```www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS
UNIT -2
QUANTUM MECHANICS
Heisenberg’s uncertainty principle:
At the atomic scale of quantum mechanics, measurement of physical parameters becomes very
difficult. At any instant, the position and momentum of a classical body can be measured with very high
accuracy. However, in the case of a quantum particle, there are uncertainties associated with the position and
momentum of the wave packet, which represent the particle.
Statement: It is impossible to determine precisely and simultaneously the values of both the members of a
pair of physical variables which describe the motion of an atomic system. Such pairs of variables are called
canonically conjugate variables.
ħ
Examples: 1) Position and momentum i.e., xp ≥
ħ
2) Energy and time i.e., Et ≥
3) Angular momentum and angular position i.e., Lθ ≥

ħ

where ħ =  ; h is Planck’s constant
Physical significance of Heisenberg’s uncertainty principle:
According to classical mechanics, it is possible for a particle to occupy a fixed position and have a
definite momentum and we can predict exactly its position and momentum at any time later. But according
to uncertainty principle, it is not possible to determine accurately the simultaneous values of position and
momentum of a particle at any time. Heisenberg’s principle implies that in physical measurements,
probability takes the place of exactness and as such phenomena which are impossible according to classical
ideas may find a small but finite probability of occurrence.
Application of uncertainty principle-Non-existence of electron in the nucleus:
The radiation emitted by a radioactive nuclei consists of ,  and  rays out of which  rays are
identified to be electrons. We apply uncertainty principle to find whether electrons are coming out of the
nucleus.
ħ
According to the uncertainty principle, Δxp ≥ 2
where x is the uncertainty or imprecision (standard deviation) of the position measurement
p is the uncertainty of the momentum measurement.
h
or p ≥ 4x -------(1)
Diameter of the nucleus is of the order of 10-14 m. If an electron is to exist inside the nucleus, then the
uncertainty in position x must not exceed the size of the nucleus.
i.e., x  10-14 m.
Substituting this in equation (1),
h
i.e.,
p ≥ 4x
6.626 × 10−34
≥ 4×3.14×10−14
≥5.3 × 10-21 kgms-1
Since energy E = mc = pc, minimum energy of the emitted electron,
E = pc
= 5.3 × 10-21 × 3 ×108
= 15.9 × 10-13 J
= 9.94 MeV
( Note: In some text books x value is taken as 5×10-15 m which is actually nuclear radius.
Then E = 19.88 MeV)
2
Notes Compiled by: Dr. Santhosh D Shenoy, M.Sc., Ph.D.
www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS
www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS
Thus if an electron exist within the nucleus, emitted electron must possess an energy of about 9.9
MeV. But  decay studies recommend a kinetic energy below 4 MeV for an electron emitted from the
nucleus. Hence we can conclude that electrons cannot exist inside the nucleus. During  decay, neutron
transforms into a proton producing an electron and neutrino.
Wavefunction:
The probability that a particle will be found at a given place in space at a given instant of time is
characterized by the function ψ(x,y,z,t). It is called wave function. This function can be real or complex.
Thus the quantity with which quantum mechanics is concerned is the wavefunction  of a particle. The
quantity, probability density p=2dV or *dxdydz is proportional to the probability of finding the particle
in the volume element dxdydz about the point (x,y,z).
Physical significance of wavefunction – Probability density and normalization of wavefunction:
Wavefunction  is a function of position and time. It has no direct physical significance, as it is not
an observable quantity. Generally the wave function Ψ is a complex function, but the probability must be
real. Therefore to make it real quantity Ψ is to be multiplied by its complex conjugate Ψ *. Thus the square of
the absolute value of the wavefunction (2) or * is related to the moving particle and is known as the
Probability density. The quantity 2dV or *dxdydz is proportional to the probability of finding the
particle in the volume element dxdydz about the point (x,y,z). This interpretation was first given by Max
Born.
∞
If ∫−∞ ψ∗ ψdxdydz =0, the particle does not exist.
∞
If ∫−∞ ψ∗ ψdxdydz = ∞, the particle is everywhere simultaneously
Since the particle exist somewhere at all times,
∞
∫−∞ ψ∗ ψdxdydz = 1
The wavefunction  satisfying the above condition is called normalized wavefunction.
Properties of Wavefunction:
1. The wavefunction  must be continuous, finite and single valued everywhere.
2. ∂/∂x, ∂/∂y and ∂/∂z must also be continuous, finite and single valued everywhere.
3.  can be normalized.
Setting up of a one dimensional, time independent Schrödinger wave equation
Schrödinger wave equation is a differential equation with the variable . It is the fundamental
equation of Quantum mechanics in the same sense that the second law of motion is the fundamental
equation of Newtonian mechanics.
