UNIT
1
Modern Physics
1.1
CLASSICAL PHYSICS
Newton’s laws of motion are the basis of the most elementary principles of classical physics.
Equations based on these laws are the simplest and they are suitable for solution of simple
dynamical problems, such as the motion of macroscopic bodies, Lagrange’s equations, Hamilton’s
equations and Hamilton’s principle are also fundamental principles of classical mechanics, because
they are consistent with each other and with Newton’s laws of motion. Lagrange’s and Hamilton’s
equations are useful for solving many complicated dynamical problems. In principle, the properties
of bulk matter must be deducible from the properties of electrons and atomic nuclei of which it is
composed. However, it is found that many the observed properties of matter cannot be explained
on the assumption that the particles obey the laws of classical mechanics.
At the end of 19th century and in the beginning of 20th century, many new phenomena such as
photoelectric effect, x-rays, line spectra, nuclear radiation were discovered which wanted
explanation on the basis of classical physics. Laws of classical mechanics failed to explain the above
said newly observed properties of matter.
Therefore the need of new concepts was felt in many areas of physical sciences. The concepts
developed led to a new mechanics called quantum mechanics. Another form of quantum mechanics
is called wave mechanics. The mathematical theory of this mechanics was developed by Erwin
Schroedinger in 1926. Numerous problems of atomic physics have been solved by the application of
quantum mechanics. To understand the development of wave mechanics, we begin with brief
account of black body radiation, which could not be explained by classical mechanics. This is
followed by description of some phenomena like the photoelectric effect, the Compton effect, etc.
Explanations of these phenomena are based on Planck’s quantum hypothesis.
1.1.1
Black Body Radiation
A body which completely absorbs radiations of all wavelengths incident on it is called a black body,
and the radiation emitted by such a body is called black body radiation or full radiation. The nearest
approach to a black body is shown in Fig 1.1(a). It consists of a porcelain sphere, having a small
opening.
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6000 K
(a )
Intensity
(b)
4500 K
3000 K
v
2
4
6
8
10
12
in 10
14
Hz
Fig. 1.1 (a) A black body, (b) Spectral energy distribution
The inner surface is coated with lamp black. Any radiation which enters the sphere through the
opening suffers a few reflections. At each reflection about 98% of the incident radiation is absorbed.
Thus after a few reflections at the inner surface, the radiation is completely absorbed. If the area of
the opening is very small, the radiation cannot be reflected out of the sphere again. The sphere also
emits radiant energy through the opening. To study the distribution of radiant energy over different
wavelengths, the black body is maintained at a constant temperature. By means of an infrared
spectrometer and a bolometer the emissive powers of the black body for different wavelengths (or
frequency) are measured. The results of the experiment conducted are illustrated in Fig. 1.1(b). The
inference is that at a given temperature the radiation energy density initially increases with
frequency, then peaks at around a particular frequency and after that decreases finally to zero at
very high frequencies.
Black body radiation is an important phenomenon because its properties have a universal
character, being independent of the properties of any particular material substance. The other
conclusions are:
(i) The area under a curve which measures the total energy of radiation at that temperature,
increases according to the fourth power of the absolute temperature. Thus, Stefan’s law is
verified.
(ii) The maximum energy peak shifts towards the shorter wavelength side with the increase in
temperature of the body. This confirms Wien’s displacement law.
(iii) Wien’s energy distribution formula developed in 1983 agrees with these curves for short
wavelengths only while Rayleigh-Jean’s formula in 1900 agrees for longer wavelengths.
1.2
PHOTOELECTRIC EFFECT
Photoelectric effect is probably one well-established phenomenon with sound experimental
verification of the existence of particle-like properties in radiations. In 1887 Hertz showed that a
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!
metallic surface could emit electrons if illuminated with light of very short wavelength. In 1905
Einstein interpreted this phenomenon in terms of the energy relationship,
hw =
mv 2
1
mv 2 + ef =
+W
2
2
(1.1)
for which he was awarded the Nobel prize. The above equation can be written as
h
1
(2pn) = hn = mv 2 + ef
2p
2
(1.2)
where W = e f is the work function of the metal corresponding to the height of a potential barrier at
the surface of the metal that electrons in the metal must overcome in order to escape into vacuum.
1.2.1
Experimental Arrangement for Observing Photoelectric Effect
Here two electrodes are closed in an evacuated quartz bullbs (or glass). The quartz container does
not absorb ultraviolet light. The cathode C is made of a photosensitive metal (zinc, sodium, lithium
or cesium). A potential divider is used to apply a p.d between anode A and cathode C. This applied
p.d say V volt is measured with the help of a voltmeter. The current in the circuit is detected with the
help of a sensitive galvanometer or micro ammeter. There are three basic experimental variables in
the photoelectric experiment. They are
(a) (i) Intensity of light: The graph between the photoelectric current and intensity is a straight
line. This shows that the photoelectric current increases linearly with intensity.
(ii) No detectable time lag has been measured between the switching of light and emission
of photoelectrons.
(b) (i) Stopping potential: There exists a minimum potential V 0 below which there is no
photoelectric emission and hence no photoelectric current. This potential which is just
Light
S
–
–
–
C
A
V
A
( )
Fig. 1.2 Millikan’s apparatus for studying photoelectric effect
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sufficient to stop the photoelectric current is called stopping potential or the cut off
potential.
(ii) The stopping potential V0 is independent of the intensity of the incident light.
(c) (i) Threshold frequency: There exists a minimum frequency n0 at which the stopping potential
is zero. This means that no electrons are emitted if the frequency of the incident light is
below n0. The minimum frequency n0 below which there is no photoelectric current is
called threshold frequency.
