1
UNIT – 1
QUANTUM PHYSICS
Unit-01/Lecture-01
Concept of matter waves
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 = h
ν
(1)
where ν is the frequency associated with the radiation.
Einstein’s mass-energy relation states that
E = mc
2 (2)
Combining the two equations, it can be written as
E = h
ν = mc
2
Hence, the momentum associated with the photon is given by
P = mc = h
ν/c = h/λ
Extending this to particles, he suggested that any particle having a momentum p is
associated with a wave of wavelength given by
λ = h/p
(3)
This is called
de Broglie’s hypothesis of matter waves and
λ
is called the de Broglie
wavelength.
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)mv
2
= 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
2
λ = h/p = h/(2meV)
1/2
(4)
This equation suggests that, at a given speed, the de Broglie wavelength associated with
the particle varies inversely as the mass of the particle.
Definitions [Rgpv June 2011, Dec 2011 (7)]
Wave Packet
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.
Phase velocity
The velocity of a individual wave of a wave packet is known as Phase velocity.
Group velocity
Group velocity is the velocity with which the wave packet travels.
Q. Derive the formula of Phase velocity and Group velocity and also find relation
between them?
A wave is represented by the formula
y = A cos (
ωt – kx)
(1)
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 v
p
= dx/dt =
ω/k
= 2
πν/(2π/λ) = λν
(2)
v
p
is called the ‘wave velocity’ or
‘phase velocity’.
For group velocity, consider the combination of two waves represented by the
formula
y
1
= A cos (
ωt-kx)
y
2
= A cos {(
ω+∆ω)t – (k+∆k)x}
3
The resultant displacement is given by
y
= y
1
+ y
2
=
2A cos {(
ω+ω+∆ω)t–(k+k+∆k)x} cos (∆ωt-∆kx)
2
2
≈ 2A cos(ωt–kx).cos(∆ωt/2-∆kx/2)
(3)
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,
2A cos(
∆ωt/2-∆kx/2) = constant
or
(
∆ωt/2-∆kx/2) = constant
or x =
constant + (
∆ωt/∆k)
Group velocity v
g
= dx/dt = (
∆ω/∆k)
(4)
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
v
g
= d
ω
(5)
d
k
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.
S.NO
RGPV QUESTIONS
Year
Marks
Q.1
What is eave packet? Define group velocity and
phase velocity. Derive an expression for the de
Broglie wavelength associated with an electron
accelerated by the electric potential V.
Dec 2011
14
Q.2
Derive an expression for the group velocity and
phase velocity. Also find relation between them.
June 2011
14
4
Unit-01/Lecture-02
Relation between phase velocity and group
velocity: [Rgpv June 2013(7)]
The mathematical relation for phase velocity given by
v
p
=
ω/k or ω = k.v
p
The group velocity v
g
is given by
v
g
= d
ω = d(k.v
p
) dk
dk
=
v
p
+ k.dv
p
dk
=
v
p
+ (2
π/λ). dv
p
d(2
π/λ)
= v
p
+ (2π/λ).(-λ
2
/2
π).dv
p
d
λ
= v
p
–
λ. dv
p
(6)
d
λ
In the above expression, if (dv
p
/d
λ) = 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, (dv
p
/d
) is
positive and hence the group velocity is less than the phase velocity.
Relation between group velocity and particle velocity (Velocity of de
Broglie waves):
The phase velocity of waves depends 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 = 2πmc
2
/h =
2
πm
0
c
2
(1)
h(1-v
2
/c
2
)
1/2
and k = 2
π/λ = 2πmv/h =
2
πm
0
v
(2)
h(1-v
2
/c
2
)
1/2
The group velocity of de Broglie waves is given by
V
g
= d
ω/dk = dω/dv
dk/dv
5
d
ω/dv = (2πm
0
c
2
/h).d(1-v
2
/c
2
)
1/2
= 2
πm
0
v
(3)
dv
h(1-v
2
/c
2
)
3/2
dk/dv = ____2
πm
0_____
(4)
h(1-v
2
/c
2
)
3/2
From equations 3 and 4 we get,
v
g
= 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,
V
p
=
ω/k = c
2
/v = c
2
/v
g
(5)
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.