The motion of a free particle can be described by the equation,
ψ = Aei(kx- ωt)
Differentiating above eq. w.r.t ‘t’
dψ
 dt = -i Aei(kx- ωt)
And again differentiating above eq. w.r.t ‘t’
d2 ψ
= i22 Aei(kx- ωt)
= -2Aei(kx-ωt)
= -2ψ ------- (1)
The expression for a traveling wave is given by,
dt2
d2 y
1 d2 y
= v2 dt2
dx2
Replacing y by ψ,
d2 ψ
dx2
1 d2 ψ
= v2
dt2
Substituting the value of
d2 ψ
dt2
from equation (1),
Notes Compiled by: Dr. Santhosh D Shenoy, M.Sc., Ph.D.
www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS
www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS
d2 ψ
dx2
d2 ψ
dx2
1
= v2 -2ψ
=
−2
ψ=
v2
−(2πν)2
ψ=
(νλ)2
−(4π)2
(λ)2
ψ ---------------(2)
p2
Thus Total energy of the particle E = Kinetic energy + Potential energy = 2m+ V
2
Or, E-V = 2 =
h
( )2
λ
2m
2m(E−V)
1

= h2
λ2
Substituting this in equation (2),
Then
d2 ψ
dx2
−4 π2 2m(E−V)
=[
=[

h2
−8 π2 m(E−V)
h2
]ψ
]ψ
(−)
Or
+[
] =0

This is known as Schrödinger’s time independent wave equation in one dimension which is widely used
in Quantum mechanics.
Its three dimensional form is   ψ+ [
(−)

] =0
Eigen values and Eigen functions:
By solving the Schrödinger equation, we obtain the possible set of  functions. In case of bound
particles, the acceptable solutions for the differential equations are possible only for certain specified values
of energy. These descrete values of energy E1,E2,........,En are called energy eigen values of the particle. The
solutions 1,2,...........,n corresponding to the eigen values are called eigenfunctions.
Applications of Schrödinger wave equation
1. Energy eigen values for a free particle
For a free particle, potential energy is zero. Hence Schrödinger time independent wave equation for a free
particle can be written as,
d2 ψ
dx2
8 π2 mE
+[
Let K2=
d2 ψ
]ψ =0
h2
8 π2 mE
h2
2
(since V=0)
----------------(1)
Then dx2 + K ψ =0
The solution of this equation is
 = AsinKx + BcosKx
Since we are not having any boundary conditions to apply, solving the constants A and B pose some
difficulties.
Thus E=
h2 K 2
8mπ2
(From eq (1))
We can see there is no quantization of energy in the case of free particle and hence we can conclude that a
free particle is a ‘classical entity’.
2. Energy Eigen values of a particle inside a potential well of infinite height
Notes Compiled by: Dr. Santhosh D Shenoy, M.Sc., Ph.D.
www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS
www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS
Consider the motion of a particle confined to move inside a potential well of infinite height at x = 0
and x = L. The width of the well is L. Assuming there is no interaction between the walls of the well and
particle, the potential energy V of the particle is taken to be zero. Schrödinger time independent wave
equation in this case is
i.e,
d2 ψ
dx2
8 π2 mE
+[
]ψ =0
h2
8 π2 mE
Let K2=
h2
2
d2 ψ
----------------(1)
Then dx2 + K ψ =0
The solution of this equation is
 = AsinKx + BcosKx ------(2)
Since the particle is inside the well,
ψ = 0 at x 0 and also at x  L.
This is possible only if ψ = 0 at x = 0 and x = L as demanded by the continuity condition.
These are the boundary conditions of this problem.
At x = 0, ψ = 0
Therefore equation (2) becomes,
0 = Asin0 + Bcos0
Or B = 0
Therefore ψ = AsinKx ----------(3)
At x = L, ψ = 0
Now equation (2) becomes,
0 = AsinKL
(Since B=0)
But A≠ 0 (because if A=0 then eq(2) becomes zero which is not possible)
Hence sin KL=0
Therefore KL = n
nπ
Or K = ---------(4)
L
nπx
Therefore ψ = A sin( L )------ (5)
By applying normalized wave condition, for one dimensional case,
L
∫0 ψ ψ* dx = 1.