1
mv2max, this
2
means that the maximum velocity of emitted electrons increases linearly with frequency.
(ii) The stopping potential increases linearly with the frequency. Since V0 =
1.2.2
Laws of Photoelectric Emission
The results of the experiment conducted by Millikan are
(i) The photoelectric current increases linearly with intensity of light.
(ii) The maximum velocity of photoelectrons depends only on the frequency of incident light
and is independent of its intensity
(iii) The photoelectric effect does not occur below a certain frequency called the threshold
frequency. This frequency depends on the metal used in photo cathode.
(iv) The emission of photoelectrons is an instantaneous process.
1.3
EINSTEIN’S PHOTOELECTRIC EQUATION
When light of frequency n shines on a metal, an electron instantly absorbs a photon of energy hn. If
hn is greater than the binding energy the absorbed photon will eject the electron from the metal
surface and will appear as a photoelectron. The energy of a photon is used in two ways.
(i) The part of absorbed photon energy is used in releasing the electron from the metal surface
(ii) The balance of energy appears as the kinetic energy of the electron. Thus we can write
Energy of incident photon = Binding energy of electron + K.E of electron
i.e.,
hn =
1
mv2 + W
2
Thus
1
mv 2 = h n - W
2
When W = W 0, the kinetic energy of emitted electrons will be maximum. Thus
1
mv2max = h n - W 0
2
...(1.3)
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Suppose a photon of frequency n = n0 has just sufficient energy to remove the least bound
electron from the metal. Then the maximum kinetic energy of emitted electron is zero. Using this
fact we have
0 = hn0 - W 0
or,
h n0 = W0
...(1.4)
The frequency n0 determines the threshold frequency. It will depend on the work function of a
particular material. Substituting the value of W 0 from Eqn. (1.4) in Eqn. (1.3) we obtain
1
mv2max = hn - hn0
2
...(1.5)
This equation is called Einstein’s photoelectric equation.
1.3.1
Conclusions
1. When we increase the intensity of incident light more photons will strike photocathode. As a
result of it the electrons emitted will increase. Thus an increase in intensity causes an increase
in photoelectric current. This is the first law of photoelectric emission.
2. It is evident from Eqn. (1.5) that the maximum velocity vmax of photoelectrons depends only
on the frequency of incident light and is independent of its intensity. This is precisely the
second law of photoelectric emission.
3. Equation (1.5) tell us that if frequency of light n is less than n0, the kinetic energy becomes
negative. So no electron emission can occur. This is nothing but the third law of photoelectric
emission.
4. According to Einstein’s theory the light energy incident on the photo-cathode is supplied in
concentrated bundles of photons. These photons are immediately absorbed by some atoms.
As a result the immediate emission of photoelectron starts. This is exactly the fourth law of
photoelectric emission.
1.4
DUALISM
The optical phenomena like reflection, refraction, interference, diffraction and polarization of light
could be easily explained by wave theory of light. On the other hand, the wave theory of light failed
completely to explain the photoelectric effect, Compton effect, absorption and emission of radiation
by substances. These phenomena could be easily explained by quantum theory of light. It appears
as if light presents itself in such a form as to support wave theory at one hand and quantum theory
on the other hand. This complex nature of light is said to be dual nature. In quantum theory of light,
the energy is thought to shoot out from the source in the form of energy packet, called photons. The
energy carrying particle ‘photon’ itself exhibits diffraction effects, thereby showing that photon is
though a particle, yet it possesses wave like characters. The diffraction effects due to single photon
were experimentally observed by a Cambridge student G.I. Taylor. The details of the
experiment have been discussed in the next section. But this leads to an important conclusion that
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light possesses dual characters, i.e., at the same time it behaves wave – and particle – like.
1.4.1
Compton Effect
When a monochromatic beam of x-rays is scattered by a block of paraffin or graphite the scattered
x-rays consist of two components:
One component having the same wavelength as that of the incident x-rays and the second
component having a greater wavelength. This is known as Compton effect. The former is called the
unmodified radiation and the latter the modified radiation. The classical electromagnetic theory
explained the unmodified radiation but it totally failed to explain the presence of the modified
radiation. Compton, however, gave a satisfactory explanation for the modified radiation on the
basis of quantum theory.
1.4.2
General Theory
Consider a photon of energy E = hn colliding with an electron. It transfers some of its energy to the
electron. Thus the scattered photon will have less energy than the incident photon. Since energy of
the photon is directly proportional to frequency, the frequency of scattered photon, will be smaller
than the frequency of incident photon. This implies that wavelength, l¢, of the scattered photon will
be greater than the wavelength l of the incident beam. The energy associated with the scattered
photon E¢ = hn¢. Let q be the angle of scattering of the photon after the collision. As a result of
r
collision, the electron acquires some velocity say v . Let us say that the electron recoils at an angle f
to the direction of the incident beam. If m is the relativistic mass of the electron, then momentum of
the electron after collision is
r
pe = m v
...(1.6)
The total energy and the total momentum of the system will remain conserved as the collision
between photon and electron is elastic.
