Quantum Chemistry: Classical to Computational 9781032789897

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Quantum Chemistry Classical to Computational

Quantum Chemistry Classical to Computational

Dr. Amita Dua M.Sc., Ph.D. Assistant Professor, Department of Chemistry Dyal Singh College University of Delhi, Delhi  Dr. Chayannika Singh M.Sc., M.Phil., Ph.D. Assistant Professor, Department of Chemistry Deen Dayal Upadhyaya College University of Delhi, Delhi

First published 2025 by CRC Press 4 Park Square, Milton Park, Abingdon, Oxon, OX14 4RN and by CRC Press 2385 NW Executive Center Drive, Suite 320, Boca Raton FL 33431 CRC Press is an imprint of Informa UK Limited © 2025 Manakin Press Pvt. Ltd. The right of Amita Dua and Chayannika Singh to be identified as authors of this work has been asserted in accordance with sections 77 and 78 of the Copyright, Designs and Patents Act 1988. All rights reserved. No part of this book may be reprinted or reproduced or utilised in any form or by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying and recording, or in any information storage or retrieval system, without permission in writing from the publishers. For permission to photocopy or use material electronically from this work, access www. copyright.com or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. For works that are not available on CCC please contact [email protected] Trademark notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Print edition not for sale in South Asia (India, Sri Lanka, Nepal, Bangladesh, Pakistan or Bhutan). British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library ISBN13: 9781032789897 (hbk) ISBN13: 9781032789903 (pbk) ISBN13: 9781003490135 (ebk) DOI: 10.4324/9781003490135 Typeset in Times New Roman by Manakin Press, Delhi

Detailed Contents

1. Classical Mechanics

1–44

Dalton’s Atomic Theory What are Classical Mechanics and Quantum Mechanics? J.J. Thomson Model of Atom Rutherford’s Nuclear Model of Atom—Discovery of Nucleus Developments Leading to the Bohr Model of Atom 1.5.1 Dual Nature of Electromagnetic Radiation 1.5.2 Emission and Absorption Spectra 1.6 Bohr Model of Atom 1.7 Sommerfeld Theory

1 2 4 4 6 7 21 28 43

1.1 1.2 1.3 1.4 1.5

2. Towards Quantum Mechanics 2.1 2.2 2.3 2.4

45–58

Reasons for the Failure of Classical Model of Atom or Bohr Model of Atom Developments Leading to Quantum Mechanical Model of Atom de-Broglie’s Dual Nature of Matter Heisenberg’s Uncertainty Principle

3. Introduction to Quantum Mechanics

59–120

Necessity of Quantum Mechanics Schrodinger Wave Equation Derivation of Time Independent Schrodinger Wave Equation Physical Significance of Wavefunction (ψ) and Probability Density (ψ2) Concept of Atomic Orbital Quantum Mechanical Model of Atom Eigen Value and Eigen Wavefunction Normalised, Orthogonal and Orthonormal Wavefunction Operators Postulates of Quantum Mechanics Derivation of Time Independent Schrodinger Wave Equation on the Basis of Postulates of Quantum Mechanics 3.12 Steady State Schrodinger Wave Equation 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11

4. Particle in a Box: Quantisation of Translational Energy 4.1 Application of Postulates of Quantum Mechanics to Simple System 4.2 Operation of Quantum Mechanics

45 45 46 52

59 60 61 64 66 68 68 71 76 112 115 117

121–195 121 121

vi

Detailed Contents

4.3 Introduction to Translational Motion of a Particle 4.4 Particle in One Dimensional Box: Quantisation of Translational Energy 4.4.1 Solution of Schrodinger Wave Equation 4.4.2 Conclusions from the Study of a Particle in One-dimensional Box 4.4.3 Solution of Properties in One Dimensional Box 4.4.4 Application of Particle in a One-dimensional Box 4.5 Particle in Two-dimensional Box 4.5.1 Two Dimensional Rectangular Box 4.5.2 Two Dimensional Square Box (Concept of Degeneracy) 4.6 Particle in Three Dimensional Box 4.6.1 Three Dimensional Cuboidal Box 4.6.2 Three dimensional Cubical Box (Concept of Degeneracy) 4.7 Free Particle 4.7.1 Solution of Schrodinger Wave Equation 4.7.2 Quantisation of Energy Levels: Difference between Particle in Free State and Under Bound State

5. Rigid Rotator: Quantisation of Rotational Energy

194

197–250

5.1 Introduction 5.2 Classical Treatment of Rigid Rotator 5.2.1 Kinetic Energy of Rigid Rotator 5.3 Quantum Mechanical Treatment: Schrodinger Wave Equation for Rigid Rotator 5.3.1 Deriving Schrodinger Equation by Reducing Angular Momentum of Kinetic Energy Operator in Polar Co-ordinates. 5.3.2 Deriving Schrodinger Equation by Reducing Laplacian of Kinetic Energy Operator in Polar Coordinates 5.4 Wavefunction of Rigid Rotator 5.4.1 Solving the F Part of Wavefunction 5.4.2 Solving the Θ Part of Wavefunction 5.4.3 Spherical Harmonics, Y(θ, φ) 5.5 Rotational Energy of the Rigid Rotator 5.6 Rotational spectra

6. Linear Harmonic Oscillator: Quantisation of Vibrational Energy

123 124 124 129 138 165 171 171 175 180 181 185 191 191

197 197 198 201 201 215 228 230 231 238 240 241

251–295

6.1 Introduction 251 6.2 Classical Treatment of Linear Harmonic Oscillator 252 6.2.1 Frequency of Linear Harmonic Oscillator 253 6.2.2 Potential Energy of Linear Harmonic Oscillator 253 6.2.3 Total Energy of the Linear Harmonic Oscillator 254 6.3 Quantum Mechanical Treatment: Schrodinger Wave Equation for Linear Harmonic Oscillator 256

Detailed Contents

6.4 Solution of Schrodinger Wave Equation: 6.4.1 Factorization Method 6.4.2 Power Series Method 6.5 Vibrational Energy Levels for Linear Harmonic Oscillator 6.6 Wavefunction Plots for Linear Harmonic Oscillator 6.7 Probability Plots for Linear Harmonic Oscillator 6.8 Symmetry of the Vibrational Wavefunction 6.9 Calculation of Properties of Linear Harmonic Oscillator 6.10 Orthonormal Sets of Wavefunction 6.11 Virial Theorem 6.12 Vibrational Spectra

7. “Hydrogen Atom”: Quantisation of Electronic Energy

vii

257 257 272 278 278 280 281 281 284 285 288

297–438

7.1 Necessity of Replacing Bohr Theory 297 7.2 Setting of Schrodinger Equation for Hydrogen Atom 297 7.2.1 Solving the F Part of Wavefunction 306 7.2.2 Solving the Θ Part of Wavefunction 307 7.2.3 Solving the Radial Part, R, of Wavefunction 314 7.3 Quantum Numbers 336 7.3.1 Significance of Quantum Numbers 336 7.3.2 Importance of Quantum Numbers in Explaining Hydrogen Spectrum: 339 7.4 Degenerate and Non-Degenerate Orbitals 340 7.5 Degeneracy of Energy Levels 341 7.6 Wavefunction of the Hydrogen Atom 345 7.6.1 Radial Wavefunction and Radial Plots 346 7.6.2 Angular Wavefunction and Shape of Orbitals 362 7.6.3 Contour Diagram or Contour maps 380 7.6.4 Total Wavefunction of Hydrogen Atom 386 7.7 Calculation of Properties of Hydrogen Atom 389 389 7.7.1 Probability density (ψ2) 7.7.2 Most Probable Distance (r) or Position of Maximum Probability Using Radial Probability Density Function 389 7.7.3 Average Value of Position or Expectation Value of Position or Mean Radius 〈r〉 394 7.7.4 Expectation value of Energy 400 7.8 Magnetic Properties: Angular Momentum and Magnetic Moment 403 7.8.1 Classical Expression of Magnetic Moment due to Orbital Motion of Electron 403 7.8.2 Quantum Mechanical Expression of Magnetic Moment Due to Orbital Motion of Electron 404 7.8.3 Potential Energy Due to Orbital Motion of an Electron in a Magnetic Field: Larmor Precession 406

viii

Detailed Contents

7.8.4 7.8.5 7.8.6 7.8.7 7.8.8 7.8.9

Zeeman Effect Anomalous Zeeman Effect Explanation of Anomalous Zeeman Effect: Spin Quantum Number Magnetic Moment Due to Spinning Motion of Electron Potential Energy Due to Spinning Motion of Electron in a Magnetic Field Experimental Demonstration of Electron Spin: Stern-Gerlach Experiment

7.9 Spin Orbit Coupling and Term Symbols

8. Multielectron System and Approximate Methods

409 410 410 411 412 414 419

439–465

8.1 Introduction

439

8.2 Pertubation Method

440

8.2.1

Solution of Pertubation Method

441

8.2.2

Application of Pertubation Method on Helium Atom

446

8.3 Variation Method

449

8.3.1

Solution of Variation Method

450

8.3.2

Application of Variation Method on Helium Atom

451

8.3.3

Application of Variation Method for Various Other Systems

453

8.4 Self-Consistent Field Method

9. Chemical Bonding

464

467–549

9.1 Introduction

467

9.2 Born-Oppenheimer Approximation

468

9.3 Approximate Methods to Solve Schrodinger Equation of a Molecule : Valence Bond Approach and Molecular Orbital Approach

471

9.3.1

Valence Bond (VB) Approach

472

9.3.2

Molecular Orbital (MO) Approach

474

9.4 Approximation of Linear Combination of Atomic Orbitals: LCAO–MO Treatment

481

9.5 LCAO–MO Treatment of Hydrogen Molecule Ion (H2+)

484

9.6 Hydrogen Molecule: Qualitative Treatment

493

9.6.1

Pauli’s Exclusion Principle w.r.t. Hydrogen Molecule

497

9.6.2

Hund’s Rule of Maximum Multiplicity w.r.t. Hydrogen Molecule

499

9.6.3

Aufbau’s Principle w.r.t. Hydrogen Molecule

500

9.7 LCAO–MO Treatment of Hydrogen Molecule

500

Configuration Interaction

506

9.7.1

9.8 Valence Bond Treatment (VBT) of Hydrogen Molecule

507

9.9 Comparison of VBT and MOT

515

9.10 LCAO–MO Treatment of Homonuclear Diatomic Molecules

516

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9.11 LCAO–MO Treatment of Heteronuclear Diatomic Molecules

523

9.12 LCAO–MO Treatment of Triatomic Molecules

530

10. Hückel Molecular Orbital Theory 10.1 Introduction 10.2 Application of HMO Theory of Conjugated Systems 10.2.1 Ethylene 10.2.2 1,3-Butadiene 10.2.3 Cyclobutadiene 10.2.4 Cyclopropene 10.2.5 Benzene 10.3 Delocalization Energy

11. Basics of Computational Chemistry 11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8 11.9

Introduction Potential Energy Surface (PES) Stationary Point Geometry Optimization Molecular Mechanics (MM) Method 11.5.1 Applications of MM Method Ab-initio Method 11.6.1 Applications of ab-initio Method Semi-Empirical Method 11.7.1 Applications of Semi-Empirical Method Density Functional Theory (DFT) Method 11.8.1 Applications of Density Functional Theory Method Basis Set

Appendices Appendix A1: Appendix A2: Appendix A3: Appendix A4: Appendix A5: Appendix A6: Appendix A7: Appendix A8: Appendix A9:

Index

551–563 551 552 552 554 555 556 557 560

565–591 565 566 568 569 570 571 572 579 579 581 582 586 587

593–611 Fundamental Physical Constants Standard Integrals Important Mathematical Formulae Selected Derived Units Energy Conversion Factors Unit Prefixes Logarithms Table Antilogarithms Table How to Take Log and Antilog

593 596 599 600 601 602 603 606 609

613–620

Preface This book has evolved from our constant efforts, real time teaching of quantum mechanics and addressing the chemical problems at University of Delhi. All major developments of quantum mechanics from classical to computational chemistry have been comprehensively discussed in this book. This work is a very fine blend of classical, quantum and computational chemistry. A robust knowledge bank has been included in this book which makes it very user friendly. The knowledge bank includes exhaustive derivations, mathematical proofs and theorems. A series of solved numerical have been inserted after each topic to have a better understanding and absorption of the subject. This book is need of every science student from undergraduate level to the professors at universities. It will also be used as a help book for students appearing in competitive examinations. This book will surely prove to be an asset for students and academicians. We welcome your constructive suggestions for any further improvement in this work. Authors

1 Classical Mechanics 1.1

DALTON’S ATOMIC THEORY

John Dalton in 1808 put forward a theory known as Dalton’s atomic theory. The main points of this theory are: 1. Matter is made up of extremely small indivisible particles called atoms. 2. Atoms of the same element are identical in all respects i.e., size, shape and mass. 3. Atoms of different elements have different masses, sizes and chemical properties. 4. Atoms of the same or different elements combine together to form molecules. 5. When atoms combine with one another to form molecules, they combine in simple whole number ratios. 6. Atoms of two elements may combine in different ratios to form more than one compound. Like C combine with O to form CO and CO2. 7. Atom is the smallest particle that takes part in a chemical reaction. 8. An atom can neither be created nor be destroyed. Dalton’s atomic theory was the first achievement towards the smallest structure of matter called ‘atom’. This theory was able to explain the law of conservation of mass, law of constant composition and law of multiple proportion. However, it failed to explain the results of many other experiments. This leads to the discovery of many different types of sub-atomic particles in the beginning of the twentieth century.

2

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Table 1.1: Discoveries of various sub-`atomic particles

Discoveries Electron/cathode rays Charge to mass ratio of electron

Discoverers Faraday (1879) J.J. Thomson (1897)

Charge on electron (oil drop experiment)

R.A. Millikan (1917)

Mass of electron

R.A. Millikan (1917)

Result

e = 1.76 × 1011 C/kg m e = –1.6 × 10–19 C e = 4.8 × 10–10 esu me = 9.11 × 10–31 kg [It is 1/1837th of H-atom]

Positive rays/Proton rays/ Canal rays/Anode rays (Discharge tube experiment) Charge on proton Mass of proton Neutron Charge on neutron

Mass of neutron

Radioactivity [phenomena of spontaneous emission of radiations by certain elements like uranium etc.]

1.2

Goldstein (1886)

Ernest Rutherford (1917) Ernest Rutherford (1917) Chadwick (1932) Ernest Rutherford (1920) and confirmed by Chadwick (1932) Ernest Rutherford (1920) and confirmed by Chadwick (1932) Henri Bequeral and Marie Curie (1896)

+1.6 × 10–19 C mp = 1.67 × 10–27 kg [same as that of H-atom] Zero

mn = 1.67 × 10–27 kg

After the discovery of electron and proton, it was well-established that atom is divisible and is made of charged particles. This was further confirmed by the phenomena of radioactivity.

WHAT ARE CLASSICAL MECHANICS AND QUANTUM MECHANICS?

Classical Mechanics is the Newtonian mechanics, whose foundation was laid by the Newton’s laws of motion. The term “Mechanics” means study of motion and Issac Newton gave the laws of motion. Issac Newton in 1687 gave three laws of motion: 1. First Law: An object either remains at rest or continues to move at constant velocity until or unless acted upon by any external force. This law is called as “Law of Inertia”.

1.2 What are Classical Mechanics and Quantum Mechanics?

3

2. Second Law: Rate of change of momentum is directly proportional to the force. Mathematically, F∝

dp dt

F ∝ m⋅

(p = momentum = mv) dv dt

F=m·a

(a = acceleration)

3. Third Law: Every action has an equal (in magnitude) and opposite (in direction) reaction. These are called as Newton’s laws of motion. Newton’s laws explain all the major aspects of a moving body and its motion. They describe the relationship between an object and the forces acting upon it and its motion in response to the said forces. However, later it was observed that Newton’s laws were well-followed by the macroscopic objects (which are visible by naked eye e.g.: falling stone, moving car, recoil of gun etc.) but they failed to explain the motion of microscopic objects (which are not visible by naked eye and we need high resolution microscopes to see them e.g., electron, atom, molecule, etc.). This was observed because macroscopic objects essentially possess particle nature, that is why Netwon’s laws provide extremely accurate results as long as the area of study is restricted to large objects. But, when the moving object becomes sufficiently small i.e., microscopic objects then they have essentially a wave nature due to which Newton’s laws fail to explain the mechanics of microscopic objects. The failure of classical mechanics to explain the laws of motion for microscopic objects gave birth to “Quantum Mechanics”. Quantum mechanics gives the laws of motion which microscopic objects obey, and they are different from the Newton’s laws of motion. Quantum mechanics was introduced in 1900 which was brought to account for the dual nature of microscopic objects i.e., particle nature and wave nature. S. No. Particle nature Wave nature 1. Particle is localized in space i.e. 1. Wave is delocalized in space i.e. a wave is spread out in a particle occupies a well-defined space. position in space. e.g., sound waves. e.g., a cricket ball. 2. Two particles cannot occupy the 2. Two or more waves can co-exist same space simultaneously and do in the same region of space not interfere. and hence waves interfere.

4

Chapter 1

Classical Mechanics

• Actually macroscopic and microscopic objects both have particle nature as well as wave nature. Now, macroscopic objects have essentially particle nature since the wave attached to a macroscopic object has a small wavelength which is not significant. The classical mechanics treats the moving objects as particle, therefore macroscopic objects are studied under classical mechanics. Whereas, microscopic objects have essentially wave nature because the size of microscopic particle is too small, thus the wavelength of the wave attached to it is large enough and thus significant. The quantum mechanics treats the moving object as wave, therefore microscopic objects are studied under quantum mechanics. • Earlier (before 1900) when no such quantum mechanics was introduced, all the moving objects (macroscopic or microscopic) were considered to have particle nature and studied under classical mechanics. Thus, the various models of atom, such as Thomson model of atom, Rutherford nuclear model of atom and Bohr model of hydrogen atom are studied under classical mechanics. Since in all these models, the sub-atomic particles (like electron, neutron, proton, atom) are considered to have particle nature, thus all these above models are studied under classical mechanics.

1.3

J.J. THOMSON MODEL OF ATOM

J.J. Thomson in 1914, proposed that an atom was a sphere of positive charge in which number of electrons were embedded to neutralise the positive charge. The atom is stable because there is a balance between the inter-electronic repulsive forces and their attraction towards the positive charge. This model is compared with cake or pudding in which raisins are embedded or with a watermelon in which seeds are embedded randomly. Thus, this model is also known as plum pudding model or raisin pudding model or watermelon model. This model was able to explain the overall neutrality of the atom, but it could not satisfactorily explain the results of scattering experiments, carried out by Rutherford in 1911 and was rejected.

1.4

RUTHERFORD’S NUCLEAR MODEL OF ATOM— DISCOVERY OF NUCLEUS

Rutherford in 1911, performed scattering experiments in which he bombarded thin foils of metals like Au, Ag, Pt or Cu with a beam of fast moving α-particles. (as shown in Fig. 1.1).

1.4 Rutherford’s Nuclear Model of Atom—Discovery of Nucleus

Beam of -particles

5

Thin gold foil Undeflected -particle

Deflected -particle by small angle

Deflected -particle by large angle Fig. 1.1. View of gold foil in Rutherford a-particle scattering experiment.

From the results of the scattering experiments, the Rutherford model was given. According to Rutherford model of atom, the atom consists of two parts: 1. Nucleus which is very small, carries positive charge and in which the entire mass is concentrated. The mass of the atom is mainly due to protons and neutrons, since electrons have negligible mass. Hence, protons and neutrons must be present in the nucleus. The presence of positively charged protons in the nucleus also accounts for the positive charge on the nucleus. It is present at the centre of the atom. 2. Extra-nuclear part is the space around the nucleus in which the electrons are distributed. Table 1.2: Atomic numbers and Atomic masses

Atomic number (Z) Atomic mass (A) Nucleons Isotopes

Number of protons or Number of electrons Number of protons + Number of neutrons Collectively protons and neutrons are known as nucleons Atoms of the same element having same atomic number but different mass number (or same number of protons but different number of neutrons) 37 e.g., 11H, 12H, 13H; 35 17Cl, 17Cl

Isobars

Atoms of the different elements having different atomic number but same mass number (or different number of protons and different number of neutrons). 40Ar, 40 K, 40 Ca e.g., 18 19 20

6

Chapter 1

Classical Mechanics

Isotones

Atoms of the different elements having different atomic number and different mass number (or different number of protons but same number of neutrons). e.g., 146C, 157N, 168O

Isoelectronics

The species (atoms or ions) containing the same number of electrons. e.g., O2–, F–, Na+, Mg2+, Al3+, Ne, each of them contains 10 electrons.

• Drawbacks of Rutherford’s Nuclear Model of Atom 1. Unable to explain the stability of atom: According to Rutherford model, electrons are revolving around the small and heavily positively charged nucleus. According to Maxwell’s electromagnetic theory, whenever a charged particle revolves in a field of force (like electron revolves around the nucleus) it undergoes acceleration due to change in direction even if the speed remains constant and it starts loosing energy in the form of electromagnetic radiations (Fig. 1.2). Thus, the orbit of the revolving electron will keep on becoming smaller and smaller and ultimately the electron should fall into the nucleus.

+

Fig. 1.2. Loss of electromagnetic radiations by revolving electron.

But, this actually does not happen and the atom is quite stable. Thus, Rutherford model could not explain the stability of the atom. 2. Unable to explain distribution and energies of electron. 3. Unable to explain the line spectra of the elements.

1.5

DEVELOPMENTS LEADING TO THE BOHR MODEL OF ATOM

It was seen that all the elements give characteristic atomic spectra or line spectra which could not be explained on the basis of Rutherford nuclear model of atom. In order to understand atomic spectra or line spectra, it is essential to understand the nature of radiation. The nature of radiation was first explained on the basis of “Electromagnetic Wave Theory” (Classical Theory) and then by “Planck’s Quantum Theory” (Quantum Theory).

1.5 Developments Leading to the Bohr Model of Atom

7

Thus, two developments played a major role in the formulation of Bohr model of atom. 1. Dual nature of electromagnetic radiation. 2. Atomic spectra or Line spectra After these two developments, a new model of atom, called Bohr model of atom was put forward. This model of atom was able to explain the stability of atom and main lines in the spectra of hydrogen atom. • Now, we shall take up a brief discussion of the above developments made one by one, before discussing the Bohr model of atom.

1.5.1 Dual Nature of Electromagnetic Radiation A. Classical theory of radiation or Electromagnetic wave theory (radiation has wave nature) B. Quantum theory of radiation or Planck’s quantum theory (radiation has particle nature) A. Classical Theory of Radiation or Electromagnetic Wave Theory James Clark Maxwell in 1864, put forward the electromagnetic wave theory. The main points of this theory are: 1. When electrically charged particles move under acceleration, alternating electric field and magnetic field are produced and transmitted. These fields are transmitted in the form of waves called Electromagnetic waves (EM waves) or Electromagnetic radiations. 2. The oscillating electric and magnetic fields produced by oscillating charged particles are perpendicular to each other and both are perpendicular to the direction of electromagnetic radiations. 3. The radiations possess wave nature and travel with the velocity of light (c = 3 × 108 ms–1). 4. Unlike sound waves or water waves, these electromagnetic radiations do not require any medium for propagation and can move in vaccum. Some important characteristics of a wave: Crest



Crest

a

 Trough

Trough

8

Chapter 1

Classical Mechanics

1. Wavelength (λ): It is defined as the distance between any two consecutive crests or troughs. Unit: Å or m or cm or nm or pm. 1 Å = 10–10 m; 1 nm = 10–9 m; 1 pm = 10–12 m; 1 cm = 10–2 m. 2. Frequency (v): It is defined as the number of waves passing through a point in one second. Unit: Hertz (Hz) or cycles/sec (cps) or sec–1 1 Hz = 1 cps = 1 sec–1 3. Velocity (c): It is defined as the distance travelled by the wave in one second. Unit: cms–1 or ms–1 In vacuum, all types of electromagnetic waves, regardless of wavelength, travel at same speed i.e., 3 × 108 ms–1. This is called speed of light (c). c = 3 × 108 m s–1 4. Amplitude (a): It is defined as the height of the crest or depth of the trough. Unit: Å or m or cm or nm or pm. 5. Wave number ( v ) : It is defined as the number of waves present per unit length. It is equal to the reciprocal of the wavelength. v =

1 λ

Unit: cm–1 or m–1. There are many types of electromagnetic radiations which differ from one another in wavelength and hence in frequencies. When these electromagnetic radiations are arranged in order of their increasing wavelengths or decreasing frequencies, the complete arrangement obtained is called electromagnetic spectrum. (as shown in Fig. 1.3).

1.5 Developments Leading to the Bohr Model of Atom

9

Decreasing wavelength or Increasing frequency -6 10 nm -5 10 nm -4 10 mn -3 10 nm -2 10 nm -1 10 nm 1 nm 10 nm

100 nm

103 nm

10 m 100 m 1000 m 10 mm 10 cm 100 cm 10 m 100 m 1000 m 10 km 100 km 1 Mm 10 Mm 100 Mm

1Å

1 m

1 mm 1 cm

1m

1 km

Gamma-Rays

X-Rays

Blue

Violet Indigo

UHF VHF HF MF LF

Visible Light ~400 nm - ~700 nm

Ultravoilet

Visible light

Near Infrared

Far Infrared

Microwave

Radio Audio

Green

Yellow Orange Red

Fig. 1.3. The Electromagnetic spectrum.

Limitations of Classical theory of radiation or Electromagnetic wave theory: Some of the experimental phenomena such as reflection, refraction, diffraction, interference, polarization etc. can be explained by the wave nature of electromagnetic radiations, but it could not explain the following observations: 1. Black body radiation (the nature of emission of radiation from hot bodies). 2. Photoelectric effect (ejection of electrons from metal surface when radiation strikes it). 3. Compton effect (Scattering of X-rays) 4. Variation of heat capacity of solids as a function of temperature. 5. Line spectra of atoms with special reference to hydrogen.

10

Chapter 1

Classical Mechanics

These phenomena could be explained only if electromagnetic radiations are supposed to have particle nature instead of wave nature. Thus to explain the above facts, Quantum theory of radiation was given by Max Planck. B. Planck’s Quantum Theory or Quantum Theory of Radiation or Corpuscular Theory of Radiation. To explain the phenomena such as black body radiation, photoelectric effect, compton effect, variation of heat capacity of solids and line spectra of atoms, Max Planck in 1900 put forward Planck’s quantum theory. This theory was further extended by Einstein in 1905. The main points of this theory are: 1. The radiant energy is emitted or absorbed not continuously but discontinuously in the form of small discrete packets of energy. Each such packet of energy is called quanta. In case of light, the packet of energy is called photon. 2. Each wave packet is associated with a definite amount of energy. The energy of each quanta is directly proportional to the frequency of the radiation E∝v as,

E = hv c = vλ

hc λ where, h is a proportionality constant called Planck’s constant. h = 6.626 × 10–27 erg sec = 6.626 × 10–34 J sec. 3. A body can emit or absorb energy in terms of the integral multiples of quanta E = Nhv Nhc or E= λ where, N is any integral number. • Dual Nature of Radiation: Some properties of radiation like reflection, refraction, diffraction, interference, polarization etc. can be explained only if we consider radiation to have wave nature, whereas some properties of radiation like black body radiation, compton effect, photoelectric effect etc. can be explained if we consider radiation to have particle nature. Thus, Einstein in 1905 suggested that radiation has dual nature i.e. wave nature as well as particle nature. E=

1.5 Developments Leading to the Bohr Model of Atom

11

Problems based on Planck’s Quantum theory 1. Calculate the wave number and frequency of radiation having wavelength 3 × 106 nm. Ans. λ = 3 × 106 nm = (3 × 106) × 10–9 m = 3 × 10–3 m 1 1 (i) Wave number = v = = = 0.33 × 103 m–1 λ 3 × 10−3 c 3 × 108 ms −1 (ii) Frequency = v = = = 1011 s–1. λ 3 × 10−3 m 2. Calculate the wavelength of the electromagnetic radiation having frequency of 1367 kHz (kilohertz). Which part of the electromagnetic spectrum does it belong to? Ans. v = 1367 kHz = 1367 × 103 Hz = 1367 × 103 s–1 (1 Hz = 1 s–1) c = 3 × 108 ms–1, c 3 × 108 ms −1 = = 219.5 m v 1367 × 103 s −1 It belongs to the radiowave region of the electromagnetic spectrum. 3. Calculate the frequency and energy of a photon of light having wavelength 3000 Å. Ans. λ = 3000 Å = 3000 × 10–10 m c = 3 × 108 ms–1 h = 6.62 × 10–34 Js λ=

(i) (ii)

3 × 108 ms −1 c = = 1015 s–1 v= 3000 × 10−10 m λ

Energy of a photon = E = hv E = (6.626 × 10–34 Js) × (1015 s–1) E = 6.6 × 10–19 J. 4. Calculate the energy of one mole of photons of light having frequency 3 × 1014 s–1. Ans. v = 3 × 1014 s–1 h = 6.626 × 10–34 Js Energy of one photon = E = hv = (6.626 × 10–34) × (3 × 1014) = 19.8 × 10–20 J Number of atoms in 1 mole = NA = 6.023 × 1023

12

Chapter 1

Classical Mechanics

Energy of one mole of photons = NAhv E = (6.023 × 1023) × (6.626 × 10–34) × (3 × 1014) E = 119.25 × 103 J. 5. A 100 Watt bulb emits light of wavelengths 6000 Å. Calculate the number of photons emitted by the bulb in one minute. Ans.

