CHAPTER 1: ELEMENTS OF WAVE MECHANICS
1.1 Introduction. 1.2 Black body radiation. 1.2.1 Experimental observation of black body radiation. 1.2.2 Laws of black body radiation. 1.2.3 Stefan – Boltzmann radiation law. 1.2.4 Wien’s law. 1.2.5 Rayleigh – Jean’s law. 1.2.6 Planck’s radiation law. 1.2.7 Derivation of Wien’s law from Planck’s law. 1.2.8 Derivation of Rayleigh – Jean’s law from Planck’s
law. 1.3 Photoelectric effect. 1.4 Compton effect. 1.5 Matter waves and de Broglie hypothesis
1.5.1 Davisson and Germer experiment.
1.5.2 G.P.Thomson experiment
1.5.3 Wave packet and de Broglie wave
1.6 Phase and group velocities
1.6.1 Relation between phase velocity and group velocity
1.6.2 Relationbetween group velocity and particle
velocity (Velocity of de Broglie waves)
1.6.3 Derivation of de Broglie relation
1.7 Uncertainty principle.
1.7.1 Origin and nature of the principle.
1.7.2 An illustration of uncertainty principle.
1.7.3 Physical significance of uncertainty principle.
1.8 Wave mechanics.
1.8.1 Characteristics of wave function.
1.8.2 Physical significance of wave function.
1.8.3 Schrodinger’s wave equation.
1.8.4 Eigen values and Eigen functions.
1.9 Applications of Schrodinger’s equation.
1.9.1 Case of a free particle.
1.9.2 Particle in a box.
1.9.3 Finite potential well
1.9.4 Tunnel effect
1.9.5 Examples of tunneling across a finite barrier.
1.9.6 Theoretical interpretation of tunneling
1.9.7 Harmonic oscillator.
1.9.8 Practical applications of Schrodinger’s wave
equation.
Numerical examples
1.1 INTRODUCTION:
The nineteenth century was a very eventful period as
far as Physics is concerned. The pioneering work on
dynamics by Newton, on electromagnetic theory by
Maxwell, laws of thermodynamics and kinetic theory were
successful in explaining a wide variety of phenomena.
Even though a majority of experimental evidence agreed
with the classical physics, a few experiments gave
results that could not be explained satisfactorily.
These few experiments led to the development of modern
physics. Modern physics refers to the development of
the theory of relativity and the quantum theory.
Inability of the classical concepts to explain certain
experimental observations, especially those involving
subatomic particles, led to the formulation and
development of modern physics. Early twentieth century
saw the development of modern physics. The pioneering
work of Einstein, Planck, Compton, Roentgen, Born and
others formed the basis of modern physics. The dual
nature of matter proposed by de Broglie was confirmed
by experiments. The wave mechanics and quantum
mechanics were later shown to be identical in their
mathematical formulation. The validity of classical
concepts was explained to be the result of an
extrapolation of modern theories to classical
situations. In the present chapter, experimental
observations of three important phenomena – black body
radiation, photoelectric effect and Compton effect –
considered as the beginning of modern physics, are
briefly described.
1.2 BLACK BODY RADIATION:
When radiation is incident on material objects, it
is either absorbed, reflected or transmitted. These
processes are dependent on the radiation and the object
involved. An object that is capable of absorbing all
radiation incident on it is called a black body.
Practically, we cannot have a perfect black body but
can have objects that are only close to a black body.
For example, a black body can be approximated by a
hollow object with a very small hole leading to the
inside of the object. Any radiation that enters the
object through the hole gets trapped inside and will be
reflected by the walls of the cavity till it is
absorbed. Objects that absorb a particular wavelength
of radiation are also found to be a good emitter of
radiation of that particular wavelength. Hence, a
black body is also a good emitter of all radiations it
has absorbed.
Emissions from objects depend on the temperature
of the object. It has been observed that the energy
emitted from objects increases as the temperature of
the object is increased. Laws of radiation have been
formulated to explain the emission of energy by objects
maintained at specific temperatures.
1.2.1 Experimental observation of black body radiation:
Experiments have been carried out to study the
distribution of energy emitted by a practical blackbody
as a function of wavelength and temperature. Figure 1.1
.
1.1 Distribution of emitted energy as a function of wavelength and temperature for a black body.
shows the distribution curves in which the energy
density Eλ is plotted as a function of wavelength at
different temperatures of the black body. Energy
density is defined as the energy emitted by the black
body per unit area of the surface. The important
features of these distribution curves may be summarized
as follows:
(i) The energy vs wavelength curve at a given
temperature shows a peak indicating that the emitted
intensity is maximum at a particular wavelength and
decreases as we move away from the peak.
(ii) An increase in temperature results in an increase
in the total energy emitted and also the energy emitted
at all wavelengths.
(iii) As the temperature increases, the peak shifts to
lower wavelengths. In other words, at higher
temperatures, maximum energy is emitted at lower
wavelengths.
1.2.2 Laws of black body radiation:
The initial attempts to explain black body
radiation were based on classical theories and were
found to be limited in application. They could not
explain the entire spectrum of the radiation
satisfactorily.
1.2.3 Stefan Boltzmann radiation law:
It states that the total energy density Eo of
radiation emitted from a black body is directly
proportional to the fourth power of its absolute
temperature T. Energy density E0 is defined as the
total of all the energy emitted at all wavelengths per
unit area of the emitter surface.
Eo ∝ T4
or Eo = σ T4 (1.1)
where σ is a constant called Stefan’s constant. It
has a numerical value equal to 5.67 x 10-8 watt m-2 K-4 .
This law was suggested empirically by Stefan and later
derived by Boltzmann on thermodynamic considerations.
The law agrees well with the experimental results.
1.2.4 Wien’s Laws:
Wien’s displacement law states that the wavelength
λm corresponding to the maximum emissive energy
decreases with increasing temperature.
i.e. λm ∝ 1/T or λm T = b (1.2)
where b is called the Wien’s constant and is equal to
2.9 x 10-3 mK. The energy density emitted by a black
body in the wavelength range λ and λ + dλ is given by
Eλ dλ = c1 λ-5 exp(-c2/λT) dλ (1.3)
where c1 and c2 are constants. This is known as Wien’s
distribution law. This law holds good for smaller
values of λ but does not fit the experimental curves
for higher values of λ (fig 1.2).
1.2 Comparison of experimental distribution curve with Wien’s law.
1.2.5 Rayleigh – Jean’s law:
According to this law, the energy density emitted
by a black body in the wavelength range λ And λ + dλ is
given by
Eλ dλ = 8πkT λ4
dλ (1.4)
This equation does not show any peak in the energy
value but the energy goes on increasing with decrease
in wavelength. The total energy emitted is infinite
for all temperatures above 0 K.
1.3 Comparison of experimental distribution curve with Rayleigh-Jean’s law.
This is not at all in agreement with the experimental
observation. The law holds good only for large values
of wavelength (fig 1.3). At lower wavelengths, the
energy density increases and becomes very large for
wavelengths in the ultra violet region. Such a large
increase in the energy emitted at low wavelength does
not occur experimentally. This discrepancy is known as
“Ultraviolet catastrophe” of classical physics.
All the above laws are based on classical
thermodynamics and statistics. They are insufficient
to explain the black body radiation satisfactorily.
1.2.6 Planck’s radiation law:
This law is based on quantum theory. Max Planck
proposed that atoms or molecules absorb or emit
radiation in quanta or small energy packets called
photons. Energy of each photon can be expressed as
E = hν
where ν is the frequency of the radiation corresponding
to the energy E, h is a constant called Planck’s
constant and is equal to 6.63 x 10-34 Js. Light quanta
are indistinguishable from each other and there is no
restriction on the number of quanta having the same
energy. In other words, Pauli’s exclusion principle is
not applicable to them. The quantum statistics
applicable to photons is Bose-Einstein statistics.
Considering all the energy emitted by the black body in
the form of photons of different energy, Planck applied
Bose - Einstein statistics to obtain the energy
distribution of photons. Accordingly, the energy
density emitted in the wavelength range λ and(λ + dλ)
is given by
Eλ dλ = 8πhc _____1 ___ λ5 (ehc/λkT –1)
dλ (1.5)
This distribution agrees well with the
experimental observation of black body radiation and is
valid for all wavelengths. Further, it reduces to
Wien’s law for lower wavelength region and to Rayleigh
– Jean’s law for higher wavelength region.
1.2.7 Derivation of Wien’s law from Planck’s law:
When λ is small, we can consider
ehc/λkT > 1
∴ [ehc/λkT - 1 ] ≈ ehc/λkT
Substituting in equation (1.5), we get
Eλ dλ = 8πhc . 1__ λ5 (ehc/λkT)
. dλ
i.e. Eλ dλ = c1 λ-5 . exp(-c2/λT) dλ (1.6) where c1 = 8πhc and c2 = hc/k
Equation (1.6) is the Wien’s law.
1.2.8 Derivation of Rayleigh – Jean’s law from Planck’s
law.
When λ is large, hc λkT
< 1.
∴ [ehc/λkT - 1 ] ≈ hc / λkT Substituting in equation (1.5), we get
Eλ dλ = 8πhc . λkT λ5 hc
. dλ
i.e. Eλ dλ = 8πkT λ4
dλ (1.7)
Equation (1.7) is the Rayleigh – Jean’s law.
1.3 PHOTOELECTRIC EFFECT:
Emission of electrons from a metal surface when
light of suitable energy falls on it is called
Photoelectric effect. The experimental setup for
observing photoelectric effect consists of a pair of
metal plate electrodes in an evacuated tube connected
to a source of variable voltage as shown in fig.1.4.
1.4 Experimental set up to study photoelectric effect.
When light of suitable energy is incident on the cathode,
electrons are emitted and a current flows across the tube.
