Download elements of quantum mechanics

1
1.1
ELEMENTS OF
QUANTUM
MECHANICS
INTRODUC TION
A revolution took place in physics between 1900 and 1930. This was the era
of a new and more general scheme called quantum mechanics. This new
approach was highly successful in explaining the behaviour of atoms, molecules and nuclei. Moreover, the quantum theory raduces to classical physics
when applied to macroscopic systems.
The basic ideas of quantum theory were first introduced by Max Planck,
but most of the subsequent mathematical developments and interpretations
were made by a number of distinguished physicists including Einstein, Bohr,
Schrödinger, de Broglie, Heisenberg, Born and Dirac.
This chapter is simply an introduction to the underlying ideas of quantum theory and the wave-particle nature of matter. We shall also discuss
some simple applications of quantum theory.
1.2
WA
VE-PARTICLE DU
ALIT
Y
WAVE-PARTICLE
DUALIT
ALITY
It is well recognised that the same light beam that can eject photoelectrons
from a metallic surface can also be diffracted by a grating. In other words, the
photon concept and the wave theory of light complement each other. Thus,
all forms of electromagnetic radiation can be described from two points of
view. In one extreme electromagnetic waves describe interference and diffraction pattern formed by a large number of photons while in other extreme,
the photon description is natural when we deal with highly energetic photon
of very short wavelength.
Light and other electromagnetic radiation sometimes act like waves and
sometimes like particles. Interference and diffraction demonstrate wave
behaviour while emission and absorption of photons demonstrate the particle behaviour. Besides waves that sometimes acts like particles, quantum
mechanics extend the concept of wave-particle duality to include particles
that sometimes show wavelike behaviour. In these situations, a particle is
modelled as an inherently spread-out entity that cannot be described as a
point with a perfectly definite position and velocity.
1
2
LASER SYSTEMS AND APPLICATIONS
1.3
de BR
OGLIE W
AVES
BROGLIE
WA
de Broglie postulated that a free particle with mass m moving with speed v
should have a wavelength λ related to its momentum exactly the same way
as for a photon. A photon of light of frequency ν has the momentum.
h
.
p =
c
Since λν = c so, the wavelength of a photon is specified by its momentum
according to the relation
h
.
λ =
...(1.1)
p
de Broglie suggested that eqn. (1.1) is a completely general one that applies
to material particles as well as to photons. The momentum of a particle of
mass m and velocity v is p = mv and its de Broglie wavelength is accordingly.
h
.
mv
In eqn. (1.2) m is the relativistic mass
λ =
m =
m0
1  v2 c2
...(1.2)
.
The wave and particle aspects of moving bodies can never be observed
at the same time. In certain situations a moving body exhibits wave properties and in others it exhibits particle properties. Which set of properties is
most conspicuous depends upon how its de Broglie wavelength compares
with its dimensions and the dimensions of whatever it interacts with.
de Broglie Wave Velocity
If we call the de Broglie wave velocity as W, we may write
W = νλ.
Where λ is the de Broglie wavelength and ν is the frequency. To find the
frequency, we equate the quantum expression E = hν with the relativistic
formula for total energy E = mc2
hν = mc2
mc 2
.
h
The de Broglie wave velocity is therefore
ν =
 mc2   h  c 2
W = νλ = 
 .

 h   mv  v
Since particle velocity v must be less than the velocity of light c, de
Broglie waves always travel faster then light. In order to understand this
unexpected result, we must understand distinction between phase velocity
and group velocity. Phase velocity is the velocity of wave.
ELEMENTS OF QUANTUM MECHANICS
1.4
3
WA
VE EQU
ATION
WAVE
EQUA
The equation of the wave travelling in the direction of increasing x and whose
vibrations are in the y direction can be given as
x 
y = A cos 2π ν  t 

W

where A is the amplitude of vibrations (that is, their maximum displacement
on either side of the x-axis) and ν their frequency. The wave has some speed
W and travels the distance x in time t.
Since the wave speed W is given by
W = νλ.
We have wave equation
x
y = A cos 2π  t   .
...(1.3)


Equation (1.3) is more convenient to use. The most widely used description of a wave, however, is still another form with angular frequency and
wave number. Angular frequency ω and wave number k can be defined by
the formulas
2 

.

W
The unit of ω is the radian per second and that of k is the radian per
meter. In terms of ω and k eqn. (1.3) becomes
y = A cos (ωt – kx).
...(1.4)
ω = 2πν, k =
1.5
PHASE AND GR
OUP VEL
OCITIES
GROUP
VELOCITIES
The wave representation of a moving particle corresponds to a wave packet
or wave group as shown in Fig. 1.1.
Mathematically a wave group can
be described in terms of the superposition of individual waves of different
wavelengths whose interference with
one another results in the variation in
amplitude. If the velocities of the
Fig. 1.1. Wave group
waves are the same, the velocity with
which the wave group travels is the common wave velocity. If the wave
velocity varies with wavelength, the different individual waves do not proceed together and the wave group has a velocity different from those of the
waves that compose it.
Let us suppose that a wave group arises from the combination of the
following two waves that have the same amplitude A but differ in angular
frequency and wave number.
4
LASER SYSTEMS AND APPLICATIONS
y1 = A cos (ωt – kx)
y2 = A cos [(ω + dω)t – (k + dk)x].
The resultant displacement y at any time t and any position x is the sum
of y1 and y2.
y = y1 + y2 = 2A cos
cos
1
[(2ω + dω)t – (2k + dk)x]
2
1
(dωt – dkx).
2
Since dω and dk are small compared with ω and k respectively.
2ω + dω ≈ 2ω
2k + dk ≈ 2k
d
dk 
t
x
...(1.5)
y = 2A cos (ωt – kx) cos 
2 
 2
Equation (1.5) represents a wave of angular frequency ω and wave number k which has superimposed upon it a wave (the process is known as
d 
 dk 
modulation) of angular frequency 
 and wave number   . The effect
 2 
 2 
of this modulation is to produce successive ‘wave groups’.
The ‘phase velocity’ or wave velocity W is given by

k
while the group velocity u of the ‘group of waves’ is given by
W = νλ =
2
 2  

d
.
dk
The group velocity may be greater or less than the phase velocity. It
depends on the manner in which phase velocity varies with wave number.
For light waves in vacuum, the group and phase velocities are the same.
The angular frequency and wave number of the de Broglie waves associated with a particle of rest mass m0 moving with the velocity v are
Angular frequency of de Broglie waves
u =
2 m0 c 2
2 mc 2

.
h
h 1  v2 c2
Wave number of de Broglie waves
2 m0 v
2 2 mv


.
k =

h
h 1  v2 c2
ω = 2πν =
de Broglie phase velocity
 c2

k
v
which exceeds both the velocity of the particle v and the velocity of light c,
since v < c.
W =
ELEMENTS OF QUANTUM MECHANICS
5
The group velocity u of the de Broglie waves associated with the particle
is
u =
d d dv

dk
dk dv
2  m0 v
d
=
dk
h(1  v 2 c 2 )3/2
2 m0
dk
=
dv
h(1  v 2 c 2 )3/2
and
so
u = v.
The de Broglie wave group associated with a moving particle travels
with the same velocity as the particle.
The wave velocity W of de Broglie waves has no simple physical significance.
The wave velocity is a function of wavelength even in free space and
here it differs from the velocity of light which may be shown as under
p = mv and total energy E = mc2
E
v and now using the relativistic expression for the mass of a
c2
particle, the equation for its momentum can be given as
So, p =
m0
p =
v
v2
1 2
c
where m0 is the rest mass of the particle.
Eliminating v between the last two equation, we get
2
E2 p

c4
c2
h
Since we known that p =
and E = hν.