Power (P) = 100 W Time (t) = 1 min = 60 sec Energy = Power × Time E = 100 × 60 = 6000 J λ = 6000 Å = 6000 × 10–10 m E = Nhv = N=

Nhc λ

E×λ (6000) × (6000 × 10−10 ) = h×c (6.626 × 10−34 ) × (3 × 108 )

N = 1.81 × 1022 photons. 6. 3 × 1018 photons of a certain light are found to produce 1.5 J of energy. Calculate the wavelength of the light. Ans. N = 3 × 1018 E = 1.5 J c = 3 × 108 ms–1 h = 6.626 × 10–34 Js Nhc λ Nhc λ= E

E=

(3 × 1018 ) × (6.626 × 10−34 ) × (3 × 108 ) 1.5 λ = 39.78 × 10–8 m. λ=

7. Calculate and compare the energies of two radiations, one with a wavelength of 4000 Å and other with wavelength 8000 Å. Ans. Case I: λ1 = 4000 Å = 4000 × 10–10 m (6.626 × 10−34 ) × (3 × 108 ) hc = E1 = (4000 × 10−10 ) λ1 E1 = 4.95 × 10–19 J

1.5 Developments Leading to the Bohr Model of Atom

13

Case II: λ2 = 8000 Å = 8000 × 10–10 m (6.626 × 10−34 ) × (3 × 108 ) hc = E2 = (8000 × 10−10 ) λ2 E2 = 2.48 × 10–19 J On comparing the two value of energy i.e. taking ratio, we have E1 4.95 × 10−19 = E2 2.48 × 10−19 E1 = 2E2.

Explanation of Various Phenomena Using Planck’s Quantum Theory 1. Black Body Radiation: If any substance with high melting point (like an iron rod) is heated in a furnace, it first becomes dull red and then progressively becomes more and more red as the temperature increases. As this is heated further, the radiation emitted becomes yellow, then white and finally begins to glow with blue light. In terms of frequency, the radiation emitted goes from a lower frequency (red) to a higher frequency (blue) as the temperature increases. The variation of intensity with wavelength at two different temperatures is shown in Fig. 1.4. From the given figure, it is seen that on moving from right to left in the plot at a given temperature, intensity of radiation increases as the wavelength decreases (or frequency increases), reaches a maximum value at a particular wavelength, and then starts decreasing with further decrease of wavelength. If the substance being heated is a black body or ideal body (body which emits and absorbs all frequencies), then the radiation emitted by such a body is called black body radiation. T2

Intensity

T2 > T1 T1

Wavelength Fig. 1.4. Variation of intensity versus wavelength.

14

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The above observations cannot be explained by electromagnetic wave theory. According to Electromagnetic Wave Theory, the electromagnetic radiation are emitted or absorbed from a substance continuously in the form of waves. The energy of any wave is proportional to its intensity (i.e., square of amplitude) and is independent of its frequency. Thus, the energy of electromagnetic radiation is proportional to its intensity and not to its frequency. Thus, according to electromagnetic wave theory, if a substance is continuously heated, the radiation emitted by the substance should have the same colour (i.e., frequency does not change) although its intensity may vary. Thus, electromagnetic wave theory could not explain the phenomena of black body radiation. According to Planck’s Quantum Theory, when a substance is heated, the atoms present in the substance are set into oscillation and emit radiation of frequency ‘v’. On continuous heating, more and more energy is being absorbed by the atoms and they emit radiations of higher and higher frequency. As red light has minimum frequency and yellow has higher frequency, thus, the substance on heating first becomes red, then yellow and so on. 2. Photoelectric Effect: When radiations of appropriate frequency strikes the surface of certain metals (like Cs, K, Pb), the electrons are ejected from the surface of the metal. This phenomena is called photoelectric effect, [Fig. 1.5(a)] and the electrons emitted are called photoelectrons. The apparatus used for studying photoelectric effect is photoelectric cell [Fig. 1.5(b)]. Photoelectron emitted

Incident light

Electron in atom Metal Fig. 1.5. (a) Photoelectric effect.

Metal surface

Light

Anode –

Cathode + e– e– –

e– e–

+ Detector Fig. 1.5. (b) Photoelectric cell.

1.5 Developments Leading to the Bohr Model of Atom

15

Some important facts about the photoelectric effect are: (i) The electrons are ejected from the metal surface as soon as the beam of light strikes the surface i.e., there is no time lag between the striking of light beam and the ejection of electrons from the metal surface. (ii) The number of photoelectrons ejected is proportional to the intensity or brightness of incident radiation. (iii) The kinetic energy of photoelectrons ejected is proportional to the frequency of incident radiation. (iv) The electrons are ejected out of the metal surface only if the incident radiation has a certain minimum frequency. This minimum frequency is called threshold frequency (v0). If the frequency is less than v0, then no electrons are ejected. The above observations cannot be explained by electromagnetic wave theory. According to Electromagnetic Wave Theory, the electromagnetic radiations are emitted or absorbed from a substance continuously in the form of waves. Thus, it is possible to accumulate radiations of all frequencies on the surface of the metal, and thus all kind of radiations (i.e., irrespective of its frequency) should be able to eject electrons. Also according to wave theory, energy is proportional to its intensity, thus the kinetic energy of the ejected photoelectrons should depend upon the intensity of the incident radiation and not on its frequency. According to the Planck’s Quantum Theory: Using Planck’s quantum theory, Einstein in 1905 was able to give an explanation of the different points of the photoelectric effect as follows: (i) When radiation of some particular frequency falls on the surface of metal, the quanta gives its entire energy to the electron of the metal atom. The electron will be detached from the metal atom only if the energy of the quanta is sufficient to overcome the force of attraction of the electron by the nucleus. That is why photoelectrons are ejected only when the incident radiation has a certain minimum frequency (threshold frequency, v0). The minimum energy required to eject the electron (hv0 ) is called work function (w0 ). (ii) If the frequency of the incident radiation (v) is more than the threshold frequency (v0), the excess energy is imparted to the electron as kinetic energy  1 i.e., hv = hv0 + kinetic energy of photoelectron  mv 2   2

16

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Hence, greater is the frequency of the incident light, greater is the kinetic energy of the emitted electron. (iii) Increasing the intensity of radiation of a given frequency, increases the number of quantas but does not increase the energies of quantas. Hence, when the intensity of radiation is increased, more electrons are ejected but the energies of these electrons are not altered. 3. Compton Effect: When a beam of one particular wavelength (mono chromatic X-rays) of high frequency are made to fall on some light element like carbon, X-rays get scattered. These scattered X-rays have wavelength larger than the incident X-rays. In other words, the scattered X-rays have lower frequency or lower energy than the incident X-rays. This phenomena is known as Compton effect. on

ctr

(i)

Incident photon of X-rays (i) Target electron at rest

le il e

co  Sca  tte of red p X- hot ray on s Re

(s)

Scatterer Fig. 1.6. Compton effect: Scattering of X-rays.

According to Electromagnetic Wave Theory: The scattered X-rays should possess the same wavelength as the incident one. Thus scattering should be independent of the wavelength of the incident radiation. In the classical view, compton scattering is simply described as an inelastic collision process between X-rays and electrons of the scatterer. Thus, energy and momentum remain conserved even after the collision. According to Planck’s Quantum Theory: Using Planck’s quantum theory, Prof. A.H. Compton in 1921, was able to give an explanation of the scattering of X-rays. According to Compton, the phenomena of scattering might be regarded as an elastic collision between high energy photon of X-rays and the electron of the scatterer. When a photon of incident X-rays of energy ‘hv’ collide with an electron, the loosely bound electrons from the outer shell of atom or molecule, gains kinetic energy from the incident X-rays. Thus, the scattered photon of X-rays will have a smaller energy and in consequence, a lower frequency or greater wavelength than that of the incident photon. The ejected electron is known as Compton recoil electron.

1.5 Developments Leading to the Bohr Model of Atom

17

(Loss in photon energy = Gain in electron energy). It was also seen that the decrease in frequency or increase in wavelength of scattered X-rays varied with direction followed by the incident beam of X-rays. The change in the wavelength (Dλ), Compton shift, is given as Compton equation Dλ = λs – λi =

2h h θ sin 2   = (1 − cos θ) me c  2  me c

where, Dλ = compton shift or wavelength shift λi = wavelength of incident radiation λs = wavelength of scattered radiation me = mass of electron c = velocity of X-rays (same as velocity of light) θ = angle of scattering i.e., angle between incident and scattered X-rays. Depending upon angle, θ, the following two cases arise: Case I: θ = 0°, scattered radiation parallel to the incident radiation h (1 – cos 0°) = 0 [Q cos 0° = 1] me c This is the case of no wavelength shift and thus no Compton effect. Case II: θ = 90º, scattered radiation is perpendicular to the incident radiation Dλ =

Dλ =

h (1 − cos 90° ) me c

Dλ =

h = 0.0242 Å. mec

[cos 90º = 0]

Here, Dλ = 0.0242 Å is known as Compton wavelength. 4. Specific Heat of Solids: Heat appears as a rise in temperature of the system in case of heat absorbed by the system. The increase in temperature is proportional to the heat transferred q = Coefficient × DT q = C × DT

18

Chapter 1

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The coefficient C is called heat capacity and its magnitude depends on the size, composition and nature of the system. Now, q C= ∆T If C is large, a given amount of heat results in only a small temperature rise. Like: water has a large heat capacity i.e. a lot of energy is needed to raise its temperature. Heat capacity (C): It is the quantity of heat required to raise the temperature of a substance by one degree celcius (or one kelvin). Molar heat capacity (Cm): It is the quantity of heat needed to raise the temperature of one mole of a substance by one degree celcius (or one kelvin). C Cm = ; where n = number of moles of substance. n Specific heat/Specific heat capacity (S): It is the quantity of heat required to raise the temperature of unit mass of a substance by one degree celcius (or one kelvin). C S= ; where m = mass of substance. m According to Electromagnetic Wave Theory: Lavosier and Laplace had determined the specific heats of a number of metals with considerable accuracy. In 1819, Pierre Louis Dulong and Alexis Thèrèse Petit, pointed out that, using the accepted atomic weights for many metals and multiplying them by the specific heats, an approximate constant figure was obtained for each metal. The average value of this constant is 6.4. The product is termed as atomic heat. Thus, Dulong and Petit gave their law such as: “At room temperature, the product of the specific heat and the atomic weight of a solid is a constant for all solid elements, and is approximate equal to 6.4”. Atomic heat = Specific heat × Atomic weight ≈ 6.4 Success of this law was that when this law was applied to different solid substances, it was found that the atomic heat is almost constant at about 6.2 ± 0.4 cals/degree inspite increase in the atomic weights from 7 to 200. The various exceptions to Dulong and Petit’s law are: 1. According to Dulong and Petit’s law, the atomic heats at room temperature of all solids must be about 6.4. Actually, this is not true. For example, the elements Be, B, C and Si have exceptionally low atomic heats; the values at ordinary temperature being 3.5, 2.5, 1.35 and 4.7 kcals/deg respectively.

1.5 Developments Leading to the Bohr Model of Atom

19

2. As the atomic weight of an element is constant, the atomic heat must also be constant, i.e. its value should be same at all temperatures. This is contrary to the observed facts. For example, Diamond shows variation of atomic heats with temperature. Similarly, Si, the atomic heat increases rapidly with temperature and reaches fairly constant value above 200°C. From this we conclude that, Dulong and Petit’s law would apply to all substances provided the temperature is high enough. 3. Measurement of specific heats at low temperatures indicates that atomic heat decreases slowly with fall of temperature and below a certain temperature, characteristic of each element, the atomic heat decreases rapidly, tending finally to the zero value at the absolute zero of temperature. This type of change is not permitted by the law of Dulong and Petit. Thus, we can say that: “The law of Dulong and Petit is entirely accidental and the atomic heat of 6.4 cals/degree has a theoretical significance only.” According to the Planck’s Quantum Theory: The quantum theory of the heat capacity of solids was developed by A. Einstein in 1907 and P. Debye in 1912. The theory is based on the quantisation of the vibrational motion of the atoms in a crystal. Einstein’s Theory on the Heat Capacities of Solids: Einstein assumed three things when he investigated the heat capacity of solids. First, he assumed that each solid was composed of a lattice structure consisting of N atoms. Each atom was treated as moving independently in three dimensions within the lattice (3 degrees of freedom). This meant that the entire lattice’s vibrational motion could be described by a total of 3 N motions, or degrees of freedom. Secondly, he assumed that the atoms inside the solid lattice did not interact with each other and thirdly, all of the atoms inside the solid vibrated at the same frequency. The third point highlights the main difference in Einstein’s and Debye’s two models. Einstein’s first point is accurate because the experimental data supported his hypothesis, however his second point is not because if atoms inside a solid could not interact sound could not propagate through it. For example, a tuning fork’s atoms, when struck, interact with one another to create sound which travels through air to the listener’s ear. Atoms also interact is a solid when they are heated. Take for example a frying pan. If the pan is heated on one side, the heat transfers throughout the metal effectively warming the entire

20

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pan. Molecules that make up the frying pan interact to transfer heat. Much in the same way the oscillators in a solid interact when energy is added to the system. The extent of these interactions lead to the physically observed heat capacity. The internal energy of solid made up of N atoms is:

1

U = 3NK θE 

2

+

1 e θ E /T

 , − 1 

...(1)

where θE is the Einstein temperature given by θE = hn/K, h is planck’s constant and K is Boltzmann’s constant and n is oscillator frequency of atoms inside the solid. The heat capacity at constant volume is 2

eθ /T  θE   ∂U  Cv =   = 3NK  T  (eθ /T − 1) 2 ,  ∂T  v E

E

...(2)

The Einstein temperature’s accesibility of the vibrational energy inside of a solid molecule determines the heat capacity of that solid. The greater the accesibility the greater the heat capacity. If the vibrational energy is easily accesible the collisions in the molecule have a greater probability of exciting the atom into an upper vibrational level. This is displayed below. Vibrational States of Excitement

Einstein T = 1.0

Einstein T = 4.0

Einstein T = 10

Fig. 1.7. The figure illustates the effect of the Einstein temperature on the likelihood of an oscillator absorbing energy from a collision and tranferring that energy into stored heat or heat capacity. As the Einstein temperature increases, the greater the probability of an oscillator being excited to the next vibrational state.

1.5 Developments Leading to the Bohr Model of Atom

21

For temperatures very large compared with the Einstein temerature. Cv ≈ 3 NK = 3nR.

...(3)

Thus the high temperature limit of Einstein’s equation gives the value of Dulong and Petit. The failure of their law becomes evident when we examine the low temperature limit. For θE/T >> 1, 2

 θ  − θ /T Cv = 3NK  E  e E T 

...(4)

As T approaches zero, Cv also goes to zero, since the exponential decay overpowers the growth of (θE/T)2. Einstein’s theory also explains low heat capacities of some elements at moderately high for temperatures. If an element (like diamond) has a large Einstein temperature, the ratio θE/T will be large even for temperatures well above absolute zero, and Cv will be small. For such an element θE ≡ hn/K must be very large and, according, n, must be large. For an oscillator with force constant k and reduced mass m, the oscillator frequency is n=

1 2π

k m

...(5)

A large frequency value suggests a small reduced mass or a large force constant, corresponding to lighter elements and elements that produce very hard crystals. The theory correctly predicts the failure of the law of Dulong and Petit for those elemetns. As an example, the heat capacity of diamond approaches 3 NK only at extremely high temperatures (θE = 1450 K for diamond). 5. Line spectra of atoms with special reference to hydrogen (detailed explanation given in next section 1.5.2)

1.5.2 Emission and Absorption Spectra • Interaction of electromagnetic radiations with matter produces spectra. The branch of science dealing with the study of spectra is called spectroscopy. • An instrument used to separate the radiations of different wavelengths (or frequencies) is called spectroscope. A spectroscope consists of a prism or a diffraction grating for the dispersion of radiations and a telescope to examine the emergent radiations with the human eye.

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• The spectra are broadly classified into: A. Emission spectra B. Absorption spectra SPECTRA

A. Emission spectra

1. Continuous Spectra

B. Aborption Spectra

2. Line Spectra or Atomic Spectra

A. Emission Spectra: When electromagnetic radiation interacts with matter, (like by passing electric discharge through a gas at low pressure or by heating some substance at high temperature etc.), then atoms and molecules may absorb energy and reach to a higher energy state. With higher energy, these are in an unstable state. For returning to their normal energy state, the atoms and molecules emit radiations. When these emitted radiations are passed directly through the prism and then received on the photographic plate, the spectra obtained is called ‘Emission Spectra’. Depending upon the source of radiation, the emission spectra are mainly of two types: 1. Continuous Spectra: When radiations from any source such as sun, a bulb or any hot glowing body are passed directly through a prism, it splits up into seven different colours from violet to red (like rainbow). These colours are so continuous that each of them merges into the next. The spectra obtained is called Continuous Spectra. 2. Line Spectra or Atomic Spectra: When some substance (like volatile salt) is heated at high temperature or an electric discharge is passed through a gas at low pressure, the radiations are emitted. When these radiations are passed directly through a prism, it is found that no continuous spectra is obtained but some isolated coloured lines are obtained on the photographic plate separated from each other by dark spaces. The spectra obtained is called Line Spectra. Each line in the spectra corresponds to a particular wavelength.

1.5 Developments Leading to the Bohr Model of Atom

23

Each element gives its own characteristic line spectra, differing from those of all other elements. Hence, spectra of the elements are regarded as their finger prints differing from each other like the finger prints of the human beings. Since spectra of most of the elements consists of a number of bright lines separated by dark lines, that is why line spectra is also called atomic spectra. B. Absorption Spectra: When white light from any source is passed through the solution or vapours of a chemical substance, then some wavelengths are absorbed, depending upon the nature of the element. After absorption of certain wavelengths, the transmitted light is analysed by the spectroscope, then it is observed that some dark lines are obtained in the continuous spectra of the white light which is irradiated on the source. Further, it is observed that the dark lines are at the same place where coloured lines are obtained in the emission spectra of the same substance. The spectra obtained is called absorption spectra. Absorption spectra is always discontinuous spectra consisting of dark lines. • Line Emission Spectra of Hydrogen When an electric discharge is passed through gaseous hydrogen at low pressure, hydrogen molecules dissociate and the energetically excited hydrogen atoms produced emit radiations of discrete frequencies. The emitted radiations are examined with a spectroscope, and the spectra obtained is called the emission spectra of hydrogen (as shown Fig. 1.8). The hydrogen spectra consists of a large number of lines which are grouped into different series, named after the discoverers. Line spectra of hydrogen atom is simplest because hydrogen being the lightest element, the number of lines produced in the spectra is minimum. As the molecular mass of the gas increases, the spectra becomes more and more complex. Series : Lyman

Balmer

Region : Ultraviolet

Visible

Paschen

Brackett

Infrared

Fig. 1.8. (a) Line emission spectra of hydrogen.

Pfund

4340.5 Å

4101.7 Å

Classical Mechanics

4861.3 Å

Chapter 1

6562.8 Å

24

Continuum

H

H

H

H

H

Energy Fig. 1.8. (b) Hydrogen spectra in visible region (Balmer series).

Rydberg in 1890 gave theoretical equation i.e., Rydberg formula to calculate the wave number of the lines appearing in the line emission spectra of hydrogen atom (Z = 1 for hydrogen atom).  1 1  v = RH  2 − 2   n1 n2 

n2 n1

cm–1

where, RH is called Rydberg constant and has a value equal to 109677 or 1.097 × 107 m–1. For a particular series, n1 and n2 are whole numbers, n1 is constant and n2 varies. Table 1.3: Spectral series found in line emission spectra of atomic hydrogen

Spectral series

n1

n2

Regions of spectrum

Lyman series Balmer series

1 2

2, 3, 4,… 3, 4, 5, …

Ultraviolet Visible

Paschen series

3

4, 5, 6, …

Infrared

Brackett series

4

5, 6, 7, …

Infrared

Pfund series

5

6, 7, 8, …

Infrared

Humphrey series

6

7, 8, 9, …

Far Infrared

For hydrogen-like atoms (i.e., one electron system) the Rydberg formula is  1 1  2 v = RH  2 − 2  Z  n1 n2  where, Z = atomic number of the hydrogen like atoms. • Explanation of Line Emission Spectra using Planck’s Quantum Theory: Max Planck gave a theoretical explanation of the spectrum of radiation emitted by an object that glows when heated. He suggested

1.5 Developments Leading to the Bohr Model of Atom

25

that the walls of a glowing solid could be imagined to contain a series of resonators (or particles) that oscillate at different frequencies. These resonators gain energy in the form of heat from the walls of the object and loose energy in the form of electromagnetic radiation. The energy of these resonators at any moment is proportional to the frequency with which they oscillate. To explain the observed spectrum, Planck had to assume that the energy of these resonators could take only a limited number of values. In other words, the spectrum of energies for these resonators was no longer continuous. Because the number of values of energy of these resonators is limited, they are theoretically countable. Thus, energy of the resonators in this system is therefore said to be quantised. Planck introduced the notion of quantisation to explain how radiation was emitted. Albert Einstein extended Planck’s work to the light that had been emitted.

Problems based on line emission sPectra of hydrogen and hydrogen like atoms . 1. Name the first three series of lines that occur in the atomic spectra of hydrogen. Indicate the regions in the electromagnetic spectra where these lines occur? Ans. Spectral series Lyman Balmer Paschen

n1

n2

Regions of spectrum

1 2 3

2, 3, 4, … 3, 4, 5, … 4, 5, 6, …

Ultraviolet Visible Infrared

2. Calculate the wavelength and the frequency of the radiation emitted when an electron in the hydrogen atom jumps from n = 3 state to n = 1 state (ground state). In which region of the electromagnetic spectrum will this line lie. (Rydberg constant = 109677 cm–1). Ans. According to Rydberg formula  1 1  2 v = RH  2 − 2  Z  n1 n2  For H-atom, Z = 1 RH = 109677 cm–1 n1 = 1 and n2 = 3 1 1 v = 109677  2 − 2  3  1

n2

3

n1

1

26

Chapter 1

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8  1 v = 109677 1 −  = 109677 × 9 9   v = 97490.7 cm–1 1 1 = (i) wavelength = λ = = 1.03 × 10–5 cm = 1.03 × 10–7 m 97490.7 v (ii) frequency = v =

c 3 × 108 ms −1 = = 2.9 × 1015 s–1 λ 1.03 × 10−7 m

This line will lie in the Lyman series i.e., UV region of the electromagnetic spectrum. 3. In the Balmer series, the wavelength of the first line is 650 nm. Calculate the wavelength of the third line. Also calculate the limiting line in Balmer series. (R = 109677 cm–1) Ans. According to Rydberg formula v =

 1 1  2 1 = RH  2 − 2  Z λ  n1 n2 

For H-atom, Z = 1 For Balmer series, n1 = 2 (fixed) and n2 = 3, 4, 5, 6, … (i) For the first line in the Balmer series n1 = 2 and n2 = 3 1 1   1 = RH  2 − 2  650 3  2

n2

3

n1

2

1 1 1 = RH  −  650 4 9 5 RH 1 = 650 36 Similarly, for the third line in the Balmer series n1 = 2 and n2 = 5. 1  1  1 = RH  2 − 2  λ 5  2

…(1)

n2

5

n1

2

1 1 1  = RH  −  λ  4 25  21RH 1 = λ 100

…(2)

1.5 Developments Leading to the Bohr Model of Atom

27

Dividing eq. (1) by eq. (2), we have 5 RH 100 λ × = 650 36 21RH 325, 000  5 × 100  λ=   × 650 = 756  36 × 21  λ = 429.89 cm. (ii) For the limiting line in the Balmer series n1 = 2 and n2 = ∞  1 1  1 = RH  2 −  λ (∞ ) 2  2

n2



n1

2

1  1 = RH  − 0  λ  4 R 1 = H λ 4

…(3)

Dividing eq. (1) by eq. (3), we have 5 RH 4 λ × = 650 36 RH 13000  5× 4  λ=   × 650 = 36  36  λ = 361.11 cm. 4. Calculate the wavelength of the radiation emitted in the spectrum of He+ ion when the electron jumps from n = 3 to n = 1 state. (RH = 109677 cm–1) Ans. He+ is a one electron system like hydrogen atom Using Rydberg formula 1  1 1  = RH  2 − 2  Z 2  n λ  1 n2  He+

For ion, Z = 2 Here, n1 = 1 and n2 = 3 1 1 1 = 109677  2 − 2  (2) 2 λ 3  1 1  1 = 109677 1 −  × 4 λ  9

n2

3

n1

1

28

Chapter 1

Classical Mechanics

1 8 = 109677 × × 4 λ 9 9 9 λ= = = 2.56 × 10–6 cm. 109677 × 8 × 4 3509664

1.6

BOHR MODEL OF ATOM

Neil Bohr a Danish physicist in 1913 proposed a new model of atom called Bohr model of atom. This was given to overcome the drawbacks of Rutherford model of atom. This model was based on Planck’s quantum theory. The main points of Bohr’s theory of atom are: 1. An atom consists of a small, heavy, positively charged nucleus at the centre and the electrons revolve around the nucleus in circular path. 2. There are large number of circular paths arranged concentrically around the nucleus, but the electrons revolve around the nucleus only in certain selected circular path of fixed radius and fixed value of energy. These circular paths are called orbits or allowed energy levels or stationary states as shown in Fig. 1.9.

Nucleus +

K L MN n = 1 n = 2, 3 4

Fig. 1.9. Circular paths around the nucleus.

These energy levels are designated as K, L, M, N, … or numbered as 1, 2, 3, 4, … . Also, the word stationary does not mean that the electrons are stationary but it means that the energy of the electron revolving in a particular energy level is fixed and does not change with time. 3. Since the electrons revolve only in the orbits which have fixed values of energy, hence electrons in an atom can have only certain definite values

1.6 Bohr Model of Atom

29

of energy and not any value of their own. Thus, Bohr theory of the atom leads to the concept of quantisation of electronic energy. Quantisation means that a quantity cannot change continuously to have any arbitrary values but can change only discontinuously to have some specific values. Thus the energy of the electron cannot change continuously but can have only definite values. i.e., energy of an electron is quantised. 4. An electron did not radiate energy if it stayed in one orbit, and therefore did not slow down. Thus, as long as the electron remains in a particular orbit, it does not loose or gain energy. But when energy from some external source is absorbed by the electron, it may jump to some higher energy level (excited state). When the electron jumps back to the lower energy level, it radiates some amount of energy. Thus, energy is emitted or absorbed only when the electron jumps from one orbit to the other. 5. The frequency (v) of radiation absorbed or emitted when transition occurs between two energy levels that differ in energy by DE is given in the form of Bohr frequency rule. Energy changes during electronic jumps (DE) is shown in Fig. 1.10. Ef

Ei

nf

ni E

E Ei

ni

nf

Ef

Fig. 1.10. (a) When an electron jumps from lower energy to higher energy level, net energy is absorbed

Fig. 1.10. (b) When an electron jumps from higher energy to lower energy level, net energy is emitted.

n=7 n=6 n=5 n=4 n=3 n=2 n=1 +

Atom emits energy

Atom absorbs energy

Fig. 1.10. (c) Bohr model showing absorption and emission of energy.