The characteristic curves for the photoelectric emission as
shown in fig. 1.5.
1.5 Current – voltage characteristics of photocell. The Intensity of illumination increases from L1 to L3.
The important properties of the emission are as
follows:
(i) There is no time interval between the incidence of
light and the emission of photoelectrons.
(ii) There is a minimum frequency for the incident
light below which no photoelectron emission occurs.
This minimum frequency, called threshold frequency,
depends on the material of the emitter surface. The
energy corresponding to this threshold frequency is the
minimum energy required to release an electron from the
emitter surface. This energy is characteristic of the
material of the emitter and is called the work function
of the material of the emitter.
(iii) For a given constant frequency of incident
light, the number of photoelectrons emitted or the
photo current is directly proportional to the intensity
of incident light.
(iv) The photoelectron emission can be stopped by
applying a reverse voltage to the phototube, i.e. by
making the emitter electrode positive and the collector
negative. This reverse voltage is independent of the
intensity of incident radiation but increases with
increase in the frequency of incident light. The
negative collector potential required to stop the photo
electron emission is called the stopping potential.
These characteristics of photoelectron emission can
not be explained on the basis of classical theory of
light but can be explained using the quantum theory of
light. According to this theory, emission of electrons
from the metal surface occurs when the energy of the
incident photon is used to liberate the electrons from
their bound state. The threshold frequency corresponds
to the minimum energy required for the emission. This
minimum energy is called the work function of the
metal. When the incident photon carries an energy in
excess of the work function, the extra energy appears
as the kinetic energy of the emitted electron. When the
intensity of light increases, the number of
photoelectrons emitted also increases but their kinetic
energy remains unaltered. The reverse potential
required to stop the photoelectron emission, i.e. the
stopping potential, depends on the energy of the
incident photon and is numerically equivalent to the
maximum kinetic energy of the photoelectrons.
When a photon of frequency ν is incident on a metal
surface of work function Φ, then,
hν = Φ + (½ mv2)max (1.8)
where (½ mv2)max is the maximum kinetic energy of the
emitted photoelectrons. This is known as Einstein’s
photoelectric equation. Since Φ = hνo, it can also be
written as
(½ mv2)max = hν– Φ = h(ν-νo ) (1.9)
If Vo is the stopping potential corresponding to the
incident photon frequency ν, then,
(½ mv2)max = hν – Φ = eVo (1.10)
Then, by experimental determination of Vo, it is
possible to find out the work function of the metal.
The experimental observation of photoelectric
effect leads to the conclusion that the energy in light
is not spread out over wavefronts but is concentrated
in small packets called photons. All photons of a
particular frequency have the same energy. A change in
the intensity of the incident light will change the
number of photoelectrons emitted but not their
energies. Higher the frequency of the incident light,
higher will be the kinetic energy of the
photoelectrons. These observations confirm the particle
properties of light waves.
1.4 COMPTON EFFECT:
When x-rays are scattered by a solid medium, the
scattered x-rays will normally have the same frequency
or energy. This is a case of elastic scattering or
coherent scattering. However, Compton observed that in
addition to the scattered x-rays of same frequency,
there existed some scattered x-rays of a slightly
higher wavelength (i.e., lower frequency or lower
energy). This phenomenon in which the wavelength of x-
rays show an increase after scattering is called
Compton effect.
Compton explained the effect on the basis of the
quantum theory of radiation. Considering radiation to
be made up of photons, he applied the laws of
conservation of energy and momentum for the interaction
of photon with electron. Consider an x-ray photon of
energy hν incident on an electron at rest (fig. 1.6.)
After the interaction, the x-ray photon gets scattered
at an angle θ with its energy changed to a value hν’ and
the electron which was initially at rest recoils at an
angle Φ. It can be shown that the increase in
wavelength is given by
1.6 Schematic diagram of the scattering of a photon by
a stationary electron.
∆λ = _h_ moc
(1-cos θ) (1.11)
where mo is the rest mass of the electron.
When θ = 90o, ∆λ = __h__ moc
= 0.0242 A°.
This constant value is called Compton wavelength.
When θ = 180o, ∆λ = moc
2h
Experimental observation indicate that the change in
the wavelength of the scattered x-rays is indeed in
agreement with equation (1.11), thus providing further
confirmation to the photon model.
Thus, Planck’s theory of radiation, photoelectric
effect and Compton effect are experimental evidences in
favour of the quantum theory of radiation.
1.5 MATTER WAVES AND DE BROGLIE’S HYPOTHESIS
Quantum theory and the theory of relativity are the
two important concepts that led to the development of
modern physics. The quantum theory was first proposed
by Planck to explain and overcome the inadequacies of
classical theories of black body radiation. The
consequences were very spectacular. Louis de Broglie
made the suggestion that particles of matter, like
electrons, might possess wave properties and hence
exhibit dual nature. His hypothesis was based on the
following arguments:
The Planck’s theory of radiation suggests that energy
is quantized and is given by
E = hν (1.12)
where ν is the frequency associated with the radiation.
Einstein’s mass-energy relation states that
E = mc2 (1.13)
Combining the two equations, it can be written as
E = hν = mc2
Hence, the momentum associated with the photon is given
by
P = mc = hν/c = h/λ
Extending this to particles, he suggested that any
particle having a momentum p is associated with a wave
of wavelength λ given by
λ = h/p (1.14)
This is called de Broglie’s hypothesis of matter waves
and λ is called the de Broglie wavelength.
The de Broglie wavelength can be calculated for any
particle using the above relation. In case of charged
particles like electrons, a beam of high energy
particles can be obtained by accelerating them in an
electric field. For example, an electron starting from
rest when accelerated with a potential difference V,
the kinetic energy acquired by the electron is given by
(1/2)mv2 = eV
where v is the velocity of the electron. The momentum
may be calculated as
p = mv = (2meV)1/2
Using the de Broglie equation, the wavelength
associated with the accelerated electron can be
calculated as
λ = h/p = h/(2meV)1/2 (1.15)
This equation suggests that, at a given speed, the de
Broglie wavelength associated with the particle varies
inversely as the mass of the particle. This concept of
matter waves aroused great interest and several
physicists launched experiments designed to test the
hypothesis. Heisenberg and Schrodinger proceeded on to
develop mathematical theories whereas Davisson and
Germer, G.P.Thomson and Kikuchi attempted experimental
verification.
1.5.1 Davisson-Germer experiment The hypothesis of de Broglie was verified by the
electron diffraction experiment conducted by Davisson
and Germer in the United States. The experimental set
up used by them is shown in the figure 1.7.
1.7 Experimental arrangement for Davisson-Germer
experiment.
The apparatus consists of a filament heated with a
small a.c power supply to produce thermionic emission
of electrons. These electrons are attracted towards an
anode in the form of a cylinder with a small aperture
maintained at a finite positive potential with respect
to the filament. They pass through the narrow aperture
forming a fine beam of accelerated electrons. This
electron beam was made to incident on a single
crystalline sample of nickel. The electrons scattered
at different angles were counted using an ionization
counter as a detector. The experiment was repeated by
recording the scattered electron intensities at various
positions of the detector for different accelerating
potentials (Fig.1.8).
Fig.1.8. Scattered electron intensity maps at different
accelerating potentials.The vertical axis represents
the direction of the incident electron beam and ф is
the scattering angle.The radial distance from the
origin at any angle represents the intensity of
scattered electrons.
When a beam of electrons accelerated with a potential
of 54 V was directed perpendicular to the nickel
target, a sharp maximum occurred in the electron
density at an angle of 500 with the incident beam. When
the angle ф between the direction of the incident beam
and the direction of the scattered beam is 500, the
angle of incidence will be 250 and the corresponding
angle of diffraction θ will be 650. The spacing of the
planes responsible for diffraction was found to be
0.091 nm from x-ray diffraction experiment. Assuming
first order diffraction, the wavelength of the electron
beam can be calculated as
λ = 2d sin θ = 2 x 0.091 x sin 650 = 0.165 nm.
The wavelength of the electrons can also be calculated
using the de Broglie’s relation as
λ = h/(2meV)1/2
= 6.63 x 10-34/(2 x 9.1 x 10-31 x 1.6 x 10-19 x 54)1/2
= 0.166 nm.
Thus, the Davisson-Germer experiment directly verifies
the de Broglie’s hypothesis.
1.5.2 G.P.Thomson experiment At almost the same time as the Davisson-Germer
experiment, G.P.Thomson of England carried out electron
diffraction experiments independently using a thin
polycrystalline foil of aluminium metal. The
experimental set up is shown in fig. 1.9.
1.9 Experimental arrangement of G.P.Thomson experiment.
He allowed a beam of accelerated electrons to fall on
the aluminium foil and observed a diffraction pattern
consisting of a series of concentric rings around the
direction of the incident beam. This pattern was
similar to the Debye-Scherrer pattern obtained for
aluminium using x-ray diffraction. Using the data
available on aluminium, he calculated the wavelength of
the electrons using the Bragg’s equation,
nλ = 2d sin θ
He also calculated the de Broglie wavelength of the
electrons with the knowledge of accelerating potential
using the relation,
λ = h/(2meV)1/2
The value of wavelength calculated from the two
equations matched well thereby experimentally proving
the de Broglie’s relation.
A similar experiment was conducted by Kikuchi in
Japan in which he obtained electron diffraction pattern
by passing an electron beam through a thin foil of mica
to confirm the validity of de Broglie’s relation.
The wave nature of particles is not restricted to
electrons. Any particle with a momentum p has a de
Broglie wavelength equal to (h/p). Neutrons produced in
nuclear reactors possess energies corresponding to
wavelength of the order of 0.1 nm. These particles also
should be suitable for diffraction by crystals.