p¢
Scattered
photon
Incident photon
q
f
p
Electron
at rest
Recoil
electron
Fig. 1.3
Scattering of photon by an electron
pe
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1.4.3
%
Law of Conservation of Energy and Law of Conservation of Momentum
Let Ee and E¢e be the energies of the electron before and after collision. The conservation of energy
requires that
E + Ee = E¢ + E¢e
hn + Ee = hn¢ + E¢e
...(1.7)
If m0 is the rest mass and m is the relativistic mass, then Eqn. (1.7) becomes
hn + m0c2 = hn¢ + mc2
hn - hn¢ + m0c2 = mc2
c
and
l
Using n =
m=
m0
e
2
1 - v /c
2
j
...(1.8)
m0
=
1- b
2
where b =
v
c
Eqn. (1.8) becomes
LM 1 - 1 OP
Nl l¢ Q + m c
ch
0
2
m0c 2
=
1-b
...(1.9)
2
Dividing each term by c
h
LM 1 - 1 OP + m c =
N l l¢ Q
m0 c
0
1- b
...(1.10)
2
Squaring both sides and simplifying, one gets
h2
LM 1 - 1 OP
Nl l¢ Q
2
+ m20c2 + 2m0ch
LM 1 - 1 OP = m c
N l l ¢ Q e1 - b j
2 2
0
2
i.e.,
h2
2
l
+
h2
2
l¢
-
b
g
m 02 c 2
2 h 2 2m 0 ch l ¢ - l
=
- m 20 c2
+
2
ll ¢
ll ¢
1- b
e
j
m02 c 2 - m02 c 2 1 - b 2
2m 0 c h l ¢ - l
2h 2
h2
h2
+
+
=
2
ll ¢
l2
l ¢2
ll¢
1- b
b
g
e
h2
l2
+
h2
l ¢2
with b2 = v2/c2
-
2h 2
ll¢
+
b
2m 0 ch l ¢ - l
ll ¢
g=mc
2 2
0
e
j
j
- m02 c 2 + m02 c 2b 2
1 - b 2
&
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Thus
h2
2
l
1.4.4
+
h2
2
l¢
-
2h 2
ll¢
+
b
2m 0 ch l ¢ - l
ll ¢
g=
m 02 v 2
1 - v 2/c 2
e
...(1.11)
j
Conservation of Momentum
r
r
Suppose p and p ¢ are the momenta of incident photon and scattered photon respectively. The
conservation of momentum requires
r
r
r
p = p ¢ + pe
r r
r
p - p ¢ = pe
Take the dot product of both sides with themselves. This gives
r r
r r
r r
p - p ¢ ◊ p - p ¢ = pe ◊ pe
b
gb
g b
g
This yields
r r
p 2 + p ¢ 2 - 2 p ◊ p ¢ = p e2
r
r
Since the angle between p and p ¢ is q,
p2 + p ¢ 2 - 2p p ¢ cos q = m2v2
...(1.12)
Substituting for m,
p2 + p ¢ 2 - 2pp¢cos q =
1.4.5
m02 v 2
1 - v2 / c2
e
j
...(1.13)
Momentum of Photon and Compton Wavelength
The relativistic relation between energy and momentum according to Einstein’s theory is
E2 = m 02 c4 + p2c2
Since a photon always travels with the speed of light, its rest mass must be zero. That is m0 = 0.,
Hence
E2 = p2c2; E = pc
E
c
This gives energy-momentum relation for a photon. This shows that momentum of the photon
is equal to its energy divided by speed of light. Thus momentum of the incident photon is
and
p=
p=
hn
h
=
l
c
...(1.14)
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and momentum of scattered photon is
hn ¢
h
=
l¢
c
Substituting these values in Eqn. (1.13) one gets
P¢ =
...(1.14a)
m02 v 2
2h 2
h2
h2
+
cos
q
=
l2
l ¢2
ll¢
1 - v2 / c2
e
j
...(1.15)
Subtracting Eqn. (1.15) from Eqn. (1.11)
-
b
g
2m0 ch l ¢ - l
2h 2 cos q
2h 2
+
+
=0
ll¢
ll ¢
ll¢
b
2h 2 cos q 2m0 ch l ¢ - l
2h 2
=
ll¢
ll ¢
ll¢
or
g
h (1 - cos q) = m0c ( l¢ – l)
i.e.,
(l¢ – l) =
h
(1 - cos q)
m0 c
l¢ = l +
h
(1 - cos q)
m0 c
...(1.16)
l¢ gives the wavelength of the modified line. We also infer the following from Eqn. (1.16). The
wavelength of the modified line:
(i) depends on the wavelength of the incident radiation
(ii) depends on the angle of scattering q
(iii) is always greater then the wavelength of incident radiation
(iv) is independent of the nature of the scatterer.
The quantity
h
has dimension of length and is known as Compton wavelength. The
m0 c
difference in wavelength between the scattered x-rays and the incident x-rays is called Compton
shift designated byDl. Thus
Dl = l¢ - l =
1.5
h
(1 - cos q)
m0 c
...(1.17)
EXPERIMENTAL VERIFICATION OF COMPTON’S THEORY
The apparatus used by Compton to study his theory consisted of an x-ray tube having a
molybdenum target (Fig. 1.4) giving a strong Ka line. The characteristic x-rays emerging from the
tube were allowed to be incident on a small graphite piece C. The x-ray tube and the graphite piece
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are both placed inside a lead block with a slit S1 in one wall.
A further system of slits S2 placed behind S1 allows the beam scattered at a definite angles q to
pass through them. The distribution of intensity with wavelength in the x-rays scattered at various
angles q is measured by means of the Bragg’s x-ray spectrometer. By this method, spectra of the
molybdenum Ka line after being scattered by the graphite C at different angles were obtained. If the
scattered rays were of the same wavelength as the primary, the spectrum of the former should be
the same as that of the latter. Compton, however, found in the spectrum obtained not only the Ka
line but another K¢a on the longer wavelength side as illustrated in Fig. 1.5. Further with q = 90° the
wavelength of the unmodified line Ka was found to be 0.708Å, while that of the modified line K¢a is
0.730Å, the Compton wavelength. The change in wavelength was also found to increase rapidly as
the angle of scattering was increased.