30

Chapter 1

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DE = Ef – Ei hv = Ef – Ei E f − Ei

…(Bohr frequency rule) h where, v is the frequency of radiation absorbed or emitted during jumping of electron and the expression is Bohr frequency rule. 6. Like energy, the angular momentum of an electron in an atom is also quantised. The angular momentum of the electron can have certain specific values and not any value of their own. The only permissible values of angular momentum are given by nh mvr = 2π where, h is Planck’s constant; m is mass of electron; v is velocity of revolving electron; r is radius of orbit and n is any integer. In other words, only those orbits are permitted in which the angular momentum of the electron is a whole number multiple of h/2π. Thus, only certain fixed orbits are allowed. Success of Bohr Model of Atom: 1. Concept of Orbits: The stationary states of electron are numbered as n = 1, 2, 3, … . These integral numbers are known as principal quantum number. 2. Stability of Atom: According to Bohr theory, an electron cannot loose energy as long as it stays in a particular orbit. Thus, the point of losing energy continuously and falling into the nucleus does not arise. Therefore, Bohr theory was able to explain the stability of the atom. 3. Bohr Radii of Orbit: The radii (r) of the stationary state is given by the expression v=

r=

a0 n 2 0.529 × n 2 = Å Z Z

For hydrogen atom, Z = 1 If, n = 1; r = (1)2 × 0.529 Å n = 2; r = (2)2 × 0.529 Å n = 3; r = (3)2 × 0.529 Å

…[Bohr radius]

Derivation of Bohr Radii of Orbit For an electron to remain in its orbit, the electrostatic attraction between the electron and the nucleus (which tends to pull the electron towards the

1.6 Bohr Model of Atom

31

nucleus) must be equal to the centrifugal force (which tends to throw the electron away from the nucleus). For an electron of mass ‘m’, moving with a velocity ‘v’ in an orbit of radius ‘r’. Centrifugal force =

mv 2 r

If the charge on the electron is ‘e’, the number of charges on the nucleus is Z, and the permittivity of a vacuum ε0. Coloumbic attractive force =

Ze 2 4πε0 r 2

mv 2 Ze 2 = r 4πε0 r 2 Hence,

v2 =

Ze 2 4πε0 mr

…(1)

To an electron, only those orbits are perimitted in which the angular momentum of the electron is a whole number multiple of h/2π. Thus, the only permissible values of angular momentum are given by mvr = v= v2

nh 2π nh 2πmr

n2h2 = 4π2 m 2 r 2

…(2)

Equating eq. (1) and eq. (2) Ze 2 n2h2 = 4πε0 mr 4π2 m 2 r 2 r=

Put, Thus,

ε0 n 2 h 2 Ze 2 πm

ε0 h 2 = a0 = 0.529 Å e 2 πm r=

a0 n 2 Z

…(3)

in eq. (3)

32

Chapter 1

Classical Mechanics

4. Bohr Energy of Electron in an Orbit: The energy of the electron in the stationary state is given by the expression En = −

2π2 me 4 Z 2 K 2 n2h2

Substituting the value of m (mass of electron), e (charge on electron), h (Planck’s constant) and K (electrostatic force constant), we get:  1312 × 103  2  1312  En = −  2  Z 2 kJ/mole = −   Z J/mole n2  n     1.312 × 106  2 = −   Z J/mole n2   or,

 Z2  Z2 −18 En = − RH  2  J/atom = −2.18 × 10 × 2 J/atom n n  (RH = 2.18 × 10–18 J/atom)

or

 13.6  En = −  2  Z 2 eV/atom  n  (1 eV = 1.6022 ×10–19 J) For hydrogen atom, Z = 1, thus values of various energy levels in hydrogen atom can be given as: 1 E1 = −2.18 × 10−18  2  = –2.18 × 10–18 J 1   1  E2 = −2.18 × 10−18  2  = –0.545 × 10–18 J 2  −18  1  E∞ = −2.18 × 10  2  = 0 ∞ 

Thus, the first energy level (n = 1) which is close to the nucleus has lowest energy. As we move outwards starting from the first energy level, the energy of the levels increases. Thus, E1 < E2 < E3 … < E∞. When, n = ∞, E∞ = 0; this shows that at infinite distance from the nucleus, there is no force of attraction on the electron by the nucleus. Hence, at infinite distance, the electron is considered as a free electron at rest, where its energy is taken as zero. As the electron moves towards the nucleus, it starts experiencing a force of attraction by the nucleus, and starts radiating some amount of energy. Further, as the electron goes more and more close to the nucleus, the attraction increases and more energy is radiated out. Since its value was already zero, hence the energy

1.6 Bohr Model of Atom

33

of the electron becomes less and less i.e., becomes more negative. Thus, the energy of the electron decreases as we move from the outer to the inner levels.

Derivation of Bohr Energy of Electron in An orbit The atom will radiate energy only when the electron jumps from one orbit to another. The kinetic energy of an electron is given as:

1 2

E = − mv 2 Substituting the values of ‘v2’ from eq. (1) [refer derivation of Bohr radii of orbit] v2 =

Ze 2 4πε0 mr

1 Ze 2 Ze 2 m× = − 8πε0 r 2 4πε0 mr Substituting the value of ‘r’ from eq. (3) [refer derivation of Bohr radii of orbit] E= −

E= − E= −

Ze 2 Ze 2 πm × 8 πε0 ε0 n 2 h 2 Z 2e4 m 8ε02 n 2 h 2

Multiply and divide by 2π2.

Put,

E= −

Z 2 e 4 m 2π2 × 8ε02 n 2 h 2 2π2

E= −

2π2 Z 2e 4 m (16π2ε02 )n 2 h 2

E= −

1 2π2 Z 2e 4 m × (4πε0 ) 2 n2h2

1 =K 4πε0 2 E = −K ×

E= −

2π2 Z 2e 4 m n2h2

2π2 me 4 Z 2 K 2 n2h2

34

Chapter 1

Classical Mechanics

5. Bohr Velocity of Electron in an Orbit: The velocity of electron in the stationary state is given by the expression vn = v0 ×

Z n

where, v0 = 2.18 × 108 cms–1, is the velocity of the electron in the first orbit of hydrogen atom. Derivation of Bohr Velocity of Electron in an Orbit: We know that, nh mvr = 2π nh \ v= 2π mr Putting the expression of ‘r’ i.e., eq. (3) [refer derivation of Bohr radii of orbit] ε n2 h2 r= 0 2 Ze πm v=

nh Ze 2 π m × 2 π m ε0 n 2 h 2

Multiply and divide by 2π. v=

1 Ze 2 2π × × 2ε0 nh 2π

1 2π Ze 2 × v= nh 4πε0

1 =K 4πε0 2π KZe 2 v= nh Except n and Z, all are constants. Putting the value of constants, the above expression becomes Z v = (2.18 × 108) × n Z v = v0 × n where, v0 = 2.18 × 108 cms–1. For the electron in first orbit of hydrogen atom, n = 1 and Z = 1 v = v0 Thus, v0 = 2.18 × 108 is the velocity of the electron in the first orbit of hydrogen atom.

Put,

1.6 Bohr Model of Atom

35

6. Quantitative Explanation of Line Spectrum of Hydrogen Atom: Bohr theory provides a satisfactory quantitative explanation for the line spectrum of hydrogen. According to Bohr theory, as long as the electron remains in a particular orbit, it does not loose or gain energy. But when energy from some external source is absorbed by the electron, it may jump from the orbit of lower energy level to the orbit of higher energy level. Whereas, the energy is emitted if the electron moves from higher energy level to lower energy level. The frequency (v) of the quanta of radiation absorbed or emitted depends upon the energy difference (DE) of the two energy levels concerned. Corresponding to the frequency [or wavelength (λ) or wave number ( v ) ] of each quanta emitted, there appears a line in the emission spectra of hydrogen atom. The frequencies (or wavelengths or wave number) of the spectral lines calculated by Bohr was found to be in good agreement with the experimental values. Thus, Bohr theory shows a remarkable success in explaining the line spectrum of hydrogen atom and hydrogen like atoms (He+, Li2+, Be3+ etc.) Let ni is the initial orbit and nf is the final orbit, wherein the electron jumps. The energy gap between the two orbits is: ∆E = Ef – Ei  R H DE =  − 2 n f 

  R  −  − H2   ni 

  

1 1  DE = RH  2 − 2   ni n f  where, Rydberg constant, RH = 2.18 × 10–18 Joules  1  −18 1 DE = 2.18 × 10  2 − 2   ni n f  v=

∆E 2.18 × 10−18 = h 6.6 × 10−34

1 1   2 − 2  s–1  ni n f 

1 1  v = 3.29 × 1015  2 − 2  s–1  ni n f  v =

v 3.29 × 1015 1 = = λ 3 × 108 c

1 1   2 − 2  m–1  ni n f 

36

Chapter 1

Classical Mechanics

1 1  v = 1.09677 × 107  2 − 2  m–1  ni n f  7 v = 1.09677 × 10 100

1 1   2 − 2  cm–1  ni n f 

1 1  v = 109677  2 − 2  cm–1  ni n f  1 1  v = RH  2 − 2  cm–1  ni n f  where, Rydberg constant, RH = 109677 cm–1. Thus, the wave number of the spectral line calculated by Bohr is in good agreement with experimental value of wave number seen in hydrogen spectrum earlier. There are two cases: Case I: Absorption spectrum When an electron jumps from lower orbit (ni) to some higher orbit (nf), the energy gap between the orbits is given as: 1 1  DE = RH  2 − 2   ni n f 

Ef

nf E = (+ve)

Ei

ni

1 1  Since, nf > ni, the term in the brackets,  2 − 2  would be positive and  ni n f  thus DE would be positive. This indicates that energy is absorbed by the system. Case II: Emission spectrum When an electron jumps from higher orbit (ni) to some lower orbit (nf), the energy gap between the orbits is given as: 1 1  ∆E = RH  2 − 2   ni n f 

Ei

ni E = (–ve)

Ef

nf

1.6 Bohr Model of Atom

37

1 1  Since, nf < ni, the term in the brackets,  2 − 2  would be negative and  ni n f  thus DE would be negative. This indicates energy is emitted out of the system. 7. Qualitative Explanation of Line Spectrum of Hydrogen Atom: When a sample of hydrogen gas is heated to a high temperature or an electric discharge is passed, the hydrogen molecules split into hydrogen atoms. Since in a sample of hydrogen, there are large number of atoms, each containing one electron, thus the electron in different atoms absorb different amount of energies and are excited to different higher energy levels. Then from excited state, they return to some lower energy level in one or more jumps. Since large number of different types of downward transitions takes place simultaneously in a sample of hydrogen, therefore, large number of lines are obtained in emission spectrum of hydrogen (Fig. 1.11). Thus, Bohr accounts for the existence of large number of lines in atomic spectra of hydrogen atom, although hydrogen atom contains only one electron. The brightness or intensity of spectral lines depends upon the number of photons of same frequency (or wavelength) emitted (in case of emission spectrum) or absorbed (in case of absorption spectrum). n=8 n=7 n=6 n=5

Pfund series (Far Infrared)

n=4

Brackett series (Infrared)

n=3 n=2

n=1

Paschen series (Infrared) Balmer series (Visible) Lyman series (Ultraviolet)

Fig. 1.11. Electronic transitions producing different series in the hydrogen spectrum.

Limitations of Bohr Model of Atom Bohr model of atom was an improvement over Rutherford nuclear model of atom. Although Bohr model accounts for the stability of atom and line spectra of hydrogen atom and hydrogen like atoms, but could not explain the following:

38

Chapter 1

Classical Mechanics

1. Spectra of Multielectron Atoms: Bohr theory was successful in explaining the line spectra of one electron atom (i.e., hydrogen and hydrogen like atoms) but it failed to explain the spectra of atoms having more than electron. Like: Helium atoms with two electrons. 2. Finer Details of the Hydrogen Spectrum: When spectroscopic techniques with better resolving powers were used, each line in hydrogen spectrum was split into a number of closely spaced lines (doublet i.e., two closely spaces lines or triplet i.e. three closely spaced lines). This fine structure could not be explained by Bohr model of atom. 3. Splitting of Spectral Lines in the Presence of Magnetic Field and Electric field: If the source emitting the radiation and producing line spectra is placed in a magnetic field or electric field, then each fine spectral line further splits into a number of lines. The splitting of spectral lines in the presence of magnetic field is called Zeeman effect. Whereas, the splitting of fine spectral lines in the presence of electric field is called Stark effect. 4. Three Dimensional Model of Atom: The Bohr model looks like a twodimensional model of the atom because it restricts the motion of the electron to a circular orbit in a two dimensional plane. But in reality, the Bohr model is a one-dimensional model (i.e., flat model of atom), because a circle can be defined by specifying only one dimension; i.e., radius (r). As a result only one co-ordinate (r) is needed to describe the orbits in the Bohr model. Unfortunately, electrons are not particles that can be restricted to one-dimensional circular orbit. They act to some extent as waves and therefore exist in three-dimension space. Thus, atom in reality is threedimensional. 5. Ability of atoms to form molecules by chemical bonds and shapes of molecules.

Problems based on bohr model of atom 1. Energy of the electron in hydrogen atom in nth shell is given by the expression 1.312 106 En = − Z 2 J/mole n2 and for hydrogen atom, Z = 1 (i) Calculate the amount of energy required to promote electron from the first energy level to the third energy level. nf Ans. 3 energy absorbed ni

1

1.6 Bohr Model of Atom

39

DE = Ef – Ei DE = E3 – E1  1.312 × 106   1.312 × 106  DE =  −  − −  (3) 2 (1) 2     1  DE = −1.312 × 106  − 1 9  6 8 DE = +1.312 × 10 ×   = +1.166 × 106 J/mol 9 The positive value indicates that energy is absorbed by the system. (ii) What will be the ionization energy (I.E.) of hydrogen atom? Ans. Ionization energy (I.E.) is the amount of energy required to remove the electron from the ground state of neutral gaseous atom. nf

 energy absorbed 1

ni

I.E. = DE = E∞ – E1  1.312 × 106   1.312 × 106  I.E. =  −  − −  (∞ ) 2 (1) 2     I.E. = 0 + (1.312 × 106) I.E. = +1.312 × 106 J/mole The positive value indicates that energy is absorbed by the system. Energy present in the ground state (n = 1) is, E1 = –1.312 × 106 J/mole. In other words, the electron is revolving in the ground state with E1 = –1.312 × 106 J/mole. But the energy required to remove an electron from this shell (n = 1) is +1.312 × 106 J/mole. Thus, the energy required to remove an electron from a particular energy level is equal in magnitude with the energy with which it is revolving in that particular energy level. 2. Calculate using the Bohr theory of hydrogen atom, the ionization potential of H and He+ atom. Ans. According to Bohr theory of hydrogen atom, energy of the electron in hydrogen atom in nth shell is given by the expression

En = −

1.312 × 106 × Z 2 J/mole 2 n

40

Chapter 1

Classical Mechanics

(i) Ionization potential of H-atom: nf

 energy absorbed 1

ni

For hydrogen atom, Z = 1 I.E. = DE = Ef – Ei = E∞ – E1  1.312 × 106 I.E =  − (∞ ) 2 

  1.312 × 106   − −  (1) 2   

= 0 + (1.312 × 106) I.E. = +1.312 × 106 J/mole (ii) Ionization potential of He+ ion: nf

 energy absorbed 1

ni

For helium atom, Z = 2 I.E. = DE = Ef – Ei = E∞ – E1  −1.312 106   −1.312 × 106  2 × ( ) –  2 I.E. × =   × ( 2) 2  2 2 (∞ ) (1)     8 I.E. = +5.248 × 10 J/mole. 3. Calculate the first excitation energy of the electron in the hydrogen atom. nf Ans. 2 energy absorbed 1

ni

For hydrogen atom, Z = 1 DE = Ef – Ei = E2 – Ei

1.6 Bohr Model of Atom

41

 1.312 × 106   1.312 × 106  = −  − −  ( 2) 2 (1) 2      6 1 = −1.312 × 10  − 1 4    3 = +1.312 × 106 ×   = 0.984 × 106 J/mole.  4

4. Calculate the frequency and wavelength of the radiation emitted when an electron in the hydrogen atom jumps from the third orbit (n = 3) to the ground state (n = 1). In which region of the electromagnetic spectrum will this line lie. Ans. [Method I: See through problems based on line emission spectra of hydrogen and hydrogen like atoms] Method II: Using Bohr theory of hydrogen atom: ni

3 energy emitted

nf

1

(Ground state)

For hydrogen atom, Z = 1 DE = Ef – Ei = E1 – E3  1.312 × 106   1.312 × 106  = −  − −  (1) 2 (3) 2     1 6  = −1.312 × 10 1 −   9 5 DE = –11.66 × 10 J/mole Here, the negative sign shows that energy is emitted out of the system 11.66 × 105 J/mole DE = + 6.023 × 1023 mole Now, (a) Frequency,

DE = 1.93 × 10–18 J DE = hv v=

1.93 × 10 −18 J ∆E = h 6.626 × 10 −34 Js

v = 2.9 × 1015 s–1

42

Chapter 1

(b)

c = vλ

Wavelength,

λ=

Classical Mechanics

c 3 × 108 ms −1 = v 2.9 × 1015 s −1

λ = 1.03 × 10–7 m. 5. Calculate the radius of the first orbit (n = 1) of H and He+? Ans. According to Bohr theory, radius of nth orbit is given as 0.529 × n 2 Å rn = Z 0.529 × (1) 2 = 0.529 Å (i) For H-atom, Z = 1; r1 = 1 0.529 × (1) 2 (ii) For He+ ion; Z = 2; r1 = = 0.2645 Å. 2 6. Calculate the velocity of the electron in the first orbit of hydrogen atom. Ans. According to Bohr theory, velocity of electron in an orbit is given by nh 2π nh v= 2πmr

mvr =

n= h= m= r1 = r1 =

1 (first orbit) 6.626 × 10–34 (Planck’s constant) mass of electron = 9.11 × 10–31 kg radius of first orbit 0.529 × (1)2 = 0.529 Å = 0.529 × 10–10 m

v=

(1) × (6.626 × 10−34 ) 2 × (3.14) × (9.11 × 10−31 ) × (0.529 × 10−10 )

v = 2.18 × 106 ms–1. 7. Calculate the velocity in third orbit of He+ ion. The velocity of electron in first Bohr’s orbit is 2.18 × 106 ms–1. Ans. For He+ ion, Z = 2 Here, n = 3 and v0 = 2.18 × 106 ms–1 Z v = v0 × n 2 v = (2.18 × 106) × 3 v = 1.4 × 106 ms–1

1.7 Sommerfeld theory or Bohr-Sommerfeld theory

1.7

43

SOMMERFELD THEORY OR BOHR-SOMMERFELD THEORY

With Arnold Sommerfeld’s extension of the Bohr theory in 1915–1916, it was turned into a powerful tool of atomic research and adopted and further developed by German physicists in particular. Bohr model of atom was able to calculate the radii and energies of the stationary orbits around the nucleus in an atom and the calculated values of Bohr were found to be in good agreement with the experimental values. But Bohr model was not able to explain the very fine spectral lines of hydrogen spectrum. When the fine spectral lines of hydrogen spectrum was examined under a high resolution spectroscope, it was found that fine spectral line was composed of several lines close together i.e., there is further splitting of these finer lines into very fine spectral lines. The explanation of several lines originating in hydrogen spectrum was given by Arnold Sommerfeld who extended the Bohr model of circular orbits by introducing the concept of elliptical orbits. According to Sommerfeld, the stationary orbits around the nucleus in the atom are not circular but elliptical in shape. It is due to the influence of the centrally located nucleus. For the orbit closest to the nucleus. i.e., for the principal quantum number n = 1, there exists a circular orbit. For next orbit i.e., for the principal quantum number n = 2, both circular and elliptical orbits are possible. When the electron revolves in elliptical path with nucleus at one of its foci, there will be a major and minor axis of the path. To define elliptical orbit another quantum number ‘k’ was required. The shape of the ellipse is defined by the ratio of the lengths of the major and minor axes, major axis n = minor axis k where, k has integral values upto n and is known as azimuthal or subsidiary quantum number. For n = 2, n/k may be 2/2 for circular orbit and 2/1 for elliptical orbit (Q, n = 2, k = 1, 2). For n = 3, k can have three values (1, 2, 3) so that n/k becomes 3/3 for circular orbit, 3/2 for elliptical orbit and 3/1 for further narrow elliptical orbit as shown in Fig. 1.12. k=3 k=2 k=1

Fig. 1.12. Bohr-Sommerfeld orbits when n = 3.

44

Chapter 1

Classical Mechanics

With increase in value of k, the path becomes more and more elliptical and eccentric. When k = n, the path becomes circular. The original quantum number ‘k’ was later replaced by ‘l’, new quantum number where l = k – 1 so that; n 1 2 3 4

l 0 0 or 1 0 or 1 or 2 0 or 1 or 2 or 3

The new and more general Bohr-Sommerfeld theory described the atom in terms of two quantum numbers (n and k), while Bohr had originally used only one quantum number (n). With this extension the theory provided an explanation of the fine structure of the hydrogen spectrum and the Zeeman effect. (a) Bohr was not able to explain the reason for the fine spectral lines visible by high revolving power spectroscopes but Sommerfeld explained the reason for the same. He said that the energy of the stationary orbit depends not only on n but on k to some extent as well. So when a transition of electron from a higher level to a lower level occurs, it would be different from what proposed by Bohr as there may be more than one values of k. In this way Sommerfeld was able to explain the reason behind those fine spectral lines. Even the frequencies of some of those fine spectral lines came to be in well agreement with the frequencies by Sommerfeld. (b) Further Zeeman effect was also explained by Sommerfeld. The Zeeman effect was experienced under the strong external magnetic field in which further splitting of fine spectral lines was observed. The elliptical orbits can only take up certain orientations in the presence of the strong external magnetic field rather than precessing freely. Each of these orientations are associated with the third quantum number ‘m’, which can possess l, (l – 1), … 0 … (–l + 1), –l values. This is how the single line splits into (2l + 1) number of lines under the force of strong magnetic field. rrr

2 Towards Quantum Mechanics 2.1

REASONS FOR THE FAILURE OF CLASSICAL MODEL OF ATOM OR BOHR MODEL OF ATOM

The results of Bohr model can be used to explain the motion of macroscopic objects or big objects (like orbiting planets and sun, falling stone etc.). These macroscopic objects have essentially a particle nature and Newtonian laws of motion can be applied on them. But it fails when applied to explain the motion of microscopic objects or small objects (like electrons, atoms, molecules etc.). This is because: 1. According to Bohr model, an orbit is a well-defined path and this path can completely be defined only if both the position and velocity of the electron are known exactly at the same time. But this is not possible according to the Heisenberg’s uncertainty principle. Thus, Bohr model contradicts the Heisenberg’s uncertainty principle. 2. According to Bohr model, an electron is regarded as a charged particle moving in well-defined circular orbits concentric about the nucleus. The wave character of the electron is not considered. Thus, Bohr model ignores the de-Broglie dual nature of matter.

2.2

DEVELOPMENTS LEADING MECHANICAL MODEL OF ATOM

TO

QUANTUM

In view of the shortcomings of Bohr model of atom, efforts were made to develop a new model of atom which could overcome the limitations of Bohr model. The development of the new model was mainly based on the following two concepts that had been put forward:

46

Chapter 2 Towards Quantum Mechanics

1. de-Broglie concept of dual nature of matter. 2. Heisenberg uncertainty principle.

2.3

de-Broglie’s DUAL NATURE OF MATTER

In 1905, Einstein suggested that radiation has dual nature, i.e. wave nature as well as particle nature. Later in 1924, a French physicist, Louis-Victor deBroglie gave theory of wave particle duality of matter. He suggested that just as radiation possesses both wave and particle nature, all macroscopic as well as microscopic material objects (like electron, proton, atom, molecule, a piece of chalk, a piece of stone etc.) also possess dual nature i.e., wave and particle nature. The wave associated with a material particle is called matter waves or de-Broglie waves. • de-Broglie equation: He calculated the wavelength of the wave associated with a material particle of mass (m) moving with a velocity (v). The deBroglie relationship is λ=

h h = p mv

where, p = momentum of the particle λ = de-Broglie wavelength. h = Planck’s constant. • Derivation of de-Broglie relationship: Consider a photon of light having energy E. If photon is assumed to have particle nature, its energy is given by Planck’s quantum theory as E = hv …(2.1) If photon is assumed to have wave nature, its energy is given by Einstein equation as …(2.2) E = mc2 where, h = Planck’s constant c = velocity of light (3 × 108 ms–1) v = frequency of the wave m = mass of photon Equating equations (2.1) and (2.2) hv = mc2 c Now, v= λ hc 2 = mc λ h λ= mc

2.3 de-Broglie’s Dual Nature of Matter

47

This equation is applicable to any material particle. This is done by replacing the mass of the photon by mass of the material particle and the velocity ‘c’ of the photon by velocity ‘v’ of the material particle. Thus, for any material particle like electron, proton etc., we have h h or λ = λ= (de-Broglie equation) p mv 1 i.e., de-Broglie Thus, from de-Broglie equation, we conclude that, λ ∝ p wavelength of wave associated with material particle is inversely proportional to its momentum. Therefore, de-Broglie equation relates the particle character with wave character of material particles. Electromagnetic waves Matter waves/de-Broglie waves 1. When electrically charged particles 1. These waves are not associated with move under acceleration, alternating electric and magnetic fields. electric and magnetic fields are produced which are perpendicular to each other. These fields are produced and transmitted in the form of electromagnetic waves. 2. They can pass through vacuum. 2. They cannot pass through vacuum i.e., they require medium for propagation. 3. They travel with the speed of light. 3. They do not travel with the same speed as that of light. 4. The wavelength of the electromagnetic 4. The wavelength of the de-Broglie c h wave is given by λ = . wave is given by λ = . ν mv 5. They are emitted from a particular 5. They are not emitted from a source source in the form of radiation. but attached to a moving particle.

• Significance of de-Broglie equation: According to de-Broglie, every moving object whether macroscopic or microscopic have a wave character. The wavelength associated with ordinary macroscopic objects (like running car, cricket ball etc.) are so small (because of their large masses) that their wave properties cannot be detected i.e., wavelength associated with large objects is not significant. Whereas, the wavelength associated with microscopic objects, (like electrons and other subatomic particles) is large enough (because of their small masses) and thus can be detected experimentally. Since in everyday life, we come across macroscopic objects, therefore de-Broglie relationship has no significance in everyday life.

48

Chapter 2 Towards Quantum Mechanics

• Derivation of Bohr’s postulate of angular momentum from de-Broglie equation: de-Broglie applied his theory of wave particle duality to the Bohr model to explain why only certain orbits are allowed for the electron. He argued that only certain orbits allow the electron to satisfy both its particle and wave properties at the same time because only certain orbits have a circumference (2πr) that is an integral multiple of the wavelength (nλ) of the electron. In other words, according to de-Broglie concept, electron is not only a particle moving in a circular path but is instead a standing wave* (no energy radiating motion) extending around the nucleus in the circular path. In order that the wave may be completely in phase (Fig. 2.1), the circumference of the orbit must be equal to an integral multiple of wavelength (λ). i.e. 2πr = nλ …(2.3) where, r is the radius of orbit and, n is an integer h (de-Broglie equation) But, λ= mv Substituting the value of λ in eq. (2.3)  h  2πr = n    mv  nh 2π which is Bohr’s postulate of angular momentum. mvr =

(a)

(b)

Fig. 2.1. (a) Wave in phase and (b) Wave out of phase.

*Stationary or Standing Waves: If a string is stretched between two fixed points, then a wave is allowed to travel along the string in one direction such that when the wave reaches the other end, it is reflected in the opposite direction with the same wavelength, speed and amplitude. The string is also divided into a number of vibrating segments, separated by points of zero amplitude. Such a vibrating motion is called a stationary or a standing wave. The points of zero amplitude (points a) are called nodes.

2.3 de-Broglie’s Dual Nature of Matter

49

Node a

a

a

a

A

B

Fig. 2.2. Stationary or standing waves.

• Experimental verification of wave nature of electrons: In 1927, C.J. Davisson and Germer gave experimental verification for the revolutionary postulate given by de-Broglie. It was further verified by G.P. Thomson. 1. C.J. Davisson and Germer’s Experiment: In 1927, Davisson and Germer observed that when a beam of electrons (or neutron, or proton or hydorgen atoms etc.) is allowed to fall on the surface of nickel crystal and the reflected rays are received on a photographic plate, a diffraction pattern (consisting of a number of concentric rings) is obtained (Fig. 2.3). The diffraction rings formed by the electrons were similar to those formed by X-rays. Diffraction pattern

s

ted ray

Reflec Incident Beam of high speed electrons

Nickel crystal

Photographic plate Fig. 2.3. Diffraction of electron beam by nickel crystal.

Since, X-rays are electromagnetic waves i.e., they have wave character, therefore, electrons must also have wave character. Further, the wavelength determined from the diffraction pattern is in good agreement with those calculated from de-Broglie equation.

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Chapter 2 Towards Quantum Mechanics

2. Thomson’s Experiment: In 1928, G.P. Thomson observed that when an electron beam is allowed to fall on a thin foil of gold in place of nickel crystal and reflected rays are received on a photographic plate, a diffraction pattern (consisting of a number of concentric rings) is obtained (Fig. 2.4). This again confirmed the wave nature of electrons. Diffraction pattern

Incident beam of high speed electron

ed lect Ref ays r

Thin gold foil Photographic plate Fig. 2.4. Diffraction of electron beam by thin gold foil.

• Thus, de-Broglie’s prediction was confirmed experimentally when it was found that an electron beam undergoes diffraction, a phenomena characteristic of waves. This is used in making an electron microscope (used to measure the size of objects) which is based on the wave like behaviour of electrons just as an ordinary microscope utilises the wave nature of light. • Experimental verification of particle nature of electrons: The particle nature of electron can be justified on the basis of following observations. 1. When an electron strikes a screen coated with zinc sulphide, a spot of light known as scintillation is produced. One electron produce only one scintillation. Since a scintillation is localised on the screen, thus the striking electron which produces scintillation must also be localised. This suggests that electron possess particle character because localised character is possessed by the particles. Hence, electron has the particle nature. 2. The particle nature of electron can also be shown by J.J. Thomson’s experiment used for the determination of charge to mass ratio of electron and Milliken oil drop Experiment used for the determination of charge on electron.

2.3 de-Broglie’s Dual Nature of Matter

51

Thus, we conclude that everything (light and matter) in nature have the properties of both particle as well as wave. In other words, dual character is applicable to everything.

Problems based on de-broglie eQuation 1. Calculate the wavelength of a body of mass 2 mg moving with a velocity of 20 ms–1 (h = 6.626 × 10–34 kgm2s–1). Ans. m = 2 mg = 2 × 10–3 g = 2 × 10–6 kg v = 20 ms–1 λ=

h 6.626 × 10−34 kg m 2s −1 = mv (2 × 10−6 ) kg × (20) ms −1

λ = 1.65 × 10–29 m. 2. Calculate the momentum of a moving particle which has de-Broglie wavelength of 300 pm. Ans. λ = 300 pm = 300 × 10–12 m h = 6.626 × 10–34 kg. m2. s–1 h λ= p h 6.626 × 10−34 kg. m 2 . s −1 p= = λ 300 × 10−12 m p = 2.2 × 10–24 kg ms–1. 3. The mass of an electron is 9.11 × 10–31 kg and its kinetic energy is 5 × 10–25 J. Calculate its de-Broglie wavelength. 1 2 Ans. K.E. = mv 2 1 −31 2 5 × 10–25 = × (9.11 × 10 ) × v 2 v=

Now,

2 × (5 × 10−25 ) 9.11 × 10−31

v = 1.048 × 103 ms–1 h λ= mv 6.626 × 10 −34 kg m 2 s −1 λ= (9.11 × 10 −31 kg ) × (1.048 × 103 ms −1 ) λ = 6.94 × 107 m.

