sin 
(1.1)
The value of  determines whether the two waves are in phase when they arrive at
point P.
Angular positions of bright and dark fringes: If  is either zero or some integer
multiple of the wavelength, the two waves are in phase at point P and constructive
interference results. Therefore, the condition for bright fringes, or constructive
interference, at point P is,
d sin  bright  m ;
 m  0,  1,  2, ...
(1.2)
The number m is called the order number.
3
When d is an odd multiple of  / 2 , the two waves arriving at point P are 180° out of
phase and give rise to destructive interference. Therefore, the condition for dark
fringes, or destructive interference, at point P is,
1

d sin  dark   m    ;
2

 m  0,  1,  2, ...
(1.3)
Linear positions of bright and dark fringes: From the triangle OPQ in Fig. 1.3a,
y
(1.4)
tan  
L
Using this result, the linear positions of bright and dark fringes are given by
ybright  L tan  bright
ybright  L
m
d
(1.5)
(small angle approximation)
ydark  L tan  dark
ydark
(1.6)
(1.7)
1

m  
2
 L
d
(small angle approximation)
(1.8)
1.3 INTENSITY DISTRIBUTION OF THE DOUBLE-SLIT
INTERFERENCE PATTERN
Consider two coherent sources of sinusoidal waves such that they have same angular
frequency  and phase difference . The total magnitude of the electric field at
point P on the screen in Fig. 1.3a is the superposition of the two waves. Assuming
that the two waves have same amplitude E0 , we can write the magnitude of electric
field at point P due to each source as
E1  E0 sin  t
and
E2  E0 sin  t   
(1.9)
Although the waves are in phase at the slits, their phase difference  at P depends
on the path difference   r2  r1  d sin  . A path difference of  (for constructive
interference) corresponds to a phase difference of 2 radians.

2


2

d sin 
(1.10)
Using the superposition principle and Equation (1.9), we obtain the following
expression for the magnitude of the resultant electric field at point P:
4
EP  E1  E2  E0 sin  t  sin  t   
(1.11)

 

EP  2 E0 cos   sin   t  
2
2

(1.12)
This result indicates that the electric field at point P has the same frequency  as
the light at the slits but that the amplitude of the field is multiplied by the factor
2 cos( / 2) . If   0, 2 , 4 ,......the magnitude of the electric field at point P is 2E0 ,
corresponding to the condition for maximum constructive interference. Similarly, if
   , 3 , 5 ,...... the magnitude of the electric field at point P is zero.
Intensity of a wave is proportional to the square of the resultant electric field
magnitude at that point. Using Equation 1.12, we can express the light intensity at
point P as
I

 

 EP2  4 E02 cos 2   sin 2   t  
2
2

(1.13)
Most light-detecting instruments measure time-averaged light intensity, and the


time averaged value of sin 2   t   over one cycle is ½. Therefore, we can write the
2

average light intensity at point P as,
 
I  I max cos2  
2
(1.14)
where I max is the maximum intensity on the screen. Substituting the value for  from
Equation 1.10;
 d sin  
I  I max cos2 

 

Alternatively, since sin  
d
I  I max cos 2 
 L

y

(1.15)
y
for small values of  , we can write;
L
(1.16)
A plot of light intensity versus d sin  is given in Fig. 1.4. The interference pattern
consists of equally spaced fringes of equal intensity.
5
Figure 1.4 Light intensity versus d sin for a double-slit interference pattern when the screen is far
from the two slits (L >> d).
1.4 CHANGE OF PHASE DUE TO REFLECTION
Young’s method for producing two coherent light sources involves illuminating a
pair of slits with a single source. Another simple arrangement for producing an
interference pattern with a single light source is known as Lloyd’s mirror (Fig. 1.5).
Figure 1.5 Lloyd’s mirror. The reflected ray undergoes a phase change of 180°.
A point light source S is placed close to a mirror, and a viewing screen is positioned
some distance away and perpendicular to the mirror. Light waves can reach point P
on the screen either directly from S to P or by the path involving reflection from the
mirror. The reflected ray can be treated as a ray originating from a virtual source S.
As a result, we can think of this arrangement as a double slit source where the
6
distance d between sources S and S in Fig. 1.5 is analogous to length d in Fig. 1.3a.
Hence, at observation points far from the source (L >> d), waves from S and S form
an interference pattern exactly like the one formed by two real coherent sources.
But, the positions of the dark and bright fringes, however, are reversed relative to
the pattern created by two real coherent sources (Young’s experiment). Such a
reversal can only occur if the coherent sources S and S differ in phase by 180°.
In general, an electromagnetic wave undergoes a phase change of 180° upon
reflection from a medium that has a higher index of refraction than the one in which
the wave is traveling. Analogy between reflected light waves and the reflections of a
transverse pulse on a stretched string is shown in Fig. 1.6.
Figure 1.6 Comparisons of reflections of light waves and waves on strings.
The reflected pulse on a string undergoes a phase change of 180° when reflected from
the boundary of a denser string or a rigid support, but no phase change occurs when
the pulse is reflected from the boundary of a less dense string or a freely-supported
end. Similarly, an electromagnetic wave undergoes a 180° phase change when
reflected from a boundary leading to an optically denser medium, but no phase
change occurs when the wave is reflected from a boundary leading to a less dense
medium.
1.5 INTERFERENCE IN THIN FILMS
Interference effects are commonly observed in thin films, such as thin layers of oil
on water or the thin surface of a soap bubble. The varied colors observed when white
light is incident on such films result from the interference of waves reflected from
the two surfaces of the film. Consider a film of uniform thickness t and index of
refraction n. Assume light rays traveling in air are nearly normal to the two surfaces
of the film as shown in Fig. 1.7. If  is the wavelength of the light in free space and n
is the index of refraction of the film material, then the wavelength of light n in the
film is n 

n
.
7
Figure 1.7 Light paths through a thin film.
Reflected ray 1, which is reflected from the upper surface (A) in Fig. 1.7, undergoes a
phase change of 180° with respect to the incident wave. Reflected ray 2, which is
reflected from the lower film surface (B), undergoes no phase change because it is
reflected from a medium (air) that has a lower index of refraction. Therefore, ray 1
is 180° out of phase with ray 2, which is equivalent to a path difference of n/2. We
must also consider that ray 2 travels an extra distance 2t before the waves recombine
in the air above surface A. (Remember that we are considering light rays that are
close to normal to the surface. If the rays are not close to normal, the path difference
is larger than 2t). If 2t = n/2, rays 1 and 2 recombine in phase and the result is
constructive interference. In general, the condition for constructive interference in
thin film is,
1

2t   m   n
2

(m  0, 1, 2, ...)
(1.17)
(m  0, 1, 2, ...)
(1.18)
Or,
1

2nt   m   
2

If the extra distance 2t traveled by ray 2 corresponds to a multiple of n the two
waves combine out of phase and the result is destructive interference. The general
equation for destructive interference in thin films is
2nt  m
(m  0, 1, 2, ...)
(1.19)
1.6 NEWTON’S RINGS
When a plano-convex lens is placed on top of a flat glass surface as shown in Fig.
1.8a, interference fringes are formed, and these fringes can be seen under the
8
traveling microscope. With this arrangement, the air film between the glass surfaces
varies in thickness from zero at the point of contact to some value t at point P. If the
radius of curvature R of the lens is much greater than the distance r and the system
is viewed from above, a pattern of light and dark rings is observed as shown in Fig.
1.8b. These circular fringes, discovered by Newton, are called Newton’s rings.
Figure 1.8 (a) The combination of rays reflected from the flat plate and the curved lens surface gives
rise to an interference pattern known as Newton’s rings. (b) Photograph of Newton’s rings.
Expressions for radii of the bright and dark rings:
Using the geometry shown in Fig. 1.8a, we can obtain expressions for the radii of the
bright and dark rings in terms of the radius of curvature R and wavelength . For
the thin air film trapped between the two glass surfaces as shown in the figure above,
the conditions for constructive (bright rings) and destructive (dark rings)
interference are given by equations (1.18) and (1.19).
Consider the dark rings (destructive interference)
2nt

m ,
For air film, n  1 ,
m  0, 1, 2, 3...

2t  m
From the above figure, t  R  R 2  r 2
  r 2 
t  R  R 1    
  R  
12
Binomial theorem is, 1  y   1  ny 
n
n(n  1) 2
y  .......
2!
If r / R  1, using binomial theorem and neglecting higher order terms,
9
 1  r 2
 r2
t  R  R 1     ........ 
 2  R 
 2 R
(1.20)
Substituting the value of t from equation (1.20) into equation (1.19), we get
rdark  mR
(m  0, 1, 2, ...)
(1.21)
In general, for any thin film of refractive index n film , the expression for the radii of
the dark rings is given by rdark 
mR
n film
(m  0, 1, 2, ...)
(1.22)
Similarly, the expression for the radii of the bright rings is given by,
1

 m   R
2
(1.23)
rbright  
(m  0, 1, 2, ...)
n film
That is, the diameters of Newton’s dark rings are proportional to square root of the
natural numbers and the diameters of Newton’s bright rings are proportional to
square root of natural odd numbers.
1.7 MICHELSON INTERFEROMETER
The interferometer, invented by American physicist A. A. Michelson (1852–1931),
splits a light beam into two parts and then recombines the parts to form an
interference pattern. A schematic diagram of the interferometer is shown in Fig. 1.9.
A ray of light from a monochromatic source is split into two rays by mirror M0, which
is inclined at 45° to the incident light beam. Mirror M 0, called a beam splitter,
transmits half the light incident on it and reflects the rest. One ray is reflected from
M0 to the right toward mirror M1, and the second ray is transmitted vertically
through M0 toward mirror M2. Hence, the two rays travel separate paths L1 and L2.
After reflecting from M1 and M2, the two rays eventually recombine at M0 to produce
an interference pattern, which can be viewed through a telescope.
The interference condition for the two rays is determined by the difference in their
path length. When the two mirrors are exactly perpendicular to each other, the
interference pattern is a target pattern of bright and dark circular fringes. As M1 is
moved, the fringe pattern collapses or expands, depending on the direction in which
M1 is moved. For example, if a dark circle appears at the center of the target pattern
(corresponding to destructive interference) and M1 is then moved a distance /4
toward M0, the path difference changes by /2. This replaces dark circle at center by
bright circle. Therefore, the fringe pattern shifts by one-half fringe each time M1 is
moved a distance /4. The wavelength of light is then measured by counting the
number of fringe shifts for a given displacement of M1. So it can also be used to
10
detect small change in path length as in the laser interferometer gravitational-wave
observatory.
Figure 1.9 Schematic diagram of Michelson Interferometer
1.8 QUESTIONS
1. What is interference of light waves?
2. What is coherence? Mention its importance.
3. Write the necessary condition for the constructive and destructive
interference of two light waves in terms of path/phase difference.
4. Obtain an expression for intensity of light in double-slit interference.
5. Write the conditions for constructive and destructive interference of
reflected light from a thin soap film in air, assuming normal incidence.
6. Explain the formation of fringes in Michelson interferometer.
1.9 PROBLEMS
1. A viewing screen is separated from a double slit by 4.80 m. The distance
between the two slits is 0.0300 mm. Monochromatic light is directed toward
the double slit and forms an interference pattern on the screen. The first dark
fringe is 4.50 cm from the center line on the screen. (A) Determine the
11
2.
3.
4.
5.
6.
wavelength of the light. (B) Calculate the distance between adjacent bright
fringes. Ans: 562 nm and 9 cm
A light source emits visible light of two wavelengths:  = 430 nm and / = 510
nm. The source is used in a double-slit interference experiment in which L =
1.50 m and d = 0.025 0 mm. Find the separation distance between the thirdorder bright fringes for the two wavelengths. Ans: 1.81 cm
A laser beam ( = 632.8 nm) is incident on two slits 0.200 mm apart. How far
apart are the bright interference fringes on a screen 5.00 m away from the
double slits? Ans: 15.8 mm
A Young’s interference experiment is performed with monochromatic light.
The separation between the slits is 0.500 mm, and the interference pattern
on a screen 3.30 m away shows the first side maximum 3.40 mm from the
center of the pattern. What is the wavelength? Ans: 515 nm
Young’s double-slit experiment is performed with 589-nm light and a
distance of 2.00 m between the slits and the screen. The tenth interference
minimum is observed 7.26 mm from the central maximum. Determine the
spacing of the slits. Ans: 1.54 mm
Two radio antennas separated by d = 300 m as shown in figure simultaneously
broadcast identical signals at the same wavelength. A car travels due north
along a straight line at position x = 1000 m from the center point between
the antennas, and its radio receives the signals. (a) If the car is at the position
of the second maximum after that at point O when it has traveled a distance
y = 400 m northward, what is the wavelength of the signals? (b) How much
farther must the car travel from this position to encounter the next minimum
in reception? Note: Do not use the small-angle approximation in this
problem. Ans: 55.7 m and 124 m
7. Two narrow parallel slits separated by 0.250 mm are illuminated by green
light ( = 546.1 nm). The interference pattern is observed on a screen 1.20 m
away from the plane of the slits. Calculate the distance (a) from the central
maximum to the first bright region on either side of the central maximum
and (b) between the first and second dark bands. Ans: 2.62 mm and 2.62 mm
12
8. In a Young’s interference experiment, the two slits are separated by 0.150 mm
and the incident light includes two wavelengths: 1 = 540 nm (green) and
2 = 450 nm (blue). The overlapping interference patterns are observed on a
screen 1.40 m from the slits. Calculate the minimum distance from the center
of the screen to a point where a bright fringe of the green light coincides with
a bright fringe of the blue light. Ans: 2.5 cm
9. In a double slit experiment, let L = 120 cm and d = 0.250 cm. The slits are
illuminated with coherent 600-nm light. Calculate the distance y above the
central maximum for which the average intensity on the screen is 75.0% of
the maximum. Ans: 48 micrometer
10. Show that the two waves with wave functions E1  6.00 sin(100t ) and
E2  8.00 sin(100t   / 2) add to give a wave with the wave function
E R sin(100t   ) . Find the required values for ER and  . Ans: 10 and 53.1
11. Calculate the minimum thickness of a soap-bubble film that results in
constructive interference in the reflected light if the film is illuminated with
light whose wavelength in free space is  = 600 nm. The index of refraction
of the soap film is 1.33. (b) What if the film is twice as thick? Does this
situation produce constructive interference? Ans: 113 nm and No
12. Solar cells—devices that generate electricity when exposed to sunlight—are
often coated with a transparent, thin film of silicon monoxide (SiO, n = 1.45)
to minimize reflective losses from the surface. Suppose a silicon solar cell (n
= 3.5) is coated with a thin film of silicon monoxide for this purpose.
Determine the minimum film thickness that produces the least reflection at
a wavelength of 550 nm, near the center of the visible spectrum. Ans: 95 nm
13. A thin film of oil (n = 1.25) is located on smooth, wet pavement. When viewed
perpendicular to the pavement, the film reflects most strongly red light at
640 nm and reflects no green light at 512 nm. How thick is the oil film? Ans:
512 nm.
14. An oil film (n = 1.45) floating on water is illuminated by white light at normal
incidence. The film is 280 nm thick. Find (a) the wavelength and color of the
light in the visible spectrum most strongly reflected and (b) the wavelength
and color of the light in the spectrum most strongly transmitted. Explain your
reasoning. Ans: 541 nm and 406 nm
15. An air wedge is formed between two glass plates separated at one edge by a
very fine wire of circular cross section as shown in Figure 12. When the wedge
is illuminated from above by 600-nm light and viewed from above, 30 dark
fringes are observed. Calculate the diameter d of the wire. Ans: 8.7
micrometer
13
16. When a liquid is introduced into the air space between the lens and the plate
in a Newton’s-rings apparatus, the diameter of the tenth ring changes from
1.50 to 1.31 cm. Find the index of refraction of the liquid. Ans: 1.31
17. A certain grade of crude oil has an index of refraction of 1.25. A ship
accidentally spills 1.00 m3 of this oil into the ocean, and the oil spreads into a
thin, uniform slick. If the film produces a first-order maximum of light of
wavelength 500 nm normally incident on it, how much surface area of the
ocean does the oil slick cover? Assume the index of refraction of the ocean
water is 1.34. Ans: 5km2
18. In a Newton’s-rings experiment, a plano-convex glass (n = 1.52) lens having
radius r = 5.00 cm is placed on a flat plate as shown in Figure 1.8a. When light
of wavelength 650 nm is incident normally, 55 bright rings are observed, with
the last one precisely on the edge of the lens. (a) What is the radius R of
curvature of the convex surface of the lens? (b) What is the focal length of
the lens? Ans: 70.5 m and 138 m
19. Monochromatic light is beamed into a Michelson interferometer. The
movable mirror is displaced 0.382 mm, causing the interferometer pattern to
reproduce itself 1700 times. Determine the wavelength of the light. What
color is it? Ans: 449 nm, Blue
20. Mirror M1 in Figure 1.9 is moved through a displacement L. During this
displacement, 250 fringe reversals (formation of successive dark or bright
bands) are counted. The light being used has a wavelength of 632.8 nm.
Calculate the displacement L. Ans: 39.6 micrometer
21. One leg of a Michelson interferometer contains an evacuated cylinder of
length L, having glass plates on each end. A gas is slowly leaked into the
cylinder until a pressure of 1 atm is reached. If N bright fringes pass on the
screen during this process when light of wavelength  is used, what is the
index of refraction of the gas? Ans: n = 1 + (Nλ)/(2L)
14
2 DIFFRACTION PATTERNS AND
POLARIZATION
OBJECTIVES
•
•
•
•
To understand the principles of diffraction.
To explain the intensity distribution in diffraction under various
conditions.
To explain the diffraction of light waves at single, multiple slits and
circular apertures.
To understand polarization phenomena and various techniques used to
produce polarized light.
2.1 INTRODUCTION TO DIFFRACTION PATTERNS
Light of wavelength comparable to or larger than the width of a slit spreads out in
all forward directions upon passing through the slit. This phenomenon is called
diffraction. When light passes through a narrow slit, it spreads beyond the narrow
path defined by the slit into regions that would be in shadow if light traveled in
straight lines. Other waves, such as sound waves and water waves, also have this
property of spreading when passing through apertures or by sharp edges.
A diffraction pattern consisting of light and dark areas is observed when a narrow
slit is placed between a distant light source (or a laser beam) and a screen, the light
produces a diffraction pattern like that shown in Figure 2.1 (a). The pattern consists
of a broad, intense central band (called the central maximum) flanked by a series
of narrower, less intense additional bands (called side maxima or secondary
maxima) and a series of intervening dark bands (or minima).
Figure 2.1 (a) The diffraction pattern that appears on a screen when light passes through a narrow vertical
slit. (b) Diffraction pattern created by the illumination of a penny, with the penny positioned midway
between the screen and light source.
15
Figure 2.1 (b) shows a diffraction pattern associated with the shadow of a penny. A
bright spot occurs at the center, and circular fringes extend outward from the
shadow’s edge. From the viewpoint of ray optics (in which light is viewed as rays
traveling in straight lines), we expect the center of the shadow to be dark because
that part of the viewing screen is completely shielded by the penny. We can explain
the central bright spot by using the wave theory of light, which predicts constructive
interference at this point.
2.2 DIFFRACTION PATTERNS FROM NARROW SLITS
Let’s consider light passing through a narrow opening modeled as a slit and
projected onto a screen. To simplify our analysis, we assume the observing screen is
far from the slit and the rays reaching the screen are approximately parallel. In
laboratory, this situation can also be achieved experimentally by using a converging
lens to focus the parallel rays on a nearby screen. In this model, the pattern on the
screen is called a Fraunhofer diffraction pattern.
Until now, we have assumed slits are point sources of light. In this section, we
abandon that assumption and see how the finite width of slits is the basis for
understanding Fraunhofer diffraction. We can explain some important features of
this phenomenon by examining waves coming from various portions of the slit as
shown in Figure 2.2.
According to Huygens’s principle, each portion of the slit acts as a source of light
waves. Hence, light from one portion of the slit can interfere with light from another
portion, and the resultant light intensity on a viewing screen depends on the
direction . Based on this analysis, we recognize that a diffraction pattern is an
interference pattern in which the different sources of light are different portions of
the single slit.
Figure 2.2(a) Geometry for analyzing the Fraunhofer diffraction pattern of a single slit. (b) Photograph of a
single-slit Fraunhofer diffraction pattern.
16
Figure 2.3 Paths of light rays that encounter a narrow slit of width a and diffract toward a screen in the
direction described by angle .
To analyze the diffraction pattern, let’s divide the slit into two halves as shown in
Figure 2.3. Keeping in mind that all the waves are in phase as they leave the slit,
consider rays 1 and 3. As these two rays travel toward a viewing screen far to the
right of the figure, ray 1 travels farther than ray 3 by an amount equal to the path
difference (a/2) sin , where a is the width of the slit. Similarly, the path difference
between rays 2 and 4 is also (a/2) sin , as is that between rays 3 and 5. If this path
difference is exactly half a wavelength (corresponding to a phase difference of 180°),
the pairs of waves cancel each other and destructive interference results.
This cancellation occurs for any two rays that originate at points separated by half
the slit width because the phase difference between two such points is 180°.
Therefore, waves from the upper half of the slit interfere destructively with waves
from the lower half when
𝑎

sin  = ±
2
2
Dividing the slit into four equal parts and using similar reasoning, we find that the
viewing screen is also dark when
sin  = ± 2