It was also established that it is independent of l. Compton and Rose, using photographic
method found that D l for the Kb line of molybdenum was the same as for Ka line for a given angle
of scattering. Moreover changing the target in the x-ray tube should not produce, according to the
theory, an alteration in the value of D l provided q is the same. In other words different targets
produce different primaries with different values of l, yet the change in wavelength Dl was found
to be the same for a given angle of scattering. The results for 0°, 45°, 90° and 135° as scattering angles
are shown in Fig. 1.5. The study of compton effect leads to the conclusion that in its interaction with
matter, radiant energy behaves as a stream of discrete particles (photons) each of energy hn and
momentum hn/c. In other words, radiant energy is quantised. Therefore the compton effect is
considered as a decisive phenomenon in support of the quantisation of radiant energy.
Crystal
S1
q
C
S2
E
T
Fig. 1.4
Experimental set up for the study of Compton effect
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U (unmodified line)
Intensity
q=0
l
0
Dl
M (modified line)
Intensity
U
Wavelength
q = 45°
0
l
U
l¢
Wavelength
Dl
M
Intensity
q = 90°
0
l
Fig. 1.5
1.6
l¢
Wavelength
Results of Compton effect
DUAL NATURE AND DE BROGLIE’S HYPOTHESIS
Louis de Broglie in 1924 enunciated a hypothesis on matter waves. According to this concept every
moving particle has a wave-packet associated with it. The wavelength of such waves depends upon
the momentum of the particle.
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He gave an expression for the wavelength of such waves as
h
h
l=
=
mv
p
...(1.18)
where m is the mass of the particle moving with a velocity v, the product mv = p is the momentum of
the particle, h is the Planck’s radiation constant. If we consider Planck’s theory of radiation, the
energy of a quantum is given by
hc
E = hn =
...(1.19)
l
where n is the frequency, c is the velocity of light in vacuum and l is the wavelength or
hc
l=
...(1.20)
E
If m is the mass of the particle converted into energy, the equivalent energy is given by Einstein
energy-mass relation as
E = mc2
Hence Eqn. (1.20) can be written as
l=
hc
mc
2
=
h
=
mc
h
p
...(1.21)
where mc = p is the momentum associated with quantum. Hence if a body of mass m moves with a
velocity v, then mv and Eqn. (1.21) is written as
l=
h
mv
=
h
p
The above relation may also be written in terms of kinetic energy which is given as
K=
Therefore
p2
1 m2 v 2
1
mv2 =
=
2 m
2
2m
p=
2mK
l=
h
2mK
...(1.22)
When a charged particle, carrying a charge e, is accelerated through a potential difference of V
volt, then kinetic energy
K = eV
0.1227
h
=
nm
...(1.23)
2meV
V
Also, if the material particles are in thermal equilibrium at associated temperature T, then
3
K = kB T
2
l=
and
l=
h
2meV
RS 2 UV
T 3k T W
B
1
2
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or
l=
h
!
...(1.24)
3mk B T
where kB =1.38 ¥ 10 - 23 joule/kelvin
1.6.1
Experiments of Davisson and Germer
The experiments of Davisson and Germer were the first experimental evidence in support of matter
waves. These two American physicists performed experiment on the diffraction of electron waves
by a nickel target.
The electron beam from an electron gun is accelerated and collimated to strike a nickel crystal.
C is an ionisation chamber for receiving the electron after they have been scattered by a nickel
crystal. The ionisation chamber can be moved along a graduated circular scale so that it is able to
receive the scattered electrons at all angles between 20° to 90° and their intensity is measured by the
galvanometer current. The whole assembly is placed in a very high vacuum. Graphs are drawn at
various voltages and the pronounced maximum obtained for 54 volt at f = 50°. (Fig. 1.6b)
f + q + q = 180°
2q = 180 - 50 = 130°
q = 65°
The interplanar distance for nickel is 0.1 nm.
Thus
2d sin q = l
l = 2 ¥ 0.1 ¥ sin 65° = 0.167 nm
...(1.25)
Ionisation
chamber
Electron
gun
G
C
Primary
electron
beam
q
f
q
40 V
44 V
48 V
54 V
60 V
64 V
Diffracted
beam
Bragg’s plane
Nickel single crystal
(a)
Fig. 1.6
(b)
Davisson-Germer experiment on diffraction of electron waves
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By de Broglie’s hypothesis Eqn. (1.23)
l =
0.1227
54
nm = 0.167 nm
...(1.26)
Thus we see 100% agreement in the value. See Eqn. (1.25) and compare with Eqn. (1.26).
1.7
1.7.1
WAVE VELOCITY AND GROUP VELOCITY FOR DE BROGLIE WAVES
Wave Velocity
A wave is a disturbance from equilibrium condition that travels or propagates with time from one
region of space to another The original displacement gives rise to an elastic force in the material
adjacent to it, then the next particle is displaced and then the next and so on. Thus the motion is
handed over from one particle to the next. Therefore, every particle begins its vibration a little later
than its predecessor. Thus there is a progressive change of phase from one particle to the next. The
phase relationship of these particles is termed as wave and the velocity with which planes of
constant phase propagate through the medium is known as wave velocity or phase velocity.