Neutrons from a nuclear reactor are slowed down to
thermal energy of the order of kT and used for
diffraction and interference experiments. The results
agree well with the de Broglie relation. Since neutrons
are uncharged particles, they are particularly useful
in certain situations for diffraction studies. Neutron
beams have also been used as probes to investigate the
magnetic properties of nuclei.
1.5.3 Wave packet and de Broglie waves We have seen that moving particles may be
represented by de Broglie waves. The amplitude of these
de Broglie waves does not represent any parameter
directly describing the particle but is related to the
probability of finding the particle at a particular
place at a particular time. Hence, we cannot describe
de Broglie waves with a simple wave equation of the
type,
y = A cos(ωt-kx) (1.16)
Instead, we have to use an equation representing a
group of waves. In other words, a wave packet
consisting of waves of slightly differing wavelengths
may represent the moving particle. Superposition of
these waves constituting the wave packet results in the
net amplitude being modified, thereby defining the
shape of the wave group. The phase velocity of
individual waves depends on the wavelength. Since the
wave group consists of waves with different
wavelengths, all the waves do not proceed together and
the wave group has a velocity different from the phase
velocities of the individual waves. Hence, de Broglie
waves may be associated with group velocity rather than
the phase velocity.
1.5.4 Characteristics of matter waves 1. Matter waves are associated with a moving body.
2. The wavelength of matter waves is inversely
proportional to the velocity with which the body
is moving. Hence, a body at rest has an infinite
wavelength and the one traveling with a high
velocity has a lower wavelength.
3. Wavelength of matter waves also depends on the
mass of the body and decreases with increase in
mass. Due to this reason, the wavelike behaviour
of heavier bodies is not very evident whereas wave
nature of subatomic bodies could be observed
experimentally.
4. A wave is normally associated with some quantity
that varies periodically with the frequency of the
wave. For example, in a water wave, it is the
height of the water surface; in a sound wave it is
the pressure and in an electromagnetic wave, it is
the electric and magnetic fields that vary
periodically. But in matter waves, there is no
physical quantity that varies periodically. We use
a wave function to define matter waves and this
wave function is related to the probability of
finding the particle at any place at any instant,
which varies periodically.
5. Matter waves are represented by a wave packet made
up of a group of waves of slightly differing
wavelengths. Hence, we talk of group velocity of
matter waves rather than the phase velocity. The
group velocity can be shown to be equal to the
particle velocity.
6. Matter waves show properties similar to other
waves. For example, a beam of accelerated
electrons produces interference and diffraction
effects similar to an electromagnetic wave of
same wavelength.
1.6 PHASE AND GROUP VELOCITIES:
A wave is represented by the formula
y = A cos (ωt – kx) (1.16)
where y is the displacement at any instant t, A is the
amplitude of vibration, ω is the angular frequency
equal to 2πν and k is the wave vector, equal to (2π/λ).
The phase velocity of such a wave is the velocity with
which a particular phase point of the wave travels.
This corresponds to the phase being constant.
i.e., (ωt – kx) = constant
or x = constant + ωt/k
Phase velocity vp = dx/dt = ω/k
= 2πν/(2π/λ) = λν (1.17)
vp is called the ‘wave velocity’ or ‘phase velocity’.
The de Broglie waves are represented by a wave packet
and hence we have ‘group velocity’ associated with
them. Group velocity is the velocity with which the
wave packet travels. In order to understand the concept
of group velocity, consider the combination of two
waves represented by the formula
y1 = A cos (ωt-kx)
y2 = A cos {(ω+∆ω)t – (k+∆k)x }
The resultant displacement is given by
y = y1 + y2 = 2A cos {(ω+ω+∆ω)t–(k+k+∆k)x} cos 2 2
(∆ωt-∆kx)
≈ 2A cos(ωt–kx).cos(∆ωt/2-∆kx/2) (1.18)
The velocity of the resultant wave is given by
the speed with which a reference point, say the maximum
amplitude point, moves. Taking the amplitude of the
resultant wave as constant, we have
2A cos(∆ωt/2-∆kx/2) = constant
or (∆ωt/2-∆kx/2) = constant
or x = constant + (∆ωt/∆k)
Group velocity vg = dx/dt = (∆ω/∆k) (1.19)
Instead of two discrete values for ω and k,
if the group of waves has a continuous spread from ω to
(ω+∆ω) and k to (k+∆k), then, the group velocity is
given by
vg = dω (1.20) dk
It can be shown that the group velocity of the wave
packet is equal to the velocity of the particle with
which the wave packet is associated.
1.6.1 Relation between phase velocity and group
velocity:
We have the mathematical relation for phase velocity
given by
vp = ω/k or ω = k.vp
The group velocity vg is given by
vg = dω dk dk
= d(k.vp)
= vp + k.dvp dk
= vp + (2π/λ). dvp d(2π/λ)
= vp + (2π/λ).(-λ2/2π). dλ
dvp
= vp - λ. dvp dλ
(1.21)
In the above expression, if (dvp/dλ) = 0, i.e., if the
phase velocity does not depend on wavelength, then the
group velocity and phase velocity are equal. Such a
medium is called a non-dispersive medium. In a
dispersive medium, (dvp/dλ) is positive and hence the
group velocity is less than the phase velocity.
1.6.2 Relation between group velocity and particle
velocity (Velocity of de Broglie waves):
The phase velocity of waves depend on the
wavelength. This is responsible for the well known
phenomenon of dispersion. In the case of light waves in
vacuum, the phase velocity is same for all wavelengths.
In the case of de Broglie waves, we have,
ω = 2πν = 2πmc2/h = 2πm0c2 h(1-v2/c2)1/2
(1.22)
and k = 2π/λ = 2πmv/h = 2πm0v h(1-v2/c2)1/2
(1.23)
The group velocity of de Broglie waves is given by
Vg = dω/dk = dω/dv dk/dv
dω/dv = (2πm0c2/h).d(1-v2/c2)1/2 = 2πm0v dv h(1-v2/c2)3/2
(1.24)
dk/dv = ____2πm0__ h(1-v2/c2)3/2
___ (1.25)
From equations 1.24 and 1.25 we get,
vg = v
Thus, the group velocity associated with de
Broglie waves is just equal to the velocity with which
the particle is moving. If we try to calculate the
phase velocity,
Vp= ω/k = c2/v = c2/vg (1.26)
Since the group velocity or the particle velocity is
always less than c, the phase velocity of de Broglie
waves turn out to be greater than c. This only
indicates that we cannot talk of phase velocity of de
Broglie waves since they are made up of a group of
waves. Phase velocity has no physical significance for
de Broglie waves.
1.6.3 Derivation of de Broglie relation: The de Broglie relation may be derived as follows.
If we assume a particle having a kinetic energy equal
to mv2/2 to have a de Broglie wavelength λ, we can
write
hν = mv2/2 (assuming the energy of the particle
to be purely kinetic)
or ν = _m .v2 (1.27) 2h Differentiating with respect to λ,
dν = m . 2v. dv dλ 2h dλ
(1.28)
But we have vg = v = dω = 2πdν = -λ2
dk 2πd(1/λ) dλ dν
∴ dν = - v_ dλ λ2
(1.29)
Substituting in eqn.1.28, we get mv . dv h dλ λ2
= - v
Rewriting this, we have dv dλ mλ2
_ = - h_ (1.30)
Integrating with respect to λ,
v = h _ mλ
+ c
where c is the constant of integration. By applying the boundary condition that the wavelength tends to infinity as the velocity tends to zero, we find that the constant of integration has to be zero. Hence, we get
λ = _h_ (1.31) mv which is the de Broglie relation.
1.7 HEISENBERG’S UNCERTAINTY PRINCIPLE:
1.7.1 Origin and nature of the Principle:
When we assign wave properties to particles there
is a limitation to the accuracy with which we can
measure the properties like position and momentum.
1.10 A wave packet with an extension ∆x along x-axis.
Consider a wave packet as shown in fig.1.10. The
particle to which this wave packet corresponds to may
be located anywhere within the wave packet at any
instant. The probability density suggests that it is
most likely to be found in the middle of the wave
packet. However, there is a finite probability of
finding the particle anywhere within the wave packet.
If the wave packet is smaller in extension, the
position of the particle can be specified more
precisely. But the wavelength of the waves will not be
well defined in a narrow wave packet. Since wavelength
is related to momentum through de Broglie’s relation,
the momentum is not precisely known. On the otherhand,
a wave packet with large extension can have a more
clearly defined wavelength and hence momentum at the
cost of the knowledge about the position. This leads
to the conclusion that it is impossible to know both
the position and momentum of an object precisely at the
same time. This is known as Uncertainty principle.
For a wave packet of extension ∆x with an
uncertainty in the wave number ∆k assuming the
uncertainties to be the standard deviation in the
respective quantities, it may be shown that a minimum
value of the product of such deviations is given by
∆x . ∆k = ½ (1.32)
This minimum value of the product of uncertainties is
for the case of a gaussian distribution of the wave
functions. Since the wave packets in general do not
have gaussian forms, the uncertainty relation becomes
∆x . ∆k ≥ ½ (1.33)
But we have
k = 2π/λ (1.34)
Also λ = h/p (1.35)
Hence, k = 2πp/h
∆k = 2π
. ∆p (1.36) h
Substituting in equation (1.33), we get
∆x . ∆p ≥ h_ 4π
or ∆x. ∆p ≥ ħ (1.37) 2
This equation states that the product of
uncertainty ∆x in the position of an object at some
instant and the uncertainty in the momentum in the x-
direction at the same instant is equal to or greater
than ħ/2.