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Chapter 2 Towards Quantum Mechanics

4. Calculate the mass of the photon with which a wave is associated of wavelength 2.5 Å. Ans. λ = 2.5 Å = 2.5 × 10–10 m v = Velocity of photon = Velocity of light = 3 × 108 ms–1 h λ= mv 6.626 × 10 −34 kg m 2 s −1 h = m= λv (2.5 × 10 −10 m) × (3 × 108 ms −1 ) m = 0.88 × 10–32 kg. 5. An electron beam is accelerated by a potential difference of 10,000 eV. What is the wavelength of the wave associated with electron beam. Ans. m = mass of electron = 9.11 × 10–31 kg K.E. = 1000 eV K.E. = 10,000 × 1.6 × 10–19 J (Since, 1 eV = 1.6 × 10–19 J) K.E. = 1.6 × 10–15 J 1 2 Now, K.E. = mv 2 1 −31 2 1.6 × 10–15 J = × (9.11 × 10 kg ) × v 2 v= Now,

2.4

2 × (1.6 × 10 −15 J ) = 5.9 × 107 ms–1 (9.11 × 10 −31 kg )

h 6.626 × 10−34 kg m 2 s −1 = λ= mv (9.11 × 10−31 kg ) × (5.9 × 107 ms −1 ) λ = 0.123 × 10–10 m λ = 0.123 Å.

HEISENBERG’S UNCERTAINTY PRINCIPLE

Werner Heisenberg stated uncertainty principle which is the consequence of dual nature of matter and radiation. All moving objects that we see in daily life i.e., macroscopic objects (like moving car, ball thrown in air etc.) have definite paths or trajectories along which they move. The definite trajectories indicates that their position and velocity can be measured accurately at any moment of time. But, such accurate measurements of position and velocity cannot be done for sub-atomic particles i.e. microscopic objects (like electron, proton, atom etc.). There are always some uncertainties

2.4 Heisenberg’s Uncertainty Principle

53

in measurement of position and velocity for small moving particles. Thus, for such subatomic particles, the existence of definite paths or trajectories is ruled out. Thus in microscopic world, it is found that however refined our instrument there is a fundamental limitation to the accuracy with which the position and velocity of microscopic particle can be known simultaneously. In 1927, a German physicist, Werner Heisenberg gave a principle about the uncertainties in simultaneous measurements of position and momentum of subatomic particles, which is known as Heisenberg’s uncertainty principle. This principle states that: It is impossible to determine simultaneously the exact position and momentum (or velocity) of small moving particles with accuracy or certainty. If one quantity is known then the determination of the other quantity will become impossible Mathematically, h (2.4) Dx · Dpx ≥ 4π h Dx · mDvx ≥ 4π where, Dx = uncertainty in position Dpx = uncertainty in momentum in x-direction Dvx = uncertainty in in velocity in x-direction m = mass of the particle h = Planck’s constant Also,

Dx ∝

1 ∆v x

This implies that the position and velocity of a particle cannot be determined simultaneously with accuracy. If the position of the particle is known with high degree of accuracy (Dx is small), then velocity of particle will be uncertain (Dvx is large) and vice versa. The uncertainty principle also applies for the same axis of position and momentum. Like if Dx is along the y-axis, then Dp must also be along the y-axis. Uncertainty in Time and Energy: Similar to uncertainty in position there is another principle uncertainty which limits the accuracy in the measurement of time i.e., if DE is the energy uncertainty in time Dt then we have an expression similar to equation (2.4). DE × Dt ≥ h/4π ...(2.5) • Justification of Heisenberg Uncertainty Principle: To understand this principle, let us first know how the position of an object is determined. A light phenomena is used. When a beam of visible light falls on an object, the photons of light gets scattered at the object and the reflected

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photons enter the retina of our eyes which produces the sensation of sight. If the object is of sufficient size, its position and velocity will not be changed by the energy of the photons of light falling on it. As such it will be possible to know both the position and the velocity of the object at the same time. But in case of microscopic objects (such as an electron), an entirely different situation exists due to very small size of the particle to be located. According to principle of optics, the accuracy with which the position of a particle can be located depends upon the wavelength (λ) of light used and the position cannot be located more accurately than ± λ. Therefore, due to the very small size of the electron its position can be determined with sufficient accuracy only by choosing light of very small wavelength. But shorter wavelength of light means that its photons would possess higher energy. When such a high energy photon falls on the electron, part of the energy of the photon falls on the electron and part of the energy of the photon would be transferred to the electron. As a result, the electron would undergo considerable change in its velocity (therefore, momentum) [Fig. 2.5]. Thus we find that if a photon of smaller wavelength is used there will be large inaccuracy in the determination of momentum of the electron. On the other hand, if a photon of larger wavelength is used, there will be large uncertainty in the location of its position. Microscope

New path and velocity

e– Init i al

pa t

h

d te n In

ath ed p

Incident photon

Fig. 2.5. The momentum of electron changes on collision with photon

• Significance of Heisenberg Uncertainty Principle: 1. Although Heisenberg’s uncertainty principle holds good for all the objects but its effect is significant only for motion of microscopic objects and is negligible for that of macroscopic objects. Since in everyday life, we deal with macroscopic objects, thus Heisenberg uncertainty principle has no significance in everyday life. The reason why the uncertainty principle is of no importance in our daily life is that planck’s constant ‘h’ is so small (6.625 × 10–34 joule-seconds) that the uncertainties in position and momentum of even quiet small

2.4 Heisenberg’s Uncertainty Principle

55

(not microscopic objects) objects are far too small to be experimentally observed. For microscopic phenomena such as atomic processes, the displacements and momentum are such that the uncertainty relation is critically applicable. 2. It rules out the existence of definite path or trajectories of small microscopic particles. This is because position and velocity of a moving particle fix its trajectory. Since for microscopic object, it is impossible to determine simultaneously the position and velocity at any given instant with accuracy. Thus, it is not possible to talk of the trajectory of the small moving microscopic particles. 3. The principle describes the incompleteness of Bohr’s atomic theory. This is the reason that Bohr’s concept of fixed circular path with definite position and momentum of the electron (subatomic particle) have been replaced by stating that the electron has the probability of having a given position and momentum. This latter on, forms the basis of quantum or wave mechanical model of atom.

Problems based on heisenberg’s uncertainty eQuation In the numerical problems, the mathematical expression for the Heisenberg uncertainty principle is simply written as: h Dx · Dpx = 4π Dx · m(Dvx) =

h 4π

1. State and explain Heisenberg’s uncertainty principle. Ans. Heisenberg’s uncertainty principle states that it is impossible to determine simultaneously the exact position and momentum (or velocity) of small moving particle with accuracy or certainty. Mathematically, h Dx · Dpx ≥ 4π Dx · D(mvx) ≥

h 4π

Dx · mDvx ≥

h 4π

56

Chapter 2 Towards Quantum Mechanics

where,

Also, Dx ∝

Dx = uncertainty in position Dp = uncertainty in momentum in x-direction Dvx = uncertainty in velocity in x-direction. 1 , which implies that the position and velocity of a particle ∆v x

cannot be determined simultaneously with accuracy. If the position of the particle is known with high degree of accuracy (Dx is small), then velocity of particle will be uncertain (Dvx is large) and vice versa. 2. According to Heisenberg’s principle of uncertainty why is it impossible to measure simultaneously the exact position and momentum of a fast moving electron? Ans. [Refer the topic: “Justification of Heisenberg Uncertainty Principle”] 3. The orbital picture of atom accommodates Heisenberg’s uncertainty principle. Explain. Ans. The position and velocity of a moving particle fix its trajectory. According to Heisenberg’s uncertainty principle, for microscopic objects (like electron) it is impossible to determine simultaneously the position and velocity at any given instant with accuracy. Thus, it is not possible to talk of the trajectory of the small moving microscopic particles. This is the reason that Bohr’s concept of fixed circular path with definite position and momentum of the electron have been replaced by stating that the electron has the probability of having a given position and momentum. This latter on, forms the basis of orbital picture of atom or quantum mechanical model of atom. 4. The uncertainty in the momentum of a particle is 6 × 10–2 kg ms–1. Calculate the uncertainty in its position. Ans. Dp = 6 × 10–2 kg m s–1 h = 6.626 × 10–34 kg m2 s–1 Dx × Dp = Dx =

h 4π

h 1 6.626 × 10−34 kg m 2 s −1 × = 4π ∆p 4 × 3.14 × (6 × 10−2 kg m s −1 )

Dx = 8.76 × 10–34 m. 5. Calculate the uncertainty in the velocity of a cricket ball of mass 150 g, if the uncertainty in the position is of the order of ± 1 Å?

2.4 Heisenberg’s Uncertainty Principle

Ans.

57

Dx = 1 Å = 10–10 m m = 150 g = 150 × 10–3 kg h = 6.6 × 10–34 kg m2 s–1 h Dx × Dp = 4π Dx × mDv =

h 4π

Dv =

1 h × 4π m × ∆x

Dv =

(6.626 × 10 −34 kg m 2 s −1 ) 4 × 3.14 × (150 × 10 −3 kg ) × (10 −10 m)

Dv = 3.5 × 10–24 ms–1. 6. The uncertainty in the position and velocity of a particle are 10–2 m and 5.27 × 10–24 ms–1 respectively. Calculate the mass of the particle. Ans. Dx = 10–2 m Dv = 5.27 × 10–24 ms–1 m=? h = 6.626 × 10–34 kg m2 s–1 h 4π h Dx × mDv = 4π Dx × Dp =

m=

h 1 × 4π ∆x × ∆v

m=

(6.626 × 10−34 kg m 2 s −1 ) (4 × 3.14) ×

(10

−2

1 m) × (5.27 × 10−24 ms −1 )

m = 0.1 kg. rrr

3 Introduction to Quantum Mechanics 3.1

NECESSITY OF QUANTUM MECHANICS

In the 17th century, it was thought that motion of macroscopic and microscopic objects could be well-expressed by using the laws of classical mechanics. But towards the end of 19th century, experimental studies showed that classical mechanics failed when applied to microscopic objects. This was mainly because classical mechanics did not take into account the concept of de-Broglie’s dual nature of matter and Heisenberg’s uncertainty principle. Also, classical mechanics does not put any restrictions on the values of dynamical properties (likes position, momentum, energy etc.) calculated for microscopic object. According to classical mechanics, microscopic objects can have any value of dynamical properties, (exact, large or small is permissible). Also any number of properties can be calculated simultaneously for a system. However, experimental measurements for microscopic objects show that there exists a well-defined set of energy levels. Like, an electron moving around the nucleus of an atom have a discrete set of energy levels known as quantised energy levels. So to understand such behaviour of microscopic objects, it was necessary to postulate a new system of mechanics called Quantum mechanics. Quantum mechanics deals with the study of motion of matter at microscopic level. It is a new branch of science which takes into account the dual nature of matter. According to classical mechanics, it is possible to write down equations which tells exact values of position and momentum of a system at any time. The state of a given system (containing a number of particles) can be completely explained by knowing the position and momentum of every particle in the system

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at any particular time. Whereas, the state of a given system (containing a number of particles) in quantum mechanics is not given merely by its position and momentum but by something called “wavefunction”. By knowing the state of a system i.e., wavefunction, does not enable to predict the results of measurements with certainty, but rather gives a set of probabilities for the possible outcomes of any measurement. Thus, the word exactness in classical mechanics has been replaced by the word probability in quantum mechanics. When quantum mechanics is applied to macroscopic objects, the results are the same as those from the classical mechanics.

3.2

SCHRODINGER WAVE EQUATION

The two important principles laid the foundation of quantum mechanics, the de-Broglie dual nature of matter and Heisenberg uncertainty principle. Quantum mechanics, developed by Erwin Schrodinger in 1926, is based on the wave motion associated with the small moving objects i.e., microscopic objects. For the wave motion of the electron in the three dimensional space around the nucleus, Schrodinger put forward an equation, known after his name as Schrodinger wave equation, which is considered as the heart of quantum mechanics. This equation takes into account the dual nature of matter and is given as: ∂ 2 ψ ∂ 2 ψ ∂ 2 ψ 8π 2 m + 2 ( E − V )ψ = 0 + + ∂x 2 ∂y 2 ∂z 2 h where,

ψ = amplitude of the electron wave x, y, z = coordinates of the electron E = total energy of the electron V = potential energy of the electron m = mass of the electron h = Planck’s constant

∂ 2ψ = second derivative of ψ w.r.t. x. ∂x 2 The solution of Schrodinger wave equation for an electron give the values of E and ψ. The values of E represent the possible values of energy which the electron in the atom can occupy. The corresponding values of ψ are called wavefunctions. The wavefunction corresponding to an energy state contains all the information about an electron which is present in that state.

3.3 Derivation of Time Independent Schrodinger Wave Equation

61

Classical Mechanics: Deals with the study of behaviour of large objects under the action of forces. Quantum Mechanics: Deals with the study of behaviour of small objects under the action of forces. Quantum Chemistry: The quantum mechanics when applied to chemistry. Wave Mechanics: Deals with the mathematical treatment of the behaviour of microscopic object (like electron). Statistical Mechanics: Deals with the mathematical treatment of the behaviour of macroscopic object.

3.3

DERIVATION OF TIME INDEPENDENT SCHRODINGER WAVE EQUATION

Schrodinger equation was proposed by the Austrian phyicist Erwin Schrodinger in 1926. Schrodinger developed a mathematical wave equation to describe the wave motion of the electron in a atom. He considered that the nucleus is surrounded by a vibrating electron wave which may be mathematically compared with stationary or standing wave formed by Fig. 3.1. A standing wave. a vibrating string fixed between two points as shown in Fig. 3.1. The electron in an atom is under constraint imposed by the attraction of the nucleus. Therefore, there must be a wave equation which describe the wave motion of electron analogous to that used to describe a “standing or stationary” wave system (wave appearing in stretched string) whose solution in one dimension is given by the sine wave equation (3.1) 2πx ψ = A sin …(3.1) λ where, ψ is the amplitude of the wave of wavelength λ at a distance x from the origin and A is a constant. On double differentiation with respect to x, we get

∂ψ 2π 2πx ⋅ cos = A⋅ λ λ ∂x ∂ 2ψ ∂x

2

∂2ψ ∂x 2

= A⋅ =

2π 2π  − sin 2πx  ⋅ ⋅  λ λ  λ 

− 4π2  2πx  A sin  2  λ  λ 

 d  dx (sin x) = cos x   d  dx (cos x) = − sin x 

…(3.2)

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Chapter 3

Substituting ψ = A sin

Introduction to Quantum Mechanics

2πx from eq. (3.1) into eq. (3.2), we get λ ∂ 2ψ 4π2 − (ψ ) = λ2 ∂x 2

∂ 2 ψ 4π2 + 2 ψ =0 or ∂x 2 λ This is the equation for one dimensional standing wave which can be extended to describe motion of an electron in three dimensions by a second order partial differential equation (3.3) ∂ 2ψ ∂ 2ψ ∂ 2ψ 4π 2 + + + 2 ψ =0 …(3.3) ∂x 2 ∂y 2 ∂z 2 λ where, ψ is a function of cartesian coordinates x, y, z. This may be written more precisely as ∇ 2ψ +

4π 2 ψ =0 λ2

…(3.4)

 ∂2 ∂2 ∂2  where, ∇2 =  2 + 2 + 2  is the Laplacian operator (discussed latter in ∂y ∂z   ∂x the topic operators). Incorporating de-Broglie relationship into eq. (3.4) for an electron of wavelength λ, h λ= , mv 4π 2 m 2 v 2 ∇2ψ + ψ =0 we get …(3.5) h2 Considering the electron as a particle, the total energy (E) of the system is sum of kinetic (1/2mv2) and potential (V) energies, i.e., 1 E = mv 2 + V 2 2( E − V ) v2 = …(3.6) m In order to calculate values for the energy states, v2 from eq. (3.6) is substituted into eq. (3.5) to obtain a fundamental form of Schrodinger wave equation for a single particle in three dimensions. ∇ 2ψ +

8π 2 m ( E − V )ψ = 0 h2

…(3.7)

3.3 Derivation of Time Independent Schrodinger Wave Equation

∂ 2 ψ ∂ 2 ψ ∂ 2 ψ 8π 2 m + 2 (E − V )ψ = 0 + + ∂x 2 ∂y 2 ∂z 2 h

or

63

…(3.8)

If the electron is at rest at an infinite distance from the nucleus, its potential energy with respect to the nucleus at a distance r is given by V=

−Ze 2 r

where,

Z = atomic number +Ze = charge on the nucleus –e = charge on the electron. Consequently, the equation (3.8) for one electron atomic system in three dimensions may be written as, ∂ 2 ψ ∂ 2 ψ ∂ 2 ψ 8π 2 m  Ze 2  + + + + E ψ = 0  r  ∂x 2 ∂y 2 ∂z 2 h 2  Eq. (3.7) is known as Schrodinger’s fundamental wave equation with respect to space. The Schrodinger equation is valid for a particle executing simple harmonic motion (SHM). The permitted solutions of Schrodinger equation are known as wavefunction (ψ) which denotes the definite energy state. It is a second order differential equation. When it is solved, it gives infinite solutions of ψ, represented as ψ1, ψ2, ψ3, ψ4, … etc. with corresponding energies E1, E2, E3, E4, … etc. But all the solutions are not acceptable. Only those solutions are acceptable which have some special properties i.e., only well behaved ψ’s are acceptable. The eq. (3.7) may also be expressed as ∇ 2ψ +

8π2 mEψ 8π2 mV ψ =0 − h2 h2 −∇ 2 ψ +

Dividing throughout by −

8π2 mE 8π2 mV ψ ψ = h2 h2 8π 2 m we get: h2

h2 ∇ 2ψ + V ψ = Eψ 2 8π m

 h2  2  − 2 ∇ + V  ψ = Eψ  8π m 

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Introduction to Quantum Mechanics

Hˆ ψ = Eψ

…(3.9)

2

h [where, Hˆ = − 2 ∇ 2 + V is the Hamiltonian operator, (discussed latter in 8π m the topic operators] Total energy of electron = K.E. + P.E. Hˆ = Tˆ + Vˆ \

(Tˆ + Vˆ ) = Eψ

Problem based on schrodinger Wave eQuation Q.1. Write down Schrodinger wave equation and explain the terms involved in it. ∂ 2 ψ ∂ 2 ψ ∂ 2 ψ 8π 2 m + 2 ( E − V )ψ = 0 + + h ∂x 2 ∂y 2 ∂z 2 where, ψ = amplitude of the wave associated with the electron x, y, z = cartesian coordinates of the electron m = mass of the electron h = Planck’s constant E = total energy of the electron V = potential energy of the electron.

Ans.

3.4

PHYSICAL SIGNIFICANCE OF WAVEFUNCTION (Ψ) AND PROBABILITY DENSITY (Ψ2)

The wavefunction ψ in the Schrodinger wave equation is the amplitude of the electron wave or amplitude of vibration at any point surrounding the nucleus. The wavefunction for an electron in an atom has no physical significance by itself. It can neither be measured nor be observed. It has only mathematical significance. It is a mathematical function whose value depends upon the coordinates of the electron in an atom, ψ(x, y, z). A complete description of the state of the system can be described by wavefunction ψ. In the classical theory of light or sound, the square of the amplitude of the wave at any point gives the intensity of the sound or light at that point. Similarly, very similar concept was suggested by Max Born in quantum mechanics, according to which the square of the amplitude of the electron wave i.e., ψ2 at any point gives the intensity of the electron wave at that point. Thus, ψ2 at any point gives the probability of finding the electron.

3.4 Physical Significance of Wavefunction (ψ) and Probability Density (ψ2)

65

According to Max Born, a German physicit, probability concept of the uncertainty principle should be utilized to get an idea of distribution of the electron in the atom in terms of the wavefunction ψ. When the amplitude of the wave is large, the probability of the event is large and when the amplitude is small, the probability of the event is small. But the wavefunction itself cannot represent probability because probability of finding an electron at a given point in space is always real and positive. Whereas, wavefunction, ψ may be positive, negative and in some cases a complex quantity i.e., contains imaginary number (i = −1) . Thus, the square of wavefunction i.e., ψ2 is proposed. By taking square of the wavefunction (ψ2 = ψ · ψ*), the imaginary or negative value vanishes (Q, ψ* is conjugate of ψ). Thus, the ψ2 obtained has a real and positive value. If ψ has imaginary value, then it can be expressed as ψ = a + ib where a and b are real numbers and i (known as iota) = −1 . In such a case, ψ is made real by multiplying it by its own complex conjugate (ψ*) which is expressed as ψ* = a – ib The product obtained by multiplying ψ with ψ* is ψψ = (a + ib)(a – ib) = a2 – (i2b2) (Q i = −1 and i 2 = 1) = a2 – (–1 × b2) = a2 + b2 However, if ψ is real, then ψ = ψ*, then ψψ* = ψ2. The square of the wavefunction ψ2 has physical significance and it represents intensity. Greater the intensity of wavefunction at a particular point, greater is the probability of locating the electron at that point. Thus, ψ2 is interpreted being proportional to electron density. When ψ2 is high, electron density is high, that is, the probability of finding an electron is high. If ψ2 is low, probability of finding an electron is low. If ψ extends over a finite volume in space dx · dy · dz (= dτ), then the electron would be found in that volume. Thus, the probability of finding electron in the small volume element dx · dy · dz (= dτ) situated at a point in space is given as: Probability density (ψ2) = ψψ * dx dy dz = ψψ * d τ = ψψ * dV . The probability of finding the electron is finite which means ψψ*dτ must have real value.

66

3.5

Chapter 3

Introduction to Quantum Mechanics

CONCEPT OF ATOMIC ORBITAL

Taking into consideration the wave nature of electron and the concept of probability, the movement of the electron around the nucleus is like that of bee around the hive. Sometimes, it may come close to the nucleus and sometimes it may move away from it. Further, it does not move in one plane (as wrongly postulated by Bohr) but moves in all directions and in all planes around the nucleus. Thus, in the light of uncertainty principle, Bohr model of atom postulating that electrons move along definite circular paths and hence their position and momentum could be determined with absolute accuracy became unacceptable. Now, we speak in terms of ‘probability’ of finding the electron at any particular position about the nucleus at any instant of time. The region of space around the nucleus which describes the probability of finding an electron of given energy is called an electron cloud. Also the probability of finding an electron is never zero. Even at large distance from the nucleus, there is a finite, though small, probability of finding an electron of a given energy. This means that electron clouds do not have sharp boundaries (Fig. 3.2).

Fig. 3.2. Electron cloud for the only electron of the hydrogen atom.

However, for the sake of pictorial clarity and for convenience of representation, a boundary surface (Fig. 3.3) may be drawn which connects points of equal probability and encloses a certain volume of the space around nucleus within which the probability of finding an electron of given energy is maximum (say upto 90%). This is called atomic orbital. Thus, atomic orbital may be defined as the three dimensional space around the nucleus within which the probability of finding an electron of given energy is maximum (say upto 90%).

3.5 Concept of Atomic Orbital

67

Fig. 3.3. Boundary surface for maximum probability density.

By finding ψ2 at different points around the nucleus in an atom, we can predict the region of space around the nucleus within which the probability of finding the electron with a definite value of energy is maximum. This space around the nucleus which represents the electron density at different points is called an atomic orbital. That is why, the wavefunction for an electron in an atom is called orbital wavefunction or atomic orbital. Orbit Orbital 1. An orbit (proposed by Bohr) is a 1. An orbital is a quantum mechanical circular path around the nucleus in concept and it refers to the three which an electron moves. dimensional space around the nucleus within which the probability of finding an electron is maximum (upto 90%). 2. It represents the planar motion of 2. It represents the three dimensional electron around the nucleus. motion of an electron around the nucleus. 3. The concept of orbit is not in accordance 3. The concept of an orbital is in with the wave nature of electrons and accordance with the wave character of uncertainty principle. Thus, a precise electrons and uncertainty principle. description of this path of the electron is impossible according to Heisenberg uncertainty principle. 4. Bohr orbits have no real meaning 4. It is a quantum mechanical concept and and their existence can never be refers to the one electron wavefunction demonstrated experimentally. ψ in an atom. The value of ψ depends on the coordinates of the electron. Also, ψ itself has no physical significance but ψ2 has a physical significance. 5. The maximum number of electrons in 5. The maximum number of electrons an orbit is given by 2n2, where, n is the present in any orbital is two. number of the orbit.

68

3.6

Chapter 3

Introduction to Quantum Mechanics

QUANTUM MECHANICAL MODEL OF ATOM

Just as Bohr model of atom was developed on the basis of classical mechanics, the quantum mechanical model of atom has been developed on the basis of quantum mechanics i.e., Schrodinger wave equation. This quantum mechanical model of atom takes into account the two concepts i.e., de-Broglie dual nature of matter and Heisenberg uncertainty principle. Thus, quantum mechanical model takes wave nature of electron into consideration. The Bohr concept of fixed circular path with definite position and momentum of the electron have been replaced in quantum mechanical model by stating that the electron has the probability of having a given position and momentum. According to this model, an electron does not move in one plane but moves in all directions and in all planes around the nucleus. There is a finite, though small, probability of finding an electron of a given energy around the nucleus. The probability of finding an electron is never zero. The electrons in an atom have only quantised values of energy. These quantised values of energy are obtained from the solution of Schrodinger wave equation. The corresponding values of the wavefunction ψ are also obtained from the solution of Schrodinger wave equation. This wavefunction ψ is simply a function of coordinates of the electron and has no physical significance as such. However, ψ2 gives the probability of finding the electron at that point i.e., electron density at that point. The concept of probability is justified in view of Heisenberg’s uncertainty principle. By finding ψ2 at different points around the nucleus in an atom, we can predict the region of space around the nucleus within which the probability of finding the electron with a definite value of energy is maximum. This space around the nucleus is called orbital. This wavefunction ψ is called orbital wavefunction or atomic orbital. Since an electron can have many wavefunctions, therefore, there are many atomic orbitals in an atom. The wavefunction ψ contains all the information about an electron in an atom.

3.7

EIGEN VALUE AND EIGEN WAVEFUNCTION

Schrodinger wave equation is a second order differential equation. When it is solved, it gives infinite solutions of ψ represented as ψ1, ψ2, ψ3, … etc. with corresponding energies E1, E2, E3, … etc. But all the solutions are not acceptable. Only those values of ψ are acceptable which correspond to some definite values of the total energy E. The acceptable values of the wavefunction ψ are called eigen wavefunction or acceptable wavefunction. The value of E given by a particular eigen wavefunction is called eigen value or proper value or characteristic value. Thus, every value of wavefunction is not an eigen wavefunction.

3.7 Eigen Value and Eigen Wavefunction

69

The properties of an eigen function are: S, C, F, B, N 1. ψ must be single valued (S) ψ must be single valued at any given point. 1

Like:

(x)

2

3

4 A x

Four values are permitted at a point A, thus here ψ(x) is not single valued. 2. ψ must be continuous (C) ψ must vary smoothly without breaks i.e., there should not be any gap or discontinuity in the function. Like: sine and cosine plots are continuous however tan plot is not continuous. 3. ψ must have finite value (F) ψ must have finite values and it should not possess any imaginary values Like: iota (i = −1 ) values. 4. ψ must be bounded in the space (B) ψ must have boundary conditions in the space. 5. ψ must be normalised (N) ψ being normalised means that the probability of finding the particle inside the bounded space is unity (i.e., sure success in probability). +∞

∫−∞ ψ*ψ d τ = 1 The function which possess all the above properties is called well behaved wavefunction or acceptable wavefunction.

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Problem based on eigen function 1. Which of the following functions are acceptable in quantum mechanics? Ans. ψ = sin x; π for 0 ≤ x ≤ 2 ψ = cos x + sin x; π for 0 ≤ x ≤ 2 ψ = tan x; π for 0 ≤ x ≤ 2 ψ = cosec x; π for 0 ≤ x ≤ 2 –ax ψ=e ; for x ≥ 0 ψ = xe–ax; for x ≥ 0 ψ = e–ax; for x ≤ 0 2

acceptable when x lies between 0 and

π 2

acceptable when x lies between 0 and

not acceptable because ψ → ∞ as x →

π 2 π 2

not acceptable because ψ → ∞ as x → 0

acceptable not acceptable because ψ → ∞ as x → 0 acceptable not acceptable

− ax ; ψ= e for x ≤ 0 ψ=x

not acceptable because ψ → ∞ as x → ∞

ψ=

x2

not acceptable because ψ → ∞ as x → ∞

ψ = ex

not acceptable because ψ → ∞ as x → ∞

ψ=

not acceptable because ψ → ∞ as x → – ∞

e–x

−x ψ= e

2

acceptable because ψ → 0 as x → ± ∞

2. What are the properties of an acceptable wavefunction? or What are conditions that ψ must obey as a wavefunction? or What are the conditions under which a wavefunction is said to be a well behaved one?