Thus velocity of advancement of a monochromatic wave (wave of single wavelength and
frequency) through a medium is called the wave velocity. For example, the equation of plane
progressive wave is given by
y = a sin (wt – k x)
2p
2p
= 2pn is the angular frequency and k =
is the propagation constant. The term
l
T
(wt –kx) represents the phase of the wave motion. Hence the planes of constant phase are defined by
where w =
(wt – k x) = constant
Differentiating this equation with respect to time,
w–k
or
dx
=0
dt
w
dx
=
=u
k
dt
dx
is the phase velocity or wave velocity. Thus phase velocity (or wave velocity) is the
dt
ratio of angular frequency w to the propagation constant k and is the velocity with which a plane
progressive wave front travels forward. If l is the wavelength and n the frequency of the wave, then
phase velocity
where u =
u = n l; n =
u
l
...(1.27)
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If E is the energy of the wave, then its frequency n is given by
E = h n; n =
E
h
Also from de-Broglie theory, the wavelength of material particle
l=
h
mv
Therefore, the phase velocity of the associated de-Broglie wave
u = nl =
LM OP
N Q
E
E h
=
h mv
mv
But from Einstein’s mass energy relation E = mc2,
Therefore,
u=
mc 2
c2
=
mv
v
...(1.28)
Since c >>v, Eqn. (1.28) implies that the phase velocity of the associated wave is greater than c,
the velocity of light. It indicates that the associated wave with the particle travels faster than the
particle itself. Thus the particle will be left far behind. Obviously, a monochromatic de Broglie wave
can not transport a particle or carry energy. The phase velocity is thus a purely mathematical
concept and represents merely the rate at which a given phase of a monochromatic wave train
advances.
1.7.2
Group Velocity–The Concept of Wave Packet
The phase velocity of a wave associated with a particle comes out to be greater than the velocity of
light. This difficulty can be overcome by assuming each moving particle of matter to consist of a
group of waves or a wave packet, rather than a single wave train. A wave group corresponding to a
certain wavelength l consists of a number of component waves of slightly different wavelengths in
the neighbourhood of l superimposed upon each other. The mutual interference between
component waves results in the variation of amplitude that defines the shape of the wave packet.
The component waves interfere constructively over only a small region of space, outside of which
they interfere destructively and hence the amplitude reduces to zero rapidly. Thus the resultant
wave pattern consists of points of maximum amplitude and points of minimum amplitude.
Between any two consecutive minima, say A and C, there is a position of maximum amplitude i.e.,
at B (mid way between A and C) as shown in Fig. 1.7. The dotted loop represents a group of waves
or a typical wave packet (Fig.1.7). This group of waves ( or the wave packet ) moves forward in the
medium with a velocity called the group velocity. Thus group velocity is the velocity with which the
slowly varying envelope of the modulated pattern due to a group of waves travels in a medium.
The importance of the group velocity lies in the fact that it is velocity with which the energy in the
wave group is transmitted.
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B
A
C
Fig. 1.7
1.7.3
Wave packet and phase velocity
Expression for Group Velocity
To derive an expression for group velocity, consider a group of waves which consists of only two
components of equal amplitude a but slightly different angular frequency w1 and w2 and
propagation constants k1 and k2. Their separate displacements may be represented by the equations
y1 = a sin (w1t - k1x)
y2 = a sin (w2t - k2x)
The resultant amplitude due to superposition is given by
y = y1+ y2 = a sin (w1t - k1x) + a sin (w2t - k2x)
or
i.e.,
or
y = a[sin (w1t - k1x) + sin (w2t - k2x)]
LM (w + w ) t - (k + k ) xOP ¥ cos êé (w - w )t - ( k - k ) x úù
2
2
2
N 2
Q
ë
û
L cos (w - w ) t - (k + k ) xOP sin LM (w + w )t - ( k + k ) x OP
y = 2a cos M
2
2
N 2
Q N 2
Q
y = 2 a sin
1
2
1
1
2
1
2
1
2
2
1
1
2
1
2
2
...(1.29)
This equation represents a wave system of amplitude
A = 2a cos
LM (w
N
1
- w 2 ) t ( k1 - k2 ) x
2
2
OP
Q
which is modulated both in space and time. Eqn. (1.29) can be written in modified form as
y = 2a sin (wt – k x) cos
LM D w t - D k xOP
N2 2 Q
...(1.30)
Modern Physics
where
w=
and
(w 1 + w 2 )
,
2
k=
%
k1 + k 2
2
D w = w1 - w2, D k = k1 - k2
The resultant wave (Eqn.1.30) thus has two parts,
(i) A wave of frequency w, propagation constant k and velocity
u=
2 pn
w
=
=nl
2 p/l
k
which is the phase velocity or wave velocity.
(ii) Another wave frequency
Dw
Dw
Dk
, propagation constant
and velocity G =
,
Dk
2
2
superimposed upon the first wave. It represents a very slowly moving envelope of frequency
Dw
(w 1 - w 2 )
D k (k1 - k 2 )
=
and propagation constant
=
. This envelope is represented by
2
2
2
2
the dotted curve in Fig.1.7 and moves with a velocity
G=
w1 - w2
Dw
=
Dk
k1 - k 2
known as the group velocity. If a group contains a number of frequency components in a very small
frequency interval, then the above expression may be written as
G=
G=
∂w
∂k
∂n
2p∂n
= - l2
1
∂l
2p∂
l
FG IJ
H K
This is the expression for group velocity.
