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2020 - CONCISE QUANTUM MECHANICS
CONCISE QUANTUMMECHANICS
[FOR MSc PHYSICS & MSc CHEMISTRY STUDENTS OF ALL INDIAN UNIVERSITIES]
BY
Dr. C.SEBASTIAN ANTONY SELVANASSISTANT PROFESSOR
R. V. GOVT.ARTS COLLEGECHENGALPATTU
9444040115DEC 2020
Web: http://www.csaslearningcenter.com/
1. CLASSICAL MECHANICS
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Planc’s Quantum Theory Of Radiation
Photo Electric Effect Compton Effect: Inadequacy Of Classical Theory(Failure Of Classical Mechanics): Black Body Radiation Stephen – Boltzman LawOrigin Of Hydrogen Spectrum De-Brogilie Equation ( Wave Particle Duality) Hisenberg Uncertainty Principle Zeeman EffectBohr’s Correspondence Principle Quantum Tunnelling
3INTRODUCTION TO QUANTUM CHEMISTRY
3.1 Postulates of quantum mechanics:
3.2 Operators - position ,differential, momentum, kinetic energy, total energy Hamiltonian operatorVector operator, Laplacian operator,Angular momentum operator3.3 Operator algebra-Addition and subtraction of operators, multiplication of operators, linear property . commutative property Hermitian property3.4 eigen functions and eigen values
4. SCHRODINGER EQUATION &
ITS APPICATIONS TO SIMPLE SYSTEMS
4.1 Schrodinger Wave Equation (Time independent Time dependent)4.2 Applications of Schrodinger wave equation:4.2.1 .particle in a one 4.2.2 particle in a two 4.2.3 particle in a three dimensional box4.2.4 , particle in a ring
5. SCHRODINGER EQUATION APPLICABLE TO COMPLEX SYSTEMS
5.1 Harmonic oscillator( vibrational motion)
5.2 Rigid rotator with free axis
5.3 Hydrogen atom
6 APPROXIMATION 6.1 Perturbation method application to hydrogen and
helium atom
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METHODS 6.2 Variation method -Secular equation:
Application to particle in one dimensional box,
hydrogen and helium atom
6.3 Born – Oppenheimer Approximation
6.4 Russell – Saunders Coupling ( LS –Coupling)
6.5 Huckel Molecular Orbital Theory- Application to
ethylene, allyl systems ,butadiene and benzene
7. CHEMICAL BONDING 7.1 Hybridisation sp , sp 2 and .sp3 hybridization
7.2 Slater Type Orbitals and Slater Rules
7.3 Linear Combination Of Atomic Orbital- Molecular
Orbital (LCAO-MO)-treatment of diatomic molecules
hydrogen molecule ion, hydrogen molecule
7.4 Valence Bond Theory For Hydrogen Molecule
(Heitler – London Theory)
7.5 Hartree – Fock Self Consistent Field Method.
(HFSCF –Method)
7.6 Semi Emprical Methods
7.7 Paulis Anti -symmetric Principle
8 APPENDIX
I CLASSICAL MECHANICSCHAPTER -1- CLASSICAL MECHANICS
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1 Planc’s Quantum Theory Of Radiation
2 Compton Effect
3 Wave Particle Duality (De-Brogilie Equation)
5 Inadequacy Of Classical Theory(Failure Of Classical Mechanics):
6 Black Body Radiation
7 Stephen – Boltzman Law
8 Origin Of Hydrogen Spectrum
9 Photo Electric Effect
10 Hisenberg Uncertainty Principle
11 Zeeman Effect
12 Bohr’s Correspondence Principle
13 Quantum Tunnelling
113PLANC’S (1900) QUANTUM THEORY OF RADIATION
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[ Max Planck was a German theoretical physicist who discovered the quantum of action, now known
as Planck's constant, h, in 1900. This work laid the foundation for quantum theory, which won him the Nobel Prize for Physics in 1918.]
According to this theory,
1. A black body can not absorb or emit energy in a continuous manner. It can absorb or emit energy in the
multiples of small units called quanta
Thus light radiations obtained from excited molecules are not continuous waves.
2. Each quanta is associated with definite amount of energy given by
E= hϑ Where ϑ - frequency of radiation h - Planc’s constant = 6.62 × 10 -34 JS
3. An atom or molecule can emit or absorb energy only in whole number multiples of quantum
1hϑ , 2hϑ , 3hϑ ....
4.It can never emit or absorb 1.5 hϑ , 2.4 hϑ , 3.2 hϑ ...
5. The energy density in the wavelength range λ and λ + dλ is given by
E = 8 π hc
λ5 (¿ eh ϑKT−1 )¿
dλ
6. The energy density in terms of frequency in the range ϑ and ϑ + dϑ is given by
E = 8 π hc
λ5 (¿ eh ϑKT−1 )¿
ϑ= cλ
λ ¿ cϑ
dλ = - cϑ2 dϑ
substituting in the above equation we get
E = 8 π hc
( cϑ)
5 (¿ehϑKT−1)
¿ ×(-
cϑ2 )dϑ
= - 8 π ϑ 3
c3 × h
(eh ϑKT−1)
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For long wavelengths:
λ is very high ehϑKT = 1+ hϑ
KT . Therefore the above equation becomes
E= 8πhc
λ5(1+ hϑKT−1)
= 8 πhc
λ5( hϑKT )
= 8 πhcKT
λ5h ( cλ)
E = 8πKT
λ4 which is Rayleigh – Jeans law
[ in 1900, the British physicist Lord Rayleigh derived the λ−4 dependence of the Rayleigh–Jeans law based on
classical physical arguments and empirical facts A more complete derivation, which included the proportionality
constant, was presented by Rayleigh and Sir James Jeans in 1905].
In terms of frequency
E = 8 πhc
( cϑ)
5
( hϑKT ) ×(-
cϑ2 )dϑ
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E = −8 πKT ϑ 2
c3
For short wavelengths:
λ is very low and ϑ = cλ is high e
hϑKT is also very high when compared to 1
. Therefore the above equation becomes
E= 8πhc
λ5 ehϑKT
Which is Wein’s law[ Wien's approximation (also sometimes called Wien's law or the Wien distribution law) is a law of physics used
to describe the spectrum of thermal radiation (frequently called the blackbody function). This law was first derived
by Wilhelm Wien in 1896. The equation does accurately describe the short wavelength (high frequency) spectrum
of thermal emission from objects, but it fails to accurately fit the experimental data for long wavelengths (low
frequency) emission]
PLANCK’S RADIATION LAW ( Application of BE statistics to photon gas )
Planck's law describes the spectral density of electromagnetic radiation emitted by a black body in thermal
equilibrium at a given temperature T.
Eϒ dγ = 8 πh γ3
c3 d γ
(e hγKT−1)
This equation is known as Planck’ energy distribution law in terms of 𝛄.
DERIVATION:
An assembly of bosons ( indistinguishable particles with zero or integral spin ) is termed as photon gas.
Consider an enclosure containing electromagnetic radiation. If the enclosure is maintained at
temperature T , it will emit and reabsorb photons. After certain time there will be a thermodynamic
equilibrium.. This electromagnetic radiation within the enclosure is called black body radiation.
Energy of one photon = hγ
If there are ‘n’ photons , total energy = n hγ
Iv ‘V represents the volume of the container then Energy density = n h γV
Energy density in the given frequency interval between γ and γ + d γ , in volume V is given by
Eϒ dγ = h γ nV -----------------------------1
h - planc constant, γ – frequency, V – total volume, n- number of photonsPage 7 of 419
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According to Bose – Einstein distribution law,
n = g
eα+β ∈−1 ------------------------------2
where g is the number of degeneracies
Photons are indistinguishable from one another. Every process of such emission creates photons and
every process of absorption results in the annihilation of photons. In this condition their number in the system
is not constant. ∑ dni = 0 is invalid .Therefore the undetermined multiplier α = 0 , put β = 1
KT
n = g
e∈
KT−1 ---------------------------------3
To find the number of degeneracies( g)
number of eigen states (g)= Degeneracy × V shell
V element -------------------4
where V shell – volume of shell, V element - volume of element
= ( 2S +1) × V shell
V element [ Degeneracy = 2S +1]
= ( 2 (12 ) +1) ×
V shell
V element [For photon s = ½ ]
g = ( 2) × V shell
V element ----------------5
According to quantum idea , the element of volume in the momentum space
V element = h3
V
The volume of the shell having their momenta between p and p+dp
= 43 π ( p + dp ) 3 -
43 π ( p ) 3
= 43 π [ p 3 + 3 p2 dp + 3 p (dp) 2 + (dp ) 3 ] -
43 π p 3
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= 43 π [ p 3 + 3 p2 dp + 3 p (dp) 2 + (dp ) 3 - p 3 ]
= 43 π [ 3 p2 dp + 3 p (dp) 2 + (dp) 3 ]
= 43 π [ 3 p2 dp ] [ neglecting the higher terms of dp]
V shell = 4 π p2 dp ---------------------------6
For a photon, p = hλ ---------------------7
= hγc ------------------8
Differentiating
∴ dp = hdγc -----------------------9
Substituting in equation 5 we get
number of eigen states (g) = 2 × 4 π ( hγ
c)
2
h3
V
×hdγc [above equation]
g = 8π V γ 2
c3 d γ ------------------------10
Substituting in 3 we get
n =
8 π V γ 2c3
e h γKT−1
d γ
Substituting the value of ‘n’ in 1 we get
Eϒ dγ = hγV ×
8 π V γ 2c3
e h γKT−1
d γ
= 8 πh γ3
c3 d γ
(e hγKT−1)
This equation is known as Planck’ energy distribution law in terms of 𝛄.
In terms of λ,
put γ = cƛ
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∴ dγ = - ( cλ2 ) dƛ
Energy density in terms of λ isE λ d λ = 8πh( c
ƛ)
3
c3
1
e h γKT−1 × - (
cλ2 ) dƛ
= - 8πhC
λ5 1
e hCƛ KT
−1 dƛ
This equation is known as Planck’ energy distribution law in terms of λ,
213 COMPTON EFFECT:
The Compton effect was observed by Arthur Holly Compton in 1923 at Washington University in St. Louis and
further verified by his graduate student Y. H. Woo in the years following. Compton earned the 1927 Nobel Prize in Physics for the discovery.
When X –rays are allowed to fall on light element, the scattered X-rays have wavelengths larger than the incident rays. This increase in wavelength of X- rays after scattering from the surface of an object is known as Compton effect.
Compton effect is conceived to be the result of the collision between the incident photon and the
electrons of the scatterer. If a photon collides with an electron at rest a part of energy of photon will be
imparted to the electron.therefore the energy of photon decreases and hence its wavelength increases. The
photon after collision travels in some other direction. The angle ' θ ' made by the direction of motion after
collision with its direction before collision is called angle of scattering. The electron having obtained some
energy during collision escapes the scatterer. It is called recoil electron.
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The increase in wavelength is called Compton shift which is given by
∆λ = 2 hmc sin 2 θ/2 This can also be written as ∆λ =
hmc ( 1- cos θ )
Where m is the mass of electron, c is the velocity of light, θ is the angle between the incident and scattered X- rays.
EXPRESSION FOR COMPTON SHIFT- DERIVATION:
It is derived on the basis of
1. Conservation of energy
2. Conservation of momentum and
3. Theory of relativity
Let ‘p’ denotes momentum Applying Law of conservation of momentum to X-axis
Momentum before collision = Momentum after collision
1. X-axis
before collision = hϑc
after collision = hϑ '
c cos θ + p cos φ
According to Law of conservation of momentum
hϑc =
hϑ 'c cos θ + p cos φ ----------------------1
Rearranging
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p cos φ = hϑc -
hϑ 'c cos θ -------------------------2
Squaring the above equation
( pcosφ)2 = ( hϑc−h ϑ '
ccosθ)
2
-----------------3
Applying Law of conservation of momentum to Y-axis
2.Y-axis
before collision = 0
after collision = hϑ '
csinθ−p sin φ
According to Law of conservation of momentum
0 = hϑ '
c sin θ - p sin φ ---------------------4
Rearranging
p sin φ = hϑ '
c sin θ -------------------5
Squaring the above equation
( p sin φ)2 = ( hϑ '
csin θ)
2
--------------------------6
adding 3 and 6 we get
( pcosφ)2 + ( p sin φ)2 = ( hϑc−hϑ '
ccosθ)
2
+ ( h ϑ '
csin θ)
2
p2 cos2 φ + p2 sin2 φ = ( hϑc−hϑ '
ccosθ)
2
+ ( hϑ '
csin θ)
2
p2 ¿¿ + sin2 φ¿ = ( hϑc−h ϑ '
ccosθ)
2
+ ( h ϑ '
csin θ)
2
¿¿ + sin2 φ = 1]
p2 = ( hϑc−h ϑ '
ccosθ)
2
+ ( h ϑ '
csin θ)
2
------------------------6a ¿ = a2 - 2ab + b2]
= ( hϑc )
2
+ ( h ϑ '
ccosθ)
2
- 2 ( hϑc
× hϑ '
ccosθ ¿+ ( h ϑ '
csin θ)
2
-------------7
Multiplying by ‘c2’
p2 c2 = (hϑ )2 + (hϑ ' cosθ )2 - 2 (hϑhϑ ' cosθ ¿+ (hϑ ' sin θ )2 --------------8a
= h2 [ϑ2 + (ϑ ' cosθ )2 - 2 (ϑϑ ' cosθ ¿+ (ϑ ' sin θ )2 ----------------8
= h2 [ϑ2 + (ϑ ' )2[cos¿¿2θ ]¿- 2 (ϑϑ ' cosθ ¿+ (ϑ ' )2[sin¿¿2 θ]¿
= h2 [ϑ2 - 2 (ϑϑ ' cosθ ¿+ (ϑ ' )2[cos¿¿2θ+sin2θ ]¿ ¿¿ + sin2 φ = 1]
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= h2 [ϑ2 - 2 ϑϑ ' cosθ + ϑ ' 2 ] ---------------9
By a relativistic formula the kinetic energy ‘T’ of the recoil electron is given by
p2 c2 = T 2+2T m0 c2 ---------------------10
Comparing 9 and 10 we get
T 2+2T m0 c2 = h2 [ϑ2 + ϑ ' 2 - 2 ϑϑ ' cosθ ¿ ------------------12
Applying Law of conservation of energy
T = hϑ - hϑ ' ---------------14
Substituting in 12 we get
¿ +2( hϑ – hϑ ’) m0 c2 = h2 [ϑ2 + ϑ ' 2 - 2 ϑϑ ' cosθ ¿ [ (a−b )2 = a2 - 2ab + b2]
h2 [ϑ2 + ϑ ' 2 - 2 ϑϑ ' ¿ +2h( ϑ – ϑ ’) m0 c2 = h2 [ϑ2 + ϑ ' 2 - 2 ϑϑ ' cosθ ¿
h2 ϑ2 +h2 ϑ ' 2 - 2h2 ϑϑ ' +2h( ϑ – ϑ ’) m0 c2 = h2 ϑ2 +h2 ϑ ' 2 - 2 h2 ϑϑ ' cosθ¿
- 2h2 ϑϑ ' +2h( ϑ – ϑ ’) m0 c2 = - 2 h2 ϑϑ ' cosθ
Dividing by (-2h2 )
ϑϑ ' - 1h ( ϑ – ϑ ’) m0 c2 = ϑϑ ' cosθ
Rearranging
ϑϑ ' - ϑϑ ' cosθ = m0c2
h( ϑ – ϑ ’)
Dividing by ‘ϑϑ ' '
1 - cosθ = m0c2
h (ϑ – ϑ ’)
ϑϑ '
Rearranging
h
m0c2 (1- cosθ¿ = (ϑ – ϑ ’)
ϑϑ '
(ϑ – ϑ ’)
ϑϑ ' =
hm0c2 (1- cosθ¿
ϑ
ϑϑ ' - ϑ ’ϑϑ ' =
hm0c2 (1- cosθ ¿
1ϑ ' -
1ϑ =
hm0c2 ( 1-cosθ )
Multiplying by ‘c’
cϑ '−
cϑ =
hm0c ( 1-cosθ )
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λ ' - λ = h
m0c ( 1-cosθ )
dλ = h
m0c ( 1-cosθ )
This is the expression for Compton shift.
ALITER
EXPRESSION FOR COMPTON WIDTH
It is derived on the basis of conservation of energy and momentum.
Energy before collision = Energy of electron+ Energy of photon
Energy before collision = m0 c2 + hϑ
Energy after collision = m c2 + hϑ '
According to Law of conservation of energy,
Energy before collision = Energy after collision
m0 c2 + hϑ = mc2 + hϑ '
mc2 = m0 c2 + hϑ - hϑ '
mc2 = m0 c2 + h¿ - ϑ ' ) ---------------------------1
Momentum along X-axis before collision = Momentum along X-axis of photon + Momentum along X-axis of
electron
Momentum along X-axis before collision = hϑc + 0
Momentum along X-axis after collision = hϑ 'c cos θ + mv cos φ
According to Law of conservation of momentum,
momentum before collision = momentum after collision
hϑc =
hϑ 'c cos θ + mv cos φ
Multiplying by ‘c’
hϑ = hϑ ' cos θ + mvc cos φ
Rearranging ,
mvc cos φ = hϑ - hϑ ' cos θ
mvc cos φ = h¿ - ϑ ' cos θ ) -----------------2
Momentum along Y-axis before collision = Momentum along Y-axis of photon + Momentum along Y-axis of
electron
Momentum along Y-axis before collision = 0 + 0Page 14 of 419
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Momentum along X-axis after collision = hϑ 'c sin θ - mv sin φ
According to Law of conservation of momentum,
0 = hϑ '
c sinθ - mv sin φ
Multiplying by ‘c’
0 = hϑ ' sin θ - mvc sin φ
Rearranging ,
mvc sin φ = hϑ ' sin θ -------------------3
squaring and adding equation 2 and 3
m2 v2 c2 = h2 (ϑ−ϑ ' cosθ)2 +(hϑ ' sin θ)2
= h2 [ (ϑ−ϑ ' cosθ)2 +(ϑ ' sin θ)2 ]
m2 v2 c2 = h2 [ϑ2 +ϑ ' 2 - 2 ϑ ϑ ' cosθ ] -----------------------4
From equation 1 , mc2 = m0 c2 + h¿ - ϑ ' )
squaring on both sides
m2 c4 = [m0 c2+h (ϑ−ϑ ' ) ]2 = mo
2 c4 + h2 [ϑ2 +ϑ ' 2 - 2 ϑ ϑ ' ] +2 m0 c2h (ϑ−ϑ ' )----------5
Equation 5 – 4 gives
m2 c4 - m2 v2 c2 = mo2 c4 + h2 [ϑ2 +ϑ ' 2 - 2 ϑ ϑ ' ] +2 m0 c2h (ϑ−ϑ ' ) - h2 [ϑ2 +ϑ ' 2 - 2 ϑ ϑ ' cosθ ]
m2 c2 ( c2- v2 ) = mo2 c4 - 2h2 ϑ ϑ ' +2 m0 c2h (ϑ−ϑ ' ) + 2h2 ϑ ϑ ' cosθ ]
= mo2 c4 - 2h2 ϑ ϑ ' ( 1-cosθ ) +2 m0 c2h (ϑ−ϑ ' )
From the theory of relativity the variation of mass with velocity is
m2 c2 ( c2- v2 ) = mo2 c4
Substituting in the above equation and cancelling the common term we get
2h2 ϑ ϑ ' ( 1-cosθ ) = 2 m0 c2h (ϑ−ϑ ' )
Dividing by ‘2h’ we get
hϑ ϑ ' ( 1-cosθ ) = m0 c2 (ϑ−ϑ ' )
rearranging
(ϑ−ϑ ' )ϑ ϑ ' =
hm0c2 ( 1-cosθ )
1ϑ -
1ϑ ' =
hm0c2 ( 1-cosθ )
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Multiplying by ‘c’
cϑ -
cϑ ' =
hm0c ( 1-cosθ )
λ - λ ' = h
m0c ( 1-cosθ )
dλ = h
m0c ( 1-cosθ )
This is the expression for Compton shift.
ALITER
Expression for Compton shift:
Psin φ = p’ sin θ,-----------------------1
P cos φ = p-p’ cos θ -------------2
Squaring and adding 1 and 2
P2 ( sin2 θ +cos 2θ ) = p’2sin2 θ+( p-p’cosθ )2
Expanding P2 = p’2sin2 θ+ p2 + p’2cos 2θ - 2 pp’cosθ [sin2 θ +cos 2θ = 1]
= p’2 ( sin2 θ +cos 2θ ) + p2 - 2 pp’cosθ
= p’2 + p2 - 2 pp’cosθ ---------------3
By energy conservation, we have
? cp + mc2 = cp’ + ( m2c4 + c2p2 ) ½
rearranging cp - cp’ + mc2 = ( m2c4 + c2 P2 ) ½
c (p - p’) + mc2 = ( m2c4 + c2P2 ) ½
Squaring on both side ( c (p-p’) + mc2 ) 2 = ( m2c4 + c2P2 )
Expanding ( c2 (p-p’) 2 + m2c4 + 2 c (p-p’) mc2 = ( m2c4 + c2P2 ) -------4
Using 3 , (c2 (p-p’)2+ m2c4 +2c(p-p’) mc2 = (m2c4 +c2 (p’2 + p2 - 2 pp’cosθ)
Expanding c2 p 2 + c2 p’ 2 - 2c2pp’ + m2c4 + 2 p mc3 - 2 p’ mc3
= ( m2c4 + c2 p’2 + c2 p2 - 2 c2 pp’cosθ)
- 2c2pp’ + 2 p mc3 - 2 p’ mc3 = - 2 c2 pp’cosθ
2 p mc3 - 2 p’ mc3 = 2c2pp’ - 2 c2 pp’cosθ
2 mc3 ( p- p’ ) = 2c2 pp’ ( 1 - cosθ)
mc ( p- p’ ) = pp’ (1 - cosθ) [ divide by 2c2]
dividing by pp’ mc ( 1p '−1
p ) = (1 - cos θ)
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λ = hp ∴ p =
hλ substituting in the above equation
mc ( λ 'h− λ
h ) = (1 - cos θ)
mch ( λ’ – λ ) = (1 - cos θ)
λ’ – λ = h
mc (1 - cos θ)
This is the expression for Compton shift
λ’ - λ = ∆λ =2 hmc
sin 2 θ2
=h
mc ( 1- cos θ )
Where m is the mass of electron, c is the velocity of light, θ is the angle between the incident and
scattered X- rays.
Case 1: when θ = 0
∆λ =h
mc ( 1- cos 0 )
= h
mc ( 1- 1)
= 0
i.e there is no change in wave length.
Case 2: when θ = 90
∆λ =h
mc ( 1- cos 90 )
= h
mc ( 1- 0 )
= h
mc
= 0.0242 A
Case 3 : when θ = 180
∆λ =h
mc ( 1- cos 180 )
= h
mc ( 1- (-1) )
= 2hmc
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= 0.0484 A
Problem1 When x-rays are scattered through 90°, the Compton shift was found to be 0.0242, if the scattering
angle is 60°, find the Compton shift
Solution:
When θ = 90, ∆λ = 0.024
0.024 =h
mc ( 1- cos 90 )
0.024 = h
mc ( 1- 0 )
hmc = 0.024
When θ = 60, ∆λ = ?
∆λ=h
mc ( 1- cos 60 )
= h
mc ( 1- ½ )
= h
2mc
= 0.024
2
= 0.012
Problem2 .When x-rays are scattered through certain angle , the Compton shift was found to be 0.048, if the
value of h
mc is 0.024 the scattering angle will be
Solution:
θ = ?
∆λ = 0.048
h
mc = 0.024
0.048 =h
mc ( 1- cos θ )
0.048 = 0.024 ( 1- cos θ )
0.0480.024 = ( 1- cos θ )
2 = ( 1- cos θ )
cos θ = 1-2
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= -1
θ = 180
Problem3 When x-rays of 5000A was scattered by tungsten plate the Scattered rays have wavelength 5002
A. Find the Compton shift
Solution:
∆λ=λ’- λ
= 5002 – 5000
= 2
Problem4 The Compton shift of an electron was found to be equal to twice the value of h
mc . Find the
scattering angle
Solution:
Given ∆λ=2 hmc
∆λ=h
mc ( 1- cos θ )
2 hmc =
hmc ( 1- cosθ )
1- cos θ =2
cos θ = 1-2
= -1
∴θ = 180
. Problem5 When x-rays are scattered through certain angle , the Compton shift was found to be 0.0242, the
scattering angle will be [h
mc = 0.024 ×10−11 m ]
Solution:
Given ∆λ= 0.0242
0.0242 = h
mc ( 1- cosθ )
= 0.24 ×10−11( 1- cosθ )
= 0.24 ×10−1( 1- cosθ ) A
0.0242 = 0.0242 ( 1- cosθ ) A
1- cosθ = 1
cosθ = 0
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Problem 6When x-rays are scattered through 90°, the Compton shift was found to be 0.0242, if the scattering
angle is 30°, find the Compton shift
Solution:
When θ = 90, ∆λ = 0.024
0.024 =h
mc ( 1- cos 90 )
0.024 = h
mc ( 1- 0 )
hmc = 0.024
When θ = 30, ∆λ = ?
∆λ=h
mc ( 1- cos 30 )
= h
mc ( 1- √32
)
= h
mc ( 1- 1.732
2 )
= h
mc ( 1- 0. 866 )
= h
mc ( 0.134 )
= 0.0242× 0.134
= 0.00324
EXPERIMENTAL VERIFICATION OF COMPTON EFFECT:
A beam of monochromatic X-rays of wave length λ is allowed to fall on a scattering material. The
scattered X- rays are received by a Bragg spectrometer.
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Diagram
The intensity of scattered X-rays is measured for various scattering angle. A graph is plotted as intensity versus
wavelength. It is found that the curves have two peaks one corresponding to unmodified radiation and the
other corresponding to scattered radiation. The difference between the two peaks on the wavelength axis
gives the Compton shift.
The change in wave length 0.0243A at 𝛉 = 90 is found to be good agreement with the theoretical value . thus
Compton effect is experimentally verified.
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Problem:X-rays of 1 o A are scattered from a carbon block. Find the wavelength of the scattered beam in a
direction making 90 with the incident beam
2.1. ORIGIN OF HYDROGEN SPECTRUM.
According to Bhor, the spectrum arises when the electron in the higher stationary orbit to the lower stationary orbit. The difference of energies associated with these orbitals is emitted as photon of frequency ϑ .
ORIGIN OF HYDROGEN SPECTRUM.
According to Bhor, the spectrum arises when the electron in the higher stationary orbit to the lower stationary orbit. The difference of energies associated with these orbitals is emitted as photon of frequency ϑ . Accordingly
ϑ = RH [ 1n f
2−1ni
2 ] where RH is Rydeberg’s constant. Using this equation, the wave numbers of photons of the
various spectral series are as follows.
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In these series, the spectral lines corresponds to the transition of the electrons from some higher energy to the first state.
ϑ = RH [ 112−
1ni
2 ] ni = 2,3,4... . These lines are found to be in the ultra violet region of electro magnetic
spectrum.
2. Balmer series
In these series, the spectral lines corresponds to the transition of the electrons from some higher energy to the second state.
ϑ = RH [ 122−
1ni
2 ] ni = ,3,4... . These lines are found to be in the visible region of electro magnetic spectrum.
3.Paschen series
ϑ = RH [ 132−
1ni
2 ] ni = 4,5,6... . These lines are found to be in the infra red region of electro magnetic
spectrum.
4.Bracket series.
ϑ = RH [ 142−
1ni
2 ] ni = 5,6,7... . These lines are found to be in the infra red region of electro magnetic
spectrum.
5.Pfund series:
ϑ = RH [ 152−
1ni
2 ] ni = 6,7,8... . These lines are found to be in the infra red region of electro magnetic
spectrum.
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2.2 PHOTO ELECTRIC EFFECT:
[ The photoelectric effect was discovered in 1887 by the German physicist Heinrich Rudolf Hertz. In
connection with work on radio waves, Hertz observed that, when ultraviolet light shines on two metal
electrodes with a voltage applied across them, the light changes the voltage at which sparking takes place]
Einstein was awarded the 1921 Nobel Prize in Physics for "his discovery of the law of the photoelectric effect"
and Robert Millikan was awarded the Nobel Prize in 1923 for "his work on the elementary charge of electricity and on the photoelectric effect".
Definition:
When light of certain frequency , strikes the surface of a metal, electrons are ejected from the metal.
This phenomenon is known as photo electric effect. The ejected electrons are called photo electrons
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Experimental set – up:
The apparatus consists of two photo sensitive surfaces A and C enclosed in an evacuated quartz bulb.
The plate C is connected to the negative terminal of a Battery while the plate A is connected to the positive
terminal through a Ammeter A .In the absence of light ,no current flows through the Ammeter . But when
monochromatic light is allowed to fall on plate C , a current starts flowing in the circuit, which is indicated by
the Ammeter This current is known as photo current
STOPPING POTENTIAL:
The voltage required to stop electrons from moving between plates and creating a current in the
photoelectric experiment is known as stopping voltage (or stopping potential)
1. Stopping potential does not depend on the intensity of incident radiation.
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For three different intensities ( I 1 , I 2 , I3) a graph is drawn between anode potential and photo electric
current In all the cases it was found out that the Stopping potential does not depend on the intensity of
incident radiation.
2. Higher the frequency of incident light higher the value of stopping potential
For three different frequencies ( ϑ1 , ϑ2 , ϑ3) a graph is drawn between anode potential and photo
electric current .It was found out that when frequency of incident light increases ( ϑ 1 to ϑ 3) the stopping
potential increases.
Einstien’s photo electric equation
The kinetic energy of emitted electrons is given by = h ϑ−hϑ 0. Where ϑ - frequency of light falling on
the metal. hϑ 0 is the threshold energy or work function. This equation is known as Einstien’s photo electric
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1. In an experiment tungsten cathode which has a threshold wavelength of 2400 A is irradiated by UV light of wavelength 1200 A. Calculate the maximum energy of emitted photo electrons
Solution: Wave length of incident radiation λ = 1200 AThreshold wavelength λ0 = 2400 A Energy E = h ϑ−h ϑ 0
= h ( cλ -
cλ0
)
= h ×c ×( 1λ -
1λ0
)
= 6.62 ×10−34 ×3× 108 ( 1
1200− 1
2400 )
= 19 .86 ×10−26× 0.5 = 9.93 ×10−26 J
2. Calculate the velocity of photo electron if the work function of the target material is 1.24 eV and the wavelength of incident light is 400 A
Solution:
Wave length of incident radiation λ = 400 AWork function of the target material(h ϑ 0 ) = 1.24 eV = 1.24 × 1.6 ×10−19 J
½ m V2 = h ϑ−h ϑ 0
∴ V2 = 2m (
hcλ - 1.24× 1.6 ×10−19 )
= 2
9.11×10−31 ¿ - 1.24×1.6 ×10−19
=
V = 7.43 × 10 5 m/s
In the photoelectric effect, the maximum kinetic energy of electrons emitted from a metal is 1.6 × 10-19 J, when
the frequency of radiation is 7.5 × 1014 Hz. Calculate the threshold frequency of the metal and stopping
potential of the electrons.
Solution:
qV = Kmax
where V is stopping potential, qe is the charge on the electron
Kmax = h ( f – f0)
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where h is the Planck constant and ϑ f is the frequency of the incident photon, f0 is the threshold frequency
Kmax = 1.6 × 10-19 J,
f = 7.5 × 1014 Hz
Kmax = h ( f – f0)
( f – f0) = Kh
threshold frequency f0 = f - Kh
= 7.5 × 1014 - 1.6× 10−19
6.62× 10−34
stopping potential of the electron V = Kq
= 1.6 × 10−19
1.602×10−19
= 1 volt
INADEQUACY OF CLASSICAL THEORY
Classical mechanics explains satisfactorily, the motion of objects which can be observed by microscope
but it cannot be applied to atomic phenomena( motion of electron)
a.Classical mechanics does not explain stability of atom.
Rutherford stated that atom consists of positive charged nucleus at the centre and electrons are revolving
around the nucleus. But according to Maxwells theory, revolving electrons radiate energy and hence the atom
will come closer and closer to the nucleus and finally it collapses with the nucleus.
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This is contradictory to the observed fact of stability of atoms.
b. Classical mechanics could not explain the spectrum of hydrogen.
According to classical mechanics, the spectrum of hydrogen will be continuous but it is found out that
the hydrogen spectrum consists of discrete set of lines given by
c. Classical mechanics and black body radiation:
Rayleigh Jeans radiation law derived from classical considerations, stated that the energy emitted by a
black body, increases with frequency and becomes infinite at large frequencies..
But experimentally observed radiation curve is in disagreement with this conclusion
d. Difficulties with classical theory of specific heat of solids:
According to classical mechanics, the specific heat of all solids should be constant and is independent
of temperature. Debye ‘s theory explains, the observed variation of specific heat of solids.
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2.5. BLACK BODY RADIATION
Any object, which absorbs the radiations of all wavelength , incident on it is called black body and when such a body is placed inside an isothermal enclosure at high temperature , it emits radiations of all wavelengths.
This heat radiation emitted by a black body is known as black body radiation.
Energy distribution:
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1. The energy is not uniformly distributed2. At a given temperature , the intensity of radiations increases with increase in wavelength to a
maximum and then decreases.3. With increase in temperature , the wavelength at which maximum emission of energy decreases..
2.6 STEPHEN –BOLTZMAN LAW
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Laws of black body radiation:
2.7 PLANC’S QUANTUM THEORY OF RADIATION
1. PLANC’S QUANTUM THEORY OF RADIATION
The foundation of Planc’s theory is theory of black body radiation
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According to Planc the quantity ,energy is quantised . If energy is quantized then frequency and wavelength
is also quantized
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2.8. de-BROGILIE EQUATION ( WAVE PARTICLE DUALITY)
de –Broglie suggested that electron has a dual character i.e it behaves as particles as well as wave.
Einstein’s mass energy relation is E = mc2 -----------1
But E = h ϑ -----------------------2
From 1 and 2
h ϑ = mc2 -------------------3
but ϑ = cƛ substituting in the above we get,
h cƛ = mc2
therefore λ = h
mc
replacing c by the velocity of electron u we get
λ = h
mu
= hp
where p is the momentum of the particle.
This is known as de- Broglie equation.
de-Brogilie equation in terms of energyE = ½ mc2 Multiplying by ‘m’mE = ½ m2c2 rearranging2mE = m2c2
mc = √2mE
substituting in the above equation we get λ = h
√2 mEde Broglie wavelength of an electron
When an electron of mass m and charge e is accelerated through a potential difference V, then the
energy eV is equal to kinetic energy of the electron
. 12 mc2 = eV
. 12 m2c2 = eVm [ multiplying by ‘m’ on both sides]
m2c2 =
mc = √2 eVm
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The de Broglie wavelength is ,
therefore λ = h
√2Vam
Substituting the known values in equation
λ = 12.27√V
If V = 100 volts, then λ = 1.227 Å i.e., the wavelength associated with an electron accelerated by 100 volts is
1.227 Å.
1. de – Broglie wave equation is λ = h
mc
2. if momentum (p) is given λ = hp
3. if .energy is given λ = h
√2mE
4. If potential of (V) is given λ = h
√2 meV
= 6.62× 10−34
√2× 9.11×10−31× 1.602× 10−19× V
= 12.25√V
Significance of de-Broglie waves
The wave nature of matter, however, has no significance for objects of ordinary size because
wavelength of the wave associated with them is too small to be detected. This can be illustrated by the
following examples.
i) Suppose we consider an electron of mass 9.1 10-31 kg and moving with a velocity of 107 ms-1. Its de-Broglie wavelength will be
λ = 7.27 × 10-11 m
This value of λ can be measured by the method similar to that for the determination of wave length of X-rays.
ii) Let us now consider a ball of mass 10-2 kg moving with a velocity of 102 ms-1. Its de-Broglie wave length
will be;
λ = 6.62 × 10 -34 m
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This wavelength is too small to be measured, and hence de-Broglie relation has no significance for such a large
object. Thus, de-Broglie concept is significant only for sub-microscopic objects in the range of atoms,
molecules or smaller sub-atomic particles.
Problem1: Calculate the de-Broglie wavelength of a particle moving with a velocity 6.62 × 10 7 m/sec
whose mass is 0.5 × 10 -27Kg
Solution:
Given v = 6.62 × 10 7 m/sec
Mass m = 0.5 × 10 -27 Kg
λ = h
mu
= 6.62 ×10−34
6.62× 107× 0.5 ×10−27
= 2 × 10 -14 m
Problem2: Find the energy of a particle whose mass 6.62 ×10 -34 Kg , de-Broglie wavelength is 1 A
Solution:
Given mass m = 6.62 × 10 -34 Kg
Wavelength λ = 1 A
= 10−10 m
squaring and rearranging 2mE = h2
λ2
∴ E = h2
2m λ2
= (6.62 ×10−34)2
2× (6.62× 10−34 ) (10−10 )2
= 6.62× 10−34
2 × (10−10)2
= 3.31 ×10−14 J
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Problem3: Find the energy of particle of mass 6.62 10−48 Kg whose de Broglie wavelength is 1A
Solution:
λ = h
√2 mE
Squaring on both sides
λ2 = h2
2mE
E = h2
2m λ2
E = (6.62×10−34 )2
2 (6.62 ×10−48 )(10−10)2
= 6.62 ×10−34
2× 10−14(10−10)2
= 3.31 J
Problem4.: What is the de-Broglie wavelength of an electron which has been accelerated from rest through
a potential difference of 100 V
Solution: λ = 12.25√V
= 12.25√100
= 1.225 A
Experimental verification of de-Broglie equation :(Davisson and Germer experiment The apparatus consists of electron gun where electrons are produced and obtained as fine pencil of
known velocity using slit. The electrons are accelerated in the electric field of known potential difference V .To measure the angular distribution there is an electron detector (Faraday cylinder) which can move on a circular graduated scale between 20 o to 90 o to receive the reflected electrons
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Diagram:
The beam of electrons is directed to fall on a large single crystal of Nickel. The electrons acting like wave are diffracted in different directions.. The Faraday cylinder was moved on the circular scale and for a given accelerating voltage 54 Volts the scattering direction was found to be 50 o .
Since the inter atomic distance for nickel crystal is 2.15 A The inter planar distance = 2.15 sin 25 = 0.09AUsing Bragg’s formula 2 d sin θ = n λ 2 ( 0.09) sin ( 90-25) = 1 λ ∴ λ = 1.65 Awhen an electron of charge ‘e’ is accelerated by a potential of V volts , its
kinetic energy = eV
eV = ½ mc2 Multiplying by ‘m’
meV = ½ m2c2 rearranging
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2meV = m2c2
mc = √2 meV
substituting in the above equation we get λ = h
√2 meV
= 6.62× 10−34
√2 × 9.11×10−31× 1.602× 10−19× V
= 12.25√54
= 1.67 A
As the two values are in good agreement, confirms the de-boglie concept of matter waves.
1. Compute the de-Broglie wavelength of 10 KeV neutron whose mass is 1.675 ×10 -27 Kg
Solution: ½ mv2 = 10 KeV = 10 × 1000 × 1.602 ×10 -19 J = 1.602 ×10 -15 J
∴ v2 = 2× 1.602×10−15
1.675 ×10−27
v = ?
λ = h
mv
= 6.62× 10−34
1.675× 10−27 × ( 2× 1.602×10−15
1.675 ×10−27 ) ½
= 2.86 × 10 -13 m
HISENBERG’S UNCERTAINITY PRINCIPLE
According to this principle, “ it is impossible to specify precisely and simultaneously
Both the position and momentum of a microscopic particle “
It means that if there is accuracy in measuring position there will be uncertainty in measuring its
momentum. If ∆x is the uncertainty in determining the position of the particle. ∆p is the uncertainty in
determining the momentum of the particle then according to this principle
∆x ×∆p = h
Derivation:
Consider the case of measurement of the position and momentum of electron in the microscope. Let a
beam of particle with momentum ‘p’ travelling in the ‘y’ direction. Let the beam fall on a narrow slit behind
this slit there is a photographic plate.
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Particle that passes through the slit of width ‘w’ have an uncertainty ∆ x
Since the microscopic particles have wave properties, they are diffracted by the slit producing a diffraction
pattern as shown below.
Sin r = BCAC
= px
p
∆ px = p Sin r ∆ x = w∆ x × ∆ px = w p Sin r -------------------1
Condition for I order minimum :
Sin r = BCAB
BC = AB Sin r
= 12w sinr
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Path difference = 12w sin α
Path difference should be equal to 12 λ
12w sin α =
12 λ
w sin α = λ substituting in equation 1 we get ∆ x × ∆ px = p × λ
= p× hp
= h
In terms of energy
∆ x × ∆ px = h
Dividing and multiplying by ‘v’ we get
∆ xv × v ∆ px = h
Dividing and multiplying by ‘2’ we get
∆ xv ×2 × 1
2v ∆ px = h
Since momentum p = mv, we get
∆ xv ×2 × 1
2v ∆(mv) = h
∆ xv ×2× 1
2∆ m v2 = h -----------------2
Velocity = distance
time
∴time = distancevelocity
∆ t=¿ ∆ xv
substituting in equation 2 we get ∆ t ×2 ∆ E = h
∆ t × ∆ E = h2
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Problem1: If the uncertainty in position of an electron is 4 A calculate the uncertainty in its momentum.
Solution:
∆x = 4 ∆p = ?
∆x ×∆p = h
∴ ∆p =h
∆ x
= 6.62× 10−34
4 ×10−10
= 1.65 ×10−24 Kg m/sec
Problem2: An electron has a speed of 1000 m/s with an accuracy of 0.001 % calculate the certainty with
which we can locate the position of electron.
Solution:
speed of electron = 1000 m/s
accuracy = 0.001 %
mass of electron = 9.11 ×10−31 Kg
momentum of electron = mv
= 9.11 ×10−31 × 1000 ×0.001100
= 9.11 ×10−31 ×1
100
= 9.11 ×10−33
∆x = ?
∆x ×∆p = h
∴ ∆x =h
∆ p
= 6.62× 10−34
9.11×10−33
= 662× 10−1
911
= 0.7266 ×10−1
= 7.266 ×10−2
Problem3:. A particle of mass 5 ×10−10 Kg has a speed of 100 times of Planc constant with an accuracy of
0.005 % .The certainty with which we can locate the position of electron is
Solution:
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speed of electron = 100× h [ given 100 times of Planc constant]
accuracy = 0.005 %
mass of particle = 5 ×10−10Kg
∆x ×∆p = h
∆x = h
∆ p
= h
5× 10−10× 100×h ×5×10−3
= 1
25× 10−11
= 0.04 ×10+11 m
= 4 ×109 m
Problem4: The momentum of particle measured with a accuracy of 1% is 19.86× 10−25 the uncertainity in
position is 6.67 A [ TRB]
Solution:
∆p = 19.86× 10−25 × 0.01
∆x = ?
∆x = h
∆ p
= 6.62× 10−34
0.1986 ×10−25
= 13 ×10−9
= 0.333 ×10−9
= 0.333 ×10−10 ×101
= 3.33 ×10−10
= 3.33 A
Problem5: The momentum of particle measured with a accuracy of 1% is 3.32 × 10−25 Kg-m/s2 the
uncertainity in position is
Solution: ∆p = 3.32 × 10−25 × 0.01 ∆x = ?
∆x = h
∆ p
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= 6.62× 10−34
3.32× 10−27
= 2 ×10−7 m = 2 ×10−7 ×10−3 ×10+3
= 2 ×10+3
= 2000 A
. ZEEMAN EFFECT
1. The splitting of spectral lines, in the presence of applied magnetic field is called Zeeman effect.
2.This effect arises as a result of interaction between B and the magnetic moment associated with electron spin.
3. The field splits the line with a given J value in to( 2J+1) levels.
for example if J = 1 the line will be split up in to [( 2 (1) + 1 = 3]three
for example if J = 2 the line will be split up in to [( 2 (2) + 1 = 5] five
4. Transition of electrons in these levels is known as Zeeman transition.
5. If the term is singlet , it is called normal Zeeman effect.
6. If it is multiplet ( doublet, triplet) it is called anomalous Zeeman effect.
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2.11 BOHR’S CORRESPONDENCE PRINCIPLE.
When the energy difference between quantized levels are very small, Quantum mechanics is in
agreement with classical physics.
This is known as Bohr’s correspondence principle.
Bohr's correspondence principle demands that classical physics and quantum physics give the same
answer when the systems become large.
... Bohr provided a rough prescription for the correspondence limit: it occurs when the quantum numbers
describing the system are large.
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What is quantum mechanical tunneling? Explain any two evidences.
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2.12 QUANTUM TUNNELLING
Quantum tunnelling or tunneling is the quantum mechanical phenomenon where a
wavefunction can propagate through a potential barrier.
The transmission through the barrier can be finite and depends exponentially on the barrier
height and barrier width.
Quantum tunneling plays an essential role in physical phenomena, such as nuclear fusion. It has
applications in the tunnel diode, quantum computing, and in the scanning tunneling microscope.
A tunneling current occurs when electrons move through a barrier that they classically shouldn't be able to
move through. In classical terms, if you don't have enough energy to move “over” a barrier, you
won't. ... Tunneling is an effect of the wavelike nature.
the quantum mechanical phenomenon sometimes exhibited by moving particles that succeed in passing from
one side of a potential barrier to the other although of insufficient energy to pass over the top.
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1. It has important applications to modern devices such as the tunnel diode,
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2. The scanning tunnelling microscope
A scanning tunneling microscope (STM) is an instrument for imaging surfaces at the atomic level.
STM is based on the concept of quantum tunneling. When a conducting tip is brought very near to the surface
to be examined, a bias (voltage difference) applied between the two can allow electrons to tunnel through the
vacuum between them. The resulting tunneling current is a function of tip position, applied voltage, and
the local density of states (LDOS) of the sample.[4] Information is acquired by monitoring the current as the tip's
position scans across the surface, and is usually displayed in image form
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3.1 POSTULATES OF QUANTUM MECHANICS:
1. The physical state of a system at time’t‘ is described by the wave function Ψ ( x, t)
2. The wave function should be continuous ,finite and single valued for all values of x.
3. The wave function is should be normalised.
4. The condition for normalization is
∫−∞
∞
Ψ Ψ∗¿ dx = 1 where Ψ∗¿ is the complex conjugate of Ψ
5. The wave function Ψ is the solution of the equation HΨ=E Ψ
This equation is known as eigen value equation. Ψ is an eigen function, E is eigen value
6. The average value ( expectation value) of an operator A is < A > = ∫−∞
∞
Ψ A Ψ ¿dx
∫−∞
∞
Ψ Ψ ¿ dx
7 .For every observable there will be an operator in quantum mechanics.
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Quantum mechanical symbol
1. Position operator along X direction x
along Y direction y
along Z direction z
2. Differential operator ddx
3. Linear momentum operator along X direction px = hi
∂∂ x
along Y direction py = hi
∂∂ y
along Z direction pz = hi
∂∂ z ,
4. Angular momentum operator
L = Lx i + Ly j + Lz k
r→
= x i + yj + z k
p→
= pxi + py j + pz k
along X direction Lx = r × px
along Y direction Ly = r × py
along Z direction Lz = r × pz
5. Potential energy operator V
6. Kinetic energy operator −h2
2m∇2
6. Hamiltonian operatorH = −h2
2m∇2 + V
7. Vector operator ∇= ∂∂ x i +
∂∂ y j +
∂∂ z k
8. Laplacian operator ∇2 = ∂2
∂ x2 + ∂2
∂ y2 + ∂2
∂ z2
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3.2. OPERATORSAn operator may be defined as, a mathematical term, which is used to transform one function into
another function.
If P is an operator applied to a function f1 then P[f1] = f2 where f2 is the transformed function
For example :ddx ( sin x ) = cos x
Here ddx is the operator, Sinx is the function called operand, cosx is transformed function.
Various operators:
1. Position operator : x
2. Differential operator : ddx
3. Momentum operator: px = hi
∂∂ x
py = hi
∂∂ y
pz = hi
∂∂ z
4. Angular momentum operator: Lx = r × px ,
Ly = r × py
Lz = r × pz
5. kinetic energy operator - h2
2m∂2
∂ x2
6. Hamiltonian operator: H = −h22 m
∂ 2∂ x 2 + V(x)
7. Hamiltonian operator in 3- dimension : −h 22m
¿ ∂ 2
∂ x 2 +∂2
∂ y2 + ∂2
∂ z 2 ) + V(x,y,z)
8. Vector operator( del operator) ∇ (del) = ∂
∂ x i + ∂
∂ y j + ∂
∂ z k .
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9. Laplacian operator ∇ 2 (square of this vector operator) = ∂ 2
∂ x 2 +∂2
∂ y2 + ∂2
∂ z 2
1.Position operator : (x) It multiplies the given function by x
Problem 1 The position operator(x) operates on a function 3x2 . Find the resultant.
Solution:
(x) [3x2 ] = x× 3x2
= 3x3
Problem :The position operator( y) operates on a function 3x2 . Find the resultant.
Solution:
( y) [3x2 ] = y × 3x2
= 3x2y
2.Differential operator : ddx or
∂∂ x It differentiates the given function with respect to x
Problem The differential operatorddx operates on a function 3x2 . Find the resultant.
Solution:
ddx [3x2 ] = 3 × 2 x
= 6x
Problem .The differential operator∂
∂ x operates on a function 3y2 . Find the resultant.
Solution:
∂∂ x [3y2 ]= 0
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3. Momentum operator:
The momentum operator is defined as , the operator which operating on the wave function , reproduces
the wave function multiplied by the momentum
It differentiates the given function with respect to x and multiplies the result by hi .It is represented
as px = ћi
∂∂ x
Problem: 3.2.1 Find the expression for linear momentum operator
Solution:
The wave function of a particle moving along x- direction is
Ψ = A e
iћ ( px−Et ) where ћ =
h2 π
∂Ψ∂ x
= A e
iћ ( px−Et ) (
iћ p)
= iћ p Ψ --------------------------------------------1
∴ ћi
∂Ψ∂ x
= p Ψ
Deleting the function , we get
p = ћi
∂∂ x
This is the expression for momentum operator.
Problem.The momentum operator along x- direction operates on a function 3x2 . Find the resultant.
Solution: px(3x2 ) =ћi
∂∂ x (3x2)
=ћi
× 3× ∂∂ x (x2)
=ћi
× 3× 2x
=6 xћ
i
ANGULAR MOMENTUM OPERATOR:
The angular momentum is given by the vector product of position vector and linear momentum. i.e L
= r × p
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where L = Lx i + Ly j + Lz k -----------------1
r→
= x i + yj + z k and
p→
= pxi + py j + pz k
r × p = ⌈i j kx y zpx p y pz
⌉
= i(ypz – zpy) + j ( zpx- xpz ) + k ( xpy – ypx )
Comparing with 1, we get
Lx = ypz - zpy
Ly = zpx – xpz
Lz = xpy – ypx
Since px= -ih∂
∂ x , we get
Lx= h
2 πi ( y∂
∂ z - z ∂
∂ y )
Ly= h
2 πi ( z∂
∂ x - x ∂
∂ z ),
Lz= h
2 πi ( x∂
∂ y - y ∂
∂ x )
KINETIC ENERGY OPERATOR.
kinetic energy operator K = - ћ2
2m ∂2
∂ x2
Problem: 3.2.2 Find the expression for kinetic energy operator
Solution:
We know that the momentum operator with a function is
p Ψ = ћi
∂ Ψ∂ x
differentiating with respect to x,
p ∂Ψ∂ x = ћ
i∂2Ψ∂ x2
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substituting the value of ∂Ψ∂ x from equation 1
p (iћ p Ψ ) =
ћi ∂
2Ψ∂ x2
p2 Ψ = - ћ 2 ∂2Ψ
∂ x2
dividing by 2m,
p2
2m Ψ = - ћ2
(2m ) ∂
2Ψ∂ x2
= - ћ2
2m ∂
2Ψ∂ x2
[ ћ = h
2 π ]
But kinetic energy K = p2
2m∴ the above equation becomes,
KΨ = - ћ2
2m ∂
2Ψ∂ x2
Deleting the function , we get K = - ћ2
2m ∂2
∂ x2
This is the expression for kinetic energy operator.
1. Total energy operator:
Total energy operator is E = i ћ ∂Ψ∂t
Problem: 3.2.3 Find the expression for total energy operator
Solution:
The wave function of a particle moving along x- direction is
Ψ = A e
iћ ( px−Et )
Differentiating with respect to time
∂Ψ∂t
= A e
iћ ( px−Et ) (
iћ ) ( - E)
= iћ ( - E) Ψ
= - iћ E Ψ
i ћ ∂Ψ∂t = E Ψ [ multiplying by i ћ on both sides]
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∴ E = i ћ ∂∂ t [ deleting the function]
This is the expression for total energy operator.
It is also given by E = p2
2m + V
E = −h2
2m ∇2 + V( x,y,z)
6. HAMILTONIAN OPERATOR (TOTAL ENERGY OPERATOR)
Hamiltonian is a form of total energy , in terms of position and momentum co-ordinates.
. It is denoted by H. It is a differential operator which operating on a wave function, reproduces the same wave
function multiplied by the total energy. Hamiltonian is a form of total energy , but in the former energy is in
terms of position and momentum co-ordinates.
Hamiltonian operator H = kinetic energy operator + potential energy operator
= −h2
2m∂2
∂ x2 + V
7. VECTOR OPERATOR( DEL OPERATOR) :
It is denoted by ( del)∇ = ∂
∂ x i + ∂
∂ y j + ∂
∂ z k .
8. LAPLACIAN OPERATOR
The square of this vector operator ( ∇2 ) is called “ Laplacian operator”
∇2 = ∇. ∇
= (∂
∂ x i + ∂
∂ y j + ∂
∂ z k ) . ( ∂
∂ x i + ∂
∂ y j + ∂
∂ z k )
= (∂
∂ x i ).( ∂
∂ x i + ∂
∂ y j + ∂
∂ z k ) + ( ∂
∂ y j) .( ∂
∂ x i + ∂
∂ y j + ∂
∂ z k )+ (∂
∂ z k ) .( ∂
∂ x i + ∂
∂ y j + ∂
∂ z k )
= ∂2
∂ x2 + 0 +0 + ∂2
∂ y2 +0 +0 + ∂2
∂ z2 +0+0 [ i.j = 0, i.k = 0]
= ∂2
∂ x2 + ∂2
∂ y2 + ∂2
∂ z2
In spherical co-ordinates,
∇ 2 = 1
r 2 ∂
∂ r ( r 2 ∂
∂ r ) + 1
r 2 sinθ ∂
∂θ ( sin θ ∂
∂θ ) + 1
r 2sin 2 θ ∂ 2
∂ φ 2
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Problem Find the resultant when ∇2 operator operates on the fuction sin2xsin2ysin2z
Solution:
∇2 (sin2xsin2ysin2z) = ∂2
∂ x2 +∂2
∂ y2 + ∂2
∂ z2 (sin2xsin2ysin2z)
= ∂2
∂ x2 (sin2xsin2ysin2z) + ∂2
∂ y2 (sin2xsin2ysin2z)+ ∂2
∂ z2 (sin2xsin2ysin2z)
= sin2ysin2z ∂2
∂ x2 (sin2x) + sin2xsin2z ∂2
∂ y2 (sin 2y)+ sin2xsin2y ∂2
∂ z2 (sin2z)
= sin2ysin2z∂
∂ X (2cos2x) + sin2xsin2z ∂
∂ y (2 cos2y)+ sin2xsin2y ∂
∂ z (2cos2z)
= sin2ysin2z (-4 sin2x) + sin2xsin2z (-4sin2y)+ sin2xsin2y ( -4sin2z)
= -4 [ sin2ysin2z sin2x + sin2xsin2z sin2y+ sin2xsin2y sin2z]
= -12 sin2ysin2z sin2x
3.3 OPERATOR ALGEBRA.ADDITION AND SUBTRACTION OF OPERATORS:
The addition or subtraction of operators yields new operators.
( A + B) f( x) = A f( x) + B f( x)
( A - B) f( x) = A f( x) - B f( x)
Example :
Consider two operators A = ‘log’ and B = ‘d/dx’ and a function “ x2 “
The addition of the two operators is
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( A + B) ( x2 ) = [ log + ddx ] ( x2 )
= log ( x2 ) + ddx ( x2 )
= 2log x + 2x [ddx ( x2 ) = 2x ]
= A ( x2 ) + B ( x2 )
Problem 3.3.1 The operators ‘x’ and ddx are added and operates on a function ‘sinx ‘ .Find the resultant.
Solution:
( x + ddx ) ( sin x ) = x sin x +
ddx sin x
= x sin x + cos x [ ddx sin x = cos x]
Problem 3.3.2 The operators ‘x’ and ddx are added and operates on a function ‘siny .Find the resultant.
Solution:
( x + ddx ) ( sin y ) = x sin y +
ddx sin y [
ddx sin y = 0]
= x sin x + 0
= x sin x
Problem3.3.3 :The operator ddy is subtracted from another operator ‘y’ and operates on e2y. Find the resultant
Solution:
( y - ddy ) (e2y ) = y e2y -
ddy e2y
= y e2y - 2 e2y [ddy e2y = 2 e2y ]
= e2y (y – 2 )
Problem 3.3.4: The operator ddy is subtracted from another operator ‘y’ and operates on e2x. Find the resultant
Solution:
( y - ddy ) (e2x ) = y e2x -
ddy e2x
= y e2y - 0 [ddy e2x = 0 ]
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= ye2y
MULTIPLICATION OF OPERATORS:
Multiplication by two operators means, operations by the two operators one after the other, the order
of operation being from right to left.
For example AB f(x) means the function f(x) is first operated by B to yield a new function g(x) which
is then operated by A to yield the final function h(x)
AB f(x) = A[ B (f(x) ]
= A [ g(x) ]
= h (x)
Consider two operators A = ‘x’ and B = ‘ddx ’ and a function “ x2 “. The multiplication of the two
operators is A B ( x2 ) = x ddx [ x2]
= x ( 2x) [ddx [ x2] = 2x
= 2x2
Problem 3.3.5 : The operators ‘x’ and ddx are multiplied and operates on a function ‘sin x ‘ .Find the
resultant.
Solution:
x ddx ( sin x)
= x ( cos x) [ ddx ( sin x) = cos x]
= x cosx
Problem3.3.6 : The operators ‘y’ and ddx are multiplied and operates on a function ‘sin x ‘ .Find the resultant.
Solution:
y ddx ( sin x)
= y ( cos x) [ ddx ( sin x) = cos x]
= y cosx
Problem3.3.7 The operators ‘y’ and ddx are multiplied and operates on a function ‘sin y ‘ .Find the resultant.
Solution:
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y ddx ( sin y)
= y (0) [ ddx ( sin y) = 0]
= 0
,3. LINEAR PROPERTY ( OPERATOR) : An operator( P) is said to be linear if it satisfies the following condition.
P(f1 +f2 ) = P f1+ Pf2
For example
if ddx ( 2e5x + 3 e5xx ) =
ddx (2e 5x) +
ddx (3 e5xx) , the operator
ddx is said to be linear
Problem 3.3.8 Verify the differential operator is linear with respect to the functions 2x and 3x.
Solution: Differential operator is “ ddx ”
Function-1 = 2x
Function- 2 = 3x
LHS :ddx ( 2x + 3x ) =
ddx ( 5x )
= 5
RHS: ddx ( 2x) = 2
ddx ( 3x) = 3
ddx ( 2x) +
ddx ( 3x ) = 2+3
= 5 .
ddx ( 2x + 3x ) =
ddx ( 2x) +
ddx ( 3x )
∴ ddx is a linear operator.
Problem 3.3.9 Check ‘log’ operator and square root (√) operator are linear or not
Solution:
1. Log operator
log ( A+ B ) ≠logA + log B
2. Square root (√) operator
√9+16= √25
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= 5
√9 + √16 = 3 + 4
= 7
√9+16≠√9 + √16
Therefore ‘log’ operator and square root operators are not linear
Problem 3.3.10 Verify the linear momentum operator along y direction is linear with respect to the functions
π4 y and
π2 y.
Solution: Linear momentum operator along y direction = ћi
∂∂ y
LHS : ћi
∂∂ y (
π4 y +
π2 y.) =
ћi
∂∂ y (
3π4 y )
= ћi (
3 π4 ) [
∂∂ y (
3 π4 y ) =
3π4 ]
RHS: ћi
∂∂ y (
π4 y ) +
ћi
∂∂ y (
π2 y) =
ћi (
π4 ) +
ћi (
π2 )
= ћi (
π4 +
π2 )
= ћi (
3 π4 )
∴ Linear momentum operator along y direction is linear .
4. COMMUTATIVE PROPERTY ( COMMUTATOR)
If two operators are such that the result of their successive applications is the same, irrespective of the
order of operations, then the two operators are said to be commutative.
Two operators P and Q are said to be commutative if it satisfies the following condition.
P Q [f] = Q P [f]
If two operators do not commutate , the difference between the co-efficient of the two transformed
function is called commutator and is represented as [ P Q]. i.e
[P,Q] = P Q [f] - Q P [f]
Problem 3.3.11 Check the operators ddx and 3commute or not.
Solution:Consider an arbitrary function [f].
ddx
3 [f] = ddx [ 3f ] [3 multiplies the given function by 3 ]
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= 3 ddx [ f ]
= 3 f’ ddx [ f ] = f ’
3 ddx [ f] = 3 [ f’]
= 3 f’ [3 multiplies the given function by 3 ]
i.e ddx
3[ f] = 3ddx [f]
Thus they commute each other.
Problem. 3.3.12 Find the commutator of the operators x and ddx with respect to the function x2
Solution:
xd
dx [x2] = x[ 2x] [ ddx (x2)=2x]
= 2x2 [x multiplies the given function by x ]
ddx
x[x2] = ddx [ x3] [x multiplies the given function by x ]
= 3x2
xd
dx [x2] ≠ ddx
x [x2 ]∴They do not commutate each other. Commutator = 3x2 - 2x2
= 1x2.
= 1 [deleting the function]
Problem 3.3.13 Show that the position operator and momentum operator do not commute with each other with
respect to the function e x and Find the commutator.
Solution:
Position operator:x (it multiplies the given function by x)
Momentum operator: hi
∂∂ x
x hi
∂∂ x [ ex ] = x h
i [ ex ] [ ∂∂ x ( e x ) = e x ]
= hi [x e x ] [ xmultiplies the given function by x]
hi
∂∂ x
x [ex] = hi
∂∂ x [x e x] [ xmultiplies the given function by x]
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= hi [ x
∂∂ x¿e x ) + e x ∂
∂ x (x) ] [ d (UV) = U dV + V dU ]
= hi [x e x + e x] [
∂∂ x¿e x ) = e x ,
∂∂ x (x)=1]
x hi
∂∂ x [ex ] ≠
hi
∂∂ x
x [ex]
∴They do not commute each other.
commutator = xhi
∂∂ x [e x ] -
hi
∂∂ x
x[e x]
= hi x ex -
hi [x e x + e x]
= hi x ex -
hi x e x --
hi e x]
= - -hi ex
= - -hi [ Deleting the function]
= - hi × i
i [multiplying
Nr and Dr by i]
= - hii2
= - hi
(−1 ) [i 2 = -1]
= h i
Problem 3.3.14 Find [z3, ddz ]
Solution:
Consider an arbitrary function f (z)
[z3 ddz ] f = z3 d
dz [f ] - ddz z3 f
z3 ddz [f ] = z3 f’ [
ddz [f ] = f’ ]
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ddz z3 f =
ddz ( z3 f )
= z3 ddz [ f ] + f
ddz [ z3 ] [
ddz (UV) = U
ddz V + V
ddz U ]
= z3f’ + 3z2 f [ ddz [z3]= 3 z2 ]
z3 ddz [f ] -
ddz z3 f
= z3f’ - [ z3f + 3z2f ]
= - 3z2f
= - 3z2 [ Deleting the arbitrary function’ f’ ]
[z3, ddz ] = - 3z2
Problem 3.3.15 Find [z3, ddz ]
Solution:
Consider an arbitrary function f (z)
[z3, ddz ] f (z) = z3 d
dz [f (z)] - ddz z3f (z)
z3 ddz [f (z)] = z3 f’ (z) [
ddz [f (z)] = f’ (z)]
= z3 f’ (z)
ddz z3f (z) =
ddz z3f (z)
= z3 ddz [ f(z) ] + f(z)
ddz [ z3 ] [
ddz (UV) = U
ddz V + V
ddz U ]
= z3f’ (z)} + 3z2f (z) [ ddz [z3]= 3 z2 ]
z3 ddz [f (z)] -
ddz z3f (z)
= z3f’(z) - [ z3f’(z)} + 3z2f (z)]
= - 3z2f (z)
= - 3z2 [ Deleting the arbitrary function f (z)]
[z3, ddz ] = - 3z2
Problem 3.3.16 Find the commutator of px, x
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Solution:
Commutator of px, x = ( px,x ) f - ( x px )
( px,x ) f(x) = h
2πi ( ∂
∂ x ) x [f(x)] [ px = h
2 πi∂
∂ x and f (x) a function]
= h
2 πi ( ∂
∂ x ) [x f(x)] [ x multiplies the given function by x]
= h
2 πi [x ∂
∂ x f(x) + f (x) ∂
∂ x (x )] [ d ( UV) = UdV + VdU ]
= h
2πi [xf ‘(x) + f(x) (1)] [∂
∂ x f(x) = f ‘(x) and ∂
∂ x x = 1]
(xpx ) f(x) = xh
2 πi∂
∂ x [f(x)] [px =h
2 πi∂
∂ x ]
= xh
2πi [f’(x)] [∂
∂ x f(x) = f’(x) ]
= x × h2πi [f’(x)] [ x multiplies the given function by x]
= h
2 πi [x f’(x)]
px ,x - x px = h
2 πi [xf ‘(x) + f(x)] - h
2πi [ x f’(x)]
commutator = h
2 πi f(x)]
= h
2 πi [Deleting the arbitrary function f (x)]
≠ 0
∴ px and x do not commute each other.