𝑎
Likewise, dividing the slit into six equal parts shows that darkness occurs on the
screen when
sin  = ± 3

𝑎
Therefore, the general condition for destructive interference is
17
sin dark  m

a
m  1,  2,  3, ...
(2.1)
2.3 INTENSITY OF SINGLE-SLIT DIFFRACTION PATTERNS
Analysis of the intensity variation in a diffraction pattern from a single slit of width
‘a’ shows that the intensity is given by
I  I max
 sin  a sin  /   


  a sin  / 

2
(2.2)
where Imax is the intensity at  = 0 (the central maximum) and  is the wavelength
of light used to illuminate the slit. Intensity variation plot and photograph of the
pattern are shown below.
Figure 2.4 A plot of light intensity I versus (/)a sin  for the single-slit Fraunhofer diffraction
pattern. (b) Photograph of a single slit Fraunhofer diffraction pattern.
2.4 INTENSITY OF TWO-SLIT DIFFRACTION PATTERNS
When more than one slit is present, we must consider not only diffraction patterns
due to the individual slits but also the interference patterns due to the waves coming
from different slits. Intensity due to combined effect is given by
  d sin 
I  I max cos 


2
  sin  a sin  /   


   a sin  / 

2
(2.3)
18
Above equation represents the single-slit diffraction pattern (the factor in square
brackets) acting as an “envelope” for a two slit interference pattern (the cosinesquared factor).
We have seen that angular position of interference maxima is given by d sin  = m,
where d is the distance between the two slits. Also, the first diffraction minimum
occurs when a sin  = , where a is the slit width. Dividing interference equation by
diffraction equation,
𝑑
=𝑚
𝑎
In this case, mth interference maximum coincides with first diffraction minimum.
Figure 2.5 The combined effects of two-slit and single-slit interference.
2.5 RESOLUTION OF SINGLE-SLIT AND CIRCULAR
APERTURES
The ability of optical systems to distinguish between closely spaced objects is limited
because of the wave nature of light. To understand this limitation, consider Figure
2.6, which shows two light sources far from a narrow slit of width a. The sources can
be two noncoherent point sources S1 and S2; for example, they could be two distant
stars. If no interference occurred between light passing through different parts of
the slit, two distinct bright spots (or images) would be observed on the viewing
screen. Because of such interference, however, each source is imaged as a bright
central region flanked by weaker bright and dark fringes, a diffraction pattern. What
is observed on the screen is the sum of two diffraction patterns: one from S1 and the
other from S2.
19
Figure 2.6 Two-point sources far from a narrow slit each produce a diffraction pattern. (a) The
sources are separated by a large angle. (b) The sources are separated by a small angle.
When the central maximum of one image falls on the first minimum of another
image, the images are said to be just resolved. This limiting condition of resolution
is known as Rayleigh’s criterion.
From Rayleigh’s criterion, we can determine the minimum angular separation min
subtended by the sources at the slit in Figure 2.6 for which the images are just
resolved. Equation 2.1 indicates that the first minimum (m = 1) in a single-slit
diffraction pattern occurs at the angle for which

sin  = 𝑎
(2.4)
where a is the width of the slit. According to Rayleigh’s criterion, this expression
gives the smallest angular separation for which the two images are resolved. Because
 << a in most situations, sin  is small and we can use the approximation sin   .
Therefore, the limiting angle of resolution for a slit of width a is
𝑚𝑖𝑛 =

𝑎
(2.5)
where min is expressed in radians. Hence, the angle subtended by the two sources
at the slit must be greater than /a if the images are to be resolved.
Many optical systems use circular apertures rather than slits. The diffraction pattern
of a circular aperture as shown in the photographs of Figure 2.7 consists of a central
circular bright disk surrounded by progressively fainter bright and dark rings. Figure
2.7 shows diffraction patterns for three situations in which light from two point
sources passes through a circular aperture. When the sources are far apart, their
images are well resolved (Fig. 2.7a). When the angular separation of the sources
satisfies Rayleigh’s criterion, the images are just resolved (Fig. 2.7b). Finally, when
the sources are close together, the images are said to be unresolved (Fig. 2.7c) and
the pattern looks like that of a single source. Analysis shows that the limiting angle
of resolution of the circular aperture is
20
𝑚𝑖𝑛 = 1.22

𝐷
(2.6)
where D is the diameter of the aperture. This expression is similar to Equation 2.4
except for the factor 1.22, which arises from a mathematical analysis of diffraction
from the circular aperture.
Figure 2.7 Individual diffraction patterns of two-point sources (solid curves) and the resultant
patterns (dashed curves) for various angular separations of the sources as the light passes through a
circular aperture. In each case, the dashed curve is the sum of the two solid curves.
2.6 DIFFRACTION GRATING
The diffraction grating, a useful device for analyzing light sources, consists of
many equally spaced parallel slits. A transmission grating can be made by cutting
parallel grooves on a glass plate with a precision ruling machine. The spaces between
the grooves are transparent to the light and hence act as separate slits. A reflection
grating can be made by cutting parallel grooves on the surface of a reflective
material. The reflection of light from the spaces between the grooves is specular,
and the reflection from the grooves cut into the material is diffuse. Therefore, the
spaces between the grooves act as parallel sources of reflected light like the slits in
a transmission grating.
21
Figure 2.8 Side view of a diffraction grating. The slit separation is d, and the path difference between
adjacent slits is d sin.
A plane wave is incident from the left, normal to the plane of the grating. The
pattern observed on the screen far to the right of the grating is the result of the
combined effects of interference and diffraction. Each slit produces diffraction, and
the diffracted beams interfere with one another to produce the final pattern. The
waves from all slits are in phase as they leave the slits. For an arbitrary direction 
measured from the horizontal, however, the waves must travel different path
lengths before reaching the screen. Notice in Figure 2.8 that the path difference 
between rays from any two adjacent slits is equal to d sin . If this path difference
equals one wavelength or any integral multiple of a wavelength, waves from all slits
are in phase at the screen and a bright fringe is observed. Therefore, the condition
for maxima in the interference pattern at the angle bright is
d sin bright  m
m  0,  1,  2,  3, ...
(2.7)
22
Figure 2.9 Intensity versus sin  for a diffraction grating. The zeroth-, first-, and
second-order maxima are shown.
The intensity distribution for a diffraction grating obtained with the use of a
monochromatic source is shown in Figure 2.9. Notice the sharpness of the principal
maxima and the broadness of the dark areas compared with the broad bright fringes
characteristic of the two-slit interference pattern.
Figure 2.10 Diagram of a diffraction grating spectrometer.
A schematic drawing of a simple apparatus used to measure angles in a diffraction
pattern is shown in Figure 2.10. This apparatus is a diffraction grating spectrometer.
The light to be analyzed passes through a slit, and a collimated beam of light is
incident on the grating. The diffracted light leaves the grating at angles that satisfy
Equation 2.7, and a telescope is used to view the image of the slit. The wavelength
can be determined by measuring the precise angles at which the images of the slit
appear for the various orders. The spectrometer is a useful tool in atomic
spectroscopy, in which the light from an atom is analyzed to find the wavelength
components. These wavelength components can be used to identify the atom.
23
2.7 DIFFRACTION OF X-RAYS BY CRYSTALS
In principle, the wavelength of any electromagnetic wave can be determined if a
grating of the proper spacing (on the order of ) is available. X-rays, discovered by
Wilhelm Roentgen (1845–1923) in 1895, are electromagnetic waves of very short
wavelength (on the order of 0.1 nm). It would be impossible to construct a grating
having such a small spacing by the cutting process. The atomic spacing in a solid is
known to be about 0.1 nm, however. In 1913, Max von Laue (1879–1960) suggested
that the regular array of atoms in a crystal could act as a three-dimensional
diffraction grating for x-rays. Subsequent experiments confirmed this prediction.
The diffraction patterns from crystals are complex because of the three-dimensional
nature of the crystal structure. Nevertheless, x-ray diffraction has proved to be an
invaluable technique for elucidating these structures and for understanding the
structure of matter.
Figure 2.11 Crystalline structure of sodium chloride (NaCl).
The arrangement of atoms in a crystal of sodium chloride (NaCl) is shown in Figure
2.11. Each unit cell (the geometric solid that repeats throughout the crystal) is a cube
having an edge length a. A careful examination of the NaCl structure shows that the
ions lie in discrete planes (the shaded areas in Fig. 2.11). Now suppose an incident xray beam makes an angle  with one of the planes as in Figure 2.12. The beam can
be reflected from both the upper plane and the lower one, but the beam reflected
from the lower plane travels farther than the beam reflected from the upper plane.
The effective path difference is 2dsin. The two beams reinforce each other
(constructive interference) when this path difference equals some integer multiple
of . The same is true for reflection from the entire family of parallel planes. Hence,
the condition for constructive interference (maxima in the reflected beam) is
2d sin   m
m  1, 2, 3, ...
(2.8)
This condition is known as Bragg’s law, after W. L. Bragg, who first derived the
relationship. If the wavelength and diffraction angle are measured, Equation 2.8 can
be used to calculate the spacing between atomic planes.
24
Figure 2.12 A two-dimensional description of the reflection of an x-ray beam from two parallel
crystalline planes separated by a distance d.
2.8 POLARIZATION OF LIGHT WAVES
An ordinary beam of light consists of many waves emitted by the atoms of the light
source. Each atom produces a wave having some orientation of the electric field
vector ⃗𝑬, corresponding to the direction of atomic vibration. The direction of
polarization of each individual wave is defined to be the direction in which the
electric field is vibrating. In Figure 2.13, this direction happens to lie along the y axis.
⃗⃗ vector
All individual electromagnetic waves traveling in the x direction have an 𝑬
parallel to the yz plane, but this vector could be at any possible angle with respect
to the y axis. Because all directions of vibration from a wave source are possible, the
resultant electromagnetic wave is a superposition of waves vibrating in many
different directions. The result is an unpolarized light beam, represented in Figure
2.14a. The direction of wave propagation in this figure is perpendicular to the page.
The arrows show a few possible directions of the electric field vectors for the
individual waves making up the resultant beam. At any given point and at some
instant of time, all these individual electric field vectors add to give one resultant
electric field vector.
⃗ vibrates in
A wave is said to be linearly polarized if the resultant electric field ⃗𝑬
the same direction at all times at a particular point as shown in Figure 2.14b.
(Sometimes, such a wave is described as plane-polarized, or simply polarized.) The
plane formed by ⃗𝑬 and the direction of propagation is called the plane of polarization
of the wave. If the wave in Figure 2.14b represents the resultant of all individual
waves, the plane of polarization is the xy plane. A linearly polarized beam can be
obtained from an unpolarized beam by removing all waves from the beam except
those whose electric field vectors oscillate in a single plane.
25
Figure 2.13 Schematic diagram of an electromagnetic wave propagating at velocity c in the x direction.
The electric field vibrates in the xy plane, and the magnetic field vibrates in the xz plane.
Figure 2.14 (a) A representation of an unpolarized light beam viewed along the direction of
propagation. The transverse electric field can vibrate in any direction in the plane of the page with
equal probability. (b) A linearly polarized light beam with the electric field vibrating in the vertical
direction.
2.9 POLARIZATION BY SELECTIVE ABSORPTION
The most common technique for producing polarized light is to use a material that
transmits waves whose electric fields vibrate in a plane parallel to a certain direction
and that absorbs waves whose electric fields vibrate in all other directions. Polaroid,
that polarizes light through selective absorption. This material is fabricated in thin
sheets of long-chain hydrocarbons. The sheets are stretched during manufacture so
that the long-chain molecules align. After a sheet is dipped into a solution
containing iodine, the molecules become good electrical conductors. Conduction
takes place primarily along the hydrocarbon chains because electrons can move
easily only along the chains.
If light whose electric field vector is parallel to the chains is incident on the material,
the electric field accelerates electrons along the chains and energy is absorbed from
26
the radiation. Therefore, the light does not pass through the material. Light whose
electric field vector is perpendicular to the chains passes through the material
because electrons cannot move from one molecule to the next. As a result, when
unpolarized light is incident on the material, the exiting light is polarized
perpendicular to the molecular chains. It is common to refer to the direction
perpendicular to the molecular chains as the transmission axis. In an ideal polarizer,
⃗ parallel to the transmission axis is transmitted and all light with 𝑬
⃗
all light with 𝑬
perpendicular to the transmission axis is absorbed.
Figure 2.15 Two polarizing sheets whose transmission axes make an angle  with each other. Only a
fraction of the polarized light incident on the analyzer is transmitted through it.
Figure 2.15 represents an unpolarized light beam incident on a first polarizing sheet,
called the polarizer. Because the transmission axis is oriented vertically in the figure,
the light transmitted through this sheet is polarized vertically. A second polarizing
sheet, called the analyzer, intercepts the beam. In figure, the analyzer transmission
axis is set at an angle  to the polarizer axis. We call the electric field vector of the
⃗ 𝟎 . The component of 𝑬
⃗⃗ 𝟎 perpendicular to the analyzer axis
first transmitted beam 𝑬
is completely absorbed. The component of ⃗𝑬𝟎 parallel to the analyzer axis, which is
transmitted through the analyzer, is E0 cos . Because the intensity of the
transmitted beam varies as the square of its magnitude, we conclude that the
intensity I of the (polarized) beam transmitted through the analyzer varies as
I  I max cos2 
(2.9)
where Imax is the intensity of the polarized beam incident on the analyzer. This
expression, known as Malus’s law.
2.10 POLARIZATION BY REFLECTION
When an unpolarized light beam is reflected from a surface, the polarization of the
reflected light depends on the angle of incidence. If the angle of incidence is 0°, the
27
reflected beam is unpolarized. For other angles of incidence, the reflected light is
polarized to some extent, and for a particular angle of incidence, the reflected light
is completely polarized.
Figure 2.16 (a) When unpolarized light is incident on a reflecting surface, the reflected and refracted
beams are partially polarized. (b) The reflected beam is completely polarized when the angle of
incidence equals the polarizing angle p, which satisfies the equation n2/n1 = tan p. At this incident
angle, the reflected and refracted rays are perpendicular to each other.
Now suppose the angle of incidence 1 is varied until the angle between the reflected
and refracted beams is 90° as in Figure 2.16b. At this angle of incidence, the reflected
beam is completely polarized (with its electric field vector parallel to the surface)
and the refracted beam is still only partially polarized. The angle of incidence at
which this polarization occurs is called the polarizing angle p. Using Snell’s law
of refraction
𝑛2
𝑛1
=
sin 𝑝
sin 2
2.10
But, 2 = 90 - p. So, we can write,
tan 𝑝 =
𝑛2
𝑛1
2.11
This expression is called Brewster’s law, and the polarizing angle p is sometimes
called Brewster’s angle, after its discoverer, David Brewster. Because n varies with
wavelength for a given substance, Brewster’s angle is also a function of wavelength.
28
2.11 POLARIZATION BY DOUBLE REFRACTION
In certain class of crystals like calcite and quartz, the speed of light depends on the
direction of propagation and on the plane of polarization of the light. Such materials
are characterized by two indices of refraction. Hence, they are often referred to as
double-refracting or birefringent materials. When unpolarized light enters a
birefringent material, it may split into an ordinary (O) ray and an extraordinary
(E) ray. These two rays have mutually perpendicular polarizations and travel at
different speeds through the material. There is one direction, called the optic axis,
along which the ordinary and extraordinary rays have the same speed.
Figure 2.17 Unpolarized light incident at an angle to the optic axis in a calcite crystal splits into an
ordinary (O) ray and an extraordinary (E) ray
Figure 2.18 Point source S inside a double-refracting crystal (calcite) produces a spherical wave front
corresponding to the ordinary (O) ray and an elliptical wave front corresponding to the extraordinary
(E) ray.
Some materials such as glass and plastic become birefringent when stressed.
Suppose an unstressed piece of plastic is placed between a polarizer and an analyzer
so that light passes from polarizer to plastic to analyzer. When the plastic is
unstressed, and the analyzer axis is perpendicular to the polarizer axis, none of the
polarized light passes through the analyzer. In other words, the unstressed plastic
has no effect on the light passing through it. If the plastic is stressed, however,
29
regions of greatest stress become birefringent and the polarization of the light
passing through the plastic changes. Hence, a series of bright and dark bands is
observed in the transmitted light, with the bright bands corresponding to regions of
greatest stress. Engineers often use this technique, called optical stress analysis, in
designing structures ranging from bridges to small tools. They build a plastic model
and analyze it under different load conditions to determine regions of potential
weakness and failure under stress.
Figure 2.19 The pattern is produced when the plastic model is viewed between a polarizer and analyzer
oriented perpendicular to each other. Such patterns are useful in the optimal design of architectural
components
2.12 POLARIZATION BY SCATTERING
Figure 2.20 The scattering of unpolarized sunlight by air molecules.
30
When light is incident on any material, the electrons in the material can absorb and
reradiate part of the light. Such absorption and reradiation of light by electrons in
the gas molecules that make up air is what causes sunlight reaching an observer on
the Earth to be partially polarized. An unpolarized beam of sunlight traveling in the
horizontal direction (parallel to the ground) strikes a molecule of one of the gases
that make up air, setting the electrons of the molecule into vibration. These
vibrating charges act like the vibrating charges in an antenna. The horizontal
component of the electric field vector in the incident wave results in a horizontal
component of the vibration of the charges, and the vertical component of the vector
results in a vertical component of vibration. If the observer in Figure 2.20 is looking
straight up (perpendicular to the original direction of propagation of the light), the
vertical oscillations of the charges send no radiation toward the observer. Therefore,
the observer sees light that is completely polarized in the horizontal direction as
indicated by the orange arrows. If the observer looks in other directions, the light is
partially polarized in the horizontal direction.
2.13 OPTICAL ACTIVITY
Many important applications of polarized light involve materials that display optical
activity. A material is said to be optically active if it rotates the plane of polarization
of any light transmitted through the material. The angle through which the light is
rotated by a specific material depends on the length of the path through the material
and on concentration if the material is in solution. One optically active material is a
solution of the common sugar dextrose. A standard method for determining the
concentration of sugar solutions is to measure the rotation produced by a fixed
length of the solution.
2.14 QUESTIONS
1. Explain the term diffraction of light.
2. Discuss qualitatively, the Fraunhofer diffraction at a single-slit.
3. Draw a schematic plot of the intensity of light in single slit diffraction against
phase difference.
4. Explain briefly diffraction at a circular aperture.
5. State and explain Rayleigh’s criterion for optical resolution.
6. Effect of diffraction is ignored in the case of Young’s double slit interference.
Give reason.
7. Discuss qualitatively, the diffraction due to multiple slits.
8. What is diffraction grating? Write the grating equation.
9. Briefly explain x-ray diffraction and Bragg’s law.
10. Distinguish between unpolarized and linearly polarized light.
11. Explain Malus’s law.
12. How to produce linearly polarized light by (a) selective absorption, (b)
reflection, (c) double refraction, (d) scattering ? Explain.
31
2.15 PROBLEMS
1. Light of wavelength 540 nm passes through a slit of width 0.200 mm. (a) The
width of the central maximum on a screen is 8.10 mm. How far is the screen
from the slit? (b) Determine the width of the first bright fringe to the side of
the central maximum. Ans: (a) 1.5 m (b) 4.05 mm
2. Helium–neon laser light ( = 632.8 nm) is sent through a 0.300-mm-wide
single slit. What is the width of the central maximum on a screen 1.00 m from
the slit? Ans: 4.22 mm
3. A screen is placed 50.0 cm from a single slit, which is illuminated with light
of wavelength 690 nm. If the distance between the first and third minima in
the diffraction pattern is 3.00 mm, what is the width of the slit?
Ans: 2.3x10-4 m
4. A beam of monochromatic light is incident on a single slit of width 0.600
mm. A diffraction pattern forms on a wall 1.30 m beyond the slit. The distance
between the positions of zero intensity on both sides of the central maximum
is 2.00 mm. Calculate the wavelength of the light. Ans: 462 nm
5. A diffraction pattern is formed on a screen 120 cm away from a 0.400-mmwide slit. Monochromatic 546.1-nm light is used. Calculate the fractional
intensity I/Imax at a point on the screen 4.10 mm from the center of the
principal maximum. Ans: 0.0162
6. Yellow light of wavelength 589 nm is used to view an object under a
microscope. The objective lens diameter is 9.00 mm. (a) What is the limiting
angle of resolution? (b) Suppose it is possible to use visible light of any
wavelength. What color should you choose to give the smallest possible angle
of resolution, and what is this angle? (c) Suppose water fills the space
between the object and the objective. What effect does this change have on
the resolving power when 589-nm light is used? Ans: (a) 79.8 x 10-6 rad (b)
400nm, 54.2 x 10-6 rad (c) Resolving power will improve with minimum resolvable
angle 60 x 10-6 rad
7. The angular resolution of a radio telescope is to be 0.100° when the incident
waves have a wavelength of 3.00 mm. What minimum diameter is required
for the telescope’s receiving dish? Ans: 2.1 mm
8. White light is spread out into its spectral components by a diffraction grating.
If the grating has 2000 grooves per centimeter, at what angle does red light
of wavelength 640 nm appear in first order? Ans: θ = 7.35o
9. Light of wavelength 500 nm is incident normally on a diffraction grating. If
the third-order maximum of the diffraction pattern is observed at 32.0°, (a)
what is the number of rulings per centimeter for the grating? (b) Determine
the total number of primary maxima that can be observed in this situation.
Ans: 3530 rulings/cm (b) 11
32
10. If the spacing between planes of atoms in a NaCl crystal is 0.281 nm, what is
the predicted angle at which 0.140-nm x-rays are diffracted in a first-order
maximum? Ans: θ = 14.4o
11. The first-order diffraction maximum is observed at 12.6° for a crystal having
a spacing between planes of atoms of 0.250 nm. (a) What wavelength x-ray is
used to observe this first-order pattern? (b) How many orders can be
observed for this crystal at this wavelength? Ans: (a) 0.109 nm (b) 4
12. Plane-polarized light is incident on a single polarizing disk with the direction
of E parallel to the direction of the transmission axis. Through what angle
should the disk be rotated so that the intensity in the transmitted beam is
reduced by a factor of (a) 3.00, (b) 5.00, and (c) 10.0? Ans: (a) 54.70 (b) 63.40
(c) 71.60
13. Unpolarized light passes through two ideal Polaroid sheets. The axis of the
first is vertical, and the axis of the second is at 30.0° to the vertical. What
fraction of the incident light is transmitted? Ans: 0.375
14. The angle of incidence of a light beam onto a reflecting surface is continuously
variable. The reflected ray in air is completely polarized when the angle of
incidence is 48.0°. What is the index of refraction of the reflecting material?
Ans: 1.1
15. The critical angle for total internal reflection for sapphire surrounded by air is
34.4°. Calculate the polarizing angle for sapphire. Ans: 60.5o
33
3 QUANTUM PHYSICS
OBJECTIVES:
 To learn certain experimental results that can be understood only by
particle theory of electromagnetic waves.
 To learn the particle properties of waves and the wave properties of the
particles.
 To understand the uncertainty principle.
3.1 BLACKBODY RADIATION AND PLANCK’S
HYPOTHESIS
A black body is an object that absorbs all incident radiation. A small hole cut into a
cavity is the most popular and realistic example. None of the incident radiation
escapes. The radiation is absorbed in the walls of the cavity. This causes a heating of
the cavity walls. The oscillators in the cavity walls vibrate and re-radiate at
wavelengths corresponding to the temperature of the cavity, thereby producing
standing waves. Some of the energy from these standing waves can leave through
the opening. The electromagnetic radiation emitted by the black body is called
black-body radiation.
Figure 3.1 A physical model of a blackbody
•
•
•
The black body is an ideal absorber of incident radiation.
A black-body reaches thermal equilibrium with the surroundings when the
incident radiation power is balanced by the power re-radiated.
The emitted "thermal" radiation from a black body characterizes the equilibrium
temperature of the black-body.
34
•
The nature of radiation from a blackbody does not depend on the material of
which the walls are made.
Basic laws of radiation
(1) All objects emit radiant energy.
(2) Hotter objects emit more energy (per unit area) than colder objects. The total
power of the emitted radiation is proportional to the fourth power of temperature.
This is called Stefan’s Law and is given by
P =  A e T4
(3.1)
where P is power radiated from the surface of the object (W), T is equilibrium
surface temperature (K), σ is Stefan-Boltzmann constant (= 5.670 x 10−8 W/m2K4 ),
A is surface area of
the object (m2) and e is emissivity of the surface (e =1 for a perfect blackbody).
(3) The peak of the wavelength distribution shifts to shorter wavelengths as the
black body temperature increases. This is Wien’s Displacement Law and is given
by
λm T = constant = 2.898 × 10−3 m.K , or λm  T−1
(3.2)
where λm is the wavelength corresponding to peak intensity and T is equilibrium
temperature of the blackbody.
Figure 3.2 Intensity of blackbody radiation versus wavelength at two temperatures
(4) Rayleigh-Jeans Law: This law tries to explain the distribution of energy from
a
black body. The intensity or power per unit area I (,T)d, emitted in the
wavelength interval  to +d from a blackbody is given by
2  c kB T
(3.3)
I(  ,T ) 
4