1.7.4
Relation Between Phase Velocity and Group Velocity
If u is the phase (wave) velocity, then, since u =
But
w
, the group velocity is given by
k
G=
dw
du
d
=
(uk ) = u + k
dk
dk
dk
k=
2p
,
l
\ dk = - (2p/l2)d l
...(1.30a)
&
Engineering Physics
l
k
=dl
dk
Hence
Therefore, group velocity is given by
LM l OP du
N dlQ
LduO
G = u - M P ◊l
Nd lQ
G =u+ -
...(1.31)
The relation shows that group velocity G is less than the phase velocity u in a dispersive
medium i.e., when u is function of l. In a non-dispersive medium, waves of all wavelength travel
with the same speed i.e.,
du
= 0 and then G = u. This is true for electromagnetic waves in vacuum
dl
and elastic waves in homogeneous medium.
1.7.5 Relation Between Group Velocity and Particle Velocity
Consider a material particle of rest mass m0. Let its mass be m when moving with a velocity v. Then
its total energy E is given by
m0 c 2
E = mc2 =
LM1 - v OP
N cQ
2
2
Its momentum is given by
m0 v
p = mv =
...(1.32)
2 1/2
LM1 - v OP
N cQ
2
The frequency of the associated de Broglie wave is given by
n=
\
E
=
h
m0 c 2
h
R| F v
S| GH1 - c
T
2
2
I U|V
JK |
W
2 pm0 c 2
w = 2pn =
h
R| F v
S| GH1 - c
T
2
2
I U|V
JK |
W
Modern Physics
Or
2 pm 0
dw =
2 3/2
L v OP
h M1 N cQ
(vdv )
'
...(1.33)
2
The wavelength of the associated de Broglie wavelength is given by (See Eqn. 1.32)
2 1/2
L vO
h M1 - P
c Q
h
l= = N
2
...(1.34)
p
m0 v
2p
=
l
2 pm0 v
Hence, propagation constant
k=
LM
N
h 1-
v2
c2
OP
Q
1/2
LM L v O
MM MN1 - c PQ
MM
N
2 1/2
dk =
2 pm 0
h
2
2 p m0
dk =
h
2
2
- 1/2
2
2
2 pm 0
h
F
GH
v2
v2
dv + 2 1- 2
c
c
- 3 /2
2
dk =
2
dv
2 -1
+
LM
N
v2
v2
1
c2
c2
2
dk =
2 pm0
LM
N
h 1-
v2
c2
OP
Q
3/2
dv
2
LMF v I
MNGH 1 - c JK
LM L v O
MM MN1 - c PQ
MM L1 - v O
MN MN c PQ
2
2 - 1/2
F v I LM1 - v OP
dv + v
Hc K N c Q
LM1 - v OP
N cQ
R| F
S| GH
T
¥ dv 1 -
I
JK
F
GH
I
JK
- 3/2
OP
OP dvPP
Q PP
PQ
-3/2
v2
v2
2 + dv
c
c2
OP
PP
PP
Q
I U|V
JK W|
dv
OP
PQ
Engineering Physics
or
dk =
2 pm0 dv
L v OP
h M1 N cQ
2
...(1.35)
3/2
2
Since group velocity, G =
velocity, G =
dw
, therefore dividing Eqn. (1.33) by Eqn. (1.35), one gets group
dk
dw
= v, the particle velocity. Thus the wave group associated with a moving material
dk
particle travels with the same velocity as the particle. It proves that a material particle in motion is
equivalent to a group of waves or a wave packet.
Table 1.1
Photoelectric threshold frequency and work function for some metals.
Photoelectric work
Metal
function, W0
Calcium
Threshold frequency
n0 =
Threshold wavelength
W0
h
l0 =
c
n0
(in eV)
in 1015 Hz
in nm
1.90
0.458
655
Gold
4.90
1.182
253.8
Iron
4.48
1.081
277.5
Nickel
4.01
0.968
310.0
Platinum
6.30
1.520
197.5
Rhodium
4.57
1.103
272.0
Silver
4.74
1.144
266.2
Sodium
2.00
0.483
621.1
Tantalum
4.05
0.977
307.1
Tungsten
4.58
1.105
271.5
Table 1.2
Wavelengths of electrons under selected voltages
Voltage applied in volt
Wavelength, l in nm
44
0.18
54
0.17
74
0.14
100
0.1227
Modern Physics
OBJECTIVE QUESTIONS
1.1 The photoelectric effect occurs with
(a) only free electrons
(b) only bound electrons
(c) both bound and free electrons
1.2 The maximum velocity of photoelectrons
(a) depends on the frequency of the incident radiation
(b) depends on the intensity of the incident radiation
(c) independent of the wavelength of the incident radiation
(d) all are true
1.3 The velocity of photon in the visible region is
(a) c2
(b)
c
2
(c)
1
c
(d) c
where c is the velocity of light
1.4 The photoelectric effect is observed only if the wavelength of light is
(a) above threshold wavelength
(b) below threshold wavelength
(c) zero
(d) equal to threshold wavelength
1.5 Quartz bulb is used for irradiating the photosensitive cathode because quartz
(a) absorbs ultraviolet rays easily
(b) does not absorb ultraviolet rays
(c) reflects light radiation easily
(d) is a polarizer
1.6 At what angle of scattering the wavelength of the scattered photon will be minimum?
(a) 0°
(b) 30°
(c) 60°
(d) 90°
1.7 At what angle of scattering the wavelength of the scattered photon will be maximum?
(a) 90°
(b) 40°
(c) 60°
(d) 0°
1.8 The de-Broglie wavelength of a particle at rest is
(a) zero
(b) infinite
(c)
h
p
(d)
h
v
1.9 The phenomenon which points toward the particle nature of electromagnetic radiation is
(a) diffraction
(b) interference
(c) Compton effect
(d) none of these
1.10 Can the wavelength of de Broglie wave of a particle of mass m and energy E be represented
by l =
h
2mE
(a) yes
(b) no
Engineering Physics
1.11 In wave mechanics the group velocity is equal to particle velocity
(a) true
(b) false
1.12 An electron, neutron and a proton have the same de Broglie wavelength. Which particle has
greater velocity?