Problem 3.3. 17.Do the following operators commute. px, z
Solution:
[ px, z ] = px z - z px
( px , z) [ f(x)] = h
2 πi ( ∂
∂ x )z [ f(x)] [ px = h
2 πi∂
∂ x ]
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= h
2 πi ( ∂
∂ x ) z f(x)] [z multiplies the given function by z]
= hz
2 πi ( ∂
∂ x )f(x)]
= hz
2 πi f’(x)] [ ∂
∂ x f(x) = f’(x) ]
z px = z [ h
2πi∂
∂ x f(x)]
= z h
2 πi f’(x)] [ ∂
∂ x f(x) = f’(x) ]
= h
2πi z f’(x)] [ z multiplies the given function by z]
px , z - z px = hz2 πi f’(x)] -
h2 πi z f’(x)]
= 0
∴ px and z commute with each other.
Problem 3.3.18.Prove that [Lz, z ] = 0
Proof :
[ Lz, z] = Lz, z - z Lz
= h
2πi ( x∂
∂ y - y ∂
∂ x )z - z [ h
2πi ( x ∂
∂ y - y ∂
∂ x )] [ Lz = h
2 πi (x∂
∂ y - y ∂
∂ x
) ]
Consider any function Ψ
Lz, z- z Lz (Ψ) = h
2 πi ( x ∂Ψ∂ y - y
∂Ψ∂ x )z - z [
h2πi ( x
∂Ψ∂ y - y
∂Ψ∂ x )]
= h
2 πi [xz ∂Ψ∂ y - yz
∂Ψ∂ x - z x
∂Ψ∂ y - zy
∂Ψ∂ x ] [z multiplies the given function by z]
= 0
Similarly we can prove [ Lx,x ] = 0 ,
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[ Ly, y ] = 0
[ Lx,y ] = h
2 πi z ,
[ Ly,z ] = h
2πi x,
[ Lx,z ] = h
2πi y
[ Ly,x ] = −h2 πi z ,
[ Lz,y ] = - h
2 πi x,
[ Lz,x ] = - h
2 πi y
Lx and Ly , Ly and Lz, Lx and Lz do not commute each other.
i.e [ Lx , Ly ] ≠ 0 , [ Ly , Lz ] ≠0 , [ Lz , Lx ] ≠ 0 ∴They can not be measured simultaneously.
.
Problem 3.3.19 : Show that square of angular momentum and its y component can be specified
simultaneously
.Proof :
To prove [ L2 , Ly ] = 0
To prove L2^ Ly^ - Ly ^L 2 ^ = 0
We know that
L 2 = Lx 2 + Ly 2 + Lz 2
[ L2 , Ly ] = [ ( Lx 2 + Ly 2 + Lz 2 ) , Ly )
= [ Lx 2 Ly ] + [ Ly 2 , Ly ] + [ Lz 2 , Ly ] -----------1
Let us take the first term
[ Lx 2 Ly ] = Lx 2 Ly - Ly Lx 2
= Lx Lx Ly - Ly Lx Lx
= Lx Lx Ly - Ly Lx Lx + Lx Ly Lx - Lx Ly Lx [ adding and subtracting Lx Ly Lx ]
= Lx (Lx Ly - Ly Lx ) + (Lx Ly - Ly Lx ) Lx
= Lx [Lx^ Ly ^] + [Lx^ Ly ^] Lx
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We know that [Lx Ly ] = ih2 π L z
∴ [ Lx 2 Ly ] = Lx (ih2 π L z ) + (
ih2 π L z ) Lx
= ih2 π ( Lx L z + L z Lx)
Similarly we can prove [ Lx 2 Lz ] = - ih2 π ( Lx L y + L y Lx)
[ Lx 2 Lx ] = Lx 2 Lx - Lx Lx 2
= Lx ( Lx Lx - Lx Lx )
= 0∴ [ Lx 2 Ly ] + [ Ly 2 , Ly ] + [ Lz 2 , Ly ]
= ih2π ( Lx L z + L z Lx) -
ih2π ( Lx L y + L y Lx) + 0
= 0
This means square of angular momentum and its x component commute with each other and can be specified simultaneously. Similarly L2 also commute with other components.
ALITER:
In polar co ordinates, L2 = - h2
4 π2 [ 1
sinθ ∂
∂θ ( sin θ ∂
∂θ ) + 1
sin2θ ∂2
∂ φ2
Lx = -i h
2 π ∂
∂ φ
L2 Lx = - h2
4 π2 [ 1
sinθ ∂
∂θ ( sin θ ∂
∂θ ) + 1
sin2θ ∂2
∂ φ2 [ -i h
2π ∂
∂ φ ]
= -i h3
8π3 [ 1
sinθ ∂
∂θ ( sin θ ∂2
∂θ ∂ φ ) +
1sin2θ
∂3
∂ φ3 ]
Lx L2 = [ -i
h2π
∂∂ φ ] [ - h2
4 π2 [ 1
sinθ ∂
∂θ ( sin θ ∂
∂θ ) + 1
sin2θ ∂2
∂ φ2 ]
= -i h3
8π3 [ 1
sinθ ∂
∂θ ( sin θ ∂2
∂ φ ∂ θ ) +
1sin2θ
∂3
∂ φ3 ]
Since ∂2
∂θ ∂ φ= ∂2
∂ φ ∂θ
L2 Lx = Lx L2
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L2 Lx - Lx L2 = 0
[ L2 Lx ] = 0
This shows that square of angular momentum and its x component commute with each other
Problem 3.3.20 Show that [ Lx , Ly ] ≠ 0 and find the commutator.
To prove Lx^ Ly^ - Ly ^Lx ^ ≠ 0
LxLy = h
2 πi ( y ∂
∂ z - z ∂
∂ y ) × h
2 πi ( z ∂
∂ x - x ∂
∂ z )
= −h24 π 2 { y
∂∂ z (z
∂∂ x ) - y
∂∂ z (x
∂∂ z ) - z
∂∂ y (z
∂∂ x ) + z
∂∂ y (x
∂∂ z )
= −h24 π 2 ( y[ z
∂ 2∂ z ∂ x +
∂∂ x (1) ) – y[x
∂2∂ z 2 +
∂∂ x (0) )] – z [z
∂ 2∂ y∂ x +
∂∂ x (0) )] + z[x
∂2∂ y∂ z +
∂∂ z (0) )
= −h24 π 2 ( y[ z
∂ 2∂ z ∂ x +
∂∂ x ) – y[x
∂2∂ z 2 ] – z [z
∂ 2∂ y∂ x ] + z[x
∂2∂ y∂ z ]
= −h24 π 2 ( y z
∂ 2∂ z ∂ x + y
∂∂ x – yx
∂2∂ z 2 – z 2
∂ 2∂ y∂ x + zx
∂2∂ y∂ z ) ----------1
Ly ^Lx ^ = h
2 πi (z ∂
∂ x - x ∂
∂ z ) × h
2 πi (y ∂
∂ z - z ∂
∂ y )
= −h 24 π 2 { z
∂∂ x (y
∂∂ z ) - z
∂∂ x (z
∂∂ y ) - x
∂∂ z (y
∂∂ z ) + x
∂∂ z (z
∂∂ y )
---- Comparing 1 and 2, it is clear that
Lx^ Ly^ ≠ Ly ^Lx ^
i.e Lx^ Ly^ - Ly ^Lx ^ ≠ 0 i.e [ Lx , Ly ] ≠ 0
∴ Lx and Ly do not commute.
To find the commutator:
[Lx Ly ] = Lx^ Ly^ - Ly ^Lx ^
Thus the commutator [Lx Ly ] = ih2 π L z Similarly [Ly Lz ] =
ih2 π L x
[Lx Lz ] = ih2π L y Note: [Lx Lz ] = - [Lz Lx ]
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Operator - 1 Operator -2 commutator
px x h2 πi
px y 0
px z 0
Lz, z 0
HERMITIAN PROPERTY OF OPERATORS
An operator is said to be Hermition if∫Ψ ¿(A φ) dx = ∫(AΨ )¿φ dx WhereΨ and φ are two
different functions and A is the operator
Problem Verify that the operator d2
dx2 is Hermitian or not
Solution: Consider two functions Ψ = eix and φ = sinx. and let A= d2
dx2
The condition for Hermition property is ∫Ψ ¿(A φ) dx = ∫(AΨ )¿φ dx
LHS
∫Ψ ¿(A φ)dx= ∫ e(ix )∗¿( d2
dx 2 sin x)¿ dx
= ∫ e(−ix)( d2
dx2 sin x)dx [e(ix )∗¿=e−ix ¿
= ∫ e(−ix)( ddx
cos x)dx [ ddx¿
= ∫ e(−ix )(−sin x) dx [ ddx¿
=( - ) ∫ e(−ix )(sin x) dx
RHS
∫(AΨ )¿φdx = ∫¿¿¿dx
= ∫ d2
dx2 e (−ix ) (sin xφ )dx [e(ix )∗¿=e−ix ¿
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= ∫ ddx(−i e−ix) (sin xφ )dx
= ∫(−i)(−i e−ix) (sin xφ )dx
= ∫+i2 e−ix (sin x φ )dx
= ∫(−1)e−ix (sin x φ )dx [ i2 = -1]
=( - ) ∫ e(−ix )(sin x φ) dx
= LHS
d2
dx2 is Hermitian
Properties of Hermition operator:
1. Eigen values of Hermition operator are real
2. Eigen functions of a Hermition operator are orthogonal
1. Eigen values of Hermition operator are real
Proof:
consider the eigen equation
AΨ = aΨ
Multiplying both sides by Ψ * and integrate over space
∫Ψ ¿A Ψ dτ = ∫Ψ ¿aΨ dτ
= a ∫Ψ ¿Ψ dτ [Since ‘a’ is a number ]
The complex conjugate of AΨ = aΨ is written as
A¿Ψ ¿ = a¿ Ψ ¿
Multiplying both sides by Ψ and integrate over space
∫Ψ A¿Ψ ¿ dτ = ∫Ψ a¿Ψ ¿ dτ
= a¿ ∫Ψ ¿Ψ dτ [Since ‘a’ is a number ]
Since the operator A is Hermition
∫Ψ ¿A Ψ dτ = ∫Ψ A¿Ψ ¿ dτ
Therefore
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a ∫Ψ ¿Ψ dτ = a¿ ∫Ψ ¿Ψ dτ
a = a¿
a number and its conjugate will be equal only if the number is real thus the Hermition operator will give real
eigen values.
2. Eigen functions of a Hermition operator corresponding to different eigen values are orthogonal
Proof: Consider an eigen function Ψ 1 with eigen value ‘a1’ . Let ‘A’ be a Hermition operator. The eigen equation is
AΨ 1 = a1Ψ 1 ------------------------1
Consider another eigen function Ψ 2 with eigen value ‘a2’ . Let ‘A’ be a Hermition operator. The eigen
equation is
AΨ 2 = a2Ψ 2 -----------------------2
The condition for orthogonality is
∫Ψ 1Ψ 2 dτ = 0 OR ∫Ψ 2¿ Ψ 1dτ = 0 OR ∫Ψ 1
¿ Ψ 2dτ = 0
Multiplying by Ψ 2¿ on both sides and integrate over space
∫Ψ 2¿ A Ψ 1 dτ = ∫Ψ 2
¿ a1Ψ 1dτ
= a1 ∫Ψ 2¿ Ψ 1dτ ----------------------3
Since A is a Hermition operator
∫Ψ 2¿ AΨ 1dτ = ∫Ψ 1 ( A Ψ 2 )
¿ dτ
= ∫Ψ 1 (a2Ψ 2 )¿ dτ
= a2¿∫Ψ 1 (Ψ 2 )
¿ dτ
Using equation (3) we can write
a2¿∫Ψ 1 (Ψ 2 )
¿ dτ = a1 ∫Ψ 2¿ Ψ 1dτ
Rearranging
a2¿∫Ψ 1 (Ψ 2 )
¿ dτ - a1 ∫Ψ 2¿ Ψ 1dτ = 0
( a2¿ - a1 ) ∫Ψ 2
¿ Ψ 1dτ = 0
a2¿ - a1 ≠ 0. Therefore ∫Ψ 2
¿ Ψ 1dτ must be equal to zero
∫Ψ 2¿ Ψ 1dτ = 0
This proves that Ψ 1 and Ψ 2 are orthogonal
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Problem 3.3.22 .Show that the momentum operator is HermitianSolution:
Momentum operator is hi
∂∂ x
To prove ∫Ψ ( hi
∂∂ x ) φ dx = ∫φ (
hi
∂∂ x ) Ψ dx
LHS:
∫Ψ ( hi
∂∂ x ) φ dx = ∫Ψ (
hi
∂ φ∂ x ) dx
= ( hi ) ∫Ψ
∂ φ∂ x dx let u = Ψ , du = dΨ, dv =
∂ φ∂ x dx ∴ v = φ
= ( hi ) [ Ψφ - ∫φdΨ dx ]
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Hermitian:
If ‘f’ is a function with f ¿ as its conjugate , and ‘A ‘ is an operator then
∫ f A f ¿ = ∫ f ¿ A¿ f
A is said to be Hermitian
Problem 1: check ddx is Hermitian or not
Solution:
Let us consider a function e ix
f = e ix then f ¿ = e−ix
∫ f A f ¿ = ∫ eix ddx e−ix dx
= ∫ eix (-i) e−ix dx [ ddx ¿) = (-i) e−ix]
= -i ∫ eix e−ix dx
= -i ∫ dx
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= -ix
∫ f ¿ A¿ f = ∫ e−ix ddx e ix dx
= ∫ e−ix (i)e ix dx [ ddx ¿) = (i) e−ix]
= i ∫ eix e−ix dx
= i ∫ dx
= ix
≠ ∫ f A f ¿
Therefore ddx is not Hermitian
Problem 2: check −ih ddx ( momentum operator) is Hermitian or not
Solution:
Let us consider a function e ix
f = e ix then f ¿ = e−ix
∫ f A f ¿ = ∫ eix (−ih ddx ) e−ix dx
= -ih ∫ eix ddx e−ix dx [
ddx ¿) = (-i) e−ix]
= -ih ∫ eix (-i)e−ix dx
= +i2 h ∫ dx
= +i2 h x
∫ f ¿ A¿ f = ∫ e−ix (−ih ddx )
¿
e ix dx
= ∫ e−ix (+ih ddx¿ e ix dx
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= +ih ∫ e−ix ddx e ix dx [
ddx ¿) = (i) e−ix]
= +ih ∫ eix (i)e−ix dx
= +i2 h ∫ dx
= +i2 h x
= ∫ f A f ¿
Therefore momentum operator is Hermitian
Problem 2: check −h2
2m d2
d x2 ( Kinetic energy operator) is Hermitian or not
Solution:
Let us consider a function e ix
f = e ix then f ¿ = e−ix
∫ f A f ¿ = ∫ eix (−h2
2m d2
d x2 ) e−ix dx
= (−h2
2m¿ ∫ eix d2
d x2 e−ix dx [ d2
d x2 ¿) = - e−ix]
= (−h2
2m¿ ∫ eix (- e−ix ) dx
= (+h2
2m¿ ∫ eix e−ix dx
= (+h2
2m¿ ∫ dx
= (+h2
2m¿ x
∫ f ¿ A¿ f = ∫ e−ix ¿¿ e ix dx
= ∫ e−ix ((−h2
2m d2
d x2 ¿ eix dx [ d2
d x2 ¿) = - e ix]
= (−h2
2m¿ ∫ e−ix (- e ix ) dx
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= (+h2
2m¿ ∫ e−ix e ix dx
= (+h2
2m¿ ∫ dx
= (+h2
2m¿ x
= ∫ f A f ¿
Therefore Kinetic energy operator is Hermitian
Similarly ‘x’ and ‘ iddx ‘ are hermitian
‘ i d2
d x2 ‘ and ‘ xddx ‘ are not hermitian
3.4. EIGEN FUNCTIONS AND EIGEN VALUES If a function( Ψ) , by a result of operation with an operator H, gives the same function, then it is
called eigen function and the coefficient( E) is called eigen value. .[ the German word eigen means
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For example consider a function sin 2x and the operator d2
d x2
d2
d x2 ( sin 2x ) = -4 (sin 2x)
∴- 4 is the eigen value.
Problem 3.4.1 Find the eigen value of cos 5x with the operator d2
d x2 .
d2
d x2 ( cos 5x ) = ddx d
dx (cos 5x )
= ddx ( -5 sin 5x )
ddx (cos 5x ) = - 5 sin 5x
= - 5 ddx ( sin 5x )
ddx ( sin 5x ) = 5 cos 5x
= - 5 ( 5cos 5x )
= - 25 (cos 5x)∴ - 25 is the eigen value.
Problem 3.4.2 Which of the functions are eigen functions of d2
d x2 . Find the eigen values.
a. Sin 2x, b. 6 cos 3x, c. 5 x2, d. log x e e -2x
Solution:
a. d2
d x2 ( sin 2x) = ddx
ddx ( sin2 x )
= ddx ( 2 cos2x )
ddx (sin2 x ) = 2cos 2x
= 2 ddx (cos 2x )
ddx ( cos2x ) = -2sin2x
= 2 (- 2sin2x )
= - 4sin 2x
Since there is ‘sin 2x‘ term in the result, it is an eigen function . - 4 is the eigen value.
b. d2
d x2 ( 6 cos 3x)
= ddx
ddx (6 cos 3x )
= ddx (- 18 sin3 x ) [
ddx (cos3 x ) = -3 sin3x]
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= - 18 ddx ( sin 3x )
= -18 ( 3cos3 x ) [ ddx ( sin3x ) = 3 cos3x ]
= - 54 cos 3x
Since there is ‘cos 3x ‘ term in the result, it is an eigen function with the eigen value -54
c. d2
d x2 (5 x2 ) = 20
= ddx
ddx (5 x2 )
= ddx ( 10 x ) [
ddx ( x2 ) = 2x]
= 10 [ddx ( x ) = 1]
Since there is no ‘x2 ‘ term in the result, it is not an eigen function.
d. d2
d x2 ( log x)
= ddx
ddx ( log x )
= ddx (
1x ) [
ddx ( log x ) =
1x ]
= −1x2 [
ddx (
1x ) =
−1x2 ]
Since there is no ‘log x ‘ term in the result, it is not an eigen function.
e. d2
d x2 ( e -2x )
= ddx
ddx ( e -2x )
= ddx ( - 2 e -2x ) [
ddx (e -2x ) = - 2 e -2x ]
= + 4 e -2x [ddx (e -2x ) = - 2 e -2x ]
Since there is ‘e -2x ‘ term in the result, it is an eigen function with the eigen value -4
Problem 3.4.3 : Find the eigen value of ∇2 when it operates on the function sin 2xsin2ysin2z
Solution:
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∇2 = ∂2
∂ x2 + ∂2
∂ y2 + ∂2
∂ z2
∇2 (sin2xsin2ysin2z) = ∂2
∂ x2 +∂2
∂ y2 + ∂2
∂ z2 (sin2xsin2ysin2z)
= ∂2
∂ x2 (sin2xsin2ysin2z) + ∂2
∂ y2 (sin2xsin2ysin2z)+ ∂2
∂ z2 (sin2xsin2ysin2z)
= sin2ysin2z ∂2
∂ x2 (sin2x) + sin2xsin2z ∂2
∂ y2 (sin 2y)+ sin2xsin2y ∂2
∂ z2 (sin2z)
= sin2ysin2z∂
∂ X (2cos2x) + sin2xsin2z ∂
∂ y (2 cos2y)+ sin2xsin2y ∂
∂ z (2cos2z)
= sin2ysin2z (-4 sin2x) + sin2xsin2z (-4sin2y)+ sin2xsin2y ( -4sin2z)
= -4 [ sin2ysin2z sin2x + sin2xsin2z sin2y+ sin2xsin2y sin2z]
= - 4[ 3 sin2ysin2z sin2x]
= - 12[ sin2ysin2z sin2x]∴∇2 Ψ = - 12Ψ
eigen value = -12
Problem 3.4.4: Find the eigen function of the operator ddx
Consider a function Ψ
ddx [Ψ ] = k Ψ [ condition for eigen function)
dΨΨ = k dx
Upon integration ln Ψ = k x + c
Taking exponential Ψ = ekx × e c
= K ekx This is the eigen function of the operator ddx
Problem 3.4.5: Find the eigen function of the operator d2
d x2
Solution: Consider a function ‘f’ then
d2
d x2 [f ] = K f [ K is a constant]
d2 f
d x2 - K f = 0
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this is a second order differential equation which can be written as
( D2 – K ) f = 0
auxiliary equation is m2 - K = 0
m = ± √K
Therefore solution is f = A e+√K x + B e−√ K x
This is the the eigen function of the operator d2
d x2
Problem 3.4.6 Show that the wave function Ψ = x e− x2
2 is an eigen function of the operator H = - d2
d x2 + x2
and find the eigen valueProof:
HΨ = (- d2
d x2 + x2) ( x e− x2
2 )
= - d2
d x2 ( x e− x2
2 ) +( x2) ( x e− x2
2 )
= - d2
d x2 ( x e− x2
2 ) + ( x3 e− x2
2 )
= - ddx ( x e
− x2
2 ¿) + e− x2
2 (1) ) + ( x3 e− x2
2 ) [d (UV) = UdV + VdU ]
= - ddx (- x2 e
− x2
2 + e− x2
2 ) + ( x3 e− x2
2 )
= - { (- x2 e− x2
2 ¿) + e− x2
2 (-2x) + e− x2
2 ¿ ) } + ( x3 e− x2
2 )
= - { ( x3 e− x2
2 - 2x e− x2
2 - x e− x2
2 } + ( x3 e− x2
2 )
= - { ( x3 e− x2
2 - 3x e− x2
2 } + ( x3 e− x2
2 )
= - x3 e− x2
2 + 3x e− x2
2 + x3 e− x2
2
= 3 x e− x2
2
= 3 Ψ Therefore this is an eigen function . The eigen value = 3
Problem 3.4.7 Calculate the eigen value, if the function 1π sin ( 3.5 x) is an eigen function of the operator
H = - h2
8 π 2m d2
d x2
Solution:
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HΨ = - h2
8 π 2m d2
d x2 [ 1π sin ( 3.5 x) ]
= - 1π
× h2
8π 2m d2
d x2 [ sin ( 3.5 x) ]
= - 1π
× h2
8π 2m
ddx [ 3.5 ×cos ( 3.5 x) ] [
ddx [ sin (A x) = A cos Ax
= - 1π
× h2
8π 2m [ - (3.5)2 sin (3.5 x) ] [
ddx [ cos (A x) = - A sin Ax]
= [- h2
8π 2m [ - (3.5)2 ] [
1π sin (3.5 x) ]
=[+ h2
8π 2m(3.5)2] [
1π sin (3.5 x) ]
eigen value = + h2
8π 2m(3.5)2
4. SCHRODINGER EQUATION &ITS APPICATIONS TO SIMPLE SYSTEMS
1.1 .1Time independent Schrodinger equation
Schrodinger gave a wave equation to describe the behaviour of electron waves in atoms and molecules.
He shared the 1933 physics Nobel prize with P.A.M . Dirac for the discovery of new forms of atomic theory.
Consider a system of stationary waves to be associated with the particle. Let ( x,y,z) be the co-ordinates
of the particle and Ψ be the wave function at any time t. The differential equation of the wave motion of the
particle is
∂ 2
∂ x 2 + ∂ 2
∂Y 2 +∂ 2
∂ Z 2 = 1c2
∂ 2∂ t 2
Considering x – direction alone the equation becomes
d2 Ψdx2 =
1c2
d 2Ψdt 2 ------------------------1
A wave can be represented as Ψ = A sin w t
Differentiating with respect to t dΨdt = Aw cos w t
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Again differentiating with respect to t d2Ψ
d t 2 = ( -w2 ) A sin w t
= -w2 Ψ
= - ( 2πγ)2 Ψ [ w = 2πγ]
= - 4 π2γ2 Ψ
= - 4 π2 ( cλ )2 Ψ [γ =
uλ ]
= −4 π 2c2
λ2 Ψ
Substituting in equation 1 we get
d2Ψ
d x2 = 1
c2 (−4 π 2 c2
λ 2) Ψ
= −4 π 2
λ2 Ψ
d2Ψd x2 + 4 π 2
λ2 Ψ = 0
d2Ψd x2 + 4 π 2 (mv ) 2
h 2 Ψ = 0 [ λ = h
mv ] ---------------1
Kinetic energy (KE) = ½ mv2
2 KE = mv2
Multiplying by m on both sides,
2mKE = (mv)2
If the total energy of the system is represented by E and potential energy as V then
KE = E – V
The above equation bwcomes
2m (E – V) = (mv)2
Substituting in equation 1 we get
d 2Ψdx2 +
4 π 2 (2m(E−V ))h2
Ψ = 0
d2Ψdx2 + 8π 2m
h2 ( E-V) Ψ = 0
This is known as time independent schrodinger equation.
4.1.2.Time dependent Schrodinger wave equation :
From time independent Schrodinger equation , we know that
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d 2Ψdx2 +
8π 2mh 2 ( E-V) Ψ = 0 Put ħ =
h2 π
d 2Ψdx2 +
2mħ2 ( E Ψ -V Ψ) = 0
2mħ2 ( E Ψ -V Ψ) = - d 2Ψ
dx2
( E Ψ -V Ψ) = - ħ2
2m
d2 Ψdx2
E Ψ = - ħ2
2m
d 2 Ψdx2 + VΨ
= - ħ2
2m ∇2 Ψ + VΨ ------------------1
The solution of basic wave equation is Ψ = Ψ0 e−iwt
∂Ψ∂t = - iw Ψ0 e−iwt
= - i ( 2πυ) Ψ [ w = 2πυ ]
= - i ( 2π Eh ) Ψ [ E = hυ ]
= - i Eħ Ψ [ ħ =
h2π ]
∴ EΨ = - ħi
∂Ψ∂t
= i ħ ∂Ψ∂t (multiplying both Nr and Dr by i, i2 = -1] ------2
Comparing 1 and 2 we get iħ∂Ψ∂t = - ħ2
2m 𝛁 2 Ψ + VΨ
This is known as Schrodinger time dependent wave equation
1.2 APPLICATION OF SCHRODINGER WAVE EQUATIONTO PARTICLE IN A one – dimensional box
Consider the one dimensional motion of a particle of mass m in a hollow rectangular box having
perfectly rigid walls along X- axis . Let the origin be at one corner of the box. Let L be the distance between
the walls so that the motion along x- axis is confined between x= 0 and x= L. i.e 0 < x < L The particle can
not exist outside the box and therefore for x ≤ 0 and x ≥ L the wave function is zero.
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The time independent Schrodinger equation is
d2Ψd x2 + 8 π 2m
h2 ( E-V) Ψ = 0 -------------------1
Inside the box , no force acts on the particle . Therefore
potential energy inside the box = 0, and the above equation becomes
d2Ψd x2 + 8π 2m
h2 E Ψ = 0 - -------------------2
LET K 2 = 8π 2mh2 E
d2Ψ
d x2 + K2Ψ = 0 ----------------------3
( D2 + K2 ) Ψ = 0
The general solution for second order differential equation is
Ψ = A sin K x + B cos K x ----------------------4
We have the boundary conditions
1. At x = 0 , Ψ = 0
2. At x = L , Ψ = 0
Applying condition 1 in equation 4, we get
0 = A sin K (0) + B cos K (0)
0 = 0 + B (1) [ sin 0 = 0, cos 0 = 1]
∴ B = 0,
substituting in equation 4 we get Ψ = A sin K x --------------------------5
Applying condition 2 in equation 5, we get
0 = A sin K (L)
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There are two possibilities . either A = 0 or sin K(L) = 0 . A can not be zero because already we have proved
that B = 0. Therefore the only possibility is sin KL = 0. This is true only if
KL = n π [ Since sin nπ = 0]
∴ K = nπL Therefore equation 5 becomes
Ψ = A sin ( nπL ) x -----------------------6
This is the wave function for the motion in the region 0 < x< L where A is normalisation constant.
To find normalisation constant:
The condition for normalised wave function is
∫0
L
ΨΨ * dx = 1
∫0
L
A sin (nπL)x A sin( nπ
L) x dx = 1
A2 ∫0
L
sin 2( nπL)x dx = 1
A2 ∫0
L
[1- cos2(nπL) x]/2 dx = 1
A 22 ∫
0
L
[1- cos2(nπL) x] dx = 1
A 22 [ ∫
0
L
1dx - ∫0
L
cos 2( nπL)x dx] = 1
A 22 { [ L ¿ - [
sin 2( nπL)x
2 ¿0
L } = 1
A 22 [ L ¿ = 1
A2 = 2L
∴ A = √ 2L
Hence the normalised wave function is
Ψ = √ 2L sin (
nπL ) x -----------------------7
Ψ1 = √ 2L sin (
πL) x Ψ2 = √
2L sin (
2πL ) x
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Ψ3 = √ 2L sin (
3 πL ) x Ψ4 = √
2L sin (
4 πL ) x
Nodes:
The points where wave function becomes zero are called nodes.
For Ψ1,
√ 2L sin (
πL) x = 0
Divide by √ 2L
sin ( πL) x = 0
we know that sin nπ = 0
( πL) x = nπ
Put n = 0, 1,2,3 ,4
∴ ( πL) x = 0, π, 2π, 3π, 4π
1.When ( πL) x = 0
the value of x is x = 0
2. When ( πL) x = π
the value of x is x = L
3.When ( πL) x = 2π
( 1L) x = 2
x = 2L
( This is impossible because the length of the box is ‘L’ only]
These two points are called extremities. Not nodes.
Therefore there are no nodes for Ψ1
For Ψ2.
√ 2L sin (
2 πL ) x = 0
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Divide by √ 2L
sin ( 2 πL ) x = 0
we know that sin nπ = 0
( 2 πL ) x = nπ
Put n = 0, 1,2,3 ,4
∴ ( 2 πL ) x = 0, π, 2π, 3π, 4π
1.When ( 2 πL ) x = 0
the value of x is x = 0
2.When ( 2 πL ) x = π
( 2L) x = 1
x = L2
3.When ( 2 πL ) x = 2π
( 1L) x = 1
x = L
4.When ( 2 πL ) x = 3π
( 2L) x = 3
x = 3 L2
( This is impossible because the length of the box is L only]
Therefore there is only one node for Ψ2 i.e L2
For Ψ3
√ 2L sin (
3 πL ) x = 0
Divide by √ 2L
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sin ( 3πL ) x = 0
we know that sin nπ = 0
( 3 πL ) x = nπ
Put n = 0, 1,2,3 ,4
∴ ( 3 πL ) x = 0, π, 2π, 3π, 4π
1.When ( 3 πL ) x = 0
the value of x is x = 0
2.When ( 3 πL ) x = π
( 3L) x = 1
x = L3
3.When ( 3 πL ) x = 2π
( 3L) x = 2
x = 2 L3
4.When ( 3 πL ) x = 3π
( 1L) x = 1
x = L
5.When ( 3 πL ) x = 4π
( 3L) x = 4
x = 4 L3
( This is impossible because the length of the box is L only]
Thus the wave function Ψn has ( n- 1) nodes.
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MAXIMA’S IN WAVE FUNCTION:
For Ψ1,
MAXIMUM value of sin( πL) x
sin ( πL) x = 1
sin π2 = 1 sin
3π2 = 1 ,sin
5 π2 = 1
When ( πL) x =
π2
the value of x is x = L2
( πL) x =
3π2 ,
x = 3L2
( This is impossible because the length of the box is L only]
Therefore there is only one maxima for Ψ1,
For Ψ2.
sin ( 2 πL ) x = 1
∴ ( 2 πL ) x =
π2 ,
3 π2 ,
5 π2 , ...
When ( 2 πL ) x =
π2 , the value of x is x =
L4
( 2 πL ) x =
3π2 , the value of x is x = -
3 L4
2 πL ) x =
5 π2 x =
5L4 ( This is impossible because the length of the box is L only]
Therefore there is one positive maxima and one negative ,maxima for Ψ1,
For Ψ3
Similarly for Ψ3 , there are two positive maxima and one negative maxima.
The wave function is represented as
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Nodes: The points where wave function becomes zero are called nodes.
( x= 0 and x= L are the extremities. These are not nodes)
no node for Ψ1,
there is only one node for Ψ2
there are two nodes for Ψ3
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i.eL3 ,
23 L are the two nodes for Ψ3,
Thus the wave function Ψn has ( n-1) nodes.
Eigen values of energy:
From equation 3
K 2 = 8π 2m
h2 E
∴ E = K 2h 28π 2m substituting the value of K,
E = n2 π 2h28π 2m L 2
= n2 h2
8 m L2 This represents the eigen values of the energy.
EIGEN VALUES OF ENERGY
Eigen values of the energy.E = n2 h2
8 m L2
1. If n= 0 , then energy will be zero, Therefore ‘ n’ begins from one ( n = 1,2,3,4 .....)
2. The lowest energy of the particle is called the ground state energy or zero point energy
E1 = h2
8m L2
E2 = 4 h2
8 m L2
= 4 E1
= 22 E1
E3 = 9 h2
8 m L2
= 9 E1
= 32 E1
E4 = 16 h2
8 m L2
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= 16 E1
= 42 E1
In general energy of n th level in terms of ground state energy is En = n2 E1
This shows the energy of the particle is discrete.
3. The spacing between second and first level is
E2- E1 = 4 h2
8 m L2 - h2
8 m L2
= 3 h2
8 m L2
The spacing between third and second level is
E3- E2 = 9 h2
8 m L2 - 4 h2
8 m L2
= 5 h2
8 m L2
In general the spacing between nth energy level and next higher ( n+1 th level is given by
[ ( n+1) 2 – n2 ] E1
= [ n2 + 2n +1 – n2] E1
= [ 2n + 1] E1
Note: If Ψ1 and Ψ2 of the particle in a box of length L are √ 2L sin (
πL) x and √
2L sin (
2 πL ) x ,
Then the value of ∫0
L
Ψ 2Ψ 1 dx = 0
The orthogonal wave function of a particle in one- D box is given by
Ψn = √ 2L sin (
nπL ) x and Ψ m = √
2L sin (
mπL ) x
Then the value of ∫0
L
Ψ m× Ψ n dx = 0
Problem 1 :( 18 of online test) Find the ground state energy of the particle with mass 0.25 units in a one
dimensional box of length 6.62 × 10 – 34 m
Solution:
n = 1 ( ground state)
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L = 6.62 × 10 – 34m
m = 0.25 units
E = n2 h2
8 m L2
= 1× (6.62× 10−34 )2
8 ×0.25 (6.62 ×10−34 )2
= 1
8 ×0.25
= 0.5 J
Problem3 :( 19 of online test) Find the lowest energy of the particle with mass m , in a one dimensional box,
whose length is equal to that of Planc’s constant [ TRB]
Solution:
n = 1 (lowest state)
L = h
E = n2 h2
8 m L2
= 12× h2
8m h2
= 1
8 m
Problem4:( 20 of online test) A particle in 1-D box has a minimum energy of 2.5 eV. Find the next higher
energy it can have .
Solution
E1 = 2.5 eV
E2 = 22 E1
= 4 ( 2.5 )
= 10 Ev
Problem 5. :( 21 of online test)
A particle is confined to 1D box of length ‘2a’ the energy difference between n= 2 and n= 3
Solution:
E2 = 4 h2
8 m (2 a )2
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= 4h2
8 m× 4a2
= 4 h2
32m a2
E3 = 9 h2
8 m (2a )2
= 9h2
8 m× 4a2
= 9h2
32m a2
E3 - E2 = 9h2
32m a2 - 4 h2
32m a2
= 9
5 h2
32m a2
Problem 6 :( 22 of online test) Find the lowest energy of electron confined to move in a one D box of
length 1 Ao. mass of electron = 9.11× 10 -31 Kg, Planc’s constant = 6.62 × 10 -34 JS,
Solution:
n = 1 ( lowest value of n = 1)
L = 1 Ao
= 1 ×10−10 m
E = n2 h2
8 m L2
= 1× (6.62 ×10−34 )2
8 ×9.11×10−31× (10−10 )2
= (6.62 )2×10−68× 10+31×10+20
72. 88
= 43.7 ×10−17
72.88
= 0. 6 × 10 – 17 J
Problem 7 .Find the lowest energy of electron confined to move in a one D box of 1 Ao. mass of electron =
9.11× 10 -31 Kg, Planc’s constant = 6.62 × 10 -34 JS, 1 eV = 1.6 ×10 -19 J
Solution:
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n = 1 ( lowest value of n = 1)
L = 1 Ao
= 1 ×10−10 m
E = n2 h2
8 m L2
= 1× (6.62×10−34 )2
8 ×9.11×10−31×10−10
= 6 × 10 – 18 J
= 6 ×10−18
1.6× 10−10
= 37.5 eV.
Problem 8 If Ψ1 and Ψ2 of the particle in a box of length L are √ 2L sin (
πL) x and
√ 2L sin (
2πL ) x , find the value of ∫
0
L
Ψ 2Ψ 1 dx
Solution:
Ψ1 = √ 2L sin (
πL) x
Ψ2 = √ 2L sin (
2 πL ) x
∫0
L
Ψ 2×Ψ 1dx
= ∫0
L
√ 2L
sin( 2 πL ) x × √ 2
Lsin ( π
L) x dx
= 2L∫0
L
sin( 2 πL )x ×sin( π
L) x dx [ √
2L
× √ 2L =
2L ]
= 2L
× 12∫0
L
cos¿¿¿ dx [ sin A sin B = ½ [ cos ( A-B) + cos (A-B) ]
= 1L [ ∫
0
L
cos (2−1 )( πL ) x+∫0
L
cos (2+1 )( πL ) x] dx
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= 1L [ ∫
0
L
cos ( πL ) x dx+∫
0
L
cos (3 πL ) x] dx
= 1L[sin( πL )x
π /L+
sin (3 πL ) x
3 π /L]
0
L
∫0
L
cos ( πL ) x =
sin( πL )x
π /L
= 1L ×
{sin( π
L )Lπ /L
+sin( 3 π
L )L3 π /L
} – ( sin 0 + sin 0 )
= 1L ×{ sin π
π /L+ sin 3 π
3 π /L } [ sin 0 = 0 ]
= 1L ×0 [ sin π = sin 3π = 0 ]
= 0
Problem 9 The normalised wave function of a particle in one- D box is given by
Ψn = √ 2L sin (
nπL ) x and Ψ m = √
2L sin (
mπL ) x Find the value of ∫
0
L
Ψ m× Ψ n dx
Solution: Ψ m = √ 2L sin (
mπL ) x and
Ψ n = √ 2L sin (
nπL ) x
∫0
L
Ψ m× Ψ n dx = ∫0
L
√ 2L
sin(mπL)x ×√ 2
Lsin ( nπ
L)x dx
= 2L∫0
L
sin(mπL ) x×sin ( nπ
L)x dx
= 2L ×
12 [ ∫
0
L
cos (m−n )( πL ) x+∫0
L
cos (m+n )( πL ) x] dx
= 1L[
sin((m−n) πL ) x
(m−n)π /L+
sin((m+n)πL )x
(m+n) π /L]0
L
= 1L
×{ sin((m−n) π
L )L(m−n) π /L
+sin( (m+n) π
L )L(m+n) π /L
} – { 0 +0} [ sin 0 = 0 ]
= 1L
×{ sin(m−n) π(m−n) π /L
+sin (m+n )π(m+n)π /L
}
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= 1L
×{ 0+0} [ sin (m−n )π = 0]
= 0
Problem 10 Find the length of butadiene chain which has absorption maximum at 2150 A
[ 5.71 ×10−10 m]
Solution:
λ = 2150 ×10−10 m
L = ?
The absorption maximum corresponds to HOMO( Highest Occupied Molecular Orbital) to LUMO ( Lowest
Unccupied Molecular Orbital) transition.
In butadiene E2 is HOMO and E3 is LUMO
Therefore transition from E2 to E3 gives absorption maximum
E2 = 4 h2
8 m L2
E3 = 9 h2
8 m L2
E3- E2 = 9 h2
8 m L2 - 4 h2
8m L2
∆ E = 5 h2
8 m L2
∆ E = hcλ
5h2
8m L2 = hcλ [ comparing]
L2 = 5 h2 λ8 mh c
= 5 h λ8 mc
= 5× 6.62× 1034× 2150 ×10−10
8× 9.11×10−31×3×108
=
L = √❑
= 5.71 ×10−10 m
Problem 4.2.8 Octatetatraene gives first transition absorption band at 4667 A. At what length of the molecule
this transition corresponds to?[ 11.2 A]Page 105 of 419
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Solution:
λ = 4667 ×10−10 m
L = ?
The absorption maximum corresponds to HOMO( Highest Occupied Molecular Orbital) to LUMO ( Lowest
Unccupied Molecular Orbital) transition.
In Octatetatraene E4 is HOMO and E5 is LUMO
Therefore transition from E4 to E5 gives absorption maximum
E4 = 16 h2
8 m L2
E5 = 25 h2
8 m L2
E5- E4 = 25 h2
8 m L2 - 16 h2
8m L2
∆ E = 9 h2
8 m L2
∆ E = hcλ
9h2
8m L2 = hcλ [ comparing]
L2 = 9 h2 λ8 mh c
= 9 h λ8 mc
= 9 ×6.62 ×1034 ×2150 ×10−10
8× 9.11×10−31×3×108
=
L = √❑
= 11.2 ×10−10 m
Problem 4.2.9 The length of Hexatriene molecule was found to be 8.67 A . Find the wavelength for the first
transition. [ 354 nm] OR Calculate the wavelength of π → π transition in 1,3,5 – hexatriene
Solution:
L = 8.67 ×10−10 m
λ = ?
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The absorption maximum corresponds to HOMO( Highest Occupied Molecular Orbital) to LUMO ( Lowest
Unccupied Molecular Orbital) transition.
In Hexatriene E3 is HOMO and E4 is LUMO
Therefore transition from E3 to E4 gives absorption maximum
E3 = 9 h2
8 m L2
E4 = 16 h2
8 m L2
E4- E3 = 16 h2
8 m L2 - 9 h2
8m L2
∆ E = 7 h2
8 m L2
∆ E = hcλ
7h2
8m L2 = hcλ [ comparing]
λ = hc8 m L2
7h2
= c8 m L2
7 h
= 3× 108 ×8 × 9.11× 10−31×(8.67 × 10−10)2
7 × 6.62× 1034
= 354 ×10−9 m
Problem 4.2.11 . Find the lowest energy of neutron confined to move in a one D box of length 10 -14 m.
Given mass of neutron = 1.67× 10 -27 Kg, Planc’s constant = 6.62 × 10 -34 JS, 1 eV = 1.6 ×10 -19 J
E = n2h 28 m L 2
= 2.05 × 106 Ev
Problem 4.2.12. Show that Ψ 1 and Ψ2 of the particle in a box of length L are orthogonal.
Solution:
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Ψ1 = √ 2L sin (
πL) x
Ψ2 = √ 2L sin (
2πL ) x
∫0
L
Ψ 1 ×Ψ 2 dx = ∫0
L
√ 2L
sin( πL)x ×√ 2
Lsin (2 π
L) x dx
= 2L ∫
0
L
sin ( πL )x ×sin( 2 πL)x dx
= 2L ×
12 [ ∫
0
L
cos (2−1 )( πL ) x+∫0
L
cos (2+1 )( πL ) x] dx
= 1L [ ∫
0
L
cos ( πL ) x+∫0
L
cos ( 3 πL )x] dx
= 1L [sin ( πL ) x
π /L+
sin( 3 πL ) x
3 π /L]0
L
= 0∴ Ψ 1 and Ψ2 are orthogonal
Problem 4.2.13. The normalised wave function of a particle in one- D box is given by
Ψ = √ 2L sin (
nπL ) x Show that the function is orthogonal.
Solution: Consider two wave functions
Ψ m = √ 2L sin (
mπL ) x and
Ψ n = √ 2L sin (
nπL ) x
∫0
L
Ψ m ×Ψ n dx = ∫0
L
√ 2L
sin(mπL)x ×√ 2
Lsin ( nπ
L)x dx
= 2L ∫
0
L
sin (mπL )x ×sin( nπ
L)x dx
= 2L ×
12 [ ∫
0
L
cos (m−n )( πL ) x+∫0
L
cos (m+n )( πL ) x] dx
= 1L [sin ((m−n) π
L )x(m−n) π /L
+sin ((m+n)π
L ) x(m+n)π /L
]0
L
= 0
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∴ Ψ m and Ψn are orthogonal
PROBABILITY OF LOCATION OF THE PARTICLE :
According to quantum mechanics, the probability p(x) dx that the particle be found over a small
distance is given byp(x) dx = ⌊ΨΨ ⌋2 dx [ TRB- 17]
= 2L sin2 (
nπL ) x
Problem 8 A particle is confined in a 1- D box of length ‘L’ . Show that the probability that the particle is in
the region 0≤x≤ L2 is 0.5 [ TRB]
Solution:
probability = ∫0
L2
❑Ψ 1Ψ 1 dx
= 0.5
Problem9 For a particle in a state n= 1, of a 1- D box of length ‘L’ Show that the probability that the particle
is in the region 0≤x≤ L4 is
14 -
12 π
Solution:
probability = ∫0
L4
❑Ψ 1Ψ 1 dx
= ∫0
L4
❑√ 2L sin (
πL) x √
2L sin (
πL) x dx
= 2L∫0
L4
❑ sin 2 ( πL) x dx
= 2L∫0
L4
❑ [¿¿ ] dx
= 1L [ ∫
0
L4
d x - ∫0
L4
cos ( 2πL)x dx]
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= 1L { [
L4 ] -
sin( 2πL ) L
42 πL
}
= 1L { [
L4 ] -
sin( 2πL ) L
42 πL
}
= 1L { [
L4 ] -
sin( π2 )2πL
}
= 1L { [
L4 ] -
12 πL
} [sin( π2 ) = 1]
= 1L { [
L4 ] -
L2 π }
= 14 -
12π
The most probable position of the particle is at sin2 ( nπL ) x is maximum. But sin2 (
nπL ) x is maximum, when
( nπL ) x is equal to
π2 ,
3π2 ,
5 π2 ,
7π2 ,
9π2
i.e ( nπL ) x =
π2 ,
3 π2 ,
5 π2 ,
7 π2 ,
9 π2
For the state n= 1, ( πL) x =
π2 ∴ x =
L2
For the state n= 2, ( 2 πL ) x =
π2 ∴ x =
L4
( 2 πL ) x =
3 π2 ∴ x =
3 L4
most probable positions are x = L4 and
3 L4 [ for x=
5 π2 value of x > L which is impossible]
similarly For the state n= 3 , most probable positions are L6 ,
L2 and
5 L6
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Problem 4.2.14 : Find the expectation ( average)value of energy of a particle of mass m confined to move in a
one -D box of width L and infinite height with potential energy zero inside the box. The normalised wave
function is Ψ n = √ 2L sin (
nπL ) x
Solution:
Total energy = kinetic energy + potential energy
Since potential energy is zero, the energy is wholly kinetic which is given by
E = ½ mv2
= m2v 2
2m [ multiplying and dividing by m ]
= p 22m
The operator for p2 is - ћ2 d 2dx2
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∴ The operator for E is - ћ22m
d 2dx 2 . Hence Hamiltonian H = -
ћ22m
d 2dx2
Average energy < E > = ∫0
L
Ψ n H Ψ n dx
= ∫0
L
√ 2L sin (
nπL ) x [ -
ћ22m
d 2dx 2 ] √
2L sin (
nπL ) x dx
= - ћ22m ×
2L ∫
0
L
sin ( nπL ) x
d 2dx2 [ sin (
nπL ) x ] dx
= - ћ2m L ∫
0
L
sin ( nπL ) x
ddx [
nπL cos (
nπL ) x dx
= - ћ2m L ∫
0
L
sin ( nπL ) x [ - (
nπL ) 2 cos (
nπL ) x ] dx
= + ћ2m L ( nπ
L ) 2 ∫0
L
sin 2 ( nπL ) x dx
= + ћ2m L ( nπ
L ) 2 ∫0
L
¿¿] dx
= + ћ2m L ( nπ
L ) 2 12 [ ∫
0
L
1 d x - ∫0
L
cos 2( nπL)x dx]
= + ћ2m L ( nπ
L ) 2 12 { [ L ¿ - [ sin( nπ
L)x ¿¿0
L }
= + ћ2m L ( nπ
L ) 2 12 { L -0 }
= ћ 2n 2π 2
2mL3 L
Substituting the value of ћ = h
2 π we get
Average energy < E > = h2 n2 π 2
4 π 2× 2 mL2
= h2 n28 mL2
Problem 4.2.15 Verify uncertainity principle for the particle in a one- D box of width L and infinite height.
Solution: Hisenberg uncertainity principle is ∆p x ∆ x > h2 where
∆p x 2 = < px 2 > - < px > 2 ,
∆ x 2 = < x 2 > - < x > 2
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1.To find < x>
< x> = ∫0
L
Ψ n x Ψ n dx
= ∫0
L
√ 2L sin (
nπL ) x ( x) √
2L sin (
nπL ) x dx
= 2L ∫
0
L
( x ) sin 2 ( nπL ) x dx
= 2L ∫
0
L
x [¿¿ ] dx
= 1L [ ∫
0
L
x d x - ∫0
L
x cos2( nπL) x dx]
= 1L { [
x 2¿2¿0
L - ( L❑ 2 nπ) xsin( 2 nπ x
L )+(1)(L/2nπ)2cos( 2nπ xL)¿¿0
L }
= 1L {
L22 - 0 } =
L2
2.To find < x 2 >
< x> = ∫0
L
Ψ n x2 Ψ n dx
= ∫0
L
√ 2L sin (
nπL ) x (x2) √
2L sin (
nπL ) x dx
= 2L ∫
0
L
(x2 ) sin 2 ( nπL ) x dx
= 2L ∫
0
L
x2 [¿¿ ] dx
= 1L [ ∫
0
L
x 2 d x - ∫0
L
x 2cos 2( nπL)x dx]
= 1L { [
x3¿3¿0
L - ( L2nπ ) x2sin( 2 nπ x
L )+(2 x )( L2nπ )2 cos ( 2nπ x
L )¿¿0L
+(2)( L2 nπ )3 sin( 2 nπ x
L ) ] =
L23 -
1L { (2 x )( L
2nπ )2 cos ( 2nπ xL )} [ other terms will lead to zero]
= L23 -
1L [
2 L 34n 2 π2 - 0 ]
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= L23 -
1L [
2 L 34n 2 π2 ]
= L23 -
L22n2 π2
3.To find < px >
The operator px is given by ( - i ћ d
d x )
∴ < px > = ∫0
L
Ψ n ( - i ћ d
d x ) Ψ n dx
= ∫0
L
√ 2L sin (
nπL ) x ( - i ћ
dd x ) √
2L sin (
nπL ) x dx
= −2i ћ
L ∫0
L
sin ( nπL ) x
dd x [ sin (
nπL ) x ] dx
= −2i ћ
L ∫0
L
sin ( nπL ) x (
Lnπ ) cos (
nπL ) x ] dx
= −2i ћ
nπ ∫0
L
sin ( nπL ) x cos (
nπL ) x dx put u = sin (
nπL ) x
= −2i ћ
nπ L
nπ [ 12 sin 2 (
nπL ) x ]❑O
L du = ( nπL ) cos (
nπL ) x dx
= 0 ∫udu = u22
4.To find < px 2 > The operator px 2 is given by ( - i ћ
dd x )2
∴ < px 2 > = ∫0
L
Ψ n ( - i ћ d
d x ) 2 Ψ n dx
= ∫0
L
√ 2L sin (
nπL ) x ( - i ћ
dd x ) 2 √
2L sin (
nπL ) x dx
= −2ћ2
L ∫0
L
sin ( nπL ) x
d 2d x 2 [ sin (
nπL ) x ] dx
= −2ћ2
L ∫0
L
sin ( nπL ) x
dd x { (
nπL ) cos (
nπL ) x ] }dx
= −2ћ2
L ∫0
L
sin ( nπL ) x { - (
nπL )2 sin (
nπL ) x ] }dx
= +2ћ2
L (nπL )2 ∫
0
L
sin ( nπL ) x sin (
nπL ) x dx
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= +2ћ2
L (nπL )2 ∫
0
L
sin 2 ( nπL ) x dx
= +2ћ2
L ( nπL )2 ∫
0
L
¿¿] dx
= +2ћ2
L (nπL )2 × 1
2 [ ∫0
L
1 d x - ∫0
L
cos 2( nπL)x dx]
= +2ћ2
L ( nπL )2 × 1
2 { [ L ¿ - [ sin( nπL)x ¿¿0
L }
= +2ћ2
L (nπL )2 × 1
2 { L - 0 }
= ћ 2n2π 2
L 2
= h 2
4 π 2 × n 2π 2
L 2
= n 2h 24 L 2
∆p x 2 = < px 2 > - < px > 2 =
n 2h 24 L 2
∆ x 2 = < x 2 > - < x > 2 = L23 -
L 22n2 π2 - (
L2 ) 2
= L212 -
L22n2 π2
∆ x 2 × ∆p x 2 = ( L212 -
L 22n2 π2 ) × (
n2h24 L 2 ) = (
112 -
12n 2 π 2 ) × (
n 2h 24 )
∴ ∆ x × ∆p x = nh2 × (
112 -
12n2 π2 ) ½ >
h2 This verifies Heisenberg’s uncertainty principle.
9.average value of position of a particle in a 1-D box in its ground state
< x> = ∫0
L
Ψ n x Ψ n dx
= ∫0
L
Ψ 1 x Ψ 1 dx
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= ∫0
L
√ 2L sin (
πL) x ( x) √
2L sin (
πL) x dx
= 2L ∫
0
L
( x ) sin 2 ( πL) x dx
= 2L ∫
0
L
x [¿¿ ] dx
= 1L [ ∫
0
L
x d x - ∫0
L
x cos2( πL)x dx]
= 1L { [
x 2¿2¿0
L - ( L❑ 2 nπ) xsin( 2 nπ x
L )+(1)(L/2nπ)2cos( 2nπ xL)¿¿0
L }
= 1L { L2
2 - 0 }
= L2
10 average value of px 2 of a particle in a 1-D box in
The operator px 2 is given by ( - i ћ d
d x )2
∴ < px 2 > = ∫0
L
Ψ n ( - i ћ d
d x ) 2 Ψ n dx
= ∫0
L
√ 2L sin (
nπL ) x ( - i ћ
dd x ) 2 √
2L sin (
nπL ) x dx
= −2ћ2
L ∫0
L
sin ( nπL ) x
d 2d x 2 [ sin (
nπL ) x ] dx
= −2ћ2
L ∫0
L
sin ( nπL ) x
dd x { (
nπL ) cos (
nπL ) x ] }dx
= −2ћ2
L ∫0
L
sin ( nπL ) x { - (
nπL )2 sin (
nπL ) x ] }dx
= +2ћ2
L (nπL )2 ∫
0
L
sin ( nπL ) x sin (
nπL ) x dx
= +2ћ2
L (nπL )2 ∫
0
L
sin 2 ( nπL ) x dx
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= +2ћ2
L ( nπL )2 ∫
0
L
¿¿] dx
= +2ћ2
L (nπL )2 × 1
2 [ ∫0
L
1 d x - ∫0
L
cos 2( nπL)x dx]
= +2ћ2
L ( nπL )2 × 1
2 { [ L ¿ - [ sin( nπL)x ¿¿0
L }
= +2ћ2
L (nπL )2 × 1
2 { L - 0 }
= ћ 2n2π 2
L 2
= h 2
4 π 2 × n 2π 2
L 2
= n 2h 24 L 2
3.3 APPLICATION OF SCHRODINGER WAVE EQUATIONTO PARTICLE IN A TWO dimensional box
In two dimensional box the total energy of the oscillator is sum of its components whereas total wave
function is the product of these two
Ψtotal = Ψx × Ψy
Etotal = Ex + Ey
The normalised wave function for a particle in a one dimensional box of length L in x – direction is
Ψ nx = √ 2a sin (
nπa ) x
The normalised wave function for a particle in a one dimensional box of length L in y – direction is
Ψ ny = √ 2b sin (
nπb ) y ,
Ψ xy = Ψ nx Ψny
= √ 2a sin (
nπa ) x √
2b sin (
nπb ) y
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= 2√ab
sin ( nπa ) x sin (
nπb ) y
Eigen values of energy:
If the dimensions of the particles are ‘a’ and ‘b’ then
Exy = h2
8m [
nx2
a2 + ny
2
b2 ]
Problem 1. 7 An electron is confined to a 2-D square of side 1 nm .Find its minimum energy
Solution:
nx = n y = 1 ( minimum)
a = b = 1 ( square )
Exy = h2
8m [ 1 + 1 ]
= h2
4 m
Problem 1. 7 An electron is confined to a two dimensional square of side 1 nm . Find the minimum excitation energy. Solution: nx = 1 n y = 2 ( minimum excitation ) a = b = 1 ( square )
Exy = h2
8 m [ 12 + 22 ]
= 5 h2
8 m
5. Calculate the energy of a particle in a rectangular box of sides 2a and a. Solution:
Exy = h2
8 m [ 12
(2 a)2+ 22
(a)2 ]
= h2
8m [
14 a2+
4a2 ]
= h2
8m [
1+164 a2 ]
= h2
8m [
174 a2 ]
Exy = h2
8 m [
12
(a)2+
22
(2 a)2 ]
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= h2
8m [
1a2 +
44 a2 ]
= h2
8m [
4+44 a2 ]
= h2
8m [
2a2 ]
3.4 APPLICATIONS OF SCHRODINGER WAVE EQUATIONTO PARTICLE IN A THREE dimensional box
Let a particle of mass m be in motion in a rectangular potential box with sides of length a,b, c parallel
to the x, y and z- axis respectively. Suppose there is no force acting on the particle inside the box, so that
inside the region 0 < x < a, 0 < y < b, 0 < z < c,
0 < x < a, The potential energy V = 0
0 < y < b, The potential energy V = 0
0 < z < c The potential energy V = 0
The time independent Schrodinger equation for the particle is
∂ 2Ψ∂ x2 +
∂ 2Ψ∂ y 2 +
∂ 2Ψ∂ z 2 +
8 π 2mEh2 Ψ = 0 ------------------1
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We assume that the wave function Ψ is equal to the product of three functions X,Y and Z each of which is a
function of one variable only.
Thus we have Ψ( x, y, z) = X(x) Y(y) Z(z)
i. e Ψ = X Y Z ------------------2
∂Ψ∂ x = YZ
dXdx
∂Ψ∂ y = XZ
dYdy
∂Ψ∂ z = XY
dZdz
∂ 2Ψ∂ x2 = YZ
d2 Xdx 2
∂ 2Ψ∂ y 2 = XZ
d 2Ydy2
∂ 2Ψ∂ z 2 = XY
d2 Zdz2
We have used ordinary derivative instead of partial derivatives because each of the functions X,Y, and Z is a
function of one variable only.
Substituting in equation 1, we get
YZ d 2 Xdx 2 + XZ
d 2 Ydy 2 + XY
d 2 Zdz 2 +
8 π 2 mEh 2 XYZ = 0
Dividing by XYZ
1X
d 2 Xdx 2 +
1Y
d2 Ydy 2 +
1Z
d2 Zdz 2 +
8 π 2mEh 2 = 0 -----------------------3
Since velocity and hence the kinetic energy ( E) of the particle , being a vector quantity, the energy in the
above expression can be expressed as the sum of the corresponding terms Ex Ey and Ez .Therefore the above
equation becomes
1X
d2 Xdx 2 +
1Y
d 2Ydy2 +
1Z
d 2 Zdz 2 +
8π 2 m(E x+Ey+Ez)h2
= 0
1X
d2 Xdx 2 +
8 π 2m E xh 2 +
1Y
d2 Ydy 2 +
8 π 2m Eyh2 +
1Z
d2 Zdz 2 +
8π 2 m Ezh2 = 0 --------4
This equation gives three independent equations.
1X
d 2 Xdx 2 +
8π 2 m E xh2 = 0 -------------------5
1Y d 2Y
dy2 + 8π 2 m Eyh2 = 0 --------------------6
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1Z
d 2 Zdz2 +
8 π 2 m Ezh 2 = 0 -------------------7
Equation 5 can be written as
d2 Xdx 2 +
8 π 2m E xh 2 X = 0
This is the equation for one dimensional case. The boundary condition applicable to the solution is
Ψ nx = √ 2a sin (
nπL ) x , Ψ ny = √
2b sin (
nπL ) y , Ψ nz = √
2c sin (
nπL ) z
Ψ = Ψ nx Ψ ny Ψ nz
= √ 2a sin (
nπL ) x √
2b sin (
nπL ) y √
2c sin (
nπL ) z
Ψ = √ 8√abc
sin ( nxπ
L ) x sin ( nyπL ) y sin (
nzπL ) z
The normalised wave function for a particle in a one dimensional box of length ’a’ along X- direction is
given by Ψx = √ 2a sin (
nπa ) x
breadth ’b’ along Y - direction is given by Ψ y = √ 2b sin (
nπb ) y
height ’c’ along Z- direction is given by Ψz = √ 2c sin (
nπc ) z
The normalised wave function for a particle in a three dimensional box of length ’a’, breadth ‘b’ and height ‘c’ is
Ψ = √ 2a sin (
nπa ) x × √
2b sin (
nπb ) y × √
2c sin (
nπc ) z
= √ 8√abc
sin ( nπa ) x sin (
nπb ) y sin (
nπc ) z
= √ 8V
sin ( nπa ) x sin (
nπb ) y sin (
nπc ) z [ V = abc]
Eigen values of energy:
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E xyz = h2
8 m [
nx2
a2 + ny
2
b2 + nz
2
c2 ]
For a cubic box , a = b= c = L
E xyz = h2
8m L2 [ nx2+ny
2+nz2]
PARTICLE IN 3-D BOX
Time independent Schrodinger equation is d2Ψ
dx2 + 8 π 2mh2 ( E-V) Ψ = 0
where E – is the total energy and
V potential energy
‘h’ – Plank’s constant
Ψ represents the amplitude of spherical wave
The wave function should be multiplied whereas energy should be added
The normalised wave function for a particle in a one dimensional box of
length ’a’ along X- direction is given by Ψx = √ 2a sin (
nπa ) x
breadth ’b’ along Y - direction is given by Ψ y = √ 2b sin (
nπb ) y
height ’c’ along Z- direction is given by Ψz = √ 2c sin (
nπc ) z
The normalised wave function for a particle in a three dimensional box of length ’a’, breadth ‘b’ and height ‘c’
is Page 122 of 419
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Ψn = √ 2a sin (
nπa ) x × √
2b sin (
nπb ) y × √
2c sin (
nπc ) z
= √ 8√abc
sin ( nπa ) x sin (
nπb ) y sin (
nπc ) z
For cube ‘ a= b= c’ , abc = volume of the cube V ‘ therefore the above equation becomes
Ψn = √ 8V
sin ( nπa ) x sin (
nπb ) y sin (
nπc ) z [ V = abc]
√ 8abc
is called normalization constant
Problem Find the ground state wave function of a particle in a cubic box of dimension 2 units
Solution:
nx = n y = nz= 1 [ ground state]
a = 2 , b= 2, c= 2 [ cubic box]
Ψxyz = √ 8√abc
sin ( nπa ) x sin (
nπb ) y sin (
nπc ) z
= √ 8
√(2)(2)(2) sin (π2 ) x sin (
π2 ) y sin (
π2 ) z
= sin (π2 ) x sin (
π2 ) y sin (
π2 ) z
Problem The wave function of a particle in the state (112) in a cubic box of length 3 units
Solution:
nx = n y = 1
nz= 2
a = 3 , b=3, c= 3 [ cubic box]
Ψxyz = √ 8√abc
sin ( nπa ) x sin (
nπb ) y sin (
nπc ) z
= √ 8
√(3)(3)(3) sin (π3 ) x sin (
π3 ) y sin (
2 π3 ) z
= √ 8√27
sin (π3 ) x sin (
π3 ) y sin (
2 π3 ) z
Eigen values of energy:
E xyz = h2
8 m [
nx2
a2 + ny
2
b2 + nz
2
c2 ]
For cube ‘ a= b= c = L
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E xyz = h2
8m L2 [ nx2+ny
2+nz2]
1. If n= 0 , then energy will be zero, which is not true .Therefore the possible values of n are
n = 1,2,3,4 .....
2. The lowest energy of the particle is obtained by putting nx = ny = nz= 1 in the above equation. It is
called the ground state energy or zero point energy
E111 = h2
8m L2 [ 12 + 12+ 12]
= 3 h2
8 m L2
. Zero point energy of particle in a 3-D is 3 times that of 1- D box.
3. Similarly E112 = h2
8m L2 [ 12 + 12+ 22]
= 6h2
8m L2
E121 = h2
8 m L2 [ 12 + 22+ 12]
= 6h2
8m L2
E211 = h2
8 m L2 [ 22 + 12+ 12]
= 6h2
8m L2
E112 = E121 = E211 = 6h2
8 m L2
Degeneracy:
The particle having different wave functions may have same energy in an excited state . Such states
are said to be degenerate.
For example consider the first excited state The possible values for nx ,n y and nz are
( 1,1,2), ( 1,2,1), ( 2, 1, 1). Their energies are calculated as follows.
E 112 = h2
8 m L2 [12 +12 + 22 ] = 6 h2
8 m L2
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E 121 = h2
8 m L2 [12 +22 + 12 ] = 6 h2
8m L2
E 211 = h2
8 m L2 [22 +12 + 12 ] = 6h2
8m L2
In all cases energy is same . Such states are said to be degenerate. The number of possibilities are called
degree of degeneracy.
In the above case there are three possibilities and hence degree of degeneracy is three. It is said to be triply
degenerate.
Problem .A particle is in the state (123) Find its degeneracy.
Solution
Given state (123) :
The possibilities are (123), (132),(213),(231), (321),(312)
The number of possibilities is six
Therefore for the state ( 123) degeneracy = 6 [ SIX FOLD ]
Problem A particle is in the state (12,2 ) Find its degeneracy.
Solution : state (122) :
The possibilities are (122), (2,1,2),(2,2,1)
The number of possibilities is three
Therefore for the state ( 122) degeneracy = 3
NOTE : .The maximum number of degeneracy = 6
Problem1 . ( 13 of test) The ground state energy of a particle in a 1D box of length 10 m is 5 eV . if the same
particle is placed in a cubic box of side 10 m, its ground state energy is [ TRB]
Solution:
Given h2
8 m L2 = 5 eV
3h2
8m L2 = ?