35
kB is Boltzmann’s constant, c is speed of light in vacuum, T is equilibrium blackbody
temperature. It agrees with experimental measurements only for long wavelengths.
It predicts an energy output that diverges towards infinity as wavelengths become
smaller and is known as the ultraviolet catastrophe.
Figure 3.3 Comparison of experimental results and the curve predicted by the Rayleigh–Jeans law for
the distribution of blackbody radiation
(5) Planck‘s Law:
Max Planck developed a theory of blackbody radiation that leads to an equation for
I (,T) that is in complete agreement with experimental results. To derive the law,
Planck made two assumptions concerning the nature of the oscillators in the cavity
walls:
(i) The energy of an oscillator is quantized hence it can have only certain discrete
values:
En = n h f
(3.4)
where n is a positive integer called a quantum number, f is the frequency of cavity
oscillators, and h is a constant called Planck’s constant. Each discrete energy
value corresponds to a different quantum state, represented by the quantum
number n.
(ii) The oscillators emit or absorb energy only when making a transition from one
quantum state to another. Difference in energy will be integral multiples of hf.
36
Figure 3.4 Allowed energy levels for an oscillator with frequency f
Planck’s law explains the distribution of energy from a black body which is given by,
I(  ,T ) 
2 h c 2

1
5
e
hc
λkB T
(3.5)
1
where I (,T) d is the intensity or power per unit area emitted in the wavelength
interval d from a blackbody, h is Planck’s constant, kB is Boltzmann's constant,
c is speed of light in vacuum and T is equilibrium temperature of blackbody .
The Planck‘s Law gives a distribution that peaks at a certain wavelength, the peak
shifts to shorter wavelengths for higher temperatures, and the area under the curve
grows rapidly with increasing temperature. This law is in agreement with the
experimental data.
The results of Planck's law:
 The denominator [exp(hc/λkT)] tends to infinity faster than the numerator (λ–5),
thus resolving the ultraviolet catastrophe and hence arriving at experimental
observation:
I (λ, T)  0 as λ  0.
hc
exp( hckT )  1   k T  I(  ,T )  2  c 4 k T
 For very large λ,
i.e. I (λ, T)  0 as λ  .
From a fit between Planck's law and experimental data, Planck’s constant was
derived to be h = 6.626 × 10–34 J-s.
37
3.2 PHOTOELECTRIC EFFECT
Ejection of electrons from the surface of certain metals when it is irradiated by an
electromagnetic radiation of suitable frequency is known as photoelectric effect.
A
E
V
C
Figure 3.5(a) Apparatus (b) circuit for studying Photoelectric Effect (T – Evacuated
glass/ quartz tube, E – Emitter Plate / Photosensitive material / Cathode, C –
Collector Plate / Anode, V – Voltmeter, A - Ammeter)
Experimental Observations:
Figure 3.6 Photoelectric current versus applied potential difference for two light intensities
1.
When plate E is illuminated by light of suitable frequency, electrons are emitted
from E and a current is detected in A (Figure 3.5).
38
2.
Photocurrent produced vs potential difference graph shows that kinetic energy
of the most energetic photoelectrons is,
Kmax = e Vs
3.
4.
5.
6.
(3.6)
where Vs is stopping potential
Kinetic energy of the most energetic photoelectrons is independent of light
intensity.
Electrons are emitted from the surface of the emitter almost instantaneously
No electrons are emitted if the incident light frequency falls below a cutoff
frequency.
Kinetic energy of the most energetic photoelectrons increases with increasing
light frequency.
Classical Predictions:
1. If light is really a wave, it was thought that if one shine of light of any fixed
wavelength, at sufficient intensity on the emitter surface, electrons should
absorb energy continuously from the em waves and electrons should be ejected.
2. As the intensity of light is increased (made it brighter and hence classically, a
more energetic wave), kinetic energy of the emitted electrons should increase.
3. Measurable / larger time interval between incidence of light and ejection of
photoelectrons.
4. Ejection of photoelectron should not depend on light frequency
5. In short experimental results contradict classical predictions.
6. Photoelectron kinetic energy should not depend upon the frequency of the
incident light.
Einstein’s Interpretation of electromagnetic radiation:
1. Electromagnetic waves carry discrete energy packets (light quanta called
photons now).
2. The energy E, per packet depends on frequency f: E = hf.
3. More intense light corresponds to more photons, not higher energy photons.
4. Each photon of energy E moves in vacuum at the speed of light: c = 3 x 108 m/s
and each photon carries a momentum, p = E/c.
Einstein’s theory of photoelectric effect:
A photon of the incident light gives all its energy hf to a single electron (absorption
of energy by the electrons is not a continuous process as envisioned in the wave
model) and the kinetic energy of the most energetic photoelectron
Kmax = hf −  (Einstein’s photoelectric equation)
(3.7)
39
 is called the work function of the metal. It is the minimum energy with which an
electron is bound in the metal.
All the observed features of photoelectric effect could be explained by Einstein’s
photoelectric equation:
1. Equation shows that Kmax depends only on frequency of the incident light.
2. Almost instantaneous emission of photoelectrons due to one -to –one
interaction between photons and electrons.
3. Ejection of electrons depends on light frequency since photons should have
energy greater than the work function  in order to eject an electron.
4. The cutoff frequency fc is related to  by fc =  /h. If the incident frequency
f is less than fc , there is no emission of photoelectrons.
The graph of kinetic energy of the most energetic photoelectron Kmax vs frequency
f is a straight line, according to Einstein’s equation.
Figure 3.7 A representative plot of Kmax versus frequency of incident light for three different metals
3.3 COMPTON EFFECT
When X-rays are scattered by free/nearly free electrons, they suffer a change in their
wavelength which depends on the scattering angle. This scattering phenomenon is
known as Compton Effect.
Classical Predictions: Oscillating electromagnetic waves (classically, X-rays are
em waves) incident on electrons should have two effects: i) oscillating
electromagnetic field causes oscillations in electrons. Each electron first absorbs
radiation as a moving particle and then re-radiates in all directions as a moving
40
particle and thereby exhibiting two Doppler shifts in the frequency of radiation. ii)
radiation pressure should cause the electrons to accelerate in the direction of
propagation of the waves. Because different electrons will move at different speeds
after the interaction, depending on the amount of energy absorbed from
electromagnetic waves, the scattered waves at a given angle will have all frequencies
(Doppler- shifted values).
Compton’s experiment and observation: Compton measured the intensity of
scattered X-rays from a solid target (graphite) as a function of wavelength for different
angles. The experimental setup is shown in Figure 3.8. Contrary to the classical
prediction, only one frequency for scattered radiation was seen at a given angle. This
is shown in the Figure 3.9.
The graphs for three nonzero angles show two peaks, one at o and the other at ’
>o . The shifted peak at ’ is caused by the scattering of X-rays from free electrons.
Shift in wavelength was predicted by Compton to depend on scattering angle as
λ′ − λ =
h
(1−cos θ)
mc
(3.8)
where m is the mass of the electron, c is velocity of light, h is Planck’s constant.
This is known as Compton shift equation, and the factor
Compton wavelength and 𝑚ℎ𝑐 = 2.43 pm.
ℎ
𝑚𝑐
is called the
Figure 3.8 Schematic diagram of Compton’s apparatus. The wavelength is measured with a rotating
crystal spectrometer for various scattering angles θ.
41
Figure 3.9 Scattered x-ray intensity versus wavelength for Compton scattering at  = 0°, 45°, 90°,
and 135° showing single frequency at a given angle
Derivation of the Compton shift equation:
Compton could explain the experimental result by treating the X-rays not as waves
but rather as point like particles (photons) having energy E = hfo = hc/o ,
momentum p = hf/c = h/ and zero rest energy. Photons collide elastically with free
electrons initially at rest and moving relativistically after collision.
Let o , po = h/o and Eo = hc/o be the wavelength, momentum and energy of
the incident photon respectively. ’, p’ = h/’ and E’ = hc/’ be the corresponding
quantities for the scattered photon.
We know that, for the electron, the total relativistic energy 𝐸 = √𝑝2 𝑐 2 + 𝑚2 𝑐 4
Kinetic energy K = E − m c2
1
And momentum p =  mv.
where  
2
1  vc 2
v and m are the speed and mass of the electron respectively.
Figure 3.10 Quantum model for X-ray scattering from an electron
In the scattering process, the total energy and total linear momentum of the
system must be conserved.
For conservation of energy we must have, Eo = E’ + K
42
Eo = E’ + (E − m c2)
ie,
Eo − E’ + m c2 = 𝐸 = √𝑝2 𝑐 2 + 𝑚2 𝑐 4
Or
Squaring both the sides, (𝐸𝑜 − 𝐸′)2 + 2(𝐸𝑜 − 𝐸′) 𝑚𝑐 2 + 𝑚2 𝑐 4 = 𝑝2 𝑐 2 + 𝑚2 𝑐 4
For conservation of momentum, x-component: 𝑝𝑜 = 𝑝′ 𝑐𝑜𝑠 𝜃 + 𝑝 𝑐𝑜𝑠 𝜙
y-component: 0 = 𝑝′ 𝑠𝑖𝑛 𝜃 − 𝑝 𝑠𝑖𝑛 𝜙
Rewriting these two equations
𝑝𝑜 − 𝑝′ 𝑐𝑜𝑠 𝜃 = 𝑝 𝑐𝑜𝑠 𝜙
𝑝′ 𝑠𝑖𝑛 𝜃 = 𝑝 𝑠𝑖𝑛 𝜙
Squaring both the sides and adding,
𝑝𝑜2 − 2𝑝𝑜 𝑝′ 𝑐𝑜𝑠 𝜃 + 𝑝′2 = 𝑝2
Substituting this 𝑝2 in the equation :
(𝐸𝑜 − 𝐸′)2 + 2(𝐸𝑜 − 𝐸′) 𝑚𝑐 2 = 𝑝2 𝑐 2 , one gets
(𝐸𝑜 − 𝐸′)2 + 2(𝐸𝑜 − 𝐸′) 𝑚𝑐 2 = (𝑝𝑜2 − 2𝑝𝑜 𝑝′ 𝑐𝑜𝑠 𝜃 + 𝑝′2 )𝑐 2
Substituting photon energies and photon momenta one gets
(
ℎ𝑐
𝜆𝑜
−
2
ℎ𝑐
𝜆
) + 2(
′
ℎ𝑐
𝜆𝑜
−
ℎ𝑐
ℎ𝑐
𝜆
𝜆𝑜
2
) − 2(
ℎ𝑐
) 𝑚𝑐 2 = (
′
ℎ𝑐
) 𝑚𝑐 2 = (
′
ℎ𝑐
𝜆𝑜
ℎ𝑐
2
) ( 𝜆′ ) 𝑐𝑜𝑠 𝜃 + ( ′ )
𝜆
Simplifying one gets
2
ℎ𝑐
ℎ𝑐
ℎ𝑐
ℎ𝑐
𝜆𝑜
𝜆
𝜆
2
( ) − 2 ( ) ( ′ ) + ( ′ ) + 2 ℎ𝑐 (
𝜆𝑜
i.e.,
−
ℎ𝑐
𝜆𝑜 𝜆′
OR,
+ ( 𝜆1 −
𝑜
′
1
𝜆′
) 𝑚𝑐 2 = −
(𝜆𝜆−𝜆𝜆𝑜′ ) 𝑚𝑐 2 =
𝑜
ℎ𝑐
𝜆𝑜 𝜆′
1
𝜆𝑜
−
ℎ𝑐
𝜆𝑜 𝜆′
1
𝜆
𝜆𝑜
2
) − 2(
ℎ𝑐
𝜆𝑜
ℎ𝑐
ℎ𝑐
2
) ( 𝜆′ ) 𝑐𝑜𝑠 𝜃 + ( ′ )
𝜆
𝑐𝑜𝑠 𝜃
(1 − 𝑐𝑜𝑠 𝜃)
Compton shift:
𝝀′ − 𝝀𝒐 =
𝒉
𝒎𝒄
(𝟏 − 𝒄𝒐𝒔 𝜽)
43
3.4 PHOTONS AND ELECTROMAGNETIC WAVES [DUAL
NATURE OF LIGHT]
•
•
•
Light exhibits diffraction and interference phenomena that are only explicable
in terms of wave properties.
Photoelectric effect and Compton Effect can only be explained taking light as
photons / particle.
This means true nature of light is not describable in terms of any single picture,
instead both wave and particle nature have to be considered. In short, the
particle model and the wave model of light complement each other.
3.5 de BROGLIE HYPOTHESIS - WAVE PROPERTIES OF
PARTICLES
We have seen that light comes in discrete units (photons) with particle properties
(energy E and momentum p) that are related to the wave-like properties of
frequency and wavelength. Louis de Broglie postulated that because photons have
both wave and particle characteristics, perhaps all forms of matter have wave-like
properties, with the wavelength λ related to momentum p in the same way as for
light.
de Broglie wavelength: 𝜆 =
ℎ
𝑝
=
ℎ
(3.9)
𝑚𝑣
where h is Planck’s constant and p is momentum of the quantum particle, m is mass
of the particle, and v is speed of the particle. The electron accelerated through a
potential difference of V, has a non-relativistic kinetic energy
1
𝑚 𝑣 2 = 𝑒 ∆𝑉
where e is electron charge.
2
Hence, the momentum (p) of an electron accelerated through a potential difference
of V is
𝑝 = 𝑚 𝑣 = √2 𝑚 𝑒 ∆𝑉
Frequency of the matter wave associated with the particle is
(3.10)
𝐸
ℎ
, where E is total
relativistic energy of the particle
Davisson-Germer experiment and G P Thomson’s electron diffraction experiment
confirmed de Broglie relationship p = h /. Subsequently it was found that atomic
beams, and beams of neutrons, also exhibit diffraction when reflected from regular
crystals. Thus de Broglie's formula seems to apply to any kind of matter. Now the
dual nature of matter and radiation is an accepted fact and it is stated in the
44
principle of complementarity. This states that wave and particle models of either
matter or radiation complement each other.
3.6 THE QUANTUM PARTICLE
Quantum particle is a model by which particles having dual nature are represented.
We must choose one appropriate behavior for the quantum particle (particle or
wave) in order to understand a particular behavior.
To represent a quantum wave, we have to combine the essential features of both an
ideal particle and an ideal wave. An essential feature of a particle is that it is localized
in space. But an ideal wave is infinitely long (non-localized) as shown in Figure 3.11.
Figure 3.11 Section of an ideal wave of single frequency
Now to build a localized entity from an infinitely long wave, waves of same
amplitude, but slightly different frequencies are superposed (Figure 3.12).
Figure 3.12 Superposition of two waves Wave 1 and Wave 2
If we add up large number of waves in a similar way, the small localized region of
space where constructive interference takes place is called a wave packet, which
represents a quantum particle (Figure 3.13).
45
Figure 3.13 Wave packet
Mathematical representation of a wave packet:
Superposition of two waves of equal amplitude, but with slightly different
frequencies, f1 and f2, traveling in the same direction are considered. The waves are
written as
𝑦1 = 𝐴 𝑐𝑜𝑠(𝑘1 𝑥 − 𝜔1 𝑡)
and 𝑦2 = 𝐴 𝑐𝑜𝑠(𝑘2 𝑥 − 𝜔2 𝑡)
where 𝑘 = 2𝜋/𝜆 ,
The resultant wave
𝜔 = 2𝜋𝑓
y = y 1 + y2
𝛥𝑘
𝑦 = 2𝐴 [𝑐𝑜𝑠 ( 2 𝑥 −
𝛥𝜔
𝑡)
2
𝑘1 +𝑘2
𝑥
2
𝑐𝑜𝑠 (
−
𝜔1 +𝜔2
𝑡)]
2
where k = k1 – k2 and  = 1 – 2.
Figure 3.14 Beat pattern due to superposition of wave trains y1 and y2
The resulting wave oscillates with the average frequency, and its amplitude envelope
(in square brackets, shown by the blue dotted curve in Figure 3.14) varies according
to the difference frequency. A realistic wave (one of finite extent in space) is
characterized by two different speeds. The phase speed, the speed with which wave
crest of individual wave moves, is given by
𝑣𝑝 = 𝑓 𝜆
or
𝑣𝑝 =
𝜔
𝑘
(3.11)
The envelope of group of waves can travel through space with a different speed than
the individual waves. This speed is called the group speed or the speed of the wave
packet which is given by
46
(𝛥𝜔
)
2
𝑣𝑔 =
(𝛥𝑘
)
2
𝛥𝜔
=
(3.12)
𝛥𝑘
For a superposition of large number of waves to form a wave packet, this ratio is
𝑣𝑔 =
𝑑𝜔
𝑑𝑘
In general these two speeds are not the same.
Relation between group speed (vg) and phase speed (vp):
𝜔
𝑣𝑃 =
But
𝑑𝜔
𝑣𝑔 =