m02 =
So,
m0 =
h 2
1

c c2  2
For de Broglie wave W =
Hence
m0 =
νλ.
h W2
1

c c 2 2  2
W = c 1
m02c2
2
h2
This equation shows that for a particle of rest mass m0 > 0, the wave
velocity W is always greater than c and also the wave velocity of de Broglie
waves is a function of λ even in free space.
From which
6
LASER SYSTEMS AND APPLICATIONS
Now consider electromagnetic wave as a special case of de Broglie wave
and so, these de Broglie waves are propagated with a velocity W = c. The
velocity of the associated particle i.e., photon is also equal to c. If W = c is
substituted in the last equation given above, we find that rest mass of the
photon m0 = 0 i.e., there is no such thing as a photon at rest, photon always
moves with the velocity c.
1.6
DA
VISSON AND GERMER EXPERIMENT
DAVISSON
In 1927, Davisson and Germer predicted
experimentally the electron waves predicted by de Broglie. Davisson and
Germer were studying the scattering of
electrons from a nickel target using an
apparatus like that sketched in Fig. 1.2.
The energy of the electrons in the
primary beam, the angle at which they
are incident upon the target and the
position of the detector could all be varied. The nickel target was subjected to
such a high temperature treatment that
the crystal was transformed into a group
Fig. 1.2
of crystals. In this case the reflection became anomalous and the reflected intensity showed striking maxima and minima
instead of a continuous variation of scattered electron intensity with angle.
The position of the maxima and minima observed depended upon the
electron energy. Then, they suspected that the beam of electron might be
diffracted from the crystals like X-rays. This shows that electrons behave like
waves under certain circumstances. Typical polar graphs of electron intensity
after the heat treatment are shown in Fig. 1.3.
Fig. 1.3
ELEMENTS OF QUANTUM MECHANICS
7
The method of plotting is such that the intensity at any angle is proportional to the distance of the curve at that angle from the point of scattering.
If the intensity were the same at all scattering angles, the curves would be
circles centered on the point of scattering.
de Broglie hypothesis suggested the interpretation that the electron waves
were being diffracted by the target, much as X-rays are diffracted by the
planes of atoms in a crystal. This idea was confirmed when it was realised
that the effect of heating a block of nickel at high temperature is to cause
many small individual crystals of which it is normally composed to form into
a single large crystal all of whose atoms are arranged in a regular lattice.
To verify whether de Broglie waves are responsible for the findings of
Davisson and Germer, an analysis of the observation should be made.
For the beam of electrons falling normally
on the surface of the crystal, the current observed in detector is a measure of the intensity of the diffracted beam. Several curves
were obtained for different voltage electrons
when graphs were plotted between the colatitude (angle between the incident beam and
the beam entering the detector) which are
shown in Fig. 1.4.
It is observed that a bump begins to appear in the curve at 44 volt electrons. This
bump moves upward for 54 volts at colatiFig. 1.4
tude of 50°. Above 54 volts the bump again
diminishes. The bump at 54 volts offers the evidence for the existence of
electron waves.
The angles of incidence and scattering relative to the family of Bragg
plane shown in Fig. 1.4 are both 65°. The spacing of the planes in this family,
which can be measured by X-ray diffraction is 0.091 nm. The Bragg equation
for maxima in the diffraction pattern is
nλ = 2d sin θ.
Here d = 0.091 nm, θ = 65°. For n = 1, the de Broglie wavelength λ of the
diffracted electrons is
λ = (2) (0.091) (sin 65°) = 0.165 nm,
We use de Broglie formula to calculate the expected wavelength of the
electrons. The electron kinetic energy of 54 eV is small compared with its rest
energy m0c2 of 0.51 MeV so we can ignore relativistic considerations.
Since
K =
1
mv 2 .
2
8
LASER SYSTEMS AND APPLICATIONS
The electron momentum mv is
mv =
2mk  (2) (9.1  10 31 ) (54) (1.6  10 19 )
= 4.0 × 10–24 kg m/s.
The electron wavelength is therefore
λ =
6.63  1034
h

= 1.66 × 10–10 m = 0.166 nm
mv
4.0  10 24
is in excellent agreement with the observed wavelength. The Davisson-Germer
experiment thus provides direct verification of de Broglie’s hypothesis of the
wave nature of moving bodies.
SOL
VED EXAMPLES
SOLVED
EXAMPLE 1: Calculate the de Broglie wavelength associated with a proton mov1
th of the velocity of light.
20
SOLUTION: Velocity of proton
ing with a velocity equal to
3  10 8
= 1.5 × 107 m/s
20
Mass of the protonm = 1.67 × 10–27 kg
v =
λ =
6.6  10 34
h

= 2.634 × 10–14 m.
mv 1.67  10 27  1.5  107
EXAMPLE 2: Calculate the de Broglie wavelength of neutron of energy 12.8 MeV.
Given h = 6.62 × 10–34 J.sec, m = 1.67 × 10–27 kg.
SOLUTION:
m0c2 =
1.507  1010
= 941.87 MeV
1.6  1019
Since 12.8 MeV is very small compared to rest mass energy hence relativistic consideration may be ignored
λ =
λ =
h

mv
h
E = Ve where V is voltage in volts
2mE
6.62  10 34
2  1.67  10 27  12.8  106  1.6  1019
= 8.0 × 10–5 Å.
9
ELEMENTS OF QUANTUM MECHANICS
EXAMPLE 3: Show that the de Broglie wavelength for a material particle of rest
mass m0 and charge q accelerated from rest through a potential difference of V volts
relativistically is given by
h
λ =

qV
2m0 qV  1 

2m0 c 2




.
SOLUTION: Kinetic energy Ek = Vq.
Ek ≠ mv2 because v is relativistic velocity and so, we cannot find momentum directly from Ek. Now, we use relativistic formula
E2 = p2c2 + m02c4
E = Ek + m0c2 = Vq + m0c2
P2c2 = E2 – m02c4 = (Vq + m0c2)2 – m02c4
P2c2 = V2q2 + 2Vq m0c2
P2 =

V 2 q2
Vq
 2Vq m0  2m0 Vq  1 
2

c
2m0 c 2


Vq
2m0 Vq  1 

2m0 c2

P =
∴ de Broglie wavelength λ =






1

Vq
2m0 Vq  1 

2m0 c 2




EXAMPLE 4: Calculate the wavelength associated with an electron accelerated to
a potential difference of 1.25 keV.
SOLUTION: If E is the K.E. of the electron, the de Broglie wavelength of the
wave associated with the electron is
λ =
h