E = 3 ( 5)
= 15 eV
Problem 2. ( 14 of test) Find the ground state energy of a particle with mass ‘m’ in a 3-D box of
dimensions1,2 and 2 units
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Solution:
nx = n y = nz= 1 [ ground state]
a = 1 , b= 2, c= 2
E xyz = h2
8 m [
nx2
a2 + ny
2
b2 + nz
2
c2 ]
E 122 = h2
8m [
112 +
122 +
122 ]
= h2
8 m [1 +
14 +
14 ]
= h2
8 m [
64 ]
= 3h2
16 m
Problem3 ( 15 of test) The ground state energy of a particle with mass ‘ 10 ’ unit present in a cubic box of
dimensions 3 units is
Solution:
nx = n y = nz= 1 [ ground state]
a = 3 , b= 3, c= 3 [ cubic box]
E xyz = h2
8 m L2 [ nx2+ny
2+nz2]
E 111 = h2
8 (10 )(3)2 [ 1+1+1 ]
= h2
240
Problem 4 ( 16 of test) Find the energy required for the transition of electron with mass “m” in a cubic box
of dimension of 3 units, from ground state to first excited state
Solution:
For ground state
nx = n y = nz= 1
a = 3 , b= 3, c= 3 [ cubic box]
E xyz = h2
8 m L2 [ nx2+ny
2+nz2]
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E 111 = h2
8 m(3)2 [ 1+1+1 ]
= h2
24 m
For first excited state nx = n y = 1, nz= 2
a = 3 , b= 3, c= 3 [ cubic box]
E xyz = h2
8 m L2 [ nx2+ny
2+nz2]
E 112 = h2
8 m(3)2 [12 +12 + 22 ]
= 6 h2
72m
= h2
12m
Energy required for the transition of electron from ground state to first excited state
= E 112 - E 111
= h2
12m - h2
24 m
= h2
24 m
Problem 5( 17 of test) A particle in a 3-D box has energy 14 h2
8 m L2 Find its degeneracy.
Solution:
E = 14 h2
8m L2
Exyz = h2
8m L2 [ x2 + y2+ z2]
= 14 h2
8 m L2
∴ x2 + y2+ z2 = 14
put x = 1(assuming) then y2+ z2 = 14 – 1
= 13
y2 = 13 - z2
possible values of ‘z’ and ‘y’ are 2 and 3
The energy state is (123)Page 127 of 419
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Possible values are (123), (132), (231),(213), (312), and (321)
Therefore degeneracy is 6
Problem 6. ( 18 of test) A particle in a 3-D box has energy 9 h2
8 m L2 Find its degeneracy
Solution: . E= 9 h2
8 m L2
Exyz = h2
8 m L2 [ x2 + y2+ z2]
= 9 h2
8 m L2
∴ x2 + y2+ z2 = 9
put x = 1 then y2+ z2 = 9 – 1 = 8
y2 = 8 - z2
possible values of ‘z’ and ‘y’ are 2 and 2
The energy state is (122) ,
Possible values are (122), (221), and (212) Therefore degeneracy is 3
Problem 7. ( 19 of test) .Find the degeneracy of the excited state of a particle in a 3D cubic box with energy
three times of its ground state energy
Solution: For a cubical box Ground state energy E = 3 h2
8 m L2
Given energy = three times of Ground state energy = 3 × 3 h2
8 m L2
= 9 h2
8 m L2
x2 + y2+ z2 = 9put x = 1 then y2+ z2 = 9 – 1 = 8 y2 = 13 - z2
possible values of ‘z’ and ‘y’ are 2 and 2 The energy state is (122),Possible values are (122), (221), and (212) Therefore degeneracy is 3
Problem 8. ( 20 of test) Find the lowest energy of neutron confined to move in a cubical box each side 6.62
Ao. Assume mass of neutron = 1.7× 10 -27 Kg, Planc’s constant = 6.62 × 10 -34 JS, 1 eV = 1.6 ×10 -19 J
Solution: For a cubical box
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E xyz = 3h2
8m L2 [ nx2+ny
2+nz2]
for lowest energy nx = ny = nz = 1
∴ E = 3 h2
8 m L2
= 3 × (6.62× 10−34 )2
8 ×1.7 ×10−27× (6.62× 10−34 )2
= 3
13.6×1027 J
= 30
136×1027 J
= 0. 22 ×1027J
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5. SCHRODINGER EQUATION APPLICABLE TO COMPLEX SYSTEMS
5.1. HARMONIC OSCILLATOR( vibrational motion)
Quantum mechanical treatment of harmonic oscillator gives different results from those obtained from
classical treatment. Simple harmonic motion is a back and forth motion along the same path in which the
displacement varies with time.
5.1.1. Classical Treatment of Harmonic Oscillator
Harmonic oscillator is one, whose oscillation obeys the following conditions
(1) When it oscillates, the restoring force should be proportion to the displacement.
(2) The restoring force should be directed towards the mean position.
Consider an oscillator whose displacement is given by,
x = a sin ωt
It’s potential energy (PE) = ½ k x2
where γ = 1
2 π √ km
, k – force constant , m- mass
Squaring γ2 = = 1
4 π2 km , ∴ k = γ2 4π2 m
Substituting in2 PE = ½ × γ2 4π2 m x2
= 2γ2 π2 m x2
= 2mγ2 π2 a2 sin2 wt [ using 1] -------------------3
KE = ½ m v2 = ½ m ( dxdt ) 2 -----------------------------4
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From 1, dxdt = aw cos wt
( dxdt ) 2 = a2 w2 cos2 wt
= a2 (2πγ)2 cos2 wt
= 4γ2 π2 a2 cos2 wt
From4, KE = ½ m×4γ2 π2 a2 cos2 wt
= 2m γ2 π2 a2 cos2 wt -------------------4
TE = PE + KE
= 2mγ2 π2 a2 sin2 wt + 2m γ2 π2 a2 cos2 wt
= 2mγ2 π2 a2 ( sin2 wt + cos2 wt )
= 2mγ2 π2 a2
k = γ2 4π2 m
k2= 2mγ2 π2 Total Energy = ½ k a2
Quantum Mechanical Treatment:
When a particle oscillates about its mean position under the action of force , and if the
restoring force is
1. proportional to the displacement and
2. directed towards the mean position,
then the motion of the particle is said to be simple harmonic and the particle is called harmonic oscillator.
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.When x is the displacement from 0, the restoring force is given by
F = - K x where K = mw2 --------------------1
Let the particle moves a small distance dx
Potential energy = work done to bring a particle from infinity to small distance dx.
= Force × displacement [ F ∝ - displacement]
= - ∫x
∞
k xdx [ F = - k x]
= - k x2
2¿¿x
∞ [∫0
∞
xdx= x2
2 ]
= - k [ ∞ - x2
2 ]
Potential energy = k x2
2
Potential energy harmonic oscillator V = k x2
2. where ‘k’ is called force constantPotential energy curve of
harmonic oscillator is parabola.
Problem 3.1 Find the potential energy harmonic oscillator
Solution
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Potential energy = k x2
2
WAVE FUNCTION and ENERGY
The time independent Schrodinger equation is
d2Ψ
d x2 + 8 π 2mh2 ( E-V) Ψ = 0 --------------------1
the potential energy harmonic oscillator is k x2
2 where ‘k’ is force constant = m ω2 substituting in the above
equation
d2Ψ
dx2 +8 π 2mh2 (E - ½ kx2 ) Ψ = 0 ------------2
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Expanding
d2Ψ
dx2 + [ 8 π 2mh2 E - 8 π 2m
h2 ( ½ kx2 )] Ψ = 0
d2Ψ
dx2 + [ 8 π 2mh2 E - 4 π2 m
h2 kx2 ] Ψ = 0
d2Ψ
dx2 + [ 8π 2mh2 E - 4 π2 m2w2
h2 x2 ] Ψ = 0 [ k = mw2 ]
let α 4 = 4 π2 m2w2
h2 ,and β = 8 π 2mEh2 [ why not α =
4 π2 m2w2
h2 ? see appendix]
then the above equation becomes,
d2Ψ
d x2 + [β – α 4 x2 ] Ψ = 0
d2Ψ
d x2 + [β – α 2 α 2 x2 ] Ψ = 0
d2Ψ
d x2 + [β – α 2 (αx )2 ] Ψ = 0
The above second order differential equation can not be solved easily due to the coefficient α 4 . In order to
eliminate α 4 :
Let y = αx then d2Ψ
d x2 = α 2 d2Ψ
d y2 *
∴ α 2 d2Ψd x2 + [β – α 2 y2 ] Ψ = 0
d2Ψ
d y2 + [β
α 2 – y2 ] Ψ = 0 [Dividing by α 2 ]
Let λ = β
α 2 , then
d2Ψ
d y2 + [λ– y2 ] Ψ = 0
------------------------------------------------------------------------------------------------Note * y = αx
dydx
= α
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By chain rule, we have d Ψdx
= dΨdy
× dydx
= dΨdy
× (α )
Again differentiating with respect to x
d2Ψd x2 =
ddx
(α dΨdy
) = ddy
( α dΨdy
) × dydx
[ by chain rule]
= ddy
( α dΨdy
) × (α )
= α 2 d2Ψd y2
-------------------------------------------------------------------------------------------
When λ≪ y d2Ψ
d y2 – y2 Ψ = 0
The general solution of the equation is Ψ = e− y2
2 or e+ y2
2
Since Ψ has to remain finite for all values of y, since , e - ∞ = 0 and e + ∞ = ∞ , the acceptable solution is Ψ
= e− y2
2
∴ The general solution is Ψ = H × e− y2
2 where H is a finite polynomial in y
To find H:
Ψ = H ×e− y2
2
dΨdy = H (e
− y2
2 ) ( - y) + (e− y2
2 ) H ‘ [differentiating with respect to y ]
= e− y2
2 [ H ‘ – y H]
d2Ψ
d y2 = e− y2
2 [ H ‘‘– y H’– H ]+[ H ‘– y H]( ½ (-2y) e− y2
2 [again differentiating by UV model]
= e− y2
2 [ H ‘‘ – y H’ – H ] + [ H ‘ – y H] (-y) e− y2
2
= e− y2
2 [ H ‘‘ – y H’ – H ] + [ -y H ‘ + y 2 H] e− y2
2
= e− y2
2 [ H ‘‘ – y H’ – H -y H ‘ + y 2 H]
= e− y2
2 [ H ‘‘ –2 y H’ – H + y 2 H]
Substituting in d2Ψ
d y2 + [λ– y2 ] Ψ = 0 we get
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(e− y2
2 ) [ H’’– 2y H’– H+ y2 H ] + [λ– y2 ] H e− y 2
2 = 0
[ H’’ – 2y H’ + y2 H – H] + [λ– y2 ] H = 0 [ Dividing by e− y2
2 ]
H’’ – 2y H’ + y2 H – H + λ H – y2 H = 0
H’’ – 2y H’ – H + λ H = 0
H’’ – 2y H’ + ( λ - 1 ) H = 0
This resembles Hermite differential equation which is
d2Ψ
d y2 - 2 x dydx + ( λ - 1 ) y = 0
Therefore the solution of the above equation is that of Hermite equation which is
y = N e− y2
2 Hn ( y) here N is normalisation constant = ¿ ( απ)
14
Hn( y) is Hermite polynomial , Hn = ( -1) n e y2
dn
d yn (e− y2
)
∴ The total wave function is
Here n can have minimum value zero. Therefore
To find energy:
The Hermite differential equation is d2 H
d y2 - 2 y dHdy + ( λ - 1 ) H = 0 -----1
OR H’’ – 2yH’ + ( λ - 1 ) H = 0
According to Frobenius method, the solution of this equation will be the following form
Let H = ∑n=0
∞
An y n
Differentiating H’ = ∑n=0
∞
An n y n-1
Again differentiating H’’ = ∑n=0
∞
An n ( n-1) y n-2
To convert this term into term with yn , we expand H’’, by putting the values for ‘n’ from ‘0’
H’’ = A0 (0) y -2 + A1(0) y -1 + A2 (2) (1) y 0 +A3 (3) (2) y 1 +A4 (4) (3) y 2 +…
= 0 + 0 + A2 (2) (1) y 0 +A3 (3) (2) y 1 +A4 (4) (3) y 2 +…
To convert y n-2 in to y n this can be generalized as belowPage 136 of 419
Ψ =¿ ( απ)
14 × e
− y2
2 × ( -1) n e y2
dn
d yn (e− y2
)
2020 - CONCISE QUANTUM MECHANICS
H’’ = ∑n=0
∞
(n+2)(n+1) An+2 y n [ why not we expand H’?, see appendix]
Substituting in equation 1 we get
∑n=0
∞
(n+2)(n+1) An+2 y n -2y ∑n=0
∞
An n y n-1 +∑n=0
∞
An y n = 0
∑n=0
∞
(n+2)(n+1) An+2 y n -2 ∑n=0
∞
An n y n +∑n=0
∞
An y n = 0 [y (y n-1 ) = y n ]
Dividing by ∑0
∞
yn we get
(n+2 ) (n+1 ) A n+2 – 2 n An + ( λ - 1 ) An = 0
Taking An as common for second and third term
(n+1 ) (n+2 ) A n+2 - [ 2 n - ( λ - 1 ) ] An ¿ = 0
(n+1 ) (n+2 ) A n+2 = [ 2 n - ( λ - 1) ] An
rearranging A n+2 = [2 n−(λ−1)](n+1 ) (n+2 ) An
This is known as recursion formula. For certain value of ‘n’ the numerator becomes zero.
2n - ( λ - 1 ) = 0
2n = λ - 1
∴ λ = 2 n + 1
Substituting the values of λ = β
α 2 in equation we get
βα2 = 2n + 1
but β = 8π 2mEh2
and , α 4 = 4 π2 m2w2
h2 , ∴ α 2 = 2πmw
h
the above equation becomes
8 π 2mEh2 ×
h2πmw = 2n + 1
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4 πEhw = 2n +1
Substituting w= 2πγ we get 4 πE
h(2πγ ) = 2n +1
2Ehγ = 2n +1
E = (2n+1)
2 hγ
= hγ (n + 12)
This is the expression for energy.
Significance of wave function
Here n can have minimum value zero. Therefore
Ψ0 = ¿ ) e− y2
2 H0
= (απ ) ¼
e− y2
2 This is the ground state wave function.
Ψ1 = ¿ ( απ ) ¼
e− y2
2 H1
= ¿ ( απ ) ¼
e− y2
2 × 2y
= ¿ ( απ ) ¼
e− y2
2 × 2y
= 1√2
× ( απ ) ¼
e−y2
2 × √2 × √2 y [ 2 = √2 × √2 ]
= ( απ ) ¼
e− y2
2 ×√2 y
Ψ2 = ¿ ( απ ) ¼
e− y2
2 H2
= 1
2√2 (
απ ) ¼
e− y2
2 ×2(2 y2 -1)
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= 1√2
( απ ) ¼
e− y2
2 ×(2 y2 -1)
Similarly Ψ3 = ( απ ) ¼
e− y2
2 (y 2 - 2 )
Values of Normalisation constant
n= 0( α
π)
14
n= 1 1√2( α
π)
14
n= 2 1√8( α
π)
14
Physical interpretation of wave function:
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HERMITE POLYNOMIAL
Hermite differential equation is d2 Hd y2 - 2 y
dHdy + (λ -1) H = 0
Its solution is known as Hermite polynomial
Hn( y) is Hermite polynomial , Hn = ( -1) ne y2 dn
d yn (e− y2
)
When n= 0 H0 = 1 similarly H1 = 2y, H2 = 4y2 – 2 H3 = 8y3 -12y
Problem 3.2 Find the value of H0 andH1 Solution
Hn = ( -1) ne y2 dn
d yn (e− y2
)
H0 = ( -1) 0 e y2 × e – y2 )
= 1
H1= ( -1) e y2 ddy (e−y2
)
= -e y2
( - 2ye− y2
)
= 2y
Problem 3.3 Find the value of H2 Solution
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Hn = ( -1) ne y2 dn
d yn (e− y2
)
H2 = ( -1) 2e y2 d2
d y2 (e− y2
)
= + e y2 ddy (−2 y e− y2
)
= - 2e y2 ddy ( ye− y2
)
= - 2e y2
{ y¿) + e− y2
(1) }
= - 2e y2
{ −2 y2e− y2
+ e− y2
}
= 4 y2 -2
= 2(2 y2 -1)
Values of Hermite polynomial
H 0 1
H1 2y
H 2 4y2 -2
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Significance of energy
energy of the oscillator is given by E = hγ (n + 12)
where ‘n’ is vibrational quantum number.
1. The lowest energy of the oscillator , is called ground state energy or zero point energy and is obtained by
putting n = 0, in the above expression.
E0 = 12 h γ
but according to classical mechanics , zero point energy of oscillator is zero
2. All the energy levels of the oscillator are non-degenerate.
E1 = hγ (1 + 12)
= 32 h γ,
E2 = hγ (2 + 12)
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= 52 h γ,
E3 = hγ (3 + 12)
= 72 h γ
3. The difference in energies of successive energy levels are
E1 – E0 = 32 h γ -
12 h γ
= h γ
E2 – E1 = 52 h γ -
32 h γ
= h γ
E3 – E2 = 72 h γ -
52 h γ
= h γ .
This shows the successive energy levels are equally spaced.
The separation between two adjacent energy levels being hγ.
Values of energy
E0 12 hϑ
E1 32 hϑ
E2 52 hϑ
difference in energies of successive
energy levels
E1 – E0 hϑ
E2 – E1 hϑ
E3 – E2 hϑ
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1.Show that the ground state wave function of harmonic oscillator is normalized
Solution:
Ψ0 = (απ ) ¼
e− y2
2
We have to prove ∫−∞
∞
Ψ 0 × Ψ 0 dx = 1
∫−∞
∞
Ψ 0 × Ψ 0 dx = ∫−∞
∞
(απ )¼
e− y2
2 × (απ )¼
e− y2
2 dx
= (απ )12 ∫−∞
∞
e− y2
dx
Since y ¿√α x, dy =√αdx . Therefore dx = dy√α
substituting we get
= (απ )12 ×2 × ∫
0
∞
e− y2
dy√α
[ ∫−∞
∞
e− y2
dy = 2 ∫0
∞
e− y2
dy ]
= (απ )12 ×2× 1
√α ∫
0
∞
e− y2
dy
= √α√π
× 2 × 1√α
× [ 12
×√ π ] ∫0
∞
e−α x2
dx = 12
×√ πα
]
= 1
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2.Show that the wave function Ψ1 = ( απ ) ¼ e
− y2
2 ×√2 y where y = (√α x) is normalized
Solution:
We have to prove ∫−∞
∞
Ψ 1 × Ψ 1 dx = 1
∫−∞
∞
Ψ 1 × Ψ 1 dx
= ∫−∞
+∞ [(απ )¼ e− y2
2 ×√2 y ]×[(απ )¼ e− y2
2 × √ 2 y¿]¿ dx
= ∫−∞
+∞ [(απ )¼ ×(απ )¼
× e− y2
2 × e− y2
2 ×√2 y×√2 y ] dx
= ∫−∞
+∞
[(απ )12 × e− y2
× 2 y2] dx
= 2 ×∫0
+∞
[( απ )
12 ×e− y2
×2 y2] dx [ ∫−∞
∞
e− y2
dy = 2 ∫0
∞
e− y2
dy ]
= 4 ×(απ )12∫
0
+∞
[e− y 2
y2 ] dx
Since y ¿√α x, dy =√αdx . Therefore dx = dy√α
substituting we get
=2 ∫0
+∞
[e− y2
y2 ] dy√α
= 4 (απ )12 × 1√α
×∫0
+∞
[e− y2
y2 ]dy
= 4 (απ )12 × 1√α
× [ 14
×√ π ] [ ∫0
∞
e−α x2
x2 dx = 14
×√ πα3 ]
= √α√π
×1√α
×√π
= 1
This shows that wave function Ψ1 is normalized
3. Show that the wave function Ψ0 and Ψ1 of harmonic oscillator are orthogonal
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Solution:
We have to prove∫−∞
∞
Ψ 0 × Ψ 1 dx = 0
Ψ0 = ( απ ) ¼ e
− y2
2
Ψ1 = 1√2
( απ ) ¼ e
− y2
2 ×2 y
∫−∞
∞
Ψ 0 × Ψ 1 dx = ∫−∞
∞
¿¿ 1√2
(απ )¼
e− y2
2 ×2 y ] dx
= (απ )12 ∫−∞
∞
¿¿ 1√2
×2 y ] dx
= (απ )12 × √2 ∫
−∞
∞
¿¿ y ] dx
Since y ¿√α x, dy =√αdx . Therefore dx = dy√α
substituting we get
= (απ )12 × √2 ∫
−∞
∞
¿¿ y ] dy√α
= (απ )12 × √2 1
√α ∫−∞
∞
¿¿ y ] dy
= (απ )12 × √2 1
√α [ 0] ∫
−∞
∞
¿¿ y ] dy is an odd function. So its value is zero
4. Show that the wave function Ψ1 and Ψ2 of harmonic oscillator are orthogonal
Solution:
We have to prove∫−∞
∞
Ψ 1 × Ψ 2 dy = 0
Ψ1 = ( απ ) ¼ e
− y2
2 ×2 y,
Ψ2 = 1√2 (απ )
¼
e− y2
2 × 4 y2−2¿¿
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∫−∞
∞
Ψ 1 × Ψ 2 dx = ∫−∞
+∞ [(απ )¼ e− y2
2 × 2 y ]×[ 1√2 (απ )
¼
e− y2
2 ׿¿dx ]¿
= (απ )12∫−∞
∞
[e− y2
×2 y (4 y2−2) ] dx
= (απ )12 {∫−∞
∞
[ e− y2
×8 y3¿ ] dx - ∫−∞
∞
[ e− y2
× 4 y¿ ] dx }
= (απ )12 {8∫
−∞
∞
[ e− y2
y3¿ ] dx - 4× ∫−∞
∞
[ e− y2
× y¿ ] dx }
= (απ )12 {8(0) - 4 (0)}
∫−∞
∞
[ e− y2
y3¿ ] and ∫−∞
∞
[ e−y2
× y¿ ] are odd functions. So their value s are zero
5. Show that the function Ψ = e−A x2
2 is the eigen function of harmonic oscillator and find the eigen value.
Where A = 4 π2 ϑ mh
Solution:
The Hamiltonian of harmonic oscillator is H = - h2
8π 2m d2
d x2 + ½ kx2
Where k = 4 π2ϑ 2m
H Ψ = [- h2
8π 2m d2
d x2 + ½ kx2 ] e−A x2
2
= [- h2
8π 2m d2
d x2 [e−A x2
2 ¿ +[ ½ kx2 ] e−A x2
2
= [- h2
8π 2m
ddx [ −A (2x )
2e−A x2
2 ¿ +[ ½ kx2 ] e−A x2
2
= [+ A h2
8 π 2m
ddx [ x e
−A x2
2 ¿ +[ ½ kx2 ] e−A x2
2
= [+ A h2
8 π 2m [ x¿ +[ ½ kx2 ] e
−A x2
2
= [+ A h2
8 π 2m [ x¿ +[ ½ kx2 ] e
−A x2
2
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= [+ h2
8 π 2m [ - A2 x2+A ¿e
−A x2
2 +[ ½ kx2 ] e−A x2
2
= [+ h2
8 π 2m [ - ( 4 π2 ϑ m
h)
2
x2+ 4 π2ϑmh
¿e−A x2
2 +[ ½ (4 π2ϑ 2m)x2 ] e−A x2
2
= [ - 2 π2ϑ 2m x2 + hϑ2 ] e
−A x2
2 +[ 2π 2ϑ2mx2 ] e−A x2
2 [ k = 4 π2ϑ 2m]
= [ hϑ2 ] e
−A x2
2
So the function e−A x2
2 is the eigen function
Eigen value = hϑ2
6. Show that the function Ψ = x e−A x2
2 is the eigen function of harmonic oscillator and has the eigen value
3hϑ4 . Where A = 4 π2 ϑ m
h
7. For a diatomic oscillator show that the frequency is given by ϑ = 1
2 π √ kμ
where k – force constant, μ –
reduced mass of diatomic molecule.
Proof :
The asymptotic solution of harmonic oscillator is Ψ = e−B x2
This should satisfy Schrodinger wave equation for harmonic oscillator.
d2Ψ
d x2 + 8 π 2mh2 ( E - k x2
2) Ψ = 0 ---------------------------1
Ψ = e−B x2
Ψ ‘ = - (B2x) e−B x2
Ψ ‘’ = -2B [ x(-2Bx e−B x2
) + e−B x2
(1) ]
= -2B [ (-2Bx2 +1)] e−B x2
= (4B2 x2 -2B) e−B x2
-----------------2
Substituting in equation 1 we get
(4B2 x2 -2B) e−B x2
+ 8π 2mh2 ( E - k x2
2) e−B x2
= 0 ----------3
Collecting all the x2 terms
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[ (4B2 - 4 π2 mkh2 )x2 +( - 2B+ 8 π 2mE
h2 ) e−B x2
= 0
e−B x2
can not be zero. Therefore the individual terms may be equal to zero
4B2 - 4 π2 mkh2 = 0 -------------------4
and - 2B+ 8 π 2mEh2 = 0 ------------------5
from equation 4
4B2 = 4 π2 mkh2
∴B2 = π2m kh2 -------------------6
from equation 5
2B = 8 π 2mEh2
∴B = 4 π2 mEh2
Squaring this
B2 = 16 π 4 m2 E2
h4 ----------------7
Comparing with equation 6 we get
π 2m kh2 = 16 π 4 m2 E2
h4
k = 16 π2 m E2
h2
= 16 π2 mh2 (
hϑ2 )2
= 4 π 2m ϑ2
ϑ2 = k
4 π2 m
ϑ = 12 π √ k
m
Problem 3.4 Find the energy difference between fourth and first level of harmonic oscillator.
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Solution
E3 – E0 = 72 h γ -
12 h γ
= 3 h γ .
HARMONIC OSCILATOR
Write down Schrodinger wave equation for simple harmonic oscillator and solve it for its wave function and
energy
Write down the expression for i. Hermite polynomial ii. Hermite differential equation
Find the value of Ψ0 , Ψ1 and Ψ3 for simple harmonic oscilator
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Find the value of H0 , H1 and H2
Show that in harmonic oscillator the successive energy levels are equally spaced
Find the separation between two adjacent energy levels
Show that the ground state wave function of harmonic oscillator is normalized
Show that the wave function Ψ1 = ( απ ) ¼ e
− y2
2 ×√2 y where y = (√α x) is normalized
Show that the wave function Ψ0 and Ψ1 of harmonic oscillator are orthogonal
Show that the wave function Ψ1 and Ψ2 of harmonic oscillator are orthogonal
Show that the function Ψ = e−A x2
2 is the eigen function of harmonic oscillator and find the eigen value. Where
A = 4 π2 ϑ mh
Show that the function Ψ = x e−A x2
2 is the eigen function of harmonic oscillator and has the eigen value 3 hϑ
4 .
Where A = 4 π2 ϑ mh
For a diatomic oscillator show that the frequency is given by ϑ = 1
2 π √ kμ
where k – force constant, μ –
reduced mass of diatomic molecule
The infrared spectrum of 75Br19F consists of an intense line at a frequency of 1.14× 1013 s-1.
Calculate the force constant of 75Br19F
The force constant for H79Br is 392 Nm-1. Calculate the fundamental vibrational frequency and zero point
energy of H79Br
TWO DIMENSIONAL HARMONIC OSCILLATOR
In two dimensional harmonic oscillator The total energy of the oscillator is sum of its components
whereas total wave function is the product of these two
Ψtotal = Ψx × Ψy
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Etotal = Ex + Ey
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1.3 APPLICATION OF SCHRODINGER WAVE EQUATIONTO PARTICLE IN A RING (ELECTRON IN A RING)
Consider a particle of mass ‘m’ rotating in a circle of radius ‘r’ in xy plane. The potential energy is zero
.
Schrodinger equation is given d2Ψ
dx2 + 8π 2mh2 EΨ = 0
converting in to polar co-ordinates by substituting
x = r cosϕ y = r sinϕ where ϕ varies from 0 to 2π and solving, we get real as well as imaginary roots.
Thus the normalized real set of wave functions are
Ψ = 1√π
sinKφ Ψ’ =
1√π
cosKφ Ψ’’ = N e±imφ where ‘N’ is normalization constant
These are called ‘CIRCULAR HARMONICS’
From the results, it is clear that, there is no barrier to the particle’s motion as long as it is on the ring.
if K = 0, Ψ = N sin 0
= 0
But Ψ = N cos 0
This is the expression for energy of a particle in a ring
For K = 0, E = 0
Therefore, the rotating particle does not have zero point energy.
Problem 4.5.1 Find the normalisation constant of the function φ m = A sinKx where x varies from 0 to 2π and
A is normalization constant
Solution:
The condition for normalization is ∫0
2 π
ΨΨ * dx = 1
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∫0
2π
A sin K x A sin K x dx = 1
A2 ∫0
2π
sin2 K x dx = 1
A2 ∫0
2 π
(1−cos2K x )
2 dx = 1 [ sin2 K x=(1−cos 2 K x)
2]
A2
2 ∫
0
2 π
[1- cos2K x] dx = 1
A2
2 [ ∫
0
2π
1 dx - ∫0
2 π
cos2 K x dx] = 1
A2
2 { [x¿ - [ (sin 2 K x )
2K¿0
2 π } = 1 [ ∫0
2π
cos2 K x dx = (sin 2 K x )2 K
]
A2
2 { [2 π ¿ - ¿¿ } = 1 [ sin 0 = 0]
A2
2 [2π ¿ = 1 [sin K2π = 0]
A2 [π ¿ = 1
A = 1√π
Problem 4.5.2 Find the normalisation constant of the function φm = N e±imφ where N is normalization constant Solution:
The condition for normalization is
∫0
2π
ΨΨ * dx = 1
N e+imφ N e- imφ dφ = 1
N2 dφ = 1 [e+imφ × e- imφ = e0
= 1]
N2 [φ ¿¿02π = 1 [ ∫ dφ=¿φ ¿ ]
N 2 [2 π−0 ] = 1
N 2 [2 π ] = 1
N = 1√2 π
Expression for energy
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Eigen values of the energy.E = K 2 h2
8 I π2
5.2 RIGID ROTATOR : Definition:
The system consisting of two spherical particles attached together, separated by finite fixed distance
and capable of rotating about an axis passing through the centre of mass and normal to the plane containing
the two particle constitutes the rigid rotator.
EXPRESSION FOR WAVE FUNCTION
The expression is obtained by solving the Schrodinger equation
Schrodinger equation in Cartesian co-ordinates is ∂2Ψ∂ x2 + ∂2Ψ
∂ y2 +∂2Ψ∂ z2 + 8 π 2m
h2 ( E – V) Ψ = 0
For a rigid rotator potential energy is zero. Therefore the above equation becomes
∂2Ψ∂ x2 + ∂
2Ψ∂ y2 +∂2Ψ
∂ z2 + 8 π 2m Eh2 Ψ = 0
This can be converted into spherical polar co-ordinates using the following transformation.
x = r sin θ cosφ
y = r sin θ sinφ
z = r cos θ
Here ‘r’ varies from ‘0’ to ‘∞ '
' θ’ varies from ‘0’ to ‘π
' φ’ varies from ‘0’ to ‘2π '
and the transformed equation is
1r2
∂∂ r ( r2
∂Ψ∂ r ) +
1r2 sinθ
∂
∂θ ( sin θ ∂Ψ∂ θ ) +
1r2sin2θ
∂2Ψ
∂ φ2 +8π 2m Eh2 Ψ = 0
In this equation the wave function Ψ is a function of r, θ,φ
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For a rigid rotator r = 1 . Therefore [∂
∂ r ( r2 ∂Ψ∂ r ) = 0] and hence the above equation becomes
0 + 1
sinθ ∂
∂θ ( sin θ ∂Ψ∂ θ ) +
1sin2θ
∂2Ψ
∂ φ2 +8π 2m Eh2 Ψ = 0
1
sinθ ∂
∂θ ( sin θ ∂Ψ∂ θ ) +
1sin2θ
∂2Ψ
∂ φ2 + 8 π 2m Eh2 Ψ = 0
1
sinθ [ sin θ ∂2Ψ∂ θ2 +
∂Ψ∂ θ cos θ ]+
1sin2θ
∂2Ψ
∂ φ2 +8 π 2m Eh2 Ψ = 0
∂2Ψ
∂ θ2 + cosθsin θ
∂Ψ∂ θ +
1sin2θ
∂2Ψ
∂ φ2 + 8π 2m Eh2 Ψ = 0 ---------------1
Separation of variables.
This equation can be separated into two equations , each involving a single independent variable. For
this purpose, we assume the wave function Ψ ( θ, φ) to be a product of two functions.
Ψ ( θ, φ) = X (θ) ×Y (φ) -------------- 4
Where X depends θ only and Y depends φ only. This equation is solved by separating into two equations
as
Ψ = XY
∂Ψ∂ θ = Y
∂ X∂ θ ,
∂Ψ∂ φ = X
∂Y∂ φ
∂2Ψ
∂ θ2 = Y ∂2 X
∂ θ2 ∂2Ψ
∂ φ = X ∂2Y
∂ φ2
Substituting in 1 we get,
Y ∂2 X∂ θ2 +
cosθsin θ Y
∂ X∂ θ +
1sin2θ
X ∂2Y∂ φ2 +β XY = 0 [ 8π 2m E
h2 = β ]
1X ∂
2 X∂ θ2 +
cosθsin θ
1X
∂ X∂ θ +
1Y sin 2θ
∂2Y
∂ φ2 +β = 0 [Dividing by XY]
sin2θX
∂2 X
∂ θ2 + sin θ cosθ 1X
∂ X∂ θ +
1Y ∂
2Y∂ φ2 +β sin2 θ = 0 [Multiplying by sin2 θ]
sin2θX
∂2 X
∂ θ2 + sin θ cosθ 1X
∂ X∂ θ +β sin2 θ = -
1Y ∂
2Y∂ φ2
In this equation RHS is function of φ only and LHS is a function of θ only.
If we change θ keeping φ constant , since LHS = RHS , LHS will also be constant and vice versa. Therefore
both side can be equated to a constant ( say m2)
- 1Y ∂
2Y∂ φ2 = m2 (azimuthal wave equation) ----------5
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sin2θX
d2 X
d θ2 + sin θ cosθ 1X
dXdθ +β sin2 θ = m2 (polar wave eqution) --------6
Where m and β are constants. In these equations, we have used ordinary derivatives instead
of partial derivatives because each function depends on only one variable
. Equations 5 and 6 are collectively called angular wave function.
1.Solution of Azimuthal wave equation.
Azimuthal wave equation is - 1Y d
2Ydφ2 = m2
d2Y
d φ2 = - m 2 Y [ rearranging]
d2Y
d φ2 + m 2 Y = 0 [ rearranging]
D2 Y+ m2 Y = 0 [D2 Y = d2Ydφ2 ]
( D2 + m2 )Y = 0
The auxiliary equation is ( K2 +m2 ) = 0 [ put D= K and divide by Y]
∴ K2 = - m2
K = √−m2
= √ i2m2 [ i2 = - 1 ]
= ± i m
[ if roots are complex like ( a +ib ) then solution is y = e±bx
The above solution can be written as K = 0 ± i m
Solution of azimuthal wave equation is Y = N e ±imφ where N is normalization constant
To find N:
∫0
2π
ΨΨ * dx = 1
N e+imφ N e- imφ dφ = 1
N2 dφ = 1 [e+imφ × e- imφ = e0
= 1]
N2 [φ ¿¿02 π = 1 [ ∫ dφ=¿φ ¿ ]
N 2 [2 π−0 ] = 1
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N = 1√2 π
∴ Complete solution of azimuthal wave equation is Ym = 1√2 π
e±imφ
2.Solution for polar wave eqution:
The polar equation is ,
sin2θX
d2 X
d θ2 + sin θ cosθ 1X
dXdθ +β sin2 θ = m2
d2 X
d θ2 + cosθSinθ
dXdθ + β X = m2
Xsin2θ
[ Multiply by X
sin2θ ]
d2 X
dθ2 + cosθSinθ
dXdθ + ( β - m2
sin2θ¿ X = 0
Put u = cos θ
d2 Xdθ2 = sin2 θ d2 X
du2 - cos θ dXdu **
-------------------------------------------------------------------------------------------------------------
** Put u = cos θ
dudθ = - sin θ
dXdθ =
dXdu ×
dudθ
= dXdu × (- sin θ)
= - sin θ× dXdu
d2 X
d θ2 = ddθ (
dXdθ )
= d
du ( dXdθ )(
dudθ ) [ by cyclic rule]
= d
du (- sin θ dXdu ) × (- sin θ)
= [ (- sin θ) d2 X
∂ d +
dXdu
ddu (- sin θ) ) ] (- sin θ)
= [ (- sin θ) d2 X
d u2 + dXdu
ddθ ×
dθdu (- sin θ) ) ] (- sin θ)
= [ (- sin θ) d2 X
∂ d +
dXdu
ddθ (- sin θ)
dθdu ) ] (- sin θ)
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= [ (- sin θ) d2 X
d u2 + dXdu (- cos θ)¿ ) ] (- sin θ)
= [ (- sin θ) d2 X
d u2 + dXdu ¿ ) ] (- sin θ)
= ( sin2 θ) d2 X
du2 + dXdu ( - cos θ)
= sin2 θ d2 X
du2 - cos θ dXdu
-------------------------------------------------------------------------------------------
Substituting the values in the above equation we ge
sin2 θ d2 Xdu2 - cos θ
dXdu +
cosθSinθ (- sin θ×
dXdu ) + ( β - m2
sin2θ¿ X = 0
sin2 θ d2 Xd u2 - cos θ
dXdu - cos θ
dXdu + ( β - m2
sin2θ¿ X = 0
sin2 θ d2 Xd u2 -2 cos θ
dXdu + ( β - m2
sin2θ¿ X = 0
( 1- u2) d2 X
d u2 -2u dXdu + ( β - m2
(1−u2)¿ X = 0 [u = cosθ ]
( 1- u2) d2 X
du2 -2u dXdu + (8π 2m E
h2 - m2
(1−u 2)¿ X = 0 [β =
8 π 2m Eh2 ]
This resembles associated Legendre differential equation which is
( 1-x2)∂2 y∂ x2 - 2x
dy∂ x + ( J (J+1) - m2
(1−x 2)¿ X = 0
solution of Associated Legendre differential equation is taken as the solution of polar wave equation.
Solution is X l,m = N 1
2l . l ! ×(1−x2 )
m2 × dm+ l
d xm+l ( x2 – 1)l ] [ ‘N’ – normalization constant]
= √ 2 l+12
× ( l−m ) !( l+m ) !
× 12l . l !
(1−x2 )m2 × dm+ l
d xm+l ( x2 – 1)l [ N =√ 2 l+1
2× ( l−m ) !
(l+m ) !
Complete solution of Schrodinger equation for rigid rotator is
Ψ = Solution of polar equation × Solution of azimuthal equation
= X l,m × Ym
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= √ 2 l+12
× (l−m ) !(l+m ) !
× 1
2l . l ! (1−x2 )
m2 × dm+ l
d xm+l [( x2 – 1)l ] ×
1√2 π
e±imφ
= √ 2 l+12
× (l−m ) !(l+m ) !
× 1
2l . l ! ×
1√2 π
(1−x2 )m2 × dm+ l
d xm+l [( x2 – 1)l ] e±imφ
This is the solution of Schrodinger equation of the rigid rotator
Values of Normalisation constant
l= 0, m= 0 1√2
l=1, m= 0 √ 32
l= 1, m=1
EXPRESSION FOR ENERGY:
To find eigen values of energy: n ( n+ 1 ) = β but β = 8 π 2m Eh2
8π 2m Eh2 = J(J + 1 ) [ n is replaced by l]
∴ E = h2
8π 2m J (J+1)
Energy of Rigid Rotator
Eigen values of energy of Rigid Rotator is E = h2
8π 2m J (J+1)
‘ J ‘is rotational quantum number
The lowest possible value of J = 0
E0 = h2
8π 2m 0 (0+1)
= 0
E1 = h2
8π 2m J (J+1)
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= h2
8 π 2m (1) (1+1) [ J = 1]
= 2h2
8 π 2m
= h2
4 π2 m
E2 = h2
8π 2m (2) (2+1) [ J = 2]
= 6 h2
8 π 2m
The difference in energy levels E2 and E1 is
E2 – E1 = 6 h2
8 π 2m - h2
4 π2 m
= 6 h2−2h2
8 π2 m
= 4 h2
8 π 2m
= h2
2 π2 m
E3 – E2 = 12 h2
8 π 2m - 6h2
8π 2m
= 6 h2
8 π 2m
Values of energy
E0 0
E1 h2
4 π2 m
E2 6h2
8 π 2m
LEGENDRE POLYNOMIAL Page 162 of 419
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Legendre differential equation is d2 Pd θ2 +
cosθSinθ
dPdθ + βP = M 2 P
sin2θ
Its solution is known as Legendre polynomial Pl(x) = 12l . l !
d l
d x l ( x2 – 1) l
Where x = cos θ and ‘ l’ is an integer including 0
Problem 5.2 Find the expression for the Legendre polynomial P0 and P1
Solution:
P0 = 120.0 !
d0
d x l ( x2 – 1) 0
= 1
P1 = 12l . l !
d l
d x l ( x2 – 1) 1
= 12
× ddx ( x2 – 1)
= 12
×(2x – 0 )
= x
= cos θ [ x = cos θ]
Similarly P2 = 3 cos 2 θ - 1
Values of Legendre polynomial
P0 1
P1 cos θ
P2 3 cos 2 θ - 1
ASSOCIATED LEGENDRE POLYNOMIAL:
Associated Legendre differential equation is
( 1-x2)d2 yd x2 - 2x
dy∂ x + ( n (n+1) -
m2
(1−x2)¿ X = 0
Its solution is known as Associated Legendre polynomial Plm(x)
Plm(x) = (1−x2 )
m2 dm
d xm × Pl(x)
= (1−x2 )m2 × dm
d xm1
2l .l !d l
d xl ( x2 – 1) l
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= 1
2l . l ! (1−x2 )
m2 dm+ l
d xm+l [( x2 – 1)l
Problem 5.3 Find the expression for the Associated Legendre polynomial P00
and P10
Solution:
P00 = ( 1-x2) 0 × d0
d x0 [( x2 – 1)0
= 1
P10 = 1
2l . l !d l
d x l ( x2 – 1) 1
= 12
× ddx ( x2 – 1)
= 12
×(2x – 0 )
= x
= cos θ [ x = cos θ]
Similarly
P20 = ½ ( 3 cos 2𝛉 – 1)
P11 = sin𝛉
P21 = 3sin𝛉 cos𝛉
Values of Associated Legendre
polynomial
P00
1
P10
cos θP1
1
SPHERICAL HARMONICS.
The solution of Schrodinger equation for the rigid rotator ( particle in a sphere) is known as
spherical harmonics. It is given by
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Ψ l ,m = √ 2 l+12
× ( l−m ) !(l+m ) !
× (1−x2 )m2 × dm
d xm [ 1
2l . l ! d l
d x l ( x2 – 1)l ] ×
1√2 π
e±imφ
Problem 5.1 Find the expression for First harmonic:
Solution:
For the first harmonic l= 0 and m = 0
Ψ 0,0 =√ 2(0)+12
× 0 !(0 ) !
× ( 1-x2) 0 × d0
d x0 [( x2 – 1)0 × 1√2 π
e 0
= 1√2
× 1√2 π
= 1
2√ π
Problem 5.4 The solution of polar equation is Θ l,m = √ 2 l+12
× ( l−m ) !( l+m ) !
× Plm ( cos θ )
Find the normalisation constant for N 0,0 is
Solution:
Given: l= 0 and m = 0
N 0,0 =√ 2(0)+12
× 0 !(0 ) !
= 1√2
Problem 5.5 The angular wave function is Θ l,m = √ 2 l+12
× ( l−m ) !(l+m ) !
×1
2l× l! ( 1- cos 2 θ ) m/2 (
dd (cosθ) )
l+m
[cos 2 θ -1 ) l 1√2 π
e imφ The value for Θ 0,0 is
Solution:
Given: l= 0 and m = 0
Θ 0,0 =√ 2(0)+12
× 0 !(0 ) !
× ( 1- cos 2 θ ) 0 ×d0
d (cosθ)0 [(cos 2 θ – 1)0 ×
1√2 π
e 0
= 1√2
× 1√2 π
= 1
2√ π
Problem 5.6. The Rodrigue formula for associated Legendre polynomial is
Plm ( cos θ ) =
12l× l!
( 1- cos 2 θ ) m/2 ( d
d (cosθ) ) l+m [cos 2 θ -1 ) l The value of P0
0 is
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Solution:
P00
=√ 2(0)+12
× 0 !(0 ) !
× ( 1- cos 2 θ ) 0 ×d0
d (cosθ)0 [(cos 2 θ – 1)0
= 1√2
Problem 5.7. The wave function of rigid rotator is Θ l,m = √ 2 l+12
× (l−m ) !(l+m ) !
× Plm ( cos θ )
The normalisation value corresponding to l=1 and m= 0 is
Solution:
Given: l= 1 and m = 0
N 1,0 =√ 2(1)+12
× (1−0) !(1+0)!
= √ 32
Problem 5.8 Find the eigen value of Ψ10= cos θ with ( L2 operator) Hamiltonian operator (- h2
8 Iπ 2 ( 1
sinθ∂
∂ θ (sin θ
∂∂θ )+
1sin2θ
∂
∂ φ 2 ) is
Solution:
E = (- h2
8 Iπ 2 ( 1
sinθ∂
∂θ (sin θ ∂
∂θ )+1
sin2θ
∂∂ φ 2 ) (cos θ)
= (- h2
8 Iπ 2 ( 1
sinθ∂
∂ θ (sin θ ∂
∂θ (cos θ )+1
sin2θ
∂∂ φ 2 (cos θ)
= (- h2
8 Iπ 2 ( 1
sinθ∂
∂ θ (sin θ ( - sin θ )+1
sin2θ (0) [
∂∂θ (cos θ) =- sinθ , ∂
∂ φ 2 (cos θ) = 0]
= (- h2
8 Iπ 2 ( 1
sinθ∂
∂ θ (- sin2 θ )
= (- h2
8 Iπ 2 ( 1
sinθ (-2 sin θ cos θ ) [ ∂
∂θ (sin2 θ ) = 2 sin θ cos θ] = (- h2
8 Iπ 2 (-2 cos θ )
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= + h2
4 Iπ 2 cos θ
S.no l m
1 0 0 12√ π
2 1 0 cosθ3 1 +1 sinθe iφ
4 1 -1 sinθe−iφ
5 2 0 cos2θ
.COMPLETE SOLUTION OF SCHRODINGER EQUATION
Complete solution of Schrodinger equation for rigid rotator is
Ψ = solution of polar equation × solution of azimuthal equation
= Norm. constant × Associated Legendre polynomial × solution of azimuthal equation
= √ 2 l+12
× ( l−m ) !( l+m ) !
× 12l . l !
(1−x2 )m2 × dm+ l
d xm+l [( x2 – 1)l ] ×
1√2 π
e imφ
3.NORMALIZATION CONSTANT
√ 2 l+12
× (l−m ) !(l+m ) !
is normalization constant
1. When l= 0, m= 0
√ 2 l+12
× (l−m ) !( l+m ) !
= √ 2(0)+12
× (0−0 )!(0+0 )!
= √ 12
× 0!0!
= √ 12
2. When l=1, m= 0
√ 2 l+12
× (l−m ) !( l+m ) !
= √ 2(1)+12
× (1−0 )!(1+0 )!
= √ 32
× 1!1!
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= √ 32
3. When l=1, m= 1
√ 2 l+12
× ( l−m ) !( l+m ) !
= √ 2(1)+12
× (1−1 )!(1+1 ) !
= √ 32
× 0 !2!
= √ 34
Values of Normalisation constant l= 0, m= 0 1
√2l=1, m= 0 √ 3
2l= 1, m=1 √ 3
4
Problem The solution of polar equation is Θ l,m = √ 2 l+12
× (l−m ) !(l+m ) !
× Plm ( cos θ )
Find the normalisation constant for N0,0 is
Solution:
Given: l= 0 and m = 0
N0,0 =√ 2(0)+12
× 0 !(0 ) !
= 1√2
Problem The Rodrigue formula for associated Legendre polynomial is
Plm ( cos θ ) =
12l× l!
( 1- cos 2 θ ) m/2 ( d
d (cosθ) ) l+m [cos 2 θ -1 ) l The value of P0
0 is
Solution:
P00 = ( 1-cos 2 θ ) 0 ×
d0
d (cosθ)0[(cos 2 θ – 1)0
= 1
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Problem The wave function of rigid rotator is Θ l,m = √ 2 l+12
× (l−m ) !( l+m ) !
× Plm ( cos θ )
The normalisation value corresponding to l=1 and m= 0 is
Solution:
Given: l= 1 and m = 0
N1,0 =√ 2(1)+12
× (1−0) !(1+0)!
= √ 32
4. LEGENDRE POLYNOMIAL:
Legendre differential equation is d2 Pd θ2 +
cosθSinθ
dPdθ + βP = M 2 P
sin2θ
where P is Legendre polynomial Pl = 12l . l !
d l
d x l ( x2 – 1) l
Where x = cos θ and ‘ l’ is an integer including 0
Problem Find the expression for the Legendre polynomial P0 and P1
Solution:
P0 =1
20 .0 !d0
d x l ( x2 – 1) 0
d0
d y0 means need not differentiate
= (1) ( x2 – 1) 0
= 1
P1 =1
2l . l !d l
d x l ( x2 – 1) 1
=12
× ddx ( x2 – 1)
=12
×(2x – 0 ) [ddx ( x2 ) = 2x ,
ddx ( 1 ) = 0]
= x
= cos θ [ x = cos θ]
P2 =1
22 .2 !d2
d x2 ( x2 – 1) 2
=18
× d2
d x2 [ x4 - 2x2 +1)
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=18
× ddx [ 4 x3 – 4x )
=18
×[ 12 x2 – 4 )
=18
× 4[ 3 x2 – 1 )
=12[ 3 x2 – 1 )
= 12 (3 cos 2 θ - 1) [ x = cos θ]
Values of Legendre polynomial
l=0 P0 1
l=1 P1 cos θ
l=2 P2 12 (3 cos 2 θ - 1)
5.ASSOCIATED LEGENDRE POLYNOMIAL:PLM
Plm(x) = 1
2l . l !(1−x2 )
m2 dm+l
d xm+l [( x2 – 1)l
Problem Find the expression for the Associated Legendre polynomial P00 and P1
0
Solution:
P00 = ( 1-x2) 0 × d0
d x0 [( x2 – 1)0
= 1
P10 = 1
2l . l !d l
d x l ( x2 – 1) 1
=12
× ddx ( x2 – 1)
= 12
×(2x – 0 )
= x
= cos θ [ x = cos θ]
P20 = ( 1-x2) 0 × 1
22 .2 !d2
d x2( x2 – 1) 2
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=18
× d2
d x2 [ x4 - 2x2 +1)
=18
× ddx [ 4 x3 – 4x )
=18
×[ 12 x2 – 4 )
=18
× 4[ 3 x2 – 1 )
=12[ 3 x2 – 1 )
= 12 (3 cos 2 θ - 1) [ x = cos θ]
P11 = ( 1-x2) 1/2 1
2l . l !d2
d x2 ( x2 – 1) 1
= ( 1-x2) 1/2 12× d2
d x2 ( x2 – 1)
= ( 1-x2) 1/2 12
×(2 )
= ( 1-x2) 1/2
= (1- cos 2 θ )1/2
= (sin 2 θ )1/2
P11 = sin𝛉
and P21 = 3sin𝛉 cos𝛉
Values of Associated Legendre
polynomial
l=0,m=0 P00 1
l=1,m=0 P10 cos θ
l=1,m=1 P11 sin𝛉
ENERGY OF RIGID ROTATOR
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Eigen values of energy of Rigid Rotator is E = h2
8π 2m J (J+1)
‘ J ‘is rotational quantum number
The lowest possible value of J = 0
E0 =h2
8π 2m 0 (0+1)
= 0
E1 = h2
8 π 2m J (J+1)
= h2
8 π 2m (1) (1+1)
= 2 h2
8 π 2m
= h2
4 π2 m
E2 = h2
8 π 2m (2) (2+1)
= 6h2
8 π 2m
Values of energy
E0 0
E1 h2
4 π2 m
E2 6h2
8π 2m
Problem : Calculate the energy difference between the energy level 2 and one
Solution
E2 – E1 = 6 h2
8 π 2m - h2
4 π2 m
= 6 h2−2h2
8 π2 m
= 4 h2
8 π 2m
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= h2
2π2 m
Problem : Calculate the energy difference between the energy level 4 and one
Solution:
E4 – E1 = 20 h2
8 π 2m - h2
4 π2 m
= 20 h2−2 h2
8 π2 m
= 18 h2
8 π 2m
SPHERICAL HARMONICS.
The solution of Schrodinger equation for the rigid rotator ( particle in a sphere) is known as spherical
harmonics. It is given by
Ψ l ,m = √ 2 l+12
× ( l−m ) !(l+m ) !
×1
2l× l! (1−x2 )
m2 × dm
d xm [ 12l . l !
d l
d x l ( x2 – 1)l ] ×
1√2 π
e imφ
Problem Find the expression for First harmonic:
Solution:
For the first harmonic l= 0 and m = 0
Ψ 0,0 =√ 2(0)+12
× 0 !(0 ) !
×1
20× 0 ! ( 1-x2) 0 × d0
d x0 [( x2 – 1)0× 1√2 π
e 0
= 1√2
× 1√2 π
= 1
2√ π
Problem Find the expression for second harmonic
Solution:
For the first harmonic l= 1 and m = 0
Ψ 1,0 =√ 2(1)+12
× (1−0) !(1+0 ) !
× 1
2l× l! ×( 1-x2) 0 × d1
d x1 [( x2 – 1)1× 1√2 π
e 0
= √ 32
× (2x) ×12
1√2 π
= √32
× 1√π
x
= √32√ π
cos θ
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5.3 HYDROGEN ATOMSchrodinger equation for H-atom
Schrodinger equation in Cartesian co-ordinates is given by
∂2Ψ
∂ x2 + ∂2Ψ
∂ y2 +∂2Ψ∂ z2 + 8π 2m
h2 ( E – V) Ψ = 0
This can be converted into spherical polar co-ordinates using the following transformation.
x = r sin θ cosφ
y = r sin θ sinφ
z = r cos θ
Here ‘r’ varies from ‘0’ to ‘∞ '
' θ’ varies from ‘0’ to ‘π
' φ’ varies from ‘0’ to ‘2π '
the transformed equation is
1r2
∂∂ r ( r2
∂Ψ∂ r ) +
1r2 sinθ
∂
∂θ ( sin θ ∂Ψ∂ θ ) +
1r2sin 2θ
∂2Ψ
∂ φ2 + 2 μh ( E +
Ze24 π∈r ) Ψ = 0 ---1
In this equation the wave function Ψ is a function of r, θ, φ
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Separation of variables.
This equation can be separated into three equations , each involving a single independent variable. For
this purpose, we assume the wave function Ψ ( r, θ,φ) , to be a product of three functions.
Ψ ( r, θ, φ) = R(r) × Θ(θ) ×Φ (φ) ---------------------------------------- 2
Where R(r) is the radial function which depends r only, Θ(θ), depends θ only and Φ (φ) depends φ only.
This equation is solved by separating into three equations as
- 1Y ∂
2Y∂ φ2 = m2 (azimuthal wave equation) ----------5
d2 X
dθ2 + cosθSinθ
dXdθ + ( β - m2
sin2θ¿ X = 0 (polar wave equation)
1r2
ddr ( r2
dRdr ) -
βr 2 R +
2μh ( E +
Ze 24 π∈0 r ) R = 0
1r2
ddr ( r2
dRdr ) + K2 r2 = l(l+1) (radial wave equation) ------------5
Where m and β are constants. In these equations, we have used ordinary derivatives instead of partial
derivatives because each function depends on only one variable. Equations 3 and 4 are collectively called
angular wave function.
1.Solution for azimuthal wave equation.
φ m = 1√2 π
e imφ
2.Solution for polar wave equation:
Θ l,m = √ 2 l+22
× ( l−m) !(l+m ) !
× ( 1-x2) m/2 × dm
d xm [1
2l . l ! d l
d x l ( x2 – 1)l ]
3.Solution for radial wave equation:
The radial wave equation is
1R
ddr ( r2
dRdr ) + K2 r2 = l(l+1) -------1 where K2 = 8 π 2m
h2 ( E + Z e2
4 πεr )
Multiplying by Rr2 on both sides
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1r2
ddr ( r2
dRdr ) + K2 R = l(l+1)
Rr2 ----------2
1r2 ( r2 d
2 Rdr 2 +
dRdr (2r) ) +[ K2−¿
l(l+1)r 2 ] R = 0 -----------3
d2 R
d r 2 + 2r
dRdr + ( K2 -
l(l+1)r 2 ) R = 0 -------------4
d2 R
d r 2 + 2r
dRdr + (8 π 2m
h2 ( E + Z e2
4 πεr ) -
l(l+1)r 2 ) R = 0 -------------5
d2 R
d r 2 + 2r
dRdr + (8 π 2mE
h2 + 2 πmZ e2
εrh2 ) - l(l+1)
r 2 ) R = 0 ----------------------6
Let P = 2 zrna
Substituting in the above equation we get
d2 R
d P2 + 2P
dRdP + (
nP -
14 -
l(l+1)P2 ) R = 0 ---------------7
Case 1: For large value of P
When P is large dRdP ,
nP and
l(l+1)P2 terms can be neglected. Therefore the above equation becomes
d2 R
d P2 - 14 R = 0 ---------------------8
The auxiliary equation is
D2 - 14 = 0
Solution of this equation is
R = A e−−P
2 + B e−+P
2 --------------9
Where A and B are constants.
Since R must be finite for P→ ∞ , the solution becomes
R = A e−−P
2 ------10 [ since e∞=∞, e−∞=¿0]
[ note : for harmonic oscillator , alse the acceptable solution is Ψ = e− y2
2
Case 2: For low value of P
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When P is low nP and
14 terms can be neglected. Therefore the above equation becomes
d2 R
d P2 + 2P
dRdP -
l(l+1)P2 ) R = 0 -----------------------11
Solution of this equation is
R = Pl --------------------12
Combining 10 and 12 we get
R = e−−P
2 Pl L(p) ----------------13
where L(p) is Associated Lagurie polynomial which is given by
L(p) = ( ddρ ) p [ e r (
ddρ ) k ( r k e –r) ]
Thus the solution of radial equation is
R = N e−−P
2 Pl ( ddρ ) p [ e r (
ddρ ) k ( r k e –r) ]
And N is normalization constant which is given by √( 2 zna 0 )
3
× (n− l−1 )!2n [ (n+l )! ]3
Thus the solution of radial equation becomes
R = √( 2 zna 0 )
3
× (n−l−1 )!2n [ (n+l )! ]3
e−−P
2 Pl ( ddρ ) p [ e r (
ddρ ) k ( r k e –r) ] ----------14
R n,l = √( 2 zna 0 )
3
× (n− l−1 )!2n [ (n+l )! ]3
× e –( zrna 0 ) × ( 2 zr
na 0 )l
( ddρ ) p [ e r (
ddρ ) k ( r k e –r) ]
.1. Find the angular eigen function of 1s orbital of H- atom in atomic mass unit.
Solution:
The angular wave function is
Θ l,m = √ 2 l+12
× (l−m ) !(l+m ) !
× ( 1-x2) m/2 × dm
d xm [1
2l . l ! d l
d x l ( x2 – 1)l ] ×
1√2 π
e imφ
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The total wave function of H-atom is Ψ n,l,m = √( 2
a 0 )3
×(1−0−1 )!
2 (1 )[ (1+0 )! ]3 × ( 2 zr
na 0 ) l × e –( zrna 0 ) × (
dd r ) p [ e r (
ddr ) k ( r k e –r) ]
× √ 2 l+12
× (l−m ) !(l+m ) !
× ( 1-x2) m/2 × dm
d xm [1
2l . l ! d l
d x l ( x2 – 1)l ]
× 1√2 π
e imφ
where p = 2l+1
k = n +l
2020 - CONCISE QUANTUM MECHANICS
GROUND STATE ENERGY EIGEN VALUE:
1r2
ddr ( r2
d Ψdr ) +[ 8 π 2m
h2 ( E + e2
4 πεr ) -
l(l+1)r 2 )] Ψ = 0 ----------1
For ground state , n= 1 , l= 0 . therefore the above equation becomes
1r2
ddr ( r2
d Ψdr ) +[ 8 π 2m
h2 ( E + e2
4 πεr ) ] Ψ = 0 -------------2
But Ψ = 2
a3 /2 e−ra
d Ψdr =
2a3 /2 [e
−ra (−1a ) ]
= −2a5 /2 [e
−ra ]
Substituting in 2 we get
1r2
ddr ( r2 (
−2a5 /2 [e
−ra ] ) +[ 8 π 2m
h2 ( E + e2
4 πεr ) ]
2a3 /2 e
−ra = 0
−2a5 /2
1r2
ddr ( r2 e
−ra ) +[ 16 π2 m
a3/2h2 ( E + e2
4 πεr ) ] e
−ra = 0
−2
r2 a5/ 2 [ ( r2 ( −1a
e−ra ) + e
−ra (2 r ) ]+ [ 16 π2 m
a3/2h2 ( E + e2
4 πεr ) ] e
−ra = 0
Divide by e−ra the above equation becomes
−2
r2 a5/ 2 [ ( r2 ( −1a ) + (2 r ) ]+ [ 16 π2 m
a3/2h2 ( E + e2
4 πεr ) ] = 0
−2a5 /2 [ (
−1a ) + ( 2r ) ]+ [ 16 π2 m
a3/2h2 ( E + e2
4 πεr ) ] = 0
+2a7 /2 - ( 4
r a5 /2 ) ]+ 16 π2 mEa3/2h2 + 16 π2 me2
a3 /2 h2 4 πεr ) ] = 0
+2a7 /2 - ( 4
r a5 /2 ) ]+ 16 π2 mEa3/2h2 + 4 πme2
a3 /2 h2 εr ) ] = 0
Rearranging
{ +2a7 /2 + 16 π2 mE
a3/2h2 } + 1r { -
4a5 /2 + 4πm e2
a3 /2 h2 ε } = 0
Each paranthesis must be equal to zero . Therefore
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Equating the second term to zero
- 4
a5 /2 + 4 πm e2
a3 /2 h2 ε = 0
4
a5 /2 = 4 πm e2
a3 /2 h2 ε
1a = πm e2
h2 ε
a = ε h2
πm e2
This is the expression for Bohr’s radiusEquating the first term to zero
+2a7 /2 + 16 π2 mE
a3/ 2h2 = 0
+2a7 /2 = - 16 π2 mE
a3/ 2h2
Multiplying by ‘a3/2 ‘
+2a2 = - 16 π2 mE
h2
E = −2 h2
16 π2 m a2
= −h2
8 π 2m a2
Substituting the value of ‘a’
E = −h2
8π 2m ( πm e2
ε h2 )2
= −h2
8π 2m × π
2m2 e4
∈2 h4
E = −m e4
8∈2 h2
This is the expression for the lowest energy state of the electron of hydrogen
atom.
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Energy E = −2 π2 me4
n2 h2 ?
Bhor’s radius r = n2h2
4 π2 me2 ?
Problem:5.3.1 Find the Angular eigen function of 1s orbital of H- atom in atomic mass unit
Solution:
For 1s orbital l=0, m=0
∴ Θ l,m = √ 2(0)+12
× 1
20× l! ( 1- x2 ) 0 (
ddx ) 0 [x2 -1 ) 0 ×
1√2 π
e 0
= √ 12
× 1√2 π
= 1
2√ π
Problem:5.3.2 Find the radial eigen function of 1s orbital of H- atom in atomic mass unit.
Solution: The radial wave function is given by
Rn,l = √( 2 zna0 )
3
×(n−l−1 )!
2n[ (n+l ) !]3
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× ( 2 zrna 0 ) l × e
− zrna0 × (
dd r ) p [ e r (
dd r ) k ( r k e –r) ]
For 1s orbital n=1, l=0 ∴
p = 2l +1 = 0 +1
= 1
k = n +l = 1 +0
= 1
∴ R 10 = - √( 2 z(1 )a 0 )
3
× (1−0−1 )!2 [(1+0 )! ]3
× ( 2 zra 0 ) 0 × e
− zra0 × (
ddr ) 1 [ e r(
ddr ( r e –r ) ]
= - √8 ×( za0 )
3
× 12
× e− zr
a0 × ddr [ e r [r(- e –r + e –r (1) ]
= - √4×( za0 )
3
× e− zr
a0 × ddr [ - r + 1 ]
= - 2 ×√( za0 )
3
× e− zr
a0– × [ - 1]
= +2 ( za0)
32 × e
− zra0
For Hydrogen atom in atomic units , Z= 1 and a0 = 1
R 10 = 2 e –r
Problem 5.3.3 Find the total eigen function of 1s orbital of H-atom in atomic units.
Solution:
The total wave function of 1s orbital is
Ψ 100 = radial function × angular function
= 2 e−r × 1
2 √ π
= 1√π
e –r
Problem 5.3.4 Find the angular eigen function of 2s orbital of H-atom.
Solution:
The angular wave function is
Θ l,m = √ 2 l+12
× (l−m ) !(l+m ) !
× ( 1-x2) m/2 × dm
d xm [1
2l . l ! d l
d x l ( x2 – 1)l ] ×
1√2 π
e imφ
For 2s orbital n = 2 , l=0, m=0
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∴ Θ l,m = √ 2(0)+12
× 1
20× l! ( 1- x2 ) 0 (
ddx ) 0 [x2 -1 ) 0 ×
1√2 π
e 0
= √ 12
× 1√2 π
= 1
2√ π
Problem 5.3.5 Find the radial eigen function of 2s orbital of H- atom in atomic mass units.