3.8 Normalised, Orthogonal and Orthonormal Wavefunction

71

Ans. Properties or Conditions of a wavefunction to be acceptable are: S = Single valued, ψ must be single valued at any given point C = Continuous, ψ must vary smoothly without breaks F = Finite, ψ must have finite values (imaginary values not allowed) B = Boundary conditions, ψ must be bounded in the space N = Normalised i.e., probability of finding the particle in the bounded space must be equal to unity. 3. What is the significance of ψ and ψ2. Ans. The wavefunction ψ in the Schrodinger equation is the amplitude of the electron wave. It has no physical significance by itself but has only mathematical significance. It is a mathematical function whose value depends upon the coordinates of the electron in an atom. The wavefunction can be positive, negative or a complex quantity. Whereas, the square of the wavefunction ψ2 i.e., square of the amplitude of the electron wave, at any point give the intensity of electron wave at that point. It gives the probability of finding the electron at a given point in space. The ψ2 have always real and positive values. 4. Explain as to why ‘ψ’ should be finite, single valued and continuous. Ans. When Schrodinger equation is solved, it gives infinite solutions of ψ with corresoponding energies. But all the solutions are not acceptable. Only those values of ψ are acceptable which fulfills certain properties like it must be finite, single valued and continuous.

3.8

NORMALISED, ORTHOGONAL AND ORTHONORMAL WAVEFUNCTION

1. Normalised Wavefunction: • The square of wavefunction (ψψ*) is proportional rather than equal to the probability of finding the electron in a given volume element dx · dy · dz (= dτ or = dV). For most of the purposes, it is taken as equal to the probability which implies that the electron is in a given volume element dτ. In such a case, the integration of ψψ* dτ over whole of the configuration space, which gives the total probability in a given volume element must be equal to unity +∞

∫−∞ ψψ* d τ

=1

or

+∞

∫−∞ ψ

2

dτ = 1

The wavefunction which satisfies the above relation is said to be normalised. • But very often, ψ is not normalised. Thus, ψ is multiplied by a constant N to give a new wavefunction, Nψ, which is also a solution of the wave

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equation. The value of N used to make the new wavefunction a normalised wavefunction can be calculated as follows: +∞

∫−∞ ( N ψ)( N ψ*) d τ = 1 N2

+∞

∫−∞ ψψ* d τ = 1  N=   

1/ 2

  +∞ ψψ* d τ  −∞  1

∫

The factor N thus introduced is called Normalisation Constant and is independent of the coordinates (x, y, z). The function Nψ is called Normalised Wavefunction. • The solution of Schrodinger equation may provide more than one wavefunction i.e., each wavefunction satisfies the Schrodinger equation with its own characteristic eigen value. If ψi and ψj are the two different eigen or acceptable wavefunction of a given system, then these functions will be normalised if they meet the requirement +∞

+∞

∫−∞ ψi ψ*i d τ = 1 and ∫−∞ ψ j ψ*j = 1. 2. Orthogonal Wavefunction: Let ψi and ψj are the two independent eigen wavefunctions obtained on solving the Schrodinger equation of a given system and each wavefunction satisfies the Schrodinger equation with its own characteristic eigen values. If two such wavefunctions satisfy the expression +∞

∫−∞ ψiψ j d τ = 0 then the two functions are said to be orthogonal to each other. Orthogonality implies the independence of energy states. 3. Orthonormal Wavefunction: (Orthonormal = Normalised + Orthogonal) If the solutions of Schrodinger equation satisfy the expression δij = 0, if i ≠ j   δ = 1, if i = j    ij then the two solutions (ψi and ψj) form orthonormal set of wavefunction. The symbol δij is called Kronecker’s delta. +∞

∫−∞ ψiψ j d τ = δij

3.8 Normalised, Orthogonal and Orthonormal Wavefunction

73

Problems based on normalised and orthogonal Wavefunction Q.1. How do you justify the normalisation condition. Ans. The square of the wavefunction ψ2 accounts for the probability of the particle in the bounded space. The probability equal to unity (ψ2 = 1) means that finding the particle inside the bounded space is sure (i.e., complete success case in probability). The conjugate of wavefunction is multiplied by the wavefunction while squaring it, in order to absorb imaginary parts. ψ*ψ = ψ2 = 1 Normalisation condition applies with the application of boundary conditions. +∞

∫−∞ ψ*ψ d τ = 1 The above condition is the condition of normalisation. Q.2. Perform the normalisation of the given wavefunction, ψ = x2 over the interval 0 ≤ x ≤ K and calculate the normalisation constant. +∞

Ans. For a function to be normalised,

∫−∞ ψψ* d τ = 1

Here, ψ = x2 and ψ* = x2 +K

∫0

2

+K

∫0

2

x ⋅ x dx =

+K

 K5  K5 − 0 = ≠1 =  5  5 

 5 x dx =  x   5 0 4

Thus, ψ = x2 is not a normalised function. To calculate the normalisation constant N, let the new normalised function be Nψ K

∫0

( N ψ )( N ψ ) dx = 1 K

∫0

N 2 x 4 dx = 1

N2

∫0

K

x 4 dx = 1

 K5  N2   =1  5  1/ 2

5 Normalisation constant = N =  5  K 

.

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Q.3. Calculate the normalisation constant N1 and N 2 over the interval –a ≤ x ≤ a for the given function respectively. Ans. (i) ψ1 = N1(a2 – x2) +a

∫−a N12

+a

∫−a

N12 (a 2 − x 2 ) 2 dx = 1

(a 4 + x 4 − 2a 2 x 2 ) dx = 1 +a

N12

 4 x5 2 2 3  − a x  =1 a x + 5 3   −a

      a5 2 2 ( − a )5 2 2 N12   a 4 (a ) + − a ⋅ (a3 )  −  a 4 ⋅ (−a) + − a (−a3 )   = 1 5 3 5 3       a 5 2 5   5 a 5 2 5     N12   a 5 + − a  + a + − a  = 1 5 3   5 3     2 4  N12 ⋅ a 5  2 + −  = 1 5 3  16  N12 ⋅ a 5   = 1 15   15  N12 =    16a 5  1/ 2

N1 =  15   16a 5  (ii) ψ2 = N2x(a2 – x2) +a

∫−a N 22 N 22

+a

∫−a

+a

∫−a

N 22 x 2 (a 2 − x 2 ) dx = 1

x 2 (a 4 + x 4 − 2a 2 x 2 ) dx = 1 (a 4 x 2 + x 6 − 2a 2 x 4 ) dx = 1 +a

N 22

 4  x3  x 7  x5   − 2a 2    = 1  a   +   3  7  5   − a

  (a 3 ) a 7  − a 5     a 5    4 ( − a 3 ) ( − a )7 + − 2a 2  N 22   a 4 + − 2a 2    −  a    = 1 3 7 3 7  5     5     

3.8 Normalised, Orthogonal and Orthonormal Wavefunction

75

  a 7 a 7 2 7   a 7 a 7 2 7   − a + + N 22   + − a  = 1 7 5  3 7 5     3 2 2 4 N 22 ⋅ a 7  + −  = 1 3 7 5

 16  N 22 ⋅ a 7  =1 105  1/ 2

 105  N2 =    16a 7 

.

(iii) ψ3 = N4(a4 – x4) 1/ 2

 45  N3 =   .  64a 9  Q.4. What are orthogonal wavefunctions? What is the physical significance of orthogonality? Ans. Let ψi and ψj are the two independent eigen wavefunction obtained on solving the Schrodinger equation of a given system. If two such Similarly,

+∞

wavefunctions satisfy the expression

∫−∞ ψiψ j = 0, then the two functions

are said to be orthogonal to each other. Physical significance of orthogonality: Orthogonality implies the independence of the energy states. Q.5. Prove that the two functions, ψ1 = x and ψ2 = x2 are orthogonal to each other over the interval –K ≤ x ≤ K. +∞

Ans. For a function to be orthogonal, Here, ψ1 = x and ψ2 = x2 +K

∫− K

ψ1ψ 2 dx =

+K

∫− K

2

x ⋅ x dx =

+K

∫− K

∫−∞ ψ1ψ 2 d τ = 0 +K

K4 K4   x4  − x dx =   =   =0 4 4  4   −K 3

Thus, the two functions, ψ1 and ψ2 are orthogonal to each other over the given range. Q.6. Show that the two functions, ψ1 = N1(a2 – x2) and ψ2 = N2x(a2 – x2) are orthogonal to each other over the interval –K ≤ x ≤ K. Ans. For a function to be orthogonal,

+∞

∫−∞ ψ1ψ 2 d τ = 0

Here, ψ1 = N1(a2 – x2) and ψ2 = N2 · x(a2 – x2)

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Chapter 3

+∞

+a

∫−∞ ψ1ψ 2 d τ = ∫−a =

N1 (a 2 − x 2 ) ⋅ N 2 ⋅ x(a 2 − x 2 ) dx

+a

∫−a

Introduction to Quantum Mechanics

N1N 2 ⋅ x(a 2 − x 2 ) 2 dx

= N1N 2 = N1N 2

+a

∫−a

+a

∫−a

x(a 4 + x 4 − 2a 2 x 2 ) dx (a 4 x + x5 − 2a 2 x3 ) dx +a

 a 4 x 2 x6 2 a 2 x 4  = N1N 2  + −  6 4  −a  2

  a 6 a 6 a 6   a 6 a 6 a 6   − − + −  = N1N 2   + 2 2 2 6 2     2 = N1N2{0} = 0 Thus, the two functions, ψ1 and ψ2 are orthogonal to each other over the given range. 1/2

1/2

1 1 Q.7. Show that the functions ψ1 =   and ψ2 =   cos nx are π  2π  orthogonal to each other over the interval 0 ≤ x ≤ 2π, where n is an integer. Ans. For a function to be orthogonal, 2π

∫0

ψ1ψ 2 dx =

2π

∫0

1/ 2

 1     2π 

+∞

∫−∞ ψiψ j d τ = 0 1/ 2

1 ⋅  π

cos nx dx

2π 1 1 [sin nx]02 π = 0 cos nx = = 0 2π 2π

∫

Thus, the two functions, ψ1 and ψ2 are orthogonal to each other over the given range.

3.9

OPERATORS

An operator is a symbol for a mathematical calculation/procedure/command which changes one function into another. Operator (Function) = New Function An operator written alone has no significance. The left hand side of the above equation does not mean that the function is multiplied with the operator. The function on which the operation is carried out is called an operand.

3.9 Operators

77

Let, (x2 + a) is a function, then Operation (i) Multiply by C (ii) Take the square (iii) Differential (iv) Integration

(v) Addition of x

Operator C × (x2 + a) (x2 + a)2

Result Cx2 + Ca x4 + a2 + 2ax2

d 2 ( x + a) dx

∫

( x 2 + a ) dx

x + (x2 + a)

2x x3 + ax 3 x + x2 + a

An operator is used with the symbol (  ) over it i.e. a CARAT is put over the symbol of the operator. Thus, P operator is represented as Pˆ and Q operator is represented as Qˆ . • algebra of Operators 1. Addition and Subtraction of Operators: If Aˆ and Bˆ are two different operators and f (x) is the function, then ˆ ( x) + Bf ˆ ( x) ( Aˆ + Bˆ ) f ( x) = Af ˆ ( x) + Bf ˆ ( x) ( Aˆ − Bˆ ) f ( x) = Af 2. Multiplication of Operators: If Aˆ and Bˆ are two different operators and f (x) is the function, then order of the operation is very important.

ˆ ˆ [ f ( x)] = Aˆ [ Bf ˆ ( x)] = Aˆ [ f ′( x)] = f ″(x) AB (where Bˆ f (x) = f ′(x)) This above expression implies that first operator Bˆ operates on function f (x) to get the result, say f ′(x) and then operator Aˆ operates on function f ′(x) to get the final result say f ″(x). When we are dealing with operators, the order of operation should be maintained. Usually the order of operation is from RHS to LHS. d and f (x) = x2 Like: Aˆ = x, Bˆ = dx ˆ ˆ [( x)] = Aˆ [ Bf ( x)] = x ⋅  d ( x 2 )  = x(2x) = 2x2 ABf  dx  Also, if the same operation is to be done a number of times in succession, it is shown by a power of operator. ˆ ˆ ( x) = Aˆ 2 f ( x) . AAf

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d Like: Aˆ = and f (x) = sin x dx 2 d2 Aˆ 2 f ( x) =  d  sin x = 2 (sin x) dx  dx 

d d d  (sin x)  = [cos x] = –sin x. dx  dx dx  3. Commutative Property: 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. ˆ ˆ [( x)] = BAf ˆ ˆ [( x)] , then the two operators Aˆ and Bˆ commute with If, ABf each other i.e. the normal product of two operators is equal to their product in reverse order. ˆ ˆ − BA ˆ ˆ ] is called commutator and is represented as The expression [ AB [ Aˆ ⋅ Bˆ ] . =

ˆ ˆ [ f ( x)] − BA ˆ ˆ [ f ( x)] . Commutator = [ Aˆ ⋅ Bˆ ] = AB If the two operators commute with one another, then the value of commutator is zero i.e. ˆ ˆ [ f ( x)] − BA ˆ ˆ [ f ( x)] = 0 [ Aˆ ⋅ Bˆ ] = AB d2 d Like: Aˆ = , Bˆ = 2 and f (x) = sin x dx dx 2 ˆ ˆ [ f ( x)] = Aˆ [ Bf ˆ ( x)] = d  d sin x  = –cos x AB  dx  dx 2  2 ˆ ( x)] = d  d (sin x)  = –cos x ˆ ˆ [ f ( x)] = Bˆ [ Af BA  2

dx  dx



ˆ ˆ [ f ( x)] − BA ˆ ˆ [ f ( x)] = (–cos x) – (–cos x) = 0 \ [ Aˆ ⋅ Bˆ ] = AB Thus, Aˆ and Bˆ commute with each other because value of commutator is zero. On the other hand, the two operators are said to be non-commutative if ˆ ˆ [ f ( x)] ≠ BA ˆ ˆ [ f ( x)] . Also if the two operators do not commute with AB one another, then the commutator has some value i.e., non-zero. ˆ ˆ [ f ( x)] − BA ˆ ˆ [ f ( x)] ≠ 0 Commutator = [ Aˆ ⋅ Bˆ ] = AB d Like: Aˆ = x, Bˆ = and f (x) = x2 dx

3.9 Operators

79

ˆ ˆ [ f ( x)] = Aˆ [ Bf ˆ ( x)] = x ⋅  d ( x 2 )  = x(2x) = 2x2 AB  dx  ˆ ( x)] = d [ x ⋅ ( x 2 )] = d ( x3 ) = 3x2 ˆ ˆ [ f ( x)] = Bˆ [ Af BA dx dx ˆ ˆ [ f ( x)] − BA ˆ ˆ [ f ( x)] = 2x2 – 3x2 = –x2 ≠ 0 [ Aˆ ⋅ Bˆ ] = AB \ Thus, Aˆ and Bˆ do not commute with each other because value of commutator is non-zero. • Significance of Commutation: If the two operators commute i.e., ˆ ˆ [ f ( x)] − BA ˆ ˆ [ f ( x)] = 0 [ Aˆ · Bˆ ] = AB then, the two observable properties represented by operators Aˆ and Bˆ can be simultaneously and exactly determined. However, if the operators do not commute i.e. ˆ ˆ [ f ( x)] − BA ˆ ˆ [ f ( x)] ≠ 0 [ Aˆ · Bˆ ] = AB then, the two observable properties represented by operators Aˆ and Bˆ cannot be simultaneously and exactly determined. • Eigen function, Eigen value and Eigen value equation: If an operator Aˆ operates on a well behaved function f (x), and give the same function multiplied by a constant. Then the function f (x) is called the eigen function and the constant is called the eigen value of the operator. The equation formed is called as Eigen value equation i.e., eq. (3.10). ˆ ( x) = af ( x) Af …(3.10) ↓ eigen function

Like: (i)

↓ eigen value

d (e ax ) = ae ax ↓ dx ↓ eigen function

eigen value

(ii) Schrodinger equation is an important eigen value equation, Hˆ ψ = Eψ. • Some of the common operators used in quantum mechanics: 1. Laplacian operator (∇2): It is an important differential operator used in quantum mechanics. ∂ ∂ ∂ ∇ = iˆ + ˆj + kˆ ∂x ∂y ∂z ˆ ˆ ˆ where, i , j and k are unit vectors along x, y and z axes respectively ∇2 =

∂2 ∂2 ∂2 + + ∂x 2 ∂y 2 ∂z 2

(iˆ ⋅ iˆ = ˆj ⋅ ˆj = kˆ ⋅ kˆ = 1)

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2. Momentum Operator ( pˆ ) : For an electron wave, the wavefunction may be represented by the function  2πix  ψ = A exp    λ  Differentiating w.r.t. x

∂ψ 2πi  2πi  2πix   ψ A exp  =  =  λ λ  ∂x  λ  Using de-Broglie relationship

…(3.11)

h px 1 p = x h λ

λ=

…(3.12)

Putting (3.12) in (3.11), we have ∂ψ p  = 2πi  x  ψ ∂x  h  On rearranging, we have: h ∂ψ or, pˆ x ψ = 2πi ∂x But ψ is a function which on being removed from the equation reduces it to a differential operator, h ∂  ∂ ∂  h  pˆ x = where,  = = = −i   2πi ∂x i ∂x ∂x  2π  ∂ h ∂  ∂ pˆ y = Similarly, = −i  = ∂y i ∂y 2πi ∂y h ∂  ∂ ∂ pˆ z = = = −i  i ∂z ∂z 2πi ∂z ˆ p Thus, the momentum operator , is given as  ∂ ∂  ∂ pˆ = pˆ x + pˆ y + pˆ z = –iħ  + +   ∂x ∂y ∂z  pˆ = –iħ∇ 2 3. Square of Momentum Operator ( pˆ ) : 2   ∂   ∂  2 ∂ pˆ x2 =  −  =   i ∂x   i ∂x  ∂x 2 2 pˆ 2y =   ∂    ∂  = −  2 ∂  i ∂y   i ∂y  ∂y 2

(i ×i = i2 = –1)

3.9 Operators

81

2 pˆ z2 =   ∂    ∂  = −  2 ∂  i ∂z   i ∂z  ∂z 2 2 2   2 pˆ 2 = pˆ x2 + pˆ 2y + pˆ z2 = − 2  ∂ + ∂ + ∂   ∂x 2 ∂y 2 ∂z 2   

pˆ = –2∇2 4. Angular Momentum Operator ( Lˆ ) : Angular momentum is given by →

→

the vector product of position ( r ) and linear momentum ( p ) →

r = xiˆ + y ˆj + zkˆ →

p = px iˆ + p y ˆj + pz kˆ

where, iˆ, ˆj and kˆ are unit vectors along x, y and z axes respectively. →

→

→

L = r×p

Now,

→

L = ( xiˆ + yjˆ + zkˆ) × ( px iˆ + p y ˆj + pz kˆ) ˆj iˆ kˆ =

x px

y py

z pz

→

L = iˆ( ypz − zp y ) − ˆj ( xpz − zpx ) + kˆ( xp y − ypx )

→

L = iˆ( ypz − zp y ) + ˆj ( zpx − xpz ) + kˆ( xp y − ypx ) …(3.13)

Also, by definition: →

L = Lx iˆ + Ly ˆj + Lz kˆ By definition, where Lx, Ly and Lz are the three components of L. Comparing equations (3.13) and (3.14). Lˆ x = ypˆ z − zpˆ y = y

∂   ∂  ∂  ∂ = y −z  −z ∂y  i ∂z i ∂y i  ∂z

∂   ∂  ∂  ∂ = z − x  Lˆ y = zpˆ x − xpˆ z = z −x ∂z  i ∂x i ∂z i  ∂x

…(3.14)

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∂   ∂  ∂  ∂ = x − y  −y Lˆ z = xpˆ y − ypˆ x = x ∂x  i ∂y i ∂x i  ∂y 5. Kinetic Energy Operator ( Tˆ ): Classically, kinetic energy is given as 1 2 mv 2 p = mv

T= Now,

…(3.15)

p2 = m2v2 Dividing throughout by 2m: m 2 v2 p2 = 2m 2m

p2 1 = mv 2 2m 2 On comparing equations (3.15) and (3.16), we have \

…(3.16)

p2 T= 2m 1 ( pˆ x2 + pˆ 2y + pˆ z2 ) Tˆ = 2m Tˆ =

1  2 ∂ 2   2 ∂ 2   2 ∂ 2    −    +  −  +  − 2m  ∂z 2   ∂y 2   ∂x 2  

Tˆ = −

∂2 ∂2  2  ∂ 2 + +   2m  ∂x 2 ∂y 2 ∂z 2 

2 2 ∇ Tˆ = − 2m 6. Hamiltonian Operator ( Hˆ ) : The total energy of a system is the sum of kinetic and potential energies. The operator corresponding to the total energy is called Hamiltonian operator ( Hˆ ) . The total energy of a system containing a particle of mass ‘m’ is given as E = K.E. + P.E. Hˆ = Tˆ + Vˆ 2 2 ∇ + V ( x, y , z ) Hˆ = − 2m

3.9 Operators

83

Suppose, there is a system containing two particles, each of mass m1 and m2. The total energy of the system is given as 2 2 2 2 Hˆ = − ∇1 − ∇ 2 + V(x1, y1, z1; x2, y2, z2) 2m1 2m2 In general for a large number of particles say i Hˆ = −

2 2

∑

1 2 ∇i + V ( x, y, z ) mi

• In quantum mechanics we deal with only Linear Hermitian operators 1. Linear Operators: By linear operator we mean that if it is applied on the sum of two functions, the result is equal to the sum of the operations on the two functions separately. Let operator Aˆ operates on two functions of f and g, then ˆ + Ag ˆ Aˆ [ f + g ] = Af

…(3.17)

Like: Differential and Integral are linear operators. d d d ( f + g) = ( f ) + (g) dx dx dx

∫ [C1 f1( x) + C2 f2 ( x)] dx = C1 ∫

f1 ( x) dx + C2

∫

f 2 ( x) dx

But Square and Square root are non-linear operators. [C1f1(x) + C2f2(x)]2 ≠ C12[f1(x)]2 + C2[f2(x)]2 + 2C1C2f1(x)f2(x) C1 f1 ( x) + C2 f 2 ( x) ≠

C1 f1 ( x) + C2 f 2 ( x)

2. Hermitian Operators: Suppose ψ and φ are the two eigen functions of the operator Aˆ , and if +∞

∫−∞

ψ*(Aˆ φ) d τ =

+∞

∫−∞ φ(Aˆ ψ)* d τ

then the operator Aˆ is called Hermitian operator. d2 is a Hermitian operator. dx 2 Let ψ = eix and φ = sin x are the two eigen functions.

Like: Show that

If ψ = eix, then ψ* = e–ix.

…(3.18)

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Chapter 3

∫

ψ* (Aˆ φ) dx =

∫

φ( Aˆ ψ )* dx = =

Introduction to Quantum Mechanics

∫

  d2 e − ix  2 (sin x) dx = − ∫ e − ix sin x dx   dx

∫

  d2 sin x  2 (e −ix )  dx   dx

∫ sin x (−e

− ix

∫

) dx = − sin x ⋅ e −ix dx









d2 Thus, the two integrals are same; therefore 2 is hermitian operator. dx • Two Properties of Hermitian Operator: (i) Eigen Values of a Hermitian Operator are Real (Positive or Negative): Let  operator be a Hermitian operator and ψ its eigen function, then ‘a’ would be the real eigen value  = a hermitian eigen real operator function eigen value

Proof: Let  be a Hermitian operator, ψ its eigen function and ‘a’ the eigen value. Âψ = aψ …(3.19) and (Âψ)* = a*ψ* …(3.20) Multiply eq. (3.19) by ψ* and on integration

∫ ψ*(Aˆ ψ) d τ = ∫ ψ*(aψ) d τ ∫ ψ*(Aˆ ψ) d τ = a ∫ ψ*ψ d τ

…(3.21)

∫ ψ (Aˆ ψ )* d τ = ∫ ψ(a*ψ*) d τ ∫ ψ(Aˆ ψ)* d τ = a*∫ ψψ * d τ

…(3.22)

Multiply eq. (3.20) by ψ and on integration

LHS of eq. (3.21) and eq. (3.22) is equal, since  is Hermitian, ψ*(Aˆ ψ ) d τ = ψ (Aˆ ψ )* d τ i.e.

∫

∫

Therefore, RHS of eq. (3.21) and eq. (3.22) is also equal i.e.

∫

∫

a ψ*ψ d τ = a* ψψ* d τ

Therefore, i.e. a is real.

a = a*

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85

(ii) Eigen Functions of a Hermitian Operator Corresponding to Different Eigen Values are Orthogonal: Let ψ1 and ψ2 are two eigen functions of a Hermitian operator  corresponding to two eigen values a1 and a2 respectively. The condition of orthogonality is

or or

∫ ψ1ψ 2 d τ = 0 ∫ ψ1ψ*2 d τ = 0 ∫ ψ1*ψ 2 d τ = 0.

Proof: The eigen value equations are Âψ1 = a1ψ1 Âψ2 = a2ψ2

…(3.23) …(3.24)

Multiply eq. (3.23) by ψ2* and on integration

∫ ψ*2 ( Aˆ ψ1) d τ = ∫ ψ*2 (a1ψ1) d τ ∫ ψ*2 ( Aˆ ψ1) d τ = a1 ∫ ψ*2ψ1 d τ

…(3.25)

Since  is Hermitian, LHS of eq. (3.25) becomes

∫ ψ*2 ( Aˆ ψ1) d τ = ∫ ψ1( Aˆ ψ 2 )* d τ = ψ1 (a2ψ 2 )* d τ ∫ = a*2 ∫ ψ1ψ*2 d τ ∫ ψ*2 ( Aˆ ψ1) d τ = a2 ∫ ψ1ψ*2 d τ

[using eq. (3.24)]

…(3.26)

[a*2 = a2 because Hermitian operators have real value] Equating RHS of eq. (3.25) and eq. (3.26), we have:

∫ ∫ (a1 − a2 ) ∫ ψ1ψ*2 d τ = 0 But a1 ≠ a2 then, ∫ ψ1ψ*2 d τ = 0

a2 ψ*2 ψ1 d τ = a2 ψ1ψ*2 d τ

i.e., ψ1 and ψ2 are orthogonal if a1 ≠ a2 However, if a1 = a2 then

∫ ψ1ψ*2 d τ ≠ 0

\, ψ1 and ψ2 may not be orthogonal if a1 = a2.

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Problems based on oPerators 1. Operate the following operator  on the function f(x) i.e., Âf(x) d2 and f(x) = 2x dx 2 d2 (2 x) = 0 Âf (x) = dx 2 d2 d (ii)  = 2 + 2 + 3 and f (x) = x2 dx dx (i)  =

 d2  2 d d2 2 d 2 2 + 2 + 3 x ( ) = 2 ( x ) + 2 ( x ) + 3x Âf (x) =  2  dx dx dx  dx  = 2 + 4x + 3x2 d and f (x) = ax2 (iii)  = dx d (ax 2 ) = 2ax Âf (x) = dx (iv)  = a and f(x) = x2 + C Âf (x) = a(x2 + C) = ax2 + aC. d ˆˆ 2. What is the value of ABf(x) , if  = 4x2, Bˆ = and f(x) = ax3? dx ˆ ˆ [ f ( x)] = Aˆ [ Bf ˆ ( x)] = 4 x 2 ⋅  d (ax3 )  Ans. AB   dx   = 4x2 · (3ax2) = 12ax4. 3. What are the expressions for the following operators:  d (a)  dx + x   

2

Ans. Let ‘f’ be the arbitrary function 2

  d d d   + x  f =  dx + x   dx + x  f     dx 

   df d + xf  + x     dx  dx

= 

3.9 Operators

87

  d2 f df + x 2 f + f  =  2 + 2 x dx   dx   d2 d 2 =  2 + 2 x + x + 1 f dx   dx 2   d2 d  d + x + x 2 + 1  =  2 + 2 x  dx   dx   dx

d  (b)  x 

2

 dx  2

 d   d   df   d  d  x  f = x x  f =x x   dx   dx   dx   dx   dx 

Ans.

 d 2 f df   2 d 2 d  + x  f = x  x 2 +  =  x 2 dx   dx dx   dx (c)  d x   

2

 dx 

d d  d  d  d  x  ( xf ) x f =  x  x f =    dx   dx  dx  dx   dx  2

Ans.

d =   dx 2 = x

 d  df    df + f  =  x2 + xf  x  x  dx  dx    dx

d2 f dx

2

+ 2x

df df + + f dx dx

  d2 d d =  x 2 2 + 2 x + + 1 f dx dx   dx 4. What are operators? Give their significance in quantum mechanics. Ans. An operator is a symbol for mathematical procedure which changes one function into another Operator (function) = New function. Significance of Operators in Quantum Mechanics: According to third postulate of quantum mechanics, for every observable in quantum mechanics (like position, velocity, momentum, energy etc.), there corresponds a

88

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mathematical operator in quantum mechanics. The physical properties of the observable can be infered from the mathematical properties of its associated operator. [Refer postulate III of quantum mechanics] 5. Show that the operator corresponding to x-component of the momentum − d . of a particle of given by pˆ x = i dx or Derive an expression for the operator pˆ x . Ans. For an electron wave, the wavefunctions may be represented by the function.  2πix  ψ = A exp.    λ  Differentating w.r.t. x:

2πi  ∂ψ  2πix   2πi A exp.  ψ =  =  λ  ∂x  λ  λ

…(1)

Using de Broglie relationship h px 1 p = x λ h Putting (2) in (1), we have λ=

…(2)

∂ψ p  = 2πi  x  ψ ∂x  h  On rearranging, we have: h ∂ψ or px ψ = 2πi ∂x But ψ is a function which on being removed from the equation reduces it to a differential operator h ∂  ∂ ∂ h = − i = (where,  = ) i ∂x 2πi ∂x ∂x 2π 6. Write Hamiltonian operator for a particle executing simple harmoic motion (SHM) at any instant. pˆ x =

Ans.