2mE
6.6  10 34
2  9.1  10 31  1.25  10 3  1.6  10 19
= 3.46 × 10–11.
EXAMPLE 5: What will be the kinetic energy of an electron if its de Broglie
wavelength equals the wavelength of the yellow line of sodium (5896 Å). The rest
mass of electron is m0 = 9.1 × 10–31 kg and h = 6.63 × 10–34 J. sec.
SOLUTION: de Broglie wavelength
λ =
h
mv
or
v=
h
.
m
10
LASER SYSTEMS AND APPLICATIONS
In the absence of relativistic consideration m = m0
λ =
h
1
, Kinetic energy K = m0 v 2
m0 v
2
K =
 h 
1
h2
.
m0 


2
2m0  2
 m0  
2
Putting the values of h, m0 and λ
K =
(6.63  10 34 )2
6.95  10 25

2  9.1  1031  (5896  10 10 )2
1.6  1019
= 4.34 × 10–6 eV.
EXAMPLE 6: A particle of rest mass m0 has a kinetic energy K. Show that its de
Broglie wavelength is given by
hc
.
λ =
K (K  2m0 c 2 )
Hence, calculate the wavelength of an electron of kinetic energy 1 MeV. What
will be the value of λ if K << m0c2?
SOLUTION: According to de Broglie’s concept of matter wave, the wavelength
λ =
1/2

v2 
1  2 
c 

1
m02
m2
=
m0
m
=
v2
c2
v2
=
c2
or
m2v2 =
or
mv =
Substituting this value
or
λ =
or
h
mv
λ =
m=
m0
(1  v 2 c 2 )1/2
m2  m02
m2
(m – m02)c2
c(m2 – m02)1/2.
of mv in equation of wavelength, we get
2
h
c m  m02
2
hc
c
2
m2  m02
c2(m2 – m02)1/2 = [c4(m – m0) (m + m0)]1/2
= [c2(m – m0) {(m + m0)c2}]1/2
= [(m – m0)c2 {(m – m0)c2 + 2m0c2}]1/2
or
c2(m2 – m02)1/2 = [K(K + 2m0c2)]1/2.
11
ELEMENTS OF QUANTUM MECHANICS
λ =
Therefore
hc
.
[K (K  2m0 c 2 ]1/2
For an electron m0c2 = 9.1 × 10–31 × (3 × 108)2 joule
=
For K = 1 MeV,
81.9  1015
eV = 0.51 × 106 eV = 0.51 MeV
1.6  1019
λ =
hc

1(1  (2  0.51))
hc
2.02
λ =
6.62  10 34  3  108
m
2.02  1.6  10 19  106
= 8.78 × 10–13 m = 8.78 × 10–3 Å.
If K << m0c then K + 2m0c2 = 2m0c2
2
λ =
hc
2m0 Kc
2
h
.
2m0 K

EXAMPLE 7: What is the de Broglie wavelengths of any electron which has been
accelerated from rest through a potential difference of 100 V.
SOLUTION:
∴
12.25
Å
V
V = 100 volts
λ =
λ =
12.25
= 1.225 Å.
100
EXAMPLE 8: Can a photon and an electron of the same momentum have the same
wavelength? Compare their wavelengths if the two have the same energy.
S OLUTION : Using de Broglie concept of matter wave, momentum
of the electron may be written as pe =
as pph =
h
h

.
c
p
h
and momentum of photon
e
So, if the electron and photon have same momentum, then we have λe = λp.
Thus, the photon and electron of the same momentum have the same
wavelength.
The de Broglie wavelength of electron is given by
h
and
mv
where E is the energy of the electron
λe =
So,
λe =
1
2mE
1
mv 2 = E or mv =
2
2mE
12
LASER SYSTEMS AND APPLICATIONS
The de Broglie wavelength of photon is given by
h
hc
but E = hν =
= pc
p

hc
=
E
λph =
λph
 ph
 2mc 2 
hc 2mE
2m

c
 
.
E
h
E
e
 E 
EXAMPLE 9: Find the energy of the neutron in units of electron volt whose de
Broglie wavelength is 1 Å.
Now
=
SOLUTION: Given mass of the neutron = 1.674 × 10–27 kg
Planck’s constant h = 6.60 × 10–34 joule/sec
We know that
or
where
λ =
h

mv
h
2mE
h2
2m 2
m = 1.674 × 10–27 kg
λ = 1 Å = 10–10 m
h = 6.60 × 10–34 J.s.
E =
E =
=
(6.60  1034 )2
2  1.674  1027  (10 10 )2
= 13.01 × 10–21 joules
13.01  10 21
= 8.13 × 10–2 eV.
1.6  10 19
EXAMPLE 10(a): What would be the wavelength of quantum of radiant energy
emitted, if an electron transmitted into radiation and converted into one quantum?
SOLUTION: When the energy of an electron is transmitted into radiation,
we use the following relations to get the value of λ
E = mc2 and
So
λ =
E = hν =
hc

6.6  10 34
h

mc 9.1  10 31  3  108
= 0.0242 × 10–10 m = 0.0242 Å.
EXAMPLE 10(b): A lamp of 150 W is emitting 8% of its energy as blue light
having a mean wavelength of 4500 Å. How many photons are being emitted by the
lamp per second.
SOLUTION: Power emitted by the lamp = 150  8 W  12 W
100
The energy emitted per second = 12 J
13
ELEMENTS OF QUANTUM MECHANICS
Energy contained in one photon
6.625  10 34  3  108
4500  1010
–20
= 44.2 × 10 J
= hc/λ =
No. of photons emitted per second =
1.7
12
 27.15  1018.
44.2  10 20
UNCERT
AINT
Y PRINCIPLE
UNCERTAINT
AINTY
It is impossible to know both the exact position and exact momentum of an
object at the same time.
To regard a moving particle as a wave group implies that there are some
fundamental limits to the accuracy with which we can measure position and
momentum of a particle.
Fig. 1.5
The narrower the wave group of the particle, its position can be precisely
determined (Fig. 1.5a).
The wavelength and hence particle’s momentum cannot be established
h
because λ =
and it is not well defined in a narrow packet.
mv
On the other hand, a wide wave group such as in Fig. 1.5b, the wavelength can be precisely determined but not the position of the particle.
The relationship between the distance Δx and the wave number spread
Δk depends upon the shape of the wave group and upon how Δx and Δk are
defined. If ψ(x) representes a wave group and the function g(k) describes how
the amplitudes of the waves that contribute to ψ(x) vary with wave number
k and Δx and Δk are taken as the standard deviations of the respective func1
tions ψ(x) and g(k), then the minimum value of Δx Δk = .
2
It is more realistic to express the relationship between Δx and Δk as
1
.
Δx Δk ≥
2
The de Broglie wavelength of a particle of momentum p is
h
.
λ =
p
The wave number corresponding to this wavelength is
2 2p

.
k =

h
14
LASER SYSTEMS AND APPLICATIONS
Hence an uncertainty Δk in the wave number of the de Broglie waves
associated with the particle results in an uncertainty Δp in the particle’s
momentum according to the formula
hk
.
Δp =
2
1
1
Since
Δx Δk ≥
, Δk ≥
2(x)
2
h
.
Δx Δp ≥
...(1.6)
4
Equation (1.6) states that the product of the uncertainty Δx in the position
of an object at some instant and the uncertainty Δp in its momentum compoh
. For
nent in the x direction at the same instant is equal to or greater than
4
a narrow wave group Δx is small and then Δp will be large and in a broad
wave group Δp is small and then Δx will be large.
h
is abbreviated as  (h bar)
Generally
2
h
 =
= 1.054 × 10–34 J.s.
2
Thus, in terms of , the uncertainty principle becomes

Δx Δp ≥
.
2
In the above discussion we have considered wave nature of particle
while uncertainty principle can be arrived considering particle nature of
waves.