Solution: The radial wave function is given by
Rn,l = - √( 2 zna0 )
3
× (n−l−1 )!2n [ (n+l )! ]3
× ( 2 zrna 0 ) l × e –( zr
na 0 ) × ( d
dr ) p [ e r ( d
d r ) k ( r k e –r) ]
For 2 orbital n=2, l=0
p = 2l +1 = 0 +1
= 1
k = n +l = 2 +0
= 2
∴ R 20 = - √( 2 z(2 )a0 )
3
× (2−0−1 ) !2(2)[ (2+0 )! ]3
× ( 2 zra 0 ) 0 × e –( zr
a0 )× ddr [ e r (
ddr ) 2 ( r2 e –r ) ]
= - √( za0 )
3
× 132
× e –( zra 0 )×
ddr [ e r d
dr [ r2 (- e –r ) + e –r (2r) ]
= - 1
4 √ 2 ( za0)
32 × e –( zr
a 0 )× ddr [ e r [ r2 (+ e –r ) +(- e –r ) (2r) + (2r)(- e –r ) + e –r (2) ]
= - 1
4 √ 2 ( za 0)
32 × e –( zr
a 0 )× ddr [ r 2 -2r – 2r +2 ]
= - 1
4 √ 2 ( za 0)
32 × e –( zr
a0 )×[ 2r – 4 ]
= - 1
2√ 2 ( za 0)
32 × e –( zr
a0 )×[ r – 2 ]
= + 1
2 √ 2 ( za 0)
32 × e –( zr
a 0 )×[ 2- r ]
For Hydrogen atom in atomic units , Z= 1 and a0 = 1
R 20 = 1
2 √ 2 ×[ 2- r ]
Problem 5.3.6 Find the total eigen function of 2s orbital of H-atom in atomic units.
Solution:
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The total wave function of 2s orbital is
Ψ 100 = radial function × angular function
= 1
2√ 2 ×[ 2- r ] × 1
2 √ π
= 1
4 √ 2π ×[ 2- r ]
Problem 5.3.7 Find the angular eigen function of 2p orbitals.
The angular wave function is
Θ l,m = √ 2 l+12
× ( l−m ) !( l+m ) !
× 1
2l× l! (1- cos 2 θ ) m/2 (
dd (cosθ) )
l+m [cos 2 θ -1 ) l × 1√2 π
e imφ
For 2p orbital l = 0 , m = 0 , ±1
When l=1, m=0
Θ 10 = √ 2(1)+12
× (1−0 )!(1+0 )!
× 1
21×1 ! (1- cos 2 θ )0 (
dd (cosθ) )
1+0 [cos 2 θ -1 ) 1 × 1√2 π
e 0
= √ 32
× 12 ×
dd (cosθ)
[cos 2 θ -1 ) × 1√2 π
= √ 32
× 12 × [ 2 cosθ ] ×
1√2 π
[ ddx
(x2 -1) = 2 x ]
= √32 √ π
× cosθ
This represent ‘z’ orbital because it contains only cos 𝛉When l=1, m=1
P11 =
1211 !
( 1- cos 2 θ ) 1/2 ( d
d (cosθ) ) 1+1 [cos 2 θ -1 ) 1× e iφ
= 12 ( sin2 θ ) 1/2 (
dd (cosθ) )
2 [cos 2 θ -1 ) × e iφ
Put x = cosθ , P11 =
12 ( sin θ) ( d2
d x2 ) [ x2 -1 ) × e iφ
= 12 ( sin θ) (
ddx ( 2x ) × e iφ
= 12 ( sin θ) [ 2 ] × e iφ
= ( sin θ) ( cos φ + i sin φ)
Taking real parts alone , P11 = sin θ cos φ
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∴ Θ l,m = √ 2(1)+12
× sin θ cos φ × 1√2 π
= √ 32
×1√2 π
sin θ cos φ
= √ 3
2 √ π sin θ cos φ
This represent ‘x’ orbital because it contains sin θ cos φ
when l=1, m=-1
P1−1 =
1211 !
( 1- cos 2 θ ) -1/2 ( d
d (cosθ) ) 1-1 [cos 2 θ -1 ) 1× e−iφ
= 12 ( sin2 θ ) -1/2 (
dd (cosθ) )
0 [cos 2 θ -1 ) × e−iφ
= 12 (
1sin θ ) [cos 2 θ -1 ) ×e−iφ
= 12 (
1sin θ ) (-) ( 1- cos 2 θ ) ×e−iφ
= - 12 (
1sin θ ) ( sin 2 θ ) × e−iφ
= - 12 sin θ × e−iφ
= - 12 sin θ × ( cos φ - i sin φ)
Taking real parts alone , P1−1 = -
12 sin θ × sin φ [ how sin φ] ?
∴ Θ l,m = √ 2(1)+12
× sin θ sin φ × 1√2 π
= √ 32
×1√2 π
sin θ sin φ
= √ 3
2 √ π sin θ sin φ
This represent ‘y’ orbital because it contains sin θ sin φ
The angular parts of wave function of 2s,2p orbitals are given by
Ψ 200 = 1
4 √ 2 π (1
a0) 3/2 e - r
2 a0 ( 2-
ra 0 )
Ψ 210 = 1
4 √ 2 π (1
a0) 3/2 e - r
2 a 0 ( r
a 0 ) cosθ
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Ψ 211 = 1
4 √ 2 π (1
a0) 3/2 e - r
2 a0 ( r
a 0 ) sin θ cosφ
Ψ 21 -1 = 1
4 √ 2 π (1
a 0) 3/2 e - r
2 a0 (
ra 0 ) sin θ sinφ
The angular parts of wave function of 3d orbitals are given by
Ψ 322 = ( 1516 π ) ½ ( 3 cos 2 θ – 1) ( dx 2 )
Ψ 321 = ( 1516 π ) ½ sin θ cosφ cos θ ( dxz )
Ψ 320 = ( 1516 π ) ½ sin θ sinφ cos θ ( dyz )
Ψ 32-1 = ( 1516 π ) ½ sin2 θ cos 2 φ ( dx2- y2 )
Ψ 32-2 = ( 1516 π ) ½ sin 2 θ sin2 φ ( dxy )
We know that x = r sin θ cosφ y = r sin θ sinφ z = r cos θ
sin2 θ cos 2 φ = sin2 θ ( cos 2 φ - sin 2 φ )
= sin2 θ cos 2 φ - sin2 θ sin 2 φ
= x 2 – y2
sin 2 θ sin2 φ = sin 2 θ × 2 sin φ cos φ
= 2 sin θ sin φ × sin θ cos φ
= 2 xy
Radial eigen function
In Bohr’s radius unit in atomic mass unit(amu)
n=1,
l=m=0
1s2 ( 1
a0)
32 × e
−ra0
2 e –r
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n=2,
l=m=0
2s 12√2
( 1a0)
32 ×( 2-
ra0
) e−r2 a0
12√2
×( 2- r ) × e−r2
n=2,
l=1
m=0
2p 12√6
( 1a0)
32 ׿ ) e
−r2 a0
12√6
× r × e−r2
n=3,
l=0
m=0
3s 281√3
( 1a0)
32 ׿ + 2
r 2
a02 ) e
−r3 a0
281√3
׿ + 2r2 ) e−r3
n=3,
l=1
m=± 1
3p 281√6
( 1a0)
32 ׿ )
ra0
e−r3 a0
281√6
׿ ) r e−r3
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NODES IN RADIAL PART
To find the node ,
1. Take the term containing ‘r’
2. leave the exponential term
3. put Ψ = 0
4. Find the value of ‘r’
For 1s orbital
Ψ 1∝ e−ra0 ×r 0 [Taking the term containing ‘r’only ]
∝ r0 [ leaving the exponential term]
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When Ψ 1 = 0,
r0 = 0
r=0 therefore no node
For 2s orbital
Ψ 2∝ ( 2- ra0
) e−r2a0 [Taking the term containing ‘r’only ]
∝ ( 2- ra0
) [ leaving the exponential term]
When Ψ 2 = 0,
( 2- ra0
) = 0
2 = ra0
r = 2a0
Therefore only one node occurs at 2 a0
For 2p orbital
Ψ 2∝ ¿ ) e−r2a0
∝ ra0
When Ψ 2 = 0,
ra0
= 0
r = 0
Therefore no node occur
For 3s orbital
Ψ 2∝ ¿ + 2r 2
a02 )
When Ψ 2 = 0,
27−18 ra0
+ 2r 2
a02 = 0
Multiplying by a02 on both sides
27 a02 – 18 a0 +2r2 = 0
This is a quadratic equation in ‘r’ which has two values. Therefore there are two nodes occurs
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For 3p orbital
Ψ 3∝ ( 6- ra0
) ra0
When Ψ 3 = 0,
( 6- ra0
) ra0
= 0
( 6- ra0
) = 0
r =6a0
Therefore only one node occurs at 6 a0
IDENTICFICATION OF ORBITALS FROM WAVE FUNCTION:
1. To find ‘n’
The denominator of the exponential term indicates the value of ‘n’ .For example
If the wave function is of the form Ψ = Nr e−r2 cos θ then n= 2
If the wave function is of the form Ψ = Nr e−r3 cos θ then n= 3
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2. To find ‘l’
The power of ‘r’ in the wave function gives the value of ‘r’. [ don’t see the exponential ‘r’ )
.For example
If the wave function is of the form Ψ = Nr e−r2 cos θ then l = 1 ( ‘p’ orbital)
If the wave function is of the form Ψ = Nr2 e−r3 cos θ then l= 2 ( ‘d’ orbital)
3. To designate the orbital:
a. If the wave function has l= 0 and no trigonometric term then it is ‘s’ prbital
b. If the wave function contains l= 1 and
1. if cos θ alone is present then it is ‘pz’ orbital
2. if sinθ cos φ is present then it is ‘px’’ orbital
3. if sinθ sin φ is present then it is ‘py’’ orbital
c . if the wave function contains l= 2 and
1. if cos2θ alone is present then it is ‘d z2’ orbital
2. if sin θ cosφ cos θ then it is ‘dxz’’ orbital
3. sin θ sinφ cos θ - dyz orbital
4. sin2 θ cos 2 φ - d x2− y2
5. sin2θ sin2 φ - dxy
Problem 5.3.8 Find the value of ‘n’ and ‘l’ and designate the orbital if Ψ = Nr e−r2 cos θ
Solution:
r - its power ‘1’ shows that l = 1
e−r2 - shows that n = 2
cos θ shows that it belongs to ‘z’ co ordinate.
Therefore the designation is 2 p z
Problem 5.3.9 . Find the value of ‘n’ and ‘l’ and designate the orbital if Ψ = Nr e−r2 sin θ cosφ
Solution:
r - its power ‘1’ shows that l = 1
e−r2 - shows that n = 2
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sin θ cosφ shows that it belongs to ‘x’ co ordinate.
Therefore the designation is 2 p x
Problem 5.3.10 Ψ = (6-Zr ) r e−2 r
3 cos θ, Find the value of n,l,m Solution:
Erom e−2 r
3 we can say n = 3
r 1 shows that l = 1
cos θ shows that m = 0Problem 5.3.11 The angular parts of wave function of 3d orbitals are given by
a. ( 3 cos 2 θ – 1) b. sin θ cosφ cos θ . Designate them.
Solution:
a. ( 3 cos 2 θ – 1)
This contains only cos2 θ. Therefore the orbital is designated as d z2
b. sin θ cosφ cos θ
= (sin θ cos φ ) cos θ [ x = r sin θ cosφ y = r sin θ sinφ z = r cos θ ]
= xz
Therefore the orbital is designated as d xz
Problem 5.3.12 The angular parts of wave function of 3d orbitals are given by
a. sin θ sinφ cos θ b. sin2 θ cos 2 φ c. sin2θ sin2 φ Designate them.
Solution:
a. sin θ sinφ cos θ
= (sin θ sin φ ) cos θ
= yz [[ x = r sin θ cosφ y = r sin θ sinφ z = r cos θ ]
Therefore the orbital is designated as d yz
b. sin2 θ cos 2 φ
sin2 θ cos 2 φ = sin2 θ ( cos 2 φ - sin 2 φ ) [ cos 2A = cos 2A – sin2 A]
= sin2θ cos 2 φ - sin2θ sin 2 φ
= x2 – y2
Therefore the orbital is designated as dx2- y2
c. sin2θ sin2 φ
sin2 θ sin2 φ = sin 2 θ × 2 sin φ cos φ [ sin 2A = 2 sin A cos A]
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= 2 xy
Therefore the orbital is designated as d xy
m value 2 1 0 -1 -2
angular part sin2θ cos2 φ sin θ cosφ cosθ 3 cos2θ – 1 (sinθ sin φ ) cos θ sin2 θ sin2 φ
designation x2 – y2 xz x2 yz xy
NORMALISATION
Problem 5.3.13 Find the normalization constant for e−r
Solution:
Condition for normalization is ∫ΨΨ * dτ = 1
∫Ne−r × Ne−r dτ = 1
N2∫ e−r× e−r r2 dr ∫0
π
sin θ dθ ∫0
2 π
dφ = 1 [ dτ=∫0
∞
r 2dr ∫0
π
sin θ dθ ∫0
2 π
dφ]
N2∫ e−2 r r2 dr ∫0
π
sin θ dθ ∫0
2 π
dφ = 1
N2 × 2!(2)2+1 [ 2] [ 2π ] = 1
N2 × 28 [ 2] [ 2π ] = 1
N2 × [ π ] = 1
N = 1√π
Problem 5.3.14 Find the normalization constant for r e−r2 cos θ
Solution:
Condition for normalization is ∫ΨΨ * dT = 1
∫Nr e−r2 cos θ × Nr e
−r2 cos θ dτ = 1
N2∫r 2e−r cos2θ r2 dr ∫0
π
sin θ dθ ∫0
2 π
dφ = 1
N2∫ e−r r4 dr ∫0
π
cos2θ sin θ dθ ∫0
2 π
dφ = 1
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N2 × 4 !(1)4+1 [ -
13 -
13] [ 2π ] = 1
N2 × 24 × [- 23 ] [ 2π ] = 1
N2 × [32 π ] = 1
N = 1
√32 π
Problem 5.3. 15.Show that the 1s orbital with wave function 1√π
e−r is normalized
Solution
∫ΨΨ * dτ = ∫ 1√π
e−r × 1√π
e−r dτ
= 1π ∫
−∞
∞
× e−2 r r2 dr ∫0
π
sin θ dθ ∫0
2 π
dφ [ dτ = r2 dr ∫0
π
sin θ dθ ∫0
2 π
dφ]
= 1π ∫−∞
∞
× e−2 r r2 dr ∫0
π
sin θ dθ [2π]
= 2 ∫−∞
∞
× e−2 r r2 dr ∫0
π
sin θ dθ
= 2 × ∫−∞
∞
× e−2 r r2 dr [ - cos θ] [∫sin θ dθ = - cos θ]
= 2 × ∫−∞
∞
× e−2 r r2 dr [ - ( -1-1 ) ] [cosπ = -1 , cos0 = 1]
= 4 × ∫−∞
∞
× e−2 r r2 dr
= 4 × 2 !
(2 )2+1 [ ∫0
∞
r ne−Ardr = n!
(A )n+1 ]
= 4 × 28
= 1
Problem 5.3.16 Show that the 1s orbital oh Hydrogen atom with wave function 1√π( z
a 0)
32 ×e
− zra0 is
normalized Proof:
Normalisation condition is ∫−∞
∞
Ψ Ψ ¿dτ = 1
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Ψ= 1√ π
( za 0)
32 × e
− zra0
∫−∞
∞
Ψ Ψ ¿dτ
= ∫−∞
∞1√π( z
a 0)
32 ×e
− zra0 × 1
√ π( za 0)
32 × e
−zra0 dτ
= 1π( za 0)3
∫−∞
∞
× e−2zr
a0 dτ
= 1π( za 0)3
∫−∞
∞
× e−2zr
a0 r2 dr ∫0
π
sin θ dθ ∫0
2 π
dφ [ dτ = r2 dr ∫0
π
sin θ dθ ∫0
2 π
dφ]
= 1π( za 0)3
∫−∞
∞
× e−Ar r2 dr ∫0
π
sin θ dθ ∫0
2π
dφ [ A = 2 za0
]
= 1π( za 0)3
∫−∞
∞
× e−Ar r2 dr ∫0
π
sin θ dθ [2π]
= 2 ×( za 0)
3
∫−∞
∞
× e−Ar r2 dr ∫0
π
sin θ dθ
= 2 ×( za 0)
3
∫−∞
∞
× e−Ar r2 dr [ - cos θ] [∫sin θ dθ = - cos θ]
= 2 ×( za 0)
3
∫−∞
∞
× e−Ar r2 dr [ - ( -1-1 ) ] [cosπ = -1 , cos0 = 1]
= 4 ×( za 0)
3
∫−∞
∞
× e−Ar r2 dr
= 4 ×( za0)
3
2 !
(A )2+1 [ ∫0
∞
r ne−Ardr = n !
(A )n+1 ]
= 4 ×( za0)
3
2
( 2 za0 )
3 [ A = 2 za0
]
= 4 ×( za0)
3
2
8×( za0 )
3
= 1
Problem 5.3. 17 .Show that the 2s orbital with wave function1
√32 π ×(2 - r) × e –( r2 ) is normalised
Solution:
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∫ΨΨ * dτ = ∫0
∞ 1√32 π
×(2−r )× e – ( r2 ) 1√32 π
×(2−r )× e – ( r2 )d τ
=
= 132 π∫0
∞
(2−r )2 e−r dτ
= 1
32 π ∫0
∞
(2−r )2 e−r r2dr ∫0
π
sin θ dθ ∫0
2 π
dφ [ dτ = r2 dr ∫0
π
sin θ dθ ∫0
2 π
dφ]
= 1
32 π ∫0
∞
(2−r )2 e−r r2 dr ∫0
π
sin θ dθ [2π]
= 1
16 ∫0
∞
(2−r )2 e−r r2 dr [ - cos θ] [∫sin θ dθ = - cos θ]
= 1
16 ∫0
∞
(2−r )2 e−r r2 dr [ - ( -1-1 ) ] [cosπ = -1 , cos0 = 1]
= 18 ∫
0
∞
(2−r )2 e−r r2 dr
= 18 ∫
0
∞
(4−4 r+r2)e−r r2 dr
= 18 ∫
0
∞
(4 e−r r2−4 e−r r3+e−r r4) dr
= 18 [ 4 ×
21 - 4 ×
3!1 +
4 !1 ] [ ∫
0
∞
r ne−Ardr = n !
(A )n+1 ]
= 18 [ 8 - 4(6) + 24 ]
= 1
ORTHOGONAL
Problem 5.3. 18.Show that the 1s orbital and 2s orbital of H-atom are orthogonal Solution
The wave function for 1s orbital and 2s orbitals are 1√π
e−r and 1
√32 π ×(2- r)×e
−r2 respectively.
∫Ψ 1Ψ 2 dτ = ∫ 1
√πe−r × 1
√32 π ×(2- r)×e
−r2 dτ
= 1
π √32 ∫ e−r× (2- r)×e
−r2 dτ
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= 1
π √32 ∫ e−r× (2- r)×e
−r2 r2 dr ∫
0
π
sin θ dθ ∫0
2 π
dφ [ dτ = r2 dr ∫0
π
sin θ dθ ∫0
2 π
dφ]
= 1
π √32 ∫ e
−3 r2 (2- r) r2 dr ∫
0
π
sin θ dθ [2π]
= 1√8
∫ e−3 r
2 (2- r) r2 dr ∫0
π
sin θ dθ
= 1√8
∫ e−3 r
2 (2- r) r2 dr[ - cos θ] [∫sin θ dθ = - cos θ]
= 1√8
∫ e−3 r
2 (2- r) r2 dr [ - ( -1-1 ) ] [cosπ = -1 , cos0 = 1]
= 1√2
∫ e−3 r
2 (2- r) r2 dr
= 1√2
∫ e−3 r
2 (2r2 - r3 ¿ dr
= 1√2
¿ 2r2 dr - ∫ e−3 r
2 r3 dr ]
= 1√2
[ 2× 2
( 32 )3 -
3 !
( 32 )4 ] [ ∫
0
∞
r ne−Ardr = n!
(A )n+1 ]
= 1√2
[ 2× 1627 -
6×1651 ]
= 1√2
[ 2× 1627 -
2× 1627 ]
= 0
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ASSIGNMENT-1
2. Find the number of radial and angular nodes of 4d orbital
Rn,l = - √( 2 zna0 )
3
× (n− l−1 )!2n [ (n+l )! ]3
× ( 2 zrna 0 ) l × e –( zr
na 0 ) × ( d
d r ) p [ e r ( d
d r ) k ( r k e –r) ]
n= 4
l= 3
p = 2l +1 = 7
k = n +l =7
Rn,l = ( 2 zrna 0 ) 3 × e –( zr
na 0 ) × ( d
d r ) 7 [ e r ( d
d r ) 7 ( r 7 e –r) ]
?
The angular wave function is
Θ l,m = √ 2 l+12
× (l−m ) !(l+m ) !
× 1
2l× l! (1- cos 2 θ ) m/2 (
dd (cosθ) )
l+m [cos 2 θ -1 ) l × 1√2 π
e imφ
Θ l,m = (1- cos 2 θ ) m/2 ( d
d (cosθ) ) l+m [cos 2 θ -1 ) l e imφ
l= 3
m = 0
Θ l,m = ( d
d (cosθ) ) 3 [cos 2 θ -1 ) 3
= ( d
d (x) ) 3 [x2 -1 ) 3
= ( d
d (x) ) 3 [x2 -1 ) 3
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= d3
d x3 ( x6 - 3x4 +3x2 - 1)
= d2
d x2 ( 6 x5 - 12x3 +6x)
= ddx
( 30 x4 - 36x2 +6)
= ( 120 x3 – 72x)
120 x3 = 72x
120 x2 - 72 = 0
This is quadratic equation which has 2 values. Therefore no of nodes = 2
3. Find the most probable radius of 2s orbital
R = ( 18¿½ ( 1
2 a0 )32 ( r
a0) e
−r2a0
P (r) dr = r2 R2
= r2 18
( 12 a 0 )
3
(( ra0)
2
) e−ra0
At most probable distance , ddr [ p(r) ] = 0
ddr [ p(r) ] =
18
( 12 a 0 )
3 ddr¿ e
−ra0 ]
= 18
( 12 a 0 )
3
( 1a 0 )
2 ddr¿ e
−ra0 ]
= 18
( 12 a 0 )
3
( 1a 0 )
2
¿ e−ra0 (−ra0
) + e−ra0 ¿¿ )]
ddr [ p(r) ] = 0
0 = 18
( 12 a 0 )
3
( 1a0 )
2
¿ e−ra0 (−1a0
) + e−ra0 ¿¿ )]
0 = r 4 e−ra0 (−1a0
) + e−ra0 ¿¿ )]
Divide by e−r2 a0
0 = ¿ (−1a0
) + ¿¿ )]
r 4 (1a0
) = 4 r3
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r (1a0
) = 4
r = 4a0
PROBABILITY OF FINDING THE ELECTRON
Probability of finding the electron is given by
P = r2 × Ψ ×Ψ
5.3.19 Find the radial probability of finding the electron and the most probable distance at which the 1s
electron of H-atom is to be found. The value of R(r) for 1s orbital in atomic unit (a.u) is given by R(r) = 2 z 3/2 e - ( z r ) where z is the atomic number.
Solution:
Radial probability
P = r2 × [R(r)] 2
= r2 × [ 2 z 3/2 e - ( z r ) ] 2
= r2 × 4 z3 e - ( 2 z r )
To find most probable distance :
At the maximum distance dPdr = 0
dPdr = 4 z3 [ r2 ( -2z) e - ( 2 z r ) + e - ( 2 z r ) (2 r ) ]
= 4 z3 [ r2 ( -2z) + (2 r ) ] e - ( 2 z r)
Since dPdr = 0
0 = 4 z3 [ r2 ( -2z) + (2 r ) ] e - ( 2 z r)
∴ r2 ( -2z) + (2 r ) = 0
Dividing by 2r, -( r z) +1 = 0
∴ r = 1/z
For H-atom z = 1 ∴ r = 1 a.u
which is the radius of first Bohr orbit.
Problem 5.3.20 Find the most probable value of ‘r’ for an electron in ‘1s’ orbital of H- atom. Solution:
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P = r2 × [ 2 ( z
a0 ) 3/2 e –( zra 0 ) ] 2
= r2 × 4 ( z
a0 ) 3 e –( 2 zra 0 )
To find most probable value of ‘r’ put dPdr = 0
dPdr = 4 (
za 0 ) 3 [ r2 ( -
2 za 0 ) e - ( 2 zr
a 0 ) + e - ( 2 zr
a 0 ) (2 r ) ]
= 4 ( z
a 0 ) 3 [ r2 ( -2 za 0 ) + (2 r ) ] e - ( 2 zr
a 0)
0 = 4 ( z
a 0 ) 3 [ r2 ( --2 za 0 ) ) + (2 r ) ] e - ( 2 zr
a0)r)
r2 ( --2 za 0 ) ) + 2 r = 0
divide by ‘2r’ r ( --z
a 0 ) ) = - 1
r = a0
zFor H-atom z = 1 ∴ r = a0
Problem 5.3.21 : The value of R(r) for 1s orbital is given by R(r) = 2 (Z 3
πao 3) 1/2 e - ( zr
ao) Where z is the
atomic number. Find most probable distance at which the 1s electron of H-atom is to be found.
Solution:
Radial probability P = r2 × [R(r)] 2
= r2 × [2 (Z 3
πao 3) 1/2 e - ( zr
ao) ] 2
= r2 × 4 (Z 3
πao 3) e - ( 2 zr
ao)
To find most probable distance :
At the maximum distance dPdr = 0
dPdr = 4 (
Z 3πao 3
) [ r2 ( - 2 za 0 ) e - ( 2 zr
ao) + e - ( 2 zr
ao) (2 r ) ]
= 4 (Z 3
πao 3) [ r2 ( -
2 za 0 ) + (2 r ) ] e - ( 2 zr
ao)
Since dPdr = 0
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0 = 16π (Z 3
πao 3) [ r2 ( -
2 za 0 ) + (2 r ) ] e - ( 2 zr
ao)
∴ r2 (- 2 za0 ) + (2 r ) = 0
Dividing by 2r, -( rza0 ) +1 = 0
∴ rza0 = 1
For H-atom z = 1 ∴ r = a0
which is the radius of first Bohr orbit.
Problem 5.3.22 :Calculate the probability of finding the electron in a sphere of radius r , given that 1s -
wave function of H-atom is (1
π a03 ) ½ e
ra0
Solution:
P = ∭000
r ,θφ
Ψ ×Ψ * dT
= ∭000
r ,θφ
( 1πa 03
) ½ e - ra0
×(1
πa 03) ½ e - r
a 0 dT
= ∫0
r
¿¿ ] 2 r2 dr ∫0
π
sin θ dθ ∫0
2 π
dφ
= ( 1πa03 ) ∫0
r
e−2ra 0
r2 dr ∫0
π
sin θ dθ ∫0
2π
dφ
= ( 1πa03 ) ∫0
r
e−2ra0
r2 dr ∫0
π
sin θ dθ [2π]
= 2 π( 1πa03 ) ∫0
r
e−2ra 0
r2 dr ∫0
π
sin θ dθ
= 2 π( 1πa03 ) ∫0
r
e−2ra 0
r2 dr [ - cos θ]
= 2 π( 1πa03 ) ∫0
r
e−2ra 0
r2 dr [ - ( -1-1 ) ]
=4 π( 1πa03 ) ∫0
r
e−2ra0
r2 dr
= 4
a 03 ∫0
r
e−2ra 0
r2 dr
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= 4
a 03 [ r2 [ e−2 r
a 0−2a0
] -2 r [ e−2 r
a 0+4a02
] + 2 [ e−2 r
a 0−8a 03
]
= 4
a 03 [ a02 [
e−2−2a 0
] -2 a0 [ e−2+4a02
] + 2 [ e−2−8a03
] - [ 2 [ e0−8a03
]
= e -2 × 4
a 03 { [ ao3 (-1/2) -2 a0
3 ( +1/4) + 2 a0 3 ( -1/8) ] - [ 2a0
3 ( -1/8 ] }
= e -2 × 4 [(-1/2) -2 ( +1/4) + 2 ( -1/8) ] - 4
a 03 [ 2a0 3 ( -1/8 ]
= -5 e -2 + 1
= - 1.323 +1
= 0.323
Problem5.3.23 : What is the probability of an electron being found at a distance of r = 0 and r = ½ a.u. from
the nucleus of H- atom in 1s state whose wave function is Ψ = 1
√ π e – r
Solution:Probability P = r2 × Ψ 2
= r2 × (1
√ π e – r ) 2
= r2 × 1π e – 2 r
When r = 0, P = 0
When r = ½ P = ( ½)2 × 1π e – 2 r
= 1
4 π × e – 1
Problem 5.3.24 Find the most probable value of ‘r’ for an electron in ‘1s’ orbital of H- atom
Solution:
Radial probability P = r2 × [R(r)] 2
= r2× [ 2 e−r ] 2 [ R(r)] for ‘1s’ orbital is 2 e−r ]
= r2 × 4 e−2 r
dPdr = 4 [ r2 ( -2) e - ( 2 r ) + e - ( 2 r ) (2 r ) ]
= 4 [ r2 ( -2) + (2 r ) ] e - ( 2 r)
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0 = 4 [ r2 ( -2) + (2 r ) ] e - ( 2 r) [ At the maximum distance dPdr = 0]
∴ r2 ( -2) + (2 r ) = 0
-( r ) +1 = 0 [Dividing by 2r]
r = 1 a.u which is the radius of first Bohr orbit.
Problem 5.3.25 : Find the probability of an electron in 2pz orbital of H- atom being found at a distance of one a.u. from the nucleus for θ = π/2 to θ = π/4 whose wave function is
Ψ = 1
√32 πr e– r/2 cos θ
Solution:
r =1
Probability P = r2 × Ψ 2
= (1)2 × ( 1
√32 π (1) e
−12 cos θ ) 2
= ( 1
32 π e –1 cos 2 θ )
When θ = π/2 P = ( 1
32 π e –1 × cos 2 (π2 )
= ( 1
32 π e –1 × 0 ) [ cos (π2 ) = 0 ]
When θ = π/4 P = ( 1
32 π ×e−1 × cos 2 (π4 )
= ( 1
32 π ×e−1 × ( 1√2
) 2 [cos (π4 ) =
1√2
]
= ( 1
32 π ×e−1 × ( 12 )
= e−1
64 π
Problem5.3.26 Show that probability of finding the electron in the H- like orbitals is. independent of φ
Solution:
Probability = r2 × Ψ × Ψ *
= r2 × R(r) × P ( θ) × e+imφ × R(r) × P ( θ) × e imφ
= r2 ×[ R(r) × P (θ) ] 2 This is independent of φ
Problem 5.3.27 : Show that the total probability of 2p- orbital depends r only
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Solution:
Total probability ( Ψ2 )
= Ψ px2 + Ψ py
2 +¿ Ψ pz2
= r2 e−r[ sin2θcos2φ + sin2θ sin2φ + cos2θ ] [ x =sinθcosφ y = sinθcosφ z = cosθ] = r2 e−r [ sin2θ ( cos2φ + sin2φ )+ cos2θ ] = r2 e−r [ sin2θ + cos2θ ] = r2 e−r
depends r only
Problem 5.3.28: Calculate the probability of an electron described by is wave function of Hydrogen will be
found with in Bohr radius of the nucleus
Solution:
Given : within Bohr radius. Therefore limit is ‘0’ to a0
Probability P =∫0
a0 4a0
3 r2 e−2 r
a0 dr
= 4a0
3∫0
a0
r 2 e−2 r
a0 dr
Put x = −2 r
a0 therefore r = (
−a0
2 )x
dx = −2a0
dr
∴dr = (−a0
2 ) dx
Limit : when r = 0 , x = 0
When r = a0 , x = -2
Probability P = 4a0
3∫0
−2
((−a0
2 ) x)2
ex (−a0
2 ) dx
= 4a0
3 ×((−a0
2 ))2
×(−a0
2)∫
0
−2
x2ex dx
= −12
×∫0
−2
x2 ex dx
= −12 [x2 ex – 2x ex + 2 ex ] [ Bernouilis formula ∫UV = UV’ – U 1V’’+U IIU’’’]
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= −ex
2 [x2 – 2x + 2 ]
= −e−2
2 [(−2 )2 – 2(-2) + 2 ] – e
0
2 [ 0 +0 +2]
= −e−2
2 [10 ] –
12 [ 2]
= −5 e−2 –1
? = −5 e−2 +1
Problem 5.3.29: Calculate the probability of ‘1s’ electron of Hydrogen will be found with in a distance of 2
a0 from the nucleus
Solution:
Ans: 1−13e−4
Problem 5.3.30 Calculate the radius of the sphere that encloses i) 50 % and ii) 90 % probability of finding a
‘1s’ electron of hydrogen atom
Solution:
The relation between probability(P) and distance(d) is
P = 1- e−2 d [ 2d2 + 2d +1 ]
Ans : 1.339 , 2.66
Problem 5.3.31 : Find the points of maximum probability of electron in 2px,state.
Solution
P = Ψ 2
= r2 e−r sin2 θ cos2φ [ x= sin θ cos φ ]
dpdr = 2r e−r - r2 e−r
2r e−r - r2 e−r = 0 [dpdr = 0 at the maximum point]
2r - r2 = 0 [ dividing by e−r on both sides]
r = 2
dpdθ = r2 e−r cos2φ [ 2 sin θcosθ ]
r2 e−r cos2φ [ 2 sin θcosθ ] = 0 [dpdθ = 0 at the maximum point]
sin θ cos θ = 0 [ dividing by 2r2 e−r cos2 φ on both sides]
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If sin θ = 0 , the probability in everywhere is zero. Therefore only cos θ = 0 . This is possible only if θ
= π2
dpdφ = r2 e−r sin2 θ [- 2 cosφ sinφ ]
r2 e−r sin2 θ [- 2 cosφ sinφ ] = 0 [dpdφ = 0 at the maximum point]
cosφ sinφ = 0 [ dividing by -2r2 e−r sin2θ on both sides]
If sinφ = 0 , the probability in everywhere is zero. Therefore only cos φ = 0 .This is possible only if φ = π2
r = 2, θ = π2 and φ =
π2
AVERAGE DISTANCE
Problem : Find the average value of ‘r’ for 1s electron from the nucleus of H-atom.
Solution:
Average value = ∫Ψ (r) Ψ × dτ = ∫
0
∞
( 1π a0
3 )12 e
−ra0 ( r) ( 1
π a03 )
12 e
−ra0 r2 dr ∫
0
π
sin θ dθ ∫0
2 π
dφ
= 1
π a03 ∫
0
∞
e−2 r
a0 r3 dr ∫0
π
sin θ dθ ∫0
2π
dφ -----------------1
∫0
∞
e−2 r
a0 r3 dr = 3 !
(−2a0)
4 ∫0
∞
e−ar rndr = n!
an+1 ]
= 3× 2× 1
16× a0
4
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= 38
× a04
∫0
π
sin θ = [ - cos θ]
= - [ cosπ - cos 0]
= - [ -1 -1]
= 2
∫0
2π
dφ = [φ]
= 2π
Substituting in 1
Average value = 1
π a3 [38
× a04 × 2 × 2π]
= [38
× a0 × 2 × 2]
= 32 a0
Problem : Find the average value of (1r ) for 1s electron from the nucleus of H-atom.in Bohr’s atomic
unit
Solution:Ψ = 1√π
e−r
Average value = ∫Ψ ((1r ) ) Ψ × dτ
= ∫¿¿ (1r ) ×( 1
√πe−r ) dτ
= 1π ∫¿¿ (1
r ) ∫0
∞
r 2dr ∫0
π
sin θ dθ ∫0
2 π
dφ
= 1π ∫
0
∞
r e−2 r dr ∫0
π
sin θ dθ ∫0
2 π
dφ
[ ∫0
∞
r e−2 r dr = 1 !(2)2
∫0
∞
e−ar rndr = n!
an+1 ]
= 14
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Average value = 1π [
14 ] [ 2] [2π]
= 1
Problem : Find the average value of ‘x’ for 1s electron from the nucleus of H-atom.
Solution: Ψ = 1√π
e−r Average value = ∫Ψ (x) Ψ × dτ = ∫¿¿ (x ) ×( 1
√πe−r ) dτ
= 1π∫ ¿¿(r sinθ cos φ) ×(e−r )∫
0
∞
r 2dr ∫0
π
sin θ dθ ∫0
2 π
dφ [ x = r sinθ cos φ] =
1π ∫¿¿) dr ∫
0
π
sin2θ dθ ∫0
2 π
cosφ dφ =
1π ∫¿¿) dr ∫
0
π
sin2θ dθ (0)
= 0
Problem : Find the average distance of electron in 2s state from the nucleus of H-atom.
Solution: Ψ for 2 s orbital is 1√32 π
(2−r ) e−r2
Average value = ∫Ψ (r) Ψ × dτ ∫¿¿ (r ) ×( 1
√32 π(2−r ) e
−r2 dτ
= 1
32 π ∫¿¿ (r ) ×( (2−r ) e−r2 ∫
0
∞
r 2dr∫0
π
sin θ dθ ∫0
2 π
dφ =
132 π∫ r3 (2−r )2 e−r dr ∫
0
π
sin θ dθ ∫0
2 π
dφ [e−r2 × e
−r2 = e−r]
= 1
32 π ∫ r3(4+r 2−4 r)e−r dr [ 2] [2π]
= 6
Problem Show that average distance of en electron is equal to32 times of its most probable distance
Solution:
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Average value = ∫Ψ (r) Ψ × dτ = ∫
0
∞
( 1π a0
3 )12 e
−ra0 ( r) ( 1
π a03 )
12 e
−ra0 r2 dr ∫
0
π
sin θ dθ ∫0
2 π
dφ
= 1
π a03 ∫
0
∞
e−2 r
a0 r3 dr ∫0
π
sin θ dθ ∫0
2π
dφ
= 1
π a03 [
3 !
(−2a0)
4 × (2)× (2π )]
= 1
π a03
616a0
4 × 4π [ ∫
0
∞
xn e−ax dx = n !(a)n+1 ]
= 32 a0
Most probable distance = a0
∴ Average value = 32 (Most probable distance)
Problem : Find the average distance of electron in 2pz state from the nucleus of H-atom.
Given Ψ for 2 pz orbital is 1√32 π
r e−r2 cos θ, dτ = ∫
0
∞
r 2dr ∫0
π
sin θ dθ ∫0
2 π
dφ Solution:
Average value = ∫Ψ (r) Ψ × dτ = ∫¿¿ cos θ ) (r ) ×( 1
√32 πr e
−r2 cos θ ) dτ
= 1
32 π ∫¿¿ cos θ ) (r ) ×( r e−r2 cos θ )∫
0
∞
r 2dr ∫0
π
sin θ dθ ∫0
2 π
dφ =
132 π ∫
0
∞
r 5e−r dr ∫0
π
cos2θ sin θ dθ ∫0
2 π
dφ -------------1∫0
∞
r 5e−r dr = 5 !(1 )5
= 5×4 × 3 ×2 ×1 = 120Page 214 of 419
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∫0
π
cos2θ sin θ dθ
Put t = cosθ
dt = −sin θ dθ
∫0
π
cos2θ sin θ dθ = ∫0
π
-t 2dt
= −t3
3
= [ −cos3 θ3
]
= −13 [ -1-1]
= 23
∫0
2π
dφ = [φ]
= 2π
Substituting in 1
= 1
32π× 120 ×
23 ×2π
= 5
Problem 5.3.37: Show that average distance of en electron is equal to32 times of its most probable distance
Solution:
Average value = ∫Ψ (r) Ψ × dτ = ∫
0
∞
( 1π a0
3 )12 e
−ra0 ( r) ( 1
π a03 )
12 e
−ra0 r2 dr ∫
0
π
sin θ dθ ∫0
2 π
dφ
= 1
π a03 ∫
0
∞
e−2 r
a0 r3 dr ∫0
π
sin θ dθ ∫0
2π
dφ
= 1
π a03 [
3 !
(−2a0)
4 × (2)× (2π )]
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= 1
π a03
616a0
4 × 4π [ ∫
0
∞
xn e−ax dx = n !(a)n+1 ]
= 32 a0
Most probable distance = a0
∴ Average value = 32 (Most probable distance)
6.APPROXIMATION METHODSNeed for approximation methods:
Schrodinger wave equation can be solved for systems having only one electron. But if an atom or
molecule has many interacting electrons , we can not solve Schrodinger equation. Because there will be more
than one potential energy terms. In such a case we must go for approximation methods.
6.1.PERTURBATION METHOD
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This method involves the determination of eigen functions( Ψ ‘ ) , and eigen values ( E’ ) of the of
the perturbed system , in terms of those of the unperturbed (Ψ0 and E 0 )
This method is applied if
1.The system differs only slightly from the unperturbed ( known)system.
2.Energy , wave function and Hamiltonian for the unperturbed system ( E0 , Ψ 0 , and H0 ) are known.
Example :
1.Normal Hydrogen atom is unperturbed system
When placed in an electric ( Stark effect), it causes perturbation.
When placed in magnetic field (Zeemann effect) it is perturbed system
2.A simple harmonic oscillator is an unperturbed system On the other hand an unharmonic oscillator is a
perturbed system.
There are two types
1. Time independent perturbation method
2. Time dependent perturbation method
Time independent perturbation method
Suppose we have a system with Time independent Hamiltonian H and we are unable to solve the
Schrodinger equation HΨ = E Ψ for the eigen value and eigen function.
There are two cases
1. Non – degenerate case
2. Degenerate case
Time independent non- degenerate perturbation method
Consider a system whose .Energy , wave function and Hamiltonian are known. Let them be E 0 , Ψ 0 ,
and H0 . If a small perturbation is applied, then
The Hamiltonian is decomposed in to two parts as
H = H0 + λ H’
where H0 is the perturbed part and λ H’ is the perturbation.
Let the eigen values and eigen functions of the unperturbed problem be E10 , E2
0 , E30 , and Ψ 1
0 , Ψ 20 ,Ψ 3
0
It is assumed that the eigen functions and the energies can be expressed in the form of power series as
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Ψ = Ψ n0 + λ Ψ n
1 + λ 2 Ψ n2 .........Ψ n
k
E = En0 + λ En
1 + λ2 En2 ..........En
k
Where Ψ nk and En
k are the ‘ k th ‘ order correction in wave function and energy
The Schrodinger equation is given by
HΨ = E Ψ
Substituting the values, and omitting the higher terms
{ H0 + λ H’ }{ Ψ n0 + λ Ψ n
1 + λ 2 Ψ n2 }= { En
0 + λ En1 + λ2 En
2 }{Ψ n0 + λ Ψ n
1 + λ 2 Ψ n2 . }
Expanding
H0Ψ n0
+ H0 λ Ψ n1 + H0 λ 2 Ψ n
2 + λ H’ Ψ n0 + λ 2H’ Ψ n
1 + λ 3 H’ Ψ n2
= En0 Ψ n
0 + En0λ Ψ n
1 +En0 λ2Ψ n
2 + λ En1 Ψ n
0 + λ2 En1 Ψ n
1 + λ3 En1 Ψ n
2 + λ2 En1 Ψ n
0 + λ3 En1 Ψ n
1 + λ4 En2 Ψ n
2
Neglecting the higher terms (λ3 and λ4 terms )
H0Ψ n0
+ H0 λ Ψ n1 + H0 λ 2 Ψ n
2 + λ H’ Ψ n0 + λ 2H’ Ψ n
1
= En0 Ψ n
0 + En0λ Ψ n
1 +En0 λ2 Ψ n
2 .+ λ En1 Ψ n
0 + λ2 En1 Ψ n
1 + λ2 En1 Ψ n
0
rearranging
H0Ψ n0
+ λ [ H0Ψ n1 + H’Ψ n
0 ] + λ 2 [ H0 Ψ n2 + H’ Ψ n
1 ]
= En0 Ψ n
0 + λ [ En0 Ψ n
1 + En1 Ψ n
0 ] + λ2 [ En0 Ψ n
2 .+ En1 Ψ n
1 + En1 Ψ n
0]
Comparing constant term , the coefficient of λ and coefficient of λ 2 we get the Schrodinger equation
H0Ψ n0
= En0 Ψ n
0 : unperturbed system.
H0Ψ n1 + H’Ψ n
0 = En0 Ψ n
1 + En1 Ψ n
0 : first order perturbation
H0 Ψ n2 + H’ Ψ n
1 = En0 Ψ n
2 .+ En1 Ψ n
1 +En1 Ψ n
0 : Second order perturbation
Evaluation of First order energy:
The Schrodinger equation for first order perturbation is
H0Ψ n1 + H’Ψ n
0 = En0 Ψ n
1 + En1 Ψ n
0
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The unknown function Ψ n1
can be expanded in terms o known functions
Ψ 10 , Ψ 2
0 ,Ψ 30 ,…….Ψ m
0
Ψ n 1 = ∑0
m
Cm Ψ m0
Substituting in the above equation,
H0 ∑0
m
Cm Ψ m0 + H1 Ψ n
0 = En0 ∑
0
m
Cm Ψ m0 + En
1 Ψ n0
H0 Ψ m0 ∑
0
m
Cm+ H1 Ψ n0 = En
0 Ψ m ∑0
m
Cm0 + En
1 Ψ n0
[Rearranging]
since H0 Ψ m0 = Em
0 Ψ m 0 the above equation becomes
Em0 Ψ m 0
∑0
m
Cm + H1 Ψ n0 = En
0 Ψ m ∑0
m
Cm + En1 Ψ n
0
Em0 Ψ m
0 ∑0
m
Cm - En0 Ψ m ∑
0
m
Cm + H1 Ψ n0 = En
1 Ψ n0
[Rearranging]
Ψ m0 ∑
0
m
Cm { Em0−En
0 } + H1 Ψ n0 = En
1 Ψ n0
------------------5
Multiplying the above equation by Ψ n0 from the left and integrate
∫Ψ n0Ψ m
0∑0
m
Cm {Em0−En
0 } dτ+ ∫Ψ n0 H1 Ψ n
0 dτ = ∫Ψ n0 En
1 Ψ n0
dτ [ ∫Ψ n
0 En1 Ψ n
0 =En
1 ∫Ψ n0Ψ n
0
but ∫Ψ n0 H1 Ψ n
0dτ ≠ H 1∫Ψ n0Ψ n
0dτ ]
Therefore the above equation becomes
∫Ψ n0Ψ m
0∑0
m
Cm {Em0−En
0 }) dτ + ∫Ψ n0 H1 Ψ n
0 dτ = En1∫Ψ n
0 Ψ n0 dτ
but ∫Ψ n0Ψ m
0 = 0,
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∫Ψ n0Ψ n
0 = 1
Therefore the above equation becomes
0 + ∫Ψ n0 H1 Ψ n
0 = En1
∴ En1
= ∫Ψ n0 H1 Ψ n
0
This is the expression for first order perturbed energy .This shows that the first order perturbation energy
for a non-degenerate system is just the perturbation function averaged over the corresponding
unperturbed state of the system
Thus the first order energy is
En = En0 + λ ∫Ψ n
0 H1 Ψ n0 dT
Evaluation of First order wave function:
From equation 5
Ψ m 0 ∑0
m
Cm { Em0−En
0 }+ Ψ n0 H1 = Ψ n
0 En1
Multiplying the above equation by Ψ m0 from the right and integrate
∫Ψ m0∑
0
m
Cm {Em0−En
0 }Ψ m0 dτ + ∫Ψ n
0 H1 Ψ m0 dτ = ∫ Ψ n
0 En1 Ψ m
0 dτ [ ∫Ψ n
0 En1 Ψ n
0 =En
1 ∫Ψ n0Ψ n
0 but ∫Ψ n0 H1 Ψ n
0dτ ≠ H 1∫Ψ n0Ψ n
0dτ ]
Therefore the above equation becomes
∑0
m
Cm {Em0−En
0 } ∫Ψ m0 Ψ m
0 dτ + ∫Ψ n0 H1 Ψ m
0 dτ = En1 ∫ Ψ n
0 Ψ m0 dτ
But ∫Ψ m0 Ψ m
0 = 1,
∫ Ψ n0 Ψ m
0 = 0
Therefore the above equation becomes
∑0
m
Cm {Em0−En
0 } + ∫Ψ m0 H 1 Ψ n
0 = 0
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∴ Cm = - ∫Ψ m
0 H 1Ψ n0 dT
{Em0−En
0 }
Substituting in 4, we get Ψ n = - ∑m=0
m ∫Ψ m0 H 1Ψ n
0d T
{Em0−En
0 } Ψ m0
This shows that the first order correction in wave function
1. is always negative
2. can be determined from the eigen function and eigen values of unperturbed system.
Thus the first order wave function is Ψ n = Ψ n0 - λ ∑
m=0
m ∫Ψ m0 H 1Ψ n
0 d T
{Em0−En
0 } Ψ m
0
APPLICATION OF PERTURBATION THEOREM TO HYDROGEN ATOM
Perturbation theorem is applied to hydrogen atom to find the corrected ( perturbed ) energy and the
corresponding perturbed wave function It involves the following steps.
1. Making perturbation:
Normal hydrogen atom is unperturbed but when it is kept in electric field it is subjected to
perturbation.
2.Writing the expression for wave function:
The value of Ψ for hydrogen atom is identified Ψ=( e−r
√ π )2. Identifying the perturbed Hamiltonian:
The application of electric field changes the Hamiltonian of hydrogen atom
H= H0 + H ’ . In this H’ is the perturbed Hamiltonian.
3. Calculation of first order perturbed energy.
The perturbed energy is calculated by the following formula
The first order perturbed energy is given by E1 = ∫Ψ 0 H ’Ψ 0 dτ
In spherical co-ordinates E = ∫Ψ 0 H ’Ψ 0 r2 dr∫0
π
sin θ d θ∫0
2π
dφ
4. Adding with the ground state energy:
The perturbed energy is added with the ground state energy
For example When a uniform electric field along Z- axis is applied, the perturbation term is given asPage 221 of 419
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H ‘ = Fz.
The first order perturbed energy is given by E1 = ∫Ψ 0 H ’Ψ 0 dτ
In spherical co-ordinates E = ∫Ψ 0 H ’Ψ 0 r2 dr∫0
π
sin θ d θ∫0
2π
dφ and z = r cos θ
= ∫0
∞
( e−r
√ π )(F rcosθ)( e−r
√ π )r2dr∫0
π
sin θ d θ∫0
2 π
dφ
= Fπ ∫
0
∞
e−2 r (rcos θ)r2 dr∫0
π
sin θ dθ∫0
2 π
dφ
= eFπ ∫
0
∞
e−2 r r3 dr∫0
π
cos θ sin θ d θ∫0
2 π
dφ
= eFπ ∫
0
∞
e−2 r r3 dr 12∫0
π
sin 2θ d θ∫0
2 π
dφ [ sin 2x = 2 sin x cos x ]
= eFπ [
3!24 ] [ ½ (
−cos2θ2 ) ¿0
π ( 2 π ) ∫0
∞
xn e−ax dx= n!(a)n+1
= eFπ [
3!24 ] [ ½ ( 0)
= 0
This shows that there is no first order effect.
6.1.1 APPLICATION TO HYDROGEN ATOM (STARK EFFECT)
Let H0 be the Hamiltonian of H – atom and Ψ 0 ,E0 be the wave function and energy respectively. When
a uniform electric field along Z- axis is applied, the perturbation term is given as
H ‘ = Fz.
The first order perturbed
energy is given by E1 = ∫Ψ 0 H ’Ψ 0 dτ
In spherical co-ordinates E = ∫Ψ 0 H ’Ψ 0 r2 dr∫0
π
sin θ d θ∫0
2π
dφ and z = r cos θ
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= ∫0
∞
( e−r
√ π )(F rcosθ)( e−r
√ π )r2dr∫0
π
sin θ d θ∫0
2 π
dφ
= Fπ ∫
0
∞
e−2 r (rcos θ)r2 dr∫0
π
sin θ dθ∫0
2 π
dφ
= eFπ ∫
0
∞
e−2 r r3 dr∫0
π
cos θ sin θ d θ∫0
2 π
dφ
= eFπ ∫
0
∞
e−2 r r3 dr 12∫0
π
sin 2θ d θ∫0
2 π
dφ [ sin 2x = 2 sin x cos x ]
= eFπ [
3!24 ] [ ½ (
−cos2θ2 ) ¿0
π ( 2 π ) ∫0
∞
xn e−ax dx= n!(a)n+1
= eFπ [
3!24 ] [ ½ ( 0)
= 0
This shows that there is no first order effect.
This perturbation may cause a second order effect owing to the mixing of Ψ 0 orbital with excited state orbitals.
The excited states close to Ψ 0 are 2s,2px,2py and 2pz . Therefore the perturbed energy is
E1 = ∫1 s H ’ 2 s dτ +∫1 s H ’ 2 px dτ +∫1 s H 2 py dτ +∫1 s H 2 pz dτ
The angular parts of various orbitals are
2s orbital : no angular part
2px orbital : sinθ cos φ2py orbital : sinθ sin φ2pz orbital : cosθ
Because of sin θ term the integrals 1,2 and 3 vanishes. Therefore
E = ∫1 s H 2 pz dτ
The wave function for 1s orbital = e−r
√π
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The wave function for 2 PZorbital = 1
4 √2π [ r e
−r2 cos θ ]
E = ∫0
∞
( e−r
√ π )(F rcosθ) 14√2 π
[ r e−r2 cos θ ]r2 dr∫
0
π
sin θ dθ∫0
2π
dφ
= F
4 π √2 ∫
0
∞
e−r(rcos θ)[r e−r2 cosθ ]r2 dr∫
0
π
sinθ dθ∫0
2π
dφ
= 1
4 π √2 ∫
0
∞ [e−3 r2 ] r4 dr∫
0
π
cos2θ sin θ d θ∫0
2 π
dφ
= 1
4 π √2 [
4 !
( 32)
5 ×∫0
π
cos2θ sin θ d θ∫0
2 π
dφ ∫0
∞
xn e−ax dx= n!(a)n+1
= 1
4 π √2 × ( 2
3)
5
× 24 ∫0
π
cos2θ sin θ d θ∫0
2 π
dφ [ 4! = 4 ×3× 2 ×1 = 24 ]
= 1
4 π √2 × ( 2
3)
5
× 24× (23 )× ∫
0
2 π
dφ ∫0
π
cos2θ sin θ dθ = −cos3 θ
3 = −13 (-1-1) =
23 ]
= 1
4 π √2 × ( 2
3)
5
× 24× (23 )× 2π ∫
0
2 π
dφ = 2π =
1√2
× ( 23)
5
× 8
Thus energy of H- atom in the presence of electric field = 1√2
×( 23)
5
× 8
Problem A Hydrogen atom is exposed to an electric field of strength F so that its perturbed Hamiltonian is
Fz. Show that there is no first order effect. Given Ψ 0¿ = e−r
√ π , ∫
0
∞
e−2 r r3 dr = 38
Proof: H ‘ = Fz.
The first order perturbed energy is given by E1 = ∫Ψ 0¿H ’Ψ 0 dτ
In spherical co-ordinates E = ∫Ψ 0¿H ’Ψ 0 r2 dr∫
0
π
sin θ d θ∫0
2π
dφ and z = r cos θ
= ∫0
∞
( e−r
√ π )(F rcosθ)( e−r
√ π )r2dr∫0
π
sin θ d θ∫0
2 π
dφ
= Fπ ∫
0
∞
e−2 r (rcos θ)r2 dr∫0
π
sin θ d θ∫0
2 π
dφ
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= eFπ ∫
0
∞
e−2 r r3 dr∫0
π
cos θ sin θ d θ∫0
2 π
dφ
= eFπ ∫
0
∞
e−2 r r3 dr 12∫0
π
sin 2θ d θ∫0
2 π
dφ [ sin 2x = 2 sin x cos x ]
= eFπ [
3!24 ] [ ½ (
−cos2θ2 ) ¿0
π ( 2 π )
= eFπ [
3!24 ] [ ½ ( 0)
= 0
This shows that there is no first order effect.
APPLICATION TO PARTICLE IN A BOX
Problem : An electric field of strength F is applied to an electron in a one dimensional box of length L, so
that the potential energy V = eFx , rises along the box V = 0 at x = 0 and V = e FL at x= L. Find the first
order correction to energy and wave function.
Solution:
First order correction to energy:
The perturbed Hamiltonian is H’ = eFx
The wave function is given by Ψ = √ 2L sin (
πL) x
The first order perturbed energy is given by
E1 = ∫Ψ 0¿H ’Ψ 0 dτ
Substituting the values, E1 = ∫{√ 2L
sin ( πL )x }{eFx } {√ 2L
sin ( πL )x } dx
= 2eF
L ∫{sin( πL ) x }{x }{sin( π
L ) x } dx
= 2eF
L ∫ x sin2( πL )x dx
Put y = πxL ∴ x =
yLπ
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dy = πL dx , ∴ dx =
Lπ dy
limit : when x=0 y = 0, when x = L y = π
E1 = 2eF
L ∫ yLπ
sin 2 y × Lπ dy
= 2 eF
L × L2
π 2 ∫ y sin2 y dy
= 2 eF L
π2 ∫ y ¿¿¿ dy
= 2 eF L2π 2 { ∫ y dy – ∫ y cos2 y dy }
= eF L
π2 { [ y2
2]0
π
- 2sin 2 y
2 – ( - cos2 y
2 ) ¿0π } [ Bernoulis formula]
= eF L
π2 [π 2
2– 0 ¿
= eF L
2
First order correction to wave function:
The first order perturbed wave function is given by
Ψ 1 = ∑m≠ n{∫Ψ m
0 H ' Ψ n0 dτ
En0−Em
0 }Ψ m0
= ∫Ψ 20 H ' Ψ 1
0 dτ
E10−E2
0 Ψ 20 + ∫Ψ 3
0 H ' Ψ 20 dτ
E20−E3
0 Ψ 30 + ....
E 1 = h 2
8m L 2
E 2 = 4 h2
8 m L 2
E1 – E2 = −3h 28m L 2
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Ψ 1 = ∫ {√ 2
Lsin( 2 π
L ) x }{eFx } {√ 2L
sin( πL ) x }dx
−3 h28m L 2
+ ....
=
2 eFL ∫{sin( 2π
L )x} {x } {sin( πL ) x }dx
−3h28 m L 2
= 2 eF
L × 8m L2
−3h2 ∫{sin( 2πL )x }{x }{sin ( πL )x }dx
= 16 eFm L−3 h2 ∫{sin(2 π
L )x }{x }{sin (πL )x }dx
Put y = πxL ∴ x =
yLπ
dy = πL dx , ∴ dx =
Lπ dy
limit : when x=0 y = 0, when x = L y = π
∴ Ψ 1 = 16 eFm L−3h2 ∫ {sin 2 y }( yL
π){sin y }L
πdy
= 16 eFm L−3 h2 (
Lπ ) 2 { ∫ y sin 2 y sin y dy }
Let I = ∫ y sin 2 y sin y dy
= ∫ y (2sin y cos y )sin y dy [ sin 2x = 2sinx cosx ]
= 2 ∫ y cos y sin2 y dy
= 2 ∫ y cos y ¿¿¿
= 2 { ∫ y cos y dy - ∫ y cos3 y dy }
= 2 [ I1 + I2 ]
I1 = ∫ y cos y dy
= y siny – ( 1)( -cos y) [ Bernoulis formula]
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= y sin y + cos y¿0π
= (0) + ( – 1 – 1 )
= - 2
I2 = ∫ y cos3 y dy cos 3x = 4 cos 3 x – 3 cos x ∴ cos 3 x = ¼ ( 3 cosx + cos 3x )
= ∫ y ¿¿
= ¾ ∫ y cos y dy+¿¿ ¼ ∫ y cos3 y dy
= ¾ [ y siny - (1) ( - cos y) ] + ¼ [ y sin 3 y
3 - ( 1) ( - cos3 y
9 ) ]
= ¾ [ y siny + cos y) ¿0π + ¼ [ y
sin 3 y3 +
cos3 y9 ¿0
π
= ¾ [ 0 + ( -1 – ( 1) ] + ¼ [ 0 + 19 ( -1 – ( 1) ]
= −64 +
−236
= −54−2
36
= −5636
= −14
9
I = 2 [ - 2 + −14
9 ]
Ψ 1 = 16 eFm L−3 h2 (
Lπ ) 2 { 2 [ - 2 +
−149 ]
= 0. 48 ( eFL ) ( m L2
h2 )
The perturbed wave function correct to first order is
Ψ = Ψ 0 + 0. 48 ( eFL ) ( m L2
h2 )
Problem: Find the first order correction to energy of a particle in a one dimensional box of length ‘a’ with
slanted bottom such that Vx = V 0
a x where V0 is constant.
Solution:
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The perturbed Hamiltonian is H’ = = V 0
ax
The wave function is given by Ψ = √ 2L sin (
nπL ) x
The first order perturbed energy is given by
E1 = ∫Ψ 0¿H ’Ψ 0 dτ
Substituting the values, E1 = ∫{√ 2L
sin ( nπL ) x}{V 0
ax }{√ 2
Lsin( nπ
L )x } dx
= 2a ×
V 0
a×∫ {sin ( nπ
L ) x}{x }{sin ( nπL ) x } dx
= 2V 0
a2 ∫0
a
x sin2( nπL ) x dx
= 2V 0
a2 (a2
4 )
= V 0
2
APPLICATION TO HARMONIC OSCILATOR
Problem: A harmonic oscillator is subjected to perturbation H = E x. Find the first order perturbation energy
and wave function. Given Ψ0 = ( βπ¿ ¼ e
− β2 x2
, ∫−∞
+∞
(x¿e−β x2
)¿ dx = 0
Solution:
The first order perturbed energy is given by E1 = ∫Ψ 0¿H ’Ψ 0 dτ
= ∫−∞
+∞
( βπ)¼ e
−β2 x 2
(Ex)( βπ)¼e
− β2 x2
dx
= E (βπ) ½ ∫
−∞
+∞
¿¿ dx
= E (βπ) ½ ∫
−∞
+∞
(x¿e−β x2
)¿ dx
= 0
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Problem: An an harmonic oscillator is subjected to perturbation H = ax3 +bx4. Find the first order
perturbation energy and wave function. Given Ψ0 = ( βπ¿ ¼ e
− β2 x2
Solution:
The first order perturbed energy is given by E1 = ∫Ψ 0¿H ’Ψ 0 dτ
= ∫−∞
+∞
( βπ)¼e
−β2 x2
(ax 3+bx 4)( βπ)¼e
−β2 x2
dx
= (βπ) ½ ∫
−∞
+∞
¿¿ dx
= (βπ) ½ ∫
−∞
+∞
(a x3¿e−β x2
)¿ dx + (βπ) ½ ∫
−∞
+∞
(b x4¿e−β x2
)¿ dx
= 0 + (βπ) ½ ∫
−∞
+∞
(b x4¿e−β x2
)¿ dx
= b (βπ) ½ ∫
−∞
+∞
(x4¿e−β x2
)¿ dx
= b (βπ) ½
1.322+1 √ π
β4+1 [ ∫0
+∞
(x2n¿e−β x2
)¿ dx = 1.3.5 …… (2 n−1 )
2n+1 √ πβ2n+1 ]
= b (βπ) ½
38 √ π
β5
= 3 b8 β2
6.1.2 APPLICATION OF PERTURBATION THEOREM TO HELIUM ATOM:
Helium atom contains two electrons and one Helium nucleus. In this atom, besides an attractive
interaction between electron and nucleus there is a repulsive interaction between the two electrons.
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Let the distance between nucleus and electron-1 and electron-2 be r1 and r2 respectively. Let r12
represents the inter electronic distance. The potential energy for a system of two electrons, and a nucleus of
charge +e is
V = - 2r1
- 2r2+ 1
r12
The Hamiltonian(H) for He atom (in a.u) H = - 12 ∇1
2 -
12 ∇1
2 - 2r1
- 2r2+ 1
r12
The first two terms are operator for kinetic energy , the third and fourth terms are potential energy of
attraction between electron and nucleus and the last term is the potential energy of inter electron repulsion.
Because of this term, Schrodinger equation for the atom can not be solved and we use approximation method
Perturbation method separates the Hamiltonian in to two parts namely H0 and H1
H = Ho +¿ H1
Ho = -{ 12 [ ∇1
2 + ∇1
2 ] -
2r1
- 2r2
is the Hamiltonian for unperturbed system
H1 = 1
r12
The wave function of unperturbed( zeroth order) system is given by
Ψ 0 = √ Z3
π e−Z r 1× √ Z3
π e−Z r 2
= Z3
π×e−Z r 1e−Zr 2
The corresponding zeroth order energy is E0 = - Z2 (a.u)
The first order perturbation energy is the average value of the perturbation function H’ = 1r12
, over the
unperturbed state of the system
E1 = ∫Ψ 0 H ’Ψ 0 dτ
= ∫ Z3
π× e−Z r1 e−Z r2 ×( 1
r12)× Z3
π× e−Z r1 e−Z r 2 dτ
= (Z3
π )2
∫¿¿ dτ
The volume element in spherical co ordinates is dτ = r12dr1sinθ1dφ1 r2
2dr2sinθ2dφ2
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upon integration we get E 1 = 58 Z ,
The unperturbed energy is given by E 0 = - Z2
Therefore the total energy E = E 0 + E 1
= - Z2 + 58 Z (a.u)
= - 4 + 58 ×2 (a.u)
= - 114 × (27.2 ) eV
= -74.8 eV
This is the ground state energy of Helium atom. Experimental value is -78.98 Ev
6.2. VARIATION METHODAdvantage over perturbation method:
It does not require that there should be a similar problem that has been solved previously.
6.2.1 METHOD:
This method says that with any trial function Ψ, the expectation value of energy E will be equal to
or greater than the true value E0, which is the lowest energy eigen value of the Hamiltonian of the system
E ≥ E0
i.e if we choose number of wave functions, Ψ 1, Ψ2, Ψ3 etc and calculate the values E1,E2,E3 ,
corresponding to them, then each of the values E , will be greater than the energy E0.
Steps:
1. Choose series of trial functions that depend some arbitrary parameters α, β etc called variational
parameter.