= 𝑓𝜆
𝑘
𝑑𝑘
𝑑(𝑘𝑣𝑃 )
=
𝑑𝑘
= 𝑘
𝑑𝑣𝑃
𝑑𝑘
𝜔 = 𝑘 𝑣𝑃
+ 𝑣𝑃
Substituting for k in terms of λ, we get
𝑣𝑔 = 𝑣𝑃 − 𝜆 (
𝑑𝑣𝑃
𝑑𝜆
)
(3.13)
Relation between group speed (vg) and particle speed (u):
𝜔 = 2𝜋𝑓 = 2𝜋
𝑣𝑔 =
𝑑𝜔
𝑑𝑘
𝐸
2𝜋
ℎ
2𝜋
ℎ
=
𝑘 =
and
ℎ
𝑑𝐸
𝜆
=
2𝜋
ℎ⁄𝑝
2𝜋𝑝
=
ℎ
𝑑𝐸
=
𝑑𝑝
2𝜋
𝑑𝑝
For a classical particle moving with speed u, the kinetic energy E is given by
𝐸 =
1
2
𝑚 𝑢2 =
𝑣𝑔 =
𝑑𝜔
𝑑𝑘
𝑝2
2𝑚
=
𝑑𝐸 =
and
𝑑𝐸
𝑑𝑝
=
2 𝑝 𝑑𝑝
2𝑚
or
𝑑𝐸
𝑑𝑝
=
𝑢
𝑝
𝑚
= 𝑢
(3.14)
i.e., we should identify the group speed with the particle speed, speed with which
the energy moves. To represent a realistic wave packet, confined to a finite region
in space, we need the superposition of large number of harmonic waves with a range
of k-values.
3.7 DOUBLE–SLIT EXPERIMENT REVISITED
One way to confirm our ideas about the electron’s wave–particle duality is through
an experiment in which electrons are fired at a double slit. Consider a parallel beam
of mono-energetic electrons incident on a double slit as in Figure 3.15. Let’s assume
the slit widths are small compared with the electron wavelength so that diffraction
effects are negligible. An electron detector screen (acts like the “viewing screen” of
Young’s double-slit experiment) is positioned far from the slits at a distance much
greater than d, the separation distance of the slits. If the detector screen collects
47
electrons for a long enough time, we find a typical wave interference pattern for the
counts per minute, or probability of arrival of electrons. Such an interference pattern
would not be expected if the electrons behaved as classical particles, giving clear
evidence that electrons are interfering, a distinct wave-like behavior.
Figure 3.15 (a) Schematic of electron beam interference experiment, (b) Photograph of a double-slit
interference pattern produced by electrons
If we measure the angle θ at which the maximum intensity of the electrons arrives
at the detector screen, we find they are described by exactly the same equation as
that for light: 𝑑 𝑠𝑖𝑛 𝜃 = 𝑚 𝜆 , where m is the order number and λ is the electron
wavelength. Therefore, the dual nature of the electron is clearly shown in this
experiment: the electrons are detected as particles at a localized spot on the
detector screen at some instant of time, but the probability of arrival at the
spot is determined by finding the intensity of two interfering waves.
3.8 UNCERTAINTY PRINCIPLE
It is fundamentally impossible to make simultaneous measurements of a particle’s
position and momentum with infinite accuracy. This is known as Heisenberg
uncertainty principle. The uncertainties arise from the quantum structure of
matter.
For a particle represented by a single wavelength wave existing throughout space, 
is precisely known, and according to de Broglie hypothesis, its p is also known
accurately. But the position of the particle in this case becomes completely
uncertain.
This means  = 0, p =0; but x = 
In contrast, if a particle whose momentum is uncertain (combination of waves / a
range of wavelengths are taken to form a wave packet), so that x is small, but  is
large. If x is made zero,  and thereby p will become .
48
In short
( x ) ( px) ≥ h / 4
(3.15)
where x is uncertainty in the measurement of position x of the particle and px is
uncertainty in the measurement of momentum px of the particle.
One more relation expressing uncertainty principle is related to energy and time
which is given by
( E ) ( t ) ≥
h / 4
(3.16)
where E is uncertainty in the measurement of energy E of the system when the
measurement is done over the time interval t.
3.9 QUESTIONS
1. Explain (a) Stefan’s law (b) Wien’s displacement law (c) Rayleigh-Jeans law.
2. Sketch schematically the graph of wavelength vs intensity of radiation from a
blackbody.
3. Explain Planck’s radiation law.
4. Write the assumptions made in Planck’s hypothesis of blackbody radiation.
5. Explain photoelectric effect.
6. What are the observations in the experiment on photoelectric effect?
7. What are the classical predictions about the photoelectric effect?
8. Explain Einstein’s photoelectric equation.
9. Which are the features of photoelectric effect-experiment explained by
Einstein’s photoelectric equation?
10. Sketch schematically the following graphs with reference to the photoelectric
effect: (a) photoelectric current vs applied voltage (b) kinetic energy of mostenergetic electron vs frequency of incident light.
11. Explain Compton effect.
12. Explain the experiment on Compton effect.
13. Derive the Compton shift equation.
14. Explain the wave properties of the particles.
49
15. Explain a wave packet and represent it schematically.
16. Explain (a) group speed (b) phase speed, of a wave packet.
17. Show that the group speed of a wave packet is equal to the particle speed.
18. (a) Name any two phenomena which confirm the particle nature of light.
(b) Name any two phenomena which confirm the wave nature of light.
19. Explain Heisenberg uncertainty principle.
20. Write the equations for uncertainty in (a) position and momentum (b) energy
and time.
21. Mention two situations which can be well explained by the uncertainty
relation.
3.10 PROBLEMS
1 Find the peak wavelength of the blackbody radiation emitted by each of the
following.
A. The human body when the skin temperature is 35°C
B. The tungsten filament of a light bulb, which operates at 2000 K
C. The Sun, which has a surface temperature of about 5800 K.
Ans: 9.4 μm, 1.4 μm, 0.50 μm
2 A 2.0- kg block is attached to a spring that has a force constant of k = 25 N/m.
The spring is stretched 0.40 m from its equilibrium position and released.
A. Find the total energy of the system and the frequency of oscillation
according to classical calculations.
B. Assuming that the energy is quantized, find the quantum number n for the
system oscillating with this amplitude.
C. Suppose the oscillator makes a transition from the n = 5.4 x 1033 state to the
state corresponding to n = 5.4 x 1033 – 1. By how much does the energy of
the oscillator change in this one-quantum change.
Ans: 2.0 J, 0.56 Hz, 5.4 x 1033, 3.7 x 10–34 J
3 The human eye is most sensitive to 560 nm light. What is the temperature of a
black body that would radiate most intensely at this wavelength?
Ans: 5180 K
4 A blackbody at 7500 K consists of an opening of diameter 0.050 mm, looking into
an oven. Find the number of photons per second escaping the hole and having
wavelengths between 500 nm and 501 nm.
50
5
6
7
8
9
10
11
12
13
Ans: 1.30 x 1015/s
The radius of our Sun is 6.96 x 108 m, and its total power output is 3.77 x 1026 W.
(a) Assuming that the Sun’s surface emits as a black body, calculate its surface
temperature. (b) Using the result, find max for the Sun.
Ans: 5750 K, 504 nm
Calculate the energy in electron volts, of a photon whose frequency is (a) 620 THz,
(b) 3.10 GHz, (c) 46.0 MHz. (d) Determine the corresponding wavelengths for
these photons and state the classification of each on the electromagnetic
spectrum.
Ans: 2.57 eV, 1.28 x 10–5 eV, 1.91 x 10–7 eV, 484 nm, 9.68 cm,
6.52 m
An FM radio transmitter has a power output of 150 kW and operates at a
frequency of 99.7 MHz. How many photons per second does the transmitter emit?
Ans: 2.27 x 1030 photons/s
A sodium surface is illuminated with light having a wavelength of 300 nm. The
work function for sodium metal is 2.46 eV. Find
A. The maximum kinetic energy of the ejected photoelectrons and
B. The cutoff wavelength for sodium.
Ans: 1.67 eV, 504 nm
Molybdenum has a work function of 4.2eV. (a) Find the cut off wavelength and
cut off frequency for the photoelectric effect. (b) What is the stopping potential if
the incident light has wavelength of 180 nm?
Ans: 296 nm, 1.01 x 1015 Hz, 2.71 V
Electrons are ejected from a metallic surface with speeds up to 4.60 x 105 m/s
when light with a wavelength of 625 nm is used. (a) What is the work function of
the surface? (b) What is the cut-off frequency for this surface?
Ans: 1.38 eV, 3.34 x 1014 Hz
The stopping potential for photoelectrons released from a metal is 1.48 V larger
compared to that in another metal. If the threshold frequency for the first metal is
40.0 % smaller than for the second metal, determine the work function for each
metal.
Ans: 3.70 eV, 2.22 eV
Two light sources are used in a photoelectric experiment to determine the work
function for a metal surface. When green light from a mercury lamp ( = 546.1
nm) is used, a stopping potential of 0.376 V reduces the photocurrent to zero. (a)
Based on this what is the work function of this metal? (b) What stopping potential
would be observed when using the yellow light from a helium discharge tube ( =
587.5 nm)?
Ans: 1.90 eV, 0.215 V
X-rays of wavelength o = 0.20 nm are scattered from a block of material. The
scattered X-rays are observed at an angle of 45° to the incident beam. Calculate
their wavelength.
51
14
15
16
17
18
19
What if we move the detector so that scattered X-rays are detected at an angle
larger than 45°? Does the wavelength of the scattered X-rays increase or decrease
as the angle  increase?
Ans: 0.200710 nm, INCREASES
Calculate the energy and momentum of a photon of wavelength 700 nm.
Ans: 1.78 eV, 9.47 x 10–28kg.m/s
A 0. 00160 nm photon scatters from a free electron. For what photon scattering
angle does the recoiling electron have kinetic energy equal to the energy of the
scattered photon?
Ans: 70°
A 0.880 MeV photon is scattered by a free electron initially at rest such that the
scattering angle of the scattered electron is equal to that of the scattered photon
( = ).
(a) Determine the angles  & . (b) Determine the energy and
momentum of the scattered electron and photon.
Ans: 43°, 43°, 0.602 MeV, 3.21 x 10–22 kg.m/s, 0.278 MeV, 3.21 x 10–22 kg.m/s
Calculate the de- Broglie wavelength for an electron moving at 1.0 x 107 m/s.
Ans: 7.28 x 10–11 m
A rock of mass 50 g is thrown with a speed of 40 m/s. What is its de Broglie
wavelength?
Ans: 3.3 x 10–34 m
A particle of charge q and mass m has been accelerated from rest to a
nonrelativistic speed through a potential difference of V. Find an expression for
its de Broglie wavelength.
Ans: λ =
h
√2 m q Δv
20 (a) An electron has a kinetic energy of 3.0 eV. Find its wavelength. (b) Also find
the wavelength of a photon having the same energy.
Ans: 7.09 x 10–10 m, 4.14 x 10–7 m
21 In the Davisson-Germer experiment, 54.0 eV electrons were diffracted from a
nickel lattice. If the first maximum in the diffraction pattern was observed at =
50.0°, what was the lattice spacing a between the vertical rows of atoms in the
figure?
Ans: 2.18 x 10–10 m
22 Consider a freely moving quantum particle with mass m and speed u. Its energy is
E= K= mu2/2. Determine the phase speed of the quantum wave representing the
particle and show that it is different from the speed at which the particle
transports mass and energy.
Ans: vGROUP = u ≠ vPHASE
52
23 Electrons are incident on a pair of narrow slits 0.060 m apart. The ‘bright bands’
in the interference pattern are separated by 0.40 mm on a ‘screen’ 20.0 cm from
the slits. Determine the potential difference through which the electrons were
accelerated to give this pattern.
Ans: 105 V
24 The speed of an electron is measured to be 5.00 x 103 m/s to an accuracy of
0.0030%. Find the minimum uncertainty in determining the position of this
electron.
Ans: 0.383 mm
25 The lifetime of an excited atom is given as 1.0 x 10-8 s. Using the uncertainty
principle, compute the line width f produced by this finite lifetime?
Ans: 8.0 x 106 Hz
26 Use the uncertainty principle to show that if an electron were confined inside an
atomic nucleus of diameter 2 x 10–15 m, it would have to be moving relativistically,
while a proton confined to the same nucleus can be moving nonrelativistically.
Ans: vELECTRON  0.99996 c, vPROTON  1.8 x 107 m/s
27 Find the minimum kinetic energy of a proton confined within a nucleus having a
diameter of 1.0 x 10–15 m.
Ans: 5.2 MeV
53
4 QUANTUM MECHANICS
OBJECTIVES:
 To learn the application of Schrödinger equation to a bound particle and
to learn the quantized nature of the bound particle, its expectation values
and physical significance.
 To understand the tunneling behavior of a particle incident on a potential
barrier.
 To understand the behavior of quantum oscillator.
4.1 AN INTERPRETATION OF QUANTUM MECHANICS
Experimental evidences proved that both matter and electromagnetic radiation
exhibit wave and particle nature depending on the phenomenon being observed.
Making a conceptual connection between particles and waves, for an
electromagnetic radiation of amplitude E, the probability per unit volume of finding
a photon in a given region of space at an instant of time as
PROBABILITY
𝑉
∝ 𝐸2
Figure 4.1 Wave packet
Taking the analogy between electromagnetic radiation and matter-the probability
per unit volume of finding the particle is proportional to the square of the amplitude
of a wave representing the particle, even if the amplitude of the de Broglie wave
associated with a particle is generally not a measurable quantity. The amplitude of
the de Broglie wave associated with a particle is called probability amplitude, or the
wave function, and is denoted by .
In general, the complete wave function  for a system depends on the positions of
all the particles in the system and on time. This can be written as
(r1,r2,…rj,…,t) = (rj) e–it
54
where rj is the position vector of the jth particle in the system.
For any system in which the potential energy is time-independent and depends only
on the position of particles within the system, the important information about the
system is contained within the space part of the wave function. The wave function
 contains within it all the information that can be known about the particle. | |2
is always real and positive, and is proportional to the probability per unit volume, of
finding the particle at a given point at some instant. If  represents a single particle,
then ||2 - called the probability density - is the relative probability per unit
volume that the particle will be found at any given point in the volume.
One-dimensional wave functions and expectation values: Let  be the wave
function for a particle moving along the x axis. Then P(x) dx = ||2dx is the
probability to find the particle in the infinitesimal interval dx around the point x.
The probability of finding the particle in the arbitrary interval a ≤ x ≤ b is
𝑏
𝑃𝑎𝑏 = ∫𝑎 |𝜓|2 𝑑𝑥
(4.1)
The probability of a particle being in the interval a ≤ x ≤ b is the area under the
probability density curve from a to b. The total probability of finding the particle is
one. Forcing this condition on the wave function is called normalization.
+∞
∫−∞ |𝜓|2 𝑑𝑥 = 1
(4.2)
Figure 4.2 An arbitrary probability density curve for a particle
All the measurable quantities of a particle, such as its position, momentum and
energy can be derived from the knowledge of . e.g., the average position at which
one expects to find the particle after many measurements is called the expectation
value of x and is defined by the equation
+∞
⟨𝑥⟩ ≡ ∫−∞ 𝜓 ∗ 𝑥 𝜓 𝑑𝑥
(4.3)
55
The important mathematical features of a physically reasonable wave function
(x) for a system are
 (x) may be a complex function or a real function, depending on the
system.
 (x) must be finite, continuous and single valued everywhere.
 The space derivatives of, must be finite, continuous and single valued
everywhere.
  must be normalizable.
4.2 THE SCHRÖDINGER EQUATION
The appropriate wave equation for matter waves was developed by Schrödinger.
Schrödinger equation as it applies to a particle of mass m confined to move along x
axis and interacting with its environment through a potential energy function U(x)
is
−
ℏ2
𝑑2 𝜓
2 𝑚 𝑑𝑥 2
+𝑈𝜓 = 𝐸𝜓
(4.4)
where E is a constant equal to the total energy of the system (the particle and its
environment) and ħ = h/2.This equation is referred to as the one dimensional, timeindependent Schrödinger equation.
Application of Schrödinger equation:
1.
2.
3.
4.
Particle in an infinite potential well (particle in a box)
Particle in a finite potential well
Tunneling
Quantum oscillator
56
4.3 PARTICLE IN AN INFINITE POTENTIAL WELL
(PARTICLE IN A “BOX”)
Figure 4.3 (a) Particle in a potential well of infinite height, (b) Sketch of potential well
Consider a particle of mass m and velocity v, confined to bounce between two
impenetrable walls separated by a distance L as shown in Figure 4.3(a). Figure
4.3(b) shows the potential energy function for the system.
U(x) = 0,
for
0 <x<L,
U (x) =  ,
for x≤ 0, x≥L
Since U (x)=  , for x< 0, x>L , (x) = 0 in these regions. Also (0) =0 and (L) =0.
Only those wave functions that satisfy these boundary conditions are allowed. In
the region 0 <x<L, where U = 0, the Schrödinger equation takes the form
𝑑2 𝜓
2𝑚
+
𝐸 𝜓 = 0
𝑑𝑥 2
ℏ2
Or
𝑑2 𝜓
𝑑𝑥 2
= − 𝑘2 𝜓 ,
where 𝑘 2 =
2𝑚𝐸
ℏ2
or
𝑘 =
√2𝑚𝐸
ℏ
The most general form of the solution to the above equation is
(x) = Asin(kx) + B cos(kx)
where A and B are constants determined by the boundary and normalization
conditions.
Applying the first boundary condition,
i.e.,
at x = 0,  = 0
leads to
0 = A sin 0 + B cos 0
or B = 0 ,
And at x = L ,  = 0 ,
57
0 = A sin(kL) + B cos(kL) = A sin(kL) + 0 ,
Since A  0 ,
k L = n π ; ( n = 1, 2, 3, ……….. )
sin(kL) = 0 .
𝜓𝑛 (𝑥) = 𝐴 𝑠𝑖𝑛 (
Now the wave function reduces to
𝑛𝜋𝑥
𝐿
)
To find the constant A, apply normalization condition
+∞
∫−∞ |ψ|2 dx = 1
𝐿1
𝐴2 ∫0
2
[1 − 𝑐𝑜𝑠(
2𝑛𝜋𝑥
)] 𝑑𝑥
𝐿
2
We get,
√2𝑚𝐸
ℏ
∴
𝑛𝜋𝑥
𝐿
2
)] 𝑑𝑥 = 1 .
= 1
2
Thus 𝜓𝑛 (𝑥) = √𝐿 𝑠𝑖𝑛 (
𝑘 =
𝐿
∫0 𝐴2 [𝑠𝑖𝑛 (
𝐴 = √𝐿
Solving we get
Since
or
𝑛𝜋𝑥
𝐿
)
√2𝑚𝐸
and
ℏ
𝐿 =
is the wave function for particle in a box.
kL = nπ
𝑛𝜋.
ℎ2
𝐸𝑛 = ( 8 𝑚 𝐿2) 𝑛2 ,
n = 1, 2, 3,
. . . . .
(4.5)
Each value of the integer n corresponds to a quantized energy value, En .
The lowest allowed energy (n = 1),
𝐸1 =
ℎ2
8 𝑚 𝐿2
.
This is the ground state energy for the particle in a box. Excited states correspond
to n = 2, 3, 4,…which have energies given by 4E1 , 9E1 , 16E1…. respectively. Energy level
diagram, wave function and probability density sketches are shown in Figure 4.4
and 4.5 respectively. Since ground state energy E1 ≠0, the particle can never be at
rest.
Figure 4.4 Energy level diagram for a particle in potential well of infinite height
58
Figure 4.5 Sketch of (a) wave function, (b) Probability density for a particle in potential well of infinite
height
4.4 A PARTICLE IN A POTENTIAL WELL OF FINITE
HEIGHT
Figure 4.6 Potential well of finite height U and length L
Consider a particle with the total energy E, trapped in a finite potential well of height
U such that
U(x) = 0 ,
0 <x<L,
U(x) = U ,
x≤ 0, x≥L
Classically, for energy E<U, the particle is permanently bound in the potential well.
However, according to quantum mechanics, a finite probability exists that the
particle can be found outside the well even if E<U. That is, the wave function is
generally nonzero in the regions I and III. In region II, where U = 0, the allowed wave
functions are again sinusoidal. But the boundary conditions no longer require that
the wave function must be zero at the ends of the well.
59
Schrödinger equation outside the finite well in regions I & III
𝑑2 𝜓
𝑑𝑥 2
=
2𝑚
ℏ2
(𝑈 − 𝐸) 𝜓
𝑑2 𝜓
or
𝑑𝑥 2
= 𝐶 2 𝜓 where
𝐶2 =
2𝑚
ℏ2
(𝑈 − 𝐸)
General solution of the above equation is
(x) = AeCx + B e−Cx
where A and B are constants.
A must be zero in Region III and B must be zero in Region I, otherwise, the
probabilities would be infinite in those regions. For solution to be finite,
I = AeCx
for x≤ 0
III = Be-Cx
for x≥L
This shows that the wave function outside the potential well decay exponentially
with distance.
Schrodinger equation inside the square well potential in region II, where U = 0
𝑑2 𝜓𝐼𝐼
𝑑𝑥 2
+ (
2𝑚
ℏ2
2𝑚𝐸
𝐸) 𝜓𝐼𝐼 = 0 ,
ℏ2
= 𝑘2
General solution of the above equation
𝜓𝐼𝐼 = 𝐹 𝑠𝑖𝑛[𝑘𝑥] + 𝐺 𝑐𝑜𝑠[𝑘𝑥]
To determine the constants A, B, F, G and the allowed values of energy E, apply the
four boundary conditions and the normalization condition:
𝑑𝜓
At x = 0 , I(0) = II(0) and [ 𝑑𝑥𝐼 ]
𝑥=0
At x = L , II(L) = III(L)
=
[
[
and
𝑑𝜓𝐼𝐼
𝑑𝑥
𝑑𝜓𝐼𝐼
𝑑𝑥
]
]
𝑥=0
𝑥=𝐿
=
[
𝑑𝜓𝐼𝐼𝐼
𝑑𝑥
]
𝑥=𝐿
+∞
∫
|𝜓|2 𝑑𝑥 = 1
−∞
Figure 4.7 shows the plots of wave functions and their respective probability
densities.
60
Figure 4.7 Sketch of (a) wave function, (b) Probability density for a particle in potential well of finite
height
It is seen that wavelengths of the wave functions are longer than those of wave
functions of infinite potential well of same length and hence the quantized energies
of the particle in a finite well are lower than those for a particle in an infinite well.
4.5 TUNNELING THROUGH A POTENTIAL ENERGY
BARRIER
Consider a particle of energy E approaching a potential barrier of height U, (E<U).
Potential energy has a constant value of U in the region of width L and is zero in all
other regions. This is called a square barrier and U is called the barrier height. Since
E<U, classically the regions II and III shown in the figure are forbidden to the particle
incident from left. But according to quantum mechanics, all regions are accessible
to the particle, regardless of its energy.
Figure 4.8 Tunneling through a potential barrier of finite height
61
By applying the boundary conditions, i.e. and its first derivative must be
continuous at boundaries (at x = 0 and x = L), full solution to the Schrödinger
equation can be found which is shown in figure. The wave function is sinusoidal in
regions I and III but exponentially decaying in region II. The probability of locating
the particle beyond the barrier in region III is nonzero. The movement of the particle
to the far side of the barrier is called tunneling or barrier penetration. The
probability of tunneling can be described with a transmission coefficient T and a
reflection coefficient R.
The transmission coefficient represents the probability that the particle penetrates
to the other side of the barrier, and reflection coefficient is the probability that the
particle is reflected by the barrier. Because the particles must be either reflected or
transmitted we have, R + T = 1.
An approximate expression for the transmission coefficient, when T<< 1 is
T ≈ e−2CL , where 𝐶 =
√ 2 𝑚 (𝑈−𝐸)
ℏ
.
(4.6)
4.6 THE SIMPLE HARMONIC OSCILLATOR
Consider a particle that is subject to a linear restoring force 𝐹 = −𝑘𝑥, where k is a
constant and x is the position of the particle relative to equilibrium (at equilibrium
position x=0).
Classically, the potential energy of the system is,
𝑈=
1 2 1
𝑘𝑥 = 𝑚𝜔2 𝑥 2
2
2
where the angular frequency of vibration is 𝜔 = √𝑘/𝑚.
The total energy E of the system is,
1
1
𝐸 = 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 + 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 = 𝐾 + 𝑈 = 𝑘𝐴2 = 𝑚𝜔2 𝐴2
2
2
where A is the amplitude of motion. In the classical model, any value of E is allowed,
including E= 0, which is the total energy when the particle is at rest at x=0.
A quantum mechanical model for simple harmonic oscillator can be obtained by
1
substituting 𝑈 = 2 𝑚𝜔2 𝑥 2 in Schrödinger equation:
ℏ2 𝑑 2  1
−
+ 𝑚𝜔2 𝑥 2  = 𝐸
2𝑚 𝑑𝑥 2 2
The solution for the above equation is
 = 𝐵𝑒 −𝐶𝑥
2
62
1
where 𝐶 = 𝑚𝜔/2ℏ and 𝐸 = 2 ℏ𝜔. The constant B can be determined from
normalization condition.
In quantum model, the energy levels of a harmonic oscillator are quantized. The
energy of a state having an arbitrary quantum number n is given by
1
𝐸𝑛 = (𝑛 + 2) ℏ𝜔;
𝑛 = 0, 1, 2 ….
(4.7)
1
The state n = 0 corresponds to the ground state, whose energy is 𝐸0 = 2 ℏ𝜔 the state
3
n = 1 corresponds to the first excited state, whose energy is 𝐸1 = 2 ℏ𝜔 and so on. The
energy-level diagram for this system is shown in Figure 4.9. The separations between
adjacent levels are equal and given by ∆𝐸 = ℏ𝜔.
Figure 4.9 Energy-level diagram for a simple harmonic oscillator, superimposed on the potential
energy function.
4.7 QUESTIONS
1 What is a wave function ? What is its physical interpretation ?
2 What are the mathematical features of a wave function?
3 By solving the Schrödinger equation, obtain the wave-functions
for a particle of mass m in a one-dimensional “box” of length
L.
4 Apply the Schrödinger equation to a particle in a onedimensional “box” of length L and obtain the energy values of
the particle.
5 Sketch the lowest three energy states, wave-functions, probability
densities for the particle in a one-dimensional “box”.
6 The wave-function for a particle confined to moving in a onedimensional box is
63
ψ(x) = A sin(nπx
) . Use the normalization condition on  to
L
show that 𝐴 = √2𝐿 .
7 The wave-function of an electron is ψ(x) = A sin(nπx
) . Obtain
L
an expression for the probability of finding the electron between
x = a and x = b.
8 Sketch the potential-well diagram of finite height U and length
L, obtain the general solution of the Schrödinger equation for
a particle of mass m in it.
9 Sketch the wave-functions and the probability densities for the
lowest three energy states of a particle in a potential well of
finite height.
10 Give a brief account of tunneling of a particle through a
potential energy barrier.
11 Give a brief account of the quantum mechanical treatment of a
simple harmonic oscillator.
4.8 PROBLEMS
1 A particle wave function is given by the equation  (x) = A 𝑒 −𝑎𝑥 2
(A) What is the value of A if this wave function is normalized?
(B) What is the expectation value of x for this particle?
Ans: A = (2a/π)¼ ,
x = 0
2 A free electron has a wave function ψ(x) = 𝐴 exp[𝑖(5.0 × 1010 )𝑥]
where x is in meters. Find (a) its de Broglie wavelength, (b) its momentum, and (c)
its kinetic energy in electron volts.
Ans: 1.26 x 10–10m, 5.27 x 10–24kg.m/s, 95.5 eV
3 An electron is confined between two impenetrable walls 0.20 nm apart. Determine
the energy levels for the states n =1 ,2 , and 3.
Ans: 9.2 eV, 37.7 eV, 84.8 eV
4 A 0.50 kg baseball is confined between two rigid walls of a stadium that can be
modeled as a “box” of length 100 m. Calculate the minimum speed of the baseball.
If the baseball is moving with a speed of 150 m/s, what is the quantum number of
the state in which the baseball will be?
Ans: 6.63 x 10–36 m/s, 2.26 x 1037
5 A proton is confined to move in a one-dimensional “box” of length 0.20 nm. (a)
Find the lowest possible energy of the proton. (b) What is the lowest possible
64
6
7
8
9
10
energy for an electron confined to the same box? (c) Account for the great
difference in results for (a) and (b).
Ans: 5.13 x 10–3 eV, 9.41 eV
(A) Using the simple model of a particle in a box to represent an atom, estimate
the energy (in eV) required to raise an atom from the state n =1 to the state n
=2. Assume the atom has a radius of 0.10 nm and that the moving electron
carries the energy that has been added to the atom.
(B) Atoms may be excited to higher energy states by absorbing photon energy.
Calculate the wavelength of the photon that would cause the transition from
the state n =1 to the state n =2.
Ans: 28.3 eV, 43.8 nm
A 30-eV electron is incident on a square barrier of height 40 eV. What is the
probability that the electron will tunnel through the barrier if its width is (A) 1.0
nm? (B) 0.10 nm?
Ans: 8.5 x 10–15, 0.039
An electron with kinetic energy E = 5.0 eV is incident
on a barrier with thickness L = 0.20 nm and height U =
10.0 eV as shown in the figure. What is the probability
that the electron (a) will tunnel through the barrier?
(b) will be reflected?
Ans: 0.0103, 0.990
A quantum simple harmonic oscillator consists of an electron bound by a restoring
force proportional to its position relative to a certain equilibrium point. The
proportionality constant is 8.99 N/m. What is the longest wavelength of light that
can excite the oscillator?
Ans: 600nm
A quantum simple harmonic oscillator consists of a particle of mass m bound by a
restoring force proportional to its position relative to a certain equilibrium point.
The proportionality constant is k. What is the longest wavelength of light that can
𝑚
excite the oscillator? Ans: 2𝜋𝑐√ 𝑘
65
5 ATOMIC PHYSICS
OBJECTIVES:






To know about the quantum model of H-atom and its wave functions.
To understand more about Visible and X ray spectra
To explain basic interactions of radiation with matter.
To understand the basic principles and requirements for working of laser.
To recognize the various applications of laser.
To apply and evaluate the above concepts by solving numerical problems
5.1 THE QUANTUM MODEL OF THE HYDROGEN ATOM
The formal procedure for solving the problem of the hydrogen atom is to substitute
the appropriate potential energy function into the Schrödinger equation, find
solutions to the equation, and apply boundary conditions as we did for the particle
in a box.
The potential energy function for the H-atom is
𝑈(𝑟) = −
𝑘𝑒 𝑒 2
(5.1)
𝑟
where ke = 1/40= 8.99 x 109 N.m2/C2 Coulomb constant and r is radial distance
of electron from H-nucleus. The mathematics for the hydrogen atom is more
complicated than that for the particle in a box because the atom is threedimensional, and U depends on the radial coordinate r.
The time-independent Schrödinger equation in 3-dimensional space is
−
ℏ2
2𝑚
𝜕2 𝜓
( 𝜕𝑥 2 +
𝜕2 𝜓
𝜕𝑦 2
+
𝜕2 𝜓
𝜕𝑧 2
) +𝑈𝜓 = 𝐸𝜓
(5.2)
Since U has spherical symmetry, it is easier to solve the Schrödinger equation
in spherical polar coordinates (r, , ) where 𝑟 = √𝑥 2 + 𝑦 2 + 𝑧 2 ,
⃗ .  is the angle between the x-axis and
 is the angle between z-axis and 𝒓
⃗⃗ onto the xy-plane.
the projection of 𝒓
66
Figure 5.1 Spherical polar coordinate system
It is possible to separate the variables r, θ,  as follows:
(r, , ) = R(r) f() g()
By solving the three separate ordinary differential equations for R(r), f(), g(),
with conditions that the normalized  and its first derivative are continuous
and finite everywhere, one gets three different quantum numbers for each
allowed state of the H-atom. The quantum numbers are integers and
correspond to the three independent degrees of freedom.
The radial function R(r) of  is associated with the principal quantum number
n. Solving R(r), we get an expression for energy as,
𝑘 𝑒2
𝐸𝑛 = − ( 2𝑒𝑎 )
𝑜
1
𝑛2
= −
13.606 𝑒𝑉
𝑛2
,
n = 1, 2, 3,
. . .
(5.3)
which is in agreement with Bohr theory.
The polar function f() is associated with the orbital quantum number . The
azimuthal function g() is associated with the orbital magnetic quantum
number m .
The application of boundary conditions on the three parts of 
leads to important relationships among the three quantum numbers:
n can range from 1 to ,
 can range from 0 to n–1 ; [n allowed values].
m can range from –to + ; [(2+1) allowed values].
All states having the same principal quantum number are said to form a shell.
All states having the same values of n and  are said to form a subshell:
 K shell
=0  s
subshell
n = 2  L shell
=1  p
subshell
n=1
67
n = 3  M shell
=2  d
subshell
n = 4  N shell
=3  f
subshell
n = 5 
O shell
=4  g
subshell
n = 6  P shell
=5  h
subshell
..
..
..
..
..
..
..
..
5.2 WAVE FUNCTIONS FOR HYDROGEN
The potential energy for H-atom depends only on the radial distance r between
nucleus and electron. Therefore some of the allowed states for the H-atom can
be represented by wave functions that depend only on r (spherically symmetric
function). The simplest wave function for H-atom is the 1s-state (ground
state) wave function (n = 1,  = 0):
𝜓1𝑠 (𝑟) =
1
√𝜋 𝑎𝑜3
𝑒𝑥𝑝 (−𝑎𝑟𝑜 ) where ao is Bohr radius (0.0529 nm).
(5.4)
|1s|2 is the probability density for H-atom in 1s-state:
|𝜓1𝑠 |2 =
1
𝜋 𝑎𝑜3
𝑒𝑥𝑝 (− 2𝑎𝑜𝑟 )
(5.5)
The radial probability density P(r) is the probability per unit radial length
of finding the electron in a spherical shell of radius r and thickness dr.
P(r)dr is the probability of finding the electron in this shell.
P(r) dr = ||2 dv = ||2 4r2 dr