Substituting the value of  from equation (5),
L
nπx
nπx
∫0 A sin( L ) A sin( L ) dx = 1

nπx
i.e., ∫0 A2 sin2(
Or,
A2
L

2nπx
∫0 {1 − cos(
2
L
or
or,
A2
2
A2
2
A2 L
2
L
)} dx =1
{since sin2θ =
(1-cos2θ)
2
}
2nπx
But ∫0 cos(
i.e.,
) dx = 1
L
) dx = 0
L
∫0 dx =1
[X]L0 = 1
=1
2
or A =√L
Substituting this value in equation (5),
Notes Compiled by: Dr. Santhosh D Shenoy, M.Sc., Ph.D.
www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS
www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS
2
nπx
i.e., ψn = √L sin(
L
) -------------(6)
Since ‘n’ can take all possible integral values, there is more than one wave function. Equation (6)
represents the Eigen functions of the particle inside the potential well. For each Eigen function we associate
an Eigen value i.e. the allowed energies associated with that particle.
Substituting the value of K from equation (4) in equation (1),
nπ
K2 = ( L )2 =
8 π2 mE
i.e., E n =
h2
n2 h2
8mL2
Since ‘n’ is restricted, particle energy is restricted to certain values. ‘n’ is called quantum number.
The energy values En are called eigen values. Lowest energy of the particle is called zero point energy or
ground state energy and is given by,
h2
E 1 = 8mL2
Case 1: When n = 1 (Ground state):
The Eigen function is ψ1 =
2
πx
√ sin( )
L
L
Plot of ψ1 versus x and
versus x are as shown in the
|ψ1 | 2
figure
Case 2: When
2
√ sin(
L
2πx
L
n = 2 (First excited state):
The Eigen function is ψ2 =
)
Plot of ψ2 versus x and
are as shown in the figure
|ψ2 | 2 versus x
Case 3: When n = 3 (Second excited state):
2
3πx
The Eigen function is ψ3 = √L sin(
L
)
Plot of ψ3 versus x and |ψ3 | 2 versus x are as shown in the figure
Notes Compiled by: Dr. Santhosh D Shenoy, M.Sc., Ph.D.
www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS
www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS
****************
VTU Model question paper
2 a I The product of uncertainty between energy and time is
A) ≥h/2
B) ≥h/4
C) h/2
D) ≤h/4
2
II According to Max Born approximation,  represents
A) Energy density
B) Particle density
C) Probability density
D) Charge density
III The first permitted energy level is called
A) Excited energy level
B) Zero point energy
C) Ground level
D) None of these
IV The wavefunction associated with a material particle is
A) Finite
B) Continuous
C) Single valued
D) All the above
b. Set up time independent one dimensional Schrödinger wave equation.
c. What are eigen values and eigen function
d. Compute first three permitted energy values for an electron in a box of width 4Å.
(4+6+6+4)
Dec08/Jan09
2 a 1) The product of uncertainty between angular momentum and angular displacement is
A) ≥h/2
B) ≥h/4
C) h/2
D)≤h/4
2) Kinetic energy of electron acclerated by a voltage 50 Volts
A) 50 eV
B) 10 eV
C) 5 eV
D) 15 eV
3) The energy of the lowest state in one dimensional potential box of length ‘a’ is
2h2
h2
h
A) Zero
B) 8ma2
C) 8ma2
D) 8ma2
4) The wavefunction for the motion of particles in one dimensional potential box of length ‘a’ is given by
n = D sin nx/a where ‘D’ is normalization constant. The value of ‘D’ is
1
A) a
2
B) √
C) a

D) √2
(04 Marks)
b. Set up time independent one dimensional Schrödinger wave equation. (06 Marks)
c. Write the physical significance of wavefunction.
(04 Marks)
d. A quantum particle confined to one dimensional box of width ‘a’ is in its first excited state. What is the

probability of finding the particle over an interval of ( 2) marked symmetrically at the centre of the box?
(06 Marks)
June-July
2009
2 a i) According
2 represents
A) Energy
C)
D)
Charge
ii)
An
to Max Born approximation,
density B) Particle density
Probability density
density
electron has a speed of 100
Notes Compiled by: Dr. Santhosh D Shenoy, M.Sc., Ph.D.
www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS
www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS
m/s, accurate to 0.005%. The uncertainty in its position is
A) 0.01 m
B) 0.0115 m
C) 0.024 m
D) 0.04 m
iii) An electron is moving in a box of length ‘a’. If ‘1’ is the wavefunction at x = a/4 with n = 1 and 2 at
x = a for n = 2, then 2/1 is
a
√2
A)
B) √2
a
D) ∝
C) 0
iv) The lowest quantized energy of a particle of mass ‘m’ in a box of length ‘L’ is given by
h2
2h2
2h2
A) Zero
B) 8mL2
C) 8mL2
D) 8mL2
(04 Marks)
b. Explain Heisenberg’s uncertainty principle. Give its physical significance. (06 Marks)
c. Set up time independent one dimensional Schrödinger wave equation. (06 Marks)
d. A quantum particle confined to one dimensional box of width ‘a’ is in its first excited state. What is the

probability of finding the particle over an interval of ( 2 ) marked symmetrically at the centre of the box?