It is worth mentioning that the lower limit of
for Δx Δp is rearely
2
attained; more usually Δx Δp ≈  or even Δx Δp ≈ h.
Another form of the uncertainty principle is sometimes useful. Let energy emitted E be measured during the time interval Δt in an atomic process.
The limited time available imposes restriction on the accuracy with which we
can determine the frequency ν of the waves.
The uncertainty Δν in our frequency measurement is
1
Δν ≥
.
t
Because the minimum uncertainty in the number of waves we count in
a wave group is one wave and to get uncertainty Δν, it is divided by the time
inverval.
The corresponding energy uncertainty is
ΔE = hΔν
h
and so
ΔE ≥
or ΔE Δt ≥ h.
t
A more precise calculation changes this result to

ΔE Δt ≥
.
2
15
ELEMENTS OF QUANTUM MECHANICS
SOL
VED EXAMPLES
SOLVED
EXAMPLE 11: Calculate the smallest possible uncertainty in the position of an
electron moving with velocity 3 × 107 m/s.
SOLUTION: (Δx)min (Δp)max = h
2
(Δp)max = p (momentum of the electron)
mv =
(Δx)min =
m0 v
(1  v 2 c 2 )
h 1  v2 c2
2 m0 v
h
= 1.05 × 10–34 J.s., v = 3 × 107 m/s, m0 = 9 × 10–31 kg, c = 3 × 108 m/s
2

(3  107 )2
 1 
(3  108 )2

9  10 31  3  107
1.05  10 34
(Δx)min =



= 3.8 × 10–12 m = 0.038 Å.
EXAMPLE 12: An electron is confined to a box of length 10–8 meter; calculate the
minimum uncertainty in its velocity. Given m = 9 × 10–31 kg;  = 1.05 × 10–34 joule
second.
SOLUTION:
(Δx)max = 10–8 meter
(Δp)min =
1.05  10 34

kg-m/s

( x )max
10 8
= 1.05 × 10–26 kg-m/s
(Δp)min = m(Δv)min = 1.05 × 10–26
(Δv)min =
1.05  10 26 1.05  10 26

= 1.17 × 104 m/s.
m
9  10 31
EXAMPLE 13: Find the uncertainty in the momentum of a particle when its
position is determined within 0.01 cm. Find also the uncertainty in the velocity of
an electron and an α-particle respectively when they are located within 5 × 10–8 cm.
SOLUTION: According to uncertainty principle
h
2
h
Δp =
2 x
(Δx) (Δp) =
or
16
LASER SYSTEMS AND APPLICATIONS
h
= 1.05 × 10–34 J.s
2
Δp =
Δx = 0.01 × 10–2 meter
1.05  10 34
= 1.05 × 10–30 kg m/s
0.01  102
Δp = mΔv
∴
Δv =
h
.
2m x
Uncertainty in the velocity of electron
m = 9 × 10–31 kg
Δx = 5 × 10–10 m
Δv =
1.05  10 34
9  1031  5  10 10
= 2.33 × 105 m/sec.
Uncertainty in the velocity of α-particle, mass of α-particle = 4 × mass of
proton
= 4 × 1.67 × 10–27 = 6.68 × 10–27 kg
Δx = 5 × 10–10 m
Δv =
1.05  10 34
6.68  1027  5  10 10
= 31.4 m/sec.
EXAMPLE 14: An electron has a speed 4 × 105 m/s within the accuracy of 0.01
per cent. Calculate the uncertainty in the position of the electron. Given h = 6.625 ×
10–34 J.s. Mass of electron = 9.11 × 10–31 kg.
SOLUTION: Uncertainty in velocity = 0.01 × 4 × 105 = 40 m/s
100
h
(Δx) (Δp) =
2
(Δx) =
=
1.055  10 34
h

m v
2 ( p)
1.055  10 34
= 2.895 × 10–6 m.
9.11  10 31  40
EXAMPLE 15: An excited atom has an average lifetime of 10–8 sec. That is during
this period it emits a photon and returns to the ground state. What is the minimum
uncertainty in the frequency of this photon?
SOLUTION:
h
2
E = hν
ΔE Δt ≥
17
ELEMENTS OF QUANTUM MECHANICS
or
ΔE = hΔν
h
2
1
Δν ≥
2 t
hΔν Δt ≥
or
Δt = 10–8 sec
Δν ≥
1
= 15.92 × 106 sec–1.
2  3.14  10 8
EXAMPLE 16: If an excited state of hydrogen atom has a lifetime of 2.5 × 10–14
sec. What is the minimum error with which the energy of this state can be measured?
SOLUTION: The uncertainty in the energy of the photon is equal to the
uncertainty in the energy of the excited state of the atom due to energy
conservation.
According to uncertainty principle
ΔE Δt ≥
ΔE ≥
=
h
2
6.63  10 34
h

= 4.22 × 10–21 J
2 t 2  3.14  (2.5  10 14 )
4.22  10 21
= 0.026 eV.
1.6  1019
EXAMPLE 17: Using the uncertainty relation ΔE.Δt = h , calculate the time
2
required for the atomic system to retain rotation energy for a line of wavelength 6000
Å and width 10– 4 Å.
SOLUTION: We know that E = hν = hc

hc

2
where Δλ is the width of the spectral lines. Here Δλ = 10– 4 × 10–10 m.
Using uncertainty relation
ΔE =
ΔE . Δt =
Δt =
h
, we have
2
h

2  E
Putting the value we get Δt =
2
h

2 c 
 hc 
2  2  
 
(6  107 )2
2  3.14  3  108  1014
= 1.9 × 10– 8 second.
18
LASER SYSTEMS AND APPLICATIONS
EXAMPLE 18: A nucleon is confined to a nucleus of diameter 5 × 10–14 m.
Calculate the minimum uncertainty in the momentum of the nucleon. Also calculate
the minimum kinetic energy of the nucleon.
SOLUTION: We have the relation
(Δp)min (Δx)max =
(Δp)min =
h
2
6.626  1034
h