(a) proton
(b) neutron
(c) electron
1.13 The group velocity is less than the phase velocity
(a) true
(b) false
SHORT QUESTIONS
1.1 Explain blackbody radiation spectrum.
1.2 What are the laws of photoelectric effect?
1.3 Explain the meaning of the following terms:
(i) Work function
(ii)
Stopping potential
1.4 What are matter waves?
1.5 Explain de Broglie’s hypothesis.
1.6 Explain phase velocity and group velocity.
1.7 What is Compton effect?
1.8 Explain the results one infers from the study of Compton effect.
1.9 Give the basic principle of Davisson-Germer experiment.
1.10 State the most important experimental observations about the photoelectric effect.
REVIEW QUESTIONS
1.1 What is photoelectric effect? Give an account of the photoelectric emission of the electrons.
Give Einstein’s interpretation for the same.
1.2 Mention the important experimental observations observed in the photoelectric emission of
electrons and explain these on the basis of Einstein’s theory.
1.3 Give the laws of photoelectric emission. Derive Einstein’s photoelectric equation and show
how this equation explains the laws.
1.4 Write down Einstein’s photoelectric equation and use it to explain the following.
In an experiment on photoelectric effect of measuring the energy and number of
photoelectrons what happens if:
(i) The frequency of incident light is changed, target material and intensity of light being
kept constant;
(ii) The target material is changed, the frequency and intensity of light being kept constant;
Modern Physics
!
(iii) The intensity of light is changed, the frequency of light and target material being kept
constant.
1.5 Write a note on black body radiation. Describe an experiment to demonstrate photoelectric
effect.
1.6 What is Compton scattering? How does it confirm the corpuscular nature of radiation?
1.7 What is Compton effect? Derive an expression for the wavelength of scattered photon. At
what angle of scattering the wavelength will be maximum?
1.8 Describe an experiment to study Compton effect. Obtain an expression for Compton shift.
1.9 Discuss the dual nature of matter waves. Derive an expression for the de Broglie wavelength.
1.10 Explain de Broglie hypothesis. Describe and explain Davisson and Germer experiment for
the confirmation of de Broglie hypothesis.
1.11 Explain clearly the wave velocity and the group velocity.
Show that the phase velocity of the wave is greater than the speed of light c while the group
velocity of the electron is equal to electron’s velocity.
1.12 Discuss the concept of group velocity. Obtain the expression for group velocity.
PROBLEMS AND SOLUTIONS
1.1 Calculate the number of photons emitted in 3 hours by a 60 watt sodium lamp. Given
l = 589.3nm
Solution
Energy of photon = h n =
=
6.62 ¥ 10 -34 ¥ c
l
6.62 ¥ 10 -34 ¥ 3 ¥ 10 8
589.3 ¥ 10 -9
= 3.374 ¥ 10-19 J
Number of photons emitted by the sodium lamp in one second =
60
3.374 ¥ 10 -19
\ The number of photons emitted by the sodium lamp in 3 hours is
60 ¥ 3 ¥ 3600
3.374 ¥ 10 -19
= 1.9 ¥ 1024
1.9 ¥ 1024
Ans.
1.2 Light of wave length 4047Å falls on a photoelectric cell with a sodium cathode. It is found
that the photoelectric current ceases when a retarding potential of 1.02 volt is applied.
Calculate the work function of the sodium cathode.
"
Engineering Physics
Let the stopping potential be V volt and the work function be f. It is given that the photoelectric
current ceases when the retarding potential is 1.02 volt
i.e.,
Vs = 1.02 volt
Maximum kinetic energy of photoelectrons = 1.02 eV
We have
hn = f + Emax
But
hn =
hc
6.62 ¥ 10 -34 ¥ 3 ¥ 10 8
=
l
4047 ¥ 10 -10
hn = 5 ¥ 10-19 J =
5 ¥ 10 -19
1.6 ¥ 10 -19
= 3.12 eV
f = hn – Emax = 3.12 – 1.02 f = 2.1 eV
Ans.
1.3 A photon of energy h n is scattered through an angle q by a free electron originally at rest.
Show that the ratio of kinetic energy of the recoil electron to the energy of incident photon is
a (1 - cos q)
1 + a(1 - cos q)
where a =
hn
m0 c 2
◊ m0 is the rest mass of the electron and c is the velocity of light.
Solution
The kinetic energy of the recoil electron is the difference between the incident photon energy and
scattered photon energy. We, therefore, have
K = hn - h n¢
FG 1 - 1 IJ
H l l¢K
R l ¢ - l UV
K = ch S
T ll ¢ W
= ch
Substituting (l' - l ) from Eqn. (1.17) and l' from Eqn. (1.16) in the above equation,
h
1 - cos q
m0 c
b
K = ch
RS
T
l l+
g
h
1 - cos q
m0 c
b
gUV
W
Modern Physics
#
R| h - q U|
b1 cos g |
ch O |
m c
L
K = M PS
N l Q | LM1 + h b1 - cos qgOP l V|
|T N m cl
Q |W
U
R h
F K I = ||S m c l (1 - cos q) ||V where 1 = n
GH ch/l JK | 1 + bh/m c lgb1 - cos qg |
l
c
|W
|T
0
0
or
0
0
R| h n (1 - cos q) U|
K
| mc
|V
= S
hn
+
q
n
h
m
c
(
cos
)(
/
)
1
1
||
||
T
W
2
0
2
0
It is assumed
hn
a=
m0 c 2
a (1 - cos q)
K
=
hn
1 + a (1 - cos q)
Ans.