(04 Marks)
Dec 09/ Jan 10
2 a i) The normalization of wavefunction is always possible, if

C) 
A)






 * dx = infinite
B)
 * dx =0
D) All of those

 * dx = finite
ii) Schrödinger’s wave equation is applicable for the particles with
A) Constant energy B) Variable energy
C) Only constant potential energy
D) All of these
iii) The ground state energy of an electron in an infinite well is 5.6 meV. If the width of the well is
doubled, the ground state energy is
A) 9.92×10-23 J
B) 4.48× 10-22 J
C) 2.24×10-22 J
D) None of these
iv) The wavefunction is acceptable if it is
A) Finite everywhere
B) Continuous everywhere
C) Single valued everywhere
D) All of these
(04 Marks)
b. State Heisenberg’s uncertainty principle and discuss its physical significance. (06 Marks)
c. Solve the Schrodinger’s wave equation for allowed energy values in case of a particle in a potential box.
(10 Marks)
May/June 2010
2 a i) If free electron exists in a nucleus, its energy value must have a minimum energy of about
A) 4 MeV
B) 20 MeV
C) 20 KeV
D) 10 KeV
2
ii) According to Max Born approximation,  represents
A) Energy density
B) Particle density
C) Probability density
D) Charge density
iii) If E1 is the energy of the lowest state of a one dimensional potential box of length ‘a’ and E2 is the
energy of the lowest state when the length of the box is halved, then
A) E2 = E1
B) E2 = 2E2
C) E2 = E1/2
D) E2 = 4E1
iv) The wavefunction for the motion of particles in one dimensional potential box of length ‘a’ is given by
n = A sin nx/a where ‘A’ is normalization constant. The value of ‘A’ is
A)
1
√
B)
2
√
2
C) √

D) √2
(04 Marks)
b. State and explain Heisenberg’s uncertainty principle and prove that nuclei do not contain electron.
(08 marks)
c. Discuss the wavefunctions and probability density for particle in an infinite potential well for first two
states.
(04 Marks)
d. An electron is bound in one dimensional potential well of width 0.18 nm. Find the energy value in eV of
the second excited state.
(04 Marks)
Notes Compiled by: Dr. Santhosh D Shenoy, M.Sc., Ph.D.
www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS
www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS
January 2011
2 a i) The uncertainty in the determination of position of an electron is h/3. Then, the uncertainty in the
determination of its momentum is
A) ¾
B) ¼
C) 4/3
D) 3
ii) The probability of locating a particle is maximum
A) at the centre of the wavepacket
B) at the nodes of the wavepacket
C) cannot be determined
D) none of these
iii) In Davisson and Germer experiment, when 54 volts was applied to electrons, the pronounced
scattering direction was found to be at
A) 900
B) 1200
C) 500
D) none of these
iv) The ground state energy of an electron in an one dimensional infinite potential well of width 2Å is 16
eV. Its energy in third excited state is
A) 32 eV
B) 64 eV
C) 144 eV
D) 256 eV
(04 Marks)
b. State and explain Heisenberg’s uncertainty principle
(04 Marks)
c. Find the eigen value and eigen functions for an electron in one dimensional potential well of infinite
height.
(08 Marks)
d. Estimate the time spent by an atom in the excited state during the excitation and de-excitation processes,
when a spectral line of wavelength 546 nm and width 10-14 m is emitted.
(04 marks)
June/July 2011
2 a. i) An electron is moving in a box of length ‘a’. If ‘1’ is the wavefunction at x = a/4 with n = 1 and 2
at x = a for n = 2, then 2/1 is
A)
√2
a
a
B) √2
C) 0
D) ∝
ii) For a particle in an infinite potential well in its 1st excited state, the probability of finding the particle
at the center of box is
A) 0
B) 0.25
C) 0.5
D) 0.1
iii) To become a nuclear constituent, the K.R of e must be of the order of
A) 20MeV
B) 2MeV
C) 20eV
D) Zero
iv)
An electron has speed of 100ms-1 accurate to 0.05%. The uncertainty in its position is
A) 0.01m
B) 0.0115m
C) 0.024m
D) 0.04m
(04 Marks)
b. What is a wave function? Explain the properties of a wave function.
(04 Marks)
c. Derive the expression for energy Eigen value for an electron in potential well of infinite depth.
(06 Marks)
d. A quantum particle confined to one dimensional box of width ‘a’ is in its first excited state. What is the

probability of finding the particle over an interval of ( 2 ) marked symmetrically at the centre of box.
(06 marks)
**************
Notes Compiled by: Dr. Santhosh D Shenoy, M.Sc., Ph.D.
www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS
```