Another form of uncertainty principle relates
energy and time. In the atomic process, if energy E is
emitted as an electromagnetic wave during an interval
of time ∆t, then, the uncertainty ∆E in the measured
value of E depends on the duration of the time interval
∆t according to the equation,
∆E . ∆t ≥ ħ/2 (1.38)
It may be mentioned that these uncertainties
are not due to the limitations of the precision of the
measuring methods or measuring instruments but due to
the nature of the quantities involved.
1.7.2 An illustration of uncertainty principle:
We have the following ‘Thought experiment’ to
illustrate the uncertainty principle. Imagine an
electron being observed using a microscope (fig.1.11).
1.11 Schematic diagram of experimental set up to study uncertainty principle.
The process of observation involves a photon of
wavelength λ incident on the electron and getting
scattered into the microscope. The event may be
considered as a two-body problem in which a photon
interacts with an electron. The change in the velocity
of the photon during the interaction may be anything
between zero( for grazing angle of incidence) and 2c
( for head-on collision and reflection). The average
change in the momentum of the photon may be written as
equal to (hν/c) or (h/λ). This difference in momentum
is carried by the recoiling electron which was
initially at rest. The change or uncertainty in the
momentum of the electron may thus be written as (h/λ).
At the same time, the position of the electron can be
determined to an accuracy limited by the resolving
power of the microscope, which is of the order of λ.
Hence, the product of the uncertainties in position and
momentum is of the order of h. This argument implies
that the uncertainties are associated with the
measuring process. This illustration only estimates
the accuracy of measurement, the uncertainty being
inherent in the nature of the moving particles
involved.
1.7.3 Physical significance of uncertainty principle:
Uncertainty principle is a consequence of the wave
particle duality. It states that it is impossible to
know both the position and momentum of an object
exactly and at the same time. Mathematically, it can
be shown that the product of uncertainties in the
position and momentum measured simultaneously will have
a value greater than ħ/2, ie (h/4π). If ∆x is the
uncertainty in the measurement of the position x of an
object and ∆px is the uncertainty in the measurement
of momentum px , then, at any instant,
∆x . ∆px > ħ/2
We can try to estimate the product of the uncertainties
with the help of illustrations as the one mentioned
above. The principle is based on the assumption that a
moving particle is associated with a wave packet, the
extension of which in space accounts for the
uncertainty in the position of the particle. The
uncertainty in the momentum arises due to the
indeterminacy of the wavelength because of the finite
size of the wave packet. Thus, the uncertainty
principle is not due to the limited accuracy of
measurement but due to the inherent uncertainties in
determining the quantities involved. But we can still
define the position where the probability of finding
the particle is maximum and also the most probable
momentum of the particle.
1.7.4 Applications of uncertainty principle:
The uncertainty principle has far reaching
implications. In fact, it has been very useful in
explaining many observations which cannot be explained
otherwise. A few of the applications of the uncertainty
principle are worth mentioning.
(a) Diffraction of a beam of electrons: Diffraction of
a beam of electrons at a slit is the effect of
uncertainty principle. As the slit is made narrower,
thereby reducing the uncertainty in the position of the
electrons in the beam, the beam spreads even more
indicating a larger uncertainty in its velocity or
momentum.
1.12 Diffraction at a single slit.
Figure 1.12 shows the diffraction of an electron beam
by a narrow slit of width ∆x. The beam traveling along
OX is diffracted along OY through an angle θ. Due to
the wave nature of the electron, we observe Fraunhoffer
diffraction on the screen placed along XY. The accuracy
with which the position of the electron is known is ∆x
since it is uncertain from which place in the slit the
electron passes. According to the theory of
diffraction, we have
λ = ∆x.sin θ or ∆x = λ/ sin θ
Further, the initial momentum of the electron along XY
was zero and after diffraction, the momentum of the
electron is p. sin θ where p is the momentum of the
electron along the incidence direction. Hence, the
change in momentum of the electron along XY is p. sin θ
or (h/λ). sin θ. Assuming the change in the momentum as
representative of the uncertainty in momentum, we get
∆x. ∆px = λ h.sin sin θ λ
θ = h
(b) Nuclear beta decay: In beta decay, electrons are
emitted from the nucleus of the radioactive element.
Assuming the diameter of the nucleus to represent the
uncertainty in the position of electron inside the
nucleus, the uncertainty in the momentum can be
calculated as follows:
Radius of the nucleus = r = 5 x 10-15 m
∆x = 2r = 10-14 m.
∆p = h/2π∆x = 6.62x10-34/(2x3.14x10-14)
= 1.055x10-20 kg m s-1
Assuming that the electron was at rest before
its emission, the change in momentum can be taken as
equal to its momentum. This magnitude of change in
momentum indicates large velocity for the electron.
Hence, the energy of the emitted electron will be
E = pc = 1.055x10-20 x 3x108 = 3.165 x 10-12 J
= 19.8 MeV.
This indicates that the electrons inside the nucleus
must have kinetic energy of 19.8 MeV. But the electrons
emitted during beta decay have kinetic energy of the
order of 1 MeV. This indicates that electrons do not
exist in the nucleus of the atom but are ‘manufactured’
by the nucleus at the time of decay.
(c)Binding energy of an electron in an atom: In a
hydrogen atom, the electron revolves round the nucleus
in an orbit of radius 5 X 10-11 m. Assuming this as the
maximum uncertainty in position, we can calculate
the minimum uncertainty in the momentum as
(∆p)min = h/2π(∆x)max = 2.1 X 10-24 kg m s-1.
Assuming this as the momentum of electron, the kinetic
energy of the electron will be equal to
K.E. = p2/2m = 2.45 X 10-18 J = 15.3 eV.
Thus, the binding energy of an electron in hydrogen
atom is nearly 15 eV which is found to be correct
experimentally.
(d) Nitrogen doping of silicon: The laws of
conservation of energy and momentum restrict the
generation and recombination processes in
semiconductors. Silicon, which is an indirect band gap
semiconductor, has low efficiency as a material for
photo diode or light emitting diode. Nitrogen doping of
silicon will bind the free electrons to the lattice
thereby restricting the value of uncertainty in
position. This results in a larger uncertainty in
momentum thereby increasing the probability for
generation or recombination process.
1.8 WAVE MECHANICS:
Quantum theory is based on the quantization
of energies. It deals with the particle nature of
radiation. It implies that addition or liberation of
energy will be between discrete energy levels. It
assigns particle status to a packet of energy by
calling it ‘quantum of energy’ or ‘photon’ and treats
the interaction of radiation with matter as a two-body
problem. On the other hand, de Broglie’s hypothesis and
the concept of matter waves led to the development of a
different formulation called ‘Wave mechanics’. This
deals with the wave properties of material particles.
It was shown later that the quantum mechanics and the
wave mechanics are mathematically identical and lead to
the same conclusion.
1.8.1 Characteristics of wave function:
Waves in general are associated with quantities
that vary periodically. For example, water waves
involve the periodic variation of the height of the
water surface at a point. Similarly, sound waves are
associated with periodic variations of the pressure.
In the case of matter waves, the quantity that varies
periodically is called ‘wave function’. The wave
function, represented by ψ, associated with matter
waves has no direct physical significance. It is not
an observable quantity. But the value of the wave
function is related to the probability of finding the
body at a given place at a given time. The square of
the absolute magnitude of the wave function of a body
evaluated at a particular time at a particular place is
proportional to the probability of finding the body at
that place at that instant.
The wave functions are usually complex. The
probability in such a case is taken as ψ∗ψ, i.e. the
product of the wave function with its complex
conjugate. Since the probability of finding the body
somewhere is finite, we have the total probability over
all space equal to certainty.
i.e.∫ ψ∗ψ dV = 1 (1.39)
Equation (1.39) is called the normalization condition
and a wave function that obeys the equation is said to
be normalized. Further, ψ must be single valued since
the probability can have only one value at a particular
place and time. Since the probability can have any
value between zero and one, the wave function must be
continuous. Momentum being related to the space
derivatives of the wave function, the partial
derivatives ∂ψ/∂x, ∂ψ/∂y and ∂ψ/∂z must also be
continuous and single valued everywhere. Thus, the
important characteristics of wave function are as
follows:
(1) ψ must be finite, continuous and single valued
everywhere.
(2) ∂ψ/∂x, ∂ψ/∂y and ∂ψ/∂z must be finite, continuous
and single valued everywhere.
(3) ψ must be normalizable.
1.8.2 Physical significance of wave function:
We have already seen that the wave function has no
direct physical significance. However, it contains
information about the system it represents and this can
be extracted by appropriate methods. Even though the
wave function itself is not directly an observable
quantity, the square of the absolute value of the wave
function is intimately related to the moving body and
is known as the probability density. This probability
density is the quantum mechanical method of finding the
body at a particular position at a particular time. The
wave function carries information about the particle’s
wave-like behaviour. It also provides information about
the momentum and energy of the particle at any instant
of time.
1.8.3 Schrodinger’s wave equation:
The motion of a free particle can be described
by the wave equation.
ψ = A exp{-i(ωt –kx)} (1.40)
But ω = 2 πν = 2π (E/h) = (E/ħ) and k = 2π/λ = 2π (p/h) = (p/ħ) where E is the total energy and p is the momentum of
the particle. Substituting in the equation (1.40), we
get,
ψ = A exp{-i ħ
(Et-px)} (1.41)
Differentiating equation (1.41) with respect to
x twice, we get,
∂2ψ = -p2 ψ or p2ψ = - ħ2 . ∂2ψ ∂x2 ħ2 ∂x2
(1.42)
Differentiating equation (1.41) with respect to t, we get, ∂ψ = - iE . ψ or E ψ = - ħ . ∂ψ ∂t ħ i ∂t
(1.43)
The total energy of the particle can be written as
E = p2
2m + U (1.44)
where U is the potential energy of the particle.