S.NO
RGPV QUESTIONS
Year
Marks
Q.1 Define group velocity and phase velocity. Prove
that for a relativistic particle and non- relativistic
particle, phase velocity is not equal to particle
velocity.
June 2013
7
6
Unit-01/Lecture-03
Heisenberg Uncertainty Principle[ Rgpv Dec 2012(7)]
This equation states that the product of uncertainty ∆x in the position of an object at some
instant and the uncertainty in the momentum ∆p in the x-direction at the same instant is equal to
or greater than ħ/2.
∆x. ∆p ≥ ħ
(1)
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
(2)
Consider the combination of two waves represented by the formula
y
1
= A cos (
ωt-kx)
y
2
= A cos {(
ω+∆ω)t – (k+∆k)x}
The resultant displacement is given by
y
= y
1
+ y
2
= 2A cos {(
ω+ω+∆ω)t–(k+k+∆k)x} cos (∆ωt-∆kx)
2
2
≈ 2A cos(ωt–kx).cos(∆ωt/2-∆kx/2)
(3)
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,
2A cos(
∆ωt/2-∆kx/2) = constant
or
(
∆ωt/2-∆kx/2) = constant
for maximum amplitude cos(∆ωt/2-∆kx/2)=0
Thus,
(
∆ωt/2-∆
kx/2)= nπ/2.
Let the displacement of two successive nodes be x1 and x2, then
(
∆ωt/2-∆kx1/2)=
π/2
(
∆ωt/2-∆kx2/2)=
3π/2
On solving, we get
∆k(x2-
x1)= π
7
∆ x=
π/∆k, where ∆k=2 π/λ
∆ x=λ/2
=h/2
∆p
∆ x. ∆p= h/2
Or
∆ x. ∆
p≈ h
This is the required principle.
S.NO
RGPV QUESTIONS
Year
Marks
Q.1
Explain the concept of wave packet and give the
mathematical proof of Heisenberg’s uncertainty
principle?
Dec 2012
7
8
Unit-01/Lecture-04
Applications of uncertainty principle: [Rgpv June 2013(7)]
(a). 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
We have the following ‘Thought experiment’ to illustrate the uncertainty principle. Imagine an
electron being observed using a microscope.
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 uncertainty is associated with the measuring process. The illustration only
estimates the accuracy of measurement, the uncertainty being inherent in the nature of the
moving particles involved.
9
b). 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.
Figure (2) shows the diffraction of an electron beam by a narrow slit of width ∆x. The beam
travelling 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). Sinθ.
Assuming the change in the momentum as representative of the uncertainty in
momentum, we get
∆x. ∆p
x
=
λ .h.sin θ = h
sin
θ λ
(c). Electron cannot reside in nucleus: 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
10
∆x = 2r = 10
-14
m.
∆p = h/2π∆x = 6.62×10
-34
/(2×3.14×10
-14
) =
1.055×10
-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 = 1.055×10
-20
x 3×10
8
= 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.
S.NO
RGPV QUESTIONS
Year
Marks
Q.1
State Heisenberg’s uncertainty principle and derive it
from hypothetical gamma ray microscope?
June 2013
7
11
Unit-01/Lecture-05
Compton Effect {Rgpv June 2012(7), Dec 2013 (7)]
He discovered that “when a beam of monochromatic radiation (X-ray) of sharply defined frequency were
incident on a material of low atomic number (like carbon), the ray suffered a change of frequency on
scattering”. The scattered beam contains two wavelengths. In addition to the expected incident
wavelength, there exists a line of longer wavelength. The change of wavelength is due to the loss of
energy of the incident rays. This phenomenon is known as
Compton Effect.