2(x)max 5  10 14  2  3.14
= 0.21 × 10–20 kg m/sec.
Since p cannot be less than (Δp)min
Emin =
( Pmin )2
2m
Putting the values of
Pmin = 0.21 × 10–20 kg . m/sec and m = 1.675 × 10–27 kg.
Emin =
=
(0.21  1020 )2
= 0.063 × 10–13 J
2  1.675  1027
0.063  10 13
= 0.039 MeV.
1.6  10 19
EXAMPLE 19: An electron is confined to a box of length 1.1 × 10–8 m. Calculate
the minimum uncertainty in its velocity. Given m = 9.1 × 10–31 kg and
× 10–34 J.s.
h
= 1.05
2
SOLUTION: We know from uncertainty principle that
Δx . Δp ≥
h
2
Δp =
h
2 x
Let Δv be the uncertainty in the velocity of a particle of mass m, so we
have
Δp = mΔv
or
Δv =
Δx = 1.1 × 10–8 m,
Δv =
p
m
h
= 1.05 × 10–34 J.s. and m = 9.1 × 10–31 kg
2
1.05  1034
h

2 m x 9.1  10 31  1.1  10 8
= 1.06 × 104 m/sec.
19
ELEMENTS OF QUANTUM MECHANICS
EXAMPLE 20(a): What is the minimum uncertainty in the frequency of a photon
whose life time is about 10–8 sec?
SOLUTION: From uncertainty principle, we have
h
2
h
h Δν Δt ≥
2
ΔE Δt ≥
But E = hν
or
Δν ≥
or ΔE = h Δν
1
2 t
1
= 15.92 × 106/sec.
2  3.14  108
EXAMPLE 20(b): A typical atomic nucleus is about 5 × 10–15 m in radius. Use
the uncertainty principle to place a lower limit on the energy an electron must have
if it is to be part of a nucleus.
Putting Δt = 10–8 sec, Δν =
SOLUTION: Let
Δx = 5 × 10–15 m
1.054  10 34


2x 2  5  10 15
≥ 1.1 × 10–20 kg . m/s
If this is the uncertainty in a nuclear electron’s momentum, the momentum p itself must be at least comparable in magnitude. An electron with such
a momentum has kinetic energy many times greater than its rest energy m0c2.
Thus we have E = pc
E = pc ≥ (1.1 × 10–20) (3 × 108)
≥ 3.3 × 10–12 J
Since 1 eV = 1.6 × 10–19 J, the energy of an electron must exceed 20 MeV
if it is to be on nuclear constituent. Experiments indicate that even the electrons associated with unstable atoms never have more than a fraction of this
energy and we conclude that electrons are not present within nuclear.
Δp ≥
1.8
WA
VE FUNCTION AND WA
VE EQU
ATION
WAVE
WAVE
EQUA
The quantity with which quantum mechanics is concerned is the wave function Ψ of a body. The square of its absolute magnitude | Ψ |2 evaluated at a
particular place at a particular time is proportional to the probability of
finding the body there at that time. The momentum, angular momentum and
energy of the body are other quantities that can be established from Ψ. In
quantum mechanics we determine Ψ for a body when its freedom of motions
is limited by the action of external forces.
If the wave function Ψ is complex with both real and imaginary parts, the
probability density | Ψ |2 is written as ΨΨ*. Where Ψ* is complex conjugate of
Ψ. A complex wave function Ψ can be written in the form
Ψ = A + iB where A and B are real functions
Ψ* = A – iB
Ψ*Ψ = A2 + B2 [i2 = –1].
20
LASER SYSTEMS AND APPLICATIONS
Thus ΨΨ* is always a positive real quantity. The condition for an acceptable wave function is that the integral of | Ψ |2 over all space must be finite,
if | Ψ |2 is equal to P, then it must be true that




2

 PdV
dV = 1 since
=1

is the mathematical statement that the particle exists somewhere at all times.
A wave function that obeys above equation is said to be normalised. Every
acceptable wave function can be normalised by multiplying by an appropriate constant.
Besides being normalisable, Ψ must be single valued, since P can have
only one value at a particular place and time and continuous. Momentum
  
,
,
be finite conx y z
tinuous and single valued. The wave functions with all these properties can
yield physically meaningful results when used in calculations.
considerations require that the partial derivatives
For a particle restricted to motion in the x direction, the probability of
finding it between x1 and x2 is given by
Probability =

x2
x1

2
dx
Schrödinger’s equation which is the fundamental equation of quantum
mechanics is a wave equation in the variable Ψ.
1.9
SC
HR
ÖDINGER’S EQU
ATION: TIME-DEPENDENT
SCHR
HRÖDINGER’S
EQUA
The wave function of a freely moving particle moving in x direction can be
represented by a wave function Ψ as
Ψ = Ae
x
 i t  


...(1.15)
Replacing ω in the above formula by 2πν and v = νλ, we get
Ψ = Ae–2πi(νt – x/λ)
...(1.16)
Which is convenient since we know ν in terms of the total energy E and
momentum p of the particle being described by Ψ.
Since
E = hν = 2π ν
and λ =
h 2

p
p
we have
Ψ = Ae–(i/)(Et – px).
...(1.17)
Equation (1.17) is a mathematical description of the wave equivalent of
an unrestricted particle of total energy E and momentum p moving in the
+ x direction. The expression for the wave function Ψ given by eqn. (1.17) is
correct only for freely moving particles.
ELEMENTS OF QUANTUM MECHANICS
21
Schrödinger’s equation which is the fundamental equation of quantum
mechanics is a wave equation in the variable Ψ. Differentiating eqn. (1.17)
twice with respect to x, we get
p2
 2

=
2
x 2
and differentiating eqn. (1.17) once w.r.t. t, we get
...(1.18)

iE
.
=
...(1.19)
x

At speeds small compared with that of light, the total energy E of a
particle is the sum of its kinetic energy p2/2m and its potential energy V,
where V is in general a function of position x and time t.
E =
p2
 V.
2m
...(1.20)
Multiplying both sides of eqn. (6) by the wave function Ψ gives
p2
 V .
2m
From eqns. (1.18) and (1.19), we see that
EΨ =
EΨ =
and
...(1.21)
 
i t
 2
...(1.22)
x 2
Substituting these expressions for E Ψ and p2 Ψ into eqn. (1.21), we get
p2 Ψ =   2

2  2 
= 
...(1.23)
 V .
t
2m x 2
Equation (1.23) is the time dependent form of Schrödinger’s equation.
In three dimensions the time-dependent form of Schrödinger’s equation is
i
i

2   2   2   2  


= 

  V .
t
2m  x 2
y 2
z 2 
...(1.24)
Where the particle’s potential energy V is some function of x, y, z
and t.
The restrictions on particle’s motion affect the potential energy funciton
V. Once V is known Schrödinger’s equation may be solved for the wave
function Ψ of particle from which its probability density | Ψ |2 may be determined for a specified x, y, z, t. The extension of Schrödinger’s equation from
the special case of freely moving particle (V = constant) to the general case
of a particle subject to arbitrary forces that vary in time and space [V = V(x,
y, z, t)] is entirely plausible.
22
LASER SYSTEMS AND APPLICATIONS
Thus, Schrödinger’s equation is postulated, solve it for a variety of physical situations and compare the results of the calculations with the results of
experiments. If they agree, the postulate is valid and if disagree it may
be discarded and some other approach would have to be explored.
Equation (1.24) can be used for non-relativistic problems and in practice it
has turned out to be accurate in predicting the results of experiments.
1.10
SC
HR
ÖDINGER’S EQU
ATION: STEAD
Y-ST
ATE
SCHR
HRÖDINGER’S
EQUA
STEADY
-STA
FORM
It has been observed that in many situations, the potential energy of a particle does not depend on time and vary with the position of the particle only.
When this is true, Schrödinger’s equation may be simplified by removing
time.
The one-dimensional wave function Ψ of an unrestricted particles may be
written as
Ψ = Ae–(i/)(Et – px) = Ae–(iE/)t e+(ip/)x
...(1.25)
–(iE/)t
Thus, Ψ is the product of a time-dependent function e
and a position dependent function Ψ. (Ψ = Ae(ip/)x)
Substituting the Ψ of eqn. (1.25) into the time-dependent form of
Schrödinger’s equation, we obtain
2
 2 (iE/ )t  
e
 V e (iE/ )t
2
2m
x
Dividing throughout by the common exponential factor
E ψ e–(iE/)t = 
 2  2m
 2 (E  V )  = 0.
...(1.26)
x 2