Multiplying both sides of the equation by ψ
E ψ = p2
2m ψ + Uψ (1.45)
Substituting for Eψ and p2ψ from equation (1.42) and (1.43) - ħ ∂ψ = - ħ2 ∂2ψ i ∂t 2m ∂x2
+ Uψ (1.46)
This is known as Schrodinger’s time dependent equation
in one dimension.
The wave function ψ in equation (1.41) may also be
written as
ψ = A exp{-i (Et-px)} = A exp(-iEt) . exp(ipx ħ ħ ħ
)
ψ = Φ exp (-iEt ħ
) (1.47 )
where Φ is a position dependent function. Substituting
this form of ψ in equation (1.45),
EΦ exp(-iEt) = p2 Φ exp(-iEt) + UΦ exp(-iEt ħ 2m ħ ħ
)
or EΦ exp(-iEt) = - ħ2 . ∂2Φ . exp(-iEt) + UΦ exp(-iEt ħ 2m ∂x2 ħ ħ
)
or ∂2Φ exp(-iEt) + 2m (E-U)Φ exp(-iEt ∂x2 ħ ħ2 ħ
) = 0
or ∂2ψ + 2m ∂x2 ħ2
(E-U)ψ = 0 (1.48)
This is the Schrodinger’s wave equation in one
dimension. In three dimensions, the above equation may
be written as
∂2ψ + ∂2ψ + ∂2ψ + 2m(
E-U)ψ = 0 ∂x2 ∂y2 ∂z2 ħ2
or ∇2ψ + 2m( ħ2
E-U)ψ =0
This equation is known as the steady state or time
independent Schrodinger wave equation in three
dimensions.
1.8.4 Eigen values and eigen functions:
These terms come from the German words and mean
proper or characteristic values or functions
respectively. The values of energy for which the
Schrodinger’s equation can be solved are called ‘Eigen
values’ and the corresponding wave functions are called
‘Eigen functions’. The eigen functions possess all the
characteristics properties of wave functions in
general (see section 1.8.1).
1.9 APPLICATIONS OF SCHRODINGER’S EQUATION:
1.9.1 Case of a free particle: A free particle is defined as one which is not
acted upon by any external force that modifies its
motion. Hence, the potential energy U in the
Schrodinger’s equation is a constant and does not
depend on position or time. For convenience, the
potential energy may be assumed to be zero. Then, the
Schrodinger’s equation for the particle becomes
∂2ψ + 2m ∂x2 ħ2
Eψ = 0 (1.49)
where E is the total energy of the particle which is
purely kinetic. This is of the form,
∂2ψ ∂x2
+ k2ψ = 0
where k2 = 2mE/ħ2. The solution of this equation may be
written as
ψ = A cos kx + B sin kx
Solving for the constants A and B pose some
difficulties because we cannot apply any boundary
conditions on the wave function as it represents a
single wave which is not localized and not
normalizable. Since the solution has not imposed any
restriction on the value of k, the free particle is
permitted to have any value of energy given by the
equation,
E = ħ2k2/2m
Since the total energy is purely kinetic, the momentum
of the particle would be p = ħk or h/λ. This is just
what we would expect, since we have constructed the
Schrodinger equation to yield the solution for the free
particle corresponding to a de Broglie wave.
1.9.2 Particle in a box:
The simplest problem for which Schrodinger’s
time independent equation can be applied and solved is
the case of a particle trapped in a box with
impenetrable walls.
Consider a particle of mass m and energy E
travelling along x-axis inside a box of width L. The
particle is thus restricted to move inside the box by
reflections at x=0 and x=L (Fig. 1.13).
1.13 Schematic for a particle in a box. The height of the wall extends to infinity.
The particle does not lose any energy when it collides
with the walls and hence the total energy of the
particle remains constant. The potential energy of the
particle is considered to be zero inside the box and
infinite outside. Since the total energy of the
particle cannot be infinite, it is restricted to move
within the box. The example is an oversimplified case
of an electron acted upon by the electrostatic
potential of the ion cores in a crystal lattice.
Since the particle cannot exist outside the box,
ψ = 0 for x < 0 and x ≥ L (1.50)
We have to evaluate the wave function inside the box.
The Schrodinger’s equation (1.48) becomes
∂2ψ + 2m∂x2 ħ2
Eψ = 0 for 0 < x < L (1.51)
ψ = A sin (2mE)1/2 x + B cos (2mE ħ2 ħ2
)1/2 x (1.52)
where A and B are constants.
Applying the boundary condition that ψ=0 at x = 0,
equation 1.52 becomes
A sin 0 + B cos 0 = 0 or B = 0.
Again, we have ψ = 0 at x = L. Then,
A.sin(2mE ħ2
)1/2.L=0
If A = 0, the wavefunction will become zero
irrespective of the value of x. Hence, A cannot be
zero.
Therefore, sin(2mE ħ2
)1/2.L=0
or (2mE ħ2
)1/2L=nπ where n=1,2,3 .. (1.53)
From (1.53), the energy eigen values may be written as
En = n2π2 ħ2
2mL2 where n = 1,2,3,… … (1.54)
From this equation, we infer that the energy of the
particle is discrete as n can have integer values. In
other words, the energy is quantized. We also note that
n cannot be zero because in that case, the wave
function as well as the probability of finding the
particle becomes zero for all values of x. Hence, n = 0
is forbidden. The lowest energy the particle can
possess is corresponding to n = 1 and is equal to
E1 = π2ħ2
2mL2
This is called ‘ground state energy’ or ‘zero point
energy’. The higher excited states will have energies
like 4E1, 9E1, 16E1, etc. This indicates that the energy
levels are not equally spaced.
The wave functions or the eigen functions are
given by
ψn = A. Sin 2mEn ħ2
1/2 x
or ψn = A. Sin nπ L
x (1.55)
Applying the normalization condition,
i.e. ∫ A2 Sin2 nπx L
. dx = 1 (1.56)
Since the wave function is non-vanishing only for
0 < x < L, it can be shown that
∫ Sin2 nπx L 2
dx = (L ) (1.57)
Substituting in equation (1.56), we have A2 (L 2 L
) = 1 or A = ( 2 )1/2 (1.58)
The eigen function or wave functions in equation (1.55) becomes ψn = ( 2 )½ sin (2mEn L ħ2
)½ x
ψn = ( 2 )½ sin nπx L L
(1.59)
1.14 Variation of wave function associated with an electron confined to a box in its ground state & excited states.
Fig. 1.14 shows the variation of the wave function
inside the box for different values of n and Fig.1.15
shows the probability densities of finding the particle
1.15 Probability function as a function of position.
at different places inside the box for different
values of n. Thus, wave mechanics suggests that the
probability of finding any particle at the lowest
energy level is maximum at the centre of the box which
is in agreement with the classical picture. However,
the probability of finding the particle in higher
energy states is predicted differently by the two
formulations.
1.9.3 Finite Potential well: In real life situations, the potential energy is
never infinite. The box with impenetrable walls has no
physical significance. However, we come across
situations where the potential energy is finite. Let us
try to solve the case of an electron in a finite
potential well. We can consider two different cases
corresponding to the following situations:
(i) the total energy E being greater than the
potential energy U, and
(ii) the total energy E being less than the
potential energy U.
Fig 1.16 Schematic for a particle in a potential well of finite depth ( E greater than U).
The first case may be represented by the figure
1.16. Consider the particle with total energy E inside
a potential well of height U. In the region II, where
the particle is not influenced by the potential (U =
0), the solution of the Schrodinger’s equation is of
the form,
ψ = A cos kx + B sin kx
where k = (2mE/ħ2)1/2. This particle may be represented
by a wave of wavelength λ = 2π/k. When the particle is
in region I and III, its wavelength changes to λ’=
2π/k’ where k’ = [2m(E-U)/ħ2]1/2. In other words, the
effect of the potential energy step is to reduce the
kinetic energy of the particle as evident from an
increase in the value of the wavelength.
In the second case, the total energy of the
particle is less than the potential energy. Under this
condition, classically, the particle cannot propagate
beyond the step since this amounts to the kinetic
energy being negative. But, wave mechanically, a
different solution results. Let U be greater than the
total energy E of the electron but finite. To analyze
this case, we have to consider the three regions
separately.
In region II, since U = 0, the electron is free and the
Schrodinger’s equation is
d2ψ + 2m dx2 ħ2
Eψ = 0 (1.60)
In regions I and III, we have
d2ψ + 2m dx2 ħ2
(E – U) ψ = 0 (1.61)
The solutions for these equations can be assumed to be ψI = A ei
βx + B e-iβx in region I (1.62) ψII = C ei
αx + D e-iαx in region II (1.63)
and ψIII = F eiβx + G e-iβx in region III (1.64)
where α = [(2mE)/ħ2 ]1/2 (1.65)
β = [ 2m (E-U)/ħ2 ]1/2 (1.66)
since E is less than U, (E-U) is (-)ve and β is
imaginary.
Let us define a new constant
ϒ = -iβ (1.67)
Then the equations (1.62) and (1.64) become
ψI = A e-ϒx + B eϒx (1.68)
ψIII = F e-ϒx + G eϒx (1.69)
To evaluate the constants, we consider the boundary
condition in the region I where the wave function
should reduce to zero as x → -∞. Then eqn (1.68)
becomes
0 = A . ∞ + B .0 or A = 0.
∴ψI = B eϒx (1.70)
Similarly, in region III, since the wave function
should reduce to zero as x → ∞, eqn (1.69) becomes
0 = F . 0 + G . ∞ or G = 0.