Let a photon of energy collides with an electron at rest. During the collision it gives a small fraction
of energy to the frequency of electron. The electron gains kinetic energy and recoils.
Before collision
i.
Energy of incident photon =
ii.
Momentum of incident photon=
iii.
Rest mass of free electron=
iv.
Momentum of rest electron=0
After collision
i.
Energy of scattered photon =
ii.
Momentum of scattered photon=
iii.
Energy of electron=
iv.
Momentum of recoil electron=mv
ν
h
c
h
ν
2
0
c
m
‘
ν
h
c
h ‘
ν
2
mc
ν
h
12
Where
Energy of system before collision=
Energy of system after collision=
Momentum before collision= momentum after collision
Where, h is Planck’s constant.
Before the scattering event, the electron is treated as sufficiently close to being at rest that its
total energy consists entirely of the mass-energy equivalence of its rest mass
m
:
2
mc
E
=
Squaring and adding above eqs.
According to the principle of conservation of energy,
𝐸𝐸 = ℎ(
ν
− ′) +
2
0
c
m
From relativistic mechanics, 𝐸𝐸 = �
2
2
c
p
+ m
0
2
c
4
On comparing above eq. ,we get �ℎ(
ν
−
′
) +
2
0
c
m
�
2
=
2
2
c
p
+ m
0
2
c
4
+
−
+
)
‘
2
‘
(
2
2
2
νν
ν
v
h
m
0
2
c
4
+ 2h(
ν
−
ν
′)
2
0
c
m
=
2
2
c
p
+ m
0
2
c
4
2
2
c
p
=
+
−
+
)
‘
2
‘
(
2
2
2
νν
ν
v
h
2h(
ν
−
ν
′)
2
0
c
m
)
‘
)(
(
2
‘
)
cos
)
‘
)(
(
2
‘
2
2
2
2
2
2
2
2
ν
ν
ν
θ
ν
ν
ν
h
h
h
v
h
h
h
h
v
h
−
+
=
−
+
+ 2h
ν
2
0
c
m
− 2h
ν
′
2
0
c
m
(ℎ𝜈𝜈)(ℎ𝜈𝜈
′
)𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = (ℎ𝜈𝜈)(ℎ𝜈𝜈
′
) − ℎ𝜈𝜈𝑚𝑚
0
𝑐𝑐
2
+ ℎ𝜈𝜈′𝑚𝑚
0
𝑐𝑐
2
Dividing the above eq. by (ℎ𝜈𝜈)(ℎ𝜈𝜈
′
), we get
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 1 −
1
ℎ𝜈𝜈′ 𝑚𝑚
0
𝑐𝑐
2
+
1
ℎ𝜈𝜈 𝑚𝑚
0
𝑐𝑐
2
1
ℎ𝜈𝜈′ 𝑚𝑚
0
𝑐𝑐
2
−
1
ℎ𝜈𝜈 𝑚𝑚
0
𝑐𝑐
2
= 1 − 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
1
𝜈𝜈′ −
1
𝜈𝜈 =
ℎ
𝑚𝑚
0
𝑐𝑐
2
(1 − 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐)
ν
ν
2
0
c
m
h
+
ν
2
‘ mc
h
+
ν
2
2
0
‘ mc
h
c
m
h
+
=
+
ν
ν
2
2
0
1
c
v
m
m
−
=
φ
θ
ν
ν
cos
cos
‘
0
mv
c
h
c
h
+
+
=
+
θ
ν
φ
cos
‘
cos
h
hv
pc
−
=
θ
ν
φ
sin
‘
sin
h
pc
=
2
2
2
2
)
sin
‘
(
)
cos
‘
(
θ
ν
θ
ν
h
h
hv
c
p
+
−
=
)
cos
)
‘
)(
(
2
‘
sin
‘
cos
‘
cos
‘
2
2
2
2
2
2
2
2
2
2
2
2
2
2
θ
ν
ν
ν
θ
ν
θ
ν
θ
νν
h
h
h
v
h
h
h
h
v
h
−
+
=
+
+
−
=
E
h
c
m
h
+
=
+
‘
2
0
ν
ν
13
𝑐𝑐 = 𝜈𝜈
λ
𝑐𝑐𝑜𝑜 𝜈𝜈 =
𝑐𝑐
λ
𝑎𝑎𝑎𝑎𝑎𝑎 𝜈𝜈′ =
𝑐𝑐
λ
′
λ
′
𝑐𝑐 −
λ
𝑐𝑐 =
ℎ
𝑚𝑚
0
𝑐𝑐
2
(1 − 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐)
λ
′
−
λ
=
ℎ
𝑚𝑚
0
𝑐𝑐 (1 − 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐)
△
λ
=
ℎ
𝑚𝑚
0
𝑐𝑐 (1 − 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐)
∆ λ
is known as
Compton shift.