Equation (1.26) is the steady-state form of Schrödinger’s equation. In
three-dimensions it is
 2   2   2  2m
 2  2  2 (E  V )  = 0.
x 2
y
z

...(1.27)
Schrödinger’s steady-state can be solved only for certain values of the
energy E. To solve Schrödinger’s equation we obtain a wave function ψ which
fulfils the requirements for an acceptable wave function, i.e., its derivatives
be continuous, finite and single valued. If there is no such wave function, the
system cannot exist in a steady-state.
The energy quantisation in wave mechanics is a natural element of the
theory and is a characteristic of all stable systems.
Particle in a Box
The simplest quantum-mechanical problem is that of a particle trapped in a
box with infinitely hard walls. We may specify the particle’s motion along the
ELEMENTS OF QUANTUM MECHANICS
23
x-axis between x = 0 and x = L. A particle does not lose energy when collides
with infinitely hard walls so that its total energy remains constant. From the
formal point of view of quantum mechanics, the potential energy V of the
particle is infinite on both sides of the box while V is constant say 0 for
convenience on the inside. The particle cannot have infinite amount of energy. So, the wave function Ψ is zero for x ≤ 0 and x ≥ L. Thus, our aim is to
find Ψ between x = 0 and x = L (Fig. 1.6).
Schrödinger’s equation for the condition specified (x = 0 and x = L)
becomes
 2  2m
 2 E  = 0 [V = 0]
...(1.28)
x 2

Equation (1.28) has the solution
2mE
2mE
x  B cos
x


where A and B are constants to be evaluated.
Applying the boundary conditions that
Ψ = 0 for x = 0 and for x = L. Since cos θ =
Fig. 1.6. Potential well which
1, the second term cannot describe the parcorresponds to a box with
ticle because it does not vanish at x = 0.
infinitely hard walls
Hence, B = 0. Since sin θ = 0, the sine term
always yields ψ = 0 at x = 0 but ψ will be 0 at x = L only when
ψ = A sin
2mE
L = nπ, n = 1, 2, 3 ...
...(1.29)

This is true because the sines of the angles π, 2π, 3π, ... are all 0.
From eqn. (1.29) it is clear that the energy of the particle can have only
certain values known as eigenvalues. These eigenvalues constituting the energy
levels of the system are
n2 2  2
where n = 1, 2, 3
...(1.30)
2mL2
The wave function of a particle in a box whose energy is En is
En =
2mEn
x

Substituting eqn. (1.30) for En, we get
ψn = A sin
...(1.31)
nx
.
...(1.32)
L
where ψn is the eigenfunctions corresponding to the energy eigenvalues En.
ψn = A sin
It can be verified that these eigenfunctions meet all the requirements. For
each quantum number n, ψn is a single valued function of x and ψn and ∂ψn/
∂x are continuous. The integral of | ψn |2 over all space is finite. By integrating
| ψn |2 dx from x = 0 and x = L.
24
LASER SYSTEMS AND APPLICATIONS



n
2
dx =

L
0
n
2
dx  A2

L
0
sin 2
nx
dx
L
L
.
...(1.33)
2
2
To normalise ψ we assign a value of A such that | ψn | dx is equal to the
probability P dx of finding the particle between x and x + dx, rather than
merely proportional to P. If | ψn |2 dx is equal to P dx, then it must be true that
= A2



n
2
dx = 1 since



P dx = 1.
...(1.34)
is the mathematical way of stating that the particle exists somewhere at all
times. Comparing eqns. (1.33) and (1.34), we find that the wave function of
a particle in a box are normalised if
2
L
.
= 1, A =
A2
L
2
The normalised wave functions of the particle are therefore
2
nx
sin
ψn =
n = 1, 2, 3.
L
L
The normalised wave functions ψ1, ψ2 and ψ3 together with the probability densities | ψ1 |2, | ψ2 |2 and | ψ3 |2 are plotted in Fig. 1.6(a). While ψn may
be negative as well as positive, | ψn |2 is always positive and, since ψn is
normalised, its value at a given x is equal to the probability density P of
finding the particle there. In every case | ψn |2 = 0 at x = 0 and x = L, the
boundaries of the box.
Fig. 1.6(a). Wave functions and probability densities of a particle
confined to a box with rigid walls
At a particular place in the box the probability of the particle being
present may be very different for different quantum numbers. For instance
| ψ1 |2 has its maximum value of 2/L in the middle of the box while | ψ2 |2 =
0. A particle in the lowest energy level of n = 1 is most likely to be in the
middle of the box, while a particle in the next higher state of n = 2 is never
there. Classical physics suggests the same probability for the particle being
anywhere in the box.
25
ELEMENTS OF QUANTUM MECHANICS
SOL
VED EXAMPLES
SOLVED
EXAMPLE 21: Show that ψ(x) = eicx where c is some finite constant is acceptable
eigenfunctions. Also normalise it over the region – a ≤ x ≤ a.
SOLUTION: The wave function ψ(x) can be written as
ψ(x) = eicx = cos cx + i sin cx.
Its derivative is given by
d ( x )
= ic eicx = ic (cos cx + i sin cx)
dx
= – c sin cx + ic cos cx.
The following points may be observed:
(i) sin cx and cos cx are periodic functions with maximum value 1 and
d
( x) are finite for all values of
c is finite constant. Thus ψ(x) and
dx
x.
d
( x) are single-valued because cos cx and
dx
sin cx are also continuous for all values of x.
Hence ψ(x) is an acceptable form of the eigenfunction. To normalise, the
wave function we may write ψ(x) as
ψ(x) = Aeicx.
Now we have to determine the value of A and we may write
(ii) The function ψ(x) and

a
a
A2

 * ( x )( x ) dx = 1
a
a
e icx e icx dx = 1
a
A2  x  a = 1
A2(2a) = 1
or
A=

.
2a
Hence normalised wave function is
ψ(x) =
 icx
e .
2a
EXAMPLE 22: A particle is moving in one-dimensional potential box (of infinite
height) of width 25 Å. Calculate the probability of finding the particle within an
interval of 5 Å at the centres of the box when it is in its state of least energy.
SOLUTION: The wave function of the particle can be written as
ψ(x) =
2
nx
sin
a
a
26
LASER SYSTEMS AND APPLICATIONS
For the particle in the least energy state n = 1 and hence
 2  sin x
 
a
a
ψ(x) =
At the centre of the box x =
a
, the probability of finding the particle in
2
the interval Δx is given as
P = | ψ(x) |2 Δx
2
 2
( a 2) 
2
2
2 
| ψ(x) |2 = 
sin
  a sin 2  a
a
a