2. Calculate E in each case, by postulate of quantum mechanics.
E = ∫Ψ H Ψ ¿dτ
∫Ψ Ψ ¿ dτ
3. Minimize E with respect to the parameter used.
4. Pick up the lowest value that would be closest to the true value .
5. The trial function corresponding to that value will be the best function.Page 232 of 419
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Proof:
Consider the wave function Ψ, which is a linear combination of normalised and orthogonal eigen
functions φ1 and φ2 with normalization constants a1 ,a2 respectively.
Ψ = a1 φ1 +a2 φ2
From postulates of Quantum mechanics,
E = ∫Ψ H Ψ ¿dτ
∫Ψ Ψ ¿ dτsubstituting the values of Ψ , we get
E = ∫(a 1 φ 1+a 2 φ 2)H ¿¿¿
Since Ψ is real function, Ψ * can be replaced by Ψ itself∴ E = ∫(a 1φ 1+a2 φ 2)H ¿¿¿
= a1
2∫φ 1 H φ 1dτ+a22∫φ 2 H φ2 dτ+a1 a2∫φ 1H φ 2dτ+a1 a2∫φ 2 H φ 1dτ
a12∫φ1
2d τ+a22∫φ2
2d τ+2a1 a2∫(φ1φ 2)d τ
∫φ1 H φ 1dτ = E1
∫φ 2 H φ 2dτ = E2
∫φ1 H φ 2dτ = 0
∫φ 2 H φ 1dτ = 0
∫φ12 dτ = 1
∫φ22 dτ = 1 ,
∫φ1 φ 2 dτ = 0
Substituting in the above equation E = a1
2 E1+a22 E 2
a12+a2
2
Subtracting E0 on both sides E - E0 = a1
2 E1+a22 E 2
a12+a2
2 - E0
= a1
2 E1+a22 E 2−E0(a1
2+a22)
a12+a2
2 [ taking LCM]
Taking only Denominator, E - E0 = a12 E 1+a2
2 E 2−E0(a12+a2
2)
= a12 E 1+a2
2 E 2−E0 a12−E0 a2
2¿
= a12 ( E1 - E0 ) + a2
2 ( E2 - E0 )
Since E0 is the lowest energy of the system, ( E1 - E0 ) and ( E2 - E0 ) are positive . Therefore
E - E0 > 0
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E > E0
This shows that the variation method provides an upper bound to the ground state energy of the
system.
6.2.2 SECULAR EQUATION:
Consider the wave function Ψ, which is a linear combination of normalised and orthogonal eigen
functions φ1 and φ2 with normalization constants a1 ,a2 respectively.
Ψ = a1 φ1 +a2 φ2
From postulates of Quantum mechanics,
E = ∫Ψ H Ψ ¿dτ
∫Ψ Ψ ¿ dτsubstituting the values of Ψ , we get
E = ∫(a 1 φ 1+a 2 φ 2)H ¿¿¿ ---------------1
Since Ψ is real function, Ψ * can be replaced by Ψ itself∴ E = ∫(a 1φ 1+a2 φ 2)H ¿¿¿
= a1
2 H11+a22 H22+2 a1 a2 H 12
a12 S11+a2
2 S22+2 a1 a2 S12
E ( a12 S11+a2
2 S22+2a 1 a 2 S12¿ = a12 H11+a2
2 H22 +2 a1 a 2H12
Differentiating with respect to a1
∂ E∂ a1
( a12 S11+a2
2 S22+2 a1 a2 S12¿ + E ( 2a1 S11+0+2a2 S12 ¿
= 2a1 H11+0 +2 a 2 H12
∂ E∂ a1
= 0
Therefore the above equation becomes
E ( 2a1 S11+2a2 S12¿ = 2a1 H11+2 a 2H 12
E (a1 S11+a2 S12 ¿ = a1 H 11+a2 H12 [ dividing by 2 ]
Ea1 S11+E a2 S12 = a1 H 11+a 2 H 12
a1¿ ) + a2¿¿ - E S12¿ = 0
Similarly, a1¿ ) + a2¿¿ - ES22¿ = 0
These are called secular equations.
This shows that a trial function depends linearly on the variation parameter leads to secular equation
and secular determinant.
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6.2.3 Application of variation method to particle in ‘1D’ box
Problem :Apply variation theorem to the probability of finding the particle in one dimensional box of length ‘l’
using the trial wave function, ѱ = x (l – x) and compare your result with the true value.
OR .Calculate the energy for a particle in a one dimensional box of length L, choosing the trial wave function
Ψ = N x ( L- x) which is finite, continuous and single valued for all values of , x, using variation theorem.
Show that the calculated value is 10π2 times of actual value
Solution :
The Hamiltonian for the particle in a box is H = −h2
8 π 2md
d x2
E = ∫Ψ H Ψ ¿dτ
∫Ψ Ψ ¿ dτ
= ∫ {x (L−x ) }{ −h2
8 π2 md2
dx 2}{x (L−x ) }dx
∫ x (L−x)N x (L−x)dx
= −h2
8π 2m ∫ {x (L−x ) }{ d2
dx2}{x (L−x ) }dx
∫ x2 (L−x )2 dx
= −h2
8π 2m ∫
0
L
{x (L−x ) }{ d2
dx2}{Lx−x2}dx
∫0
∞
x2 (L−x )2 dx
= −h2
8π 2m ∫
0
L
{x (L−x ) }{ ddx(L−2 x)dx
∫0
∞
x2 (L−x )2 dx
= −h2
8π 2m ∫0
L
{x (L−x ) }(−2)¿dx ¿
∫0
∞
x2 (L−x )2 dx
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= +2 h2
8π 2m ∫0
L
{x (L−x ) }dx
∫0
∞
x2 (L−x )2 dx
Nr = ∫0
L
{x (L−x ) } dx
= ∫0
L
{Lx−x2 } dx
= [L x2
2− x3
3]0
L
= L3
2 - L3
3
= L3
6
Dr = ∫0
∞
x2 (L−x )2 dx
= ∫0
∞
x2(L¿¿2−2 Lx+ x2)¿ dx
= ∫0
∞
x2 L2−2 L x3+x4 ¿¿ dx
= ¿
= L5
3 - 2 L5
4+ L5
5
= (10−15+6 )L3
30
= L5
30
E = h2
4 π2 m
L3
6L5
30
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= +h2
4 π2 m × L3
6 ×
30L5
= 5h2
4 m π 2 L2
. Comparison:
Ecalculated = 5h2
4 m π 2 L2
EActual = h2
8 m L2
Ecalculated = 5
4 π2 [ h2
m L2 ]
= 8 ×54 π2 [ h2
8 m L2 ]
= 10π2 [ EActual ]
Problem: Apply variation principle to show that the upper bound to the ground state energy of the particle in
a 1D box of length ‘a’ using the trial wave function Ψ = x2 (a- x) is 7h2
4 π2 ma2 . Show that the calculated value
is 14π2 times of actual value
Solution:
The Hamiltonian for the particle in a box is H = −h2
8 π 2md2
d x2
E = ∫Ψ H Ψ ¿dτ
∫Ψ Ψ ¿ dτ
¿ ∫
0
a
x2 (a−x ) −h2
8 π 2md2
d x2 x2(a−x)dx
∫0
a
x2 (a−x )× x2 (a−x )dx
numerator:
∫0
a
x2 (a−x ) −h2
8 π2 md2
d x2 x2(a− x)dx
= −h2
8 π 2m ∫
0
a
¿ d2
d x2 ¿ dx
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= −h2
8 π 2m ∫
0
a
¿ ddx¿dx
= −h2
8π 2m ∫
0
a
¿¿ dx
= −h2
8π 2m ∫
0
a
¿ dx
= −h2
8π 2m [2 a2 x
3
3
−2 a x4
4−6 a x4
4+6 x5
5¿
= −h2
8π 2m [2 a2 x
3
3
−8 a x4
4+6 x5
5¿
= −h2
8π 2m [2 a2 a
3
3
−2 aa4+6 a5
5¿
= −h2
8π 2m [2 a
3
5
−2a5+6 a5
5¿
= −h2
8π 2m [ (10−30+18)a5
15]
= −h2
8π 2m × ( -
215 a5 )
= +h2
60 π2 m × (a5 )
Denominator:
∫0
a
x2 (a−x )× x2 (a−x )dx
¿ ∫0
a
x4 (a−x)2 dx
¿ ∫0
a
x4 (a¿¿2−2ax+x2)dx ¿
¿ ∫0
a
(a¿¿2 x¿¿4−2 x5+x6)dx¿¿
¿ a2 x5
5 - 2 x6
6 + x7
7
¿ a2 x5
5 - x6
3 + x7
7
¿ a7(21−35+15)
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¿ a7
105
E = +h2
60π2 m × (a5 )×
105a7
= 7 h2
4 π2 ma2
Comparison:
Ecalculated = 7 h2
4 π2 ma2
EActual = h2
8ma2
Ecalculated = 7
4 π2 [ h2
ma2 ]
= 8 ×74 π2 [ h2
8 m L2 ]
= 14π2 [ EActual ]
APPLICATION OF VARIATION THEOREM TO HYDROGEN ATOM
Variation theorem is applied to hydrogen atom to find the minimum ground state energy and the
corresponding acceptable wave function It involves the following steps.
1. Selection of wave function
Many acceptable wave functions , in terms of variation parameter are chosen to calculate the energy
2. Writing Hamiltonian:
Hamiltonian operator of hydrogen atom in spherical co-ordinates is
H = −12r 2
ddr¿ ) -
1r ( in atomic units).
3. Calculation of energy in terms of variation parameter.
The expectation value is calculated by quantum mechanics
E = ∫Ψ H Ψ ¿dτ
∫Ψ Ψ ¿ dτ
4. Estimation of variation parameter:
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After getting energy in terms of variation parameter, the expression should be to minimised. To do this
the function should be differentiated with respect to the parameter
and equated to zero. From this the value of the parameter can be obtained.
5. Calculation of energy:
The value of parameter should be substituted in the energy expression to obtain the exact value of
energy.
For example the acceptable wave functions for hydrogen atom with ‘a’ as variation parameter are
Ψ = e−ar, Ψ=e−ar2
, Ψ = r e−ar
Hamiltonian operator of hydrogen atom in spherical co-ordinates is
H = −12r 2
ddr¿ ) -
1r ( in atomic units).
The expectation value is calculated by quantum mechanics
E = ∫Ψ H Ψ ¿dτ
∫Ψ Ψ ¿ dτ
= ∫
0
∞
Ψ H Ψ r2 dr∫0
π
sinθ dθ∫0
2π
dφ
∫Ψ Ψ ¿r2 dr∫0
π
sin θ dθ∫0
2 π
dφ d𝛕 = [r2 dr∫
0
π
sin θ d θ∫0
2 π
dφ¿
= ∫
0
∞
(e−ar )(−12 r2
ddr (r2 d
dr )−1r )(e−ar ) r2 dr∫
0
π
sin θ d θ∫0
2 π
dφ
∫ (e−ar ) (e−ar ) r2 dr∫0
π
sinθ dθ∫0
2 π
dφ
Similarly for other functions also energy is calculated.
For the wave functions Ψ=e−ar , the expectation value of energy is E=¿is ( a2
2 - a )
For Ψ = e−ar 2
is E=¿ 3∝2 -√ 8 α
π where α is the parameter
For Ψ = r e−ar the expectation value of energy is E=α2
6 -
α2 .
To find the parameter ’a’ , E should be minimized .
Its first order derivative with respect to the parameter ’a’ is
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dEda = a – 1
This should be equal to zero.
dEda = 0
0 = a – 1 ∴ a = 1
Substituting the value of a , we get
E = ½ - 1
= - 0.5
Similar treatment should be done for other energies and the function which gives lowest energy is the best
wave function.
Problem : Use the functions Ψ = e−ar to calculate the∇2Ψ of Hydrogen atom
Solution:
Ψ = e−ar
In spherical co ordinates
∇2 = 1r2
dd r (r2 d
d r )
∇2Ψ = 1r2
dd r (r2 d
dr (e−ar ¿
= 1r2
ddr (r2 ( -a e−ar ¿ [
dd r (e−ar ¿ = -a e−ar ]
= −ar2
dd r (r2 e−ar ¿
= −ar2 [ (r2 (-ae−ar ¿ + e−ar (2r)]
= [ (a2 e−ar ¿ – 2ar e−ar ]
= e−ar (a2 – 2 ar )
Problem : Use the functions Ψ = e−ar to calculate the Hamiltonian of Hydrogen atom
Solution:
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H = −h2
8 π 2m∇2 - e
2
r
Ψ = e−ar
In spherical co ordinates
∇2 = 1r2
dd r (r2 d
d r )
∇2Ψ = 1r2
dd r (r2 d
dr (e−ar ¿
= 1r2
ddr (r2 ( -a e−ar ¿ [
dd r (e−ar ¿ = -a e−ar ]
= −ar2
dd r (r2 e−ar ¿
= −ar2 [ (r2 (-ae−ar ¿ + e−ar (2r)]
= [ (a2 e−ar ¿ – 2ar e−ar ]
= e−ar (a2 – 2 ar )
H = −h2
8 π 2m∇2 - e
2
r
Substituting the value of ∇2
H = −h2
8 π 2m(a2 e−ar) – 2a
re−ar - e
2
r
Problem : Use the functions Ψ = e−ar to calculate the ∫ΨH Ψ of Hydrogen atom
Solution:
H = −h2
8 π 2m∇2 - e
2
r
Ψ = e−ar
In spherical co ordinates
∇2 = 1r2
dd r (r2 d
d r )
∇2Ψ = 1r2
dd r (r2 d
dr (e−ar ¿
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= 1r2
ddr (r2 ( -a e−ar ¿ [
dd r (e−ar ¿ = -a e−ar ]
= −ar2
dd r (r2 e−ar ¿
= −ar2 [ (r2 (-ae−ar ¿ + e−ar (2r)]
= [ (a2 e−ar ¿ – 2ar e−ar ]
= e−ar (a2 – 2 ar )
HΨ = −h2
8 π 2m∇2Ψ - e
2
rΨ
Substituting the value of ∇2
H = −h2
8 π 2m[(a2 e−ar)– 2a
re−ar ] - e
2
r
∫ΨH Ψ dτ = ∫ e−ar¿¿ - e2
r ] e−ar r2 dr
= ∫ e−ar¿¿ - e2
r e−ar] r2 dr
= ∫ −h2 a2
8 π2 m(e−2 ar ) r2 dr – ∫−h2 a2
8 π2m¿ r2 dr - ∫ e2
re−2ar r2dr
= −h2 a2
8π 2m ∫ e−2 ar r2 dr+¿ h2 a2
8π 2m ∫2 ar e−2 ar dr - e2 ∫r e−2ar dr
= −h2 a2
8π 2m ∫ e−2 ar r2 dr+¿ h
2 a2× 2a8 π 2m
∫r e−2 ar dr - e2 ∫r e−2 ar dr
= −h2 a2
8π 2m [
2(2 a )3
¿+¿ h2 a2× 2 a8 π 2m
[ 1
(2a )2 ] - e2 [
1(2 a )2
]
= −h2 a2
8π 2m [
28 a3 ¿+¿
h2 a2× 2a8 π 2m
[ 1
4 a2 ] - e2 [ 1
4 a2 ]
= −h2 a32 π2 m
+¿ h2 a16 π2 m
- e2 [ 1
4 a2 ]
= −h2 a32 π2 m
+¿ h2 a16 π2 m
- e2 [ 1
4 a2 ]
= −h2 a+2h2 a32 π2m
- e2
4 a2
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= h2 a32 π2 m
- e2
4 a2
Problem : Use the functions Ψ = e−ar to calculate the energy of Hydrogen atom in terms of parameter ‘a’
Solution:
E = ∫ΨH Ψ d τ
∫Ψ Ψ d τ
H = −h2
8 π 2m∇2 - e
2
r
Ψ = e−ar
In spherical co ordinates
∇2 = 1r2
dd r (r2 d
d r )
∇2Ψ = 1r2
dd r (r2 d
dr (e−ar ¿
= 1r2
ddr (r2 ( -a e−ar ¿ [
dd r (e−ar ¿ = -a e−ar ]
= −ar2
dd r (r2 e−ar ¿
= −ar2 [ (r2 (-ae−ar ¿ + e−ar (2r)]
= [ (a2 e−ar ¿ – 2ar e−ar ]
= e−ar (a2 – 2 ar )
HΨ = −h2
8 π 2m∇2Ψ - e
2
rΨ
Substituting the value of ∇2
H = −h2
8 π 2m[(a2 e−ar)– 2a
re−ar ] - e
2
r
∫ΨH Ψ dτ = ∫ e−ar¿¿ - e2
r ] e−ar r2 dr
= ∫ e−ar¿¿ - e2
r e−ar] r2 dr
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= ∫ −h2 a2
8 π2 m(e−2 ar ) r2 dr – ∫−h2 a2
8 π2m¿ r2 dr - ∫ e2
re−2ar r2dr
= −h2 a2
8π 2m ∫ e−2 ar r2 dr+¿ h2 a2
8π 2m ∫2 ar e−2 ar dr - e2 ∫r e−2 ar dr
= −h2 a2
8π 2m ∫ e−2 ar r2 dr+¿ h
2 a2× 2a8 π 2m
∫r e−2 ar dr - e2 ∫r e−2 ar dr
= −h2 a2
8π 2m [
2(2 a )3
¿+¿ h2 a2× 2 a8 π 2m
[ 1
(2a )2 ] - e2 [
1(2 a )2
]
= −h2 a2
8π 2m [
2(2 a )3
¿+¿ h2 a2× 2 a8 π 2m
[ 1
(2a )2 ] - e2 [
1(2 a )2
]
= −h2 a2
8π 2m [
28 a3 ¿+¿
h2
8π 2a m [
14 ] - e2 [
14 a2 ]
= −h2
32 π2 am +¿ h2
16 π2 a m - e2 [
14 a2 ]
= −h2
32 π2 m +¿ h2
16 π2 a m - e2 [
14 a2 ]
= −h2+2 h2
32 π 2am - e2
4 a2
= h2
32 π2 a m - e2
4 a2
∫Ψ Ψ d τ = ∫ e−2 ar r2 dr
= 2
(2 a )3
= 1
4 a3
E = ∫ΨH Ψ dτ
∫Ψ Ψ d τ
=
h2
32 π 2a m− e2
4 a2
14 a3
= h2 a2
8π 2m−e2a
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Problem : Use the functions Ψ = e−ar to calculate the energy of Hydrogen atom
Solution:
E = ∫ΨH Ψ d τ
∫Ψ Ψ d τ
H = −h2
8 π 2m∇2 - e
2
r
Ψ = e−ar
In spherical co ordinates
∇2 = 1r2
dd r (r2 d
d r )
∇2Ψ = 1r2
dd r (r2 d
dr (e−ar ¿
= 1r2
ddr (r2 ( -a e−ar ¿ [
dd r (e−ar ¿ = -a e−ar ]
= −ar2
dd r (r2 e−ar ¿
= −ar2 [ (r2 (-ae−ar ¿ + e−ar (2r)]
= [ (a2 e−ar ¿ – 2ar e−ar ]
= e−ar (a2 – 2 ar )
HΨ = −h2
8 π 2m∇2Ψ - e
2
rΨ
Substituting the value of ∇2
H = −h2
8 π 2m[(a2 e−ar)– 2a
re−ar ] - e
2
r
∫ΨH Ψ dτ = ∫ e−ar¿¿ - e2
r ] e−ar r2 dr
= ∫ e−ar¿¿ - e2
r e−ar] r2 dr
= ∫ −h2 a2
8 π2 m(e−2 ar ) r2 dr – ∫−h2 a2
8 π2m¿ r2 dr - ∫ e2
re−2ar r2dr
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= −h2 a2
8π 2m ∫ e−2 ar r2 dr+¿ h2 a2
8π 2m ∫2 ar e−2 ar dr - e2 ∫r e−2 ar dr
= −h2 a2
8π 2m ∫ e−2 ar r2 dr+¿ h
2 a2× 2a8 π 2m
∫r e−2 ar dr - e2 ∫r e−2 ar dr
= −h2 a2
8π 2m [
2(2 a )3
¿+¿ h2 a2× 2 a8 π 2m
[ 1
(2a )2 ] - e2 [
1(2 a )2
]
= −h2 a2
8π 2m [
28 a3 ¿+¿
h2
8π 2a m [
14 ] - e2 [
14 a2 ]
= −h2
32 π2 am +¿ h2
16 π2 a m - e2 [
14 a2 ]
= −h2
32 π2 m +¿ h2
16 π2 a m - e2 [
14 a2 ]
= −h2+2 h2
32 π 2am - e2
4 a2
= h2
32 π2 a m - e2
4 a2
∫Ψ Ψ d τ = ∫ e−2 ar r2 dr
= 2
(2 a )3
= 1
4 a3
E = ∫ΨH Ψ dτ
∫Ψ Ψ d τ
=
h2
32 π 2a m− e2
4 a2
14 a3
= h2 a2
8π 2m−e2a ---------------------------1
By variation theorem
∂ E∂ a = 0
Therefore
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∂∂ a ( h2 a2
8 π 2m−e2a) = 0
h22a8 π 2m
−e2 = 0
h2a4 π2 m
=e2
a = 4 π2 me2
h2
substituting in 1 we get
E = h2 a2
8π 2m−e2a
= h2
8 π 2m× 16 π4 m2e4
h4 −e2 × 4 π2 me2
h2
= 2π2 me4
h2 −4 π2 m e4
h2
= −2 π2 me4
h2
This is the expression for ground state energy of hydrogen atom
6.2.4 APPLICATION OF VARIATION METHOD TO HYDROGEN ATOM]
The ground state energy of H- atom can be calculated using variation method. Consider the wave
function is Ψ = e−ar, where ‘a’ is the variation parameter. The Hamiltonian for H- atom in spherical co ordinate
is
H = (−12 r2
ddr (r2 d
dr )−1r )
The expectation value is calculated by quantum mechanics
E = ∫Ψ H Ψ ¿dτ
∫Ψ Ψ ¿ dτ
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= ∫
0
∞
Ψ H Ψ r2 dr∫0
π
sinθ dθ∫0
2π
dφ
∫Ψ Ψ ¿r2 dr∫0
π
sin θ dθ∫0
2 π
dφ d𝛕 = [r2 dr∫
0
π
sin θ dθ∫0
2 π
dφ¿
= ∫
0
∞
(e−ar )(−12 r2
ddr (r2 d
dr )−1r )(e−ar ) r2 dr∫
0
π
sin θ d θ∫0
2 π
dφ
∫ (e−ar ) (e−ar ) r2 dr∫0
π
sinθ dθ∫0
2 π
dφ
Since the 𝛉 term and 𝛗 term are constants, they can be cancelled. Therefore the above equation becomes, E
= ∫
0
∞
(e−ar )(−12 r2
ddr (r2 d
dr )− 1r )(e−ar ) r2 dr
∫ (e−ar ) (e−ar ) r2 dr
= a2
2 - a [Upon integration]
To find the parameter ’a’ , E should be minimized .
Its first order derivative with respect to the parameter ’a’ is
dEda = a – 1
This should be equal to zero.
dEda = 0
0 = a – 1 ∴ a = 1
Substituting the value of a , we get
E = ½ - 1
= - 0.5 this is the true value.
Problem: Use the functions Ψ = e−ar to calculate the ground state energy of H-atom by variation method.
Compare the result with the true value. Hamiltonian in spherical co-ordinates is H = −12r 2
ddr¿ ) -
1r ( in
atomic units). .
Solution:
The given wave function is Ψ = e−ar
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By postulate of quantum mechanics E = ∫Ψ H Ψ ¿dτ
∫Ψ Ψ ¿ dτ
In spherical co-ordinates, E = ∫
0
∞
Ψ H Ψ r2 dr∫0
π
sinθ d θ∫0
2 π
dφ
∫Ψ Ψ ¿r2 dr∫0
π
sin θ d θ∫0
2π
dφ
substitutingthe values = ∫
0
∞
(e−ar )(−12 r2
ddr (r2 d
dr )−1r )(e−ar ) r2 dr∫
0
π
sin θ d θ∫0
2 π
dφ
∫ (e−ar ) (e−ar ) r2 dr∫0
π
sinθ d θ∫0
2 π
dφ
Since the 𝛉 term and 𝛗 term are constants, they can be cancelled. Therefore the above equation becomes,
E = ∫
0
∞
(e−ar )(−12 r2
ddr (r2 d
dr )−1r )(e−ar ) r2 dr
∫ (e−ar ) (e−ar ) r2 dr
Let us calculate HΨ
HΨ = (−12 r2
ddr (r2 d
dr )−1r ) (e−ar )
= (−12 r2 [ d
dr (r2 d (e−ar )dr )]− (e−ar )
r ) = ¿
= ( +a2 r2 [ d
dr(r 2e−ar )]− (e−ar )
r ) = ¿
Numerator Nr = ∫0
∞
(e−ar ) HΨ r2 dr substituting
Nr = ∫0
∞
(e−ar )¿¿
Taking e−ar as common factor,
= ∫0
∞
(e−2 ar ){( +a2 r2 [ ( r2 (– a )+(2 r ) ) ]− (1 )
r )}r2 dr
= ∫0
∞
(e−2ar ){(−a2
2+ a
r−
(1 )r )}r2dr
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= ∫0
∞
e−2 ar (−a2 r2
2 ) dr + ∫
0
∞
e−2 ar ( ar) dr - ∫0
∞
e−2 ar r dr
= −a2
2 ∫
0
∞
e−2 ar r2 dr + a ∫0
∞
e−2 ar r dr - ∫0
∞
e−2 ar r dr
= −a2
2 [
2 !(2 a)3
] + a [1 !(2 a)2
] - 1 !(2 a)2
[ ∫0
∞
xn e−ax dx = n!
an+1 ]
= −18 a +
14 a -
14 a2
= +18a -
14 a2
Nr = a−28 a2
Denominator = ∫ (e−ar ) (e−ar ) r2 dr
= ∫ (e−2 ar ) r2dr Formula : ∫0
∞
xn e−ax dx = n!
an+1
= 2 !(2 a)3
= 2
8 a3
E = a−28 a2 × 8 a3
2
E = a2
2 - a
dEda = a – 1
To find the minimum value, the first derivative must vanish.
dEda = 0
∴ a = 1 Substituting the value of a , we getE = ½ - 1
= - 0.5 this is the true value.
2 Use the functions Ψ = e−a r 2
to calculate the ground state energy of H-atom by variation method. Compare the result with the true value. Hamiltonian in spherical co-ordinates is
H =−12r 2
ddr¿) -
1r given ∫
0
∞
r 4 e−2a r2
dr = 3
32 a2 √ π2a
, ∫0
∞
r 2e−2 ar2
dr = 1
8 a √ π2 a
, ,∫0
∞
r e−a r2
dr =1
4 a
Solution:
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The given wave function is Ψ = e−a r 2
By postulate of quantum mechanics E = ∫Ψ H Ψ ¿dτ
∫Ψ Ψ ¿ dτ
In spherical co-ordinates, E = ∫
0
∞
Ψ H Ψ r2 dr∫0
π
sinθ d θ∫0
2 π
dφ
∫Ψ Ψ ¿r2 dr∫0
π
sin θ d θ∫0
2π
dφ
substitutingthe values = ∫
0
∞
(e−a r2 ) (−12 r2
ddr (r2 d
dr )−1r )(e−ar 2
)r2 dr∫0
π
sin θ d θ∫0
2π
dφ
∫ (e−a r2 ) (e−ar2 )r2 dr∫0
π
sin θ d θ∫0
2π
dφ
Since the 𝛉 term and 𝛗 term are constants, they can be cancelled. Therefore the above equation becomes,
E = ∫
0
∞
(e−ar2 ) (−12 r2
ddr (r2 d
dr )−1r ) (e−a r2 )r 2dr
∫ (e−a r2 ) (e−a r 2 )r2dr
Let us calculate HΨ
HΨ = (−12 r2
ddr (r2 d
dr )−1r ) (e−ar2 )
= (−12 r2 [ d
dr (r2 d (e−a r 2)dr )]−e−a r 2
r ) = ¿ [ d(e−a r 2
¿=−2 ar e−ar2
]
= (+2 a2 r2 [ d
dr(r3 e−a r2 )]− e−ar2
r ) = ¿[Differentiating by UV model]
= (+ar2 (– 2ar 4 e−ar2
+3 r2e−a r 2 )− e−a r2
r ) [ d(e−a r 2
¿=−2 ar e−ar2
, d(r3) = 3r2]
= (– 2a2r 2e−ar2
+3a e−ar 2
− e−a r2
r ) = e−ar 2
(– 2 a2r 2+3a−1r ) [Taking e−ar 2
as common factor]
Numerator Nr = ∫0
∞
e−a r2
HΨ r2 dr substituting
= ∫0
∞
(e−ar2 )e−ar2(– 2 a2 r2+3 a−1r )r2 dr
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= ∫0
∞
(e−2 a r2 )(– 2 a2r 2+3 a−1r )r2dr
= – 2a2∫0
∞
e−2 ar2
r4 dr + 3a ∫0
∞
e−2a r 2
r2 dr - ∫0
∞
e−2a r 2
r dr
Given ∫0
∞
r 4 e−2ar2
dr = 3
32 a2 √ π2 a
,∫0
∞
r2e−2 a r2
dr = 1
8 a √ π2 a
, , ∫0
∞
r e−ar 2
dr =1
4 a
= – 2a2¿ √ π2 a
) + 3a ( 18 a √ π
2 a ) -
14 a
= –316 √ π
2a + 3
8 √ π2a
- 1
4 a
= √ π2 a
(-316 +
38 ) -
14 a
= 316 √ π
2a -
14 a
Denominator = ∫ (e−a r2 ) (e−ar 2 )r2dr
= ∫ e−2 a r 2
r2 dr Formula : ∫0
∞
r 2e−2 ar2
dr = 1
8a √ π2 a
,
, E = ¿√ π
2a( 316− 1
4 a √ 2 aπ)
18 a √ π
2a,
[ taking √ π2a
as common]
, = ¿( 3
16 )− 14 a √ 2 a
π¿
¿1
8 a
= 8a ×316 - 8a × 1
4 a √ 2 aπ
= 3a2 - 2√ 2 a
π
dEdα =
32 – 2 ( √
2π ) ½ a -1/2
To find the minimum value, the first derivative must vanish.
dEdα = 0
∴ 32 = ( √
2π ) a -1/2
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a12 =
2 √ 23 √ π
a = 8
9 π Substituting the value of a , we get
E=¿ 3 a2 - 2 (√ 2a
π¿
= 24
18 π - 2 √169 π 2
= 4
3 π - 8
3 π
= −43 π
= - 0.424, which is greater than true value ( - 0.5 )5. Use the functions Ψ = r e−ar to calculate the ground state energy of H-atom by variation method.
Compare the result with the true value. Hamiltonian in spherical co-ordinates is H = −12r 2
ddr¿ ) -
1r ( in
atomic units). Given : ∫0
∞
xn e−ax dx = n!
an+1
Solution:
The given wave function is Ψ = r e−ar
By postulate of quantum mechanics E = ∫Ψ H Ψ ¿dτ
∫Ψ Ψ ¿ dτ
In spherical co-ordinates, E = ∫
0
∞
Ψ H Ψ r2 dr∫0
π
sinθ d θ∫0
2 π
dφ
∫Ψ Ψ ¿r2 dr∫0
π
sin θ d θ∫0
2π
dφ
substitutingthe values = ∫
0
∞
(r e−ar )(−12 r2
ddr (r2 d
dr )−1r )(r e−ar )r 2dr∫
0
π
sin θ dθ∫0
2 π
dφ
∫ ( r e−ar ) ( r e−ar ) r2 dr∫0
π
sin θ d θ∫0
2 π
dφ
Since the 𝛉 term and 𝛗 term are constants, they can be cancelled. Therefore the above equation becomes,
E = ∫
0
∞
(r e−ar )(−12 r2
ddr (r2 d
dr )−1r )(r e−ar )r 2dr
∫ ( r e−ar ) ( r e−ar ) r2 dr
Let us calculate HΨ
HΨ = (−12 r2
ddr (r2 d
dr )−1r ) (r e−ar )
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= (−12 r2 [ d
dr (r2 d (r e−ar )dr ) ]− (r e−ar )
r ) = ¿ [differentiation by UV model]
= (−12 r2 [ d
dr(−a r3 e−ar+r2 e−ar )]−e−ar)
= ¿
= ¿
= ¿
= (−a2 r2
e−ar+2 a e−ar−1r
e−ar−e−ar)Numerator Nr = ∫
0
∞
(e−ar ) HΨ r2 dr substituting
Nr = ∫0
∞
( r e−ar )((−a2 r2
e−ar+2 ae−ar−1r
e−ar−e−ar))r2 dr
Taking e−ar as common factor,
= ∫0
∞
( r e−2 ar ) {(−a2 r3
2e−ar+2 a r2 e−ar−r e−ar−r2 e−ar)}dr
= −a2∫0
∞
(r e−2 ar ) r3
2dr + ∫
0
∞
( r e−2 ar )2a r2 dr - ∫0
∞
( r e−2 ar ) r dr- ∫0
∞
( r e−2 ar ) r 2dr
=−a2
2 ∫0∞
(e−2ar )r 4 dr + 2a ∫0
∞
(e−2 ar ) r3dr- ∫0
∞
(e−2 ar ) r2dr- ∫0
∞
(e−2ar ) r3dr
= - a2
2× 4 !(2 a)5
+ 2 a × 3 !(2 a)4
−¿ 2 !(2 a)3
- 3 !(2 a)4
=- a2
2×
2432 a5 + 2 a ×
616 a4−¿
28 a3 -
616 a4
= - 3
8 a3 + 3
4 a3−¿ 1
4 a3 - 3
8a4
= −3a+6a−2a−3
8 a4
= a−38 a4
Denominator = ∫ ( r e−ar ) ( r e−ar ) r2 dr
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= ∫ (e−2 ar ) r4 dr Formula : ∫0
∞
xn e−ax dx = n!
an+1
= 4 !(2 a)5
= 24
32 a5
E = a−38a4 × 32 a5
24
E = a2−3 a6
¿ a2
6 -
a2
To find the minimum value, the first derivative must vanish.
dEdα =
α3 – ½
dEdα = 0
∴ α3 = ½
α = 32
Substituting the value of α , we get
E=α2
6 - α2
= 9
24 - 34
= 9−18
24
= −38
= - 0.375 which is greater than the true value ( - 0.5 ).
4.For the selected wave function Ψ = r e−ar the expectation value of energy is E=α2
6 -
α2 . Find the value of 𝛂 and hence calculate the energy. Compare your result with the true energy and comment the result.
solution:
The expectation value of energy is E=α2
6 -
α2
dEdα =
α3 – ½
To find the minimum value, the first derivative must vanish.
dEdα = 0
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∴ α3 = ½
α = 32
Substituting the value of α , we get
E=α2
6 - α2
= 9
24 - 34
= 9−18
24
= −38
= - 0.375 which is greater than the true value ( - 0.5 ).
6.2.5 APPLICATION OF VARIATION METHOD TO HELIUM ATOM:
Helium atom contains two electrons and one Helium nucleus. In this atom, besides an attractive
interaction between electron and nucleus there is a repulsive interaction between the two electrons.
Let the distance between nucleus and electron-1 and electron-2 be r1 and r2 respectively. Let r12
represents the inter electronic distance. The potential energy for a system of two electrons, and a nucleus of
charge +e is
V = - 2r1
- 2r2+ 1
r12
The Hamiltonian(H) for He atom (in a.u) H = -{ 12 [ ∇1
2 + ∇1
2 ] -
2r1
- 2r2+ 1
r12
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The first two terms are operator for kinetic energy , the third and fourth terms are potential energy of
attraction between electron and nucleus and the last term is the potential energy of inter electron repulsion
The variational method approximation requires that a trial wavefunction with one or more adjustable
parameters be chosen.
1. Choosing Trial Wave function
The calculations begin with a wave function in which both electrons are placed in a hydrogenic orbital
with scale factor α. The wave function is given by
Ψ(1,2)=Φ(1)Φ(2)
= a3
π e−α (r 1+r2)
2. Calculation of energy in terms of variation parameter
E = ∫Ψ H Ψ dτ
= ∫Ψ [−{12[∇1
2+∇22]− 2
r1− 2
r2+ 1
r12]Ψ dτ
= −12 ∫Ψ ∇1
2Ψ dτ −12 ∫Ψ ∇2
2Ψ dτ - ∫Ψ [ 2r1 ¿¿ ]Ψ dτ - ∫Ψ [ 2
r2]Ψ dτ + ∫Ψ [ 1
r12]Ψ dτ
Substituting the values and on integration we get
E = a2 - 278 a
Where ‘a ‘ is variation parameter
3.Determination of variation parameter
E = a2 - 278 a
Differentiating with respect to ‘a’
dEda = 2a –
278
To minimize this put dEda =0
2a = 278 -
∴ a = 2716
Substituting we get
E = (2716¿¿2 – (
278 ) ×(
2716¿
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= 2716 (
2716 -
278 )
= 2716 (
−2716 )
= −729256
= - 2.8476 a.u and the experimentally determined ground state energy is Eexp=−2.903
The deviation of energy for the optimized trial wave function from the experimental value is
E(α )−EexpEexp
= −2.847+2.9032.903
= 1.93 %
Second Trial Wave function
In this trial it is assumed that each electron is in an orbital which is a linear combination of two different1s
orbitals. This assumption gives a trial wave function of the form
Ψ(1,2) =φ1 (1)φ1 (2) + φ1 (1)φ2 (2) +φ1 (2)φ2 (1) +φ2 (2)φ2 (2) +
= e−α (r 1)e−α (r2) + e−α (r 1)e−β (r2) + e− β (r 1)e−α (r2) + e− β (r 1)e−β (r2)
This trial wave function indicates that 50% of the time the electrons are in different orbitals, while for the first
trial wave function the electrons were in the same orbital 100% of the time. In this case the expression for the
energy is being minimized simultaneously with respect to two parameters (α and β) rather than just one (α).
The energy obtained after minimizing this equation and optimizing for α and β is E=−2.86035
The deviation of energy for the optimized trial wave function from the experimental value is
E(α )−EexpEexp
= −2.860+2.9032.903
= 1.49 %
Third Trial Wave function
By introducing electron correlation directly into the first trial wave function by adding a term, r12 involving the
inter-electron separation.
Ψ(1,2)= e−α (r 1+r2) (1+βr12)
The variational method energy obtained after minimizing the above equation and optimizing for α and β is
E=−2.89112
The deviation of energy for the optimized trial wave function from the experimental value is
E(α )−EexpEexp
= −2.891+2.9032.903
= 0.43 %
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Fourth Trial Wave function
Further improvement by adding Hylleraas's r12 term to the third trial wave function as shown here.
Ψ(1,2)= e−α (r 1)e−β (r2) + e− β (r 1)e−α (r2) [1+γr12]
The energy obtained after minimizing and optimizing the three parameter α, β and γ
E=−2.90143
The deviation of energy for the optimized trial wave function from the experimental value is
E(α )−EexpEexp
= −2.901+2.9032.903
= 0.07 %
Thus it is assumed as the fourth one is better wave function
Application of variation theorem to Helium atom:
Helium atom contains two electrons and one Helium nucleus. Let the distance between nucleus
and electron-1 and electron-2 be r1 and r2 respectively. Let r12 represents the inter electronic distance. The
potential energy for a system of two electrons, and a nucleus of charge +e is
v= - 2r1
- 2r2+ 1
r12 e1 r12 e2
r1 r2
+e
The Hamiltonian(H) for He atom (in a.u) H = -{ 12 [ ∇1
2 + ∇1
2 ] -
2r1
- 2r2+ 1
r12
The wave function is given by Ψ 0 = ( Z3
π ) ½ e−Z r 1 ( Z
3
π ) ½ e−Z r 2
E = ∫Ψ H Ψ dτ
[note: z is the variation parameter, to be determined, not atomic number],Substituting
E = ∫Ψ [−{12[∇1
2+∇22]− 2
r1− 2
r2+ 1
r12]Ψ dτ
= −12 ∫Ψ ∇1
2Ψ dτ −12 ∫Ψ ∇2
2Ψ dτ - ∫Ψ [ 2r1 ¿¿ ]Ψ dτ - ∫Ψ [ 2
r2]Ψ dτ + ∫Ψ [ 1
r12]Ψ dτ
= I1 + I2 + I3 + I4 + I5
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The Laplacian operator in spherical co ordinates is ∇12 =
1r2
ddr¿) [ neglecting the θ and φ terms
I1 = −12 ∫Ψ 1
r2ddr (r2 d
dr )Ψ dτ
=−12 ∫( Z3
π )1/2
e−Z r 1( Z3
π )1/2
e−Z r2[ 1r2
ddr (r 2 d
dr )]( Z3
π )1/2
e−Zr 1(Z3
π )1/2
e−Z r2
1
2
dτ
dτ = r12 sinθ1dr1dθ1dφ1r2
2 sinθ2dr2dθ2dφ2
∴ I1 = −12 (Z3
π )2
∫ e−Z r1 e−Z r2[ 1r 1
2d
d r1 (r12 dd r1 )]e−Z r 1e−Z r 2r
1
2
sinθ1dr1dθ1dφ1r22 sinθ2dr2dθ2dφ2
−12 (Z3
π )2
∫ e−Z r1 e−Z r2 [ 1r1
2d
d r1 (r12 d
d r1 )]e−Z r1 e−Z r 2r12 sinθ1dr1dθ1dφ1∫r 2
2 sinθ2dr 2 dθ 2dφ 2
but ∫0
π
sinθ dθ∫0
2 π
dφ =2× 2π, therefore the above equation becomes
I1 =−12 (Z3
π )2
∫ e−Z r1 e−Z r2 [ 1r1
2d
d r1 (r12 d
d r1 )]e−Z r1 e−Z r 2r12 dr1 ( 4π ) ∫r 2
2dφ 2 ( 4π )
= −12 (Z3
π )2
( 4π ) 2 ∫ e−Z r1 e−Z r2 [ 1r1
2d
d r1 (r12 dd r1 )]e−Z r1 e−Z r 2r1
2 dr1 ∫r 22dr 2
= - 8 Z6 ∫ e−Z r1 [ 1r1
2d
d r1 (r12 d
d r1 )]e−Z r1 r12 dr1 ∫ e−2 Zr 2r2
2 dr 2
= - 8 Z6 ∫ e−Z r1 [ 1r1
2d
d r1 (r12 d (e−Z r1)
d r1 ) ]r 12 dr1 ∫ e−2 Zr 2r2
2 dr 2
= - 8 Z6 ∫ e−Z r1 [ 1r1
2d
d r1 (r12 d (e−Z r1)
d r1 ) ]r 12 dr1
2(2 Z )3
[ ∫0
∞
xn e−ax dx = n!
an+1 ]
= - 8 Z6 ×2
(2 Z )3 ∫ e−Z r1 [ 1
r12
dd r1 (r1
2 d (e−Z r1)d r1 ) ]r 1
2 dr1
= - 2 Z3 ∫ e−Z r1 [ 1r1
2d
d r1 (r12 d (e−Z r1)
d r1 ) ]r 12 dr1
= - 2 Z3 ∫ e−Z r1 [ 1r1
2d
d r1 (r12 d (e−Z r1)
d r1 ) ]r 12 dr1
= - 2 Z3 ∫ e−Z r1 [ dd r1 (r1
2 d (e−Z r1)d r1 )] dr1
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= - 2 Z3 ∫ e−Z r1 ¿ dr1
= - 2 Z3 ∫ e−Z r1 ¿ dr1 [ d(UV ) = UdV + VdU]
= - 2 Z3 [ ∫ e−Z r1 ¿ dr1 - ∫ e−Z r1 [ (e−Z r1 )(2 Z r1)] dr1
= - 2 Z3 [ ( Z2 ) ∫ e−2 Zr 1 r12 dr1 - 2Z ∫ e−2 Zr 1r1 dr1 ]
= - 2 Z3 [( Z2 ×) 2
(2 Z )3 - 2Z ×
1(2 Z )2
]
= - 2 Z3 [ 1
4 Z - 12 Z ]
= - 2 Z3 [ 1−24 Z ]
= - 2 Z3 [ −14 Z ]
= Z2
2
Similarly I2 = Z2
2
I3 = ∫Ψ [ 2r1 ¿¿ ]Ψ dτ
= ∫(Z3
π )1 /2
e−Z r1(Z3
π )1 /2
e−Z r2[ 2r1¿¿](Z3
π )1/2
e−Z r1(Z3
π )1 /2
e−Z r2 dτ
= (Z3
π )2
∫ e−Z r1 e−Z r2[ 2r 1¿¿]e−Z r1 e−Z r2 dτ
dτ = r12 sinθ1dr1dθ1dφ1r2
2 sinθ2dr2dθ2dφ2
= (Z3
π )2
∫ e−Z r1 e−Z r2[ 2r 1¿¿]e−Z r1 e−Z r2 r1
2 sinθ1dr1dθ1dφ1r22 sinθ2dr2dθ2dφ2
but ∫0
π
sinθ dθ∫0
2 π
dφ =2× 2π, therefore the above equation becomes
= (Z3
π )2
× ( 4π ) 2 × ∫ e−2 Zr 1[ 2r1¿¿ ] r1
2 dr1 ∫ e−2 Zr 2 r22 dr2
= 16 Z6 ×2 ∫ e−2 Zr 1 r1 dr1 ∫ e−2 Zr 2r22 dr2
= 32 Z6× 1
(2Z )2× 2(2 Z)3
= 32 Z6 × 1
4 Z2 × 28 Z3
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= 2Z
Similarly I4 = 2Z
I5 = 58 Z [ It is more tedious]
E = Z2
2 + Z
2
2 -2Z- 2Z +
58 Z
= Z2 - 4Z + 58 Z
Differentiating with respect to Z
δEδZ = 2Z – 4 + 58
To minimize this put δEδZ =0
2Z = - 4 + 58
∴ Z = 2716
Substituting we get
E = (2716¿¿2 – 4 (
2716 ) +
58 (
2716¿
= - 2.8447 a.u
= - 2.8447 ( 27.21) eV [ 1 a.u = 27.21 eV ]
= -77.48 eV
This is better than first order perturbation result ( E= -74.8)
6.3 BORN OPPENHEIMER APPROXIMATION:
1. it is otherwise called Fixed nuclei approximation
2. The Schrodinger equation includes both nuclear and electronic motions.
3. Born Oppenheimer, approximation holds good for . ground state
4. Max Born and Robert Oppenheimer assumed that nucleus is assumed to be stationary ,when the
electrons move. According to Born Oppenheimer, approximation, neutron is be treated as stationary
5. In Born Oppenheimer, approximation , nuclear kinetic energy is neglected because
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a.Electronic motion is extremely fast
b. proton is 1840 times as heavy as electron
c. Nuclear motion is very small
6. The solution obtained by solving Schrodinger equation using Born Oppenheimer, approximation gives
electronic energy
This is because nuclei are 1000 times heavier than electrons.
H = nuclear kinetic energy + electronic kinetic energy + nuclear potential energy + nuclear – electron
attraction potential energy + electronic repulsion potential energy.
H = - h2
8 π 2 ∑u
N 1mu
∇u2 - h2
8 π 2 ∑i
n 1me
∇e2 + ∑
u , v
zu z v e2
(4 π∈0 ) ruv
+¿∑i ,u
zu e2
(4 π∈0 ) r iu
+¿∑i , j
e2
(4 π∈0 ) rij
¿¿
Applying BO approximation we can neglect the first term and hence Hamiltonian becomes
H = - h2
8 π 2 ∑i
n 1me
∇e2 + ∑
u , v
zu z v e2
(4 π∈0 ) ruv
+¿∑i ,u
zu e2
(4 π∈0 ) r iu
+¿∑i , j
e2
(4 π∈0 ) rij
¿¿
Using this we can solve Schrodinger equation.
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6.4 RUSSELL – SAUNDERS COUPLING ( LS –COUPLING)
It is represented by writing the spin multiplicity as superscripts to the left of the spectral state (term letters) with angular momentum quantum number (J) value as subscript to the right.
TERM SYMBOL = spin multiplicity spectral state J
1. SPECTRAL STATE (TERM LETTERS):It depends on the value of L, which is called Orbital angular momentum.which is an integer value and
equal to sum of the l values for all the electrons.
a. To find the value of L:The value of L is given by L = ( l1 + l2), ( l1 + l2 -1)…. (l1-l2) where l1 and l2 are azimuthal quantum
numbers
b. To find the term letters:
Value of ‘ L’ 0 1 2 3 4 5 6TERM LETTER S P D F G H I
For example if the value of L is 2, then the term letter is D.
2 SPIN MULTIPLICITY
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a. Total spin angular momentum (S)s = ( s1 + s2) , ( s1 + s2 -1), ….
when two vectors are anti parallel then s = ½ – ½ = 0
when they are parallel then s = ½ + ½ = 1
b. Multiplicity
It is given by S = 2s +1,
3. TOTAL ANGULAR MOMENTUM:
It is given by J = L+ S, L+S-1, L+S-2,… . L-S 2. There will be 2s +1 values of J for each value of L3. For completely filled atom J =04. For half filled J = S5. For less than half filled J= |L−S| 6. For greater than half filled J = L +S 7. It should have positive value.8. It can not have zero value
For examplea. When L= 1 and s = ½ , then
J = 1 + ½ , 1 + ½ -1 , 1+1/2 -2 , ….(1 – ½ ) = 3/2 , ½ The possible values of J are 3/2, ½ only.
b. When L= 1 and s = 1, then J = 1 + 1, 1 +1 -1 , 1+1 -2 , ….(1-1) = 2 ,1 ,0The possible values of J are 2,1 ,0 only.
c. When L= 2 and s = 1, then J = 2 + 1, 2 +1 -1 , 2+1 -2 , ….(2-1) = 3 ,2 , 1,The possible values of J are 3, 2,1 only.
1. Find the term symbol for He
a. Spin multiplicity:
s = ½ - ½ = 0
S = 2s + 1 = 2 (0) +1 = 1
b. Spectral state:
Value of ‘ L’ 0 1 2 3 4 5 6TERM LETTER S P D F G H I
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Since L=0, the spectral term is ‘S’
c. J-value:For completely filled atom J =0
d. ∴ Term symbol for He is 1 S 0
2. Find the term symbol for C
a. Spin multiplicity:
configuration : 1s2 2s2 2p2
s = ½ + ½ = 1
S = 2s + 1 = 2 (1) +1 = 3
b. Spectral state:
L = l1+l2+l3+l4
= 0 + 0+ 0+ 1 = 1
Value of ‘ L’ 0 1 2 3 4 5 6TERM LETTER S P D F G H I
Since L=1, the spectral term is ‘P’
e. J-value:For less than completely filled atom J = L-S = 1-1 = 0
f. ∴ Term symbol for C is 3 P 0 3. Find the term symbol for N
a. Spin multiplicity:
configuration : 1s2 2s2 2p3
s = ½ + ½ + ½ = 3/2
S = 2s + 1 = 2 (3/2) +1 = 4
b. Spectral state:
L = l1+l2+l3+l4
= 0 + 0+ 1+0 -1 = 0
Value of ‘ L’ 0 1 2 3 4 5 6TERM LETTER S P D F G H I
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Since L=0, the spectral term is ‘S’
c. J-value:
For half filled atom J = s = 3/2
d. ∴ Term symbol for N is 4 S 3/2
4. Find the term symbol for F
a. Spin multiplicity:
configuration : 1s2 2s2 2p5
s = 0+ 0 + ½ = ½
S = 2s + 1 = 2 (1/2 ) +1 = 2
b. Spectral state:
L = l1+l2+l3+l4
= 0 + 0+ +1-1+1-1-1 = 1
Value of ‘ L’ 0 1 2 3 4 5 6TERM LETTER S P D F G H I
Since L=1, the spectral term is ‘P’
c.J-value:
For more than half filled atom J = L+S = 1+ ½ = 3/2
d.∴ Term symbol for F is 2 P 3/2
Rules:
1. Hund’s rule should not be violated. ∴ Possible ground state terms are 3F 3P 1G 1D 1S ?2. The most stable state is with maximum multiplicity
∴ Possible ground state terms are 3F 3P
3. For a group of terms with same multiplicity, largest L value has least energy.
∴ Ground state term symbol is 3F
Excited Term symbols are 3P 1G 1D 1S
1.Find the term symbol for d1 system: Step -1: To find L:
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l1 = 2 , l2 =0 ∴ L = 2 [ since there is only one electron]Step 2: To find term letter : Term letter = D (from the table)Step 3: To find S S = ½ Step 4: To find multiplicity 2S + 1 = 2 Step 5: To find J:
Step 6: To write term symbol: Term symbol is 2D
2 .Find the term symbol for d2 system:
Step -1: To find L: l1 = 2 , l2 =2 ∴ L = (2+2), (2+2 -1),(2+2-2)…….(2-2) = 4, 3,2,1,0Step 2: To find term letter : Term letters = G,F,D ,P,S (from the table)Step 3: To find S S = ½ +1/2 = 1 S = ½ - ½ = 0Step 4: To find multiplicity When s = 0 , 2s + 1 = 1 When s = 1 , 2s + 1 = 3 Step 5: To find J: Step 6: To write term symbol: Term symbols are 3G, 3F, 3D, 3P, 3S, 1G, 1F, 1D, 1P, 1S
1. The term letters S,P,D,F,G … are used according to the value of L as 0,1,2,3 …2. The term letter is preceded by a superscript which is equal to 2s+1. It represents the multiplicity of the
energy level.3. The term letter is further followed by a subscript J, Total angular momentum4. The term symbol may be preceded, by the principal quantum number
For example 5 3D ½ should be read as ‘ five – triplet-D- one half ’ 2P 3/2 should be read as ‘ Doublet -P- three half ’
Combination of orbital angular momenta(L) and spin (S) of individual electrons is called
RS – coupling. Total electronic angular momentum (J) is given by the.
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J = L + S
Where L = total orbital angular momentum vector = ∑ Li
S = total spin angular momentum vector = ∑ Si
The value of L is given by L = ( l1 + l2), ( l1 + l2 -1)…. (l1-l2) where l1 and l2 are azimuthal quantum
numbers This series is known as Clebsch – Gordon series.
TERM SYMBOLS FOR ATOM IN THE GROUND STATE
It is represented by writing the spin multiplicity as superscripts to the left of the spectral state (term
letters) with angular momentum quantum number (J) value as subscript to the right.
TERM SYMBOL = spin multiplicity spectral state J
To find the Spectral state
Value of ‘ L’ 0 1 2 3 4 5 6
TERM LETTER S P D F G H I
For example if the value of L is 0, then the term letter is S.
if the value of L is 1, then the term letter is P and so on.
1. Find the ground state term symbol for d1 ion
Solution:
+2 +1 0 -1 -2↑ Spin multiplicity = 2s + 1
= 2 ( ½ ) + 1
= 2 ( doublet)
L = 2
J = 2 + ½
= 5/2
Therefore the ground state symbol 2 D 5/2
2.. Find the ground state term symbol for d2 ion
Solution:
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+2 +1 0 -1 -2↑ ↑ s = ½ +½
= 1
Spin multiplicity = 2s + 1
= 2 (1 ) + 1
= 3 ( Triplet)
L = 2(1) + 1(1)
= 2 + 1
= 3
Spectral state is F
J = L+S, L+S -1, L+S-2…
= 4,3,2,1
Therefore the ground state symbol 3 F
For atoms with small atomic number, where the spin – orbit interaction will be small, in comparison to inter electronic repulsion, the total electronic angular momentum (J) is given by the combination of spin and orbital angular momenta of individual electrons.
J = L + S
Where L = total orbital angular momentum vector = ∑ Li
S = total spin angular momentum vector = ∑ Si
The value of L is given by L = ( l1 + l2), ( l1 + l2 -1)…. (l1-l2) where l1 and l2 are azimuthal quantum numbers This series is known as Clebsch – Gordon series.
This is known as RS coupling or LS coupling.
For example consider the excited state of Carbon. Its configuration is 1s2 2s2 2p1 3p1
1s2 2s2 electrons do not contribute towards total L or S. For the 2p and 3p electrons
l1 = 1, l2 = 1
The vectors l1 and l2 can be combined in three possible ways so that it gives 3 value.
The possible values of J are 2,1 ,0 only.
L = ( 1+1), (1+1-1)…. ( l1 - l2 )
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= 2, 1,0
Similarly s1 and s2 combine in two different ways
when two vectors are anti parallel then s = ½ – ½ = 0
when they are parallel then s = ½ + ½ = 1
Then Land S can combine to give J as J = 3,2 and 1,
For atoms with large atomic number , the spin orbit interaction becomes much larger than the inter electronic repulsion. Under such condition, the operators L and S do not commute
6.5 HUCKEL MOLECULAR ORBITAL THEORYThis is an empirical method for organic compounds containing conjugated carbon chain linear as well
as cyclic. The theory is based on the following assumptions.
1. The wave function of Huckel Molecular Orbital(HMO) is taken as a linear combination of atomic
orbitals(AOs)
Ψi = ai1 p1 + a i2 p2 + ai3p3 +…………….
For example, Ψ1 = a11 p1 + a12 p2 + a13p3 +…………….
2. .There will be ‘n’ energy levels corresponding to ‘n’ HMOs , each being expressed in terms of α and β
3. Energy is calculated by E = ∫Ψ HΨ dT
∫ΨΨ dT
4. Variation treatment leads to ‘n’ secular equations, where ‘n’ is number of C atoms.
For example for ethylene two secular equations and for butadiene there are four secular equations.
a1 ( H11 – ES11) + a2 ( H12 – ES12) + …………………….. + an ( H1n – ES1n ) = 0
a1 ( H21 – ES21) + a2 ( H22 – ES22) + ……………………+ an ( H2n– ES2n ) = 0
……………………………………………………………………………………
……………………………………………………………………………………..
a1 ( Hn1 – ESn1) + a2 ( Hn2 – ESn2) +…………………… + an ( Hnn – ESnn) = 0
5. He introduce some approximations
a. All integrals of type H ii = α
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For example H11 = H22 = H33 = α
b. All integrals of type in which i and j are directly bonded, H ij = β
For example H12 = H21 = β, if C1 and C2 are directly bonded.
c. All integrals of type in which i and j are not directly bonded, H ij = 0
For example H14 = H41 = 0, if C1 and C4 are not directly bonded.
d. Integrals of type S ii = 1
For example S11 = S22 = S33 = 1.
e. Integrals of type S ij = 0
For example S12 = S23 = S34 = 0.
Thus for a linear conjugated chain the secular determinant takes the form
α – E(1) β 0 0
β α – E(1) β 0
0 β α – E(1) β = 0
0 0 β α – E(1)
Dividing by β
α – Eβ 1 0 0
1 α – E
β 1 0
0 1 α – E
β 1 = 0
0 0 1 α – E
β
Put x = α – E
β the above determinant becomes
x 1 0 0
1 x 1 0 = 0
0 1 x 1
0 0 1 x
5. On expanding the n× n determinant, a polynomial of n th degree in x which has ‘n’ real
roots x1,x2,x3….will be obtained.
6. The negative root corresponds to ‘ bonding level’. The positive root corresponds to ‘ anti
bonding level’. When ‘n’ is odd, Ei = α corresponds to ‘non bonding level
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7. By inserting the values of E, in the secular equations, the values of HMO coefficients are
obtained.
EXCHANGE INTEGRAL
Integral over the coordinates of two identical particles ( the interaction between a given state
and a second state in which the coordinates of the particles are exchanged).
OVERLAP INTEGRAL
The overlap of the atomic orbital of an atom A and the atomic orbital of an atom B is called their overlap
integral. It is defined as SAB=∫ψ∗AψBdr extending over all space.
If the wave functions do not overlap, then the overlap integral is zero.
The integral can also be zero if the wave functions have positive and negative aspects that cancel out.
If the overlap integral is zero, then the wave functions are called orthogonal.
the maximum value overlap integral of S = 1
COULOMB INTEGRALS
All Hamiltonian integrals Hii are called coulomb integrals and
RESONANCE INTEGRALS
All Hamiltonian integrals of type Hij, where atoms i and j are connected, are called resonance integrals.
STEPS IN HUCKEL’S MOLECULAR ORBITAL THEORY.
1.Write down the expression for atomic orbital
2. Write down the expression for molecular orbital
3. . Write down secular determinant
5. Apply Huckel’s approximation to find the energy equation
6. Solve the secular determinant to get the polynomial
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7. Solve the polynomial
8. Substitute the value in energy equation to the energy level
6.5.1 APPLICATION OF HMO TO ETHYLENE:
The skeleton framework of ethylene is C1= C2.
1. Atomic orbitals
Since there are two carbon atoms in ethylene there are two atomic orbitals ( φ1 and φ2 )
2. Molecular orbitals
The MO may be written as the linear combination of atomic orbitals
Ψ = a1 φ1+ a2 φ2
where a1 and a2 are Huckel’s coefficients
3. The secular equations
Since there are two carbon atoms there will be two secular equations.. They are
a1 ( H11 – ES11) + a2 ( H12 – ES12) = 0
a1 (H21 – ES21 ) + a2 ( H22 – ES22) = 0
4. The secular determinant
secular determinant is formed by leaving the HMO coefficient’s
¿ = 0
5. Apply Huckel’s approximation to find energy equation
H11 = H22 = α, ( Coloumb integral)
H12 = H21 = β ( exchange integral)
S11 = S22 = 1 ,
S21 = S12 = 0 ( overlap integral)
Substituting in the above determinant,
|α−E ββ α−E|=0
dividing by β
α−Eβ 1
1 α−E
β = 0
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Put α−E
β = x
This is the energy equation.
6. Solve the secular determinant find the polynomial
x 1
1 x = 0
polynomial is x 2 - 1 = 0
7. To solve the polynomial:
x 2 - 1 = 0
x 2 = 1
x = ± 1
8. Substitute the value in energy equation.
when x = +1, let E = E1
α−Eβ = x
α−E1
β=+1
α−E 1=β
∴ E1 = α- β
when x = -1, let E = E2
α−E
β = x
α−E2
β = -1
α−E 2=−β
∴ E2 = α + β
Thus the two HMO energies of ethylene are α + β , α – βThe energy level E2 is called bonding HMO( Huckel MolecularOrbital ) because it has lower
energy ( since β is negative ) and E1 is called anti bonding HMO
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π – BOND ENERGY:
π – bond energy = total energy – energy in the absence of bonding
= 2(α + β) - 2 α
= 2 β
To determine the HMO co efficients:
a1 ( H11 – ES11) + a2 ( H12 – ES12) = 0
a1 (α- E) + a2 β = 0
since α−E2
β=−1 [Take any one value]
(α- E) = - β [ Take E2 as E]
substituting in equation 1∴ a1 (- β) + a2 β = 0
a1 β = a2 β
a1 = a2
By normalization condition
a12 + a2
2 = 1
a12 + a 1
2 = 1 [a1 = a2 ]
2 a12 = 1
∴ a1 = 1
√ 2 and a2 = 1
√ 2
Thus the wave functions are
Ψ1 = 1
√ 2 φ1+ 1
√ 2 φ2
Ψ2 = 1
√ 2 φ1 - 1
√ 2 φ2
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α + β represents bonding energy level because β is negative
The antibonding wave function of ethylene is α – β
The total energy of ethylene in terms of α and β is 2α + 2β
Number of nodal plane in Ψ1 of ethylene is zer0
Number of nodal plane in Ψ2 of ethylene is one
Charge density:
The Charge density is given ρ = 1- qr = 1 – 1 = 0
Π – bond order
This is given by pij = ∑ n(Aij× Bij) Where n is number electrons, Aij –
P12 = 2 × 1
√ 2× 1
√ 2 = 1
Free valency:
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For C1: F1 = √3 - P12 = √3 - 1 = 1. 732 -1 = 0. 732
For C1: F2 = √3 - P12 = √3 - 1 = 1. 732 -1 = 0. 732
6.5.2 APPLICATION TO ALLYL SYSTEMS
The carbon skeleton of allyl system is
C1-C2-C3
There are three AO’s to be combined to form MO’s
Ψ = a1 φ1+ a2 φ2 + a3 φ1 -------------------------------1
Where φ1 , φ2, and φ3 are the atomic orbitals and the secular equations are
a1 ( H11 – ES11) + a2 ( H12 – ES12) + a3 ( H13 – ES14) = 0
a1 ( H21 – ES21) + a2 ( H22 – ES22) + a3 ( H23– ES23) = 0
a1 ( H31 – ES31) + a2 ( H32 – ES32) + a3 ( H33 – ES33) = 0
1. The Coloumb integral H11 = H22 = H33 = α
2. The exchange integral H12 = H21 = H23 = H32 = β
3. H13 = H31 = 0 since there is no direct link between C1 and C3
4. The resonance integral S11 = S22 = S33 = 1 ,
5. The overlap integral S12 = S21 = S13 = S31 = 0
Put α−E
β = x
Substituting the above values the secular determinant becomes
|x 1 01 x 10 1 x| = 0
x( x2 – 1 ) – 1 ( x – 0) = 0
x3 – x – x = 0
x3 –2x = 0
x ( x2 –2) = 0
x = 0 and x2 –2 = 0
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x = 0 and x2 = 2
x = 0 , x = + √2 and x = - √2 are the roots
when x = 0, let E = E1
α−Eβ = x
α−E1
β=0
α−E 1=0
∴ E1 = α
when x = + √2, let E = E2
α−E
β = + √2
α−E 2=+√2 β
∴ E2 = α + √2 β
when x = - √2, let E = E3
α−E
β = - √2
α−E3−√2 β
∴ E3 = α - √2 β
π – BOND ENERGY:
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= 2(α +√2 β ) - 2 α
= 2√2 β
For radical:
π – bond energy = total energy – energy in the absence of bonding
= [2(α +√2 β ) + α ] - 3α
= 2√2 β
For anion:
π – bond energy = total energy – energy in the absence of bonding
= [2(α +√2 β ) +2 α ] - 4 α
= 2√2 β
This shows that π – bond energy for all allyl system are equal
Delocalization energy = π – bond energy - 2 β
= 2√2 β - 2 β
= 2 β (√2 – 1)
= 2 β ( 1.414 – 1)
= 0.828 β
Since β is negative , this gives negative value.
The delocalization energy is same for all the three allyl systems.
