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2020 - CONCISE QUANTUM MECHANICS CONCISE QUANTUM MECHANICS [FOR MSc PHYSICS & MSc CHEMISTRY STUDENTS OF ALL INDIAN UNIVERSITIES] BY Dr. C.SEBASTIAN ANTONY SELVAN ASSISTANT PROFESSOR R. V. GOVT.ARTS COLLEGE CHENGALPATTU 9444040115 DEC 2020 Page 1 of 566

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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 )¿

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

θ = 90Page 19 of 419

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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.

1.Lyman series:Page 22 of 419

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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

equationPage 26 of 419

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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

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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’

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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

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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

)

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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

A sin K x A sin K x dx = 1

A2 ∫0

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

1 dx - ∫0

2 π

cos2 K x dx] = 1

A2

2 { [x¿ - [ (sin 2 K x )

2K¿0

2 π } = 1 [ ∫0

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

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

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

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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

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

= ( 1πa03 ) ∫0

r

e−2ra 0

r2 dr ∫0

π

sin θ dθ ∫0

= ( 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

= 1

π a03 ∫

0

e−2 r

a0 r3 dr ∫0

π

sin θ dθ ∫0

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

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

= 1π ∫

0

r e−2 r dr ∫0

π

sin θ dθ ∫0

2 π

[ ∫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

= 1

π a03 ∫

0

e−2 r

a0 r3 dr ∫0

π

sin θ dθ ∫0

= 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

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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

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

= 1

π a03 ∫

0

e−2 r

a0 r3 dr ∫0

π

sin θ dθ ∫0

= 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

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

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

= Fπ ∫

0

e−2 r (rcos θ)r2 dr∫0

π

sin θ dθ∫0

2 π

= eFπ ∫

0

e−2 r r3 dr∫0

π

cos θ sin θ d θ∫0

2 π

= 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

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

= Fπ ∫

0

e−2 r (rcos θ)r2 dr∫0

π

sin θ dθ∫0

2 π

= eFπ ∫

0

e−2 r r3 dr∫0

π

cos θ sin θ d θ∫0

2 π

= 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

= F

4 π √2 ∫

0

e−r(rcos θ)[r e−r2 cosθ ]r2 dr∫

0

π

sinθ dθ∫0

= 1

4 π √2 ∫

0

∞ [e−3 r2 ] r4 dr∫

0

π

cos2θ sin θ d θ∫0

2 π

= 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

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

= Fπ ∫

0

e−2 r (rcos θ)r2 dr∫0

π

sin θ d θ∫0

2 π

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= eFπ ∫

0

e−2 r r3 dr∫0

π

cos θ sin θ d θ∫0

2 π

= 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 .

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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

∫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

∫ (e−ar ) (e−ar ) r2 dr∫0

π

sinθ dθ∫0

2 π

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

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

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

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

∫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

∫ (e−ar ) (e−ar ) r2 dr∫0

π

sinθ dθ∫0

2 π

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

∫Ψ Ψ ¿r2 dr∫0

π

sin θ d θ∫0

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

∫ (e−ar ) (e−ar ) r2 dr∫0

π

sinθ d θ∫0

2 π

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

∫Ψ Ψ ¿r2 dr∫0

π

sin θ d θ∫0

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

∫ (e−a r2 ) (e−ar2 )r2 dr∫0

π

sin θ d θ∫0

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

∫Ψ Ψ ¿r2 dr∫0

π

sin θ d θ∫0

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

∫ ( r e−ar ) ( r e−ar ) r2 dr∫0

π

sin θ d θ∫0

2 π

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

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 = 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:

For cation:

π – bond energy = total energy – energy in the absence of bondingPage 280 of 419

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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

= 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

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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

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( √ 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

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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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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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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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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

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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

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

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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

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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

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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

1 11 21 31 41Page 360 of 419

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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

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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.

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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

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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.

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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.

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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

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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

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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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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

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

∫−∞

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

A sin K x A sin K x dx = 1

A2 ∫0

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

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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