Different cases
If
𝑐𝑐=0
o
,
then ∆ λ=0.
If
𝑐𝑐=90
o
, then ∆ λ=
ℎ
𝑚𝑚
𝑒𝑒
𝑐𝑐
=0.0242 A
o
.
This constant value is called
Compton wavelength.
If
𝑐𝑐=180
o
, then ∆ λ=
2ℎ
𝑚𝑚
𝑒𝑒
𝑐𝑐
=0.0484 A
o
.
This is the maximum wavelength.
S.NO
RGPV QUESTIONS
Year
Marks
Q.1
What is Compton effect? Explaining the Compton expression,
discuss the various possibilities of X-ray scattering?
June
2012
14
Q.2
An X ray photon of Wavelength 0.4A
o
is scattered through an
angle of 45
o
by a loosely bound electron. Find the wavelength of
the scattered photon.
DEC
2013
7
14
UNIT 1/LECTURE 6
Characteristics of wave function: [RGPV June 2013 (7)]
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)
Equation (1) is called the normalization condition and a wave function that obeys the equation
is said to be
normalized. Further, 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.
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
15
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.
Schrodinger’s wave equation: [RGPV JUNE 2013, DEC 2013 (7)]
The motion of a free particle can be described
by the wave equation.
ψ = A exp{-i(ωt –kx)}
(1)
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), we get,
ψ = A exp{-i (Et-px)}
(2)
ħ
Differentiating equation (2) with
respect to
x twice, we get,
∂
2
ψ = -p
2
ψ or p
2
ψ = –
ħ
2
.
∂
2
ψ
(3)
∂x
2
ħ
2
∂x
2
Differentiating equation (2) with respect to t, we get,
∂ψ
=
– iE
.
ψ or E ψ = –
ħ . ∂ψ
(4)
∂t
ħ
i ∂t
The total energy of the particle can be written as
E
=
p
2
+ U
(5)
2m
Where U is the potential energy of the particle. Multiplying both sides of the equation by ψ
E
ψ = p
2
ψ + Uψ
(6)
2m
Substituting for E
ψ and p
2
ψ from equation (1.42) and (1.43)
16
–
ħ ∂ψ = – ħ
2
∂
2
ψ + Uψ
(7)
i
∂t
2m
∂x
2
This is known as
Schrodinger’s time dependent equation in one dimension.
The wave function ψ in equation (2) may also be written as
ψ = A exp {-i (Et-px)} = A exp (-iEt). exp (ipx)
ħ
ħ
ħ
ψ = Φ exp (-iEt)
(8 )
ħ
where Φ is a position dependent function. Substituting this form of ψ in equation (6),
E
Φ exp(-iEt) = p
2
Φ exp(-iEt) + UΦ exp(-iEt)
ħ
2m
ħ
ħ
or E
Φ exp(-iEt) = –
ħ
2
.