P(r) = 4r2 ||2
Figure 5.2 A spherical shell of radius r and thickness dr has a volume equal to 4 r2dr
Radial probability density for H-atom in its ground state:
68
𝑃1𝑠 = (
4 𝑟2
𝑎𝑜3
) 𝑒𝑥𝑝 (− 2𝑎𝑜𝑟 )
Figure 5.3 (a) The probability of finding the electron as a function of distance from the nucleus for the
hydrogen atom in the 1s (ground) state. (b) The cross section in the xy plane of the spherical
electronic charge distribution for the hydrogen atom in its 1s state
The next simplest wave function for the H-atom is the 2s-state wave function
(n = 2,  = 0):
𝜓2𝑆 (𝑟) =
1
√32𝜋𝑎𝑜3
(2 − 𝑎𝑟𝑜 ) 𝑒𝑥𝑝 (− 𝑎𝑟𝑜 )
(5.6)
2s is spherically symmetric (depends only on r). Energy corresponding to n = 2
(first excited state) is E2= E1/4 = –3.401 eV.
Figure 5.4 Plot of radial probability density versus r/a0 (normalized radius) for 1s and 2s states of
hydrogen atom
69
5.3 MORE ON ATOMIC SPECTRA: VISIBLE AND X-RAY
These spectral lines have their origin in transitions between quantized atomic states.
A modified energy-level diagram for hydrogen is shown in the Figure 5.5.
Figure 5.5 Some allowed electronic transitions for hydrogen, represented by the colored lines
In this diagram, the allowed values of , for each shell are separated horizontally.
Figure shows only those states up to  = 2, the shells from n = 4 , upward would
have more sets of states to the right, which are not shown. Transitions for which 
does not change are very unlikely to occur and are called forbidden transitions.(Such
transitions actually can occur, but their probability is very low relative to the
probability of “allowed” transitions.) The various diagonal lines represent allowed
transitions between stationary states. Whenever an atom makes a transition from a
higher energy state to a lower one, a photon of light is emitted.
The frequency of this photon is f = E/h, where E is the energy difference between
the two states and h is Planck’s constant. The selection rules for the allowed
transitions are
  1 and m  0,  1
The allowed energies for one-electron atoms and ions, such as hydrogen (H) and
helium ion (He+), are
70

ke e 2  Z 2 
13.6 eV Z 2


En  

2a0  n 2 
n2
(5.7)
This equation was developed from the Bohr theory, but it serves as a good first
approximation in quantum theory as well. For multi-electron atoms, the positive
nuclear charge Ze is largely shielded by the negative charge of the inner-shell
electrons. Therefore, the outer electrons interact with a net charge that is smaller
than the nuclear charge.
Hence, we can write
En  
13.6 eV  Z eff2
n2
(5.8)
where Zeff depends on n and 
5.4 X-RAY SPECTRA
X-rays are emitted when high-energy electrons or any other charged particles
bombard a metal target. The x-ray spectrum typically consists of a broad continuous
band containing a series of sharp lines as shown in Figure 5.6. So, the x-ray
spectrum has two parts: continuous spectrum and characteristic spectrum.
Figure 5.6 The x-ray spectrum of a metal target. The data shown were obtained when 37-keV electrons
bombarded a molybdenum target.
An accelerated electric charge emits electromagnetic radiation. The x-rays in figure
are the result of the slowing down of high-energy electrons as they strike the target.
It may take several interactions with the atoms of the target before the electron loses
all its kinetic energy. The amount of kinetic energy lost in any given interaction can
vary from zero up to the entire kinetic energy of the electron. Therefore, the
wavelength of radiation from these interactions lies in a continuous range from
some minimum value up to infinity. It is this general slowing down of the electrons
71
that provides the continuous curve, which shows the cutoff of x-rays below a
minimum wavelength value that depends on the kinetic energy of the incoming
electrons. X-ray radiation with its origin in the slowing down of electrons is called
bremsstrahlung, the German word for “braking radiation”.
Thus the emitted x-rays can have any value for the wavelength above λMIN in
the continuous x-ray spectrum. Thus
e V  hf MAX 
MIN 
hc
MIN
hc
e V
(5.9)
λMIN depends only on ∆V
The peaks in the x-ray spectrum is the characteristic of the target element
in the x-ray tube and hence they form the characteristic x-ray spectrum. When
a high energy (K = e ∆V, ∆V = x-ray tube voltage) electron strikes a target atom
and knocks out one of its electrons from the inner shells with energy
Enf (|Enf | ≤ K, nf = integer), the vacancy in the inner shell is filled up by an
electron from the outer shell (energy = Eni, ni = integer). The characteristic xray photon emitted has the energy:
hf 
hc

 Eni  Enf
Figure 5.7 Transitions between higher and lower atomic energy levels that give rise to x-ray photons
from heavy atoms when they are bombarded with high-energy electrons.
72
A K x-ray results due to the transition of the electron from L-shell to Kshell. A K x-ray results due to the transition of the electron from M-shell to
K-shell. When the vacancy arises in the L-shell, an L-series (L, L, L) of xrays results. Similarly, the origin of M-series of x-rays can be explained.
Moseley’s observation on the characteristic K x-rays shows a relation between
the frequency (f) of the K x-rays and the atomic number (Z) of the target
element in the x-ray tube:
f  C Z  1
(5.10)
where C is a constant.
Note: Based on this observation, the elements are arranged according to their
atomic numbers in the periodic table
Figure 5.8 Moseley plot
5.5 SPONTANEOUS AND STIMULATED TRANSITIONS
There are three possible processes that involve interaction between matter and
radiation.
Stimulated Absorption: Absorption of a photon of frequency f takes place
when the energy difference E2 – E1 of the allowed energy states of the
atomic/molecular system equals the energy hf of the photon. Then the photon
disappears and the atomic system moves to upper energy state E2.
73
Figure 5.9 Stimulated absorption of a photon
Spontaneous Emission: The average life time of the atomic system in the
excited state is of the order of 10–8 s. After the life time of the atomic system
in the excited state, it comes back to the state of lower energy on its own
accord by emitting a photon of energy hf = E2– E1 .
This is the case with ordinary light sources. The radiations are emitted in different
directions in random manner. Such type of emission of radiation is called
spontaneous emission and the emitted light is not coherent.
Figure 5.10 Spontaneous Emission of a photon
Stimulated Emission: When a photon (called stimulating photon) of suitable
frequency interacts with an excited atomic system, the latter comes down to
ground state by emitting a photon of same energy. Such an emission of radiation
is called stimulated emission. In stimulated emission, both the stimulating photon
and the emitted photon (due to stimulation) are of same frequency, same phase,
same state of polarization and in the same direction. In other words, these two
photons are coherent.
74
Figure 5.11 Stimulated Emission
All the three processes are taking place simultaneously to varying degrees, in the
matter when it is irradiated by radiation of suitable frequency.
Population inversion: From Boltzmann statistics, the ratio of population of
atoms in two energy states E1 and E2 at equilibrium temperature T is,
 E  E1 
nE 2 

 exp  2
nE 1 
k
T


(5.11)
where k is Boltzmann constant, n(E1) is the number density of atoms with energy
E1 , n(E2) is the number density of atoms with energy E2 . Under normal condition,
where populations are determined only by the action of thermal agitation,
population of the atoms in upper energy state is less than that in lower energy
state (i.e. n(E2)<n(E1), Figure 5.12a).
Figure 5.12 (a) Normal thermal equilibrium distribution of atomic systems (b) An inverted population,
obtained using special techniques
We have described how an incident photon can cause atomic energy transitions
either upward (stimulated absorption) or downward (stimulated emission). The two
processes are equally probable. When light is incident on a collection of atoms, a
net absorption of energy usually occurs because when the system is in thermal
equilibrium, many more atoms are in the ground state than in excited states. If the
situation can be inverted so that more atoms are in an excited state than in the
75
ground state, however, a net emission of photons can result. Such a condition is
called population inversion.
5.6 LASER (LIGHT AMPLIFICATION BY STIMULATED
EMISSION OF RADIATION)
Laser light is highly monochromatic, intense, coherent, directional and can be
sharply focused. Each of these characteristics that are not normally found in
ordinary light makes laser a unique and the most powerful tool. Population
inversion is, in fact, the fundamental principle involved in the operation of a laser.
The full name indicates one of the requirements for laser light: to achieve laser
action, the process of stimulated emission must occur.
For the stimulated emission rate to exceed the absorption rate it is necessary
to have higher population of upper energy state than that of lower energy
state. This condition is called population inversion [n(E2)>n(E1)]. This is a nonequilibrium condition and is facilitated by the presence of energy states called
‘metastable states’ where the average life time of the atom is 10-3 s which is much
longer than that of the ordinary excited state ( 10-8s).
Suppose an atom is in the excited state E2 as in the below figure and a photon with
energy hf = E2 - E1 is incident on it. The incoming photon can stimulate the excited
atom to return to the ground state and thereby emit a second photon having the
same energy hf and traveling in the same direction. The incident photon is not
absorbed, so after the stimulated emission, there are two identical photons: the
incident photon and the emitted photon. The emitted photon is in phase with the
incident photon. These photons can stimulate other atoms to emit photons in a
chain of similar processes. The many photons produced in this fashion are the
source of the intense, coherent light in a laser.
For the stimulated emission to result in laser light, there must be a buildup of
photons in the system. The following three conditions must be satisfied to achieve
this buildup:
• The system must be in a state of population inversion: there must be more atoms
in an excited state than in the ground state. That must be true because the number
of photons emitted must be greater than the number absorbed.
• The excited state of the system must be a metastable state, meaning that its
lifetime must be long compared with the usually short lifetimes of excited states,
which are typically 10-8 s. In this case, the population inversion can be established
and stimulated emission is likely to occur before spontaneous emission.
• The emitted photons must be confined in the system long enough to enable
them to stimulate further emission from other excited atoms. That is achieved by
using reflecting mirrors at the ends of the system. One end is made totally reflecting,
76
and the other is partially reflecting. A fraction of the light intensity passes through
the partially reflecting end, forming the beam of laser light.
Figure 5.13 Schematic diagram of a laser design.
Lasing medium (active medium), resonant cavity and pumping system are the
essential parts of any lasing system. Lasing medium has atomic systems (active
centers), with special energy levels which are suitable for laser action. This
medium may be a gas, or a liquid, or a crystal or a semiconductor. The
atomic systems in this may have energy levels including a ground state (E1),
an excited state (E3) and a metastable state (E2). The resonant cavity is a pair
of parallel mirrors to reflect the radiation back into the lasing medium.
Pumping is a process of exciting more number of atoms in the ground state
to higher energy states, which is required for attaining the population
inversion.
Figure 5.14 Energy-level diagram for a neon atom in a helium–neon laser.
77
In He-Ne laser, the mixture of helium and neon is confined to a glass tube that is
sealed at the ends by mirrors. A voltage applied across the tube causes electrons to
sweep through the tube, colliding with the atoms of the gases and raising them into
excited states. Neon atoms are excited to state E3* through this process (the asterisk
indicates a metastable state) and also as a result of collisions with excited helium
atoms. Stimulated emission occurs, causing neon atoms to make transitions to state
E2. Neighboring excited atoms are also stimulated. The result is the production of
coherent light at a wavelength of 632.8 nm.
5.7 APPLICATIONS OF LASER
Laser is used in various scientific, engineering and medical applications. It is used
in investigating the basic laws of interaction of atoms and molecules with
electromagnetic wave of high intensity. Laser is widely used in engineering
applications like optical communication, micro-welding and sealing etc. In medical
field, laser is used in bloodless and painless surgery especially in treating the retinal
detachment. Also used as a tool in treating dental decay, tooth extraction, cosmetic
surgery.
5.8 QUESTIONS
1 Give a brief account of quantum model of H-atom.
Explain the origin of (i) orbital quantum number (ii) magnetic orbital
quantum number and write the relation between them
2 The wave function for H-atom in ground state is ψ1S (r) =
1
√πa3o
exp (− aro) . Obtain an expression for the radial probability density
of H-atom in ground state.
Sketch schematically the plot of this vs. radial distance.
3 The wave
function for
H-atom
in 2s
state is ψ2S (r) =
1
√32πa3o
(2 − 𝑎𝑟𝑜 ) exp (− aro) .
Write
the
expression
for
the
radial
probability density of H-atom in 2s state. Sketch schematically the plot
of this vs. radial distance.
4 Sketch schematically the plot of the radial probability density vs. radial
distance for H-atom in 1s-state and 2s-state.
5 Explain the continuous x-ray spectrum with a schematic plot of the
spectrum.
6 Explain the origin of characteristic x-ray spectrum with a sketch of xray energy level diagram.
78
7 Write Moseley’s relation for the frequency of characteristic x-rays.
sketch schematically the Moseley’s plot of characteristic x-rays.
8 Explain three types of transitions between two energy levels, when radiation
interacts with matter
9 Explain the characteristics of a laser beam
10 Explain metastable state
11 What is population inversion? explain
12 Describe the principle of a laser using necessary schematic design and
energy level diagram
13 Mention any four applications of laser.
14 Describe the three important conditions need to be satisfied to achieve
laser action
5.9 PROBLEMS
1 For a H-atom, determine the number of allowed states corresponding
to the principal quantum number n = 2, and calculate the energies of
these states.
Ans: 4 states (one 2s-state + three 2p-states), –3.401 eV
2 A general expression for the energy levels of one-electron atoms and
ions is 𝐸𝑛 = −
𝜇 𝑘𝑒2 𝑞12 𝑞22
2 ℏ2 𝑛 2
,
where ke is the Coulomb constant, q1 and q2
are the charges of the electron and the nucleus, and μ is the reduced
m m
mass, given by μ = m 1+m2 . The wavelength for n = 3 to n = 2 transition
1
2
of the hydrogen atom is 656.3 nm (visible red light). What are the
wavelengths for this same transition (a) positronium, which consists of
an electron and a positron, and (b) singly ionized helium ?
Ans: 1310 nm, 164 nm
3 Calculate the most probable value of r (= distance from nucleus) for an
electron in the ground state of the H-atom. Also calculate the average
value r for the electron in the ground state.
Ans: ao , 3 ao/2
4 Calculate the probability that the electron in the ground state of Hatom will be found outside the Bohr radius.
Ans: 0.677
5 For a spherically symmetric state of a H-atom the Schrodinger equation
in spherical coordinates is−
ℏ2
2𝑚
𝜕2 𝜓
( 𝜕𝑟 2 +
2 𝜕𝜓
𝑟
) −
𝜕𝑟
𝑘𝑒 𝑒 2
𝑟
𝜓 = 𝐸 𝜓 . Show
79
that the 1s wave function for an electron in H-atom
1
√𝜋𝑎𝑜3
𝜓1𝑆 (𝑟) =
𝑒𝑥𝑝 (− 𝑎𝑟𝑜) satisfies the Schrodinger equation.
6 The ground-state wave function for the electron in a hydrogen atom is
1
𝑟
𝜓1𝑆 (𝑟) =
𝑒𝑥𝑝 (− 𝑎 )
𝑜
√𝜋𝑎𝑜3
where r is the radial coordinate of the electron and a0 is the Bohr radius.
(a) Show that the wave function as given is normalized. (b) Find the
probability of locating the electron between r1 = a0/2 and r2 = 3a0/2.
7 What minimum accelerating voltage would be required to produce an x-ray
with a wavelength of 70.0 pm? Ans: 17.7 kV
8 A tungsten target is struck by electrons that have been accelerated from rest
through a 40.0-keV potential difference. Find the shortest wavelength of the
radiation emitted. Ans: 0.031 nm
9 A bismuth target is struck by electrons, and x-rays are emitted. Estimate (a)
the M- to L-shell transitional energy for bismuth and (b) the wavelength of
the x-ray emitted when an electron falls from the M shell to the L shell.
Ans: (a) 14 keV (b) 0.885 Å
10 The 3p level of sodium has an energy of -3.0 eV, and the 3d level has an energy
of -1.5 eV. (a) Determine Zeff for each of these states. (b) Explain the
difference. Ans: (a) 1.4 and 1.0
11 The K series of the discrete x-ray spectrum of tungsten contains wavelengths
of 0.0185 nm, 0.020 9 nm, and 0.021 5 nm. The K-shell ionization energy is
69.5 keV.
(a) Determine the ionization energies of the L, M, and N shells.
(b) Draw a diagram of the transitions.
Ans: (a) L shell = 11.8 keV ; M shell = 10.2 keV ; N shell = 2.47 keV
80
12 When an electron drops from the M shell (n = 3) to a vacancy in the K shell
(n = 1), the measured wavelength of the emitted x-ray is found to be 0.101 nm.
Identify the element. Ans: Gallium (Z=31)
13 A ruby laser delivers a 10.0-ns pulse of 1.00-MW average power. If the
photons have a wavelength of 694.3 nm, how many are contained in the
pulse? Ans: 3.49x1016 photons
14 A pulsed laser emits light of wavelength . For a pulse of duration t having
energy TER, find (a) the physical length of the pulse as it travels through space
and (b) the number of photons in it. (c) The beam has a circular cross section
having diameter d. Find the number of photons per unit volume. (a) cΔt (b)
𝜆𝑇𝐸𝑅
ℎ𝑐
4𝜆𝑇
(c) 𝑛 = 𝜋ℎ𝑐 2 𝑑𝐸𝑅
2 Δ𝑡
81
6 MOLECULES AND SOLIDS
OBJECTIVES:





To understand the bonding mechanism, energy states and spectra of
molecules
To understand the cohesion of solid metals using bonding in solids
To comprehend the electrical properties of metals, semiconductors and
insulators
To understand the effect of doping on electrical properties of
semiconductors
To understand superconductivity and its engineering applications
6.1 MOLECULAR BONDS
The bonding mechanisms in a molecule are fundamentally due to electric forces
between atoms (or ions). The forces between atoms in the system of a molecule are
related to a potential energy function. A stable molecule is expected at a
configuration for which the potential energy function for the molecule has its
minimum value. A potential energy function that can be used to model a molecule
should account for two known features of molecular bonding:
1. The force between atoms is repulsive at very small separation distances. This
repulsion is partly electrostatic in origin and partly the result of the exclusion
principle.
2. At relatively at larger separations, the force between atoms is attractive.
Considering these two features, the potential energy for a system of two atoms can
be represented by an expression of the form (Lennard–Jones potential)
U r   
A B

rn rm
(6.1)
where r is the internuclear separation distance between the two atoms and n and m
are small integers. The parameter A is associated with the attractive force and B with
the repulsive force. Potential energy versus internuclear separation distance for a
two-atom system is graphed in Figure 6.1.
82
Figure 6.1 Total potential energy as a function of internuclear separation distance for a system of two
atoms.
Ionic Bonding: When two atoms combine in such a way that one or more outer
electrons are transferred from one atom to the other, the bond formed is called an
ionic bond. Ionic bonds are fundamentally caused by the Coulomb attraction
between oppositely charged ions.
Figure 6.2 Total energy versus internuclear separation distance for Na + and Cl- ions.
A familiar example of an ionically bonded solid is sodium chloride, NaCl, which is
common table salt. Sodium, which has the electronic configuration 1s22s22p63s1, is
ionized relatively easily, giving up its 3s electron to form a Na+ ion. The energy
required to ionize the atom to form Na+ is 5.1 eV. Chlorine, which has the electronic
configuration 1s22s22p5 is one electron short of the filled-shell structure of argon. The
amount of energy released when an electro joins Cl atom to make the Cl ̶ ion, called
the electron affinity of the atom, is 3.6 eV. Therefore, the energy required to form
Na+ and Cl ̶ from isolated atoms is 5.1 - 3.6 = 1.5 eV. The total energy of the NaCl
molecule versus internuclear separation distance is graphed in Figure 6.2. At very
large separation distances, the energy of the system of ions is 1.5 eV as calculated
above. The total energy has a minimum value of - 4.2 eV at the equilibrium
83
separation distance, which is approximately 0.24 nm. Hence, the energy required to
break the Na+ ̶ Cl ̶ bond and form neutral sodium and chlorine atoms, called the
dissociation energy, is 4.2 eV. The energy of the molecule is lower than that of the
system of two neutral atoms. Consequently, it is energetically favorable for the
molecule to form.
Covalent Bonding: A covalent bond between two atoms is one in which electrons
supplied by either one or both atoms are shared by the two atoms. Many diatomic
molecules such as H2, F2, and CO—owe their stability to covalent bonds. The bond
between two hydrogen atoms can be described by using atomic wave functions for
two atoms. There is very little overlap of the wave functions ψ1(r) for atom 1, located
at r = 0, and ψ2(r) for atom 2, located some distance away (Figure 6.3a). Suppose now
the two atoms are brought close together, their wave functions overlap and form the
compound wave function ψ1(r) + ψ2(r) shown in Figure 6.3b. Notice that the
probability amplitude is larger between the atoms than it is on either side of the
combination of atoms. As a result, the probability is higher that the electrons
associated with the atoms will be located between the atoms than on the outer
regions of the system. Consequently, the average position of negative charge in the
system is halfway between the atoms.
Figure 6.3 Ground-state wave functions ψ1(r) and ψ2(r) for two atoms making a
covalent bond. (a) The atoms are far apart, and their wave functions overlap
minimally. (b) The atoms are close together, forming a composite wave function
ψ1(r) + ψ2(r) for the system.
Van der Waals Bonding: Ionic and covalent bonds occur between atoms to form
molecules or ionic solids, so they can be described as bonds within molecules. The
84
van der Waals force results from the following situation. While being electrically
neutral, a molecule has a charge distribution with positive and negative centers at
different positions in the molecule. As a result, the molecule may act as an electric
dipole. Because of the dipole electric fields, two molecules can interact such that
there is an attractive force between them. There are three types of van der Waals
forces.
The first type, called the dipole– dipole force, is an interaction between two
molecules each having a permanent electric dipole moment. For example, polar
molecules such as HCl have permanent electric dipole moments and attract other
polar molecules.
The second type, the dipole–induced dipole force, results when a polar molecule
having a permanent electric dipole moment induces a dipole moment in a nonpolar
molecule. In this case, the electric field of the polar molecule creates the dipole
moment in the nonpolar molecule, which then results in an attractive force between
the molecules.
The third type is called the dispersion force, an attractive force that occurs between
two nonpolar molecules. Two nonpolar molecules near each other tend to have
dipole moments that are correlated in time so as to produce an attractive van der
Waals force.
Hydrogen Bonding: Because hydrogen has only one electron, it is expected to form
a covalent bond with only one other atom within a molecule. A hydrogen atom in a
given molecule can also form a second type of bond between molecules called a
hydrogen bond. Let’s use the water molecule H2O as an example. In the two
covalent bonds in this molecule, the electrons from the hydrogen atoms are more
likely to be found near the oxygen atom than near the hydrogen atoms, leaving
essentially bare protons at the positions of the hydrogen atoms. This unshielded
positive charge can be attracted to the negative end of another polar molecule.
Because the proton is unshielded by electrons, the negative end of the other
molecule can come very close to the proton to form a bond strong enough to form
a solid crystalline structure, such as that of ordinary ice. The bonds within a water
molecule are covalent, but the bonds between water molecules in ice are hydrogen
bonds. The hydrogen bond is relatively weak compared with other chemical bonds
and can be broken with an input energy of approximately 0.1 eV. Because of this
weakness, ice melts at the low temperature of 0°C.
85
6.2 ENERGY STATES AND SPECTRA OF MOLECULES
Consider an individual molecule in the gaseous phase of a substance. The energy E
of the molecule can be divided into four categories: (1) electronic energy, due to the
interactions between the molecule’s electrons and nuclei; (2) translational energy,
due to the motion of the molecule’s center of mass through space; (3) rotational
energy, due to the rotation of the molecule about its center of mass; and (4)
vibrational energy, due to the vibration of the molecule’s constituent atoms:
E = Eel + Etrans + Erot + Evib
Because the translational energy is unrelated to internal structure, this molecular
energy is unimportant in interpreting molecular spectra. Although the electronic
energies can be studied, significant information about a molecule can be determined
by analyzing its quantized rotational and vibrational energy states. Transitions
between these states give spectral lines in the microwave and infrared regions of the
electromagnetic spectrum, respectively.
6.3 ROTATIONAL MOTION OF MOLECULES
Let’s consider the rotation of a diatomic molecule around its center of mass (Figure
6.4a). A diatomic molecule aligned along a y axis has only two rotational degrees of
freedom, corresponding to rotations about the x and z axes passing through the
molecule’s center of mass. If  is the angular frequency of rotation about one of
these axes, the rotational kinetic energy of the molecule about that axis can be
expressed as
1
Erot  I 2
2
(6.2)
In this equation, I is the moment of inertia of the molecule about its center of mass,
given by
 mm 
I   1 2  r2   r2
 m1  m2 
(6.3)
where m1 and m2 are the masses of the atoms that form the molecule, r is the atomic
separation, and  is the reduced mass of the molecule

m1m2
m1  m2
(6.4)
86
Figure 6.4 Rotation of a diatomic molecule around its center of mass. (a) A diatomic molecule
oriented along the y axis. (b) Allowed rotational energies of a diatomic molecule expressed as
multiples of E1 = ℏ2/I.
The magnitude of the molecule’s angular momentum about its center of mass is
L=Iω, which can attain quantized values given by,
L  J  J  1
J  0, 1, 2, ...
(6.5)
where J is an integer called the rotational quantum number. Combining
Equations 6.5 and 6.2, we obtain an expression for the allowed values of the
rotational kinetic energy of the molecule:
Erot  EJ 
2
2I
J  J  1
J  0, 1, 2, ...
(6.6)
The allowed rotational energies of a diatomic molecule are plotted in Figure 6.4b.
The allowed rotational transitions of linear molecules are regulated by the selection
rule ΔJ =±1. From Equation 6.6, the energies of the absorbed photons are given by,
Ephoton 
2
I
J
h2
4 2 I
J
J  1, 2, 3, ...
(6.7)
where J is the rotational quantum number of the higher energy state.
6.4 VIBRATIONAL MOTION OF MOLECULES
If we consider a molecule to be a flexible structure in which the atoms are bonded
together by “effective springs” as shown in Figure 6.5a, we can model the molecule
as a simple harmonic oscillator as long as the atoms in the molecule are not too far
from their equilibrium positions. Figure 6.5b shows a plot of potential energy versus
87
atomic separation for a diatomic molecule, where r0 is the equilibrium atomic
separation. For separations close to r0, the shape of the potential energy curve
closely resembles a parabola. According to classical mechanics, the frequency of
vibration for the system is given by
f 
1
2
k
(6.8)

where k is the effective spring constant and  is the reduced mass given by Equation
6.4. The vibrational motion and quantized vibrational energy can be altered if the
molecule acquires energy of the proper value to cause a transition between
quantized vibrational states. The allowed vibrational energies are
1

Evib   v   hf
2

v  0, 1, 2, ...
(6.9)
where v is an integer called the vibrational quantum number.
Figure 6.5 (a) Effective-spring model of a diatomic molecule. (b) Plot of the potential energy of a
diatomic molecule versus atomic separation distance.
88
Figure 6.6 Allowed vibrational energies of a diatomic molecule, where f is the frequency of vibration of
the molecule
Substituting Equation 6.8 into Equation 6.9 gives the following expression for the
allowed vibrational energies:
1 h

Evib   v  
2  2

k
v  0, 1, 2, ...

(6.10)
The selection rule for the allowed vibrational transitions is Δv = ±1. The photon
energy for transition is given by,
Ephoton  Evib 
h
2
k

(6.11)
The vibrational energies of a diatomic molecule are plotted in Figure 6.6. At ordinary
temperatures, most molecules have vibrational energies corresponding to the v = 0
state because the spacing between vibrational states is much greater than kBT, where
kB is Boltzmann’s constant and T is the temperature.
6.5 MOLECULAR SPECTRA
In general, a molecule vibrates and rotates simultaneously. To a first approximation,
these motions are independent of each other, so the total energy of the molecule is
the sum of Equations 6.6 and 6.9:
2
1

E   v   hf  J  J  1
2
2I

(6.12)
The energy levels of any molecule can be calculated from this expression, and each
level is indexed by the two quantum numbers v and J. From these calculations, an
energy-level diagram like the one shown in Figure 6.7a can be constructed. For each
allowed value of the vibrational quantum number v, there is a complete set of
rotational levels corresponding to J = 0, 1, 2, . . . . The energy separation between
successive rotational levels is much smaller than the separation between successive
vibrational levels. The molecular absorption spectrum in Figure 6.7b consists of two
groups of lines: one group to the right of center and satisfying the selection rules ΔJ
= +1 and Δv = +1, and the other group to the left of center and satisfying the selection
rules ΔJ = -1 and Δv = +1. The energies of the absorbed photons can be calculated
from Equation 6.12:
89
Ephoton  E  hf 
2
I
Ephoton  E  hf 
 J  1
2
I
J
J  0,1, 2, ...
J  1, 2, 3, ...
 J  1
 J  1
(6.13)
(6.14)
where J is the rotational quantum number of the initial state.
Figure 6.7 (a) Absorptive transitions between the v = 0 and v = 1 vibrational states of a diatomic
molecule. (b) Expected lines in the absorption spectrum of a molecule.
The experimental absorption spectrum of the HCl molecule shown in Figure 6.8.
One peculiarity is apparent, however: each line is split into a doublet. This doubling
occurs because two chlorine isotopes were present in the sample used to obtain this
spectrum. Because the isotopes have different masses, the two HCl molecules have
different values of I.
90
Figure 6.8 Experimental absorption spectrum of the HCl molecule
The second function determining the envelope of the intensity of the spectral lines
is the Boltzmann factor. The number of molecules in an excited rotational state is
given by
𝑛 = 𝑛0
−ℏ2 𝐽(𝐽+1)
𝑒 2𝐼𝑘𝐵 𝑇
where n0 is the number of molecules in the J = 0 state.
Multiplying these factors together indicates that the intensity of spectral lines
should be described by a function of J as follows:
𝐼𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 ∝ (2𝐽 + 1)𝑒
−ℏ2 𝐽(𝐽+1)
2𝐼𝑘𝐵 𝑇
(6.15)
6.6 BONDING IN SOLIDS
A crystalline solid consists of a large number of atoms arranged in a regular array,
forming a periodic structure.
Ionic Solids: Many crystals are formed by ionic bonding, in which the dominant
interaction between ions is the Coulomb force. Consider a portion of the NaCl
crystal shown in Figure 6.9a. The red spheres are sodium ions, and the blue spheres
are chlorine ions. As shown in Figure 6.9b, each Na+ ion has six nearest-neighbor Clions. Similarly, in Figure 6.9c, each Cl- ion has six nearest-neighbor Na+ ions. Each
Na+ ion is attracted to its six Cl- neighbors. The corresponding potential energy is 6kee2/r, where ke is the Coulomb constant and r is the separation distance between
each Na+ and Cl-. In addition, there are 12 next-nearest-neighbor Na+ ions at a
distance of √2r from the Na+ ion, and these 12 positive ions exert weaker repulsive
forces on the central Na+. Furthermore, beyond these 12 Na+ ions are more Cl2 ions
that exert an attractive force, and so on. The net effect of all these interactions is a
resultant negative electric potential energy
91
U attractive   ke
e2
r
(6.16)
where α is a dimensionless number known as the Madelung constant. The value
of α depends only on the particular crystalline structure of the solid (α = 1.747 for
the NaCl structure). When the constituent ions of a crystal are brought close
together, a repulsive force exists because of electrostatic forces and the exclusion
principle. The potential energy term B/rm in Equation 6.1 accounts for this repulsive
force. The repulsive forces occur only for ions that are very close together.
Therefore, we can express the total potential energy of the crystal as
U total   ke
e2 B

r rm
(6.17)
where m in this expression is some small integer.
Figure 6.9 (a) Crystalline structure of NaCl. (b) Each positive sodium ion is surrounded by six
negative chlorine ions. (c) Each chlorine ion is surrounded by six sodium ions
Figure 6.10 Total potential energy versus ion separation distance for an ionic solid, where U 0 is the
ionic cohesive energy and r0 is the equilibrium separation distance between ions
92
A plot of total potential energy versus ion separation distance is shown in Figure
6.10. The potential energy has its minimum value U0 at the equilibrium separation,
when r = r0.
U 0   ke
e2  1 
1  
r0  m 
(6.18)
This minimum energy U0 is called the ionic cohesive energy of the solid, and its
absolute value represents the energy required to separate the solid into a collection
of isolated positive and negative ions.
Covalent Solids: Solid carbon, in the form of diamond, is a crystal whose atoms are
covalently bonded. In the diamond structure, each carbon atom is covalently
bonded to four other carbon atoms located at four corners of a cube as shown in
Figure 6.11a. The crystalline structure of diamond is shown in Figure 6.11b. The basic
structure of diamond is called tetrahedral (each carbon atom is at the center of a
regular tetrahedron), and the angle between the bonds is 109.5°. Other crystals such
as silicon and germanium have the same structure. Covalently bonded solids are
usually very hard, have high bond energies and high melting points, and are good
electrical insulators.
Figure 6.11 (a) Each carbon atom in a diamond crystal is covalently bonded to four other carbon
atoms so that a tetrahedral structure is formed. (b) The crystal structure of diamond, showing the
tetrahedral bond arrangement
Metallic Solids: Metallic bonds are generally weaker than ionic or covalent bonds.
The outer electrons in the atoms of a metal are relatively free to move throughout
the material, and the number of such mobile electrons in a metal is large. The
metallic structure can be viewed as a “sea” or a “gas” of nearly free electrons
surrounding a lattice of positive ions (Fig. 6.14). The bonding mechanism in a metal
93
is the attractive force between the entire collection of positive ions and the electron
gas.
Light interacts strongly with the free electrons in metals. Hence, visible light is
absorbed and re-emitted quite close to the surface of a metal, which accounts for
the shiny nature of metal surfaces. Because the bonding in metals is between all the
electrons and all the positive ions, metals tend to bend when stressed.
Figure 6.12 Highly schematic diagram of a metal
6.7 FREE-ELECTRON THEORY OF METALS
Quantum based free electron theory of metals is centered on wave nature of
electrons. In this model, one imagines that the outer-shell electrons are free to
move through the metal but are trapped within a three-dimensional box
formed by the metal surfaces. Therefore, each electron is represented as a
particle in a box and is restricted to quantized energy levels. Each energy state
can be occupied by only two electrons (one with spin up & the other with spin
down) as a consequence of exclusion principle. In quantum statistics, it is shown
that the probability of a particular energy state E being occupied by an electrons is
given by
f E 
1
 E  EF 
exp 
  1
 kT 
(6.19)
where f(E) is called the Fermi-Dirac distribution function and EF is called the
Fermi energy. Plot of f(E) versus E is shown in figure 6.13.
94
Figure. 6.13 Plot of Fermi-Dirac distribution function f(E) Vs energy E at (a) T = 0K and (b) T > 0K
At zero kelvin (0 K), all states having energies less than the Fermi energy are
occupied, and all states having energies greater than the Fermi energy are vacant.
i.e. Fermi energy is the highest energy possessed by an electron at 0 K (Figure 6.13a).
As temperature increases (T > 0K), the distribution rounds off slightly due to
thermal excitation and probability of Fermi level being occupied by an electron
becomes half (Figure 6.13b). In other words, Fermi energy is that energy state at
which probability of electron occupation is half. The Fermi energy EF also depends
on temperature, but the dependence is weak in metals.
Density of states: From particle in a box problem, for a particle of mass m is
confined to move in a one-dimensional box of length L, the allowed states have
quantized energy levels given by,
2 2
h2

2
En 
n 
n2
2
2
8mL
2mL
n = 1, 2, 3 . . .
(6.20)
An electron moving freely in a metal cube of side L, can be modeled as particle in a
three-dimensional box. It can be shown that the energy for such an electron is
2
E
n 2  n y2  nz2 
2  x
2mL
2
(6.21)
where m is mass of the electron and nx, ny, nz are quantum numbers(positive
integers). Each allowed energy value is characterized by a set of three quantum
numbers (nx, ny, nz - one for each degree of freedom). Imagine a threedimensional quantum number space whose axes represent nx, ny, nz. The
allowed energy states in this space can be represented as dots located at positive
integral values of the three quantum numbers as shown in the Figure 6.14.
95
Figure 6.14 Representation of the allowed energy states in a quantum number space (dots represent
the allowed states)
Eq. 6.21 can be written as
nx2  ny2  nz2 
where Eo 
2
2
2
2mL
E
 n2
Eo
and n 
(6.22)
E
Eo
Eq. 6.22 represents a sphere of radius n. Thus, the number of allowed energy
states having energies between E and E+dE is equal to the number of points
in a spherical shell of radius n and thickness dn. In this quantum number space
each point is at the corners of a unit cube and each corner point is shared by eight
unit cubes and as such the contribution of each point to the cube is 1/8 th. Because
a cube has eight corners, the effective point per unit cube and hence unit volume is
one. In other words, number of points is equal to the volume of the shell. The
“volume” of this shell, denoted by G(E)dE.
1
G(E) dE =    4 n 2 dn 
8