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6.5.3 APPLICATION TO BUTADIENE:
The carbon skeleton of butadiene is
C1-C2-C3-C4
There are four AO’s to be combined to form MO’s
Ψ = a1 φ1+ a2 φ2 + a3 φ1+ a4 φ2 -------------------------------1
Where φ1 , φ2, φ3 and φ4 are the atomic orbitals and the secular equations are
a1 ( H11 – ES11) + a2 ( H12 – ES12) + a3 ( H13 – ES14) + a4 ( H14 – ES14 ) = 0
a1 ( H21 – ES21) + a2 ( H22 – ES22) + a3 ( H23– ES23) + a4 ( H24– ES24 ) = 0
a1 ( H31 – ES31) + a2 ( H32 – ES32) + a3 ( H33 – ES33) + a4 ( H34 – ES34) = 0
a1 ( H41 – ES41) + a2 ( H42 – ES42) + a3 ( H43 – ES43) + a4 ( H44 – ES44 ) = 0
1. The Coloumb integral H11 = H22 = H33 = H44 = α
2. The exchange integral H12 = H21 = H23 = H32 = H34= H43 = β
3. H13 = H31 = H14 = H41 = H24 = H42 = 0 since there is no direct link between C1 and C3
C2 and C4 and C1 and C4
4. The resonance integral S11 = S22 = S33 = S44 = 1 ,
5. The overlap integral S12 = S21 = S13 = S31 = S14 = S41 = S23 = S32 = 0
Using this value, the secular determinant is
α – E(1) β 0 0
β α – E(1) β 0
0 β α – E(1) β = 0
0 0 β α – E(1)
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Dividing by β
α – Eβ 1 0 0
1 α – E
β 1 0
0 1 α – E
β 1 = 0
0 0 1 α – E
β
Put x = α – E
β the above determinant becomes
x 1 0 1 1 0
x 1 x 1 - 1 0 x 1 = 0
0 1 x 0 1 x
x { [ x ( x2 -1 ) - 1( x-0) ] } - [ 1 ( x2 – 1) - 1 ( 0)] = 0
x4 – x2 - x 2– x2 +1 = 0
x4 –3 x2 +1 = 0 --------------1
put y = x2 then the above equation becomes,
y2 – 3y + 1 = 0
y = +3 ±√9−42
= +3 ±√52
The roots are y 1 = +3+√52
y 2 = +3−√52
To find x: y1 = x2
∴x2 = +3+√52
= 2¿¿ [ multiplying and dividing by 2]
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= 1+5+2√54
= ¿¿
x1 = + 1+√52
= 1. 618
x2 = - 1+√52
= - 1.618
similarly
y 2 = +3−√52
x3 = + 0.618 x 4 = - 0.618
∴ The four roots of equation 1 are x1 = + 1.618, x2 = - 1.618 , x3 = + 0.618 x 4 = - 0.618
when x = + 1.618 , let energy be E1,
x = α – E
β
1.618 = α – E1
β
E1 = α - 1.618 β
when x = - 1.618 let energy be E2 then -1.618 = α – E2
β
E2 = α + 1.618 β
when x = + 0.618 let energy be E3, then, 0.618 = α – E3
β
E3 = α - 0.618 β
when x = - 0.618 let energy be E4 , then, 0.618 = α – E3
β
E4 = α + 0.618 β
E1 = α - 1.618 β,
E2 = α + 1.618 β
E3 = α - 0.618 β,
E4 = α + 0.618 β
ALITER:
Expression for energy
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E = α + 2 β Cos ( (r+1)πN+1
) Where N – number of π electrons, r = 0,1,2…N-1
For butadiene N = 4
When r = 0 , E1 = α + 2 β Cos (π5 )
When r = 1, E2 = α + 2 β Cos (2 π5 )
When r = 2, E3 = α + 2 β Cos (3π5 )
When r = 3, E4 = α + 2 β Cos (4 π5 )
ESTIMATION OF DELOCALISATION ENERGY
Delocalization energy = Total energy – expected energy
TOTAL ENERGY:
E = ∑ no of electrons∈the state× energy of the state
In butadiene there are 4 π electrons ( 2 double bonds)
2 electrons are in α + 1.62 β state and other 2 electrons are in α + 0.62 βstate
∴Total energy = 2(α + 1.62 β ) + 2 (α + 0.62 β)
= 2(α + 1.62 β ) + 2 (α + 0.62 β)
= 2α + 3.24 β + 2 α + 1.24 β
= 4α + 4.48 β
EXPECTED ENERGY
= 4 ( energy of one double bond like in ethylene)
= 4( α + β)
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= 4 α +4β
DELOCALISATION ENERGY
Delocalization energy = Total energy – expected energy
= (4α + 4.48 β ) - (4 α +4β )
= 0.48β
WAVE FUNCTIONS:
Number of nodal plane in Ψ1 of butadiene is zero
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E1 and E2 levels corresponds to bonding MOs . E3 and E4 levels corresponds to anti bonding MOs The 4
electrons of occupy these bonding MOs of low energy
Total energy = 2 (α + 1.618 β ) + 2 (α + 0.618 β )
= 4 α + (3.236 + 1.236 )β
Since butadiene has 2 double bonds and hence the total energy should be equal to twice that
of compound ‘having one’ double bond∴ Expected energy = 2[ energy of ethylene]
= 2[2α + 2β] [ energy of ethylene = 2α + 2β]
Difference in energy = 4 α + 4.562 β - 2 [2α +2β]
= +0.562 β
To determine the HMO co efficients:
a1 ( H11 – ES11) + a2 ( H12 – ES12) + a3 ( H13 – ES14) + a4 ( H14 – ES14 ) = 0
a1 ( H21 – ES21) + a2 ( H22 – ES22) + a3 ( H23– ES23) + a4 ( H24– ES24 ) = 0
a1 ( H31 – ES31) + a2 ( H32 – ES32) + a3 ( H33 – ES33) + a4 ( H34 – ES34) = 0
a1 ( H41 – ES41) + a2 ( H42 – ES42) + a3 ( H43 – ES43) + a4 ( H44 – ES44 ) = 0
the above equations can be rewritten as
a1 (α- E) + a2 β + a3 (0) + a4 (0) = 0 since there is no direct link between 1-3 and 1-4
a1 β + a2 (α- E) + a3 β + a4 (0) = 0
a1 (0) + a2 β + a3 (α- E) + a4 β = 0
a1 (0) + a2 (0) + a3 β + a4 (α- E) = 0
dividing by β, and put x = α – E
β a1x + a2 = 0 a1 + a2x + a3 = 0
a2 + a3x + a4 = 0
a3 + a4 x = 0 By normalization condition
a12 + a2
2 + a32 + a4
2 = 1
put x = -1.618, we will get. a1 = a4 = 0.372 and a2 = a3 = 0.602
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Thus the wave function is Ψ1 = 0.372 φ1 + 0.602 φ2 + 0.602 φ2 + 0.372 φ2
put x = -0. 618, we will get, a1 = 0.602, a2 = 0.372, a3 = - 0.372 and a4 =- 0.602
Thus the wave function is ,Ψ2 = 0.602 φ1 + 0.372 φ2 - 0.372 φ2 - 0.602 φ2
put x = +0.618, we will get ,a1 = 0.602, a2 = - 0.372, a3 = - 0.372 and a4 = +0.602
Thus the wave function is ,Ψ2 = 0.602 φ1 - 0.372 φ2 - 0.372 φ2 + 0.602 φ2
put x = +1.618, we will get , a1 = 0.372 , a2 = - 0.602, a3 = + 0.602 and a4 = -0.372
Thus the wave function is ,Ψ2 = 0.372 φ1 - 0.602 φ2 + 0.602 φ2 - 0.372 φ2
Charge density:
The Charge density is given ρ = 1- qr
At all carbon charge density = 1 – 1 = 0
Π – bond order:
This is given by pij = ∑ n(Aij× Bij)
Where n is number electrons, Aij –
P12 = 2 ( 0.372 × 0.602 ) + 2 ( 0.602 × 0.372 ) = 0.896
P23 = 2 ( 0.602 × 0.602 ) + 2 ( 0. 372 × 0.372 ) = 0.448
P34 = 2 ( 0.602 × 0. 372 ) + 2 ( - 0.372 × - 0.602 ) = 0.896
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1. All the values are less than one indicates that each of the three double bonds is neither a full double bond
nor a single bond. All have partial double bond character.
2. P12 = P34 > P23 .
This shows that the terminal bonds C1-C2 and C3-C4 have more double bond character than the
central bond C2-C3.
Free valency:
Free valency for terminal carbons
For C1: F1 = √3 - P12
= √3 - 0.896
= 1. 792 - 0.896
= 0. 896
For C4: F4 = √3 - P34
= 1. 792 - 0.896
= 0. 896
Free valency for intermediate carbons
For C2:
F2 = P12 + P23
= 0.896 + 0.448
= 1.344
For C3:
F3 = P12 + P23
= 0.896 + 0.448
= 1.344
The lower value of C2 and C3 explains, that these carbons are deeply engaged in bonding and have a smaller
free valency. The terminal carbons C1andC4 have higher values which shows that these are more reactive.
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6.5.4 APPLICATION OF HMO TO BENZENE
The HMO wave function for benzene is Ψ1 = a1 p1 + a2 p2 + a3p3 + a4p4 + a5p5 + a6p6
Where p1 , p2 ,p3 ,p4,p5,p6 are the atomic orbitals and the secular equations are
a1 ( H11 – ES11) + a2 ( H12 – ES12) + a3 ( H13 – ES14) + a4 ( H14 – ES14 ) + a5 ( H15 – ES15 ) + a6 ( H16 – ES16 ) = 0
a1 ( H21 – ES21) + a2 ( H22 – ES22) + a3 ( H23– ES23) + a4 ( H24– ES24 ) + a5 ( H25 – ES25 ) + a6 ( H26 – ES26 ) = 0
a1 ( H31 – ES31) + a2 ( H32 – ES32) + a3 ( H33 – ES33) + a4 ( H34 – ES34) + a5 ( H35 – ES35 ) + a6 ( H36 – ES36 ) = 0
a1 ( H41 – ES41) + a2 ( H42 – ES42) + a3 ( H43 – ES43) + a4 ( H44 – ES44 ) + a5 ( H45 – ES45 ) + a6 ( H46 – ES46 ) = 0
a1 ( H51 – ES51) + a2 ( H52 – ES52) + a3 ( H53 – ES53) + a4 ( H54 – ES54 ) + a5 ( H55 – ES55 ) + a6 ( H56 – ES56 ) = 0
a1 ( H61 – ES61) + a2 ( H62 – ES62) + a3 ( H63 – ES63) + a4 ( H64 – ES64 ) + a5 ( H65 – ES65 ) + a6( H66 – ES66 ) = 0
The Coloumb integral H11 = H22 = H33 = H44 = H55= H66 = α
The exchange integral H12 = H21 = H23 = H32 = H34= H43= H45= H54= H56= H65 = β
All other Hij = 0
The resonance integral S11 = S22 = S33 = S44 = S55 = S66 = 1 ,
All other Sij = 0
The above secular equations become
a1 (α – E) + a2 ( β ) + 0 + 0 + 0+ a6 ( β ) = 0
a1 (β )+ a2 (α – E) + a3 ( β) +0 +0 +0 = 0
0+ a2 (β) + a3 (α – E) + a4 (β ) + 0 + 0 = 0
0 + 0 + a3 (β) + a4 ( α – E) + a5 (β) + 0 = 0
0+0 + 0 + a4 ( β ) + a5 ( α – E ) + a6 ( β ) = 0
a1 (β ) + 0 + 0 +0 + a5 (β ) + a6 ( α – E ) = 0
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let x = α – E
β
The secular equations in terms of ‘x’ are
a1x + a2 +a6 = 0
a1 + a2x + a3 = 0
a2 + a3x + a4 = 0
a3+a4x+ a5 = 0
a4+a5x+ a6 = 0
a1 +a5 + a6x =0
The corresponding determinant is
x 1 0 0 0 1
1 x 1 0 0 0
0 1 x 1 0 0
0 0 1 x 1 0
0 0 0 1 x 1
1 0 0 0 1 x
Expanding of this polynomial will lead to x6 – 6 x4 +9x2 -4 = 0
put y = x2, the above equation becomes y3 -6y2+9y-4 = 0
put y = 1, 1-6+9-4 =0 ∴ (y -1)is a factor
y=1 ia a root. To find the other roots 1 1 -6 9 -4 0 1 -5 4 -------------------------------- 1 -5 4 0
The equation becomes y2 – 5y +4 = 0
( y-4)( y-1) = 0
y= 4 and y = 1
y = 1, y=1 and y = 4∴ x2 = 1, x2 = 1 and x2 = 4
x= ± 1, x= ±1 and x = ±2
Therefore the six roots are +1, +1, -1,-1,+2,-2
When x1 = -2, E1 = α +2β bonding
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When x2 = -1, E2 = α +β doubly degenerate bonding HMO s
When x3 = -1, E3 = α +β
When x4 = +1, E4 = α –β doubly degenerate anti - bonding HMO s
When x5 = +1, E5 = α -β
When x6 = +2, E6 = α -2β anti -bonding
The corresponding determinant is
Expanding of this polynomial will lead to x6 – 6 x4 +9x2 -4 = 0
When x1 = -2, E1 = α +2β bonding
When x2 = -1, E2 = α +β degenerate bonding
When x3 = -1, E3 = α +β
When x4 = +1, E4 = α –β degenerate non- bonding
When x5 = +1, E5 = α -β
When x6 = +2, E6 = α -2β non-bonding
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ESTIMATION OF DELOCALISATION ENERGY
Delocalization energy = Total energy – expected energy
TOTAL ENERGY:
E = ∑ no of electrons∈the state× energy of the state
In benzene there are 6 π electrons ( 3 double bonds)
2 electrons are in α + 2 β state and other 4 electrons are in α + β state
∴Total energy = 2(α + 2 β ) +4 (α + β)
= 2α + 4 β + 4α +4 β
= 6α + 8 β
EXPECTED ENERGY
= 6 ( energy of one double bond like in ethylene)
= 6( α + β)
= 6 α +6β
DELOCALISATION ENERGY
Delocalization energy = Total energy – expected energy
= (6α + 8 β ) - (6α + 6β )
= 2β
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Number of nodal plane in of benzene is
Degree of degeneracy of lowest bonding molecular orbital of benzene is
Degree of degeneracy of highest anti bonding molecular orbital of benzene is
The lowest energy state in benzene is
In polyenes containing even number of π electrons the degenerate state is
In polyenes containing odd number of π electrons the degenerate state is
The lowest and highest energy levels are degenerate for polyenes containing
The intermediate energy levels are doubly degenerate for polyenes containing
HOMO of benzene is
LUMO of benzene is
The doubly degenerate energy level of benzene is
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ALITER:
Expression for energy
E = α + 2 β Cos ( 2 πrN ) Where N – number of π electrons, r = 0,1,2…N
For benzene N = 6
When r = 0
E1 = α + 2 β Cos (0)
= α + 2 β (1) [cos (0) = 1 ]
= α + 2 β
When r = 1
E2 = α + 2 β Cos (2 π6 )
= α + 2 β Cos (60) [ π= 180 ]
= α + 2 β (12 ) [cos (60) = ½
= α + β
When r = 2
E2 = α + 2 β Cos (4 π6 )
= α + 2 β Cos (120) [ π= 180 ]
= α + 2 β ( - 12 ) [cos (120) = -½
= α - β
When r = 3
E3 = α + 2 β Cos (6 π6 )
= α + 2 β Cos (π) [ π= 180 ]
= α + 2 β (-1 ) [cos (π ) = -1
= α - 2β
When r = 4
E3 = α + 2 β Cos (8 π6 )
= α + 2 β Cos (240) [ π= 180 ]
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= α + 2 β (-½ ) [cos (240 ) =-½
= α - β
When r = 5
E3 = α + 2 β Cos (10 π
6 )
= α + 2 β Cos (300) [ π= 180 ]
= α + 2 β (½ ) [cos (300 ) =½
= α + β
Thus the energy levels are
α +2β , α + β , α + β , α - β, α - β , α -2β
E1, E2 and E3 levels corresponds to bonding MOs. Two of these energy levels are degenerate.
The 6 electrons of benzene occupy these bonding MOs of low energy. E4, E5 and E6 levels corresponds to anti
bonding MOs .Two of these energy levels are also degenerate.
Total energy = 2 (α +2β) + 2 (α +β) + 2(α +β)
= 6α +8β
Since benzene has 3 double bonds the total energy of benzene should be equal to 3 times that
of compound ‘having one’ double bond∴ Expected energy of benzene = 3 (energy of ethylene)
= 3(2α + 2β) [ energy of ethylene = 2α + 2β]
= 6α + 6β)
Difference in energy = 6α +8β - 3 [2α +2β]
= 2β
This is due to delocalization of π electrons and this energy is known as delocalization energy.Page 297 of 419
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The value of β is –75 KJ/ mol. Therefore delocalization energy of benzene = –150 KJ/mol.(36 Kcal/mol) This
leads to stability of benzene.
.
Charge density:
The Charge density is given ρ = 1- qr
= 1 – 1 = 0
Π – bond order:
This is given by pij = ∑ n(Aij× Bij)
Where n is number electrons, Aij –
P12 = number of electrons (coeff of φ1 × coeff of φ1 in Ψ1 ) + number of electrons( coeff of
φ1 × coeff of φ1 in Ψ2) + number of electrons( coeff of φ1 × coeff of φ1 in Ψ3 )
= 2 (1
√ 6 × 1
√ 6 ) + 2 ( 12 ×
12 ) +2 (
1√ 12 ×
−1√ 12 )
= 26 +
24 -
212
= 4+6−2
12
= 23
P23 = number of electrons( coeff of φ1 × coeff of φ1 in Ψ1 ) + number of electrons( coeff of
φ1 × coeff of φ1 in Ψ2) + number of electrons( coeff of φ1 × coeff of φ1 in Ψ3 )
= 2 (1
√ 6 × 1
√ 6 ) + 2 ( 12 × 0 ) +2 (
−1√ 12 ×
−2√ 12 )
= 26 +
412 ? ?
= 4+412
= 23
P12 = P23 = P34 = P56 = P61 = 23
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S.NO SYSTEMS No.of π
Electron
s
Energy levels π –bond
energy
Delocalisation
energy
1 ETHYLENE 2 E1 = α - β 2 β -----------E2 = α + β
2 ALLYL
SYSTEM
CATION 2α
+ √2 βα α
- √2 β
2√2 β
Same for all
0.828 β
Same for all
RADICAL 3
ANION 4
3 BUTADIENE 4 E1 = α + 1.62β
E2 = α + 0.62β
E3 = α − 0.62β
E4 = α − 1.62β
4 BENZENE 6 E1 = α + 2β
2β
= 36 Kcal/mol
E2 = α + β
E3 = α + β
E4 = α − β
E5 = α − β
E6 = α − 2β
5 CYCLOBUTADIENE 4 E1 = α + 2β
E2 = α
E3 = α
E4 = α − 2β
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7. CHEMICAL BONDING 7.1 HYBRIDISATION
Definition:
Bond formation involves the overlapping of 2s and 2p orbitals .This linear combination of
orbitals of the same atom is called hybridization. Thus hybridization is the mixing of orbitals to
produce equivalent number of orbitals called hybrid orbitals.
A combination of one ‘s’ orbital and one ‘p’ orbital is called ‘sp’ hybridization. Similarly a
combination of one ‘s’ orbital and two ‘p’ orbitals is called ‘sp 2’ hybridization
9.1.1 s-p hybridization:
The combination of one s- orbital and one p- orbitals , giving two hybrid orbitals Ψ1 and Ψ2 may
be expressed as
Ψ1 = a1 s + b1 p ----------------1
Ψ2 = a2s + b2p ------------------2
The values of a1,b1, a2 and b2 can be determined by the following considerations.
Ψ1 and Ψ2 are normalized , orthogonal and equivalent.
Since the two hybridized orbitals are equivalent, the share of s functions is equal∴ a12 = a2
2 = ½ -------------------3
a12 = ½
∴ a1 = 1√2
a22 = ½
∴ a2 = 1√2
Ψ1 is normalized
a12 + b1
2 = 1
(12 ) + b1
2 = 1
∴ b12 =
12
b1 = 1√2
Ssince Ψ1 and Ψ2 are orthogonal
a1b1 + a2b2 = 0
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(1√2
) ( 1√2
) + ( 1√2
) (b2) = 0
12 + (
1√2
) (b2) = 0
( 1√2
) (b2) = - ½
∴ b2 = - 12 ×
√ 21
= - 1
√2√ 2 ×
√ 21
= 1√2
The wave functions are Ψ1 = 1√2
s + 1√2
p = 1√2
( s+p)
Ψ2 = 1√2
s + −1√2
p = 1√2
( s- p)
SSSDirectional characteristics:
Ψ1 = 1√2
( s+p), Ψ2 = 1√2
( s- p)
The angular function of 2s orbital = 1
√ 4 π *
The angular function of 2pz orbital.,l l=1, m=0 = √ 34 π
cos θ **
Substituting Ψ1 = 1√2
(1
√ 4 π+ √ 3
√ 4 π cos θ) = 1
√ 4 π (1+√ 3 cosθ ¿ ¿√2 )
Ψ2 = 1√2
(1
√ 4 π− √ 3
√ 4 π cos θ) = 1
√ 4 π (1−√ 3 cosθ ¿ ¿√2 )
The value of maximum in either case is 1+√ 3
√ 2 = 1.932 which is greater than that for pure 2s
orbital(f=1) and a 2p orbital ( √ 3 = 1.732)
The functions are maximum θ = 0, θ = π Therefore the angle between the two functions is 180 o
The functions are maximum θ = 0, θ = π Therefore the angle between the two functions is 180 o
Diagram
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Θ l,m =√ 2 l+12
× (l−m ) !(l+m ) !
× Plm(cos θ) ×
1√2 π
e imφ but Plm(cos θ)=
12l× l!
( 1- cos 2 θ ) m/2 ( d
d (cosθ) ) l+m [cos 2 θ -1 )
l
*angular function of s orbital (l=0,m=0)
P00 ( cos θ ) =
120× l!
( 1- cos 2 θ ) 0 ( d
d (cosθ) ) 0 [cos 2 θ -1 ) 0 = 1
∴ Θ 0,0 =√ 2(0)+12
×(1)× 1√2 π
= √ 12
× 1√2 π
= 1
√ 4 π
** angular function of p orbital( l=1, m=0)
P10 = ½ (
ddθ ) [cos 2 θ -1 ) l× e0 = ½ [ 2 cosθ ] = cosθ
∴ Θ 1,0 = √ 2(1)+12
× (1) × 1√2 π
= √ 32
×1√2 π
= 1
√ 4 π cosθ
------------------------------------------------------------------------------------
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sp 2 hybridization:
The combination of one s- orbital and two p- orbitals , giving three hybrid orbitals Ψ1 , Ψ2 and Ψ 3 may be expressed as
Ψ1 = a1 s + b1 px + c1 py ----------------1Ψ2 = a2 s + b2 px + c2 py ---------------2 Ψ3 = a3 s + b3 px + c3 py -----------------3
The values of coefficients can be determined by the following considerations.Ψ1 , Ψ2 and Ψ3 are normalized , orthogonal and equivalent. Since the three hybridized orbitals are equivalent, the share of s functions is equal
∴ a12 = a2
2 = a32 =
13 ----------4
a1 = a2 = a3 = 1
√ 3
Let c1 may be assigned to fixed direction. Therefore c1 = 0
To find Ψ1
Since Ψ1 is normalized,
a12 + b1
2 + c12 = 1
(1
√ 3 )2 + b12 = 1
(13 ) + b1
2 = 1
∴ b12 = 1 -
13
= 23
b1 = √23
Thus the wave function is
To find Ψ2
Since Ψ1 and Ψ2 are orthogonal a1a2 + b1 b2 + c1c2 = 0
(1√3
) ( 1√3
) + ( √ 2√3
) b2 = 0
13 + (
√ 2√3
) b2 = 0
( √ 2√3
) b2 = - 13
b2 = −13 ×
√ 3√ 2
= −1√3 √ 3
×
∴ b2 = -−1√ 6
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Ψ1 = 1
√ 3 s + √23 px
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Since Ψ2 is normalized a2
2 + b22 + c2
2 = 1
(1
√ 3 )2 +(−1√ 6 ) 2 + c2
2 = 1
(13 ) + (
16 ) + c2
2 = 1
(2+1
6 ) + c22 = 1
(12 ) + c2
2 = 1
∴ c22 = 1 – ½
c2 = 1√2
Thus the wave function is
To find Ψ3
Since Ψ1 and Ψ3 are orthogonal a1a3 + b1 b3 + c1c3 = 0
(1√3
) ( 1√3
) + ( √ 2√3
) b3 = 0
13 + (
√ 2√3
) b3 = 0
b3 = −13 ×
√ 3√ 2
= −1
√ 3 √ 3 × √ 3√ 2
∴ b3 = −1√ 6
Since Ψ2 and Ψ3 are orthogonal a2a3 + b2 b3 + c2c3 = 0
(1√3
) ( 1√3
) +(−1√ 6 )(
−1√ 6 ) +
1√2
c3 = 0
13 +
16 +
1√2
c3 = 0
12 +
1√2
c3 = 0
c3 = −12 ×
√ 21
= −1
√ 2 √ 2 × √ 21
∴ c3 = −1√ 2
Thus the wave function is
Directional characteristics:
Ψ1 = 1
√ 3 s + √23 px
Ψ2 = 1
√ 3 s - ( 1
√ 6 ) px + 1√2
py
Ψ3 = 1
√ 3 s - ( 1
√ 6 ) px - 1√2
py
The angular function of 2s orbital = 1
2 √ π ,
2px orbital. = (√ 3
2√ π sin θ cos φ) and that of
2py orbital. = ( √ 3
2 √ π sin θ sin φ)
Substituting
Ψ1 = 1
√ 3 ( 1
2√ π¿ + √
23 (
√ 32√ π sin θ cos φ)
Ψ2 = 1
√ 3 ( 1
2 √ π ) - ( 1
√ 6 ) (√ 3
2 √ π sin θ cos
φ) + 1√2
( √ 3
2 √ π sin θ sin φ)
Ψ2 = 1
√ 3 ( 1
2 √ π ) - ( 1
√ 6 ) (√ 3
2 √ π sin θ cos
φ) - 1√2
( √ 3
2√ π sin θ sin φ)
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Ψ2 = 1
√ 3 s - ( 1
√ 6 ) px + 1√2
py
Ψ3 = 1
√ 3 s - ( 1
√ 6 ) px - 1√2
py
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Let us assume that Ψ1 points towards x-axis. The direction of Ψ2 can be determined as follows
Ψ2 = 1
√ 3 ( 1
2√ π ) - ( 1
√ 6 ) (√ 3
2 √ π sin θ cos φ) + 1√2
( √ 3
2√ π sin θ sin φ)
= ( 1
2√ π ) [ 1
√ 3 - ( 1
√ 6 ) (√3 sin θ cos φ) + 1√2
(√3 sin θ sin φ)
If the two functions are in the xy plane , then 𝛉 = 90. Sin 90 = 1., Substituting
Ψ2 = ( 1
2 √ π ) [ 1
√ 3 - ( 1
√ 6 ) (√3 cos φ) + 1√2
(√3 sin φ)
= ( 1
2√ π ) [ 1
√ 3 - ( 1
√ 2√ 3 ) (√3 cos φ) + 1√2
(√3 sin φ)
= ( 1
2 √ π ) [ 1
√ 3 - 1
√ 2 cos φ + √ 3√2
sin φ)
Let f = ( 1
√ 3 - 1
√ 2 cos φ + √ 3√2
sin φ)
= ( 1
√ 3 - 1
√ 2 cos φ + √ 3√2
( 1- cos 2φ) ½ )
Let x = cos φ
Then f = ( 1
√ 3 - 1
√ 2 x + √ 3√2
( 1- x2 ) ½
Differentiating with respect to ‘x’
dfdx = -
1√ 2 +
√ 3√2
× ½ ( 1- x2 ) -½ ( -2x)
Put dfdx = 0
0 = - 1
√ 2 + √ 3√2
× ½ ( 1- x2 ) -½ ( -2x)
1
√ 2 = √ 3√2
× ( 1- x2 ) -½ ( - x)
1 = - √3 ( 1 – cos 2 φ) – ½ cos φ
= - √3 ( sin 2 φ) – ½ cos φ
= - √3 1
sin φ cos φ
- 1
√ 3 = cosφsin φ
Taking reciprocal
- √3 = sin φcosφ
Tan φ = - √3
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∴ φ = 120
Thus the function f2 is found to be at an angle of 120 o with respect to f1
Similarly the function f3 is found to be at an angle of 240 o with respect to f1
Θ l,m = √ 2 l+12
× ( l−m ) !(l+m ) !
× Plm(cos θ) ×
1√2 π
e imφ but Plm(cos θ)=
12l× l!
( 1- cos 2 θ ) m/2 ( d
d (cosθ) ) l+m [cos 2 θ
-1 ) l
*angular function of s orbital (l=0,m=0)
P00 ( cos θ ) =
120× l!
( 1- cos 2 θ ) 0 ( d
d (cosθ) ) 0 [cos 2 θ -1 ) 0 = 1
∴ Θ 0,0 =√ 2(0)+12
×(1)× 1√2 π
= √ 12
× 1√2 π
= 1
√ 4 π
** angular function of p orbital( l=1, m=0)
P10 = ½ (
ddθ ) [cos 2 θ -1 ) l× e0 = ½ [ 2 cosθ ] = cosθ
∴ Θ 1,0 = √ 2(1)+12
× (1) × 1√2 π
= √ 32
×1√2 π
= 1
√ 4 π cosθ
3.sp3 hybridization:
The combination of one s- orbital and three p- orbitals , giving three hybrid orbitals Ψ1 , Ψ2 Ψ 3 and Ψ 4 may be expressed as
Ψ1 = a1 s + b1 px + c1 py + d1pz ----------------1Ψ2 = a2 s + b2 px + c2 py + d2pz ---------------2 Ψ3 = a3 s + b3 px + c3 py + d3pz -----------------3Ψ4 = a4 s + b4 px + c4 py + d4pz
The values of coefficients can be determined by the following considerations.Ψ1 , Ψ2 , Ψ3 and Ψ 4 are normalized , orthogonal and equivalent. Since the four hybridized orbitals are equivalent, the share of s functions is equal
∴ a12 = a2
2 = a32 = a4
2 = 14 ----------4
a1 = a2 = a3 = a4 = 12
Let c1 may be assigned to fixed direction. Therefore c1 = 0
To find Ψ1
Since Ψ1 is normalized, a1
2 + b12 + c1
2 + d12 = 1
(12 )2 + b1
2 + c12 + d1
2 = 1
14 + b1
2 + c12 + d1
2 = 1
b12 + c1
2 + d12 = 1 -
14
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b12 + c1
2 + d12 =
34 ----------------------------------5
The first of the four hybrids can be chosen arbitrarily to lie in any direction. let it be along x- direction. along this direction c1 = 0, d1 = 0.
∴ b12 + 0 +0 =
34
b1 = √34
Thus the wave function is
To find Ψ2
Since Ψ1 and Ψ2 are orthogonal a1a2 + b1 b2 + c1c2 + d1d2 = 0
(12 ) (
12) + (
√ 3√4
) b2 + 0 +0 = 0
14 + (
√ 32 ) b2 = 0
( √ 32 ) b2 = -
14
b2 = −14 ×
2√ 3
= −12 √ 3
Since Ψ1 is normalized, a2
2 + b22 + c2
2 + d22 = 1
Since along the xy- plane , the orbital py has no contribution, c2 = 0,
(12 )2 + (
−12√ 3
¿2 + 0 + d22 = 1
14 +
112 + d2
2 = 1
412 + d2
2 = 1
13 + d2
2 = 1
d22 = 1 -
13
= 23
d2 = √ 23
Thus the wave function is
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Ψ1 = 12 s + √3
4 px
Ψ2 = 12 s - (
12 √ 3 ) px +
√ 2√3
pz
2020 - CONCISE QUANTUM MECHANICS 2017
To find Ψ3
Since Ψ1 and Ψ3 are orthogonal a1a3 + b1 b3 + c1c3 + d1d3 = 0
(12 ) (
12) + (
√ 3√4
) b3 + 0 +0 = 0
14 + (
√ 32 ) b3 = 0
( √ 32 ) b3 = -
14
b3 = - 14 ×
2√ 3
b3 = −12 √ 3
Since Ψ2 and Ψ3 are orthogonal a2a3 + b2 b3 + c2c3 + d2d3 = 0
(12 ) (
12) + (
−12 √ 3 ) (
−12√ 3 ) + 0 + (√
23 )d3 = 0
14 +
112 + (√
23 )d3 = 0
412 + (√
23 )d3 = 0
13 + (√
23 )d3 = 0
d3 = −13 ×
√ 3√ 2
= −1
√ 3 √ 3× √ 3√ 2
= −1√ 6
Since Ψ3 is normalized, a32 + b3
2 + c32 + d3
2 = 1
(12 )2 + (
−12√ 3
¿2 + c32 +(
−1√ 6 ) 2 = 1
14 +
112 + c3
2 + 16 = 1
612 + c3
2 = 1
12 + c3
2 = 1, ∴ c3 = √ 12
Thus the wave function is
To find Ψ4
Since Ψ1 and Ψ4 are orthogonal a1a4 + b1 b4 + c1c4 + d1d4 = 0
(12 ) (
12) + (
√ 3√4
) b4 + 0 +0 = 0
14 + (
√ 32 ) b4 = 0
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Ψ3 = 12 s - ( 1
2 √ 3 ) px + 1√2
py - 1
√ 6 pz
2020 - CONCISE QUANTUM MECHANICS 2017
( √ 32 ) b4 = -
14
b4 = - 14 ×
2√ 3
b4 = −12 √ 3
Since Ψ2 and Ψ4 are orthogonal a2a4 + b2 b4 + c2c4 + d2d4 = 0
(12 ) (
12) + (
−12 √ 3 ) (
−12√ 3 ) + 0 + (√
23 )d4 = 0
14 +
112 + (√
23 )d4 = 0
412 + (√
23 )d4 = 0
13 + (√
23 )d4 = 0
d4 = −13 ×
√ 3√ 2
= −1
√ 3 √ 3× √ 3√ 2
= −1√ 6
Since Ψ3 and Ψ4 are orthogonal a3a4 + b3 b4 + c3c4 + d3d4 = 0
(12 ) (
12) + (
−12 √ 3 ) (
−12√ 3 ) + (√
12 ) c4 + (
−1√ 6 )(
−1√ 6 ) = 0
14 +
112 + (√
12 ) c4 + (
16 ) = 0
612 + (√
12 ) c4 = 0
12 + (√
12 ) c4 = 0
(√ 12 ) c4 = -
12
c4 = - 12 ×
√ 21
= - 1
√2 √ 2 ×
√ 21
∴ c4 = - 1
√ 2
Thus the wave function is
Thus the wave function of hybrid orbitals are
Ψ1 = 12 s + √
34 px
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Ψ4 = 12 s - ( 1
2 √ 3 ) px - 1√2
py - 1
√ 6 pz
2020 - CONCISE QUANTUM MECHANICS 2017
Ψ2 = 12 s - (
12 √ 3 ) px +
√ 2√3
pz
Ψ3 = 12 s - (
12 √ 3 ) px +
1√2
py - 1
√ 6 pz
Ψ4 = 12 s - (
12 √ 3 ) px -
1√2
py - 1
√ 6 pz
Directional characteristics:
The angular function of 2s orbital = 1
2 √ π ,
2px orbital. = (√ 3
2√ π sin θ cos φ) , 2py orbital. = ( √ 3
2√ π sin θ sin φ) , 2pz orbital = ( √ 3
2√ π cos θ),
Substituting the values
Ψ1 = 12 (
12√ π ) + √
34 (
√ 32 √ π sin θ cos φ)
Ψ2 = 12 (
12√ π ) - (
12 √ 3 ) (
√ 32 √ π sin θ cos φ) +
√ 2√3
( √ 3
2√ π cos θ)
Ψ3 = 12 (
12√ π ) - (
12 √ 3 ) (
√ 32 √ π sin θ cos φ) +
1√2
( √ 3
2 √ π sin θ sin φ) - 1
√ 6 ( √ 3
2 √ π cos θ)
Ψ4 = 12 (
12 √ π ) - (
12 √ 3 ) (
√ 32 √ π sin θ cos φ) -
1√2
( √ 3
2 √ π sin θ sin φ) - 1
√ 6 ( √ 3
2 √ π cos θ)
Let Ψ1 be along the x- axis , the direction of Ψ2 which lies in the xz – plane can be determined as
follows
Ψ2 = 12 (
12√ π ) - (
12 √ 3 ) (
√ 32 √ π sin θ cos φ) +
√ 2√3
( √ 3
2√ π cos θ)
= (1
2√ π ) [12 - (
12 √ 3 ) (√ 3 sin θ cos φ) +
√ 2√3
√3 cos θ)]
= (1
2√ π ) [ 12 - (
12 sin θ cos φ) + √2cos θ)]
Let f = 12 -
12 sin θ cos φ + √2cos θ]
In the xz- plane φ = 0 for positive lobe and φ = 180 for negative lobe and cos 180 = -1.
Therefore for negative lobe, the above equation becomes f = 12 +
12 sin θ + √2cos θ
Differentiating with respect to ‘θ’
dfdθ =
12 cos θ + √2 ( - sin θ)
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Put dfdx = 0
12 cos θ = √2 sin θ
sinθcosθ =
12√ 2
Tan θ = 19 o 28 ‘
Therefore the function f2 makes an angle of 19 o 28 ‘ or 90 + 19 o 28 ‘ = 109 o 28 ‘
LINEAR COMBINATION OF ATOMIC ORBITAL- MOLECULAR ORBITAL (LCAO-MO)
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-TREATMENT OF DIATOMIC MOLECULES
According to LCAO theory, the molecular orbitals are formed by linear combination of atomic orbitals
(AOs) .For effective combination the following three conditions should be satisfied.
1. The AOs of matching symmetry with respect to bond axis can combine
2. The AOs of matching energy can combine
3. The AOs that can overlap can combine.
Consider a diatomic molecule AB. Let ΦA and φB be the atomic orbitals.Tthe linear combination of atomic orbitals
leads to the MO which may be written as
Ψ = a1 φA+ a2 φB -------------------------------1
Where φ1 and φ2 are the atomic orbitals and . The secular equations are
a1 ( HAA – ESAA) + a2 ( HAB – ESAB) = 0 ----------------------------------2
a1 (HBA – ESBA ) + a2 ( HBB – ESBB) = 0 ------------------------------------3
The equations have non-trivial solution only if the secular determinant is equal to zero.
HAA – ESAA HAB – ESAB
HBA – ESBA HBB – ESBB 0
HAA = α A, HBB = α B ( Coloumb integral) , HAB = HBA = β ( exchange integral)
SAA = SBB = 1 , SBA = SAB = S ( overlap integral)
Substituting in the above determinant,
α A- E β-ES
β-ES α B - E = 0
1.THE HYDROGEN MOLECULE ION¿ )
This is the only one molecular system for which Schrodinger equation has been solved. The
Schrodinger equation for H- atom is solved by separating it into spherical co-ordinates with nucleus at
the centre, while the equation for H 2+¿¿ is separated in elliptical co-ordinates with the two nuclei at the
two foci of the ellipse. The wave functions of H – atom are atomic orbitals and those of H 2+¿¿ are called
molecular orbitals. The wave function of H 2+¿¿ is given by
Ψ = a1 φ1+ a2 φ2
Where φ1 and φ2 are the atomic orbitals and rA and rB are the distances of the electron from the
nuclei A and B respectively. The secular equations are
a1 ( HAA – ESAA) + a2 ( HAB – ESAB) = 0 r A r B
a1 (HBA – ESBA ) + a2 ( HBB – ESBB) = 0 R
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HBH A
2020 - CONCISE QUANTUM MECHANICS 2017
The equations have non-trivial solution only if the secular determinant is equal to zero.
HAA – ESAA HAB – ESAB
HBA – ESBA HBB – ESBB 0
HAA = HBB = α, ( Coloumb integral) HAB = HBA = β ( exchange integral)
SAA = SBB = 1 , SBA = SAB = 0 ( overlap integral)
Substituting in the above determinant,
α- E β
β α- E = 0
dividing by β
α−Eβ 1
1 α−E
β = 0
Put α−E
β = x
x 1
1 x = 0
x 2 = 1
x =± 1
when x = +1, let E = E1
α−E1
β=+1
α−E 1=β
∴ E1 = α- β
when x = -1, let E = E2
α−E2
β=−1
∴ E2 = α + β
From the symmetry point of view , the only Molecular Orbitals ( MO) that can be constructed are
Ψ 1 = (φ1+ φ2)
Ψ 2 = (φ1- φ2 ) .
E = ∫(φ 1+φ2¿¿1)H (φ 1+φ 2 )dτ
∫ (φ1+φ 2 )(φ 1+φ 2)dτ¿ Where H = -
12 ∇2 -
1r A
- 1rB
+ - 1R
φ1 = e−r A
√ π and φ2 = e−rB
√ π
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2020 - CONCISE QUANTUM MECHANICS 2017
Ψ1 = 1
2(1+S )1 /2 (φ1+ φ2 )
Ψ2 = 1
2(1−S)1/2 (φ1 - φ2 )
2. HYDROGEN MOLECULE ( H2)
Consider the hydrogen molecule is made up of HA and HB. The wave functions of Hydrogen
molecule is given by the Linear Combination of Atomic Orbitals( LCAO). which is
Ψ = a1 SA + a2 SB ----------------------------------1
Where SA represents the s- orbital of hydrogen A and SB represents the s- orbital of hydrogen B.and a1
and a2 are normalization constants.
The secular equations are
a1 ( HAA – ESAA) + a2 ( HAB – ESAB) = 0 e1 e2
a1 (HBA – ESBA ) + a2 ( HBB – ESBB) = 0
The equations have non-trivial solution only if the secular determinant is equal to zero.
HAA – ESAA HAB – ESAB
HBA – ESBA HBB – ESBB 0
Since both electrons are indistinguishable, HAA = HBB
Since H is a Hermitian operator HAB = HBA
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2020 - CONCISE QUANTUM MECHANICS 2017
SAA = SBB = 1 , SBA = SAB = 0 ( overlap integral)
Substituting in the above determinant,
HAA – E HAB – ES
HAB – ES HAA – E 0
( HAA – E ) 2 = ( HAB – ES ) 2
∴ HAA – E = ± HAB – ES
Let E = E1 , when positive value is taken, therefore the above equation becomes
HAA – E1 = + HAB – E1S
Rearranging, HAA – HAB = E1 - E1S
= E1 ( 1- S )
∴ E1 = H AA – H AB
(1−S)
Let E = E2 , when negative value is taken, therefore the above equation becomes
HAA – E2 = - ( HAB – E2S)
Rearranging, HAA + HAB = E2 + E2S
= E2 ( 1 + S )
∴ E2 = H AA+– H AB
(1+S )
Thus the energy levels are E1 = H AA – H AB
(1−S), E2 =
H AA+– H AB
(1+S )
E2
E
E1
To find the normalization constants, let us apply the condition for normalization
∫ΨΨ dt = 1
∫(a1 S A+a2 SB)(a1 SA+a2 SB )dt=¿¿ 1
∫ (a1 S A+a2 SB )2dt=¿¿ 1
∫ (a1 S A )2 dt + ∫ (a2 SB )
2 dt + 2 ∫ a1 SA a2 SB dt=¿¿ 1
a12∫ (SA )
2 dt + a22∫ (SB )
2 dt + 2 a1a2∫S A SB dt=¿¿ 1
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∫ (S A )2 dt = 1, ∫ (SB )
2 dt = 1, ∫ SA SB dt=¿¿ S
Therefore the above equation becomes
a12+ a2
2+ 2 a1a2 S = 1
Since both are electrons, a1=a2 = A
A 2 + A2 + 2A2S = 1
2A2 +2A2S = 1
2A2 ( 1 +S) = 1
A2 = 1
2(1+S )
A = 1
√2(1+S¿)¿ substituting in 1, we get
Let Ψ + be the for the symmetric wave function
Ψ+ = 1
√2(1+S¿)¿ SA +1
√2(1+S¿)¿ SB
= 1
√2(1+S¿)¿ ( SA + SB )
Similarly the anti symmetric wave function is given by
Ψ- = 1
√2(1−S¿)¿ ( SA - SB )
Electron density:
The electron density of electron -1 at any point is obtained by adding the probabilities of electron -2 at
all locations. Therefore
Electron density( D1) = ∫ ρ12 ρ2
2 dV2
Where dV2 is spatial co-ordinates of electron -2
ρ = 1
√2(1+S¿)¿ ( SA (i) + SB (i) ) where i = 1,2
ρ1 = 1
√2(1+S¿)¿ ( SA (1) + SB (1) ) ,
ρ 2 = 1
√2(1+S¿)¿ ( SA (2) + SB (2) )
Electron density( D1) = ∫¿¿¿ ׿¿dV2
= 1
2(1+S ) (S A (1 )+SB (1 ) )¿¿2 × 12(1+S )∫ ¿¿(2) + SB(2) ] 2 dV2
= 1
2(1+S ) (S A (1 )+SB (1 ) )¿¿2 12(1+S)
¿ ∫ SA2 dV2 + ∫ SB
2 dV2 + 2∫ SA SB dV2 ]
= 1
2(1+S ) (S A (1 )+SB (1 ) )¿¿2 12(1+S)
¿ 1+1+2S ]
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= 1
2(1+S ) (S A (1 )+SB (1 ) )¿¿2 12(1+S)
¿ 2+2S ]
= 1
2(1+S ) (S A (1 )+SB (1 ) )¿¿2 12(1+S)
¿ 2(1+S ) ]
= 1
2(1+S ) (S A (1 )+SB (1 ) )¿¿2
Similarly for electron -2
Electron density( D2) = 1
2(1+S ) (S A (2 )+SB (2 ) )¿¿2
Therefore D = D1 + D2
= 1
2(1+S ) (S A (1 )+SB (1 ) )¿¿2 + 1
2(1+S ) (S A (2 )+SB (2 ) )¿¿2
Since electrons do not carry labels, D1 = D2, SA (1 ) = SA (2 ) = SA , SB (1 ) = SB (2 )= SB
D = 1
2(1+S ) (S A+SB )¿¿
2 +
12(1+S ) (
S A+SB )¿¿2
= 21
2(1+S ) (S A+SB )¿¿
2
= 1
(1+S) (SA+SB )¿¿
2
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MOLECULAR ORBITALS OF H2
.When the 1s wave functions of the two HH atoms are linearly combined, we get a sigma () bonding orbital, denoted as 1s in the diagram here. This approach is called linear combination of atomic orbitals (LCAO), in the MO approach.
In this approach, the sum of the two 1s orbitals (one for each atom) is the bonding orbital. In terms of wave mechanics, the two waves constructively interact.
The difference of the two orbitals forms the antibonding orbital, 1s*, due to destructive interference. It is interesting to note that the anti bonding orbital is at a higher energy than the 1s atomic orbital.
The energy level can be represented below:
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Each orbital accommodates two electrons, and the two electrons in H−H fill the 1s molecular orbital (MO). Obviously, as a result of the formation of the H2molecule, the energy of the system is lowered and becomes more stable.
.
1. Which one of the following statements is the best description of valence bond theory?a. an overlap of atomic orbitalsb. satisfies the octet rulec. count the number of VSEPR pairsd. the overlap of atomic orbitalse. atomic orbitals undergo a transformation
Hint: d.
Discussion: The simple version assumes that atomic orbitals do not change during the formation of chemical bonds. They merely overlap. Modern treatments of valence bond theory do accept the fact that atomic orbitals do change a little.
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VALENCE BOND THEORY FOR HYDROGEN MOLECULE(HEITLER – LONDON THEORY)
Consider the hydrogen molecule is made up of HA and HB. . Let A and B represent the two H-
atoms , 1 and 2 represent the two electrons and S represents the ‘s’ orbital.
Let us assume that electron 1 is associated with atom A and electron 2 is associated with atom B then the wave function of this state is given by Φ1 = SA(1)SB(2) -----------------1
If the electron 1 is associated with atom B and electron 2 is associated with atom A then the wave function of this state is given by Φ2 = SA(2)SB(1) --------------------2
The wave functions of Hydrogen molecule is given by the Linear Combination of Atomic Orbitals( LCAO). which is Ψ = a1 Φ1 + a2 Φ2 where a1 and a2 are normalization constants
Substituting , Ψ = a1 [SA(1)SB(2) ] + a2 [SA(2)SB(1) ] --------------3
To find the normalization constants’a1 ‘ and ‘a2’:
The condition for normalised wave function is
∫−∞
∞
ΨΨ * dτ = 1
∫−∞
∞
{a1[S A (1) S B (2)]+a2[S A (2) SB (1)] } {a1 [ S A (1 )S B (2 ) ]+a2 [S A (2 )S B (1 ) ] }dτ = 1
∫−∞
∞
a1 [S A(1)S B(2)]+a2[S A(2)S B(1)]2 dτ = 1
a12 ∫
−∞
∞
[S A(1)S B(2)]¿¿2 dτ + a22 ∫−∞
∞
[S A(2)S B(1)]¿¿2 dτ + 2 a1 a2 ∫−∞
∞
[S A(1)S B(2)]S A (2)S B(1)¿
dτ1 dτ2 = 1
∫−∞
∞
[S A(1)S B(2)] dτ1 = ∫−∞
∞
[S A(2)S B(1)] dτ2 = 1, ∫−∞
∞
[S A(1)S B(2)]S A (2)S B(1)¿ dτ1 dτ2 = S2
Therefore the above equation becomes a12 (1)+ a2
2 (1) + 2 a1 a2 S2 = 1
Since a1 = a2 , a12 + a1
2 + 2 a12 S2 = 1
2a12 + 2 a1
2 S2 = 1
2a12 ( 1 + S2 ) = 1
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a12 =
12 (1+S2)
a1 = 1
√2 (1+S2 )
Since a1 = a2 a2 = 1
√2 (1+S2 )
therefore the function becomes
, Ψ = 1
√2 (1+S2 ) [SA(1)SB(2) ] + 1
√2 (1+S2 ) [SA(2)SB(1) ]
, = 1
√2 (1+S2 ) { [SA(1)SB(2) ] + [SA(2)SB(1) ] }
This is for symmetric wave function. For antisymmetric wave function we have
, Ψ = 1
√2 (1+S2 ) { [SA(1)SB(2) ] - [SA(2)SB(1) ] }
There is only one symmetric wave function
A(1)B(2) + B(1) A(2) [ α (1)β (2) - α (2)β (1)]
This is called singlet state because S= 0, 2S+ 1 = 2(0) + 1 = 1
There are three antisymmetric wave functions
A(1)B(2) - B(1) A(2) [ α (1) α (2)]
A(1)B(2) - B(1) A(2) [ β (1) β (2)]
A(1)B(2) - B(1) A(2) [ α (1)β (2) + α (2)β (1)]
This is called triplet state because S= 1, 2S+ 1 = 2(1) + 1 = 3
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To find energy:
Variation treatment leads to two secular equations
a1 ( H11 – E) + a2 ( H12 – ES 2 ) = 0
a1 (H21 – ES2 ) + a2 ( H22 – E ) = 0
on solving the 2× 2 determinant, we get
E + = H 11+H 12
2 (1+S2 ) , E - = H 11−H 12
2 (1−S2 )
To find H11 and H12:
The Hamiltonian is given by
H = - ½ ∇12 -
1r A1
- ½ ∇22 -
1rB2
- 1
r A 2 -
1rB 1
+ 1r12
+ 1R
Where 1
r A 1,
1r A 2
, 1
rB1 ,
1rB 2
, 1
r12 and
1R represents the distances.
Let HA = - ½ ∇12 -
1r A1
, HB = - ½ ∇22 -
1rB 2
, H’ = - 1
r A 2 -
1rB1
+ 1
r12 +
1R
Therefore the above equation can be written as
H = HA + HB + H’
H11 = E HA + EHB + ∫−∞
∞
Φ1 H ' Φ1 dτ
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E HA = EHB = EH
∴ H11 = 2E H + ∫−∞
∞
Φ1 H ' Φ1 dτ
= 2E H + ∫−∞
∞
Φ1(−1r A 2− 1
r B1+ 1
r12 )Φ1 dτ+ ∫−∞
∞
Φ1( 1R )Φ1 dτ
= 2E H + ∫−∞
∞
Φ1(−1r A 2− 1
r B1+ 1
r12 )Φ1 dτ+ 1R ∫
−∞
∞
Φ1Φ1 dτ
= 2E H + ∫−∞
∞
Φ1(−1r A 2− 1
r B1+ 1
r12 )Φ1 dτ+ 1R (1) [∫
−∞
∞
Φ1 Φ1 dτ = 1]
Let ∫−∞
∞
Φ1(−1r A 2− 1
r B1+ 1
r12 )Φ1 dτ = J, then the above equation becomes
H11 = 2E H + J + 1R
Similarly
H12 = E HA + EHB + ∫−∞
∞
Φ1 H ' Φ2 dτ
E HA = EHB = EH S2
∴ H12 = 2E H + ∫−∞
∞
Φ1 H ' Φ2 dτ
= 2E H S2 + ∫
−∞
∞
Φ1(−1r A 2− 1
r B1+ 1
r12 )Φ 2 dτ+ ∫−∞
∞
Φ1( 1R )Φ2 dτ
= 2E H S2 + ∫
−∞
∞
Φ1(−1r A 2− 1
r B1+ 1
r12 )Φ 2 dτ+ 1R ∫
−∞
∞
Φ1 Φ2 dτ
= 2E H S2 + ∫
−∞
∞
Φ1(−1r A 2− 1
r B1+ 1
r12 )Φ 2 dτ+ 1R (S2) [∫
−∞
∞
Φ1 Φ2 dτ = S2]
Let ∫−∞
∞
Φ1(−1r A 2− 1
r B1+ 1
r12 )Φ1 dτ = K, then the above equation becomes
H12 = 2E H S2 + K + S2
R
E + = 2 EH+J + 1
R+2 EH S2+K+ S2
R2 (1+S2 )
= 1
2 (1+S2) { 2 EH ( 1 + S2 ) + J + K + 1R ( 1 + S2 ) }
= EH + J+K
2 (1+S2) + 1R ( ½ ?)
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Similarly,
E - = H11−H12
2 (1−S2 )
= EH + J−K
2 (1−S2 ) + 1R
E + = 2 EH + J+K(1+S2 ) +
1R
E - = 2 EH + J−K(1−S2 ) +
1R
First two terms in the right side constitute electrostatic energy and the third term is nuclear repulsion
energy.
∆ E+ = J+K(1+S2 )
It is found out that K is more negative than J
Therefore ∆ E+ is negative for al values of R. it represents the stable state.
∆ E- is positive representing repulsive state
The ∆ E+ plot shows that two H- atoms attract each other as they approach , passes through a state of
minimum energy and begin to repel when they are very close.
The plot ∆ E- never shows minimum
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SLATER TYPE ORBITALS
Hydrogen like orbitals are given by
Ψ nlm = N r F(r) e−Zr
n Y lm ( θ,φ)
Here N is the normalisation constant, Rnl is the radial function and Y lm
( θ,φ) is the angular part ( spherical harmonics)Slater suggested an orbital by taking into consideration ,the shielding effect of the nucleus by the
electrons.it is given by
Ψ’ nlm = N r n-1 e−Z ' r
n ' Y lm ( θ,φ)
Where n’ is the effective principal quantum number and Z’ is the effective nuclear charge.Functions of these oritals are called ‘Slater Type Orbitals’ (STO)
1. STO differ from H- like orbitals only in radial part.2. Unlike H-like orbitals, STOs have no radial nodes3. STO’s are not mutually orthogonal
Find the expression for STO for 2s, orbital in Nitrogen atom.. The normalisation constant is ‘N’. Compare with Hydrogen – like orbital.Solution:
Slater Type Orbitals is given by Ψ’ nlm = N’ r n-1 e−Z ' r
n ' Y lm ( θ,φ)
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For 2s orbital, n= 2 , l=0. ∴ Ψ’ 200 = N r 2-1 e−Z ' r
n ' Y lm ( θ,φ)
Y 00 ( θ,φ) =
1√4 π
To find Z’ ( effective nuclear charge of 2s electrons in N).
Electronic configuration of N : 1s2. (2s 2p ) 5 , These are grouped as (1s) (2s,2p)
There are 5 electrons in set 2
. The effective nuclear charge on one electron
= 4×(0.35) [ rule 3 for remaining 4 electrons + 2(0.85)[ rule 5 for 2 electrons]
= 3.10
Z’ = Z-S [ Z is the atomic number of N =7]
= 7- 3.1
= 3.9
n'= n=2
Substituting, we get
Ψ’ 200 = N r 2-1 e−3.9r
2 1√4 π
= N r e−3.9r
2 1√4 π
Hydrogen – like orbital: radial function of 2s orbital is given by
Rn,l = √( 2 zna 0 )3 × (n−l−1 )!
2n [ (n+l )! ]3 × ( 2 zr
na 0 ) l × e –( zrna 0 ) × Ln+l
2 l+1 ( 2 zrna 0 )
Ln+l2 l+1 = (
ddρ ) s [ e ρ (
ddρ ) r ( ρ r e –ρ ) ] where r= n +l ,s = 2l +1 and ρ = ( 2 zr
na 0 )For 2s orbital n= 2, l=0 ∴ r=n +l = 2 +0 = 2 and s = 2l +1 = 0 +1 = 1
∴ Lrs (ρ) = (
ddρ ) 1 [ e ρ ×
ddρ 2 ( ρ 2 e –ρ )) ]
= ( ddρ ) 1 [ e ρ ×
ddρ [ 2ρ e –ρ - ρ 2e –ρ ]
= ( ddρ ) 1 [ e ρ [ 2[ρ (- e –ρ
) + e –ρ (1) ] - [ 2ρ e –ρ + ρ 2 ( - e –ρ ) ]
= ( ddρ ) 1 [ -2ρ + 2 -2ρ + ρ 2 ]
= -2 +0 -2 + 2 ρ
= (2 ρ - 4 )
∴ Rn,l = N × ( 2 zrna 0 ) 0 × e –( zr
2a 0 ) × (2 ρ - 4 )
= N × e –( zr2a 0 ) × (2 ρ - 4 )
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R 1.0 = N × e –( zr2a 0 )× (2 ρ - 4 )
Substituting the value of ρ = ( 2 zrna 0 )
R 1.0 = N × e –( zr2 a0 )× 2 (( 2 zr
2a 0 ) - 2 ).
= N × (2 - -zra0 ) × e –( zr
2a 0 ) Ψ’ 200 = N × (2 - -
zra 0 ) × e –( zr
2 a 0 ) 1√4 π
a0 = 1 in atomic mass unit
∴ Ψ’ 200 = N × (2 - zr ) × e –( zr2 ) 1
√4 π
Slater Type Orbital ΨS 200 = N× r ×e−3.9r
2 × 1√4 π
Hydrogen like orbital Ψ 200 = N× (2 - 3.9 r ) × e –( 3.9 r2 ) 1
√4 π
Comparison:
There is no node in STO, but in HLO node occurs at 2
3.9
------------------------------------------------------------------------------------
Y lm ( θ,φ) = √ 2 l+1
2× (l−m ) !
( l+m ) ! × ( 1-x2) m/2 ×
12l . l !
d l
d x l ( x2 – 1) l ×
1√2 π
e imφ
First harmonic:
Y 00 ( θ,φ) =√ 2(0)+1
2× 0 !
(0 ) ! × ( 1-x2) 0 ×
120 .0 !
d0
d x0l ( x2 – 1) 0 ×
1√2 π
e 0
= 1√2
× 1√2 π
= 1√4 π
Find the expression for STO for 2px, orbital in Nitrogen atom.. The normalisation constant is ‘N’. Compare with Hydrogen – like orbital.Solution:
The angular wave function is
Θ l,m = √ 2 l+12
× (l−m ) !(l+m ) !
× Plm ( cos θ ) ×
1√2 π
e imφ
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Plm ( cos θ ) =
12l× l!
( 1- cos 2 θ ) m/2 ( d
d (cosθ) ) l+m [cos 2 θ -1 ) l
For 2pz orbital.,l l=1, m=0
P10 =
1211 !
( 1- cos 2 θ ) 0 ( d
d (cosθ) ) [cos 2 θ -1 ) 1× e0
= 12 (
dd (cosθ) )
[cos 2 θ -1 )
= 12 [ 2 cos θ ]
= cos θ
∴ Θ l,m = √ 2(1)+12
× cos θ × 1√2 π
= √ 32
×1√2 π
cos θ
= √ 3
√ 4 π cos θ
Slater Type Orbital ΨS 200 = N× r ×e−3.9r
2 × √ 3
√ 4 π cos θ
Hydrogen like orbital Ψ 200 = N× (2 - 3.9 r ) × e –( 3.9 r2 ) √ 3
√ 4 π cos θ
Comparison:
There is no node in STO, but in HLO node occurs at 2
3.9
Calculate the average value of ‘r’ for 1s electron of Lithium atom using Slater type orbitals.Solution:
Average of r = ∫Ψ rΨ dT
∫ΨΨ dT
= ∫(N ’rn−1 e−Z 'r
n ' )r ¿¿¿ STO Ψ’ nlm = N’ r n-1 e−Z ' r
n ' Y lm ( θ,φ)
= ∫r2 n−1 e
−2 Z 'rn ' r2 dr
∫ ’r2 n−2 e−2 Z ' r
n ' r2 dr [Y l
m ( θ,φ) will be cancelled]
= ∫ r2n+1 e
−2 Z 'rn ' dr
∫ ’r2n e−2Z' r
n ' dr
= (2 n+1)!¿¿
× ¿¿ Formula : ∫0
∞
xn e−ax dx = n!
an+1
= (2n+1 )× 2n !¿¿
× 1
2n! (n+1) ! = (n+1) n!
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= (2n+1 )× n2 z
For lithium atom n = 3, for is orbital Z = z’ – 1(0.3) = 3-0.3 = 2.7
Average of r = 215.4
=3.6
Calculate the first ionisation energy for Li atom on the basis of Slater rules.
Solution:
I.P of Li = E of Li + - E of Li
E= ∑ 12
z2
n2 in a.u = ∑ 13.6 z2
n2 e.v
SLATER RULES
1. The effective principal quantum number (n’) is related to the quantum number (n) as follows
2. The effective nuclear charge (Z’) is related with actual atomic number as
Z’ = Z-S, where ‘S’ is screening constant
3. Screening constant is calculated as follows.
1. The orbitals are grouped as
(1s), (2s,2p), (3s,3p), (3d),(4s,4p),(4d)
2. No contribution from electrons in groups higher than , the group in question
For example For 3s or 3p electrons , no contribution from 3d,4s,4p,4d
3. Each electron in the same group contributes 0.35
4. Electron in 1s group contributes 0.30 only
5. In case of ‘s’ or ‘p’ group, each electron in the next lower group contributes 0.85
6. For electron in ‘d’ or ‘f’ the contribution from each inner electron is 1.0
7. Each electron in groups lower by two or more numbers from the one in question contributes 1.0
1. Determine the effective nuclear charge for the 1s electron in He.
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n 1 2 3 4 5 6
n' 1 2 3 3.7 4 4.2
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Electronic configuration of He : 1s2. There are two electrons. The effective nuclear charge on one electron =
1×(0.30) [ rule 4]
S = 0.3
Z’ = Z-S [ Z is the atomic number of He =2]
= 2-0.3
= 1.7
The effective nuclear charge for the 1s electron in He = 1.7 ( no unit)
2. Determine the effective nuclear charge of 2s and 2p electrons in C.
Electronic configuration of C : 1s2. 2s2 2p 2
These are grouped as (1s) (2s,2p)
There are 4 electrons in set 2
. The effective nuclear charge on one electron = 3×(0.35) [ rule 3 for remaining 3 electrons
+ 2(0.85)[ rule 5 for 2 electrons]
= 2.75
Z’ = Z-S [ Z is the atomic number of C =6]
= 6- 2.75
= 3.25
The effective nuclear charge for the 2s electron in C = 3.25 ( no unit)
3. Determine the effective nuclear charge of 2s and 2p electrons in N.
Electronic configuration of N : 1s2. 2s2 2p 3
These are grouped as (1s) (2s,2p)
There are 5 electrons in set 2
. The effective nuclear charge on one electron = 4×(0.35) [ rule 3 for remaining 4 electrons
+ 2(0.85)[ rule 5 for 2 electrons]
= 3.10
Z’ = Z-S [ Z is the atomic number of N =7]
= 7- 3.10
= 3.90
The effective nuclear charge for the 2s and 2p electrons in N = 3.90 ( no unit)
4. Determine the effective nuclear charge of 3s and 3p electrons in S.
Electronic configuration of S : 1s2. 2s2 2p 6 3s2 3p4
These are grouped as (1s) (2s,2p), (3s,3p)
There are 6 electrons in set 3
. The effective nuclear charge on one electron = 5×(0.35) [ rule 3 for remaining 5 electrons
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+ 8(0.85)[ rule 5 for 8 electrons]
+ 2(1.0) [ rule 7 for 2 electrons
= 10.55
Z’ = Z-S [ Z is the atomic number of S =16]
= 16- 10.55
= 5.45
The effective nuclear charge for the 3s and 3p electrons in S = 5.45 ( no unit)
5. Determine the effective nuclear charge of 1s electron in F.
Electronic configuration of F : 1s2. 2s2 2p 5
These are grouped as (1s) (2s,2p)
There are 2 electrons in set 1
. The effective nuclear charge on one electron = 1×(0.30) [ rule 4 for remaining 1 electrons
+ 0)[ rule 1 for other electrons]
= 0.30
Z’ = Z-S [ Z is the atomic number of F = 9]
= 9- 0.30
= 8.70
The effective nuclear charge for the 1s electron in F = 8.70 ( no unit)
Problem1 :Find the expression for STO for 2s, orbital in Nitrogen atom.. The normalisation constant is ‘N’. Compare with Hydrogen – like orbital.Solution:
Slater Type Orbitals is given by Ψ’ nlm = N’ r n-1 e−Z ' r
n ' Y lm ( θ,φ)
For 2s orbital, n= 2 , l=0. ∴
Ψ’ 200 = N r 2-1 e−Z ' r
n ' Y lm ( θ,φ)
= N r 2-1 e−Z ' r
n ' × 1√4 π
[ Y 00 ( θ,φ) =
1√4 π
]
To find Z’ ( effective nuclear charge of 2s electrons in N).