∂
2
Φ . exp(-iEt) + UΦ exp(-iEt)
ħ
2m ∂x
2
ħ
ħ
or
∂
2
Φ exp(-iEt) + 2m (E-U)Φ exp(-iEt) = 0
∂x
2
ħ ħ
2
ħ
or
∂
2
ψ + 2m (E-U)ψ
= 0
(9)
∂x
2
ħ
2
This is the Schrodinger’s wave equation in one dimension. In three dimensions, the above
equation may be written as
∂
2
ψ +
∂
2
ψ +
∂
2
ψ + 2m(E-U)ψ = 0
∂x
2
∂y
2
∂z
2
ħ
2
or
∇
2
ψ + 2m(E-U)ψ =0
ħ
2
This equation is known as
the steady state or time independent Schrodinger wave equation in
three
dimensions.
17
S.NO
RGPV QUESTION
YEAR
MARKS
Q.1
Discuss the concept of wave function
associated with the particle. Give
examples of admissible wave function.
Why derivatives of wave function
should be continuous everywhere?
JUNE2013
7
Q.2
Derive Schrodinger’s time dependent
equation for matter wave?
DEC 2013
7
18
UNIT 1/LECTURE 7
APPLICATIONS OF SCHRODINGER’S EQUATION: [RGPV Dec2013 (7)]
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
∂
2
ψ + 2m Eψ
= 0
(10)
∂x
2
ħ
2
Where E is the total energy of the particle which is purely kinetic. This is of the form,
∂
2
ψ + k
2
ψ = 0
∂x
2
Where k
2
= 2mE/ħ
2
.
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 = ħ
2
k
2
/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.
Particle in a one dimensional potential 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).
19
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)
We have to evaluate the wave function inside the box.
The Schrodinger’s equation (1.48) becomes
∂
2
ψ + 2m Eψ = 0 for 0 < x < L
(2)
∂x
2
ħ
2
ψ = A sin (2mE)
1/2
x + B cos (2mE )
1/2
x
(3)
ħ
2
ħ
2
where A and B are constants.
Applying the boundary condition that ψ=0 at x = 0, equation 3 becomes
A sin 0 + B cos 0 = 0 or B = 0.
Again, we have ψ = 0 at x = L. Then,
A.sin(2mE)
1/2
.L=0
ħ
2
If A = 0, the wave function will become zero irrespective of the value of x. Hence, A cannot be
zero.
Therefore, sin(2mE)
1/2
.L=0
ħ
2
or (2mE)
1/2
L=n
π
Where n=1,2,3 ..
(4)
ħ
2
From (4), the energy eigen values may be written as
E
n
=
n
2
π
2
ħ
2
Where n = 1,2,3,… …
(5)
2mL
2
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
20
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
E
1
=
π
2
ħ
2
2mL
2
This is called ‘ground state energy’ or ‘zero point energy’. The higher excited states will have
energies like 4E
1
, 9E
1
, 16E
1
, etc. This indicates that the energy levels are not equally spaced.
The wave functions or the Eigen functions are
given by
ψ
n
= A. Sin
2mE
n
1/2
x
ħ
2
or ψ
n
=
A. Sin
n
π x
(6)
L
Applying the normalization condition,
i.e.
∫ A
2
Sin
2
n
πx . dx = 1
(7)
L
Since the wave function is non-vanishing only for
0 < x < L, it can be shown that
∫ Sin
2
n
πx dx
= (L )
(8)
L
2
Substituting in equation (8), we have
A
2
(L ) = 1
or
A = ( 2 )
1/2
(9)
2
L
The eigen function or wave functions in equation (9) becomes
ψ
n
= ( 2 )
½
sin (2mE
n
)
½
x
ψ
n
= ( 2 )
½
sin n
πx
L
L
21
Fig. (2) shows the variation of the wave function inside the box for different values of n and
Fig.(3) shows the probability densities of finding the particle 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.