Ĥ = Tˆ + Vˆ Ĥ=

− 2 2 ∇ + V ( x, y , z ) 2m

3.9 Operators

89

Problems based on linear oPerators 1. Which of the following operators are linear operators: (a)  = x Operators is said to be linear if  [C1 f1(x) + C2 f2(x)] = C1  f1(x) + C2  f2(x) x [C1 f1(x) + C2 f2(x)] = C1 x f1(x) + C2 x f2(x) \ Its a linear operator. (b)  =

d2 dx 2

d2 [C1f1(x) + C2f2(x)] = C1 f1″(x) + C2 f2″(x) dx 2 \ Its a linear operator. (c) Â = ( )3 (a + b)3 = a3 + b3 + 3ab (a + b) [C1f1(x) + C2f2(x)]3 = [C1f1(x)]3 + [C2f2(x)]3 + 3C1f1(x) · C1f2(x) [C1f1(x) + C2f2(x)] \ It is not a linear operator. d  (d) Â =    dx 

2

2

d   d  d f 2 ( x)  f x C C ) + • ( ( ) + ( ) C f x C f x = 1 1 2 1 1 2 2   dx  dx     dx

2

= [C1 f1′(x) + C2 f2′(x)]2 = [C1 f 1′ (x)]2 + [C2 f2′(x)]2 + 2C1C2 f1′(x) 2

2

≠ [C1 f1 ′( x) ] + [C2 f 2 ′( x) ] \ It is not a linear operator. (e) Â = x2 – 6

(x2 – 6) [C1 f1(x) + C2 f2(x)] = x2C1 f1(x) + x2C1 f1(x) – 6C1 f1(x) – 6C2 f2(x) = (x2 – 6) C1 f1(x) + (x2 – 6) C2 f2(x) \ It is a linear operator.

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2. Define linear operators. Give an example. Ans. By linear operator, we mean if it is applied on the sum of two functions, the result is equal to the sum of the operations on the two functions separately Â[f + g] = Âf + Âg Like: Differential and Integral are linear operators. d d d (g) ( f + g) = (f)+ dx dx dx

∫ ( f + g ) dx = ∫

f dx +

∫ g · dx.

Problems based on commutators 1. Show that the two operators commute with each other given d (i) Â = , Bˆ = 3+ and f(x) as function. dx Ans. The two operators commute with each other if ˆ ˆ − BA ˆ ˆ [ f ( x)] − BA ˆ ˆ ] f ( x) = AB ˆ ˆ [ f ( x)] = 0 Commutator = [ Aˆ · Bˆ ] = [ AB ˆ ˆ [ f ( x)] = Aˆ [ Bf ˆ ( x)] = d [3 + f ( x)] AB dx = 0+

df ( x) df ( x) = dx dx

ˆ ( x)] = 3 +  d f ( x)  = 3 + df ( x) ˆ ˆ [ f ( x)] = Bˆ [ Af BA   dx dx ˆ ˆ ( x) − BAf ˆ ˆ ( x) [ Aˆ ⋅ Bˆ ] = ABf Now, =

df ( x) df ( x) −3− = –3 dx dx

Since the value of commutator is not zero, thus  and Bˆ do not commute with each other. (ii)  = 3+, Bˆ = 4+ and f(x) = ax Ans.

ˆ ˆ [ f ( x)] = Aˆ [ Bf ˆ ( x)] = 3 + [4 + (ax)] = 3 + 4 + (ax) = 7 + ax AB ˆ ˆ [ f ( x)] = Bˆ [ Af ˆ ( x)] = 4 + [3 + (ax)] = 7 + ax BA

3.9 Operators

91

ˆ ˆ [ f ( x)] − BA ˆ ˆ [ f ( x)] [ Aˆ ⋅ Bˆ ] = AB [ Aˆ ⋅ Bˆ ] = (7 + ax) – (7 + ax) = 0 Since the value of commutator is zero, thus  and Bˆ commute with each other. ˆ if  = d , Bˆ = 3x2 and f(x) = sin x. 2. Evaluate the commutator [Aˆ ⋅ B] dx ˆ ˆ − BA ˆ ˆ [ f ( x) − BA ˆ ˆ ] f ( x) = AB ˆ ˆ [ f ( x)] Ans. [ Aˆ ⋅ Bˆ ] = [ AB d d (3 x 2 ⋅ sin x) − 3 x 2 ⋅ sin x = dx dx (6 x ⋅ sin x + 3 x 2 cos x ) − 3 x 2 cos x = = 6x · sin x [ Aˆ ⋅ Bˆ ] = 6xf (x). 2 ˆ if  = d and Bˆ = x. 3. Calculate the commutator [Aˆ ⋅ B] dx 2 ˆ ˆ − BA ˆ ˆ ] f ( x) = AB ˆ ˆ [ f ( x)] − BA ˆ ˆ [ f ( x)] Ans. [ Aˆ ⋅ Bˆ ] = [ AB

    d2  d2 =  2 ⋅ [ x ⋅ f ( x)]  −  x ⋅  2 f ( x)         dx  dx

 d d  dx  dx

 

 d 2 f ( x)  2   dx 

=   ( x ⋅ f ( x))   −  x

d 2 f ( x) d  df ( x)  − f ( x ) x x + =  dx  dx dx 2 d 2 f ( x) d 2 f ( x) df ( x) df ( x) x x + + − = dx dx dx 2 dx 2 df ( x) . Commutator = [ Aˆ ⋅ Bˆ ] = 2 • dx

92

Chapter 3

4. The operator  =

Introduction to Quantum Mechanics

d ˆ , B = x2, and f (x) is an arbitrary function show that: dx

(i) Â2f (x) ≠ [Âf (x)]2 2

Ans. LHS:

Â2f (x)

d  d2 =   f ( x) = 2 f ( x)  dx  dx

 d RHS: [Âf (x)]2 =  f ( x)    dx \

2

LHS ≠ RHS.

ˆ ˆ ( x) ≠ BAf ˆ ˆ ( x) (ii) ABf ˆ ( x)] = d [ x 2 f ( x)] = 2 xf ( x) + x 2 df ( x) ˆ ˆ ( x) = Aˆ [ Bf Ans. LHS: ABf dx dx ˆ ( x)] = x 2 ⋅  d f ( x)  = x 2 df ( x) ˆ ˆ ( x) = Bˆ [ Af RHS: BAf   dx dx   \ LHS ≠ RHS. ˆ ˆ [ f ( x)] − BA ˆ ˆ − BA ˆ ˆ [ f ( x)] ˆ ˆ ] f ( x) = AB (iii) Commutator [ Aˆ ⋅ Bˆ ] = [ AB =

df ( x) d 2 [ x ⋅ f ( x)] − x 2 ⋅ dx dx

2 = x

df ( x) df ( x) + 2 xf ( x) − x 2 = 2xf (x). dx dx

Commutator = [ Aˆ ⋅ Bˆ ] = 2xf(x). d and Bˆ = x, commute with each other. dx ˆ ˆ [ f ( x)] − BA ˆ ˆ [ f ( x)] ˆ ˆ − BA ˆ ˆ ] f ( x) = AB Ans. [ Aˆ ⋅ Bˆ ] = [ AB

5. Find whether the operators  =

d    d f ( x)  =  ⋅ [ xf ( x)]  −  x ⋅    dx  dx = x

df ( x) df ( x) + f ( x) − x dx dx

[ Aˆ ⋅ Bˆ ] = f (x)

3.9 Operators

93

Since the value of commutator is not zero, thus  and Bˆ do not commute with each other. d 2 and xˆ 2 = x2. 6. Find the commutator, [pˆ x ⋅ xˆ ] where pˆ x = −i dx 2 2 2 Ans. [ pˆ x ⋅ xˆ ] = [ pˆ x xˆ − xˆ pˆ x ] f ( x)

 d  d 2 2  =  −i dx ( x ) − x ⋅  −i dx   f ( x)    df ( x) d = −i [ x 2 ⋅ f ( x)] + ix 2 dx dx

df ( x)   df ( x) −i  x 2 + 2 xf ( x)  + ix 2 = dx dx   df ( x) df ( x) − i x 2 − 2 xif ( x) + ix 2 = dx dx Commutator = [ pˆ x ⋅ xˆ 2 ] = –2xif (x). 

2 7. Evaluate the commutator, Â = x2 and  x ,



d2  . dx 2 

Ans. Suppose the arbitrary function is f(x), then. ˆ ˆ − BA ˆ ˆ ] f ( x) Commutator = [ Aˆ ⋅ Bˆ ] = [ AB ˆ ˆ [ f ( x)] − BA ˆ ˆ [ f ( x)] = AB 2  d2 2 2  d = x ⋅  2 f ( x)  − 2 [ x f ( x)]  dx  dx

2 = x

d 2 f ( x) dx

2

−

d d 2   [ x • f ( x)]  dx  dx 

2

d f ( x) d  2 df ( x)  − + 2 xf ( x)  x  2 dx dx dx   2 2  d f ( x) df ( x)   df ( x)  2 d f ( x) + 2 f ( x) −  x2 + 2x = x  − 2 x 2 2 dx   dx dx dx   = x2

2 = x

d 2 f ( x) df ( x) df ( x) df ( x) − x2 − 2x − 2x − 2 f ( x) 2 2 dx dx dx dx

[ Aˆ ⋅ Bˆ ] = −4 x

df ( x) − 2 f ( x) dx

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d 8. Evaluate the commutator; [ pˆ x · xˆ ] where, pˆ x = − i and xˆ = x dx [ pˆ x · xˆ ] = [ pˆ x · xˆ − xˆ ⋅ pˆ x ] f ( x)

 d  d  ⋅ ( x) − x  −i   f ( x) =  −i dx dx    = −i 

d df ( x) [ x ⋅ f ( x)] + ix dx dx

= −i

d dx

= − i x

df ( x)   df ( x)  x dx + f ( x)  + ix dx  

df ( x) df ( x) − i f ( x ) + i x dx dx

[ pˆ x · xˆ ] = –i f (x) Since the value of commutator is not zero, thus, pˆ x and xˆ do not commute with each other. In other words, linear momentum (p) and position (x) cannot be obtained simultaneously and exactly [Heisenberg uncertainty principle]. 9. Show that if two operators  and Bˆ commute, they have same set of eigen functions. Ans. Let ψA be the eigen function of  operator, with an eigen value ‘a’. …(1) Then,  ψA = aψA ˆ ˆ = BA ˆˆ AB ( and Bˆ commute). Given Multiply both the sides by ψA ˆˆ ˆ ˆ ψ = BAψ AB A

Aˆ ( Bˆ ψ A ) = ˆ ) = Aˆ ( Bψ

A

Bˆ ( Aˆ ψ A )

Bˆ (aψA) Aˆ ( Bˆ ψ A ) = a ( Bˆ ψ A )

[using (1)]

A

This means ( Bˆ ψ A ) is also an eigen function of  operator with the eigen value ‘a’, which is possible only if Bˆ ψ A is just a multiple of ψ i.e., Bˆ ψ A = C · ψ , or ψ is also an eigen function of Bˆ . A

A

10. Determine whether or not the operators x 2 2 d2 ˆ= d . B and dx 2 dx 2 Suppose f(x) is the arbitrary function, then

2 Ans. Here, Â = x

d2 dx 2

and

d2 dx 2

commute.

3.9 Operators

95

ˆ ˆ − BA ˆ ˆ ] f ( x) = 0 [ Aˆ · Bˆ ] = [ AB  2 d2 = x 2  dx

 d2  d2  2  − 2  dx  dx

 2 d 2   x 2    f ( x)  dx  

 2 d 4  d  d  2 d 2  =  x ⋅ 4 −    x   f ( x) dx  dx  dx  dx 2    3   2 d4  d  d2 2 d  ⋅ + − 2 x x = x   f ( x)    4 dx 2 dx3    dx   dx 4   2 d4  d2 d3 d3 2 d  − 2 + 2 + 2 + x x x = x  f ( x)  dx 2 4 dx3 dx3 dx 4     dx

 d4 d2 d3 d4  =  x 2 4 − 2 2 − 4 x 3 − x 2 4  f ( x) dx dx dx dx    d2 d3  [ Aˆ · Bˆ ] =  −2 − 4 x  f ( x) 2 dx3   dx Since the value of commutator is not zero, thus operators do not commute with each other. d2 d d 11. Determine whether the operator Aˆ = and= + 2 , commute Bˆ 2 dx dx dx or not. Ans. Suppose the arbitrary function is f(x) ˆ ˆ − BA ˆ ˆ ] f(x) [ Aˆ ⋅ Bˆ ] = [ AB  d  d2 d   d2 d d ⋅ + 2 · −  2 + 2 ·   f ( x) =    2 dx   dx dx  dx   dx  dx  d 3 d2   d3 d 2  =  3 + 2 · 2  −  3 + 2 · 2   f ( x) dx   dx dx    dx

 d3 2· d2 d3 2d 2  − −  f ( x) =  3 + dx 2 dx3 dx 2   dx

[ Aˆ . Bˆ ] = 0 Since the value of commutator is zero, thus the two operators commute with each other.

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12. What do you understand by the commutation between two operators? What is the importance of this concept? (or) What important aspect is signified by the property of commutation? Ans. 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 ˆ ˆ f ( x) = BA ˆ ˆ f ( x) AB i.e., ˆ ˆ − BA ˆ ˆ ] f ( x) . Commutator = [ Aˆ ⋅ Bˆ ] = [ AB If the two operators commute with one another then the value of commutator is zero. Importance of Commutation: A physical quantity (like; energy, momentum, position etc.,) of a system can be represented in quantum mechanics by operator. If two operators commute, then it is possible to determine accurately both the corresponding quantities simultaneously. 13. Evaluate the commutator [ Lˆ x ⋅ Lˆ y ] where, Lˆx and Lˆ y are the angular momentum operator along the x and y directions respectively. And. As we know that, ∂  ∂ Lˆ x =  y • − z •  ∂y  i  ∂z Lˆ y =

∂  ∂ z• − x•   ∂z  i  ∂x

[ Lˆ x ⋅ Lˆ y ] = Lˆ x Lˆ y − Lˆ y Lˆ x ∂   ∂ ∂  ∂  Now, Lˆ x Lˆ y =  y • − z •  •  z • − x •  ∂y  i  ∂x i  ∂z ∂z  ∂  ∂ ∂  ∂  ∂  ∂  ∂  2 ∂  = −  y  z •  − y  x •  − z  z •  + z  x •   ∂z  ∂z  ∂y  ∂z   ∂y  ∂x   ∂z  ∂y  ∂2 ∂2 ∂2  ∂2 2 ∂ − yx • 2 − z 2 • + zx • = −   y + yz  ∂z∂x ∂y∂z  ∂y∂x ∂z  ∂x Again, Lˆ y Lˆ x =

∂  ∂ ∂    ∂ z⋅ − x •  y• − z  ∂y  ∂z  i  ∂z i  ∂x

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∂  ∂  ∂  ∂  ∂  ∂ ∂  2 ∂  = −  z  y  − z  z  − x  y ·  + x  z ·   ∂x  ∂y  ∂z  ∂y   ∂z  ∂z   ∂x  ∂z  ∂2 ∂2 ∂2  ∂2 ∂ 2 − xy · 2 + x · − z2 + xz · = −   zy ·  ∂x · ∂y ∂z∂y  ∂xdz ∂y ∂z  Thus,

[ Lˆ x ⋅ Lˆ y ] = [ Lˆ x Lˆ y − Lˆ y Lˆ x ] 2  ∂ ∂2 ∂2 _ 2 ∂ ∂2 2 z · + zx · − yx · 2 = −   y · + yz · ∂y∂z ∂y∂x ∂z∂x ∂z  ∂x

− zy ·

∂2 ∂2 ∂2 ∂ ∂2   + z2 · + xy 2 − x · − xz · ∂y ∂z∂y  ∂x∂z ∂x∂y ∂z 

∂  ∂  ∂  ∂ 2 = −  y − x  =  2  x · − y  x y ∂ ∂ ∂x     ∂y Multiply and divide by ‘i’:   ∂  ∂ = i   x · − y   ∂x    i  ∂y  ∂  ℏ ∂ − y  ∵ Lz =  x ∂z   i  ∂y 

[ Lˆ x · Lˆ y ] = i Lˆz Similarly, [ Lˆ y · Lˆz ] = i Lˆ x

[ Lˆ z · Lˆ x ] = i Lˆ y Thus, none of the angular momentum operator component Lˆ x , Lˆ y or Lˆ z commute with each other, hence cannot be specified exactly and simultaneously. 14. Find whether the operator Lˆ 2 and Lˆ z commute with each other.

and

2 2 2 Ans. Lˆ 2 = ( Lˆ x + Lˆ y + Lˆ z )

[ Lˆ2 · Lˆ z ] = [( Lˆx2 + Lˆ y2 + Lz2) · Lˆ z ] 2 2 2 2 2 2 = ( Lˆ x + Lˆ y + Lˆ z ) Lˆ z − Lˆ z ( Lˆ x + Lˆ y + Lˆ z )

= Lˆ2x Lˆ z + Lˆ2y Lˆ z + Lˆ2z Lˆ z − Lˆ z Lˆ2x − Lˆz Lˆ2y − Lˆ z Lˆ2z

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2 2 2 2 2 2 = ( Lˆ x Lˆ z − Lˆ z Lˆ x ) + ( Lˆ y Lˆ z − Lˆ z Lˆ y ) + ( Lˆ z Lˆ z − Lˆ z Lˆ z ) 2 2 2 = [ Lˆ x ⋅ Lˆ z ] + [ Lˆ y ⋅ Lˆ z ] + [ Lˆ z ⋅ Lˆ z ]

Now, the first terms of the above expression can be written as: [ Lˆ2 ⋅ Lˆ ] = Lˆ Lˆ Lˆ − Lˆ Lˆ Lˆ x

z

x x z

z x x

= Lˆ x Lˆ x Lˆ z − Lˆ x Lˆ z Lˆ x + Lˆ x Lˆ z Lˆ x − Lˆ z Lˆ x Lˆ x = Lˆ x ( Lˆx Lˆ z − Lˆ z Lˆx ) + ( Lˆx Lˆ z − Lˆ z Lˆx ) Lˆx = Lˆ x [ Lˆ x ⋅ Lˆ z ] + [ Lˆ x ⋅ Lˆ z ]Lˆ x = −iLˆ x Lˆ y − iLˆ y Lˆ x

(Q[Lx Lz] = – [Lz Lx])

[ Lˆ2x ⋅ Lˆ z ] = −i(Lˆ x Lˆ y + Lˆ y Lˆ x )

…(1)

Similarly, it can be shown that [ Lˆ2y · Lˆ z ] = +i(Lˆ x Lˆ y + Lˆ y Lˆ x )

…(2)

But, [ Lˆ2z ⋅ Lˆ z ] = 0 …(3) Adding eqs. (1), (2) and (3), we obtain [ Lˆ2 ⋅ Lˆ z ] = − i(Lˆ x Lˆ y + Lˆ y Lˆ x ) + i(Lˆ x Lˆ y + Lˆ y Lˆ x ) + 0 [ Lˆ2 ⋅ Lˆ ] = 0 z

This shows that the square of the angular momentum and its z-component can be specified simultaneously. Similarly, it can be shown that Lˆ2 commutes with Lˆ and Lˆ . x

y

15. There exists a set of orthogonal functions, ψ which are simultaneously eigen function of both the operators  and Bˆ . Prove that  and Bˆ must commute for this statement to be true. Ans. Since ψ is the eigen function of both the operators Aˆ and Bˆ . Let ‘a’ and ‘b’ are the respective eigen values …(1) then, Âψ = aψ ˆ …(2) Bψ = aψ Now,

ˆ ˆ = Aˆ ( Bψ ˆ ) ABψ = Â(bψ) = b(Âψ) ˆ ˆ = b(aψ) ABψ

(Using 2) (Using 1)

…(3)

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99

ˆ ˆ = Bˆ ( Aψ ˆ ) ABψ

Similarly,

= Bˆ (aψ ) ˆ ) = a ( Bψ

(Using 1)

ˆ ˆ = abψ BAψ (Using 2) …(3) From eqs. (3) and (4) ˆ ˆ = BAψ ˆˆ ABψ Therefore, the two operators Aˆ and Bˆ commute with each other.

Problems based on eigen function and eigen values 1. Show that eikx is an eigen function of the momentum operator pˆ x , where, pˆ x = –i

d and what is the eigen value? dx

Ans. f(x) = eikx and pˆ x = − i

d dx

Using eigen value equation,

pˆ x [f(x)] = –i

d ikx (e ) dx

= –i × ik × eikx = –i2k (eikx) = + k (eikx ) Eigen value = + ħk

(i2 = –1)

d2 2. Show that the function ae–x is an eigen function of the operation . 2 dx What is the eigen value? Ans. f (x) = ae–x and  =

d2 dx 2

Using eigen value equation, d2 (ae–x) = ae–x = f (x) dx 2 Thus, ae–x is an eigen function of the given operator and eigen value is (+1). Â f (x) =

3. Show that sin 2x is not an eigen function of the operator of What is the eigen value?

d2 d but of 2 . dx dx

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Ans. Using eigen value equation d (i) (sin 2x) = 2 cos 2x dx d Thus, sin 2x is not an eigen function of dx (ii)

d2 dx 2

(sin 2x) = –4 (sin 2x)

d2 and eigen value is (–4). dx 2 4. Operate the Laplacian operator on the given function and find the eigen value. Given that f(x) = sin kx · sin my · sin nz Thus, sin 2x is an function of

∇2 =

and

∂2 ∂2 ∂2 + + . ∂x 2 ∂y 2 ∂z 2

Ans. Using eigen value equation, ∇2f(x) = ∇2 [sin kx · sin my · sin nz] ∂2 ∂2 [sin kx · sin my · sin nz] + [sin kx · sin my · sin nz] ∂x 2 ∂y 2 ∂2 + 2 [sin kx · sin my · sin nz] ∂z = sin my · sin nz (–k2 sin kx) + sin kx · sin nz (–m2 sin my) + sin kx · sin my (–n2 sin nz) = (–k2 –m2 –n2) sin kx · sin my · sin nz =

−k 2 − m2 − n 2 ) = (  ↓ eigen value

f ( x)  ↓ eigen function

2

5. Under what condition is the function e–aq an eigen function of the operator

 d2 2  2 − kq  , where, k is a constant. What is the eigen value under these   dq conditions? d2 − kq 2 dq 2 Using eigen value equation, 2

Ans. f(x) = e–aq and  =

 d2 2  − aq 2 Â f (x) =  2 − kq  e  dq 

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101

=

d2 2 2 (e–aq ) – (kq2e–aq ) dq 2 2

2

= a2 × 4q2e–aq – (kq2e–aq ) 2

4a 2 q 2 − kq 2 ) (e − aq ) Â f (x) = (    ↓ eigen value

↓ eigen function

2

The function e–aq is an eigen function if: 4a2q2 – kq2 = 0 4a2q2 = kq2 k = 4a2 If k = 4a2, then function itself is the eigen value. 6. Which of the given functions are eigen function of

d2 . Also for each eigen dx 2

function, give the eigen value. (i) sin 3x Using eigen value equation, d2 (sin 3x) = –9(sin 3x) dx 2 d2 Thus, sin 3x is an eigen function of 2 and the eigen value is (–9). dx (ii) 6 cos 4x d2 (6 cos 4x) = –96(cos 4x) = –16(6 cos 4x) dx 2 d2 Thus, (6 cos 4x) is an eigen function of 2 and the eigen value is (–16). dx 2 (iii) 5x d2 (5x2) = 10 dx 2 Thus, 5x2 is not an eigen fucnction of (iv)

1 x

Thus,

d2 dx 2

2 d2  1  = 3 2   x dx  x  1 d2 . is not an eigen function of x dx 2

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(v) 3e–5x d2 (3e–5x) = 3 × 25e–5x = 25(3e–5x) 2 dx d2 Thus, 3e–5x is an eigen function of 2 and the eigen value is 25. dx (vi) ln 2x  d2 −1 d2  d2 ) (ln 2 x (ln 2x) =   = 2 2 2 2 dx x dx  dx  Thus, ln 2x is not an eigen function of

d2 . dx 2

d 7. (i) Show that eax is an eigen function of the operator and find the dx corresponding eigen value. d Ans. (i) f(x) = eax, and  = dx Using eigen value equation: d ax (e ) = ae ax  f(x) = dx d and its eigen value is ‘a’. Hence, eax is an eigen function of dx d 2 (ii) Show that eax is not an eigen function of . dx d 2 Ans. f(x) = eax , and  = dx d ax 2 (e ) = 2ax · eax2 dx Thus, the result obtained is not a constant factor multiplied with f (x) but a 2 variable factor 2ax multiplied with f(x). Hence, eax is not an eigen function d . of dx 8. If ψ1 and ψ2 are eigen wavefunctions of an atom, prove that any linear combination of these wavefunctions will also be an eigen function of the Schrödinger wave equation. Ans. ψ = C1ψ1 + C2ψ2 Ĥ ψ = Ĥ (C1ψ1 + C2ψ2) = C1Ĥψ1 + C2Ĥψ2 = C1Eψ1 + C2Eψ2 Ĥψ = E(C1ψ1 + C2ψ2) Ĥ ψ = Eψ.

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9. If a quantum mechanical system has degenerate eigen function of operator Â, then any linear combination of the eigen functions is also an eigen function  with the same eigen value. Justify. Ans. Degenerate eigen function means a series of functions means a series of functions ψ1, ψ2 ..., ψn are all eigen functions of the operator  with the same eigen value. For a linear combination, we can write n

ψ = C1ψ1 + C2ψ2 + ... Cnψn =

∑ Ci ψ i i =1

where C1, C2 etc., are constants n

Âψ = Â

∑

= Ci ψ i

i =1 n

=

∑ Ci aψi

n

∑ Ci Aˆ ψi

i =1

[because Aˆ ψi = aψi]

i=i

n

=

∑ Ci ψ i i =1

= aψ. 10. Show that if all eigen functions of two operators  and Bˆ are the same functions,  and Bˆ commute with each other. Ans. Suppose the eigen functions of the operator  and Bˆ are ψi and the corresponding eigen values are ai and bi respectively. Then eigen value equations will be and

Âψi = aiψi Bˆ ψi = biψi

ˆ ˆ can be obtained as follows The eigen function of the operator AB ˆ ˆ ψ = Aˆ ( Bˆ ψ i ) = Aˆ (bi ψ i ) = b (Âψ ) = b a ψ AB i i i i i i

ˆ ˆ can be obtained as follows The eigen function of the operator BA ˆ ˆ = Bˆ ( Aˆ ψ )= Bˆ (a ψ ) = a ( Bˆ ψ ) = a b ψ BAψ i i i i i i i i i As,

biai = aibi

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ˆ ˆ ψ = BA ˆ ˆψ this means, AB i i ˆ ˆ = BA ˆˆ or AB Hence, they commute with each other. 11. Prove that the function Aeikx (A and K are constants) is an eigen function of the operator pˆ x . What is the eigen value? Ans. pˆ x =

 ∂ and f(x) = Aeikx i ∂x

Using eigen value equation, pˆ x f ( x) =

 ∂  ikx · Aeikx = ( i k ) Ae = ħk(Aeikx) i ∂x i

Hence, function Aeikx is an eigen function of the operator pˆ x and eigen value is ħk.  df  12. If f = xn, show that x   = nf and thus that f is an eigen function of the  dx  d  operator x   .  dx  d Ans. Let,  = x and f = xn dx  = x

…(1) d n (x ) dx

Âf = x · nxn–1 Âf = nxn  d   x  f = nf  dx  \

[using (1)]

 df  x   = nf  dx 

Thus, the function ‘f ’ is an eigen function of the given operator. 13. Schrödinger equation is an eigen value equation justify. Ans. If an operator  operates on a well behaved function f (x) and give the same function multiplied by a constant. The expression involved is eigen value equation Âf (x) = a f(x)

(eigen value equation)

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105

In Schördinger equation, Ĥψ = Eψ, the Hamiltonian operator (Ĥ) operates on well behaved function, ψ and give the same function multiplied by a constant, E. Thus, such equation is an eigen value equation.  d2  d +2 + 3  , what 2 dx  dx 

14. Show that eax is an eigen function of the operator  is the eigen value? Ans.

f (x) =

eax

d2 d +3 and  = 2 + 2 dx dx

Using eigen value euqation: d 2 ax d (e ) + 2 (e ax ) + 3e ax 2 dx dx 2 ax ax = a e + 2ae + 3eax Âf (x) = (a2 + 2a + 3)eax Âf (x) = (a2 + 2a + 3)f (x) Thus, eax is an eigen function of the given operator and eigen value is (a2 + 2a + 3). −h 2 d 2 15. Show that ψ = 3 cos 2x is an eigen function of the operator . 4π2 dx 2 What is the eigen value? Âf (x) =

Ans. Here, Â =

−h 2 d 2 and f (x) = 3 cos 2x 4π2 dx 2 Âψ =

−h 2 d 2 (3 cos 2x) 4π2 dx 2

+h 2 (3 × 4 × cos 2x) = 4π 2 4 h2 (3 cos 2x) = 4 π2 Âf(x) =

h2

f(x) π2 Thus, function ψ is an eigen function of given operator  and eigen value  h2  is  2  . π  16. Show that the function  − x2  ψ(x) = Cx exp    2 

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where C is a constant, is an eigen function of the operator:   − d 2  2  2  + x   dx   What is the eigen value.   ∂ 2  2 Ans. Here, Aˆ =  − + x   2     ∂x 

and

2 ψ(x) = C · x exp  − x   2   

   ∂ 2   − x2  ˆA ψ(x) =  − 2  + x 2  ⋅ C ⋅ x ⋅ exp   2        ∂x     = −

∂2 ∂x 2

= −C⋅

  − x2  2 C ⋅ x exp    + x   2  

∂ ∂x

= −C⋅

  − x2  C ⋅ x ⋅ exp      2  

∂   − x2   − x2   3 + ⋅ C x exp   x ⋅ exp         ∂x   2   2   

 − x2   − x2   − 2 x  ∂  exp   + x ⋅ exp     ∂x   2   2   2    − x2  + C ⋅ x3 exp   2   

= −C

 − x2   − x2   − x2  ∂  2 3 exp     + Cx ⋅ exp   + x ⋅ exp  ∂x   2   2    2 

  − x2   − x2   − 2 x   C exp x − ⋅ − exp 2        =     2   2  2    − x2   − x2   − 2 x   3 Cx + ⋅ + x 2 exp  exp         2   2   2      − x2    − x2   − x2  3 − + x exp x exp 2 = C  x exp          2    2   2   − x2  + C x3 exp    2 

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=

Aˆ ψ(x) = 3ψ(x) Hence proved, ψ(x) is an eigen function of operator and eigen value is (3).