Electronic configuration of N : 1s2. (2s 2p ) 5 , These are grouped as (1s) (2s,2p)
There are 5 electrons in set 2
. The effective nuclear charge on one electron
= 4×(0.35) [ rule 3 for remaining 4 electrons + 2(0.85) [ rule 5 for 2 electrons]
= 3.10
Z’ = Z-S [ Z is the atomic number of N =7]
= 7- 3.1
= 3.9
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n'= n=2
Substituting, we get
Ψ’ 200 = N r 2-1 e−3.9 r
2 1√4 π
= N r e−3.9r
2 1√4 π
This is the expression for STO of N atom
∴ Ψ’ 200 = N × (2 - zr ) × e –( zr2 ) 1
√4 π
Slater Type Orbital ΨS 200 = N× r ×e−3.9r
2 × 1√4 π
Hydrogen like orbital Ψ 200 = N× (2 - 3.9 r ) × e –( 3.9 r2 ) 1
√4 π
Comparison:
There is no node in STO, but in HLO node occurs at 2
3.9Problem 2 Find the expression for STO for 2px, orbital in Nitrogen atom.. The normalisation constant is ‘N’. Compare with Hydrogen – like orbital.Solution:The angular wave function is
Θ l,m = √ 2 l+12
× ( l−m ) !( l+m ) !
× Plm ( cos θ ) ×
1√2 π
e imφ
Plm ( cos θ ) =
12l× l!
( 1- cos 2 θ ) m/2 ( d
d (cosθ) ) l+m [cos 2 θ -1 ) l
For 2pz orbital.,l l=1, m=0
P10 =
1211 !
( 1- cos 2 θ ) 0 ( d
d (cosθ) ) [cos 2 θ -1 ) 1× e0
= 12 (
dd (cosθ) )
[cos 2 θ -1 )
= 12 [ 2 cos θ ]
= cos θ
∴ Θ l,m = √ 2(1)+12
× cos θ × 1√2 π
= √ 32
×1√2 π
cos θ
= √ 3
√ 4 π cos θ
Slater Type Orbital ΨS 200 = N× r ×e−3.9r
2 × √ 3
√ 4 π cos θ
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Hydrogen like orbital Ψ 200 = N× (2 - 3.9 r ) × e –( 3.9 r2 ) √ 3
√ 4 π cos θ
Comparison:
There is no node in STO, but in HLO node occurs at 2
3.9
Problem 3 Calculate the average value of ‘r’ for 1s electron of Lithium atom using Slater type orbitals.Solution:
Average of r = ∫Ψ rΨ dT
∫ΨΨ dT
= ∫(N ’rn−1 e−Z 'r
n ' )r ¿¿¿ STO Ψ’ nlm = N’ r n-1 e−Z ' r
n ' Y lm ( θ,φ)
= ∫r2 n−1 e
−2 Z 'rn ' r2 dr
∫ ’r2 n−2 e−2 Z ' r
n ' r2 dr [Y l
m ( θ,φ) will be cancelled]
= ∫ r2n+1 e
−2 Z 'rn ' dr
∫ ’r2n e−2Z' r
n ' dr
= (2 n+1)!¿¿
× ¿¿ Formula : ∫0
∞
xn e−ax dx = n!
an+1
= (2n+1 )× 2n !¿¿
× 1
2n! (n+1) ! = (n+1) n!
= (2n+1 )× n
2 z
For lithium atom n = 3, for is orbital Z = z’ – 1(0.3) = 3-0.3 = 2.7
Average of r = 215.4
=3.6
Problem 4 Calculate the first ionisation energy for Li atom on the basis of Slater rules.
Solution:
I.P of Li = E of Li + - E of Li
E= ∑ 12
z2
n2 in a.u = ∑ 13.6 z2
n2 e.v
HARTREE – FOCK SELF CONSISTENT FIELD METHOD.(HFSCF –METHOD)
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Hamiltonian for’ n’ electron atom is H = - ½ ∑ ∇2 - ∑ zr + ½ ∑ ∑
1rij
The inter electronic
repulsion term (½ ∑ ∑ 1rij
) does not permit separation of Schrodinger equation in to one electron
equations. Hence the product wave function will not yield correct energy. The best possible wave
function is obtained by using a set of orbitals , which are the solutions of n simultaneous Schrodinger
equations.
Hφ = Eφ, H = - 12 ∇2 -
zr + Vi
Where Vi represents the potential energy of electron, due to repulsion by all other electrons.. To
calculate Vi, Hartree assumed that each electron in the atom moves , in a potential field of all other
electrons. He solved the above equation by iterative technique as follows
Consider the wave function as product of one electron orbitals.
Ψ = (φ1) ( φ2) ( φ3)….
Let electron-1 is moving through the cloud of other electrons. The potential energy of
interaction of electron-1 with all other electrons is given by
V1 = ∑j=2
n ∫φi2d τ
r 1 j where φ is one electron orbital.
On solving the above expression we obtain an improved orbital φ1' .
Now the electron-2 may be considered to move in the field . The new set ( φ1) is now used to
recalculate the potential energy . The potential V2 is calculated as above The potential energy of
interaction of electron-2 with all other electrons is given by
V2 = ∑j=1,3. .
n ∫φ 1' dτr 1 j
where φ is one electron orbital.
On solving the above expression we obtain an improved orbital φ2'
.. Finally we have the wave function as Ψ = (φ1’) ( φ2’) ( φ3’)…. The orbitals are denoted by ( φ) and
are now consistent with the potential field . These are called Self Consistent Field orbitals ( SCF)
Using this value the hamiltonian and hence the energy are calculated.
The cycle of iteration is continued till we get the same value as in the earlier step .
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Ψ = φ2 × φ2
Use φ2
solve
Use φ1;
NO
YES
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[ - ½ ∇12 - ½ ∇2
2 - Zr1
- Zr1
+ ∫ (φ2 )2 d τ
r12
] φ1 = E φ1
φ1; ( improved orbital)
[ - ½ ∇12 - ½ ∇2
2 - Zr1
- Zr1
+ ∫ (φ1; )2d τ
r12
] φ2 = E φ2
φ2; ( improved orbital)
Are the
values
same ?
STOP
2020 - CONCISE QUANTUM MECHANICS 2017
F( x) = x2 – 6x +5x2 – 6x +5 = 0 6x = x2 +5
x = x 2+5
6
x1= f(0) = (0 )2+56
= 0. 83
x2 = f( 0.833) = (0.83 )2+56
= 0.68+5
6 = 0.94
x2 = f( 0.94) = (0.94 )2+56
= 0.88+5
6 = 0.98
x2 = f( 0.98) = (0.98 )2+56
= 0.96+5
6 = 0.99
x2 = f( 0.99) = (0.99 )2+56
= 0.92+5
6 = 0.99
till we get value equal to previous value
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Hartree Fock ‘s theory: Hartree’s method suffers from a drawback that the orbital product wave
function is not antisymmetric. Fock introduced anti symmetrised wave function and followed the same
iterarative procedure. The orbitals obtained are called HFSCF AO ‘s
The antisymmetrised product wave function is expressed as determinant.`
Ψ = 1√n!
φ1(1) φ 1(2) φ2 (3) φ2 (4)... φ ( n)
Where φ1(1) is a spin orbital with e1 and α – spinφ 1(2) is the spin orbital with e2 and β – spin in the
same orbital φ1. He introduced Fock’s operator ( F) in place of H to include electron repulsion and
electron exchange.
The Fock operator is defined as F = - ½ ∑ ∇2 - ∑ zr + ∑ [ 2 Ji – Ki]
Ji is called coloumb operator and Ki is called exchange operator
The hartree (symbol: Eh or Ha), also known as the Hartree energy, is the atomic unit of energy,
. It is defined as 2R∞hc, where R∞is the Rydberg constant, h is the Planck constant and c is the speed of
light.
Eh = 4.359 744 650(54)×10−18 J Page 343 of 419
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= 27.211 386 02(17) eV.[1]
The hartree energy is the electric potential energy of the hydrogen atom in its ground state and, by
the virial theorem, approximately twice its ionization energy
SEMI EMPRICAL METHODS
In HFSCF method, large number of integrals need to be evaluated. For example, even in a small molecule like CH4, we have to evaluate 1035 two electron integrals. Hence semi empirical methods with drastic approximations have been developed.
1. Neglect of Valence Electron: In this method all inner shell electrons are ignored. For example in C,N, O only 2s 2p atomic orbitals are considered. This is effected by replacing the nuclear charge Zp by effective nuclear charge Z’p in the Hamiltonian operator. Hamiltonian H =
The Fock operator =
2. Zero Differential Overlap ( ZDO) : Any product of atomic orbitals within an integral is zero if they are different. Φ μa ×φ μb = 0 if μa ≠ μb Here μa and μb refer to atomic orbitals of atomic centre Aand B.This condition is known as Zero Differential Overlap ( ZDO) condition.
3. Neglect of Diatomic Differential Overlap. (NDDO) In this scheme All overlap integrals, in which μa ≠ μb and all 2-electron integrals are taken as zero.
∫ μa (1 )μb (1 ) 1r12
μa (2 ) μb (2 )dt 1dt 2 = 0
This is known as Neglect of Diatomic Differential Overlap. (NDDO)
4. Intermediate Neglect od Differential Overlap ( INDO) In this scheme S μa μb = 0 if μa ≠ μb S μa μb = 0
∫ μa μbdτ = 0
This is known as Intermediate Neglect od Differential Overlap ( INDO)5. Complete Neglect of Differential Overlap ( CNDO)
In this scheme Both Overlap integral and non-coloumbic integral are zero.
S μa μb = 0 if μa ≠ μb and This is known as Complete Neglect of Differential Overlap ( CNDO)Hartree – Fock –Roothan (HFR) method
PAULIS ANTISYYMETRIC PRINCIPLEPage 344 of 419
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particles whose wave functions which are symmetric under particle interchange have integral or zero intrinsic spin, and are termed bosons.
Particles whose wave functions which are anti-symmetric under particle interchange have half-integral intrinsic spin, and are termed fermions.
Space wave function of two particles
Symmetrical
Antisymmetrical
MULTIPLE CHOICE QUESTIONS
TEST - 1 1. Corpuscular theory does not explain a. interference b.diffraction c. both
2. Wave theory of light was given bya. Newton b. Hygen c.Planc
3. Electromagnetic theory was given bya. Hygen b. Maxwell c. Davisson
4. Quantum theory was given bya. Planc b. Hygen c. Maxwell
5. Which is not true. Classical mechanics does not explain a. stability of atom. b. the spectrum of hydrogen. c. motion of objects .
6. Black body radiation is governed bya. Weins law b. Stefans law c. both
7. Intensity of blackbody radiation a. increases b. decreases c. increases up to a maximum then decreases
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8. Distribution of energy of black body radiation isa. uniform b. not uniform c not predictable
9. As temperature increases the peak energy shifts towardsa. Shorter wavelength b. longer wavelength c. no effect
10. Wein’s law fails at a. Lower wavelength b. higher wavelength c. low temperature
11. Rayleigh Jeans law holds goods at a.Lower wavelength b. higher wavelength c. low temperature
12. Classical theory fails to explain a. Continuous emission of radiation b. Continuous absorption of radiation c. both
13. According to Planc energy spectrum is due to a. Continuous energy b. discrete energy c. both
14. Planc formula is
a. E= 8 π
λ5 ehϑKT
b. E= 8 πhc
ehϑKT
c. E= 8 πhc
λ5 ehϑKT
15. According to Planc the quantity , quantised isa. wavelength b. energy c. frequency
16. Photo electric effect is emission of a. protons b.electrons c. X-rays
17. Photo electrons are emitted when the metal surface is exposed toa. X-rays b. UV -rays c. both
18. Photo electrons are emitted froma. Anode b. cathode c. both
19. Photo electric effect occurs at a. All frequecies b.low frequencies c.threshold frequencies
20. Number of Photo electrons are emitted is proportional toa. intensity of incident radiation b. temperature c. both
21. Kinetic energy of photo electron is proportional toa. frequency of incident radiation b. temperature c. both
22. In photo electronic devices light energy is converted in to a. Mechanical energy b. potential energy c. electrical energy
23. Counting machine usesa. Photo electric effect b.compton effect c.stark effect
24. Photo electric effect is used ina.alarms b. counting machines c. both
25. Photo electric work function is given bya. W= hϑ0 b. W= 2h c. W= ϑ0
26. The expression for energy of H- atom is
a. −me4
8 n2 h2∈2 b. −m
8 n2 h2∈2 c. −me4
n2 h2∈2
27. The wave length of Hydrogen spectrum is
a.1λ = RH [ 1
n f2−
1ni
2 ] b. 1λ = RH RH [ 1
n f2 ] c.
1λ = RH RH [ 1
n f2−
1ni
3 ]
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28. In the expression 1λ = R [ 1
n f2−
1ni
2 ], R is called
a. Gas constant b. Bohr constant c. Rydberg constant29. The expression for Rydberg constant is
a. m e4
8 ch3∈2 b. −m
8 n2 h2∈2 c. −me4
n2 h2∈2
30. The relation between Energy and Rydberg constant is
a. E = Rhn2 b. E =
−Rchn2 E =
Rchn2
31. The value of Rydberg constant isa. 10.97 b. 10.97 × 10 6 m c. 10.97 × 10 6 m -1
32. Lyman series is
a.1λ = R [12− 1
ni2 ] b
1λ = R [ 11− 1
ni2 ] c.
1λ = R [13− 1
ni2 ]
33. When electron jumps from second orbit to first orbit the line obtained is a. Balmer b. Lyman c. Pfund
34. When electron jumps from fifth orbit to first orbit the line obtained is b. Balmer b. Lyman c. Pfund
35. Lyman series occurs ina. UV region b. IR region c. far IR region
36. The first member of Lyman series has wavelength
a. λ = 4
3 R b. λ = 3
4 R c. λ = 4
6R37. The second member of Lyman series has wavelength
a. λ = 4
3 R b. λ = 9
8 R c. λ = 4
6R38. The limiting member of Lyman series has wavelength
a. λ = 4
3 R b. λ = 3
4 R c. λ = R
39. The limiting member of Lyman series has wavelength, approximatelya. 2216 A b. 1526 A c. 900 A
40. The relation between wavelength of I and II member of Lyman series is
a.III =
43 R b.
III =
3227 c.
III = R
41. The relation between wavelength of I and limiting member of Lyman series is
a. First member = 43 × limiting b.
ILimiting =
3227 c.
IR = Limiting
42. The first member of Balmer series has wavelength
a. λ = 4
3 R b. λ = 9
5 R c. λ = 4
6R43. The second member of Balmer series has wavelength
a. λ = 4
3 R b. λ = 9
8 R c. λ = 163 R
44. The limiting member of Balmer series has wavelength
a. λ = 4
3 R b. λ = 4R c. λ = R
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45. The limiting member of Balmer series has wavelength, approximatelya. 1216 A b. 1026 A c. 3600 A
46. The relation between wavelength of I and II member of Balmer series is
a . III =
43 R b.
III =
2720 c.
III = R
47. Binding energy for K-shell is a. -13.6 eV b. -3.4 eV c. -1.5 Ev
48. Binding energy for L-shell is a. -13.6 eV b. -3.4 eV c. -1.5 Ev
49. Binding energy for M-shell is a. -13.6 eV b. -3.4 eV c. -1.5 eV
50. Balmer series is
a .ϑ = RH [12− 1ni
2 ] b.ϑ = RH [ 11− 1ni
2 ] c.ϑ = RH [13− 1ni
2 ]51. Balmer series occurs in
a. UV region b. IR region c. visible region52. Paschen series is
a. ϑ = RH [ 12− 1ni
2 ] b.ϑ = RH [ 11− 1ni
2 ] c.ϑ = RH [13− 1ni
2 ]53. Paschen series occur in
a..UV region b. IR region c. visible region54. Bracket series is
a. ϑ = RH [ 12− 1ni
2 ] b.ϑ = RH [ 14− 1ni
2 ] c.ϑ = RH [13− 1ni
2 ]55. Bracket series occur in
a. UV region b. IR region c. visible region56. Pfund series is
a. ϑ = RH [ 12− 1ni
2 ] b.ϑ = RH [ 11− 1ni
2 ] c.ϑ = RH [15− 1ni
2 ]57. Pfund series occur in
a. UV region b. IR region c. visible region58. 1eV is equal to
a. 1.602 × 10 -19 J b. 1.602 × 10 19 J c. 1.602 × 10 -31 J59. Relation between Rydberg constant and energy is
a. E = −R z2
n4 b.E = −R ch
n2 c. E = −Rn2
60. The ground state energy of electron in hydrogen atom isa. -13.6 eV b. -1.36 eV c. 200 eV
61. Bohr’s quantisation postulate is
a. mvr = nh2 π b. mr =
nh2 π c. vr =
nh2 π
62. Stationary orbitals are calleda. Radiating orbitals b. non- Radiating orbitals c. equilibrium orbitals
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63. When electron jumps from higher orbital to lower orbital energy is a. absorbed b. released c. remains constant
64. When electron jumps from lower orbital to higher orbital energy isa. absorbed b. released c. remains constant
65. Bohr’s frequency condition is a.Ef – Ei = hϑ b. Ef – Ei = h c. Ef – Ei = ϑ
66. Radius of stationary orbitals is directly proportional to a. Principal quantum number b. square of Principal quantum number c. cube of Principal quantum number
67. Expression for Bohr’s radius is
a. n2 h2∈πme2 b. n2∈
πm e2 c. n2 h2
πm e2
68. Bohr;s theory fails to explaina. Fine structure b. hydrogen spectrum c. both
69. Radius of first orbit of H-atom is a. 0.2 A b. 0.5 A c. 34 A
70. Radius of second orbit of H-atom is b. 0.2 A b. 0.5 A c.0.34 A
71. Interference , diffraction explains the ---------- nature of electronsa. corpuscular b. wave c. both
72. Photoelectric effect, Compton effect indicates the ---------- nature of electrons a. corpuscular b. wave c. both
73. The one which exhibit dual nature isa. electron b. proton c. both
74. De- Broglie equation is
a. λ = cp b. λ =
cmv c. both
75. The dual nature of electron was verified bya. Planc b. Davisson and Germer c. Compton
76. When an electron is accelerated by potential V then the wave length is
a. λ = h
√VEmb. λ =
h√Vm
c. λ = h√Em
77. If mass is smaller the wave length is a. longer b. shorter c. no relation
78. If velocity is small then wave length is a.longer b. shorter c. no relation
79. De- Broglie equation of an electron accelerated with potential of V is
a. λ = 12.25√V
b. λ = 12.25√Vm
c. λ = 12.25√Vc
80. An electron is accelerated by a potential of 150 V. Its wavelength isa. 2A b.1 A c. 3 A
81. Hisenberg uncertainity principle statesa. Position and momentum cannot be determined simultaneouslyb. Position and momentum cannot be determined preciselyc. Position and momentum cannot be precisely determined simultaneously
82. Hisenberg uncertainity principle statesPage 349 of 419
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a. ∆x ×∆y = h
2 π b. ∆x ×∆p = 1
2 π c . ∆x ×∆p = hπ
83. The uncertainty in momentum of an electron is 0.2 .The uncertainty in position isa. h0.4 π b.
h0.1 π c
hπ
84. If the uncertainty in momentum is low, then the uncertainty in position isa. Low b. high c. no relation
85. The uncertainty in position of an electron is “h’ times that of uncertainty in momentum. The value of uncertainty in momentum is
a.√ h
√ 2 π b. h
√ 2 π c. 1
√ 2 π86. Davision and Germer experiment is used to verify
a. Wave matter dualism b. Compton effect c. photo electric effect.87. The metal used in Davision and Germer experiment is
a. Copper b. Nickel c. Cobolt88. The law used to calculate the wavelength in Davision and Germer experiment is
a. Planc’s law b. Stefen’s law c. Bragg’s law89. The maximum number of electrons ejected at 54 V is at an angle of
a. 60o b. 50o c..70o
90. The voltage required for the ejection of maximum number of electrons b. 60 V b. 54 V c..70 V
91. Photo electric work function of a metal is 6.62 eV the threshold frequency isa. 1.6 × 10 15 b.81.6 × 10 5 c. 7.6 × 10 5
92. The radius of first Bohr’s orbit is 0.5 A. radius of 10 th orbit is a. 5A b. 10 A c. 50 A
93. Energy of electron in the n th orbit of H- atom is proportional to
a. n2 b. 1n2 c. n3
94. The distance between successive orbits with increase in ‘n’ -------a. remains same b. decreases c. increases
95. Arrangement of spectral series in H- atom is a. Lyman, Paschen, Balmer b. . ,Paschen, Balmer .c. Lyman, Balmer Paschen, . 96. Which of the following is in visible region? a.. Lyman, b.Paschen, c. Balmer 97. Rydberg constant has dimension of a. frequency b. velocity c. wave number98 . Unit of wave number is
a. m b. m -1 c. cm99. First line of Balmer series is 656 nm. Second line of this series is a. 656 b. 486 c. 1300100. Wave length of a line is 50 nm. Its wave number is a. 2 b. 2× 10 7 c. 2× 10 8
KEY
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51 C 61 A 71 A 81 C 91 A52 C 62 B 72 B 82 A 92 C53 B 63 B 73 C 83 A 93 B54 B 64 A 74 C 84 B 94 C55 B 65 A 75 B 85 C 95 C56 C 66 B 76 A 86 A 96 C57 B 67 A 77 A 87 B 97 C58 A 68 A 78 A 88 C 98 B59 B 69 B 79 A 89 B 99 B60 A 70 A 80 B 90 B 100 B
TEST
Page 351 of 419
1 C 11 B 21 A 31 C 41 A2 B 12 C 22 C 32 B 42 B3 B 13 B 23 A 33 B 43 A4 A 14 C 24 C 34 B 44 B5 C 15 B 25 A 35 A 45 C6 C 16 B 26 A 36 A 46 B7 C 17 C 27 A 37 B 47 A8 B 18 B 28 C 38 C 48 B9 A 19 C 29 A 39 C 49 C10 B 20 A 30 B 40 B 50 A
2020 - CONCISE QUANTUM MECHANICS 2017
1.Classical mechanics does not explaina. Stability of atom b. pressure of liquids c. pressure of solid
2. A wave function should be a. Continuous b. Single valued c. continuous and Single valued 3. The wave function is a function of a. position b. time c. both 4. Eigen equation is a. H = E Ψ b. HΨ = E Ψ c. HΨ = E 5. The average value of un- normalised physical quantity is
a. ∫Ψ Ψ ¿dT
∫Ψ dT b. ∫Ψ H Ψ ¿dT c.
∫Ψ H Ψ ¿dT
∫ΨΨ ¿dT6 The average value of a normalised physical quantity M is given by
a. ∫Ψ M Ψ ¿dT
∫Ψ Ψ ¿ dT b. ∫Ψ M Ψ ¿dT c.
∫Ψ Ψ ¿dT
∫Ψ ¿dT7. The position operator in x-direction is
a. x b. xy c. x2
8. The function of position operator is a. It adds with the function b. It subtracts with the function c. It multiplies with the function9.A position operator ‘x’ operates on a function ‘y’ the resultant is
a. xy b. x-y c. xy
10. The expression for momentum operator in ‘x’ direction is
a. h2
x ∂2
∂ x2 b. hx
∂∂ x c.
hi
∂∂ x
11. The resultant of momentum operator hi
∂∂ y on a function ‘y’ is
a. y b. hi c.
xy
12. Laplacian operator is
a. ∂
∂ x +∂
∂ y + ∂
∂ z b. ∂2
∂ x2+∂2
∂ y2+∂2
∂ z2 c. ∂2
∂ x2
13. Total energy operator is
a.h∂∂ t b.i h2 ∂
∂ t c.ih∂
∂ t
14. Kinetic energy operator is
a. - h2
2m ∇2 b. - h2
4 m ∇2 c.-
h2m
∇2
15. If the operator P obeys the relation P(Ψ+Φ) =PΨ+PΦ the operator is said to be a. linear b. Hermition c. commutative16.The angular momentum operator is A. r +p B. r × p C.r /p17. An operator A is said to be Hermition if. a. ∫Ψ∗(A Ψ ) dx = ∫Ψ ( A Ψ )∗¿¿ dx b. ∫Ψ Ψ ¿¿ dx = ∫Ψ (Ψ )∗¿¿
c. ∫Ψ∗(A+Ψ ) dx = ∫Ψ+ (A Ψ )∗¿¿
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18. P and Q are the operators obeying the relation PQ[Ψ] =QP[Ψ ]then the operators are A.. linear B. Hermition C. commutative 19. Expectation value of energy of particle is A ∫Ψ Ψ∗¿¿dT B ∫Ψ 1 Ψ 2 dT C ∫Ψ H ΨdT20. The commutator of [ Lx , Ly ] isa. equal to zero b.equal to one c. not equal to zero21.. The resultant of Laplacian operator on sin 2x is
a. -4 sin 2x b. 2cos 2x c.2 sin 2x + 2cos 2x+2cos 2z22. The commutator of position and momentum operator with respect to the function e x isa. -ih ex b. - ih c. + ih ex
23. The value of [ z3, ddz ] is
a.-3 z2 b. -3z c. -2z24. [ Lz, z ^] = P then the value of P is a. 0 b.1 c. -125 The eigen value of the function sin 2xsin2ysin2z with respect to ∇ 2 a.-4 b.+4 c.-1226. The commutator of [ Lx
2 , Ly ] is a. 1 b.0 c.-1
27. The commutator for ddx
and x on x2 is
A. 2x2 B. 3x2 C. 1
28. [ddx , x ] on e2x is
A. e2x B. x0 C. 2xe2x + e2x
29. [ x, p x] for the function ex is A. i h B. –i h C. –i h2
30. [Z2,ddz ] is
A. 2z B. -3z2 ` C. -3z3
31.[L z, z] is A. 1 B. 2 C.0
32. Which is not eigen function with respect to d2
d x2
A. 1x B. 5x2 C. Both
33.Which of the function is not eigen function ofd 2dx 2
a.. Sin 2x b. 6 cos 3x c. 5 x2
34.Which of the function is eigen function ofd 2dx 2
a. 1/x b. log x c. e -2x
35. The eigen value of cos 5x with respect to the operator d 2dx2 is
a. -5 b.-25 c.+25Page 353 of 419
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37.The eigen value if sin2x with respect to the operation d2
d x2 is
A. 4 B. -4 C. 0
38. The eigen function with respect to ddx is
A. eax B. logx C.1x
40.
QUANTUM MECHANICS -II KEY
1 A 11 B 21 A 31 412 C 12 B 22 32 423 B 13 C 23 33 434 C 14 A 24 34 445 C 15 A 25 35 456 B 16 B 26 36 467 A 17 A 27 37 478 C 18 C 28 38 489 A 19 C 29 39 4910 C 20 A 30 40 50
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QUANTUM MECHANICS -III( Particle in one , two and three - dimension box)
1.A particle of mass m is confined to move in a one-D box between x=0 and x= a. The normalised wave function of the particle is
a . √8a sin
nπxa b.√
2a sin
nπxa c. √
16a sin
nπxa
2.The wave function for a particle in 1-D box of length L is Ψ = A sinnπx
L The value of A is
A √8L B √
2L C √
L2
3. The wave function for a particle in a one –D box is Ψn =√ 2L
sinnπxL The normalization constant is
A. 2L B. √ 2
L C.
4L2
4. The normalization constant for particle in a box of 1-D for the wave function Ψ1 is √ 2L
. Its value for is
Ψ2 is
A. √ 2L
B. √ 2L2 C.√ 4
L5. Normalization constant for a particle in a 1-D box A. Directly proportional to length of the box B. Inversely proportional to length of the box C. Inversely proportional to square root of length6. When length of the box increases , normalization constant A. Increases B. Decreases C. Remains same 7. Nodes are the points where the wave function becomes A infinity B negative C zero8 The wave function Ψn has ----- number of nodes. A (n+1) B ( 2n +1) C (n+2)9. The nodes of Ψ 1 for a particle in a one dimensional box of length L occurs at A 0 and L B 0 and L/2 C 0 and L/310. The number of nodes in Ψ 2 is A 2 B 3 C 411. The nodes of Ψ3 are A 0, L/3, / L B 0, L/4, 2L/3, L C 0, L/3, 2L/3 , L12. Zero point energy of the particle in a one dimensional box of length a is
A n2 h2
8 ma2 B h2
8 ma2 C n2 h2
32m L2
13. The lowest value of n in the eigen energy of the particle in a one dimensional box is A 0 B 1 C 2
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14. The energy gap is inversely proportional to the ---- of the length of the box. A cube B square root C square 15.The normalization constant for a particle in a box of length 1m is A. 0.141 B. 1.41 C. 14.116.The ratio between the normalization constant of Ψ1 and Ψ2 is
A. √ xL
B. 1 C. 0
17. Energy difference between 3 rd and 4 th level of the particle in a one dimensional box is
A 9 h2
8 m L2 B 16 h2
8 m L2 C 7h2
8m L2
18. The spacing between n th level and ( n+ 1) th level of the particle in a one 1-D box is A ( 2n +1) E1 B ( n +1) E1 C ( n +2) E1
19. The energy gap between fifth and fourth level of a particle in a one- D box is – E0
A. 9 B. 25 C.1620.Energy of the particle in a one dimensional box of length 2L is
A n2 h2
32m L2 B n2 h2
2m L2 C n2 h2
8m L2
21..The ground state energy of the particle with mass 0.25 units in a one dimensional box of length 6.62 × 10 – 34 m is A 0.5 B 0.25 C 0.7522. The lowest energy of the particle with mass m , in a one dimensional box, whose length is equal to that of Planc’s constant is A 1/16m B1/8 m C1/2 m23. The probability for the particle in state one of a one dimensional box of length L , to be within 0< X<L is A 0.0908 B 0.5 C 1.024. A particle in 1-D box has a minimum allowed energy of 2.5 eV. The next higher energy it can have is a. 3.5 eV. b. 10 eV. c. 5.0 eV.25. The probability for a particle in a one dimensional box of length ‘a’ found to be in between ‘a’ and ‘a/2’ is
a. ½ b. 1/4. c. ¾ d. 1/3 26.The probability for a particle in a one dimensional box of length ‘a’ found to be in between ‘0’ and ‘ a/2’ is
a. 1/4 b. ½ c. ¾ d. 1/3 27 .The probability for a particle in a one dimensional box of length ‘a’ found to be in between ‘a/4’ and ‘a/2’ is
a. ½ b. 1/4. c. ¾ d. 1/3 28. The probability for a particle in a one dimensional box of length ‘a’ found to be in between ‘a’ and ‘a/2’ is
a. 1/3 b. 1/4. c. ¾ d. ½ 29. The normalisation constant of for the particle in a one dimensional box of length ‘L’ is
a.2
√ L b. √ 2L c. √ 2
Ld. None
30. The ground state energy of electron in a one dimensional box of length 100 nm is 4 ev. If the particle is placed in a cubic box of side 100 nm , the energy of the particle is
a. 10 b. 12 c. 16 d. 10031. The normalization constant for the wave function of particle in a 3D is
A. √ 2abc
B. √ 4abc
C. √ 8abc
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32. The normalization constant for a box of cube is
A. √ 8v
B. √ v8
C. √ 4v
33. Zero point energy of particle in a cubic box of side L is
A 3 h2
8 m L2 B 7 h2
8 m L2 C 8 h2
8 m L2
34.. The degree of degeneracy of first excited state of particle in a 3-D is A 2 B 3 C 4
35. Zero point energy of particle in a 3-D is – times that of 1- D box. A 4 B 2 C 336. The degree of degeneracy of (222) state is A 0 B 2 C 337. The degree of degeneracy of (1,1,2) state is A. 1 B. 2 C. 338. The energy level (1,1,2) is A. Non-degenerate B. Doubly degenerate C. Triply degenerate39. An electron is confined in a cubic box of dimension 2 m. Its normalized constant is A. 1 B. 0 C. ∞ 40.An electron is confined in a box of length , breath and width 2,2,4 m respectively. Its normalized constant is
A. 12 B.√ 1
2 C.
1h
41. The energy required for a transition from 111 to 112 state of H2 which is placed in cubic box of unit dimension is
a. 36 h 2 b.
38 h 2 c.
37 h 2
42. The number of nodes in Ψ3 =√ 2L
sin3 πx
L is
a. 1 b.2 c. 3 d.4 43. Which can not be energy of the particle in a one dimensional box
a 9 h2
8 ma2 c. 16 h2
8ma2 d. 7 h2
8 ma2
44 The degree of degeneracy of the level E = 14 h2
8 ma2
a 2 b. 3 c. 4 d. 5
45. The degree of degeneracy of the level E = 12h2
8 ma2
a 2 b. 3 c. 4 d. 0
46. The integral of the product Ψ3 =√ 2L
sin3 πx
L and Ψ4 =√ 2L
sin4 πx
L on the interval 0 to ‘a’ is
a 2 b. 0 c. 1
47. If ∫0
a
Ψ 1 Ψ∗¿¿ dx = 1 , the wave function is
A orthogonal B normalised C invalid48. The wave function of 3-D box along x ,y and z directions are Ψ x , Ψ y and Ψ z and energies are Ex, E y and Ey respectively. The total wave function and energies are a.. Ψ xyz = Ψ x × Ψ y × Ψ z , Exyz = Ex × Ey × Ey
b. Ψ xyz = Ψ x + Ψ y + Ψ z , Exyz = Ex + Ey× Ey
c. Ψ xyz = Ψ x × Ψ y ×Ψ z , Exyz = Ex + Ey + Ey
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QUANTUM MECHANICS -III -KEY1 B 11 C 21 A 31 C 41 B2 B 12 B 22 B 32 A 42 C3 B 13 B 23 C 33 A 43 C4 A 14 C 24 B 34 B 44 B5 C 15 B 25 A 35 C 45 C6 B 16 B 26 B 36 A 46 B7 C 17 C 27 B 37 C 47 B8 A 18 A 28 C 38 C 48 C9 B 19 A 29 C 39 A10 B 20 A 30 B 40 B
TEST
1. In Harmonic oscillator, the force is proportional to a. amplitidue b. displacement c. velocity2. The potential energy of particle following simple harmonic oscillation is
a. ½ kx2 b. kx2 c. ½ kx 3. The Schrodinger equation Harmonic oscillator is
a. d2Ψdx2 +8π 2m
h2 (E- x) Ψ =0 b. d2Ψ
dx2 +8 π 2mh2 (E - ½ Kx2) Ψ = 0 c. d
2Ψdx2 +8π 2m
h2 E = 0
4 The wave function of harmonic oscillator is Ψ=¿ ) ½e− y 2
2 (-1) n e y2
dn
d yn (e− y2
) . the value for Ψ 0
is
a. Ψ0 = ( απ ) ¼ e
− y 22 b. Ψ0 =(
απ ) ¼
e− y 2
2 ( √2 y) , c. Ψ0 = ( απ ) ¼ e
− y 22 (
2 y 2−1√ 2 )
5. The normalisation constant for harmonic oscillator is a. ¿ ) ½ b. ¿ ) ½ c. ¿ ) ½6 The normalisation constant for a harmonic oscillator is given as¿ ) ½ It depends a. Amplitude b. velocity c. none7. When period of oscillation increases, the normalisation constant
a. increases b. decrease c. remains constant.8. The energy of Harmonic Oscillator is given by
A. hγ(n+12) B. hγ(n+1) C. hγ(n+2)
9. The zero point energy of harmonic oscillator is
a. 12 h γ b.
32 h γ c.
52 h γ
10. Which is true in harmonic oscillator?Page 358 of 419
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a All the energy levels of the oscillator are non-degenerate b. successive energy levels are equally spaced c. Both
11. The spacing between successive energy levels in harmonic oscillator a. always equal b. equal only at higher level c. equal only at lower
12. The difference in energies of first and second energy levels of harmonic oscillator is
a. 12 h γ b. h γ c. 2 h γ
13. The difference in energies of second and third energy levels of harmonic oscillator is
a. 12 h γ b.
32 h γ c. h γ
14. The difference in energies of first and third energy levels of harmonic oscillator is
a. 12 h γ b.
32 h γ c. 2 h γ
15. The spacing between successive energy levels in harmonic oscillator
a. 12 h γ b. h γ c. 2 h γ
16. The energy of harmonic oscillator at third excited level is
a. 12 h γ b.
52 h γ c. 2 h γ
17. The potential energy of rigid rotator is a. 1 b. 0 c.∞
18. The solution of azimuthal wave function is 1√2 π
e imφ The normalisation constant is
a.A1√2 π
b.1√4 π
c.1√π
19. The radial wave function of H-atom is given by A. associated Laguerre polynomial B. Legendree polynomial C. Hermite polynomial
20. The angular wave function is given by A. Laguerre polynomial B. associated Legendree polynomial C. Hermite polynomial21. Radial wave function of H- atom1. Depends r only b. depends θ only c. Depends Both r and θ22. Angular wave function of H- atom2. Depends r only b. depends θ only c. Depends Both r and θ23. The total eigen function of 1s orbital of H-atom in atomic units is
a. 1√π
e –r b. 1√π
e –r cos θ c..1√π
e –r sin θ
24. The total wave function is given by a. radial function × angular function b. radial function + angular function c. none25. To convert Cartesian co-ordinates into spherical co-ordinates a. x = r sin θ ,y = r sin θ sinφ, z = r cos θ b. x = r sin θ cosφ , y = r sin θ z= r cos θ c. x = r sin θ cosφ , y = r sin θ sinφ ,z = r cos θ 26. The wave function of 2s orbital is
a. 1
4 √ 2π (1
a 0) 3/2 e - r
2a 0 ( 2-
ra 0 ) b.
14 √ 2 π (
1a 0) 3/2 e - r
2 a0 ( r
a0 ) cosθ
c. 1
4 √ 2π (1
a0) 3/2 e - r
2a 0 ( r
a0 ) sin θ cosφ
27. The radial and angular wave function of 2s orbital of H-atom, in atomic units, are
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1
2√ 2 (2 - r ) × e –( r2 ) × √ 32
and 1√2 π
then the value of Ψ 200 is
a. 1
√32 π × (2 - r ) × e –( r2 ) b.
1√32 π
× (2 - r ) c . (2 - r ) × e –( r2 )28. The solution of azimuthal wave function is
1√2 π
e imφ The normalisation constant is
A1√2 π
B1√4 π
C1√π
29. The solution of polar wave function is Θ l,m = √ 2 l+12
× (l−m ) !(l+m ) !
× Plm ( cos θ ) The
normalisation constant Θ 0,0 is
A1√2
B1√4 π
C1√π
30. The solution of radial wave function is R n,l = √( 2 zna0 )
3
× (n−l−1 )!2n [ (n+l )! ]3
× ( 2 zrna 0 ) l × Ln+l
2 l+1 ( 2 zrna 0 )
× e –( zrna 0 )The normalisation constant R1,0 for H-atom is
A ( 2
a 0 )3/2 B1√4 π
C1√π
31. The Rodrigue formula for associated Legendre polynomial is
Plm ( cos θ ) =
12l× l!
( 1- cos 2 θ ) m/2 ( d
d (cosθ) ) l+m [cos 2 θ -1 ) l × e i mφ The value of P0
0 is
A ( 2
a 0 )3/2 B 1 C1√π
32. The Rodrigue formula for associated Laguerre polynomial is
Lrs (ρ) =(
ddρ ) s [ e ρ (
ddρ ) r ( ρ r e –ρ ) ] where r= n +l ,s = 2l +1 and ρ = ( 2 zr
na 0 )The value of L0
0 is
A ( 2
a 0 )3/2 B 1 C1√π
33. The total wave function of H-atom is Ψ n,l,m = √( 2 zna 0 )
3
× (n−l−1 )!2n [ (n+l )! ]3
× ( 2 zrna 0 ) l × Ln+l
2 l+1 ( 2 zrna 0 )
× e –( zrna 0 ) × √ 2 l+1
2× (l−m ) !
(l+m ) ! × Pl
m ( cos θ ) × 1√2 π
e imφ . The value of
Ψ 1,0,0 in atomic units is
A ( e−r
√ 2 π )3/2 B e−2 r
√ 2 π C e−r
√ π
QUANTUM MECHANICS -IV -KEY
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2 12 22 32 423 13 23 33 434 14 24 34 445 15 25 35 456 16 26 36 467 17 27 37 478 18 28 38 489 19 29 39 4910 20 30 40 50
TEST1. Wave particle duality was verified by
a. de-Broglie b. Compton c. Davisson and Germer d. Sommerfeild
2. According to Hisenberg’s uncertainity principle which is true?
a. ∆x ×∆p = h b. ∆E ×∆t = h c. ∆J ×∆θ = h d. all the above
3. If a particle of mass 5 × 10 -5 Kg moves with a velocity 6.62 × 10 -34 m/sec then its
de-Broglie wavelength is
a. 5 × 10 +5 b. 5 × 10 -5 c. 15 × 10 -5 d. 15 × 10 +5 4. The de-Broglie wavelength of the particle accelerated with a voltage of 44 V is
a. 1.12 × 10 +2 b. 1.12 × 10 -5 c. 12.25 × 10 -5 d. 2.25 × 10 +5
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5. Bohr’s model was failed because I It could not explain the occurrence of fine spectral lines. II. In
Bohr’s theory, there is no place for the wave character of electron
a. I only b. II only c. both I and II d. none of the above.
6. Which is not true. I. Classical mechanics does not explain stability of atom. II. Classical
mechanics does not explain the spectrum of hydrogen. III. Classical mechanics does not explain
motion of objects .
a. I only b. II only c. III only d. all the three
7. Intensity of blackbody radiation
a. increases b. decreases c. increases up to a maximum then decreases
d not predictable
8. According to Planc the quantity , quantised is
b. wavelength b. energy c. frequency d all
9. Photo electric effect is emission of
b. protons b.electrons c. X-rays d. Gamma rays
10. Photo electric effect occurs at ----- frequecies
b. high b. low c. threshold d. all
11. Photo electric work function is given by
b. W= hϑ0 b. W= 2h c. W= ϑ0 d. W= 5h
12. The wave length of Hydrogen spectrum is
b.1λ = RH [ 1
n f2−
1ni
2 ] b. 1λ = RH RH [ 1
n f2 ] c.
1λ = RH RH [ 1
n f2−
1ni
3 ] d. none
13. Lyman series is
b.1λ = R [12− 1
ni2 ] b
1λ = R [ 11− 1
ni2 ] c.
1λ = R [13− 1
ni2 ] d. none
14. When electron jumps from second orbit to first orbit the line obtained is
c. Balmer b. Lyman c. Pfund d. Paschen
15. Pfund series occur in
b. UV region b. IR region c. visible region d. microwave region
16. Bohr’s quantisation postulate is
b. mvr = nh2 π b. mr =
nh2 π c. vr =
nh2 π d. mvr =
h2 π
17. Radius of stationary orbitals is directly proportional to
a. Principal quantum number b. square of Principal quantum number
c. cube of Principal quantum number d. none
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18. Radius of first orbit of H-atom is
c. 0.2 A b. 0.5 A c. 34 A d. 0.05A
19. When an electron is accelerated by potential V then the wave length is
b. λ = h
√VEmb. λ =
h√Vm
c. λ = h√Em
d. λ = h
√2 VEm
20. An electron is accelerated by a potential of 150 V. Its wavelength is
b. 2A b.1 A c. 3 A
21. Hisenberg uncertainity principle states
d. Position and momentum cannot be determined simultaneously
e. Position and momentum cannot be determined precisely
f. Position and momentum cannot be precisely determined simultaneously
g. only two electrons can be present in an orbital
22. Davision and Germer experiment is used to verify
b. Wave matter dualism b. Compton effect c. photo electric effect. d. Wein effect
23..A position operator ‘x’ operates on a function ‘y’ the resultant is
a. xy b. x-y c. xy d. x +y
24.. Laplacian operator is
a. ∂
∂ x +∂
∂ y + ∂
∂ z b. ∂2
∂ x2+∂2
∂ y2+∂2
∂ z2 c. ∂2
∂ x2 d. .i h2 ∂∂ t
25. If the operator P obeys the relation P(Ψ+Φ) =PΨ+PΦ the operator is said to be
a. linear b. Hermition c. commutative d. associative
26.. If ∫0
a
Ψ Ψ∗¿¿ dx = 1 , the wave function is
a orthogonal b normalised c continuous d. single valued
27. For an orthogonal wave function Ψn = √2a sin
nπxa . The value of ∫
0
a
Ψ 1Ψ 2 dx is
a 0 b 1 c ∞ d. - 1
28.The probability of finding the particle over a small distance is
a Ψ 2 b Ψ 3 c Ψ d. √Ψ
29.. The resultant of Laplacian operator on sin 2x is
a. -4 sin 2x b. 2cos 2x c.2 sin 2x d. -4 cos 2x
30. The commutator of position and momentum operator with respect to the function e x is
a. -ih ex b. - ih c. + ih ex d. none
31 The eigen value of the function sin 2xsin2ysin2z with respect to ∇ 2
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a.-4 b.+4 c.-12 d +12
32 [ x, p x] for the function ex is
a. i h b –i h c. –i h2 d. ex
33. Which is not eigen function with respect to d2
d x2
a. 1x b 5x2 c. sin x d..all
34. Linear momentum operator along y direction operates on π4 y . The resultant is
a. ћi (
3 π4 ) b.
ћi (
π4 ) c.
ћi (
3 π2 ) d.
π4 y
35. Which is not linear I ‘log’ II. root (√ ) III. Linear momentum operator
a. I and II b. II and III c. I and III d. all
36. The operators ‘x’ and ddx are added and operates on a function ‘sinx ‘ .The resultant is
a. x sin x + cos x b.x sin x - cos x c. x cos x d. none
37. The number of nodes in √ 2L
sin4 πx
L is
a. 1 b. 2 c.3 d. 4
38.A particle of mass m is confined to move in a one-D box between x=0 and x= a. The normalised
wave function of the particle is
a . √8a sin
nπxa b.√
2a sin
nπxa c. √
16a sin
nπxa d. sin
nπxa
39.The wave function for a particle in 1-D box of length L is Ψ = A sinnπx
L The value of A is
a √8L b √ 2
Lc. √
L2 d. L
40. The normalization constant for particle in a box of 1-D for the wave function Ψ1 is √ 2L
. Its value for is
Ψ2 is
a. √ 2L
b. √ 2L2 c.√ 4
L d. √2
41. Normalization constant for a particle in a 1-D box
a.. Directly proportional to length of the box b.. Inversely proportional to length of the box
c.. Inversely proportional to square root of length d. none
42. The number of nodes in Ψ 2 of particle in a one dimensional box is
a. 0 b. 1 c. 2 d. 3
43.Zero point energy of the particle in a one dimensional box of length a isPage 364 of 419
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a. n2 h2
8 ma2 b. h2
8 ma2 c. n2 h2
32m L2 d. zero
44. The degree of degeneracy of (1,1,2) state is
a. 1 b. 2 c. 3. d.6
45. The energy level (1,2,3) is
a. Non-degenerate b. Doubly degenerate c. 3-fold degenerate d. 6-fold degenerate
46. Which can not be energy of the particle in a one dimensional box
a 9 h2
8 ma2 b. 16 h2
8 ma2 c. 7 h2
8 ma2 d. 8h2
ma2
47 The integral of the product of eigen functions Ψ3 =√ 2L
sin3πx
L and Ψ4 =√ 2L
sin4 πx
L on the interval 0
to ‘L’ is
a 2 b. 1 c. 0 d. -1
48.. Which is six fold degenerate ?
a. (111) b. (122) c. (123) d. (666)
49..The normalization constant for a particle in a one dimensional box of length 1m is
a.. 0.141 b.. 1.41 c.. 14.1 d. none.
50. The normalization constant for the wave function of particle in a 3D with dimension a,b and c is
a.. √ 2abc
b. √ 4abc
c. . √ 8abc
d. √ 2L
51. Zero point energy of particle in a cubic box of side L is
a. 3 h2
8 m L2 b. 7 h2
8 m L2 c. 8 h2
8 m L2 d. zero
52 The degree of degeneracy of first excited state of particle in a 3-D is
a. 2 b. 3 c.4 d. 6
53. The spacing between n th level and ( n+ 1) th level of the particle in a one 1-D box is
a. ( 2n +1) E1 b. ( n +1) E1 c. ( n +2) E1 d. nE
54.Energy of the particle in a one dimensional box of length 2L is
a. n2 h2
32m L2 b. n2 h2
2m L2 c. n2 h2
8 m L2 d. n2 h2
4 m L2
55. A particle in 1-D box has a minimum allowed energy of 2.5 eV. The next higher energy it can have
is
a. 3.5 eV. b. 10 eV. c. 5.0 eV. d. 2.5 eV
56. The probability for a particle in a one dimensional box of length ‘a’ is found within 0< X< a/2’ is
a.½ b. 1/4. c. ¾ d. 1/3
57. The normalisation constant of for the particle in a one dimensional box of length ‘L’ is
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a. 2
√ L b. √ 2L c. √ 2
Ld. None
58. The ground state energy of electron in a one dimensional box of length 100 nm is 4 ev. If the
particle is placed in a cubic box of side 100 nm , the energy of the particle is
a.10 b. 12 c. 16 d. 100
59..An electron is confined in a box of length , breath and width 2,2,4 m respectively. Its normalized
constant is
a. 12 b.√ 1
2 c.
18 d. 16
60.. The energy required for a transition from 111 to 112 state of H2 which is placed in cubic box of
unit dimension is
a. 3
6m h 2 b. 3
6 m h 2 c. 37 h 2 d. 5 h2
24 m
61 The degree of degeneracy of the level of particle in a 3-D box 14 h2
8 ma2
a 2 b. 3 c. 5 d.6
62. The degree of degeneracy of the level of particle in a 3-D box 12h2
8ma2
a 2 b. 3 c. 1 d. 6
63. The ground state energy of a particle with mass ‘m’ in a 3-D box of dimensions1,2 and 2 units is
a 6 h2
8 mb. h2
32mc. 6h2
32m d. 9 h2
8 m
64. The ground state wave function of a particle in a cubic box of dimension 2 units is
a. sin ( xπ2 ) sin (
yπ2 ) sin (
zπ2 ) b.√8 sin (
xπ2 ) sin (
yπ2 ) sin (
zπ2 )
c. √3 sin ( xπ2 ) sin (
yπ2 ) sin (
zπ2 ) d. √6 sin (
xπ2 ) sin (
yπ2 ) sin (
zπ2 )
65.. The wave function of a particle in the state (112) in a cubic box of length 3 units is
a. sin ( xπ2 ) sin (
yπ2 ) sin (
zπ2 ) b.√ 8
27 sin (
xπ3 ) sin (
yπ3 ) sin (
zπ3 )
c. √3 sin ( xπ1 ) sin (
yπ2 ) sin (
zπ3 ) d. √8 sin (
xπ2 ) sin (
yπ2 ) sin (
zπ2 )
66. The energy required for the transition of electron with mass “m” in a cubic box of dimension of 3
units, from ground state to first excited state is
a. h2
24 mb. 3 h2
8 mc. h2
4 m d. h2
4 m67. The potential energy of particle following simple harmonic oscillation is
a. ½ kx2 b. kx2 c. ½ kx d. 0
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68. The polynomial used to solve the schrodinger equation of harmonic oscillator is
a,Laguerre polynomial b.. Legendree polynomial c.. Hermite polynomial d. none
69. The spacing between successive energy levels in harmonic oscillator
a. always equal b. equal only at higher level c. equal only at lower d. none
70. The normalisation constant for a harmonic oscillator is given as¿ ) ½ It depends
a. Amplitude b. velocity s. frequency c. none
71. The energy of Harmonic Oscillator is given by
a.. hγ(n+12) b. hγ(n+1) c. hγ(n+2) d. zero
72. The zero point energy of harmonic oscillator is
a. 12 h γ b.
32 h γ c.
52 h γ d. zero
73. Which is true in harmonic oscillator?
a. Successive energy levels are equally spaced
b. Separation between two adjacent energy levels is hγ
c. All the energy levels are non-degenerate
d. all
74. The acceptable asymptotic solution for SHO is
a. ey2
2 b. e− y2
2 c. ey3
2 d. ey2
5
75. The difference in energies of first and third energy levels of harmonic oscillator is
a. 12 h γ b.
32 h γ c. 2 h γ d. none
76. The energy of harmonic oscillator at third level is
a. 12 h γ b.
52 h γ c. 2 h γ d. none
77.. Hermite polynomial is Hn (y) = (-1) n e y2
dn
d yn (e− y2
), then H0 is
a. 1 b. 2y, c. 4y2 – 2 d. – 2y
78. The potential energy of harmonic oscillator is proportional to x b . the value of b is
a. 1 b..2 c. 3 d. -1
79. The general solution of harmonic oscillator is Ψ = Hn × e− y 2
2 Hn stands for
a.Laguerre polynomial b. Legendree polynomial c. Hermite polynomial
80. .The potential energy of rigid rotator is b. 1 b. 0 c.∞ d. ½ k x2
81 Spherical harmonics are the wave functions of
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a. harmonic oscillator b. rigid rotator c. particle in a 3-D boxd. particle in a ring
82 .The energy of rigid rotator is
a. n2 h2
8 m L2 b. h2
8 mπ 2J(J+1) c.n2 h2
m L2 d hγ(n+12)
83.The energy difference between second and third level of a rigid rotator is
a. h2
2 π2 I b..0 c. 5 h2
8 π 2 I84. The Eigen values for the wave functions r e−r , e−r and e−r2
are -0. 375 , - 0.5 and
- 0. 405 respectively. The best wave function is
a Ψ 1 b. Ψ2 c. Ψ3 d. all
85. Secular equation is the outcome of
a perturbation method b. variation method c. Born approximation d. slater rules
86. The average energy of H-atom with variation parameter ‘a’ is given by < E > = a2
2 - a. The
minimum value of average energy is
a 0 b. - 0.5 c. 2 d. -1
87 In variation theorem the energy is calculated by
a. ∫−∞
∞
Ψ dx
∫−∞
∞
Ψ Ψ ¿ dx b.
∫−∞
∞
ΨH Ψ ¿ dx
∫−∞
∞
Ψ Ψ ¿dx c. ∫
−∞
∞
ΨH Ψ ¿ dx . d. ∫−∞
∞
Ψ Ψ ¿ dx
88. Perturbation method can be applied if
a.The system differs only slightly from the unperturbed ystem.
b.Energy and wave function for the unperturbed system are known.
c Hamiltonian for the unperturbed system is known.
d. all the above
89. Example for perturbed system is
a. Hydrogen atom placed in an electric b. Hydrogen atom placed magnetic field,
c. anharmonic oscillator d. all the above
90. The expression for first order perturbed energy is
a. En1 = ∫Ψ n
0 H1 Ψ n0 dτ b. En
1 = ∫Ψ n0Ψ n
0 dτ c.En1 = ∫Ψ n
0 H1 dτ d.En1= ∫H1 dτ
91. The wave functions of Hydrogen molecule given by Valence Bond Theory is
a. Ψ = a1 [SA(1)SB(2) ] b. Ψ = a2 [SA(2)SB(1) ]
c. Ψ = a1 [SA(1)SB(2) ] + a2 [SA(2)SB(1) ] d. Ψ = a1 [SA(1)SB(2) ] a2 [SA(2)SB(1) ]
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92. The symmetric wave function of H2 molecule according to MO theory is
a. 1
√2(1+S¿)¿ ( SA + SB ) b. 1
√2(1−S¿)¿ ( SA - SB )
c. 1
√2 (1+S2 ){[SA(1)SB(2)]+[SA(2)SB(1) ] } d.1
√2 (1+S2 ){ [SA(1)SB(2) ] - [SA(2)SB(1)] }
KEY
1 2 3 4 5 6 7 8 9 10
C D B B C C C B B C
11 12 13 14 15 16 17 18 19 20
A A B B C A B B D D
21 22 23 24 25 26 27 28 29 30
C A A B A B A A A B
31 32 33 34 35 36 37 38 39 40
C B C D A C C B B B
41 42 43 44 45 46 47 48 49 50
B B B C D C C C B C
51 52 53 54 55 56 57 58 59 60
A B A A B A C B B B
61 62 63 64 65 66 67 68 69 70
D C D A B A A C A D
71 72 73 74 75 76 77 78 79 80
A A D B C B A B C B
81 82 83 84 85 86 87 88 89 90
B B D B B B B D D A
91 92 93 94 95 96 97 98 99 100
C A
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REVIEW QUESTIONS 1.MATHEMATICS FOR QUANTUM MECHANICS
1. Convert the coordinate (2, 5) into Cartesian coordinate2. Which coordinate system is explained by two distances and an angle? Mention its limits.
3.Convert the Cartesian coordinate (-1,1,-√2) into spherical polar coordinates.
4.Name the coordinate system that is applied to describe diatomic molecule and give
its limits
5.Express the Cartesian coordinates (1,0,0) in terms of spherical coordinates
6.Give the limits of spherical coordinates
7.Define the following: 1. Closed interval 2. Even function 3. Orthonormal set of functions
8. Represent the complex number (1 - i) in the Euler form.
9. How much distance is away the point (5, 120°, 60°) from the origin?
10. Represent the following complex numbers in the form of Euler formula.
(i) 1/√2 +(1/√2)i (ii) ½ +(√3/2)i
11.Show that for spherical co ordinates r = √ x2+ y2+z2
12. How much distance the point (6,4,3) away from the origin?
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13.Convert the distance (9,6,3) in to spherical and cylindrical co ordinate.
14..Convert the distance (2,π3 ,
π6 ) in to spherical and cylindrical co ordinate
15. Convert the distance ( 4,4,2) into spherical and cylindrical
16. Show that in complex number r = √ x2+ y2
17. Find the absolute value of 4+5i
18. Determine whether the following functions or odd or even xsinx, xcosx , x2 Cosx, e ix
19.Find the inner product of <x/x+1> (0,1)
20. Find the norm of <x> (0,1)
21. Find the norm of <x +2i > (0,1)
22. Consider the functions sinx and cosx. Are they orthogonal in ( 0, π2 ) and in ( 0,π )
23. Normalise the function,Ѱ = A sin (nπx
a ) for a particle in one dimensional box of length ‘a’.
24. Show that the wave function describing 1s orbital of hydrogen atom is normalized where ѱ =
25. Normalize the trial wave functions Ψg = c1[Ψ1s(A)+Ψ1s(B)] and Ψu = c2[Ψ1s(A) –Ψ1s(B)].
26. The wave function for a particle in one dimensional box is sin sin (nπx
a ) . Normalize this function in
the interval (0,a).
27. Normalize exp(ikx) for 0 ≤ x ≤ π
28. For what value of A the function Ax2is normalized for 0 ≤ x ≤ 1
29. Normalize exp(imx) for 0 ≤ x ≤ 2π
30. Show that for 0 ≤ x ≤ a , sin(2π/a)x is orthogonal to sin(3π/a)x.
31. Define the following. 1. Hermite equation 2. Hermite polynomials
32.Obtain the following Hermite polynomials for (i) n = 0 (ii) n = 1 (iii) n = 3
33. Define the following 1. Associated Legendre equation2. Associated Legendre polynomials
34. Get the following polynomial functions for a rigid rotor: (i) P00(cosθ) (ii) P1
0(cosθ)
35. Hermite polynomial is Hn = ( -1) n e y2
dn
d yn (e− y2
) .Evaluate the first three values.
36.Associated Legendre polynomial is Pl(x) = 1
2l . l ! d l
d x l ( x2 – 1)l where x = cos θ
Find first three values.
37. Associated Laugree polynomial is Ln+l2 l+1 = Lk
p (ρ) = ( ddρ ) p [ e r (
ddρ ) k ( r k e –r) ]
Find the values for n=1,l=o , n= 2,l = 0, and n=2, l=1
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38..Legendre polynomial may be derived from the generating function
Pl(x)=1
2l . l ! d l
d x l (x2 – 1)lDerive the first four polynomials as a function of cos θ, taking x = cos θ.
39. .Find the normalisation constant of the function sin nπx
l within the limits 0 to l
40. The wave function for 1s orbital and 2s orbitals are 1√π
e –r and 1
√32 π ×(2- r) × e –( r2 ) respectively.
Show that they are orthogonal.
2.CLASSICAL MECHANICS
1. Calculate the de Broglie wave length for an electron with a kinetic energy of
a. 100 eV. b. 8.01 × 10-18 J.
2. Calculate the de Broglie wavelength of the electron in the first Bohr orbit of
hydrogen atom(given: r = 0.529Å).
3. In the photoelectric effect, the maximum kinetic energy of electrons emitted from a metal is 1.6 × 10-19 J,when the frequency of radiation is 7.5 × 1014 Hz. Calculate the threshold frequency of the metal and stopping potential of the electrons
4. If the work function of chromium is 4.40 eV, then calculate the kinetic energy of electrons emitted
from the chromium surface that is irradiated with UV radiation of wavelength 200 nm. What is the
stopping potential for these electrons?
5. Show that the value of Stefan-Boltzmann constant is 5.66 × 10-8 J m-2 K-4 s-1
6. Rigel, the brightest star in constellation Orion, has approximately a blackbody radiation spectrum with
a maximum wave length of 145 nm. Estimate the surface temperature of Rigel
7.Sirius, one of the hottest known stars, has approximately a blackbody radiation
spectrum with λmax at 2600 Å. Estimate the surface temperature of Sirius.
8. Calculate the wave length in Å of the second line in Paschen series of hydrogen spectrum
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9. Find the original energy level of the electron for a line in the Lyman series of hydrogen corresponding to a wavelength of 1.03 × 10-5 cm
10. Explain quantum mechanical tunneling with two experimental evidences for it.
11. Show that the product of Δx and Δp for a particle obey uncertainty principle.
12. State Bohr’s correspondence principle
13. State Heisenberg’s uncertinity principle
14.. a.State and explain Compton effect .
b. X- ray of wavelength 2 Ao is allowed to fall on a carbon sheet, the scattered wave is found
to have the wavelength of 2.5 A. Calculate the Compton shift
15. What do you understand by the dual character of matter?
16. Discuss the origin of hydrogen spectrum and discuss the various lines in H- spectrum.
17. Derive de-Broglie’s equation. What is its significance?
18. Calculate the de-Broglie wavelength of electron moving with velocity of 6.62 m/s.
19. What is a black body?
20. What are the characteristics of black body radiation?
21. State wein’s law.
22. Write a note on the distribution of energy in a continuous spectrum black body radiation.
23. Write the plank’s black body radiation equation and explain various terms.
24. Write the postulates of plank’s quantum theory of radiation.
25. What are photoelectrons?
26.Write Einstein’s photoelectric equation and explain the meaning of the symbols.
27. State the laws of photoelectric emission.
28.Mention any two applications of photoelectric effect.
29. What is meant by Compton scattering of X-rays?
30.What is photoelectric effect? Show that photoelectric effect depends upon frequency and not on the intensity of incident radiation.
31.What is Compton effect? Deduce a mathematical expression for the Compton shift produced in a scattering.
32.Describe an experiment to verify Compton effect.
34. Briefly explain the quantum theory of radiation proposed by planck and mention its success.
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35.A monochromatic beam of X-rays of wavelength 1.24Ao is viewed at an angle of 60o to the direction of incidence. Calculate the Compton shift.
36. An X-rays photon of wavelength 0.3Ao is scattered through an angle of 45o by a loosely bound
electron. Find the wavelength of the scattered photon .
37.X-rays of wavelength 0.112nm is scattered from a carbon target. Calculate a)the wavelength of X-
rays scattered at an angle 90o with respect to the original direction, b)the energy of the scattered electron
after the collision.
38. What do you understand by the dual character of matter?
39. The kinetic energy of sub-atomic particle is 5.85 ×10-25 J. Calculate the de-Broglie wavelength of an
electron that has been accelerated from rest through a potential difference of 1 kV
40. Calculate the wavelength associated with an electron (mass 9.1 ×10-31 kg) moving with a velocity of
103 m sec-1 (h=6.626 ×10-34 kg m2 sec-1).
41. Calculate the kinetic energy of a moving electron which has a wavelength of 4.8 pm. [mass of electron
= 9.11 ×10-31 kg, h = 6.626 ×10-34 Kg m2 s-1].
42. Calculate the momentum of a particle which has a de-Broglie wavelength of 1A°. [h = 6.626 ×10-34
kg m2 s-1] [Ans. : 6.63 ×10-24 kg ms-1]
43. Calculate the uncertainty in the velocity of a wagon of mass 3000kg whose position is known to an
accuracy of ± 10 pm (Planck’s constant = 6.626 ×Kg m2 s-1.
44. Calculate the uncertainty in the position of an electron if the uncertainty in its velocity is 5.7 ×105
m/sec (h = 6.626 ×10-34 kg m2 s-1, mass of the electron = 9.1 × kg).
45. In the photoelectric effect, the maximum kinetic energy of electrons emitted from a metal is 1.6 × 10-19
J, when the frequency of radiation is 7.5 × 1014 Hz. Calculate the threshold frequency of the metal and
stopping potential of the electrons.
46. Calculate the de-Broglie wavelength of a proton moving with a velocity 1.7 × 10 7 m/sec whose
mass is 1.67× 10 -27
47. Find the energy of the neutron whose mass 1.67 ×10 -27 Kg , de-Broglie wavelength is 1 A
48. What is the de-Broglie wavelength of an electron which has been accelerated from rest through a
potential difference of 100 V
49. Compute the de-Broglie wavelength of 10 KeV neutron whose mass is 1.675 ×10 -27 Kg
50.An electron has a speed of 600 m/s with an accuracy of 0.005 % calculate the certainty with which
we can locate the position of electron
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3. INTRODUCTION TO QUANTUM CHEMISTRY
1. Show that d/dx is a linear operator whereas √ is not.
2..Derive the expression for linear momentum operator
3. Find equivalent operator for (A+B)2 if the operators A and B commute
4. Evaluate ABC[x3] if A=d2/dx2 , B= x+3 and C = d/dx
5. What are linear operators?
6. Write note on a. addition and subtraction of operators b. multiplication of operators
7. For a particle with position vector, r = 2i-3j+k in m and momentum vector, p = i+2j-2k in kg m/s, calculate the magnitude of the angular momentum
8. Show that the wave functions corresponding to two different eigenvalues of a Hermitian operator are orthogonal
9. Show that the eigenvalues of Hermitian operators are real.
10. Explain the properties of Hermitian operator.
11. Define Hermitian operator. Give an example12. Evaluate the commutator [Lz,Lx] and mention its significance13. Show that the operators of any one of the angular momentum component commute with the operator
of the square of angular momentum (L2).