1
2
   n dn
2
96
 E  1 2 


E
G ( E ) dE  12  
 d 
 
E
E

 o   o  
 E   12
 Eo
 Eo 
G ( E ) dE 
1
2

G( E ) dE 
1
4
 2 2 

2 
 2mL 
3
1
2
E
2
E
1
1
2
using the relation n 
dE 
2
1
4
3
 Eo
2
E
1
2
E
Eo
dE
dE
3
2 m 2 L3 12
G ( E ) dE 
E dE ,
2 2 3
L3  V
Number of states per unit volume per unit energy range, called density of
states, g(E) is given by
g(E) = G(E)/V
G( E )
2 m
g ( E ) dE 
dE 
V
2 2
4 2 m
g ( E ) dE 
h3
3
2
E
1
2
3
2
3
E
1
2
dE

dE
h
2
Finally, we multiply by 2 for the two possible spin states of each particle.
8 2 m
g ( E ) dE 
h3
3
2
E
1
2
dE
(6.23)
g(E) is called the density-of-states function.
Electron density: For a metal in thermal equilibrium, the number of electrons
N(E) dE, per unit volume, that have energy between E and E+dE is equal to
the product of the density of states and the probability that a state is occupied.
that is,
N(E)dE = [ g(E)dE ] f(E)
97
8 2 m
N ( E ) dE 
h3
3
2
1
E 2 dE
 E  EF 
exp 
  1
 kT 
(6.24)
Plots of N(E) versus E for two temperatures are given in figure 6.15.
Figure 6.15 Plots of N(E) versus E for (a) T = 0K (b) T = 300K
If ne is the total number of electrons per unit volume, we require that

8 2 m
ne   N ( E ) dE 
h3
0
3
2
1

E 2 dE
 E  EF 
exp 
  1
 kT 

0
(6.25)
At T = 0K, the Fermi-Dirac distribution function f(E) = 1 for E <EF and f(E) = 0 for E
>EF. Therefore, at T = 0K, Equation 5.7 becomes
8 2 m
ne 
h3
3
2
EF

0
E
1
2
8 2 m
dE 
h3
3
2
  EF
2
3
3
2
16 2  m

3 h3
3
2
3
EF 2
(6.26)
Solving for Fermi energy at 0K, we obtain
h 2  3 ne 
EF  0  


2 m  8 
2
3
(6.27)
The average energy of a free electron in a metal at 0K is Eav = (3/5)EF.
98
6.8 BAND THEORY OF SOLIDS
When a quantum system is represented by wave function, probability density ||2
for that system is physically significant while the probability amplitude  not.
Consider an atom such as sodium that has a single s electron outside of a closed



shell. Both the wave functions  S ( r ) and  S ( r ) are valid for such an atom [ S ( r )

and  S ( r ) are symmetric and anti symmetric wave functions]. As the two sodium
atoms are brought closer together, their wave functions begin to overlap. Figure 6.16
represents two possible combinations : i) symmetric - symmetric and ii) symmetric
– antisymmetric . These two possible combinations of wave functions
represent two possible states of the two-atom system. Thus, the states are
split into two energy levels. The energy difference between these states is relatively
small, so the two states are close together on an energy scale.
When two atoms are brought together, each energy level will split into 2 energy
levels. (In general, when N atoms are brought together N split levels will occur which
can hold 2N electrons). The split levels are so close that they may be regarded as a
continuous band of energy levels. Figure 6.17 shows the splitting of 1s and 2s
levels of sodium atom when : (a) two sodium atoms are brought together (b)five
sodium atoms are brought together (c) a large number of sodium atoms are
assembled to form a solid. The close energy levels forming a band are seen clearly
in (c).
Figure. 6.16 The wave functions of two atoms combine to form a composite wave function : a)
symmetric-symmetric b) symmetric-antisymmetric
99
Figure 6.17 Splitting of 1s and 2s levels of sodium atoms due to interaction between them
Some bands may be wide enough in energy so that there is an overlap
between the adjacent bands. Some other bands are narrow so that a gap may occur
between the allowed bands, and is known as forbidden energy gap. The 1s, 2s,
and 2p bands of solid sodium are filled completely with electrons. The 3s
band (2N states) of solid sodium has only N electrons and is partially full;
The 3p band, which is the higher region of the overlapping bands, is completely
empty as shown in Figure 6.18
Figure 6.18 Energy bands of a sodium crystal
100
6.9 ELECTRICAL CONDUCTION IN METALS, INSULATORS
AND SEMICONDUCTORS
Good electrical conductors contain high density of free charge carriers, and
the density of free charge carriers in insulators is nearly zero. In
semiconductors free-charge-carrier densities are intermediate between those of
insulators and those of conductors.
Metals: Metal has a partially filled energy band (Figure 6.19a). At 0K Fermi level
is the highest electron-occupied energy level. If a potential difference is applied
to the metal, electrons having energies near the Fermi energy require only a
small amount of additional energy to reach nearby empty energy states above
the Fermi-level. Therefore, electrons in a metal experiencing a small force (from a
weak applied electric field) are free to move because many empty levels are available
close to the occupied energy levels. The model of metals based on band theory
demonstrates that metals are excellent electrical conductors.
Insulators: Consider the two outermost energy bands of a material in which the
lower band is filled with electrons and the higher band is empty at 0 K (Figure 6.19b).
The lower, filled band is called the valence band, and the upper, empty band is the
conduction band. The energy separation between the valence and conduction band,
called energy gap Eg, is large for insulating materials. The Fermi level lies
somewhere in the energy gap. Due to larger energy gap compare to thermal energy
kT (26meV) at room temperature, excitation of electrons from valence band to
conduction band is hardly possible. Since the free-electron density is nearly zero,
these materials are bad conductors of electricity.
Semiconductors: Semiconductors have the same type of band structure as an
insulator, but the energy gap is much smaller, of the order of 1 eV. The band
structure of a semiconductor is shown in Figure 6.19c. Because the Fermi level is
located near the middle of the gap for a semiconductor and Eg is small, appreciable
numbers of electrons are thermally excited from the valence band to the conduction
band. Because of the many empty levels above the thermally filled levels in the
conduction band, a small applied potential difference can easily raise the energy of
the electrons in the conduction band, resulting in a moderate conduction. At T = 0
K, all electrons in these materials are in the valence band and no energy is available
to excite them across the energy gap. Therefore, semiconductors are poor
conductors at very low temperatures. Because the thermal excitation of electrons
across the narrow gap is more probable at higher temperatures, the conductivity of
101
semiconductors increases rapidly with temperature. This is in sharp contrast with
the conductivity of metals, where it decreases with increasing temperature. Charge
carriers in a semiconductor can be negative, positive, or both. When an electron
moves from the valence band into the conduction band, it leaves behind a vacant
site, called a hole, in the otherwise filled valence band.
Figure 6.19 Band structure of (a) Metals (b) Insulators (c) Semiconductors
In an intrinsic semiconductor (pure semiconductor) there are equal number
of conduction electrons and holes. In the presence of an external electric
field, the holes move in the direction of field and the conduction electrons
move opposite to the direction of the field. Both these motions correspond
to the current in the same direction (Figure 6.20).
Figure 6.20 Movement of electrons and holes in an intrinsic semiconductor
102
Doped Semiconductors (Extrinsic semiconductors): Semiconductors in their
pure form are called intrinsic semiconductors while doped semiconductors are
called extrinsic semiconductors. Doping is the process of adding impurities to a
semiconductor. By doping both the band structure of the semiconductor and
its resistivity are modified. If a tetravalent semiconductor (Si or Ge) is doped
with a pentavalent impurity atom (donor atom), four of the electrons form
covalent bonds with atoms of the semiconductor and one is left over (Figure
6.21). At zero K, this extra electron resides in the donor-levels, that lie in the
energy gap, just below the conduction band. Since the energy Ed between the
donor levels and the bottom of the conduction band is small, at room
temperature, the extra electron is thermally excited to the conduction band.
This type of semiconductors are called n-type semiconductors because the
majority of charge carriers are electrons (negatively charged).
If a tetravalent semiconductor is doped with a trivalent impurity atom
(acceptor atom), the three electrons form covalent bonds with neighboring
semiconductor atoms, leaving an electron deficiency (a hole) at the site of
fourth bond (Figure 6.22). At zero K, this hole resides in the acceptor levels
that lie in the energy gap just above the valence band. Since the energy E a
between the acceptor levels and the top of the valence band is small, at room
temperature, an electron from the valence band is thermally excited to the
acceptor levels leaving behind a hole in the valence band. This type of
semiconductors are called p-type semiconductors because the majority of
charge carriers are holes (positively charged).
103
Figure 6.21 n-type semiconductor – two dimensional representation and band structure
When conduction in a semiconductor is the result of acceptor or donor impurities,
the material is called an extrinsic semiconductor. The typical range of doping
densities for extrinsic semiconductors is 1013 to 1019 cm-3, whereas the electron
density in a typical semiconductor is roughly 1021 cm-3.
Figure 6.22 p-type semiconductor: two-dimensional representation and band structure
6.10 SUPERCONDUCTIVITY-PROPERTIES AND
APPLICATIONS
Superconductor is a class of metals and compounds whose electrical resistance
decreases to virtually zero below a certain temperature Tc called the critical
temperature. The critical temperature is different for different superconductors.
Mercury loses its resistance completely and turns into a superconductor at 4.2K.
Critical temperatures for some important elements/compounds are listed below.
Element/Compound
La
NbNi
Nb3Ga
Tc (K)
6.0
10.0
23.8
104
Figure 6.23 Plot of Resistance Vs Temperature for normal metal and a superconductor
Meissner Effect: In the presence of magnetic field, as the temperature of
superconducting material is lowered below Tc, the field lines are spontaneously
expelled from the interior of the superconductor(B = 0, Figure 6.24). Therefore, a
superconductor is more than a perfect conductor; it is also a perfect diamagnet. The
property that B = 0 in the interior of a superconductor is as fundamental as the
property of zero resistance. If the magnitude of the applied magnetic field exceeds
a critical value Bc, defined as the value of B that destroys a material’s
superconducting properties, the field again penetrates the sample. Meissner effect
can be explained in the following way.
Figure 6.24 A superconductor in the form of a long cylinder in the presence of an external magnetic
field.
A good conductor expels static electric fields by moving charges to its surface. In
effect, the surface charges produce an electric field that exactly cancels the
externally applied field inside the conductor. In a similar manner, a superconductor
expels magnetic fields by forming surface currents. Consider the superconductor
105
shown in Figure 6.24. Let’s assume the sample is initially at a temperature T>Tc so
that the magnetic field penetrates the cylinder. As the cylinder is cooled to a
temperature T<Tc, the field is expelled. Surface currents induced on the
superconductor’s surface produce a magnetic field that exactly cancels the
externally applied field inside the superconductor. As expected, the surface currents
disappear when the external magnetic field is removed.
BCS Theory: In 1957. Bardeen, Cooper and Schrieffer gave a successful theory to
explain the phenomenon of superconductivity, which is known as BCS theory.
According to this theory, two electrons can interact via distortions in the array of
lattice ions so that there is a net attractive force between the electrons. As a result,
the two electrons are bound into an entity called a Cooper pair, which behaves like
a single particle with integral spin. Particles with integral spin are called bosons. An
important feature of bosons is that they do not obey the Pauli exclusion principle.
Consequently, at very low temperatures, it is possible for all bosons in a collection
of such particles to be in the lowest quantum state and as such the entire collection
of Cooper pairs in the metal is described by a single wave function. There is an
energy gap equal to the binding energy of a Cooper pair between this lowest state
and the next higher state.. Under the action of an applied electric field, the Cooper
pairs experience an electric force and move through the metal. A random scattering
event of a Cooper pair from a lattice ion would represent resistance to the electric
current. Such a collision would change the energy of the Cooper pair because some
energy would be transferred to the lattice ion. There are no available energy levels
below that of the Cooper pair (it is already in the lowest state), however, and none
available above because of the energy gap. As a result, collisions do not occur and
there is no resistance to the movement of Cooper pairs.
Applications: Most important and basic application of superconductors is in high
field solenoids which can be used to produce intense magnetic field.
Superconducting magnets are used in magnetic resonance imaging (MRI)
technique. Magnetic levitation, based on Meissner effect, is another important
application of superconductors. This principle is used in maglev vehicles. Detection
of a weak magnetic field and lossless power transmission are some other important
applications of superconductors.
6.11 QUESTIONS
1.
Sketch schematically the plot of potential energy and its components as a
function of inter-nuclear separation distance for a system of two atoms.
106
2.
Explain briefly (a) ionic bonding, (b) covalent bonding, (c) van der Walls
bonding, (d) hydrogen bonding.
3. Obtain an expression for rotational energy of a diatomic molecule. Sketch
schematically these rotational energy levels.
4. Obtain expressions for rotational transition photon energies and frequencies.
5. Obtain an expression for vibrational energy of a diatomic molecule. Sketch
schematically these vibrational energy levels. Obtain expression for vibrational
transition photon energies.
6. Write expression for total energy (vibrational and rotational) of a molecule.
Sketch schematically these energy levels of a diatomic molecule for the
lowest two vibrational energy values, indicating the possible transitions.
Write the expressions for the energy of the photon in the molecular
energy transitions. Write the expression for the frequency separation of
adjacent spectral lines.
7. Explain the expression for the total potential energy of a crystal. Sketch
schematically the plot of the same.
8. Define (a) ionic cohesive energy (b) atomic cohesive energy, of a solid.
9. Write the expression for Fermi-Dirac distribution function.
Sketch
schematically the plots of this function for zero kelvin and for temperature
above zero kelvin.
10. Derive an expression for density-of-states.
11. Assuming the Fermi-Dirac distribution function , obtain an expression for
the density of free-electrons in a metal with Fermi energy EF, at zero K
and, hence obtain expression for Fermi energy EF in a metal at zero K. [
3
12.
13.
14.
15.
16.
17.
8 2  m 2 12
E dE ]
Given: density-of-states function g( E ) dE 
h3
Explain the formation of energy bands in solids with necessary diagrams.
Distinguish between conductors, insulators and semiconductors on the basis of
band theory
Indicate the position of (a) donor levels (b) acceptor levels, in the energy
band diagram of a semiconductor.
What is the difference between p-type and n-type semiconductors? Explain
with band diagram.
With necessary diagrams, explain doping in semiconductors.
Why the electrical conductivity of an intrinsic semiconductor increases with
increasing temperature?
107
18. What are superconductors? Draw a representative graph of Resistance Vs
Temperature for a superconductor.
19. Explain Meissner effect.
20. Explain briefly the BCS theory of superconductivity in metals.
6.12 PROBLEMS
1. A K+ ion and a Cl– ion are separated by a distance of 5.00 x 10–10 m.
Assuming the two ions act like point charges, determine (a) the force
each ion exerts on the other and (b) the potential energy of the two-ion
system in electron volts.
Ans: a) 921 pN toward the other ion b) - 2.88 eV
2. The potential energy (U) of a diatomic molecule is
potential:
A
B
U 
 6
r 12
r
where A and B are constants. Find, in terms of A and
which the energy is minimum and (b) the energy E
diatomic molecule. (c) Evaluate ro in metres and E in
molecule.
Take A = 0.124 x 10–120 eV.m12 and B = 1.488 x 10–60 eV.m6.
Ans: a) (2A/B)1/6
b) B2/4A
c) 74.2 pm, 4.46 eV
given by Lennard-Jones
B, (a) the value of r o at
required to break up a
electron-volts for the H2
3. An HCl molecule is excited to its first rotational energy level,
corresponding to J = 1. If the distance between its nuclei is ro = 0.1275 nm,
what is the angular speed of the molecule about its center of mass ?
Ans: 5.69x1012 rad/s
4. The rotational spectrum of the HCl molecule contains lines with wavelengths of
0.0604, 0.0690, 0.0804, 0.0964, and 0.1204 mm. What is the moment of inertia of
the molecule?
Ans: 2.72x10-47 kg.m2
5. The frequency of photon that causes v = 0 to v = 1 transition in the CO molecule
is 6.42 x 1013 Hz. Ignore any changes in the rotational energy. (A) Calculate the force
constant k for this molecule. (B) What is the maximum classical amplitude of
vibration for this molecule in the v = 0 vibrational state ?
Ans: A) 1.85x103 N/m
B) 0.00479nm
6. Consider a one-dimensional chain of alternating positive and negative ions.
Show that the potential energy associated with one of the ions and its
interactions with the rest of this hypothetical crystal is
108
Ur    k e 
e2
r
where the Madelung constant is  = 2 ln 2 and r is the inter-ionic
spacing. Hint: Use the series expansion
ln 1  x   x 
x2
x3
x4


 ...
2
3
4
7. Each atom of gold (Au) contributes one free-electron to the metal.
The
28
3
concentration of free-electron in gold is 5.90 x 10 /m . Compute the Fermi Energy
of gold.
Ans: 5.53 eV
8. Sodium is a monovalent metal having a density of 971 kg/m3 and a molar mass
of 0.023 kg/mol. Use this information to calculate (a) the density of charge
carriers and (b) the Fermi energy.
Ans: 2.54 x 1028/m3, 3.15 eV
9. Calculate the energy of a conduction electron in silver at 800 K, assuming the
probability of finding an electron in that state is 0.950. The Fermi energy is
5.48 eV at this temperature.
Ans: 5.28 eV
10. Show that the average kinetic energy of a conduction electron in a metal at
zero K is (3/5) EF
Suggestion: In general, the average kinetic energy is
1
E N( E ) dE
ne 
E AV 
where the density of particles

ne   N( E ) dE
0
109
3
8 2  m2
N( E ) dE 
h3
1
E 2 dE
 E  EF
exp 
 kT

  1

11. (a) Consider a system of electrons confined to a three-dimensional box.
Calculate the ratio of the number of allowed energy levels at 8.50 eV to the
number at 7.00 eV. (b) Copper has a Fermi energy of 7.0 eV at 300 K.
Calculate the ratio of the number of occupied levels at an energy of 8.50 eV
to the number at Fermi energy. Compare your answer with that obtained in
part (a).
Ans: (a) 1.10 (b) 1.46x10-25
12. Most solar radiation has a wavelength of 1 μm or less. What energy gap should
the material in solar cell have in order to absorb this radiation ? Is silicon
(Eg= 1.14 eV) appropriate?
Ans: 1.24 eV or less; yes
13. The energy gap for silicon at 300 K is 1.14 eV. (a) Find the lowest-frequency photon
that can promote an electron from the valence band to the conduction band. (b)
What is the wavelength of this photon?
Ans: 2.7x1014 Hz, 1090 nm
14. A light-emitting diode (LED) made of the semiconductor GaAsP emits red light ( λ=
650nm). Determine the energy-band gap Eg in the semiconductor.
Ans: 1.91 eV
110