Problems based on hermitian oPerators 1. Show that the operator pˆ x for linear momentum is Hermitian. Ans. If ψ and φ are the two eigen functions of the operator Aˆ , then the operator Aˆ is Hermitian if: +∞

+∞

∫−∞ ψ * ( Aˆ φ) d τ = ∫−∞ φ( Aˆ ψ) * d τ Here, Â = pˆ x =

 ∂  ∂ and pˆ *x = − i ∂x i ∂x

It is required to prove that +∞

∫−∞ ψ * ( pˆ x φ) dx

+∞

=

∫−∞ φ pˆ x * ψ * dx

Taking LHS and integrating by parts: +∞

+∞

∫−∞ ψ*pˆ x φ dx = ∫−∞ =

 i

 d  ψ * φ dx  i dx  +∞

dφ

∫−∞ ψ * dx dx

Integration by parts:

∫ udv = uv − ∫ v du Here,

u = ψ* and v = φ =

ℏ  [ψ * φ]+∞ − +∞ φ d ψ * dx −∞    ∫−∞ dx i   0 = 

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Since both the functions ψ and φ vanish at ∞, thus the first term is zero and thus we are left with: −  +∞ d ψ * φ dx = i ∫−∞ dx +∞  −  d  = ∫−∞ φ   ψ * dx  i dx  +∞

=

∫−∞ φ pˆ x *ψ*dx

= RHS

Thus, linear momentum operator ( pˆ x ) is Hermitian. 2. (a) Show that kinetic energy operator ( Tˆx ) is Hermitian. Ans. It ψ and φ are the two eigen functions of the operator Â, then the operator  is Hermitian if +∞

+∞

∫−∞ ψ * ( Aˆ φ) d τ = ∫−∞ φ ( Aˆ ψ) * d τ 2 − 2 d 2 ˆ * = + d T Here, Â = Tˆx = and x 2m dx 2 2m dx 2

It is required to prove that:

∫

+∞

−∞

ψ*(Tˆx φ) dx =

∫

+∞

−∞

φTˆx*ψ* dx

Taking LHS and integrating by parts: +∞

∫−∞

+∞

ψ*Tˆx φ dx =

 − 2 d 2  ψ * ∫  2m dx 2  φ dx   −∞

d 2φ − 2 d  dφ  − 2 ψ * dx ψ *   dx = = 2 ∫ 2m ∫ 2m dx  dx  dx Integrating by parts:

∫ u • dv = uv − ∫ vdu Here,

u = ψ* and v = −ℏ 2 = 2m

dφ dx

+∞  +∞ d φ d ψ * dφ   ψ *  − ∫ • dx −∞  dx  −∞ dx dx   =0 

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109

Since both the functions ψ and φ vanish at ∞ because they are well behaved functions, thus the first term becomes zero and we are left with: =

+2 2m +∞

+∞

∫−∞ φ

d 2ψ * dx dx 2

 +2 d 2  φ  2  ψ * dx  2m dx 

=

∫−∞

=

∫−∞ φTˆ*x ψ* dx = RHS

+∞

Thus, kinetic energy operator (Tˆx ) is Hermitian. ˆ is Hermitian. (b) Show that potential energy operator V(x) Ans. If ψ and φ are the two eigen functions of the operator Â, then the operator Â, is Hermitian if +∞

+∞

∫−∞ ψ*(Aˆ φ) d τ = ∫−∞ φ( Aˆ ψ)* d τ For the potential energy part, Â = Vˆ ( x) = V(x) +∞

+∞

∫−∞ ψ*[Vˆ ( x)φ] dx = ∫−∞ φ [Vˆ ( x) · ψ ]* dx (because V(x) just multiplies). Thus, potential energy operator Vˆ ( x) is Hermitian. (c) Prove that Hamiltonian operator (Ĥ) for the total energy of a system is Hermitian, provided that the wavefunction is well behaved. Ĥ = Tˆx + Vˆ ( x)

Ans.

Ĥ=

[in one-dimension]

−2 d 2 + V ( x) 2m dx 2

Assuming the wavefunctions to be real, it is required to prove that +∞

+∞

∫−∞ ψ * (Ĥφ) dx = ∫−∞ φ ( Hˆ ψ ) * dx For the potential energy part i.e., Vˆ ( x) +∞

+∞

∫−∞ ψ*[Vˆ ( x)φ] dx = ∫−∞ φ [Vˆ ( x)ψ]*dx [because V(x) just multiplies] − 2 in just a constant factor. Thus, both Vˆ ( x) and Tˆx are Hermitian, while 2m Thus, Ĥ is Hermitian.

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3. Explain the term Hermitian operators. Give an example of Hermitian operator. Ans. Suppose ψ and φ are the two eigen functions of the operator  and if +∞

+∞

∫−∞ ψ * (Âφ) dτ = ∫−∞ φ (Âψ)* dτ Then the operator  is called Hermitian operator. Hermitian operators give real values. Like: Linear momentum and kinetic energy operators are Hermitian operators. 4. Show that “Hermitian operators” have real eigen values. What is the significance of this property? Ans. Let  be a Hermitian operator, ψ its eigen function and ‘a’ the eigen value; then, and

Âψ = aψ (Âψ)* = a*ψ*

…(1) …(2)

Multiply eq. (1) by ψ* and on integration.

∫ ψ * (Âψ)dτ = ∫ ψ * (aψ)d τ ∫ ψ * (Âψ)dτ = a ∫ ψ *ψd τ

…(3)

Multiply eq. (2) by ψ and on integration

∫ ψ (Âψ)*dτ = ∫ ψ(a * ψ*)d τ ∫ ψ (Âψ)*dτ = a * ∫ ψψ * d τ

…(4)

LHS of eq. (3) and eq. (4) is equal, since  is Hermitian. i.e.,

∫ ψ * (Â ψ) dτ = ∫ ψ (Âψ)*dτ

Therefore, RHS of eq. (3) and eq. (4) is also equal a ∫ ψ * ψd τ = a * ∫ ψψ * d τ a = a* i.e., a is real. Significance: The real values can be applied to real systems in the universe. Imaginary values can only be applied to hypothetical system. ˆ ˆ is 5. Show that if two operators  and Bˆ are Hermitian, their product AB also Hermitian, if and only if operators  and Bˆ commute. i.e.,

ˆ ˆ = BA ˆˆ, Ans. Since AB ˆ ˆ ) φ dτ = ∫ ψ * ( BA ˆ ˆ ) φ dτ thus, ∫ ψ * ( AB = ∫ ψ * Bˆ (Âφ) dτ

3.9 Operators

111

Since Bˆ is Hermitian, the integral is equal to

∫ ( Aˆ φ)( Bˆ ψ)* d τ = ∫ ( Bˆ ψ )*( Aˆ φ) d τ = ∫ φAˆ *(Bˆ ψ)* d τ ˆˆ = ∫ φ ( AB )* ψ* d τ

(because  is Hermitian)

ˆ ˆ ) is also Hermitian. Therefore, ( AB 6. Derive the condition for the eigen functions of Hermitian operators to be orthogonal. Ans. Let ψ1 and ψ2 be two eigen functions of a Hermitian operator  corresponding to two eigen values a1 and a2 respectively. The condition of orthogonality is

∫ ψ1 ψ 2 d τ = 0 or

∫ ψ1 ψ 2 * d τ = 0

or

∫ ψ1 * ψ 2 d τ = 0

Proof: The eigen value equations are: Âψ1 = a1ψ1 Âψ2 = a2ψ2

…(1) …(2)

Multiply eq. (1) by ψ2* and on integration

∫ ψ 2 *( Aˆ ψ1)d τ = ∫ ψ 2 * (a1ψ1 ) d τ ∫ ψ 2 *( Aˆ ψ1)d τ = a1 ∫ ψ 2 * ψ1 d τ

…(3)

Since  is Hermitian, thus LHS of eq. (3) becomes

∫ ψ 2 *( Aˆ ψ1)d τ = ∫ ψ1 ( Aˆ ψ 2 )* d τ = ∫ ψ1 (a2 ψ 2 ) * d τ = a2 * ∫ ψ1ψ 2 * d τ ∫ ψ 2 *( Aˆ ψ1) d τ = a2 ∫ ψ1ψ 2 * d τ

[using eq. (2)]

…(4)

[a2* = a2 because Hermitian operators have real values] Equating RHS of eq. (3) and eq. (4): a1 ∫ ψ 2 * ψ1d τ = a2 ∫ ψ1ψ 2 * d τ (a1 − a2 ) ∫ ψ1ψ 2 * d τ = 0

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But, a1 ≠ a2 \

∫ ψ1ψ 2 * d τ

=0

i.e., ψ1 and ψ2 are orthogonal to each other.

3.10 POSTULATES OF QUANTUM MECHANICS Quantum mechanics is based on few principles or postulates. 1. Postulate I: Wavefunction or state function is an important property in quantum mechanics. To every state of a physical system, there is a function ψ which defines the state of the system. Knowledge of this function permits a complete description of the state of the system. • The wavefunction ψ (r, t) depends on the space co-ordinates of the particles, r = (x1, y1 z1, x2, y2, z2...) and on time, t. Suppose there is a dynamic system with ‘n’ particles (1, 2, 3, ...n) and their space co-ordinates are (x1, y1, z1), (x2, y2, z2) etc. Then the state of these particles can be explained by a wavefunction, ψ . ψ (r, t) = ψ (x1, y1, z1, x2, y2, z2, …, xn, yn, zn, t). Generally, the wavefunction is a complex function, which means we have to do with its complex conjugate ψ *, as well, obtained by changing the imaginary number i = −1 into –i, [Basically, this imaginary number, ‘i’, is not present in classical mechanics but is an inherent property which is present in quantum mechanics]. Thus, we calculate +∞

∫−∞

ψ (r , t ) · ψ *(r , t )d τ =

+∞

∫−∞ ψ d τ 2

By doing this, the imaginary number ‘i’ vanishes. This ψ 2 is the probability of finding the particle in a particular volume element dτ = dx1, dy1, dz1, dx2, dy2, dz2, ..., located at distance r and time, t. • If the wavefunction depends on time, t, then called time dependent wavefunction, ψ (r, t). But, if it is independent of time, t, then called time independent wavefunction, ψ(r). ψ(r , t ) = ψ(r) · φ(t). ψ(x, y, z, t) = ψ(x, y, z) · φ(t) • Actually, when we speak about evolution, the time dependent wavefunction, ψ (r, t) is taken into account. But when we speak about the state of a physical system, then time independent wavefunction, ψ(r) is taken into account.

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113

• Thus, according to this postulate, state of a physical system is described by wavefunction ψ and all the information about a state is contained inside ψ. It is actually the store house of information. 2. Postulate II: The function ψ must be single valued, finite, smoothly varying and continuous between the boundaries at which it vanishes, but at the boundary the wavefunction has to vanish. Thus, ψ vanishes at infinity i.e., probability of finding the particle at infinity is zero. The function which possess all these properties is a well behaved wavefunction. Dynamic variable: It is any property of the system. Like energy, position, velocity, momentum etc. Observable: It is that property which can be experimentally observed. Usually, dynamic variable is an observable. But spin component and angle ‘x’ are dynamic variable and not observable. 3. Postulate III: For every observable of a system in classical mechanics, (like position, velocity, momentum, energy etc.) there corresponds a mathematical operator in quantum mechanics. The form of the operator can be obtained from the classical expressions for the observables according to the definite rules. The physical properties of the observable can be infered from the mathematical properties of its associated operator. Table 3.1: Classical mechanical observable and their corresponding quantum mechanical operators [Refer Sec. 3.9: Operators] Observable Name Symbol Position x

Symbol xˆ

Operator Operation multiplication by x

x-component of momentum

px

pˆ x

Square of x-component of momentum

px2

pˆ x2

Linear momentum

p

pˆ

 ∇ i

Square of linear momentum

p2

pˆ 2

–ħ2∇2

Lx

Lˆ x

 ∂ ∂  y −z  i  ∂z ∂y 

x-component of angular momentum

 ∂ i ∂x −

∂2 ∂x 2

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pˆ 2 Tˆx = 2m

−2 2 ∇ 2m

V

Vˆ ( x, y , z )

multiplication by V

E=T+V

Hˆ= Tˆ + Vˆ

−2 2 ∇ + V ( x, y , z ) 2m

Kinetic energy

T=

Potential energy Total energy

p2 2m

In quantum mechanics we deal with Linear Hermitian operators. [Refer sec. 3.9: Operators] A Linear operator  has the property i.e., eq. (3.17) Â[f + g] = Â(f) + Â(g) A Hermitian operator  has the property i.e., eq. (3.18) +∞

+∞

∫−∞ ψ*(Aˆ φ) d τ = ∫−∞ φ(Aˆ ψ)* d τ The Hermitian property of the operator ensures that the value of any observable obtained will be real. 4. Postulate IV: For any measurement of the observable associated with the operator Â, the only value that will ever be observed is the eigen value ‘a’ which can be obtained from eigen value equation, i.e., eq. (3.10). Âψ = aψ An eigen value is just a real number and when an observable has a definite value, the value of the observable is the eigen value. An eigen value equation is that equation in which an operator  when operates upon the wavefunction ψ, it gives a constant called eigen value ‘a’ and wavefunction ψ back. While the eigen function and the operator may be real or complex, but the eigen value must be real because it represents observable physical quantities. Moreover, quantum mechanical operators of interest (like; momentum, energy etc.) are Hermitian operators which have real eigen values. The Hermitian property of the operator ensures that value of any observable obtained will be real. 5. Postulate V: If ψ is an eigen function of operator Â, then Âψ = aψ. But, it has to be noted that ψ is not necessarily an eigen function of Â. When ψ is not an eigen function of Â, then we talk of average values in quantum mechanics.

3.11 Derivation of Time Independent Schrodinger...

115

The expected average (or expectation) value of an observable of a system whose state function is ψ, is given by mean value theorem: +∞

a avergae value

=

=

expectation value

∫−∞ ψ*Aˆ ψ d τ +∞ ∫−∞ ψ*ψ d τ

…(3.27)

Also, if ψ is an eigen function of operator Â, the expectation value is equal to the eigen value ‘a’. +∞

a ==

∫−∞ ψ*(Aˆ ψ) d τ +∞ ∫−∞ ψ*ψ d τ

=

a

+∞

∫−∞ ψ*(aψ) d τ +∞ ∫−∞ ψ*ψ d τ

=

a

+∞

∫−∞ ψ*ψ d τ ∫ ψ*ψ d τ

=a

The integrals are overall permitted values of the coordinate of the system. The expectation value is defined as the prediction of the mean of a large number of measurements of the mechanical quantity.

3.11 DERIVATION OF TIME INDEPENDENT SCHRODINGER... WAVE EQUATION ON THE BASIS OF POSTULATES OF QUANTUM MECHANICS Consider an electron in an atom of mass ‘m’ moving with velocity ‘v’. Its total energy E is the sum of its kinetic energy (T) and potential energy (V), i.e. E=T+V Put,

2 2 2 1 2 p 2 px + p y + pz mv = = T= 2m 2m 2

where, p represent the total momentum of the particle. \ Total energy E =

px2 + p 2y + pz2 2m

+V

…(3.28)

This classical expression of the quantity ‘E’ can be converted to the corresponding quantum mechanical operator for energy by the following replacements h ∂ pˆ x = 2πi ∂x pˆ y = h ∂ 2πi ∂y h ∂ pˆ z = 2πi ∂z V is a function of position co-ordinates and hence the operator Vˆ is V itself.

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The operator for the energy E is the Hamiltonian operator Hˆ . Hence, equation (3.28) takes the form: 1  h ∂   h ∂   h ∂    + Hˆ =  +  2m  2πi ∂x   2πi ∂y   2πi ∂z   2

Hˆ =

2

2

 +V 

−h 2  ∂ 2 ∂2 ∂2  + +  +V  8π2 m  ∂x 2 ∂y 2 ∂z 2 

−h2 2 ∇ +V Hˆ = 8π2 m

…(3.29)

∂2 ∂2 ∂2 + + is the Laplacian operator ∂x 2 ∂y 2 ∂z 2 Further according to the fourth postulate of quantum mechanics, i.e., eq. (3.10). Âψ = aψ (eigen value equation) Therefore,we must have i.e., eq. (3.9): Hˆ ψ = Eψ ∇2 =

where,

where, E is the eigen value of the energy for the given state of the system. Substituting the value of Hˆ from equation (3.29)   h2 2  − 2 ∇ + V  ψ = Eψ   8π m which can be rearranged and written in the from −

h2 ∇ 2 ψ + V ψ = Eψ 2 8π m

h2 ∇ 2 ψ + (E − V )ψ = 0 2 8π m ∇2ψ +

8π 2 m ( E − V )ψ = 0 h2

This is the familiar fundamental form of time independent Schördinger wave equation and is same as eq. (3.7) as derived earlier in sec. 3.3.

3.12 Steady State SchrÖdinger Wave Equation

117

3.12 STEADY STATE SCHRÖDINGER WAVE EQUATION The possible state function ( ψ ) of a system evolves in time. For every system, time dependent Schrodinger equation is given as: −  ∂ψ …(3.30) Ĥψ = i ∂t where, Ĥ is the Hamiltonian operator. Ĥ is a function of position co-ordinates (x, y, z) as well as time (t). The potential energy V also depends on position co-ordinates (x, y, z) and time (t). Thus, time dependent Schrodinger equation is given as:

  − 2 2 −  ∂ψ ( x, y, z , t ) ∇ + V ( x, y , z , t )  ψ =  ∂t i  2m  Since this is a partial differential equation involving both position and time variables, it can be solved only by separating it into two differential equations, one containing position variables and other containing time variables only. This can be done if we assume that ψ (x, y, z, t) = ψ(x, y, z) · φ(t)

…(3.31)

Put eq. (3.31) in eq. (3.30): Ĥ[ψ(x, y, z) · φ(t)] =

−  ∂ [ψ ( x, y, z ) • φ(t )] i ∂t

Suppose there is a system whose potential energy, V depends only on position co-ordinates and independent on time ‘t’. Since potential energy, V is independent of time ‘t’, therefore, Ĥ is also independent of time. This is because Ĥ = Tˆ + Vˆ . Thus, for time independent system Schordinger equation becomes,

− ∂ φ(t ) ψ ( x, y , z , ) • i ∂t Dividing both the sides by ψ(x, y, z) · φ(t) φ(t) · Ĥψ (x, y, z) =

φ(t ) · Hˆ ψ ( x, y, z ) ψ ( x, y, z ) · φ(t )

=

ψ ( x, y , z ) − ∂φ(t ) • i ψ ( x, y, z ) • φ(t ) ∂t

−  1 ∂φ(t ) Hˆ ψ ( x, y, z ) • • = i φ(t ) ∂t ψ ( x, y , z ) Now the left hand side of the above equation is independent of time ‘t’ and the right hand side is independent of position co-ordinates. Let each side of this equation may be equated to a constant, say E i.e.,

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−  1 ∂φ(t ) Hˆ ψ ( x, y, z ) • • =E = i φ(t ) ∂t ψ ( x, y , z )

…(3.32)

Thus, LHS of eq. (3.32) is given as: Hˆ ψ ( x, y, z ) =E ψ ( x, y , z ) Ĥψ(x, y, z) = Eψ(x, y, z) This is the familiar fundamental form of time independent Schrödinger wave equation, also called as the amplitude equation. Similarly, RHS of eq. (3.32) is given as: −  1 ∂φ(t ) • • =E i φ(t ) ∂t Rearranging and integrating both the sides: ∂φ(t )

∫ φ(t )

=

ln φ(t) =

−iE dt  ∫ −iEt + ln N 

where, N is a constant. Taking exponential on both sides, we get φ(t) = Ne–iEt/ħ This is the time-dependent part of the Schrödinger wave equation. • Thus, the total wavefunction would be ψ( x, y, z , t ) = ψ(x, y, z) · φ(t) ψ( x, y, z , t ) = ψ(x, y, z) · Ne–iEt/ħ • The total probability density would be ψ2 = where,

+∞

∫−∞ ψ ψ * d τ

ψ = ψ(x, y, z) · Ne–iEt/ħ

ψ * = ψ*(x, y, z) · Ne+iEt/ħ

3.12 Steady State SchrÖdinger Wave Equation

ψ2 =

119

+∞

∫−∞ ψ( x, y, z ) • N • e

− iEt / 

 ψ*(x, y, z ) • N • e + iEt /   d τ   

2 ψ2 = N ∫ ψψ* d τ

Thus, total probability density ψ2, is independent of time ‘t’. • The average value or expectation value of any observable property of the system would be +∞

= a=

∫−∞ ψ * · Aˆ · ψ d τ +∞ ∫−∞ ψ * ψ d τ

+∞

a =

a =

If ψ is normalised, Thus,

+∞

· Aˆ · (ψ · N e )d τ ∫−∞ (ψ* · N e +∞ + iEt /  ) · (ψ · N e − iEt /  )dτ ∫−∞ (ψ* · N e N2 ∫

+∞

∫

− iEt / 

ψ* Aˆ ψ d τ

−∞ 2 +∞ N −∞

∫−∞ ψ*ψd τ a =

+ iEt / 

ψ* ψ d τ

=1

+∞

∫−∞ ψ* Aˆ ψd τ

Therefore, any observable property is independent of time ‘t’ factor. Such a system is said to be stationary state and all its properties can be obtained by solving the time independent Schrodinger equation i.e.,

Hˆ ψ(x, y, z) = E ψ(x, y, z)

A stationary state does not imply that the particle or particles of the system are at rest. It is stationary in the sense that the probability density, ψ2 and every physical property of the system (like; energy, charge density, dipole moment etc.) will be independent of time. Thus, the factor (e–iEt/ħ) is of no significance and the essential part of the wavefunction for a stationary state is the time-independent wavefunction ψ.

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Problems based on stationary state Q. The state of a system is described by the function ψ(r , t ) , which is a function of both position and time, but the average value of any physical quantity is independent of time, why? ψ(r , t ) = ψ(r) · φ(t) Ans. ψ(r , t ) = ψ(r) ·Ne–iEt/ħ The average value of any observable property of the system would be +∞

a =

∫−∞ ψ * · Aˆ · ψ d τ +∞ ∫−∞ ψ * · ψ d τ

a =

) · Aˆ · (ψ · N e ) dτ ∫−∞ (ψ* · N e +∞ + iEt /  ) · (ψ · N e − iEt /  ) d τ ∫−∞ (ψ* N e

+∞

a = If ψ is normalised, Thus,

N2 ∫

+ iEt / 

+∞

ψ* Aˆ ψd τ

−∞ 2 +∞ N −∞

∫

− iEt / 

ψ* ψd τ

+∞

∫−∞ ψ* ψd τ = 1 a =

+∞

∫−∞ ψ* Aˆ ψd τ

Therefore, any observable property is independent of time ‘t’ factor. Such a system is said to be stationary state. rrr

4 Particle in a Box: Quantisation of Translational Energy 4.1

APPLICATION OF POSTULATES OF QUANTUM MECHANICS TO SIMPLE SYSTEM

All the properties and motion of a particle of mass ‘m’ which moves with potential energy, V(x, y, z) can be obtained from solution of well behaved Schrodinger equation − 2 2 ∇ ψ + V ( x, y, z )ψ = Eψ 2m If we look at this equation, system depends on mass ‘m’ and potential energy, ‘V’ of the particle. So one system differ from another system on the basis of mass and potential energy. Thus, we find the solution of this Schrodinger equation for different systems by varying the mass and potential energy of the system. We can find the exact solution to this Schrodinger equation only for very simple system. In case of complex system, the exact solution cannot be determined and if it is determined, the answer is so elaborate that it is difficult to understand. So most of the complex systems are solved by using approximation method.

4.2

OPERATION OF QUANTUM MECHANICS

It involves four steps: 1. Writing the Schrodinger Equation: Hˆ ψ = Eψ (Tˆ + Vˆ )ψ = Eψ

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According to quantum mechanics, microscopic particles in an enclosure is never at rest and hence kinetic energy (Tˆ ) is never equal to zero but potential energy (Vˆ ) can be zero. Thus, first we need to define potential energy of the system and then Hamiltonian ( Hˆ ) is written for the system. Also we need to define wavefunction (ψ) for the system. Wavefunction is always system specific and is also called store house of the information of the system. Total number of terms in the Hamiltonian: Total energy = K.E + P.E where, Number of K.E. terms = Number of particles in the system (each particle has its own K.E. term) Number of P.E. terms = Attractive P.E. terms + Repulsive P.E. terms

P.E. term + electron-electron P.E. term = electron-nuclei   attractive term

repulsive term

+ nuclei-nuclei P.E. term   repulsive term

Number of electron-nuclei P.E. term = n · e …(1) e(e − 1) …(2) Number of electron-electron P.E. term = 2 n(n − 1) …(3) Number of nuclei-nuclei P.E. term = 2 Number of K.E. terms = n + e Total number of terms = K.E. terms + P.E. terms [(1) + (2) + (3)] e(e − 1) n(n − 1)   + = [n + e] +  ne + 2 2   (n + e)(n + e − 1) = ( n + e) + 2 =

(n + e)(n + e + 1) 2

where, n = Total number of nuclei present in the system and e = Total number of electron present in the system. 2. Specifying Boundary Conditions: The restriction imposed on the system are the boundaries. On applying the boundary conditions on

4.3 Introduction to Translational Motion ofa Particle

123

the wavefunction, we satisfy the necessary condition of a wavefunction to be acceptable. In other words, ψ value obtained must satisfy the boundary conditions, if it does not satisfy then that ψ is not accepted. 3. Solving the Schrodinger Equation to get ψ and E Values: Once Schrodinger equation is obtained with boundary conditions, then we solve for the general solution of the wavefunction and energy. 4. Calculation of Properties Using Wavefunction: Properties like momentum, position etc. for a system can be solved using: (a) Eigen Value Equation: This gives exact solution of the property, i.e., eigen value, using mathematical relation ˆ = aψ Aψ where,

 = operator of the property a = eigen value of the property which should be a constant value. (b) Mean Value Theorem: Some systems do not give constant value for the eigen value i.e. they cannot give exact solution for the property. In that case, average value of the property is calculated using mean value theorem ∫ ψ*Aˆ ψ d τ a = 〈a〉 = ∫ ψ*ψ d τ where, a or 〈a〉 = average value of the property. If the system is normalised, the value of denominator becomes unity i.e., ∫ ψ*ψ d τ = 1 \

4.3

a = 〈a〉 =

∫ ψ*Aˆ ψ d τ .

INTRODUCTION TO TRANSLATIONAL MOTION OF A PARTICLE

The most simple type of motion that a particle (atom or molecule) can undergo is translational or linear motion. In general, translational motion is the kinetic motion of a particle. However, an atom or molecule can execute translational, rotational, vibrational and electronic motions, but in free state they move linearly only. In this chapter we deal with the linear motion of a particle under quantum mechanical bound state (i.e., boundary conditions) and under free state (i.e., without boundary conditions).

124

4.4

Chapter 4

Particle in a Box: Quantisation of Translational Energy

PARTICLE IN ONE DIMENSIONAL BOX: QUANTISATION OF TRANSLATIONAL ENERGY

• This is the simplest application of Schrödinger wave equation to the translational motion of a particle (atom or molecule) in space. According to classical mechanics, energy of a moving particle can vary continuously i.e., can have any value. But the quantum mechanical results obtained from the solution of Schrödinger wave equation for the translational motion of a particle in space, we can explain that, why the energies are quantised i.e., can have only discrete values. • “Particle in a box” model serves as the simplest case for the treatment of bound electrons in atoms and molecules. It is a hypothetical case, in which we assume that a single particle of mass ‘m’ is restricted to move in an imaginary one dimensional box of length ‘a’ with infinitely high walls at the two ends. Thus, the boundary conditions are defined by x = 0 to x = a. As long as particle is inside the box we assume potential energy, V = 0, which implies that particle is free to move inside the box, (i.e., the particle is not under any external force and its potential does not vary inside the box). In order that the particle may remain inside the box, and does not move outside the boundary conditions, it is essential to assume that the potential energy outside the walls is very high i.e., equals to infinity, so that as soon as the particle reaches the wall, it is reflected back into the box due to high potential difference. Thus, particle is just confined to move in a box of length ‘a’. This model is also called particle in a potential well (Fig. 4.1).  V= x=0

 V=0

x

V=

x=a

Fig. 4.1. A particle in one dimensional box.