14. Evaluate the commutator for angular momentum components Lx and Lz
15. Prove the commutation relation [x, px] = ih/2π16. Verify whether the following pair of operators commute: d2/dx2 and x.17. Show that the operators L2 and Lz commute
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18. Evaluate pxx2.
19. Prove the commutation relation [p2x, x] = -2iћp.
20. Show that [x, d/dx] = -1.
21. Show that [L2,Lx] = 0. Mention its significance.
22. Prove that the angular momentum and kinetic energy of a particle can be measured simultaneously to an arbitrary precision
23. Obtain the value of [x, px2]. Mention its physical significance.
24. Show that the function cos 2xs is an eigen function of ∇2 = 2/ x2+ 2/ y2+ 2/ z2.
25.How will you write the acceptable wave function for an atom containing two electrons?
26. Check the position operator and momentum operator commute each other.
27. Obtain the expression for the following operators. a. Momentum b. Angular momentum
28. Find the eigen value of the function sin 2x for the operator d2
dx2
29. Define an operator.30. Find the expression for angular momentum operator Lx.31.Give the expression for angular momentum in polar co- ordinates 32.What do you mean by Hermitian operator.?
33.Show that the commutator [ x, ddx ] = 1
34.Evaluate [ z3, ddz ]
35. Prove that [ Lz, z ] = 036.Check the differential operator is linear with respect to the functions x2 and 3x2.37.Define eigen function and eigen value.
38.Find the eigen value of the function cos 7x with the operator ddx
39.What do you mean by normalisation constant?.40. Give the condition for orthonormal function.
41.Show that sin 2x is not an eigen function of the operator ddx but of
d 2dx 2
42.Mention the characteristics of wave function.43.Find the expression for Kinetic energy operator.44..Prove that Lx and Ly can not be precisely specified simultaneously.45. Can L and Lx be precisely specified simultaneously ? - Reasonout 46.Show that the momentum operator is Hermitian.47.Momentum of free particle commutes with the Hamiltonian operator -Justify48.Prove the following results. a. [ Lx, x ] = 0 b. [ Lx, y ] = ihz c. [ Lz, y ] = - ihx
49. Which of the functions are eigen functions ofd 2dx2 . For the eigen functions state the eigen values. Sin
2x, 6 cos 3x, 5 x2, 1/x, log x , e -2x
50. Show that the function sin 2x sin2y sin2z is an eigen function of ∇ 2 .Page 376 of 419
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51.Verify the differential operator is linear with respect to the functions 2x and 3x.
52. Check ‘log’ operator and square root (√) operator are linear or not
53.Check the operators ddx and 3commute or not
54.Find the commutator of the operators x and ddx with respect to the function x2
55.Find the eigen function of the operator ddx
56. Show that the wave function Ψ = x e− x2
2 is an eigen function of the operator H = - d2
d x2 + x2 and
find the eigen value
57. Calculate the eigen value, if the function 1π sin ( 3.5 x) is an eigen function of the operator
H = - h2
8 π 2m d2
d x2
4. SCHRODINGER EQUATION &
ITS APPICATIONS TO SIMPLE SYSTEMS1.State and explain the postulates of quantum mechanics
2. Derive time independent Schrodinger wave equation from time dependent equation
3. Derive time-dependent and time-independent Schrodinger wave equations.
4. Calculate the energy for the transition from n = 2 to n=3 state for an electron in a one dimensional box
of length 5.78 Å.
5.Determine the energy required for the transition from nx = ny = nz =1 to nx = ny = 1, nz = 2 state for an
argon atom (atomic mass = 39.95 g mol-1) in a cubic container with 1.0 cm side
6. Derive the wave function and energy for a particle in a 1 D box of length ‘a’.
7. Derive the wave function and energy for a particle in a rectangular three dimensional box
8. Determine the wave length of light absorbed when an electron in a linear molecule of 11.8 Å long makes a transition from the energy level, n = 1 to n = 2
9. Show that for a particle in a three dimensional box with lengths, lx = ly = lz/2, the
energy levels 122 and 114 are accidentally degenerate
10. Derive the expressions for wave function and energy for a particle in a rectangular box.
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11. For a particle of mass 2 × 10-26 g in one dimensional box of length 4.00 nm, calculate the wavelength of the photon emitted when the particle goes from n = 3 to n= 2 level
12. Compare zero point energy of a particle in one and three dimensional boxes of same length
13.Write an expression for the wave function of a particle confined to move in a cubical box of edge
length ‘l’ having energy 12h2/8ml2.
14. When a particle of mass 9.1×10-31 kg in a certain one dimensional box goes from n = 5 level to n=2
level, it emits a photon of frequency 6 ×1014 s-1. Find the length of the box.
15. State the order and degree of the Schrödinger equation for a particle in a one dimensional box.
16. Write expressions for the third levels for Ψnand En for a particle in 3D box.
17. Calculate the highest translational quantum number for an oxygen molecule in 1mm length to have its
thermal energy kT at 298K(k = 1.38 x 10-23Jmolecule-1).
18. Evaluate the following for a particle in 1D box: (i) <px > (ii) px2Ψ.
19. Set up the Schrodinger equation for a particle in 1D box and hence solve for its energy and
wave function
20. A particle of mass m is confined to move in a one-D box between x=0 and x= L. Write the normalised wave function of the particle.21. Find the energy gap between third and fourth level of a particle in a one- D box.22.Sketch the wave functions Ψn and Ψn
2
23. What are nodes? Predict the points where node occurs for the first four states.24. Define degeneracy and degree of degeneracy
25. Determine the degree of degeneracy of the energy level 38 h2
8 m L2 of a particle in a cubical box.
26.Show that Ψ1 and Ψ2 for a particle in a one- D box are orthogonal.27. Find the lowest energy of electron confined to move in a one D box of 1 Ao. Given mass of electron = 9.11× 10 -31 Kg, Planc’s constant = 6.62 × 10 -34 JS, 1 eV = 1.6 ×10 -19 J 28.. The normalised wave function of a particle in one- D box is given by
Ψ = √ 2L sin (
nπL ) x Show that the function is orthogonal.
29.Find the energies of the six lowest energy levels of a particle in a cubical box. Which of the levels are degenerate.?. Mention their degeneracy30.Find the lowest energy of neutron confined to move in a cubical box each side 1 Ao. Given mass of
neutron = 1.67× 10 -27 Kg, Planc’s constant = 6.62 × 10 -34 JS, 1 eV = 1.6 ×10 -19 J
31The normalised wave function for a particle in a one dimensional box is Ψn = √2a sin
nπxa . Find the
expectation value for position x .32.Find the expectation ( average)value of energy of a particle of mass m confined to move in a one -D box of width L and infinite height with potential energy zero inside the box. The normalised wave
function is Ψ n = √ 2L sin (
nπL ) x
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33. Arrive the expression for wave function for a particle in a three dimensional box and find the expression for energy of the particle. 34..Write the Schrodinger wave equation for a particle in one-D box and solve it to find the wave function. Find the normalisation constant and the energy associated with the particle.35. What is zero point energy?. Show that ZPE of 3- D box. is thrice that of 1- D box.36. Obtain the values of x , at which nodes occur in Ψ 3.
37. Write down the degenerate states for which the eigen values for energy is 9 h2
8 m L2
38. A particle of mass m is confined in a 1-D box of length L, with the conditions that the
potential inside the box is zero. Arrive the expression for
1. wave function 2. energy eigen value.
39. A particle is placed in a rectangular box of sides a, b and c. Derive the expression for
wave function and energy.
40.A particle of mass ‘m’ is confined to move in a 1-D between x = 0 and x = L. The Potential Energy of
the particle is zero between x = 0 and x = L. Write the time independent Schrödinger equation
41.Write the expression for wave function and energy of 1 – D box for first two levels
42.Show that the Eigen energy of a particle in ground state whose mass is equal to the square of the
Planck’s constant placed in a box of one meter length is 0.125.
43.Find the energy gap between second and third level for a particle in a box and show that the energy
gap is inversely proportional to square of the length of the box
44.Sketch the functions ψ n and ψ n2 for n = 1, 2, 3.
45.A particle of mass ‘m’ is confined to move in a 3-D rectangular box of dimensions Lx, Ly & Lz.
Write the expression for Eigen function and Eigen value.
46. A particle is confined to move in a cubical box of length ‘L’. Give the expression for energy and wave
functions for three levels.
47.What is a node? Determine the nodes for a particle in a 1-D box of length L for the first three levels
48.For the particle in a 1-D box of length L. Find the probability in the ground state and first excited state
that the particle is in the region 0 < x < L/4
49. Show that the probability of finding a particle in the vicinity of the centre of a 1-D box of
length one meter is less than the probability of finding it in the range 0 < x < 0.2 nm for
the state n = 1.
50. Show that the probability of finding a particle in a 1-D box of length L in the region
x = L/4 to 3L/4 is 0.5 if n is two.
51. Write the Schrödinger wave equation for a particle in 1- D box and solve it to find out wave function. Find the normalization constant and the energy associated with the particle. 52 Butadiene contains four П electrons each of which moves freely from one end of the molecule to the other. Treat the molecule as a 1-D box whose length is equal to the length of the carbon chain plus half
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the C – C bond length on either side. The average C – C bond length is 0.14 nm. Calculate the positions of maximum and minimum electron density in the molecule. 53. Show that the E corresponds to the electronic transition for HOMO to LUMO of Hexatriene is 7
h2/8ma2 and find the length corresponds to the transition. Assume C-C bond distance is 1.4 A0
54. For and electron ( m ═ 9.1 ×10-31kg) in a 3-D rectangular box of dimensions Lx = 1× 10-15m, Ly = 1.5
x 10-15 and Lz = 2 x 10-15m, (i) Calculate the ground state Energy and the wave function.
55.What do you mean by degenarcy. Find the degeneracy of a particle confined in a 3-D box for the first
four levels.
56.A particle is confined to move in a 2-D rectangular box having dimensions Lx = 2Ly. Show that there is
no degeneracy.
57. Find the length of butadiene chain which has absorption maximum at 2150 A
[ 5.71 ×10−10 m]
58. The length of Hexatriene molecule was found to be 8.67 A . Find the wavelength for the first
transition. [ 354 nm] OR Calculate the wavelength of π → π transition in 1,3,5 – hexatriene
59. Octatetatraene gives first transition absorption band at 4667 A. At what length of the molecule this
transition corresponds to?[ 11.2 A]
5. SCHRODINGER EQUATIONAPPLICABLE TO COMPLEX SYSTEMS
1. Solve the Schrodinger equation for simple harmonic oscillator and obtain its energy levels
2. Set up the Schrodinger wave equation for a simple harmonic oscillator and solve it
for the energy eigenvalues
3.Get the normalized functions for the simple harmonic oscillator for its third vibrational level.
4.Get an expression for the total energy of a simple harmonic oscillator in terms of its
amplitude and frequency
5. The infrared spectrum of 75Br19F consists of an intense line at a frequency of 1.14× 1013 s-1.
Calculate the force constant of 75Br19F
6. The force constant for H79Br is 392 Nm-1. Calculate the fundamental vibrational frequency and zero
point energy of H79Br.
7. Verify, whether the energy of a rigid rotor is quantized
8.Use the method of separation of variables to break up Schrodinger equation for a rigid rotor into
ordinary angular equations. Discuss the nature and characteristics of the solution of each.
9.Set up the Schrodinger equation for a rigid rotor and hence solve for its energy and wave functions.
10.Evaluate the spherical harmonics Y0,0
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11 Get the following polynomial functions for a rigid rotor: (i) P21(cosθ) (ii) P3(cosθ)
12.Use the method of separation of variables to break up Schrodinger equation for hydrogen atom into ordinary angular equations and write the solutions for each
13. Write the Schrodinger equation to be solved for H atom and solve it for its energy using a simple solution, which assumes the wave function to depend only on the distance r and not on the angles θ and φ.
14. Draw the radial distribution plot for 2p,3p, 3d and 4s orbitals of H-atom and indicate the nodes.
15. Obtain the point at which the probability density of 3dz2 orbital will be maximum. Given ψ320 = Cr2exp(-r/3) (3cos2ө – 1) where C contains all the constants16.Derive an expression for the energy of electron in hydrogen atom.
17.Solve the radial eigenfunction for R2,0(r).18.Solve φm equation for hydrogen atom, when m = 0.
19.Mention the importance of radial density function with an example20.Write down the Laugerre polynomial for 1s electron.
21.Solve the polar angle dependent equation for Hydrogen atom22.Show that for a 1s orbital of a hydrogen like ion, the most probable distance from the nucleus to electron is ao/Z.23.The normalized wave function for the 1s orbital of hydrogen atom is Ψ1s =1/(π)1/2(Z/a0)3/2exp(-Zr/ao).
Show that the most probable distance of the electron is a0.
24.At what distance from the nucleus is the probability of finding the electron a maximum for a 1s
electron in hydrogen?
25.Find out the most probable distance of 1s electron of hydrogen atom using the wave
function Ψ1s = 1/(π)1/2(Z/a0)3/2exp(-Zr/ao). Calculate the values for the atoms from hydrogen
to boron and offer your comments upon their ionization potentials.
26.Find the radius of the shell where there is a maximum probability of finding the electron. Given: The probability, P = 4πr2e-2ar
27.Show that for anhydrogen like atom, in its ground state, the average distance of the electron from the nucleus is 3/2 times the most probable distance. Given: Ψ1s = 1/(π)1/2(Z/a0)3/2exp(-Zr/ao).
28.Show that the energy levels of SHO are equally spaced. 29.Find the angular eigen function of 2s orbital30.Evaluate the radial function of 1s orbital.31.What is spherical harmonics?32. What is rigid rotator? Write down the expression for its energy33. Write the normalized wave function for the spherical harmonics Y 0,0 , Y 1,0 and Y 1,1
34 Write down the expression for total wave function of H-atom and find the values of Ψ 100 , Ψ 200 and Ψ 210 35.Find the angular eigen function of 2p orbitals36.Write down the Schrodinger equation for Simple harmonic oscillator and solve it to obtain the its wave function .
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37.Write down the Schrodinger equation for rigid rotator and separate in to azimuthal and polar equation and solve the azimuthal equation to obtain the solution
38. The polar equation for a rigid rotator is sin 2θ
X d2 X
d θ2 + sin θ cosθ 1X
dXdθ +β sin2 θ = m2
Solve it to obtain the solution.
39. Write down the Schrodinger equation for H-atom in Cartesian co-ordinates, convert it into spherical polar co-ordinates and separate in to different equations.
40. Determine the energy of simple harmonic oscillator by series solution method.
41. Find the expression for energy of first four states of SHO. Are the energy levels degenerate?
42.Compare the classical energy of SHO with that of quantum mechanical energy
43 Write the Schrodinger equation for H- atom in spherical co-ordinates.
44.Show that the θ equation of rigid rotator is reduced in the form of associated Legendre differential
equation,(1-x2)∂2 y∂ x2 - 2x
dy∂ x +( n (n+1) -
m2
(1−x2)¿ X = 0
45..Establish Schrodinger equation for a linear harmonic oscillator and solve it to obtain its eigen functions and eigen values. Discuss the significance of zero point energy.
46.The wave functions for the SHO are given as Ψ1=(απ )¼
e− y 2
2 √2y), Ψ2=(απ )¼ e
− y 22 (
2 y 2−1√ 2 ) Show that a.
Ψ 2 is normalized b. Ψ 1 and Ψ2 are orthogonal.
47.Write down the Schrodinger wave equation in polar co ordinates ,for a rigid rotator. Separate the equation into azimuthal and polar wave equation and solve the azimuthal wave equation. Find the normalization constant.
48.Solve the radial part of Schrodinger equation for the H-atom and obtain the energy eigen values. Draw the wave functions of first three levels
49..Sketch the probability diagram for the ground state of simple harmonic oscillator.50..Give the normalization constant value for simple harmonic oscillator.51..Write the Rodrigue’s formula for Hermite polynomials.52.. Discuss the significance of zero point energy53. Eigen energy of SHO is non-degenerate –why?54.Show that energy levels of SHO is equally spaced but that of a particle in a box is not . 55. Show that the probability of finding a simple harmonic oscillator within the classical limits , if
the oscillator is in its normal state is 84%
56. .Solve the Schrodinger equation for the linear harmonic oscillator and determine the
normalized wave functions.
57. Show that ,if the oscillator is in its normal state ,then the probability of finding the particle
outside the classical limits is 16 %
58 Write time independent Schrodinger equation for H-atom in spherical co-ordinates and separate in to three
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59. Write the radial part of the time independent Schrodinger equation for H- atom and solve it to obtain the energy
of the atom.
60. Solve the angular equation (polar wave equation) of H- atom.
61. Calculate the probability of finding the electrons in a sphere of radius r= a0 ,given that the 1s
wave function is 1/ a03 e – r/a
0. ( ANS: -0.323)
62.Show that as the atomic number(Z) increases , the most probable distance decreases and hence
calculate the most probable distance at which the electron the 1s electron is to be found.
( ANS: r = a0/z, r = a0 )
63.The 2s state of H-atom is given as = 1/ 4 2 (2-r) e – r/2. Find the most probable distance of electron from the
nucleus and nodal point.
64. Establish the value of 100 and 200 for H-atom.
61..Show that the wave function Ψ1 = ( απ ) ¼ e
− y2
2 ×√2 y where y = (√α x) is normalized
62. Show that the wave function Ψ1 and Ψ2 of harmonic oscillator are orthogonal
63.Find the angular eigen function of 1s orbital of H- atom in atomic mass unit.
64.Find the radial eigen function of 1s orbital of H- atom in atomic mass unit.
65.Find the total eigen function of 1s orbital of H-atom in atomic units.
66.Find the angular eigen function of 2s orbital of H-atom.
67. Find the radial eigen function of 2s orbital of H- atom in atomic mass units.
68.Find the total eigen function of 2s orbital of H-atom in atomic units.
69.Find the angular eigen function of 2p orbitals.
70.Find the radial probability of finding the electron and the most probable distance at which the1s electron of H-atom is to be found.The value of R(r) for 1s orbital in atomic unit (a.u) is given by R(r) = 2 z 3/2 e - ( z r ) where z is the atomic number
71.The value of R(r) for 1s orbital is given by R(r) = 2 (Z 3
πao 3) 1/2 e - ( zr
ao) Where z is the atomic
number. Find most probable distance at which the 1s electron of H-atom is to be found.
72.Calculate the probability of finding the electron in a sphere of radius r , given that 1s - wave
function of H-atom is (1
πa 03) ½ e - r
a 0
73.What is the probability of an electron being found at a distance of r = 0 and r = ½ a.u. from the
nucleus of H- atom in 1s state whose wave function is Ψ = 1
√ π e – r
74. Find the probability of an electron in 2pz orbital of H- atom being found at a distance of one a.u.
from the nucleus for θ = π/2 to θ = π/4 whose wave function isΨ=1
√32 πre– r/2 cos θ
75.Show that the 1s orbital and 2s orbital of H-atom are orthogonal
76..Show that the 1s orbital with wave function 1√π
e –r is normalised
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77..Show that the 2s orbital with wave function1
√32 π × (2 - r ) × e –( r2 ) is normalised
78.Show that the 1s orbital oh Hydrogen atom with wave function 1√π
( za0)
32 ×e
− zra0
is normalized
6. APPROXIMATION METHODS1.Obtain expressions for the energy of the molecular orbitals of hydrogen molecular ion using variation method
2.Apply variation theorem to the probability of finding the particle in one dimensional box of length ‘l’ using the trial wave function, ѱ = x(l-x).
3.State variation theorem and apply it to the probability of finding the particle in one dimensional box of length ‘l’ using the trial wave function, ѱ = ( − ) and compare your result with the true value
4. Highlight the importance of variation method in the determination of energy of MO for Hydrogen
molecular ion.
5. Apply variation principle to get an upper bound to the ground state energy of the particles in a 1D box of length a, using the trial function Ψ = x2(a-x).
6.What are variational integral and variational parameters?
7.Mention the significance of Secular determinant.
8. What is a Secular determinant? Write down the determinants for the excited state of
He atom
9. Calculate the average energy for a particle in one dimensional box of length ‘a’ using perturbation
theory if ѱ = (2/L)1/2 sin(nπx/L).
10.Evaluate the first order correction to the energy term when an electric field of strength ‘F’ is applied to a particle in a one dimensional box of length ‘l’
11.Find the first order correction to the energy term when an electric field of strength
‘F’ is applied to the electron in a one dimensional box of length L.Given: ψ = (2/L)1/2 sin(nπx/L).
12. Identify the perturbation term in the Hamiltonian of Helium atom.
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13. State the principle of Perturbation theory and use first order perturbation theory to calculate the energy
of a particle in a one dimensional box from x = 0 to x = a with a slanted bottom, such that V x = V0 (x)/a.
Given the wave function Ψ(0) = (2/a)1/2 sin (nπx/a).
14.Mention Born-Oppenheimer approximation with an example
15. How will you apply Born-Oppenheimer approximation to simplify the Hamiltonian for H2+?
16.Verify whether an energy state with term symbol 2P 5/2 can exist
17. Obtain the ground state atomic term symbol for fluorine. and carbon.
18. Deduce the atomic term symbol for boron.
19. Determine the possible electronic configuration of the element whose ground state term symbol is4S
3/2.
20. Obtain The term symbol of a particular atomic state is 6S5/2. Suggest a possible electronic configuration
21. Suggest a possible electronic configuration for the term symbol 3P2
22. Mention the conditions to apply perturbation theory
23. Give the expression for first order perturbation energy and wave function
24. Identify H0 and H1 for the following
a. An oscillator governed by potential energy V = ½ kx2 + ax3 + bx4
b. Helium atom
25. A hydrogen atom subjected to electric field has Hamiltonian H = - 12 ∇2
- 2r1
+rcosθ.Separate the
perturbed part from unperturbed part.
26. Write down the second order Schrodinger wave equation , energy and wave function.
27..Discuss the first order time – independent perturbation theory for non – degenerate stationary state.
Obtain the expression for eigen energy and eigen function.
28. A Hydrogen atom is exposed to an electric field of strength F so that its perturbed Hamiltonian is F
r cosθ. Show that there is no first order effect. Given Ψ 0¿ = e−r
√ π and ∫
0
∞
e−2 r r3 dr = 38
29. Show that the first order perturbation energy is zero in Stark effect
30. Show that the first order perturbation energy for a non-degenerate system is just the perturbation
function averaged over the corresponding unperturbed state of the system. Derive the expression for the
eigen function of the perturbed system
31. Outline the Schrodinger perturbation time –independent theory for non- degenerate levels and apply
it to explain first order Stark effect in hydrogen
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32. An uniform electric field along Z- axis is applied on H- atom . The perturbation term is given as H
‘ = F r cosθ .Calculate second order perturbation energy.
33. Discuss the first order perturbation theory for a non –degenerate level .Using this theory, solve the
Schrodinger wave equation for ground state of Helium and obtain the expression for eigen energy
34. Using the principles of time – independent perturbation theory, obtain the second order correction to
the wave function. Show that the second order correction to the energy of the normal state is always
negative.
35. What is the need for approximation methods?36. Give the principle behind variation method.37. List out the steps followed in variation method.38. Give two examples for perturbation39. Write down the expression for first order perturbation energy and wave function.
40.The expression for expectation energy of Hydrogen atom by variation method with Ψ = e−ar 2
is 3∝2
-√ 8 απ
where α is the parameter. Minimize this energy, to find the value of the parameter α and
determine the exact energy.
41 Apply the first order perturbation result to calculate the ground state energy of Helium atom.42. Use the trial wave function Ψ = e−arto find the energy eigen values for the ground state of H- atom using variation theorem. Hamiltonian in spherical co-ordinates is
H =−12r 2
ddr¿) -
1r [given ∫
0
∞
e−2 ar r2dr = 1
4 a3 ∫0
∞
e−2arr dr = 1
4 a2]
43. Consider Ψ as a linear combination of two eigen functions φ1 and φ 2 with normalization constant a1 and a2 respectively. Find the expectation value of energy and show that it is lesser than the lowest energy E0 Derive an expression for the eigen function of the first order perturbed system.
44. For the wave function Ψ = e−ar2
show that the expectation value of energy is
E=3 a2 - 2 (√ 2 a
π¿ Hamiltonian in spherical co-ordinates is H =
−12r 2
ddr¿) - 1r
Given ∫0
∞
r 4 e−2a r2
dr = 3
32 a2 √ π2 a
,∫0
∞
r2e−2 a r2
dr = 1
8a √ π2a
, and ∫0
∞
r e−a r 2
dr =1
4 a
45. In what way variation method differs from perturbation method.
46. Give the various steps in Variation theorem
47. The average energy of H –atom in terms of variation parameter ’a’ is a2
6 -
a2 . Find the variation
parameter and the true energy.
48. Show that the variation method provides an upper bound to the ground state energy of the system. 49. A trial function depends linearly on the variation parameter leads to secular equation and secular determinant. Prove. 50. Arrive the secular equation using variation theorem
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51.The expectation energy of H-atom were calculated as a2
6 -
a2 and
3 a2 - 2√ 2a
π when the wave functions
are Ψ = re−ar and Ψ = e−ar 2
respectively. a. Minimize this average energy and find the variation parameter ‘a’ b. Find the actual energyc. Which is better wave function ?. Comment your result
52. Apply variation method to Helium atom and calculate its ground state energy. Compare your result
with that obtained from variation method.
53. State variation theorem and apply it to the probability of finding the particle in one dimensional box of
length ‘l’ using the trial wave function, ѱ = x (l – x) and compare your result with the true value.
54. Make a reasonable guess for the ground state wave function for H-atom.
55. Assuming = e – r as the trial wave function for the ground state of H-atom, find the
energy using variation method. Hamiltonian in spherical co-ordinates is -(12) d2/dr2 –(1/r) d/dr – (1/r).
56. Assuming the wave function = Ne – r2
a. Find the best value of
b. The energy of first state of H-atom
c. Compare with ground state energy calculate the percentage error.
57. Show that the energy of ground state of H- atom is -0. 375 when the wave function is
= r e – r using variation theorem
58.An electric field of strength F is applied to an electron in a one dimensional box of length L, so that the potential energy V = eFx , rises along the box V = 0 at x = 0 and V = e FL at x= L. Find the first order correction to energy and wave function
59.A Hydrogen atom is exposed to an electric field of strength F so that its perturbed Hamiltonian is
Fz. Show that there is no first order effect. Given Ψ 0¿ = e−r
√ π , ∫
0
∞
e−2 r r3 dr = 38
60. A harmonic oscillator is subjected to perturbation H = E x. Find the first order perturbation energy
and wave function. Given Ψ0 = ( βπ¿ ¼ e
− β2 x2
, ∫−∞
+∞
(x¿e−β x2
)¿ dx = 0
61.An an harmonic oscillator is subjected to perturbation H = ax3 +bx4. Find the first order perturbation
energy and wave function. Given Ψ0 = ( βπ¿ ¼ e
− β2 x2
62. Use the functions Ψ = r e−ar to calculate the ground state energy of H-atom by variation method.
Compare the result with the true value. Hamiltonian in spherical co-ordinates is H = −12r 2
ddr¿ ) -
1r ( in
atomic units). Given : ∫0
∞
xn e−ax dx = n!
an+1
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63.For the selected wave function Ψ = r e−ar the expectation value of energy is E=α2
6 -
α2 . Find the
value of 𝛂 and hence calculate the energy. Compare your result with the true energy and comment the result.
.
7. CHEMICAL BONDING1. Write down the Hamiltonian for Hydrogen molecule
2.Write the Hamiltonian for helium atom and mention the terms involved.
3.Write down the Hamiltonian for Hydrogen molecular ion.
4.Write the Hamiltonian for H2+ and explain the terms involved.
5.Write the Slater determinant for the ground state of helium atom.
6.Write the Slater determinant for the ground state of Helium atom. Show that it is antisymmetric with
respect to the exchange of the two electrons
7.What are resonance and coulomb integrals? Obtain their expressions
8.What are coulomb and exchange integrals? How are they obtained?
9.How is the energy of the orbitals of hydrogen molecular ion determined through energy and overlap integrals?
10.Solve the secular determinantel equations of allyl cation and allyl anion for their delocalization energy
11.Find the Huckel molecular orbitals and energies for 1,3-butadiene
12.Find the Huckel molecular orbitals and energies for allyl radical
13.Write down the secular determinant for ethylene molecule using Hückel's method and obtain
expressions for its energy levels.
14.Give the assumptions of Huckel molecular orbital theoryPage 388 of 419
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15.Set up the secular determinant for allyl radical and obtain its energy levels
16.Apply Huckel’s method to allyl cation and obtain expressions for the energy levels.
17.What are the features that distinguish the Huckel method from other LCAO methods?
18. Calculate the wave length of π →π* transition in 1, 3, 5-hexatriene
19.Calculate the wave length of π →π* transition of the following molecule:
(Bond distances: C-C, 0.154 nm; C=C, 0.133 nm)
20.Draw the MO diagram for the π-electrons in 1, 3-butadienyl anion radical.
21.Highlight the features that distinguish the Huckel method from other LCAO methods.
22. Obtain the normalized trial functions for bonding and antibonding orbitals of H2 molecular ion.
23.Highlight the important approximations that distinguish the HMO method from other LCAO methods
24. Explain the importance of the integrals HAAand HBB obtained for the lowest energy of H2+using LCAO
method
25. Outline the salient features of VB(Heitler-London) theory as applied to Hydrogen molecule.
26. What are the three important approximations of Huckel LCAO-MO theory?
27. State Pauli’s anti symmetric principle and illustrate it for the ground state of helium atom..
28.Illustrate the Pauli Exclusion Principle for the ground state of He atom.
29.Obtain the Pauli antisymmetric wave function for the excited state He atom
30. Discuss the Pauli principle of anti-symmetric wave function
31. What is a Hartree? Give its value.
32. Using the following HMOs calculate the Π – bond order of butadiene between the adjacent
carbons in butadiene.Ψ1 = 0.372 φ1 + 0.602 φ2 + 0.602 φ2 + 0.372 φ2
Ψ2 = 0.602 φ1 + 0.372 φ2 - 0.372 φ2 - 0.602 φ2
33. Write down the secular determinant for hexa triene.
34.Which of the following HMOs are normalized?
Ψ1 = 1
√ 2 ( p1 + p2) , Ψ2 = 1
√ 3 ( p1 -2 p2) , Ψ3 = 1
√ 3 ( p1 -2 p2 + p3)
35. What do you mean by hybridization?.
36.Mention the degenerate energy levels of benzene in terms of α and β
37. Apply HMO theory for ethylene molecule and determine its energy and wave function .calculate its free valency, charge density and π- bond order.
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38. Explain Huckel Molecular Orbital theory for multi electronic atoms and mention the approximations used in his theory39. Write down the secular equations and set up the secular determinant for butadiene . Determine the energies of the HMOs in terms of 𝛂 and 𝛃 and calculate the free valencies of the terminal carbons.40.. Give the secular equations for benzene . Calculate the energies and HMO coefficients.determine its π- bond order 41. Use 2s and 2pz atomic orbitals to construct two equivalent sp hybrid orbitals and determine the angle between the hybrid orbitals . 42.. What is Screening effect ?43 Discuss the applications and limitations of Slater rule.
44. What do you mean by Effective nuclear charge. Calculate the Effective nuclear charge of
2s electron of Nitrogen and calcium
45. What are Slater type orbitals?
46. Find the term symbol for N.
47. Determine the effective nuclear charge of 2s and 2p electrons in C .
48. Give the expression for columbic, resonance exchange and overlap integral
49. Write down the secular determinant for H2 molecule.
50.. For a one electron homonuclear diatomic molecule, the values of some relevant integrals are
∫φ1 H φ 1dτ = -2 a.u, ∫φ 2H φ 2 dτ = -2 a.u, ∫φ1 H φ 2 dτ = -1 a.u ∫φ1 φ 2dτ = 0.25 Find the
energy of this system and also the corresponding normalized wave function.
51.. Explain VB theory for H2 molecule considering Ψ as a linear combination of two eigen
functions φ1 and φ 2 with normalization constant a1 and a2 respectively.
52. Explain linear combination of atomic orbital- molecular orbital for diatomic molecules
and arrive the condition for effective combination
53. Write down the secular determinant of benzene
54 What is screening constant. Calculate the screening constant of 2s electron of Carbon
55 Differentiate Slater type orbitals and Hydrogen like orbitals...
56 Give the expression for wave function and energy of ethylene.
57. Apply HMO theory to butadiene and arrive the expression for its wave functions..
58 Explain the application of Valence bond theory to Hydrogen atom.
59. Write note on 1. LCAO-MO method for H-atom 2. HFSCF method.
60. Write note on Semi Empirical methods.
61 Find the ground state term symbol for d1 ion and d2 ion
62. Mention the degenerate energy levels of benzene in terms of α and β
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.63. Apply HMO theory for ethylene molecule and determine its energy and wave function .calculate its free valency, charge density and π- bond order.
64. Explain Huckel Molecular Orbital theory for multi electronic atoms and mention the approximations used in his theory65. Write down the secular equations and set up the secular determinant for butadiene . Determine the energies of the HMOs in terms of 𝛂 and 𝛃 and calculate the free valencies of the terminal carbons.66 Give the secular equations for benzene . Calculate the energies and HMO coefficients.determine its π- bond order 67. Find the expression for STO for 2s, orbital in Nitrogen atom.. The normalisation constant is ‘N’. Compare with Hydrogen – like orbital.68.Find the expression for STO for 2px, orbital in Nitrogen atom.. The normalisation constant is ‘N’. Compare with Hydrogen – like orbital69.Calculate the average value of ‘r’ for 1s electron of Lithium atom using Slater type orbitals.70. Calculate the first ionisation energy for Li atom on the basis of Slater rules.
71. Determine the effective nuclear charge for the
A. 1s electron in He. B. 2s and 2p electrons in C.
72. Determine the effective nuclear charge of
A. 2s and 2p electrons in N. B. 3s and 3p electrons in S.
73. Determine the effective nuclear charge of 1s electron in F.
:
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APPENDIX
Apply iterative method x1= f(0) = = 0+5
6 = =
56
= 0. 8333
f(x) = x3- 6x+5 x2= f(0.83) = (0.8333 )3+56
= 0.578+5
6 =
5.5786
= 0.9296
let x3- 6x+5 = 0 5 x3 = f(0.9296) = (0.9296 )3+56
=
6x = x3+5
x = x3+56
till we get value equal to previous value
F( x) = x2 – 6x +5x2 – 6x +5 = 0 6x = x2 +5
x = x 2+5
6
x1= f(0) = (0 )2+56
= 0. 83
x2 = f( 0.833) = (0.83 )2+56
= 6.93+5
6 = 0.94
x2 = f( 0.94) = (0.94 )2+56
= 0.88+5
6 = 0.98
x2 = f( 0.98) = (0.98 )2+56
= 0.96+5
6 = 0.99
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x2 = f( 0.99) = (0.99 )2+56
= 0.92+5
6 = 0.99
till we get value equal to previous value
INTEGRATION FORMULA
1. ∫ xn dx = xn+1
n+1
∫ x dx = x2
2
∫ dx = x
2. ∫ sinax dx = −1a cos ax 4.
∫ sinx dx = - cosx
3 ∫cos ax dx = 1a sin ax
∫cos xdx = sin ax
4 . ∫ dxx+a = log( x + a )
∫ dxx = log x
5 . ∫ eax dx = 1a eax
∫ ex dx = ex
6. ∫ d x√a2−x2 = a
2
2 sin – 1
xa
7. ∫ d xa2+x2 =
1a tan – 1
xa
8. ∫ √a2−x2dx = x2 √a2−x2 +a2
2 sin – 1
xa
9. ∫ √a2+x2 dx = x2 √a2−x2 +
12 a log ( a +√a2+x2 )
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10. ∫ eax sin bxdx = eax
a2+b2 [ a cos bx + bsin bx ]
11. ∫ eax cosbx dx= = eax
a2+b2 [ a sin bx - b cos bx ]
12. ∫ f n f , dx or ∫ f '
f n dx put t = f
13. ∫UV = U V I - U I V II + U II V III - ….. [ Bernoulis formula]
14. ∫0
∞
e−ax sin bx dx = b
a2+b2
15. ∫0
∞
e−ax cosbx dx = a
a2+b2
16. ∫ xx2+b2 dx =
12 log (x2+b2 )
PROPERTIES OF INTEGRALS
1. ∫a
b
f ( x )dx = - ∫b
a
f ( x )dx
∫1
2
f ( x )dx = - ∫2
1
f ( x )dx
2. ∫1
2
f ( x )dx = ∫1
2
f (1+2−x )dx
3. ∫−a
+a
f (x )dx =2∫0
a
f ( x )dx if f(x) is even [ if f(x) = f(-x) , the function is even if not odd]
= 0 if f(x) is odd
TRIGNOMETRIC FORMULA
1. sin2x = 2 sinx cos x
2. cos2x = cos2x - sin2x
= 1- 2 sin2x
= 2 cos2x – 1Page 394 of 419
2020 - CONCISE QUANTUM MECHANICS 2017
3. cos2x = 1−cos2 x
2
4. sin2x = 1+cos2x
2
5. cos 3x = 4cos3x – 3cosx
6. cos3x = 14 cos 3x +
34 cosx
sin 𝜋 = 0, cos 𝜋 = -1
sin π2 = 1, cos
π2 = 0
cos 0 = 1, sin 0 =0
sin π4 = cos
π4 =
1√2
sin (3π2¿ = -1
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Long live scientist DoberinerAnd his Law of Triads too
Long live Scientist NewlandAnd his Law of Octaves too -2
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Lithium Sodium Potassium Rubidium Cesium Francium -2 Which are first A group metals-2 They are known as Alkali metals-2( long)
2. Beryllium Magnesium Calcium Strontium Barium Radium -2 Which are second A group metals -2 They are known as Alkaline earths -2( long)
3. Boron belongs to third A group Aluminium Gallium Indium Thalium -2 Are the members of Boron family -2 They form first group of p- block elements-2(long)
4.Carbon Silicon Germanium Stannum Plumbum form fourth A group-2 Fourth B consists of three metals-2 Titanium Zirconium Hafnium – 2( long)
5. Phosphorous Arsenic Antimony Bismuth Belong to fifth A Nitrogen family – 2 Sulphur Selenium Tellurium Polonium – 2 Belong to sixth A Oxygen family-2( long)
6. Fluorine Chlorine Bromine Iodine Astatine are the five halogens-2 They do belong to seventh A group-2 They form salt with strong bases.-2( long)
7. Helium Neon Argon Krypton Xenon Radon are the rare gases-2 They are known as noble gases -2 Which are basically inert nature-2( long)
8.Elements forming colour compounds All are found in d- block series-2 Scandium is the first member-2 Yttrium forms the first of second row -2( long)
9 Vanadium Niobium Tantalum Chromium Molybdenum hard Tungsten-2 Manganese Technecium Rhenium-2 Are the essential d- block elements-2(long)
10.Eight group consists of nine metals Which are Ferric Iron Cobalt Nickel – 2 Ruthenium Rhodium Palladium-2 Osmium Iridium Platinum-2( long)
11.Copper Silver Gold are the So called essential coinage metals-2 Zinc Cadmium Mercury are their neighbours-2 They do belong to transition elements-2 ( long)
12.Elements following Lanthanum All but fourteen are Lanthanides-2 Cerium Prasodium Neodymium-2 Are the first three lanthanides - 2( long)
13.Promithyum Samarium Europium Gadalonium Terbium Dysprocium-2 Holmium Erbium Thulium Yutterbium – 2 Luetecium are the rest of lanthanides-2. ( long)
14.Actinides are also fourteen numbers Which are following element Actinium – 2 Thorium Protactinium Uranium -2 Are the first three actinides- 2 ( long)
15.Neptunium Plutonium Americium Curium Berkelium Californium -2 Einsteinium Fermium Mendelevium-2 Nobelium Lawrencium are the actinides-2 ( long)
16. Rutherfordium Dubnium Seaborgium Bohrium Hassium Meitnerium - 2 Darmastidium Roentgenium Copernicium - 2 Are the newly found d - block elements
17. Hydrogen resembles alkali metals Halogens also same as hydrogen So we are in need of your help in Predicting the position of Hydrogen
Composed by: Dr. C.SEBASTIAN A NTONY SELVAN. ASST. PROF in CHEMISTRY, R.V GOVT.ARTS COLLEGE, CHENGALPATTU , MOB: 9444040115 . OCT - 2017
.
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PHYSICAL CONSTANTS
VALUE
1. Planc’s constant h 6.62 ×1034 Js
2. Boltzmann constant k
3. Mass of electron m 9.11× 10−31 Kg
4 Charge of electron e
5. Gas constant R
6. Velocity of light c 3×108
7.
NOV -2017
Prove that 5 e−5 x is an eigen function of second order differention . Find its eigen value.
Obtain the value of L,S and J for the term symbol 3F2
Normalise the molecular orbital equation = N[ A+B].
Show that √❑ is a non linear operator.
Page 398 of 419
2020 - CONCISE QUANTUM MECHANICS
R.V.GOVT.ARTS COLLEGE ,CHENGALPATTU -603001
DEPARTMENT OF CHEMISTRY
Class : II. M.Sc DATE :
SUB: PHYSICAL CHEMISTRY –III UNIT V : QUANTUM CHEMISTRY – IV
: ASSIGNMENT- 1
Approximation methods –perturbation and variation method –application to hydrogen ,helium atoms –
R.S. coupling and term symbols for atoms in the ground state – Slater orbital and HF –SCF methods Born
– Heimer approximation –valence bond theory for hydrogen molecule –LACO –MO theory for di and
polyatomic molecules –concept of hybridization – Huckel theory for conjugated molecules (ethylene ,
butadiene and benzene)- semi empirical methods
PART – A
1. Identify the perturbation term in the Hamiltonian of Helium atom 2. Mention the conditions to apply perturbation theory 3. Identify H0 and H1 of an oscillator governed by potential energy V = ½ kx2 + ax3 + bx
4. A hydrogen atom subjected to electric field has Hamiltonian H = - 12 ∇2 -
2r1
+rcosθ.Separate the
perturbed part from unperturbed part.
5. Write down the second order perturbed Schrodinger wave equation
6. What is the need for approximation methods?
7. List out the steps followed in variation method.
8. Give two examples for perturbation
9. In what way variation method differs from perturbation method.
10. Write down the expression for first order perturbation energy and wave function
11. Find the term symbol for N.
12. .Verify whether an energy state with term symbol 2P 5/2 can exist
13. Deduce the atomic term symbol for boron.
14. Determine the electronic configuration of the element whose ground state term symbol is4S 3/2.
15. The term symbol of a particular atomic state is 6S5/2. Suggest a possible electronic configuration
16. Suggest a possible electronic configuration for the term symbol 3P2
17. . What is screening constant?.
18. . Calculate the screening constant of 2s electron of Carbon
20. Differentiate Slater type orbitals and Hydrogen like orbitals. 21. What are Slater type orbitals?
22.Determine the effective nuclear charge of 1s electron in F.
Page 399 of 419
2020 - CONCISE QUANTUM MECHANICS
23. Find the ground state term symbol for d1 ion and d2 ion
24. What is Born – Heimer approximation ?
25. Write down the Hamiltonian for Hydrogen molecule
26.Write the Hamiltonian for helium atom and mention the terms involved.
27. What are resonance and coulomb integrals?
28.What are exchange integrals?
29. What are the three important approximations of Huckel LCAO-MO theory?
30. What do you mean by hybridization?.
31..Mention the degenerate energy levels of benzene in terms of α and β
32. Give the expression for wave function and energy of ethylene in terms of α∧β.
33. Write down the secular determinant for H2 molecule.
34. Write down the Secular determinant for the He atom
35. Write down the secular determinant of benzene in terms of α∧β
36. Give the expression for wave functions of H2 molecule by VB theory.
37. Draw the wave functions of butadiene
37. For a one electron homonuclear diatomic molecule, the values of some relevant integrals are
∫φ1 H φ 1dτ = -2 a.u, ∫φ 2 H φ 2dτ = -2 a.u, ∫φ1 H φ 2dτ = -1 a.u ∫φ1 φ 2dτ = 0.25 Find the energy of this system and also the corresponding normalized wave function
PART- B (5 marks )
1. Discuss the application of perturbation theorem to Hydrogen atom.2. Derive the expression for first order perturbation energy and wave function.3. Arrive the secular equation using variation theorem.4. Discuss the application of variation theorem to hydrogen atom.5. Show that the variation method provides an upper bound to the ground state energy of the system.
6. The average energy of H –atom in terms of variation parameter ’a’ is ( a2
6 -
a2 ) . Find the variation
parameter and the true energy
7. The expression for expectation energy of Hydrogen atom by variation method with Ψ = e−a r 2
is 3∝2 -
√ 8 απ
where α is the parameter. Minimize this energy, to find the value of the parameter α and
determine the exact energy.
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7..For the selected wave function Ψ = r e−ar the expectation value of energy is E=α2
6 -
α2 . Find the value
of 𝛂 and hence calculate the energy8. Find the expression for STO for 2s, orbital in Nitrogen atom.. The normalisation constant is ‘N’. Compare
with Hydrogen – like orbital.
9..Find the expression for STO for 2px, orbital in Nitrogen atom.. The normalisation constant is ‘N’. Compare
with Hydrogen – like orbital
10. What do you mean by Effective nuclear charge. Calculate the Effective nuclear charge of
2s electron of Nitrogen and Calcium
11. Write note on HFSCF method
12. Explain Born-Oppenheimer approximation
13. Outline the salient features of VB(Heitler-London) theory as applied to Hydrogen molecule14 Apply HMO theory for ethylene molecule and determine its energy and wave function .
15.Give the assumptions of Huckel molecular orbital theory
16. Use 2s and 2pz atomic orbitals to construct two equivalent sp hybrid orbitals and determine the angle between the hybrid orbitals.
PART- C ( 10 marks )
1.Show that the first order perturbation energy for a non-degenerate system is just the perturbation function
averaged over the corresponding unperturbed state of the system. Derive the expression for the eigen
function of the perturbed system
2. Discuss the first order perturbation theory for a non –degenerate level .Using this theory, solve the
Schrodinger wave equation for ground state of Helium and obtain the expression for eigen energy
3.A Hydrogen atom is exposed to an electric field of strength F so that its perturbed Hamiltonian is F r
cosθ. Show that there is no first order effect. Given Ψ 0¿ = e−r
√ π and ∫
0
∞
e−2 r r3 dr = 38
4. Apply variation method to Helium atom and calculate its ground state energy. Compare your result with
that obtained from perturbation method.
5. Assuming = e – r as the trial wave function for the ground state of H-atom, find the
energy using variation method. Hamiltonian in spherical co-ordinates is - ( 12 ) d2
d r 2 –( 1r )
ddr – (
1r ).
6. Assuming the wave function = Ne – r2
a. Find the best value of
b. The energy of first state of H-atom
c. Compare with ground state energy calculate the percentage error.Page 401 of 419
2020 - CONCISE QUANTUM MECHANICS
7. Show that the energy of ground state of H- atom is -0. 375 when the wave function is
= r e – r using variation theorem
8 Use the functions Ψ = r e−ar to calculate the ground state energy of H-atom by variation method.
Compare the result with the true value. Hamiltonian in spherical co-ordinates is H = −12r 2
ddr¿ ) -
1r ( in
atomic units). Given : ∫0
∞
xn e−ax dx = n!
an+1
9. Use the trial wave function Ψ = e−arto find the energy eigen values for the ground
state of H- atom using variation theorem. Hamiltonian in spherical co-ordinates is
H =−12r 2
ddr¿) -
1r [given ∫
0
∞
e−2 ar r2dr = 1
4 a3 ∫0
∞
e−2 arr dr = 1
4 a2 ]
10. Consider Ψ as a linear combination of two eigen functions φ1 and φ 2 with normalization constant a1 and a2 respectively. Find the expectation value of energy and show that it is lesser than the lowest energy E0 11. For the wave function Ψ = e−ar2
show that the expectation value of energy is
E=3 a2 - 2 (√ 2a
π¿ Hamiltonian in spherical co-ordinates is H =
−12r 2
ddr¿) -
1r
Given ∫0
∞
r 4 e−2ar2
dr = 3
32 a2 √ π2 a
,∫0
∞
r2e−2 a r2
dr = 1
8 a √ π2a
, and ∫0
∞
r e−a r 2
dr =1
4 a
12.The expectation energy of H-atom were calculated as a2
6 -
a2 and
3a2 - 2√ 2a
π when the wave
functions are Ψ = re−ar and Ψ = e−a r 2
respectively. a. Minimize this average energy and find the variation parameter ‘a’ b. Find the actual energyc. Which is better wave function ?. Comment your result
13 Explain VB theory for H2 molecule considering Ψ as a linear combination of two eigen
functions φ1 and φ 2 with normalization constant a1 and a2 respectively
14. Write down the secular equations and set up the secular determinant for butadiene . Determine the energies of the HMOs in terms of 𝛂 and 𝛃 and calculate the free valencies of the terminal carbons.
15.Give the secular equations for benzene . Calculate the energies and HMO coefficients Determine its π- bond order
16. Write note on Semi Empirical methods.
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1. MATHEMATICS FOR QUANTUM MECHANICS
1.1 Coordinate systems Cartesian, spherical ,polar, cylindrical and elliptical coordinate system1.2 Functions: Real, complex, odd, even, orthogonal and normalized functions Single ,Double, Multiple valued function 1.3 Inner Product Of Function Norm of a function1.4 Some Polynomials Hermite, Legendre Associated Legendre, Laguree and ,Associated Lagurre polynomial
1. MATHEMATICS FOR QUANTUM MECHANICS
1.1 COORDINATE SYSTEMS:
a. CARTESIAN
Relation between Cartesian and polar co ordinates
x = r cos φ , y = r sinφPage 403 of 419
2020 - CONCISE QUANTUM MECHANICS
r = √ x2+ y2, φ = tan−1( yx ¿)¿
Volume element: dτ = dxdydzb. SPHERICAL/ POLAR
A point is represented by two angles (𝛉 ,𝛗 ) and one distance ‘r’
Relation between Cartesian and polar co ordinates
x = r sinθ cos φ, y = r sinθ sinφ , z = r cosθ Limits: 0≤ r≤ ∞, 0≤ θ≤ π , 0≤ φ≤ 2πVolume element:: dτ =r2 sinθ drdθdφShow that for spherical co ordinates r = √ x2+ y2+z2
Proof:
x2+ y2+z2 = (r sinθcos φ)2+(r sinθ sinφ)2+(r cosθ)2
= r2 sin2θ ( cos2 φ + sin2φ ) + r2 cos2θ = r2 sin2θ (1) + r2 cos2θ [sin2θ + cos2θ = 1] = r2 ¿θ + cos2θ ) = r2
∴ r = √ x2+ y2+z2
θ = cos−1¿
φ = tan−1( yx ¿)¿
Problem: Convert the distance ( 4,4,2) into spherical and cylindrical
Solution:
r = √ x2+ y2+z2
= √42+42+22
= 6θ = cos−1¿
= cos−1¿
= cos−1¿
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2020 - CONCISE QUANTUM MECHANICS
φ = tan−1( yx ¿)¿
= tan−1( 44 ¿)¿
= tan−1(1¿)¿
= 45
c. CYLINDRICAL
x = p cos 𝛗, y= p sin , z = z p = √ x2+ y2, φ = tan−1( y
x ¿)¿
Volume element:dτ =ρdρdφdzd. ELLIPTICAL COORDINATE SYSTEM
If P = r A+r B
r Q = r A−rB
r then
x= r2 √(P2−1) √(1−Q2) cos φ
y= r2 √(P2−1) √(1−Q2) sin φ
Volume element: dτ = r3
8 (P2−r 2 )drdPdφ
Problem:
1. How much distance the point (6,4,3) away from the origin?
2.Convert the distance (9,6,3) in to spherical and cylindrical co ordinate.
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3.Convert the distance (2,π3 ,
π6 ) in to spherical and cylindrical co ordinate.
1.2. NUMBER SYSTEMS:
Complex number:
A complex number is of the form a± ib where ‘a’ and ‘b’ are real numbers and ‘i’ is
called the imaginary unit having the property i2 = -1. It is denoted by Z
Examples: 3+2i , 6-4i
x = r cosθy = r sinθ
Z = x +iy
= r cosθ +i r sinθ
= r (cosθ + i sinθ )
The absolute value of Z = √a2+b2
Show that in complex number r = √ x2+ y2
x +iy = r (cosθ + i sinθ )
x – iy = r (cosθ - i sinθ )
(x +iy ) (x - iy ) = r (cosθ + i sinθ )× r (cosθ - i sinθ )
x2 + y2 = r2 (cosθ + i sinθ )× (cosθ - i sinθ )
= r2 [ cos2θ + sin2θ ] = r2
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∴ r = √ x2+ y2
Find the absolute value of 4+5i
1.3 FUNCTIONS:
Single valued function
For each value of independent variable, if there is only one dependent variable then the
function is called single valued function.
for example y = 4x
here for each value of ‘x’ there will be only one value for ‘y’
Double valued function
For each value of independent variable, if there is two dependent variable then the
function is called double valued function.
for example y2 = 4x
if x = 2 ,
y = +2 or y = -2
Multiple valued function
For each value of independent variable, if there are many dependent variables then the
function is called multiple valued function.
for example y = sinx
for x = 30, 150, 390 ..
y = 0.5 only
ODD & EVEN functions
A function is said to be odd if f( -x ) = - f(x)
For example consider f(x) = sinx
f ( -x) = sin ( -x)
= - sin x
= - f(x)
Therefore it is an odd function.
A function is said to be even if f( -x ) =- f(x)
For example consider f(x) = Cos x
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f ( -x) = Cos ( -x)
= Cos x
= f(x)
therefore it is an even function.
Determine whether the following functions or odd or even
xsinx, xcosx ,x2 sinx, x2 Cosx, e ix
Problem 1.Which is acceptable wave function? a. sinx b. cos x c. e− x d. ex
Solution:
a. sinx
When x →∞ , sinx = ∞. It is not finite and hence it is not acceptable wave function
b. cos x
When x →∞ ,cos x = ∞ It is not finite and hence it is not acceptable wave function
c. e− x
When x →∞ , e− x = 0, It is finite and hence it is acceptable wave function
d. ex
When x →∞ , ex ∞, It is not finite and hence it is not acceptable wave function
INNER PRODUCT OF FUNCTION:
It is defined as the integral product of two functions within specified limits.
Find the inner product of <x/x+1> (0,1)
∫0
1
x (x+1)dx = ∫0
1
x2+x dx
= x3
3 + x2
2
= 13 +
12
= 56
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Norm of a function:
It is defined as the integral product of its own function within specified limits.
if the function is complex the integral product of the function and its conjugate is called norm
of a function.
Find the norm of <x> (0,1)
∫0
1
x xdx = ∫0
1
x2 dx
= x3
3
= 13
Find the norm of <x +2i > (0,1)
∫0
1
(x+2i)(x−2 i)dx = ∫0
1
x2+4 dx
= x3
3 + 4x
Normalization:
A function is said to be normalized if the norm of the function in a specified limit is unity
<f/ f*> (a,b) . = 1
Orthogonal :
Two functions are said to be orthogonal if the inner product of the function is zero
<f/ f*> (a,b) . = 0
Problem: Consider the functions sinx and cosx. Are they orthogonal in ( 0, π2 ) and in ( 0,π )
a. ( 0, π2 )
Solution:
∫0
π2
sinx cos x dx = 12 ∫
0
π2
sin 2 x dx
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= 12 ( -
12 cos 2x)
= - 14 cos 2x
= 12
The functions are not orthogonal in the limit ( 0, π2 )
b. a. ( 0, π )
Solution:
∫0
π
sinx cos x dx = 12 ∫
0
π
sin 2 x dx
= 12 ( -
12 cos 2x)
= - 14 cos 2x
= 0
The functions are orthogonal in the limit ( 0, π)
NORMALISATION AND NORMALIZATION CONSTANT
A wave function is said to be normalised if it satisfies the following condition
∫ΨΨ * dT = 1 ( where dT = dxdydz)
If after solving the schrodinger equation, the wave function is not normalised , it must be
multiplied by a constant factor called normalisation factor. For example if a function Ψ( x)
satisfies the schrodinger equation, and if
∫ΨΨ * d T = c ( c ≠ 1)
1c∫ΨΨ * d T = 1
∫ 1√ c
Ψ 1√ c
Ψ * d T = 1
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Then normalisation factor is 1
√ c and the normalised wave function is 1
√ c Ψ (x)
Problem 5.1 Find the normalisation constant of the function φ m = N e imφ where N is
normalization constant
Solution:
The condition for normalization is
∫0
2π
ΨΨ * dx = 1
N e+imφ N e- imφ dφ = 1
N2 dφ = 1 [e+imφ × e- imφ = e0
= 1]
N2 [φ ¿¿02 π = 1 [ ∫ dφ=¿φ ¿ ]
N 2 [2π−0 ] = 1
N = 1√2 π
The normalisation constant of the given function is 1√2 π
Problem 5.2 Find the normalisation constant of the function Ψ =ACos2x where A is
normalization constant - π2 < x> +
π2
Solution:
Normalisation condition is ∫−∞
∞
Ψ Ψ∗¿dx = 1
∫− π
2
+π2
A cos2 x × A cos2 xdx = 1
A2∫π2
+π2
cos4 xdx = 1
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2A2∫0
π2
cos4 xdx = 1 [ ∫π2
+π2
cos4 xdx = 2∫0
π2
cos4 xdx ]
2A2 × 34 ×
12 × π2 = 1 [ ∫
0
π2
cosn xdx = n−1
n × n−3n−2 × π2 ]
A2 ×3π8 = 1
A = √ 83 π
The normalisation constant of the given function is √ 83 π
Problem 5.3 Find the normalisation constant of the function Ψ = A e−ra where A is normalization
constant
Solution
Normalisation condition is ∫−∞
∞
Ψ Ψ∗¿dτ = 1 , dτ = r2dr ∫0
π
sin θ dθ ∫0
2 π
dφ
∫−∞
∞
A e−ra A e
−ra r2 dr ∫
0
π
sin θ dθ ∫0
2 π
dφ = 1
A2 ∫−∞
∞
e−2 r
a r2 dr ∫0
π
sin θ dθ ∫0
2 π
dφ = 1[ ∫−∞
∞
e−arr n dr = n!
(a )n+1 ,
[ ∫0
π
sin θ dθ = - cos θ
= -(-1 -1) = + 2, ∫0
2 π
dφ =
2π] A2 [
2!
( 2a )2+1 ×2 × 2π] = 1
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A2 [ 2 a3
8 × 4π] = 1
A2 [a3π] = 1
A2 = 1
πa3
A = √ 1π a3
The normalisation constant of the given function is √ 1π a3
Problem 5.4 Find the normalisation constant of the function A sin( nπL)x,0 < x< L, ‘n’ is an
integer
Solution:
The condition for normalization is ∫0
2 π
ΨΨ * dx = 1
∫0
2π
A sin K x A sin K x dx = 1
A2 ∫0
2π
sin2 K x dx = 1
A2 ∫0
2 π
(1−cos2K x )2
dx = 1 [ sin2 K x=(1−cos 2 K x)2
]
A2
2 ∫
0
2 π
[1- cos2K x] dx = 1
A2
2 [ ∫
0
2π
1 dx - ∫0
2 π
cos2 K x dx] = 1
A2
2 { [x¿ - [ (sin 2 K x )
2K¿0
2 π } = 1 [ ∫0
2 π
cos2 K x dx =
(sin 2 K x )2K
]
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A2
2 { [2 π ¿ - ¿¿ } = 1 [ sin 0 = 0]
A2
2 [2 π ¿ = 1 [sin K 2 π =
0]
A2 [π ¿ = 1
A = 1√π
The normalisation constant of the function = √ 2L
5.2 ORTHOGONAL A wave function is said to be orthogonal if it satisfies the following
condition .
∫Ψ 1Ψ 2dτ = 0
5.3 ORTHO NORMAL SET:
A wave function is said to be ortho normal if it satisfies the following condition .
∫Ψ iΨ jdτ = 0 if i ≠ j
= 1 if i = j
1.4 SOME POLYNOMIALS:
HERMITE POLYNOMIAL
Hermite differential equation is d2 Hd y2 - 2 y
dHdy + (λ -1) H = 0
Its solution is known as Hermite polynomial
Hn( y) is Hermite polynomial , Hn = ( -1) ne y2 dn
d yn (e− y2
)
When n= 0 H0 = 1 similarly H1 = 2y, H2 = 4y2 – 2 H3 = 8y3 -12y
Problem 3.2 Find the value of H0 andH1 Solution
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Hn = ( -1) ne y2 dn
d yn (e−y2
)
H0 = ( -1) 0 e y2 × e – y2 )
= 1
H1= ( -1) e y2 ddy (e− y2
)
= -e y2
( - 2ye− y2
)
= 2y
Problem 3.3 Find the value of H2 Solution
Hn = ( -1) ne y2 dn
d yn (e−y2
)
H2 = ( -1) 2e y2 d2
d y2 (e−y2
)
= + e y2 ddy (−2 y e− y2
)
= - 2e y2 ddy ( ye− y2
)
= - 2e y2
{ y¿) + e−y2
(1) }
= - 2e y2
{ −2 y2e− y2
+ e− y2
}
= 4 y2 -2
= 2(2 y2 -1)
LEGENDRE POLYNOMIAL
Legendre differential equation is d2 Pd θ2 +
cosθSinθ
dPdθ + βP = M 2 P
sin2θ
Its solution is known as Legendre polynomial Pl(x) = 12l . l !
d l
d x l ( x2 – 1) l
Where x = cos θ and ‘ l’ is an integer including 0
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Problem 5.2 Find the expression for the Legendre polynomial P0 and P1
Solution:
P0 = 120.0 !
d0
d x l ( x2 – 1) 0
= 1
P1 = 12l . l !
d l
d x l ( x2 – 1) 1
= 12
× ddx ( x2 – 1)
= 12
×(2x – 0 )
= x
= cos θ [ x = cos θ]
Similarly P2 = 3 cos 2 θ - 1
ASSOCIATED LEGENDRE POLYNOMIAL:
Associated Legendre differential equation is
( 1-x2)d2 yd x2 - 2x
dy∂ x + ( n (n+1) -
m2
(1−x2)¿ X = 0
Its solution is known as Associated Legendre polynomial Plm(x)
Plm(x) = (1−x2 )
m2 dm
d xm × Pl(x)
= (1−x2 )m2 × dm
d xm1
2l .l !d l
d xl ( x2 – 1) l
= 1
2l . l ! (1−x2 )
m2 dm+ l
d xm+l [( x2 – 1)l
Problem 5.3 Find the expression for the Associated Legendre polynomial P00
and P10
Solution:
P00 = ( 1-x2) 0 × d0
d x0 [( x2 – 1)0
= 1
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P10 = 1
2l . l !d l
d x l ( x2 – 1) 1
= 12
× ddx ( x2 – 1)
= 12
×(2x – 0 )
= x
= cos θ [ x = cos θ]
Similarly
P20 = ½ ( 3 cos 2𝛉 – 1)
P11 = sin𝛉
P21 = 3sin𝛉 cos𝛉
LAGUREE POLYNOMIAL:
Lk (x) = ex dk
d xk [ xke− x]
Problem 6.8. Find the value of L1(x)
Solution:
Lk (x) = ex dk
d xk [ xke− x]
L1 (x) = ex d1
d x1[ x e− x]
= ex [ x ddx (e− x ) + e− x
ddx (x) [ d (UV) = UdV +
VdU ]
= ex[ x(- e− x ) + e− x (1)] [ddx (e− x ) = - e− x and
ddx (x) =
1 ]
= ex[ -x e− x + e− x ]
= ex × e− x ( -x +1) [ taking e− x as common]
= [ -x +1 ] [ex × e− x = ex− x = e0 = 1]
= 1 – x
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Problem 6.9. Find the value of L2(x)
Solution:
Lk (x) = ex dk
d xk [ xke− x]
L2 (x) = ex d2
d x2[ x2 e− x]
= ex ddx
ddx (x2e− x )
= ex ddx [x2
ddx (e− x ) + e− x
ddx (x2) ] [ d (UV) = UdV +
VdU ]
= ex ddx [ x2(- e− x ) + e− x (2x)] [
ddx (e− x ) = - e− x and
ddx (x2) =
2x ]
= ex ddx [ -x2 e− x + 2x e− x ]
= ex [(-x2) ddx (e− x)+e− x
ddx (-x2)+2x
ddx (e− x)+e− x d
dx (2x) d (UV) = UdV + VdU for both terms]
= ex [ ( -x2) (- e− x ) + e− x (-2x) + 2x (- e− x ) +( e− x )(2)
= ex× e− x [ +x2- 4x + 2 ] [ taking e− x as common]
= [ + x -1 - 1 ] [ ex × e− x = ex− x = e0 = 1]
L2 (x) = x2- 4x + 2
Similarly L3 (x) = 6-18 x + 9 x2 – x3
ASSOCIATED LAGURRE POLYNOMIAL:
The differential equation of Associated Laguree polynomial is
ρ d2 L
d ρ2 + [ 2(l+1) – ρ] dLdρ + ( n-1-l)L =0
the solution of this equation is known as associated Laguree polynomial, which is
Lkp (x) = d p
d x p [ ex dk
d xk [ xke− x]
Problem 6.10. Find the value of L00 and L1
1
Solution:Page 418 of 419
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L00 = d0
d x0 [ ex d0
d x0 [ x e− x]
= [ ex[ x e− x] [ d0
d x0 = 1]
L00 = x [ex × e− x = ex− x = e0 = 1]
L11 = d1
d x1 [ ex d1
d x1[ x e− x]
= ddx [ ex[ x
ddx (e− x ) + e− x
ddx (x) ] [ d (UV) = UdV +
VdU ]
= ddx [ ex [ x(- e− x ) + e− x (1)] [
ddx (e− x ) = - e− x and
ddx (x) = 1
]
= ddx [ ex× e− x [ -x +1 ] [ taking e− x as common]
= ddx [ -x +1 ] [ex × e− x = ex− x = e0 = 1]
= -1 +0 [ ddx (x) = 1 and
ddx (1) =
0 ]
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