∴ψIII = F e-ϒx (1.71)
This indicates that the wave function decreases
exponentially as we move away from the potential well
on either sides. Inside the potential well the wave
function represented by the equation (1.63) varies
sinusoidally. Further, since the wave function and its
derivative are continuous at the boundaries
corresponding to x = 0 and x = L, the wave functions
are non-zero at these boundaries. The plots of the wave
functions and the probability densities are shown in
Fig. 1.17 and 1.18 respectively.
1.17 Variation of wave function of a particle in a
finite potential well.
1.18 Probability function as a function of position.
Thus, we observe that in case of a particle in a
potential well of finite height, the particle has a
finite probability of penetrating into the wall.
However, if the walls of the well are infinitely thick,
the particle will be confined to the well and performs
oscillatory motion inside the well.
1.9.4 Tunnel effect: In the previous case of a finite potential well,
even though the height of the wall was finite, the
thickness of the wall was assumed to be infinite. As a
result, the particle was trapped in the well in spite
of penetrating into the wall. Under the same condition
of the total energy being less than the potential
energy, if the thickness of the wall is reduced and
made finite, the solution of the Schrodinger’s equation
predicts a finite probability of the particle passing
through the barrier and finding itself on the other
side. Thus, a particle without the necessary energy to
pass over the barrier can still penetrate through the
barrier. This phenomenon is called “Quantum mechanical
tunneling”.
1.19 Electron tunneling across a finite potential barrier.
Consider a particle with energy E incident on a
potential barrier of height U and width L as shown in
Fig. 1.19. The potential energy is zero in the regions
I and III, but is finite and equal to U in region II.
The Schrodinger’s equation for the three regions will
be
d2ψ + 2m dx2 ħ2
Eψ = 0 in region I (1.72)
d2ψ + 2m dx2 ħ2
(E – U) ψ = 0 in region II (1.73)
d2ψ + 2m dx2 ħ2
Eψ = 0 in region III (1.74)
The solutions of these equations can be written as ψI = A ei
αx + B e-iαx in region I (1.75)
ψII = C e-ϒx + D eϒx in region II (1.76)
ψIII = F eiαx + G e-iαx in region III (1.77)
where α = [(2mE)/ħ2 ]1/2 (1.65)
β = [ 2m (E-U)/ħ2 ]1/2 (1.66)
and ϒ = -iβ (1.67)
The wavefunction in the region I is made up of two
terms as evident from Equation (1.75). The first term
with a positive exponent represents an incoming or
incident wave moving in the positive x-direction and
the second term represents a wave reflected by the
barrier moving in the negative x-direction. Similarly,
the first term in equation (1.77) represents the
transmitted wave moving in region III in the positive
x-direction. The wavefunction in the region II is given
by equation (1.76). Here, the exponents are real
quantities and hence the wavefunction does not
oscillate. The probability density ψII2 is finite and
represent the probability of finding the particle
within the barrier. Such a particle may emerge into
region III. This is called tunneling.
The transmission probability T for a particle to
pass through the barrier is given by
T = ψIII2 = FF ψI2 AA*
* ≅ e-2γL (1.78)
The above equation represents the dependence of
tunneling probability on the width of the barrier and
the energy of the particle.
1.9.5 Examples of tunneling across a finite barrier: There are a few examples of tunneling across a
thin finite potential barrier in nature. These
observations are in fact proof in favour of the theory
of quantum mechanical tunneling. Let us consider a few
of them.
(a) Alpha decay: Alpha particles are made up of two
protons and two neutrons. In radioactive decay, the
alpha particle must free itself from the attractive
nuclear force and penetrate through a barrier of
repulsive coulombic potential to be emitted out of the
nucleus(Fig.1.20). A calculation of the energy of the
particle inside the nucleus and the measurement of the
energy of the emitted alpha particle indicate that it
is not possible that the particle has surmounted the
barrier of coulombic potential but must have penetrated
through it.
1.20 Emission of an α particle in nuclear decay.
(b) Ammonia inversion: In a molecule of ammonia, the
three hydrogen atoms form a plane with the nitrogen
atom placed symmetrically at a finite distance from the
plane. It has been observed experimentally that the
nitrogen atom oscillates between two positions on
either sides of the plane(Fig.1.21).
Classical calculations show that the nitrogen atom
cannot perform such oscillation since the hydrogen
atoms form a barrier against nitrogen atom to prevent
it from moving through the plane formed by the hydrogen
atoms. However, nitrogen atom oscillates across the
plane with a frequency higher than 1010 per second. This
can be explained only on the basis of tunneling
process.
1.21 Schematic diagram of ammonia molecule. Nitrogen atom oscillates between two symmetric positions across the repulsive plane of hydrogen atoms.
(c) Zener and tunnel diodes: These are diodes made out
of heavily doped semiconductors with special
characteristics. The current-voltage characteristics of
these diodes can be explained only on the basis of
quantum mechanical tunneling process. The high speed of
operation of these devices can be explained only as due
to tunneling since the movement of charge carriers is
otherwise by diffusion which is a very slow process.
The scanning tunneling microscope is another device
operating on the principle of tunneling.
(d) Frustrated total internal reflection: Figure 1.22
shows a beam of light reflected totally from the
surface of glass. If a second prism of glass is brought
close to the first, the beam appears through the second
glass prism indicating tunneling of light through the
surfaces of glass which were otherwise acting as
barriers.
1.22 Demonstration of frustrated total internal reflection.
1.9.6. Theoretical interpretation of tunneling:
Penetration of a particle into the forbidden
region of a step or a barrier can be explained
with the help of Heisenberg’s uncertainty
principle. To enter this region, the particle must
gain an energy of atleast (U-E) and to move in
this region, it must have an additional kinetic
energy, K. This is a violation of the principle of
conservation of energy for the particle. However,
according to the uncertainty relation, we may
write
ΔE. Δt ≈ ћ (1.79)
According to this, the conservation of energy does
not apply for a time duration Δt if the change in
energy is not greater than ΔE. If we presume that
the particle borrows an energy ΔE and returns the
borrowed energy within a time interval of Δt, the
observer will still believe that the energy is
conserved. The time interval within which the
extra energy must be returned is given by
Δt = ћ/ΔE = ћ/(U-E+K) (1.80)
The particle moves with a velocity v given by
v = (2K/m)1/2 (1.81)
If the particle travels a distance Δx into the
forbidden region and returns, then, the total
distance travelled is 2Δx and hence we can write
Δx = v.Δt/2
= (1.82)
As the kinetic energy K tends to zero, the value
of Δx also tends to zero since the velocity tends
to zero. Also, as K tends to infinity, Δx tends to
zero since it is the distance travelled in a time
interval Δt tending to zero. In between these
limits, there must be a maximum value of Δx
corresponding to a particular value of K.
Differentiating Δx with respect to K in equation
1.82, we can find the maximum value of Δx as
Δxmax = (1.83)
Or Δxmax = (1/2ϒ) (1.84)
From equation 1.78, the probability of finding the
particle at a distance Δxmax from the step is
T = e-2ϒΔxmax = e-1 (1.85)
Hence, we may define the maximum penetration
distance as the distance at which the transmission
probability is (1/e).
It may be mentioned that the particle is
never observed in the forbidden region. The
particles incident on the potential energy step
will be reflected back. Some are reflected at the
step itself where as others penetrate a finite
distance before returning. If the barrier width is
small, the particle will re-emerge on the other
side of the barrier. This phenomenon is known as
quantum mechanical tunneling.
1.9.7 Harmonic oscillator:
When a body vibrates about an equilibrium
position, the body is said to be executing harmonic
motion. We have many examples of such motion which we
come across, like the vibration in a spring that is
stretched and released, vibrations of atoms in a
crystal lattice, etc. Whenever a system is disturbed
from its equilibrium position, it can come back to its
original position only under the influence of a
restoring force. Hence, the presence of a restoring
force is a necessary condition for harmonic motion. The
system oscillates indefinitely if there is no loss of
energy.
A special case of harmonic motion is simple
harmonic motion. In simple harmonic motion, the
restoring force F acting on a particle of mass m is
linear. In other words, the restoring force is
proportional to the displacement x of the particle from
its equilibrium position and is in the opposite
direction.
i.e., F = -kx (1.86)
where k is called the force constant. The relation is
called Hooke’s law. From the second law of motion, we
have,
F = ma (1.87)
∴ -kx = m dt2
d2x
or d2x dt2 m
+ k.x = 0 (1.88)
This is the equation for the simple harmonic
oscillator. The solution of this equation may be
written as
x = A cos (ωt+φ) (1.89)
where ω = 2πν = (k/m)1/2 (1.90)
ν is called the frequency of the oscillator. φ is the
phase angle and depends on the value of x at t = 0. The
potential energy U corresponding to the restoring force
F may be calculated as equal to the work done in
bringing a particle from x = 0 to x = x against the
force.
i.e., U(x) = - F(x) dx = k x dx = kx2/2 (1.91)
A plot of the potential energy U as a function of
displacement x is a parabola as shown in fig. 1.23.
This indicates that an oscillator with energy E
1.23 Energy states in a one dimensional harmonic oscillator.
vibrates back and forth with an amplitude from –a to
+a. Classically, it appears that the oscillator can
have any value of energy forming a continuous spectrum.
Let us examine the quantum mechanical modification to
this classical picture.
The Schrodinger’s equation for the harmonic
oscillator with a potential energy U equal to kx2/2 may
be written as
∂2ψ +2m ∂x2 ħ2
(E – kx2/2) ψ = 0 (1.92)
This equation may be rewritten in terms of
dimensionless quantities a and y as
∂2ψ ∂y2
+ (a – y2) ψ = 0 (1.93)
where a = 2E/hν and y2 = kmx2/ħ2.