S.NO
RGPV QUESTION
YEAR
MARKS
Q.1
Obtain an expression of energy levels
for particle trapped in one dimensional
square with infinitely deep potential
well.
Dec2013
7
UNIT 1/LECTURE 8
Q1. X-rays of wavelength 1.54 A
o
are Compton scattered at an angle of 60
o
. Calculate the
change in the wavelength.
Solution:
Change in wavelength = ∆λ = h (1-cos θ) /m
o
c
h (1-cos 60
0
)/ m
o
c
=
= 1.2 x 10
-12
m (Ans).
In a Compton scattering experiment, incident photons of energy 10 KeV are scattered at
to the incident beam. Calculate the energy of the scattered photon.
Solution:
Change in wavelength = ∆λ = h (1-cos θ) m
o
c
= 7.1 x 10
-13
m.
Wavelength of incident photon = λ = hc/eE
= 1.243 x 10
10
m.
Wavelength of scattered photon =λ’= λ + ∆λ
22
= 1.25 x 10
10
m.
Energy of scattered photon = hc/λ’
=
1.59 x 10
-15
J
=
9.93 keV (Ans).
Q3.
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
o
? What is the kinetic energy of scattered electron?
Solution:
Wavelength of incident gamma rays = λ = hc/E
=
6.62×10
-34
x3x10
8
/1.6×10
-19
x0.5×10
6
=
2.486 x 10
-12
m.
Change in wavelength =
∆λ = h (1-cos θ)
m
o
c
=
3.24 x 10
-13
m.
Wavelength of scattered photon =λ’= λ + ∆λ
2.81 x 10
-12
m.
Kinetic energy of the scattered electron = hc/λ’
= 0.442 MeV (Ans).
Q4.
X-rays of wavelength 1.5 A
o
are Compton
scattered. At what angle will be scattered x-
rays have a wavelength of 1.506 A?
Solution:
Change in wavelength = ∆λ = h (1-cos θ) m
o
c
cos
θ = (1 – m
0
c.
∆λ/h) = (1 – 0.247) =0.753
Angle of scattering,θ = 41.2
0
(Ans).
Q5.
Calculate the de Broglie wavelength associated with an electron travelling with a velocity of
10
5
ms
-1
. Assume the mass of the electron to be 9.1 x 10
-31
kg. and h = 6.62×10
-34
Js.
Solution:
De Broglie wavelength λ = h
= 6.62 x 10
-34
____
P
9.1 x 10
-31
x10
5
λ = 7.27 x 10
-9
m. (Ans.)
23
UNIT 1/LECTURE 9/ADDITIONAL TOPICS
Photoelectric effect
The photoelectric effect occurs when matter emits electrons upon exposure to
electromagnetic radiation, such as photons of light. Here’s a closer look at what the photoelectric effect
is and how it works.
The photoelectric effect is studied in part because it can be an introduction to wave-particle
duality and quantum mechanics.
When a surface is exposed to sufficiently energetic electromagnetic energy, light will be
absorbed and electrons will be emitted. The threshold frequency is different for different
materials. It is visible light for alkali metals, near-ultraviolet light for other metals, and extreme-
ultraviolet radiation for nonmetals. The photoelectric effect occurs with photons having
energies from a few electronvolts to over 1 MeV. At the high photon energies comparable to
the electron rest energy of 511 keV, Compton scattering may occur pair production may take
place at energies over 1.022 MeV.
Einstein proposed that light consisted of quanta, which we call photons. He suggested that the
energy in each quantum of light was equal to the frequency multiplied by a constant (Planck’s
constant) and that a photon with a frequency over a certain threshold would have sufficient
energy to eject a single electron, producing the photoelectric effect. It turns out that light does
not need to be quantized in order to explain the photoelectric effect, but some textbooks
persist in saying that the photoelectric effect demonstrates the particle nature of light.