P =
2  5  10 10
2
x 
a
25  10 10
 a  25 Å 
= 0.4 

 x  5 Å 
EXAMPLE 23: A particle is in motion along a line between x = 0 and x = a with
zero potential energy. At points for which x < 0 and x > a, the potential energy is
infinite. The wave function for the particle in the nth state is given by
nx
.
a
Find the expression for the normalised wave funciton.
ψn = A sin
SOLUTION: The probability of the particle between x and x + dx for the nth
state is given as
nx
dx.
a
Since the particle is found in the region x = 0 and x = a for all times, we
have
| ψn(x) |2 dx = A2 sin 2


A2

a
0
a
0
a
0
n
A2 sin 2
2
dx = 1
nx
dx = 1
a
1
2nx 
 1  cos
 dx = 1
2
a 
a
A2 
a
2nx 
sin
x 
 = 1
2 
2n
a 0
A2
a = 1 or
2
Now the normalised wave funciton is
ψn(x) =
2
nx
sin
.
a
a
A=
2
.
a
27
ELEMENTS OF QUANTUM MECHANICS
EXAMPLE 24: Find the energy of an electron moving in one dimension in an
infinitely high potential box of width 1 Å, given mass of the electron 9.11 × 10–31
kg and h = 6.63 × 10–34 Js.
n2 h 2
(n = 1, 2, 3, ...)
8ma2
For least energy of the particle n = 1.
SOLUTION: Since we know that E =
Now
(6.63  10 34 )2
h2
=
joules
2
8ma
8  9.11  10 31 (10 10 )2
= 9.1 × 10–19 joules
=
9.1  10 19
eV = 5.68 eV.
1.602  10 19
EXAMPLE 25: Show that Ψ = ψe–iωt is a wave function of a stationary state.
SOLUTION: If ψ is the wave functions of a stationary state, then the value
2
2
of  at each point must be constant, independent of time. To find  ,
we first take the complex conjugate of Ψ = ψe–iωt which is Ψ* = ψ* e+iωt. Then

2
= Ψ*Ψ = (ψ* e+iωt) (ψ e–iωt)
= ψ*ψ e0 = 
2
where ψ is not a function of time, so 
2
it has been shown that 
Ψ = ψe
–iωt
is also independent of time. Now,
2
2
2
=  , so 
is independent of time and
is a wave function of a stationary state.
EXAMPLE 26: Find the probability that a particle trapped in a box L wide can
be found between 0.45 L and 0.55 L for the ground and first excited states.
SOLUTION: This part of the box is one tenth of the box’s width and is
centered on the middle of the box. Classically, we could expect the particle
to be in this region 10 per cent of the time. The quantum mechanics gives
quite different predictions that depend on the quantum number of the particle’s
state. The probability of finding the particle between x1 and x2 when it is in
the nth state is
P =

x2
x1
2
 n dx 
2
L

x2
x1
sin 2
x
1
2nx  2
x
sin
=  
L  x1
 L 2n
x1 = 0.45 L and x2 = 0.55 L
For ground state which corresponds to n = 1,
we have
P = 0.198 = 19.8%
nx
dx
L
28
LASER SYSTEMS AND APPLICATIONS
For the first excited state n = 2
P = 0.0065 = 065%
2
This low figure is consistent with the probability density of  n = 0 at
x = 0.50 L.
EXAMPLE 27: (a) Show that y = Ax + B where A and B are constants, is a
solution to the Schrödinger equation for an E = 0 energy level of a particle in a box.
(b) Show, however, that the probability of finding a particle with this wave
function is zero.
SOLUTION: (a) The Schrödinger equation for a particle in a box is
d2
dx
2