The solution to the equation 1.93 has to satisfy the
boundary condition,
ψ = 0 as y → ∞
and the normalization condition,
∞ ∫ Ψ2 dy = 1 -∞ These conditions will be satisfied when
a = (2n+1) where n = 0,1,2,3,…..
i.e., a = 2E/hν = (2n+1)
or E = (n +1/2) hν where n = 0,1,2,3,.. (1.94)
This solution leads to the following conclusions:
(i) The allowed energies will form a discrete spectrum
and not a continuous spectrum.
(ii) The least allowed energy is not zero but a finite
minimum value.
At n = 0, E0 = hν/2
This minimum energy E0 is called the zero point
energy. It is also observed that the higher energy
levels are all equally spaced with a spacing of hν.
This is in contrast to the result obtained for the case
of a particle in a potential well of infinite depth.
1.9.8 Practical applications of Schrodinger’s wave
equation:
The real life situations are much different from
the one considered while deriving the Schrodinger’s
wave equation. This is especially true when one is
analyzing the motion of a particle like electron
traveling at velocities comparable to that of light.
Relativistic modification to the Schrodinger’s equation
and its solution are complex. Further, the boundary
condition of an infinitely high potential barrier is
never encountered. In case of metals, conduction
electrons move in crystal lattice under the influence
of finite potentials of the ion cores. The potential
energy due to the influence of external forces acting
on it may also be functions of position of the particle
and time. Incorporation of these factors while
formulating and solving the Schrodinger’s wave equation
has led to accurate prediction of the behaviour of
subatomic, atomic, molecular and other microscopic
systems.
NUMERICAL EXAMPLES:
1.1 Calculate the velocity of photoelectrons emitted
from a metal surface of work function 1.5 eV when the
metal surface is irradiated with a light beam of
wavelength 4 x 10-7 m.
Solution:
Incident energy = hν = hc = 6.62x10-34x3x108
λ 4 x 10-7
= 4.97x10-19 J
Threshold energy = hνo = Φ = 1.5eV = 1.5 x 1.6 x 10-19 J
= 2.4 x 10-19 J
Kinetic energy of electrons = (hν - hν0 )
= (4.97-2.40) x 10-19
= 2.57 x 10-19 J.
Velocity of Photoelectrons = [2(hν - hν0)/m]1/2
= 7.52 x 105 ms-1 (Ans.)
1.2 In a photoelectric effect experiment, a stopping
potential of 4.6 V was required to stop photoelectron
emission with an incident light of frequency 2x 1015 Hz
and a stopping potential of 12.9 V when the incident
light had a frequency 4 x 1015 Hz. Evaluate the Planck’s
constant.
Solution:
If V1 and V2 represent the stopping potentials
corresponding to incident frequencies ν1 and ν2, then
eV1 = hν1 - Φ
eV2 = hν2 - Φ
h = e(V2-V1) = (ν2 - ν1) (4 x 1015 – 2x1015)
1.6 x 10-19 (12.9 4.6)
h = 6.64 x 10-34 Js. (Ans.)
1.3 Calculate the maximum change in wavelength that
can take place during Compton scattering of a photon.
Solution:
Change in wavelength = ∆λ = __h moc
(1-cos θ)
This will be maximum when Cos θ =-1,i.e.,when θ = 180°.
∴ (∆λ)max = _2h = __ moc 9.1 x 10-31 x 3 x 108
2x6.62 x 10-34___
(∆λ)max = 4.85 x 10-12m (Ans.)
1.4 The material of the emitter of a photocell has a
work function of 2eV. Calculate the threshold
frequency.
Solution:
Work function Φ = hν0 = 2 x 1.6 x 10-19 J
Threshold frequency ν0 = 2x1.6x10-19/6.62x10-34
= 4.83 x 1014 Hz.
1.5 The threshold frequency for the material of the
emitter of a Photocell is 4 x 1014 Hz. What is the
stopping potential required to supress photo electrons
emission when light of frequency 6x 1014 Hz is incident
on the emitter?
Solution:
Stopping potential V0 = h(ν-ν0)/e
= 6.62x10-34(6x1014-4x1014)/1.6x10-19
= 0.829 V (Ans).
1.6 In a photocell, a stopping potential of 2.5 V is
required to stop the photo electron emission
completely. Calculate the kinetic energy of the
emitted photo electrons.
Solution:
Kinetic energy = potential energy = e.V
= 1.6 x 10-19 x 2.5 J = 2.5 eV (Ans).
1.7 In a photocell illuminated by light of frequency
5x1014Hz,a reverse potential of 2V is required to stop
the photo electron emission. Find the work function of
the material of the emitter.
Solution:
Work function Φ = h(ν-ν0) = (hν-eV)
=6.62x10-34x5x1014-1.6x10-19x2 = 0.072 eV (Ans).
1.8 X-rays of wavelength 1.54 A° are Compton –
scattered at an angle of 60°. Calculate the change in
the wavelength.
Solution: Change in wavelength = ∆λ = h moc
(1-cos θ)
= h moc
(1-cos 600) = 1.2 x 10-12m (Ans).
1.9 In a Compton scattering experiment, incident
photons of energy 10 KeV are scattered at 45° to the
incident beam. Calculate the energy of the scattered
photon.
Solution: Change in wavelength = ∆λ = h
moc (1-cos θ)
= 7.1 x 10-13m.
Wavelength of incident photon = λ = hc/eE
= 1.243 x 10-10m.
Wavelength of scattered photon =λ’= λ + ∆λ
= 1.25 x 10-10m.
Energy of scattered photon = hc/λ’
= 1.59 x 10-15 J
= 9.93 keV (Ans).
1.10 Gamma Rays of energy 0.5 MeV are scattered by
electrons. What is the energy of scattered gamma rays
at a scattering angle of 30°? What is the kinetic
energy of scattered electron?
Solution:
Wavelength of incident gamma rays = λ = hc/E
= 6.62x10-34x3x108/1.6x10-19x0.5x106
= 2.486 x 10-12m.
Change in wavelength = ∆λ = h moc
(1-cos θ)
= 3.24 x 10-13m.
Wavelength of scattered photon =λ’= λ + ∆λ
2.81 x 10-12m.
Kinetic energy of the scattered electron = hc/λ’
= 0.442 MeV (Ans).
1.11 X-rays of wavelength 1.5 A° are Compton
scattered. At what angle will be scattered x-rays have
a wavelength of 1.506 A°?
Solution: Change in wavelength = ∆λ = h
moc (1-cos θ)
cos θ = (1 – m0c. ∆λ/h) = (1 – 0.247) = 0.753
Angle of scattering,θ = 41.20 (Ans).
1.12 Calculate the de Broglie wavelength associated
with an electron travelling with a velocity of 105 ms-1.
Assume the mass of the electron to be 9.1 x 10-31kg. and
h = 6.62x10-34Js.
Solution:
De Broglie wavelength λ = h = P 9.1 x 10-31x105
6.62 x 10-34____
λ = 7.27 x 10-9 m. (Ans.) 1.13 In an electron diffraction apparatus, the
electron beam is accelerated to a potential of 25 kV.
Calculate the de Broglie wavelength associated with the
electrons. Given:m =9.1 x 10-31kg,h=6.62 x 10-34 JS.
e = 1.6 x 10-19 C.
Solution: De Broglie wavelength λ = __h___ = ___h___ (2mE)1/2 (2meV)1/2
λ = _____6.62 x 10-34
( 2x 9.1 x 10-31 x 1.6 x 10-19 x 25 x 103) _______________________
λ = 7.6 x 10-12 m. (Ans.) 1.14 Calculate the phase velocity and group velocity
associated with an electron assuming (i) non-
relativistic case and (ii) relativistic case.
Solution:
Non – relativistic case.
Phase velocity vp = ω = (E/ħ) = E = (p2/2m) = p = K (p/ħ) p p 2m 2
v
Phase velocity is half the particle velocity. Group velocity vg = dω = dE = d(p2/2m) = p_ dk dp dp m
= v
Group velocity is equal to the particle velocity.
Relativistic case:
Phase velocity vp = E p
where E = (p2c2 + mo2c4)1/2
∴ vp = c [ 1+ mo2c2
p2 ]1/2
Phase velocity is greater than c. This indicates that
we cannot talk of phase velocity of a particle since it
is represented by a group of waves or a wave packet.
Group velocity vg = dE = _d_ dp dp
[p2c2+moc4]1/2
= pc2 {p2c2 + mo2c4]-1/2 = E
pc2
= mvc2 mc2
= v assuming E = mc2
Group velocity is equal to the particle velocity.
1.15 An electron is trapped in a one dimensional
potential well of infinite depth and a width of
1x10-10 m. What is the probability of finding the
electron in the region from x = 0.09 x 10-10 m to
x = 0.11 x 10-10 m in the ground state.
Solution:
The probability of finding an electron in the region
between x1 and x2 is given by
Method I:
x2 x2 Pn = ∫ψ*ψ dx = (2/L) ∫ sin2(2nπx/L)dx x1 x1 x2
= [x/L – (1/2nπ).sin(2nπx/L)] x1 For ground state, n = 1. x2
P1 = [x/L – (1/2π).sin(2πx/L)] x1 = [0.11-(1/2x3.14)sin(2x180x0.11/1.0)]
-[0.09-(1/2x3.14)sin(2x180x0.09/1.0)]
= (0.11 – 0.10150) – (0.09 – 0.08532)
= 0.00382 (Ans).
The probability of finding an electron in the region of
width ∆x around x is given by
Method II:
Pn = |ψn|2. ∆x = (2/L) sin2(2nπx/L). ∆x
Here, we take n = 1, x = 1x10-10m and ∆x = 0.02x10-10m.
P1 = (2/1x10-10) sin2(1x180x0.1). 0.02x10-10
= 0.00382 (Ans).