1.4 X-rays of wavelength 0.1 nm from a carbon block are scattered in a direction making 60 ° with
the incident beam. How much kinetic energy is imparted to the recoiling electron?
Solution
The Compton change in wavelength is given by
l' - l =
Dl =
2h
sin2 (q/2)
m0 c
|RS 2 ¥ 6.626 ¥ 10
|T 9.1 ¥ 10 ¥ 3 ¥ 10
-34
-31
8
|UV (0.25)
|W
= 0.001214 nm
l' = l + D l = 0.101214 nm
Energy of incident X-ray photon = hn =
Energy scattered X-ray photon = hn =
hc
l
ch
l¢
$
Engineering Physics
Hence energy imparted to the recoiling electron
=
ch ch
(l ¢ - l )
–
= ch
l
l¢
ll ¢
=
ch ( D l )
ll ¢
=
3 ¥ 10 8 ¥ 6.62 ¥ 10 -34 ¥ 0.001214 ¥ 10 -9
(0.1 ¥ 10 -9 ) (0.101214 ¥ 10 -9 )
= 2.384 ¥ 10-17 J
=
2.384 ¥ 10 -17
1.6 ¥ 10 -19
= 149 eV
149 eV
1.5 A beam of x-rays of wavelength I Å is incident on a carbon target. The scattered x-rays are
detected at an angle of 60° in the direction of the incident beam. Find the wavelength of the
scattered x-rays.
Solution
l' = l +
But
Thus
h
(1 - cos q)
m0 c
6.62 ¥ 10 -34
h
=
= 0.00243 nm
m0 c
9.1 ¥ 10 -31 ¥ 3 ¥ 10 8
l' = 0.01 + 0.00243 (1 – 0.5)
l' = 0.01121
Ans.
1.6 The most rapidly moving valence electron in metallic sodium at absolute zero temperature,
has a kinetic energy 3 eV. Show that the de Broglie wavelength is 7Å.
Solution
K = 3 eV = 3 ¥ 1.6 ¥ 10 -19 J
For the electron
m0c 2 = 9.1 ¥ 10– 31 ¥ (3 ¥ 108)2 = 9.1 ¥ 9 ¥ 10-15 J
=
9.1 ¥ 9 ¥ 10 -15
1.6 ¥ 10 -19
= 5.12 ¥ 105 eV
Therefore, kinetic energy of the electron is small compared with m0 c 2.
Hence de-Broglie wavelength is given by
Modern Physics
l =
But
%
h
m0 v
1
m0v2 = eV
2
or v =
2 eV
m0
Hence
But K = 3 ¥ 1.6 ¥ 10–19 J, l =
LM h OP
Nm Q
0
1
h
=
=
2 eV
2 m 0 eV
m0
l=7Å
6.62 ´ 10
2 ´ 9.1 ´ 10
- 31
- 34
´ 3 ´ 1.6 ´ 10 - 19
Ans.
1.7 Calculate the momentum of an electron possessing the de Broglie wavelength 6.62 ¥ 10– 11 m.
Solution
l=
h
p
p=
6.62 ¥ 10 -34
h
=
= 10-23 kg.m/s
6.62 ¥ 10 -11
l
p = 10-23kg.m/s
Ans.
1.8 Find the phase and group velocities of an electron whose de-Broglie wavelength is 0.12 nm.
Solution
l = 0.12 ¥ 10-9 m
The phase velocity of the wave is given by
vp =
w
k
vp =
hw
hk
...(1)
It can be written as
Using the relations
E = h w and p = hk
Thus
vp =
E
p
Also
E=
p2
1 2
mv =
2
2m
...(2)
&
Engineering Physics
Substitute this in Eqn. (2)
vp =
Using de-Broglie relation l =
p
2m
...(3)
h
in Eqn. (3)
p
vp =
6.62 ¥ 10 -34
h
=
2 ml 2 ¥ 9.1 ¥ 10 -31 ¥ 1.2 ¥ 10 -10
vp = 3.03 ¥ 106 m/s
The group velocity is always equal to the velocity of the particle. Thus
vg = v =
p
m
Using Eqn. (3), it becomes
vg = 2vp = 6.06 ¥ 106 m/s
Ans.
EXERCISE
1.1 Each of a photon and an electron have an energy of 1000 electron volt. Calculate their
corresponding wavelength.
(Ans: lp = 12.4 Å. le ª 0.04 nm)
1.2 Electrons are emitted with zero velocity from a certain metal surface when it is exposed to
radiation of 680 nm. Calculate the threshold frequency and the work function of the metal.
(Ans: 4.412 ¥ 1014 Hz, 1.83 eV)
1.3 The wavelength of the incident photon in Compton scattering is 0.4 nm. If sin q is 0.5,
calculate the wavelength of scattered radiation. Given l of the incident photon is 4 nm.
(Ans: 0.4015 nm)
1.4 Calculate the wavelength of a 1 kg object whose velocity is 1 m/s and compare it with the
wavelength of an electron accelerated by 100 volt (l0 = 6.6 ¥ 10-34 m, le = 1.2 ¥ 10-10 m and
le = 1.8 ¥ 1023 l0).
ANSWERS TO OBJECTIVE QUESTIONS
1.1 (b)
1.6 (a)
1.11 (a)
1.2 (a)
1.7 (a)
1.12 (c)
1.3 (d)
1.8 (b)
1.13 (a)
1.4 (b)
1.9 (c)
1.5 (b)
1.10 (b)