2m
2
E = 0
Differentiating ψ = Ax + B twice with respect to x gives
d2 
= 0 for
dx 2
the left side of the Schrödinger equation. Also E = 0 gives zero of the
right side. Since 0 = 0, ψ = Ax + B is a solution to this Schrödinger
equation for E = 0.
(b) The wave function ψ equals zero outside the box (x < 0 and x > L). In
order that ψ = Ax + B may be continuous at x = 0, it must be true that
y = 0 at x = 0. So A(0) + B = 0 or B = 0.
Similarly, in order that ψ may be continuous at x = L, it must be true
that ψ = 0 at x = L, so A (L) + 0 = 0 or A = 0. With both A and B equal
to zero, ψ = Ax + B = 0. Thus the wave function equals zero inside
the box as well as outside the box and the probability of finding the
particle anywhere with this wave function is zero.
EXER
CISES
EXERCISES
1.
2.
3.
4.
5.
6.
7.
8.
Discuss the dual nature of matter and waves.
Find de Broglie wavelength of an electron of energy MeV.
What are de Broglie matter waves?
Calculate the de Broglie wavelength associated with a proton moving
with a velocity equal to 1/20 velocity of light.
Derive time dependent and time-independent Schrödinger wave
equation.
What is uncertainty principle? Apply this to prove the non-existence
of the electron in the nucleus.
Calculate the smallest possible uncertainty in the position of electron
moving with velocity 3 × 107 m/sec.
[Ans. 0.038 Å]
An electron has a speed of 1.05 × 104 m/sec within the accuracy of
0.01 per cent. Calculate the uncertainty in the position of the electron.
[Ans. 1.10–4 m]
ELEMENTS OF QUANTUM MECHANICS
29
9. Show that the de Broglie wavelength for a material particle of rest
mass m0 and charge q, accelerated from rest through a potential difference of V volts relativistically is given by λ = h/{2m0qV [1 + qV/
2m0c2]}1/2.
10. Find the uncertainty in the momentum of a particle when its position
is determined within 0.01 cm. Find also the uncertainty in the velocity of an electron and α-particle respectively when they are located
within 5 × 10–8 cm.
[Ans. 1.05 × 10–30 kg m/s, 2.33 × 105 m/s, 31.4 m/s]
11. An electron is confined to a box of length 1.1 × 10–8 meter. Calculate
the minimum uncertainty in its velocity. Given m = 9.1 × 10–31 kg,
[Ans. 1.06 × 104 m/s]
 = 1.05 × 10–34 J. sec.
12. Calculate the kinetic energy of an electron if the wavelength of electron equals the yellow line of Sodium.
[Ans. 4.34 × 10–6 eV]
13. Calculate the de Broglie wavelength of neutron of energy 28.8 eV.
Given h = 6.62 × 10–34 J.sec, m = 1.67 × 10–27 kg. [Ans. 5.3 × 10–2 Å]
14. Derive the formula for de Broglie wavelength of particle in terms of
kinetic energy and its rest energy m0c2.
15. Describe Davisson-Germer experiment to demonstrate the wave nature of particle.
16. Deduce an expression for de Broglie wavelength of helium atom
having energy at temperature T°K.
17. What are de Broglie waves and how do they help in the interpretation of the Bohr’s quantisation rule?
18. Show that the phase velocity of de Broglie wave is greater than the
velocity of light; but the group velocity is equal to the velocity of the
particle with which the waves are associated.
19. Calculate de Broglie wavelength associated with nitrogen at 3.0 atmospheric pressure and 27°C temperature.
20. Can a photon and an electron of the same momentum have the same
wavelength? Compare their wavelengths if the two have the same
energy.
21. X-rays of wavelength 0.91 Å fall on a metal plate having work function 2.0 eV. Find the wavelength associated with the emitted photoelectrons.
[Ans. 8.7 Å]
22. Calculate kinetic energy of a neutron having de Broglie wavelength
1 Å.
[Ans. 8.13 × 10–2 eV]
23. What is the minimum uncertainty in the frequency of a photon whose
lifetime is about 10–8 sec?
[Ans. 15.92 × 106 per sec.]
24. A certain excited state of hydrogen atom is known to have a lifetime
of 2.5 × 10–4 sec. What is the minimum error, with which energy of
the excited state can be measured.
[Ans. 0.026 eV]
30
LASER SYSTEMS AND APPLICATIONS
25. What is de Broglie wavelength of an electron accelerated from rest
through a potential difference of 100 volts.
[Ans. 1.23 Å]
26. Find the expression for the energy state of a particle in one-dimensional box.
OBJECTIVE QUES
T IONS
QUEST
1. A particle possesses a kinetic energy E and mass m, then its de Broglie
wavelength is
h
(a)
(b) h 2mE
2mE
1
2mE
(d)
h 2mE
h
2. An electron is accelerated through a potential difference of V volts.
The de Broglie wavelength of the electron is
12.27
12.27
(a)
(b)
V
V
(c)
(c)
12.27 V
(d) 12.27 V
3. Read the statements (A and B) and choose the answer:
(A) The de Broglie wavelength of a moving particle is inversely
proportional to its momentum
(B) Only a charged particle in motion is associated with matter waves
(a) A and B are correct
(b) A and B are wrong
(c) A is wrong and B is correct
(d) A is correct and B is wrong
4. If the momentum of a particle is doubled then its de Broglie wavelength
(a) halves
(b) doubles
(c) quadruples
(d) remains the same
5. The masses of neutron and electron are mn and me respectively. If they
have the same de Broglie wavelength, then their velocities should be
in the ratio
(a) 1 : 1
(c)
mn
me
(b)
(d)
me
mn
me2
mn2
6. An α-particle has a mass 4 mp and a proton mp. If they possess the
same kinetic energy, then the ratio of the de Broglie wavelengths is
31
ELEMENTS OF QUANTUM MECHANICS
(a)
(c)
7. The
(a)
(b)
(c)
(d)
8. The
1:4
(b) 4 : 1
2:1
(d) 1 : 2
characteristic of wave function ψ are
real function, finite and discontinuous
complex, single valued, finite and continuous function
complex, infinite and discontinuous function
complex single valued and infinite
wave function for the motion of a particle in a potential well of
 nx 
width a is given as ψn = B sin 
 , then B is
 a 
(a)
1
a
(b)
1
a
(c)
2
a
(d)
a
2
9. When the wave function is normalized then
(a) ψ ψ* = 1
(b) ψ ψ* dx =
a
1
(c)

1
  * dx  1
(d)
  * d
x
1
a
1
10. Increasing the potential difference between anode and filament in
Davisson-Germer experiment
(a) causes an increase in the wavelength of the electron-waves
(b) causes a decrease in the wave velocity
(c) causes a decrease in the wavelength of the electron-waves
(d) causes a decrease in the momentum of the electron
11. For a particle of mass m confined to a cubical box of side L, the
allowed values of energy E are given as
(a)
(c)
n2 h 2
2 mL2
h 2 L2
, n = 1, 2, 3
(b)
, n = 1, 2, 3
(d)
n2 h2
8mL2
, n = 1, 2, 3
n2 h2
, n = 1, 2, 3
mL
2mn 2
12. An electron is confined to a potential box of infinite height and width
10 Å. The probability of finding the particle in a small interval Δx at
the centre of the box for the energy state immediately above the
ground state is
(a) zero
(b) 0.5
(c) 0.9
(d) 1
32
LASER SYSTEMS AND APPLICATIONS
13. If a charged particle of mass m is accelerated through a potential
difference of V volts, the de Broglie wavelength is proportional to
(a)
1

14.
15.
16.
17.
18.
19.
20.
21.
22.
(b) V
V2
1
(d) V2
(c) V 2
The de Broglie hypothesis is associated with
(a) wave nature of α-particles only
(b) wave nature of radiations
(c) wave nature of electrons only
(d) wave nature of all material particles
Which of the following phenomena cannot be expressed by wave
nature of light?
(a) Interference
(b) Diffraction
(c) Polarization
(d) Photoelectric effect
Matter waves
(a) are latitudinal
(b) show diffraction
(c) are electromagnetic
(d) always travel with speed of light
Of the following moving with the same velocity, the one which has
largest wavelength is
(a) a photon
(b) an electron
(c) an α-particle
(d) a neutron
The uncertainty principle holds for
(a) macroscopic particles only
(b) microscopic particles only
(c) macroscopic and microscopic particles both
(d) neither macroscopic nor microscopic particles
The uncertainty principle cannot hold for the following pairs:
(a) Energy and time
(b) Position and momentum
(c) Angular momentum and angle
(d) Linear momentum and angle
Matter waves were first experimentally observed by
(a) Frank-Hertz
(b) Davisson and Germer
(c) Ster-Gerlach
(d) de Broglie
If the momentum of a particle is increased to four times, then the de
Broglie wavelength will become
(b) four times
(a) half
(c) twice
(d) one-fourth
The Schrödinger time independent wave equation for free particle is
(a)
2 2
   (E  v )   0
2m
(b)
2 2
   ( v  E)   0
2m
33
ELEMENTS OF QUANTUM MECHANICS
2 2
2 2
   E  0
   E  0
(d)
2m
2m
23. Schrödinger time dependent wave equation for free particle is
(c)
(a)

2  2

  i
t
2m x 2
2  2

  i
t
2m x 2
i 
2m  2 

 i
(d)  2
2


t
t
 x
 x
24. An α-particle and a proton have the same kinetic energy. The ratio of
their wavelength is (m∝ = 4mp)
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1
25. In Davisson-Germer experiment, nickel crystal acts as
(a) perfect absorber
(b) perfect reflector
(c) two dimensional grating
(d) three dimensional diffraction grating
26. Momentum of a photon of energy hν is
(a) hν
(b) hνc
(c) hν/c
(d) has no momentum
(c)
2m  2 
(b)
2
2

ANSWERS
1.
6.
11.
16.
21.
26.
(a)
(d)
(b)
(b)
(d)
(c)
2.
7.
12.
17.
22.
(b)
(b)
(a)
(b)
(a)
3.
8.
13.
18.
23.
(d)
(c)
(c)
(b)
(b)
4.
9.
14.
19.
24.
(a)
(d)
(d)
(d)
(c)
5.
10.
15.
20.
25.
(b)
(c)
(d)
(b)
(d)