Note: The second method is only approximate and gives
result close to the one obtained by method I only when
∆x is very small.
1.16 An electron is trapped in a one dimensional
potential well of width 1x10-10 m and infinite height.
Find the amount of energy required to excite the
electron to its first excited state. What is the
probability of finding the electron in its first
excited state between x =0.4x10-10 m and x =0.6x10-10 m?
Solution:
Energy of the electron in the nth excited state is given by En = 8mL2
n2h2
Energy required to take the electron from ground state
(n = 1) to the first excited state (n = 2) is given by
E = E2 – E1 = h2__(22-12) = 8mL2 8mL2
3h2__
= 8 x 9.1 x 10-31 x (10-10)2
3 x (6.62 x 10-34)2_______
= 1.81 x 10-17 J = 112.87 eV. (Ans.)
The probability of finding the electron in the first
excited state between x = 0.4 A° and x = 0.6 A° is
given by
x2 x2 ∫ψ*ψ dx = (2/L) ∫ sin2(2πx/L)dx x1 x1 x2
= [x/L – (1/4π).sin(4πx/L)] x1
= (0.6 – 0.076) –(0.4 + 0.076)
= 0.048 (Ans)
1.17 The velocity of an electron is measured to be
3x106 ms-1 in a particular direction. If the velocity
is measured with a precision of 1%, what is the
accuracy with which its position can be measured
simultaneously?
Solution:
Momentum of the electron (in non – relativistic
calculation) is
Px = mvx = 9.1 x 10-31 x 3 x 106 = 2.73x10-24kg.m.s-1
Uncertainty in momentum = ∆px = 1% of px
= 2.73x10-26 kg.ms-1
∴ Uncertainty in position, ∆x ≈ ∆px
ħ
= 3.86 x 10-9 m (Ans.)
1.18 In an experimental study of nuclear decay, the
emitted energy spectrum shows a peak with a spread of
energy equal to 120 MeV. Compute the life time of the
decaying nuclei.
Solution: Life time of the decaying nuclei = ∆t ≈ _ ∆E
ħ_
∆t =____6.62 x 10-34
120 x 106 x 1.6 x 10-19 x 2 x 3.14 ___________________
= 5.5 x 10-24 s. (Ans.) 1.19 Calculate the de Broglie wavelength associated
with the following:
(a) A car of mass 1000 kg moving with a velocity
of 50 ms-1
(b) A tennis ball of mass 50 gms moving at a speed
of 75 ms-1.
(c) An air molecule of average mass 2x10-26kg
moving with a velociity of 300 ms-1.
(d) An electron with an energy of 1 MeV.
Solution:
The deBroglie wavelength can be calculated as
λ = h/p
(a) λ = 6.62 x 10-34/1000x50 = 1.324 x 10-38m.This value
is too low to be observable.
(b) λ = 6.62 x 10-34/50 x 10-3x75 = 1.765 x 10-34m.
(c) λ = 6.62 x 10-34/2 x 10-26 x 300 = 1.103 x 10-10m.
(d) λ = h/(2meV)1/2
= 6.62 x 10-34/(2x9.1x10-31x1.6x10-19x106)1/2
= 1.23 x 10-12m (Ans).
1.20 Compute the accelerating potential required to
produce an electron beam of de Broglie wavelength
0.1 A°.
Solution:
V = h2/2meλ2
= (6.62x10-34)2/2x1.6x10-19x(0.1x10-10)2
= 15.1 kV (Ans).
1.21 Calculate the deBroglie wavelength associated
with a thermal neutron at 270 C.
Kinetic energy of thermal neutrons = (3/2)kT
∴λ = h/(3mkT)1/2
= 6.62x10-34/(3x1.67x10-27x1.38x10-23x300)1/2
= 1.45 x 10-10m (Ans).
1.22 A cricket ball of mass 250 gms moves with a
velocity of 100 ms-1. If its velocity is measured with
an accuracy of 1%, what is the accuracy of a
simultaneous measurement of its position?
Solution:
∆x . ∆px > ħ/2 or ∆x . ∆px ≅ ħ
∆px = 1% of 250x10-3x100 = 250x10-3 kgms-1
∆x = ħ/∆px = 6.62 x10-34/250x10-3 = 2.65 x 10-33m(Ans).
1.23 In a gamma decay process, the life time of
decaying nuclei is found to be 2 ns. Compute the
uncertainty in the energy of gamma rays emitted.
Solution:
∆E . ∆t ≥ ħ/2 or ∆E . ∆t ≅ ħ
∆E = ħ/∆t = 6.62 x 10-34/2 x 3.14 x 2 x 10-9
= 5.27 x 10-26 J (Ans).
EXERCISE
1.1 Explain photoelectric effect. Give the assumptions
needed, and the laws of photoelectric emission.
Calculate the velocity of the photoelectrons emitted
from a metal surface whose work function is 1.5 eV and
the wavelength of the incident light being 4x10-7m.
(March 99)
1.2 Describe the ultraviolet catastrophe. Explain how
the Planck’s law of radiation overcomes it. (March 99)
1.3 Explain Wien’s law and Rayleigh-Jean’s law.
Mention their drawbacks. (August 99)
1.4 Explain Einstein’s theory of photo-electric
effect. (August 99)
1.5 What is Planck’s radiation law? Show how Wien’s
law and Rayleigh-Jean’s law can be derived from it.
(August 99)
1.6 The work function of tungsten is 5.4 eV. When
light of wavelength 170 nm is incident on the surface,
electrons with energy 1.7 eV are ejected. Estimate the
Planck’s constant. (August 99)
1.7 What is the nature of black body radiation?
Explain its significance. (March 2000)
1.8 In Compton scattering, calculate the maximum
kinetic energy of the scattered electron for a given
photon energy. A metal surface has a photoelectric cut
off wavelength of 325.6 nm. What is the stopping
potential for an incident light of wavelength of 259.8
nm? (March 2000)
1.9 State Compton effect and explain the experimental
observations of Compton shift. (August 2000)
1.10 The stopping potential of 5.2 V was observed for
light of frequency 3 x 1015 Hz. What would be the
frequency of light when the stopping potential is
doubled? (August 2000)
1.11 State and explain Planck’s law of radiation. Show
that it reduces to Wien’s law and Rayleigh-Jean’s law
under certain conditions. (August 2000)
1.12 The photoelectric threshold for a certain metal is
5000 AU. Determine the maximum energy of the
photoelectrons emitted when a radiation of wavelength
3000 AU is incident on its surface. (March 01)
1.13 Explain Planck’s radiation law. Discuss Einstein’s
theory of photoelectric effect. (March 01)
1.14 Explain Wien’s Displacement law. Mention its
drawbacks. (August 01)
1.15 Explain Compton effect and give its physical
significance. (August 01)
1.16 What is ultraviolet catastrophe? Show how it can
be overcome with the help of Planck’s radiation law.
(August 01)
1.17 What is photoelectric effect? Mention the features
of photoelectric emission. (March 02)
1.18 What is Compton scattering? With necessary
equations, explain Compton scattering effect.(March 02)
1.19 Calculate the change in wavelength in Compton
scattering at an angle of 600 to the incident
direction. (March02)
1.20 Calculate the wavelength associated with an
electron carrying an energy 2000 eV. (March 99)
1.21 Set up one dimensional Schrodinger’s wave equation
for a free particle. (March 99)
1.22 Explain group velocity and phase velocity. Derive
a relation between the two. (August 99)
1.23 Explain Heisenberg’s uncertainty principle. Give
its physical significance. (August 99)
1.24 What is de Broglie’s concept of matter waves? An
electron has a wavelength of 1.66 x 10-10m. Find the
kinetic energy, phase velocity and group velocityof the
de Broglie wave. (August 99)
1.25 Explain Heisenberg’s uncertainty principle.
Give its physical significance. The position of an
electron can be measured with an accuracy of 1.5 x 10-
10m. Find the uncertainty in its position after 1 sec.
(August 2001)
1.26 What are Eigen functions and Eigen values?
Find them for a particle in one dimensional potential
well of infinite height. (August
2001)
1.27 Obtain the time independent Schrodinger’s wave
equation for a particle in one dimensional potential
well of infinite height and discuss the solution.
(August 2000)
1.28 Write a note on group velocity and phase
velocity. (March 2001)
1.29 Explain group velocity and phase velocity.
Derive relation between the two.
(August 2001)
1.30 What is a wave function? Give its physical
significance. What is normalization of a wave function?
(August 2001)
1.31 What is de Broglie concept of matter waves?
Derive an expression for de Broglie wavelength.
(August 2001)
1.32 Set up time independent Schrodinger’s wave
equation and explain Eigen functions and Eigen values.
(August 99)
1.33 What is Heisenberg’s uncertainty principle?
Discuss its consequences. (March 2000)
1.34 Explain the behaviour of a particle in a one-
dimensional infinite potential well in terms of de
Broglie waves. (March 2000)
1.35 What is the physical interpretation of wave
function, nature of Eigen values and Eigen functions.
(March 2000)
1.36 Discuss phase velocity and group velocity. (August
2000)
1.37 Explain only the conclusions drawn in
photoelectric effect and Davisson-Germer experiment
leading to deBroglie hypothesis. (Feb 2005)
1.38 Show that a free electron cannot exist in a
nucleus of an atom. (Feb 2005)
1.39 What are matter waves? Show that the electron
accelerated by a potential difference V volt is =
1.226/ √V nm for non-relativistic case.(July 2005)
1.40 Explain phase velocity and group velocity. Derive
the expression for deBroglie wavelength using the
concept of group velocity.(July 2005)
1.41 Explain only the conclusions drawn in
photoelectric effect and Davisson-Germer experiment
leading to deBroglie hypothesis. (Feb 2005)