4.4.1 Solution of Schrödinger Wave Equation To calculate energy for particle in a box, we have to first write the Hamiltonian 2 2 ∇ + V ( x, y , z ) Hˆ = − 2m

4.4 Particle in One Dimensional Box: Quantisation of Translational Energy

125

If we take a particle in one dimensional box moving only in x-direction where the limits in x-direction are from x = 0 to x = a, the Hamiltonian and thus the Schrodinger equation would be given as 2 ∂ 2 Hˆ = − + V ( x) 2m ∂x 2 Hˆ ψ = Eψ

Thus,

 2 ∂ 2  + V ( x)  ψ = Eψ − 2  2m ∂x  (i) Schrödinger equation when particle is outside the box, (V = ∞):

− +

 2 ∂ 2ψ + ∞ψ = Eψ 2m ∂x 2

 2 ∂ 2ψ + ( E − ∞ )ψ = 0 2m ∂x 2

As E 2 always The quantity on RHS is > because 4π 3 h 4π  Dx · Dpx ≥ 2 which satisfies the Heisenberg uncertainty principle.

\

Dx · Dpx ≥

6. Kinetic Energy: We solve for exact kinetic energy using eigen value equation. Tˆ = −

2 ∂ 2 2m ∂x 2

and

ψ=

2  nπx  sin   a  a

4.4 Particle in One Dimensional Box: Quantisation of Translational Energy

2 ∂2 Tˆ · ψ = − 2 m ∂x 2

2  nπx  sin   a  a

2 Tˆ ⋅ ψ = − 2m

nπx  2 ∂2  sin  2  a  a ∂x 

2 Tˆ · ψ = − 2m

2  nπ   nπx    ⋅  − sin  a a   a 

h2

149

2

2 n 2 π2  nπx  sin  ⋅  2 2 a  a  8π m a

Tˆ · ψ =

⋅

n2h2 Tˆ · ψ = 8ma 2

 2 n2h2  nπx   sin  ⋅  ⋅ (ψ )   =  a   8ma 2  a

Therefore, exact solution for kinetic energy is possible. Thus ψ is an eigen function of Tˆ and its eigen value is n2h2/8ma2.

Problems based on Particle in a one dimensional box  nπx  1. ψ = A sin  ; (n = 1, 2, 3, …), where a = length of one-dimensional box.  a  Find the value of normalisation constant ‘A’ . Use the usual assumptions when dealing with a particle in a box problem.  nπx  Ans. ψ = A sin    a  The condition for normalisation is +∞

∫−∞ ψ*ψ d τ = 1

 nπx  ψ = ψ* = A sin    a 

Here a

∫0

[Normalisation condition]

  nπx    nπx     A sin  a    A sin  a   dx = 1       A2

Put

nπx =y a

∫0

 nπx  sin 2   dx = 1  a 

⇒

nπ dx = dy a

a

⇒ dx =

a dy nπ

150

Chapter 4

Particle in a Box: Quantisation of Translational Energy

nπx = y, Using the boundary conditions on a If x = 0, y = 0 and x = a, y = nπ Thus, the above integral becomes nπ

A2

∫0

sin 2 y ⋅

A2 a nπ

∫0

nπ

a dy = 1 nπ

sin 2 y dy = 1 2 sin2 y = 1 – cos 2y 1 − cos 2 y sin2 y = 2

A2 a nπ

nπ

∫0

A2 a 1 × nπ 2 A2 a  2nπ 

nπ

∫0

 1 − cos 2 y   dy = 1  2  

nπ

∫0

dy −

(1 − cos 2 y ) dy = 1 nπ

∫0

cos 2 y dy  = 1  nπ

sin 2 y  A2 a  =1 y−  2nπ  2  0 A2 a 2nπ

  sin 2nπ   − (0 − sin 0)  = 1  nπ − 2    A2 a [ nπ ] = 1 2 nπ A2 = A=

( sin 0 = 0 and sin 2nπ = 0)

2 a 2 . a

2. Calculate the ground state energy of an electron which is moving in a onedimensional box of length 0.2 nm. Ans. The energy of a particle moving in one dimensional box is given as E=

n2h2 8ma 2

4.4 Particle in One Dimensional Box: Quantisation of Translational Energy

151

In ground state, n = 1 Mass of electron = 9.11 × 10–31 kg Length of box = a = 0.2 nm = 0.2 × 10–9 m Planck’s constant = h = 6.6 × 10–34 Js E1 =

(1) 2 (6.6 × 10−34 ) 2 8 × (9.11 × 10−31 ) × (0.2 × 10−9 ) 2

E1 = 1.506 × 10–18 Joules. 3. For a particle in a state n = 1 moving in a one-dimensional box. Calculate the probability of finding the particle anywhere in the region. a Ans. (i) 0 to 4 For particle moving in one dimensional box. 2  nπx  sin   a  a  2  πx  sin    a a ψ=

If n = 1, ψ = a/4

∫0

2 πx dx sin 2 a a 2πx   1 − cos 2 a/ 4  a  dx =   a 0  2 

ψ 2 dx =

a/4

∫0

∫

1 − cos 2 x   2 Q sin x =  2 =

2 1 a  2

a/4

∫0

dx −

1 2

a/4

∫0

2 1 a/4 1 a =  [ x ]0 − × a  2 2 2π

a/4

∫0

cos

2πx  dx  a  a/4

2πx   sin a  0 

=

π 2 1  a  a    sin − sin 0    − a  2  4  4π  2 

=

2 a a  − (1 − 0)   a  8 4π 

ψ 2 dx =

2 a a  1 1 − = − . a  8 4π  4 2π

[sin

 

π = 1 and sin 0 = 0] 2

152

Chapter 4

(ii) 0 to

Particle in a Box: Quantisation of Translational Energy

a 2 ψ2 =

a/2

∫0

2 = a

a/4

∫0

2 πx dx sin 2 a a 2πx   1 − cos a    dx 2  

a/ 2

∫0

2πx  dx  a 

=

2 a

1  2

=

2 a

a/2  1 a/2 1 a  2πx   [ ] − × sin x   0 a  0  2 2π   2

=

 2 1  a  a (sin π − sin 0)   −  a  2  2  4π 

=

 2 a a ( 0 − 0)   − a  4 4π 

ψ 2 dx =

a/2

∫0

dx −

1 2

a/2

∫0

cos

( sin π = 0, sin 0 = 0)

1 2 a × = . 2 a 4

4. Calculate the expectation value of energy for a particle in a one dimensional box. Ans. Total energy (E) for a particle in a one dimensional box is E= T+V when particle is moving inside the box, its potential energy (V) is zero. Thus

E= T

or

Ĥ = Tˆ .

For a particle in one-dimensional box, moving only in x-direction and limits in x-direction are from x = 0 to x = a. ψ=

2 ∂ 2 2  nπx  sin  and Tˆ = −   a  a 2m ∂x 2 a

〈T〉 =

∫ 0 ψ(Tˆ ) ⋅ ψ* dx a ∫ 0 ψψ* dx

4.4 Particle in One Dimensional Box: Quantisation of Translational Energy

2 nπx   2 d 2  sin ⋅ − ∫0 a a  2m dx 2  〈T〉 = a 2 nπx 2 ∫0 a sin a ⋅ a sin Solving numerator:

2 nπx dx sin a a

a

2 2 = − × a 2m

∫0

2 2 = − × a 2m

∫0

2 2 = − × a 2m

∫0

2 2 = − × a 2m

a

a

a

a

∫0

sin

nπx d 2 nπx dx ⋅ 2 sin a dx a

sin

nπx d  d  nπx   ⋅  dx  sin  a dx  dx  a  

sin

nπx d  nπ nπx  dx ⋅ ⋅ cos  a dx  a a 

sin

nπx  n 2 π2 nπx  ⋅ − sin  dx a  a a 

2  2 n 2 π2 × 2 = + × a 2m a To solve this integral, put

nπx dx a

a

∫0

sin 2

nπx dx a

nπx =y a

nπ dx = dy a a dy dx = nπ Using the boundary conditions on If,

nπx =y a

x = 0, y = 0 x = a, y = nπ

Thus, the above integral becomes: =

2 2 n2π 2 × × 2 a 2m a

=

h2 n2 π 2 a 2 1 × × × × 2 2 a 4π nπ 2m a

nπ

∫0

sin 2 y ×

a dy nπ nπ

∫0

 1 − cos 2 y    dy 2

153

154

Chapter 4

Particle in a Box: Quantisation of Translational Energy

nπ

sin 2 y  a  n2h2 × y− = 3  π 2 2  0 n 4ma  =

n2h2 a    sin 2nπ (nπ − 0) −  × − sin 0    3 2nπ  4ma   2

=

n2h2 a × × nπ 3 2 nπ 4ma

[sin 2nπ = 0 and sin 0 = 0]

n2h2 8ma 2 Solving denominator: =

a

∫0

nπx 2 nπx 2 dx = 1 sin ⋅ sin a a a a [as solved above, refer denominator of example (1) of sec. 4.4.3]

n2h2 n2h2 ÷ 1 = 8ma 2 8ma 2 5. Consider a particle of mass ‘m’ in a one dimensional box of length ‘a’. A 2  nπx  possible solution of this system is ψ = . Evaluate 〈x〉. sin   a  a 〈T〉 =

Thus,

Ans. [Refer section 4.4.3, example (1)]. 6. What is a node? Sketch the function ψn and determine the number of nodes for a particle in a one dimensional box of length ‘a’ corresponding to n = 1 to n = 3. Ans. Node is the point in the box where ψ = 0. It is the general property of wavefunction, that greater the number of nodes in a wavefunction, more is the curvature in the particle wave. 3

n = 3, node = 2 node

node

2

n = 2, node = 1 node

1 x=0

n = 1, node = 0 a x= 2

x=a

4.4 Particle in One Dimensional Box: Quantisation of Translational Energy

155

7. What is the unit of wavefunction of particle in a one-dimensional box? Ans. For a particle in one dimensional box: 2  nπx  sin   a  a

ψ=

Unit of ψ = (cm)–1/2. 8. Consider a particle of mass ‘m’ confined in a one-dimensional box of length ‘a’. A possible solution of this system is 2  nπx  sin   a  a 

ψn =

(i) Show that ψn is not an eigen function of pˆ x but is an eigen function of pˆ x2 . Calculate its eigen value. 2  nπx   ∂ sin  Ans. (a) pˆ x = and ψ =  a  a i ∂x Using eigen value equation, we get:

pˆ x · ψ =

 ∂  2 nπx  sin   i ∂x  a a 

=

 i

=

 nπx  2 nπ  × ×  cos  i a a a 

pˆ x · ψ =

2 ∂  nπx    sin a ∂x  a 

 nπ  2 nπx  cos × i a  a a 

Thus, ψ is not an eigen function of the momentum operator ( pˆ x ) since the function is changing from sin → cos. 2 2 Again pˆ x = − 

∂2 and ψ = ∂x 2 2 pˆ x2 ψ = − 

2  nπx  sin   a  a

∂2 ∂x 2

2 = − ×

 2 nπx  sin   a   a 2  n2h2 nπx  ×  − 2 sin  a  a a 

156

Chapter 4

Particle in a Box: Quantisation of Translational Energy

= + pˆ x2 · ψ =

2 n2h2 × 2 4π2 a

 2 nπx  sin   a   a

n2h2 (ψ ) 4a 2

2 \ ψ is an eigen function of pˆ x and eigen value is n2h2/4a2. (ii) Show that the wavefunctions ψ1 and ψ2 are orthogonal. Ans. Consider two wavefunctions corresponding to n = 1 and n = 2, then 2  πx  sin   ψ1 = a  a 

2  2πx  sin   a  a 

ψ2 = a

∫0

2 = a

2  πx   2πx  2 sin  sin   dx ⋅ a  a   a  a

a

∫0

ψ 2 ⋅ ψ1 dx = a

∫0

 πx   2πx  sin   ⋅ sin   dx a  a     Q  

=

2 a

=

1 a 

2 sin A sin B = cos ( A − B ) − cos ( A + B )  cos ( A − B ) − cos ( A + B )  sin A •sin B =  2 

πx πx   cos (2 − 1) a − cos (2 + 1) a  dx

a

1 2

a

πx 3πx    − cos  dx  cos a a   

∫0

∫0

a

3πx  πx  sin sin 1 a − a  =  3π  a π  a 0  a

a

3πx  πx 1  sin a − 3 × sin a  0  1 1  = ( sin π − sin 0 ) − ( sin 3 π − sin 0 )  π 3  ( sin 0 = 0, sin nπ = 0)

1 a × = a π

4.4 Particle in One Dimensional Box: Quantisation of Translational Energy

157

a

1 [ 0] = 0 π Thus, ψ1 and ψ2 are orthogonal to each other. 9. State Bohr’s correspondence principle. Ans. [Refer sec. 4.4.2, point number. (10)] 10. An electron is constrained to move in one-dimensional box of length 0.15 nm. Calculate (i) Energy of first three energy levels. Ans. For particle in one dimensional box;

∫0 ψ 2 ⋅ ψ1 dx =

n2h2 8ma 2 Mass of particle = Mass of electron = 9.11 × 10–31 kg Length of box = a = 0.15 nm = 0.15 × 10–9 m En =

n = 1; E1 =

(1) 2 × (6.6 × 10−34 ) 2 = 2.39 × 10–19 Joules 8 × (9.11 × 10−31 ) × (0.15 × 10−9 ) 2

n = 2; E2 =

(2) 2 × (6.6 × 10−34 ) 2 = 9.56 × 10–19 Joules 8 × (9.11 × 10−31 ) × (0.15 × 10−9 ) 2

n = 3; E3 =

(3) 2 × (6.6 × 10−34 ) 2 = 21.52 × 10–19 Joules. 8 × (9.11 × 10−31 ) × (0.15 × 10−9 ) 2

(ii) Enegry absorbed when electron goes transition from E1 to E2 and E2 to E3. Ans. DE = Ef – Ei

E2

Ef

h2 4h 2 − (a) DE = E2 – E1 = 8ma 2 8ma 2

E Ei

= 9.56 × 10–19 – 2.39 × 10–19

E1

(Calculated in step (i)] = 7.17 × (b) DE = E3 – E2 =

10–19

Joules

9h 2 4h 2 − 8ma 2 8ma 2

= 21.52 ×

10–19

Ef

– 9.56 ×

10–19

E

(Calculated in step (i)] = 11.96 × 10–19 Joules.

E3

Ei

E2

158

Chapter 4

Particle in a Box: Quantisation of Translational Energy

(iii) If these amount of energy are absorbed as single photons, calculate the wavelength of radiation absorbed. Ans. DE = hv (c = nλ) hc DE = λ hc λ= ∆E h = 6.6 × 10–34 Js and c = 3 × 108 ms–1

Also,

Here,

(a) For E1 → E2; DE = E2 – E1= 7.17 × 10–19 J λ=

[Calculated in step (ii)]

(6.6 × 10−34 ) × (3 × 108 ) = 2.76 × 10–19 m 7.17 × 10−19

(b) For E2 → E3; DE = E3 – E2 = 11.96 × 10–19 J

[Calculated in step (ii)]

(6.6 × 10−34 ) × (3 × 108 ) = 1.66 × 10–7 m. −19 11.96 × 10 11. What are symmetric and anti symmetric functions? Ans. The wavefunction belongs to one of the two classes: (i) symmetric wavefunction (ii) antisymmetric wavefunction The function is symmetric, if it satisfies the condition ψ(–x) = ψ(x) The function is antisymmetric, if it satisfies the condition ψ(–x) = –ψ(x). 12. What do you understand by orthonormal set of wavefunctions? Explain this by taking the wavefunction of a particle in a one dimensional box of length ‘a’ (Given that V = 0 for 0 ≤ x ≤ a and V = ∞ elsewhere). Ans. (i) For two wavefunctions ψ1 and ψ2 to be orthonormal λ=

+∞

+∞

+∞

+∞

∫−∞ ψ1ψ1* = 1 or ∫−∞ ψ 2ψ*2 = 1 and

∫−∞ ψ1ψ 2 = 0 or ∫−∞ ψ 2ψ1 = 0

The wavefunctions for a particle moving in a one dimensional box of length ‘a’ is 2  nπx  ψn = a sin  a   

4.4 Particle in One Dimensional Box: Quantisation of Translational Energy

159

The two wavefunctions corresponding to n = 1 and n = 2 are 2  πx  sin   and ψ2 = a  a 

ψ1 =

2 a

= Put

2  πx   πx  2 sin   ⋅ sin   a  a   a  a

a

a

∫0 ψ1ψ1* dx = ∫0

Now,

πx =y ⇒ a

π dx = dy a

 πx  sin 2   dx  a 

a

∫0

⇒

dx =

a dy π

πx =y a

Using the boundary conditions on If x = 0, y = 0 and x = a, y = π

Thus, the above integral becomes 2 a a 2 π × ⋅ sin 2 y dy = = a π a 0 π

∫

=

2 π

=

2 1 × π 2

=

1 π 

1 = π

\

a

 1 − cos 2 y   dy  2  

π

∫0

π

∫0

π

∫0

(1 − cos 2 y ) dy

dy −

π

∫0

π

∫0

sin 2 y dy

1 − cos 2 y   2  Q sin y = 2

cos 2 y dy   π

sin 2 y    y − 2  0

=

 sin 2π  1   − ( 0 − 0)  π −  π  2  

=

1 ( π − 0) π

=

1 ×π =1 π

∫0 ψ1ψ1* dx = 1

2  2πx  sin   a  a 

[, sin 2π = 0]

160

Chapter 4

Particle in a Box: Quantisation of Translational Energy

a

∫0 ψ 2ψ*2 dx = 1

Similarly,

\ The wavefunction ψ1 and ψ2 are normalised wavefunction. a

Again,

∫0

2  πx   2πx  2 sin  sin   dx ⋅ a a a  a   

a

ψ 2ψ1 dx =

∫0

=

2 a

a

∫0

 πx   2πx  sin   ⋅ sin   dx  a   a 

Since, 2 sin A sin B = cos ( A − B ) − cos ( A + B )   cos ( A − B ) − cos ( A + B )  sin A •cos B =   2   =

2 a

=

1 a 

πx πx   cos (2 − 1) a − cos (2 + 1) a  dx

a

1 2

a

πx 3πx    − cos  dx  cos a a   

∫0

∫0

a

 3πx  πx sin sin  1 a − a  =  a  π  3π        a   a  0

a

  πx  1  3πx   sin  a  − 3 × sin  a   0    

=

1 a × a π

=

1 1  1      sin π − sin 3π  −  sin 0 − ⋅ sin 0   π   3 3     [ sin 0° = 0 and sin nπ = 0]

\

a

∫0 ψ 2ψ1 dx = 0

Similarly,

a

∫0 ψ1ψ 2 = 0

Therefore, the wavefunction ψ1 and ψ2 are orthogonal to each other. Since, both the wavefunctions ψ1 and ψ2 are normalised as well as orthogonal, thus these wavefunctions are orthonormal because Orthonormal = Normalised + Orthogonal.

4.4 Particle in One Dimensional Box: Quantisation of Translational Energy

13. Why is n = 0 not permitted in the equation En = Ans. For particle in a one dimensional box, ψ=

161

n2h2 ? 8ma 2

n2h2 2  nπx  . sin  and E =  a 8ma 2  a 

(i) If n = 0, then ψ = 0 or E = 0. But it is not at all acceptable for a particle moving in a box. Because for n = 0, ψ = 0 and ψ cannot be equal to zero inside the box. Since the particle is existing inside the box and if ψ = 0, then we are violating what we are saying. (ii) If n = 0, ψ = 0, E = 0 and if energy which is the kinetic energy ( Potential energy is assumed to be zero), becomes zero it means  p2  momentum is zero Q K.E. = x  . However, zero momentum means 2m   Dpx = 0 which is the violation of Heisenberg’s principle. 14. Zero point energy need not be equal to zero which Comment. Ans. Zero point energy of a particle in a box is the lowest energy of a particle in a box. The minimum value of n is 1, therefore, the energy corresponding to n = 1 will be the zero point energy (E1) and E1= h2/8ma2. The lowest value of E is not zero but h2/8ma2. This shows that particle is not at rest even at absolute zero or at zero kelvin. Thus, the particle in a box will continue to move to and fro in the region ‘0’ to ‘a’ even at the temperature of absolute zero. 15. Write an energy expression for a particle of mass ‘m’ confined to move in a one-dimensional box of length ‘a’. Calculate the energy required for a particle of mass 1 × 10–30 kg to move from energy level 3 to 4, the length of the box is 973 pm. Ans. For a particle in one dimensional box. n2h2 8ma 2 Mass of particle = m = 1 × 10–30 kg Length of the box = a = 973 pm = 973 × 10–12 m Planck’s constant = h = 6.6 × 10–34 Js E=

n = 3,

E3 =

(3) 2 × (6.6 × 10−34 ) 2 8 × (1 × 10−30 ) × (973 × 10−12 ) 2 Ef

n = 4,

E4 =

(4) 2 × (6.6 × 10−34 ) 2 8 × (1 × 10−30 ) × (973 × 10−12 ) 2

Ei

E4

E3

162

Chapter 4

Particle in a Box: Quantisation of Translational Energy

∆E = Ef – Ei = E4 – E3 (42 − 32 ) × (6.6 × 10−34 ) 2 = 8 × (1 × 10−30 ) × (973 × 10−12 ) 2

∆E = 4.02 × 10–19 Joules. 16. For a particle of mass ‘m’ in one dimensional box of length ‘a’ having wavefunction  nπx  ψ = A sin  .  a  Calculate the energy of the particle.  nπx  ψ = A sin    a 

Ans.

...(1)

∂ψ nπ  nπx  ⋅ cos  = A×  ∂x a  a 

Now, Ĥψ = Eψ

∂ 2ψ n 2 π2  nπx  A − × sin  =  2 2 ∂x a  a  −

 2 ∂ 2ψ = Eψ 2m ∂x 2

…(2)

…(3)

Putting (1) and (2) in (3) 2  n2h2  nπx    nπx  − ×  − A × 2 ⋅ sin    = E × A sin   2m  a  a    a  +

2 n2h2 × 2 =E 2m a

n2h2 . 8ma 2 17. A particle of mass ‘m’ is confined to a one dimensional box with the origin a a at the centre of the box. The box extends from − to + . The P.E. function 2 2 V(x) is V(x) = 0 (inside) and V(x) = ∞ (outside) E=

Write down the Schrodinger equation for the system showing separate equation inside and outside the box. Ans. The fundamental Schrodinger equation is given as:  2 2  ∇ + V ( x, y, z )  ψ = Eψ −  2m 

4.4 Particle in One Dimensional Box: Quantisation of Translational Energy

163

(a) Inside the box: V = 0 and the particle moves in x-direction. \

∂2 ∂x 2 Thus, the Schrodinger equation becomes ∇2 =

−

(in one dimension)

 2 ∂ 2ψ = Eψ 2m ∂x 2

(b) Outside the box: V = ∞ and particle moves in x-direction . −

 2 ∂ 2ψ + (∞)ψ = Eψ 2m ∂x 2

18. The length of the one dimensional box is an integral multiple of half wavelength of allowed wavefunctions. Show that the energy of the particle

\

 h2  moving in a box is given by E = n 2  .  8ma 2  Ans. For a particle in one dimensional box, we have, nλ a= 2 n2λ 2 2 a = 4

(given)

4a 2 …(1) n2 1 2 m2 v2 p 2 = E = mv = …(2) 2 2m 2m λ = h/p [de-Broglie’s wavelength] p = h/λ …(3)

λ2 = Now, Also, Put (3) in (2):

E=

h2 2mλ 2

E=

 h2  h2 n2 × 2 = n 2  2  2m 4a  8ma 

Using (1), we have

19. The wavefunction for a particle in a one-dimensional box (V = 0 for 0 ≤ x ≤ a and V = ∞ elsewhere) are given by the expression. ψ=

2  nπx  sin   a  a

164

Chapter 4

Show that 〈x〉 =

Particle in a Box: Quantisation of Translational Energy

a a3 a2 − 2 2 ; and 〈x2〉 = 2 3 2n π 〈px〉 = 0 and 〈px2〉 =

Also show that Dpx · Dx ≥

n2h2 4a 2

h where, 4π

(Dpx )2 = 〈px2〉 – 〈px〉2 (Dx)2 = 〈x2〉 – 〈x〉2. Ans. [Refer section 4.4.3] 20. An electron is confined in a one-dimensional box of length a as long as x = 0 to x = a and potential energy V = 0, elsewhere V = ∞ (a) Write down the wave equation to this problem. (b) If the starting wavefunction is of the form ψ = A sin Kx + B cos Kx. Calculate the normalised wavefunction for the problem. (c) Calculate the energy of various energy levels. Ans. [Refer section 4.4.1] 21. Show that the electron inside the one-dimensional box is never at rest. Ans. The electron inside the one-dimensional box is never at rest. We cannot find the exact position and exact momentum using eigen value equation, however, their average values are calculated using mean value theorem. [Refer section 4.4.3]. This means for the electron (microscopic particle), the position and momentum are changing very fast to be calculated. This is in accordance to the Heisenberg uncertainty principle. 22. Show that the wavefunction which is a solution of one-dimensional box problem is not an eigen function of the linear momentum operator. Ans. [Refer Q. 8, problems based on particle in a one dimensional box.] 23. A particle of mass ‘m’ exists in a one dimensional box of length ‘a’ extending along the x-axis. a (i) Find the probability that the particle exists in the region 0 ≤ x ≤ for 4 the states n = 1, 2, 3. (ii) What inference can be drawn from the calculated values? Ans. (i) [Refer Q. 3(i), problems based on particle in a one dimensional box.] (ii) [Refer section 4.4.2, point (7)]

4.4 Particle in One Dimensional Box: Quantisation of Translational Energy

165

4.4.4 Application of Particle in a One-dimensional Box “Particle in a box” model provides satisfactory interpretation of electronic spectra (UV or visible) of conjugated organic molecules like linear conjugated polyene molecule. The experimental spectral data available for these molecules agree fairly well with the values calculated using “particle in a box” model. A linear conjugated polyene molecule may be regarded as one-dimensional box and π electrons of molecule as a particle. The π electrons move from one end of the conjugated chain to the other end just as particle in a box continue to move in a to and fro motion within the box. Although there are more than one π electron moving in a box, but the coloumbic repulsion among the π electrons is neglected and each electron is supposed to move freely (i.e., V = 0). The π electrons occupies the various energy levels in accordance with the rules of electronic distribution in atoms and molecules. The energies of π electrons in various energy levels is calculated by the expression of energy, En = n2h2/8ma2. Also, the length (a) of the box is taken equal to the sum of all the C—C bond distances of the molecule. An electron in an atom or molecule can jump from a lower energy level to a higher energy by absorption of appropriate wavelength of light, or it can jump from a higher to a lower level by emission of light. The wavelength (λ) of the light absorbed or emitted is related to the energy gap between the two levels. hc DE = λ This energy gap further depends on the mass and the length of the molecules i.e., length of the conjugated chain. If the chain length of the molecule is small, hence the mass is less (like butadine), then the energy gap would be large and the absorption takes place in ultraviolet region (lower λ). While, if the chain length of the molecule is large, hence the mass is large, then the energy gap would be small and the absorption takes place in visible region (higher λ). These above observations can be explained by using the expressions of “particle in a box” model. Let the energy gap (DE) between highest occupied molecular orbital (n) and the lowest unoccupied molecular orbital (n + 1) is: DE = Ef – Ei DE = En+1 – En DE =

(n + 1) 2 h 2 n 2 h 2 − 8ma 2 8ma 2

( n 2 + 1 + 2n − n 2 ) h 2 DE = 8ma 2

LUMO (n + 1)

Ef E Ei

HOMO (n)

166

Chapter 4

Particle in a Box: Quantisation of Translational Energy

(2n + 1)h 2 DE = 8ma 2 hv =

(2n + 1)h 2 8ma 2

(DE = hv)

v=

(2n + 1)h 8ma 2

λ=

c 8ma 2c = v (2n + 1)h

=

1 (2n + 1)h = λ 8ma 2c

v (wave number)

Example: If we take a specific case of butadiene molecule and treat the π-electrons of this conjugated system as particles moving in a one-dimensional box. Then the ground state energy, hence, the lowest absorption frequency and the wavelength of absorbed light for the butadiene molecule can be calculated as follows: 2 3 4 There are 4π electrons in butadiene molecule: CH1 2 CH—CH CH2 and these π-electrons are distributed in various energy levels by Aufbau rule, as shown in Fig. 4.8. E4

n4

E3

n3 (LUMO)

E2

n2 (HOMO)

E1

n1

Fig. 4.8. Electron distribution in butadiene molecule.

The minimum energy or minimum LUMO (n + 1) 3 frequency or maximum wavelength is required for the transition from the highest occupied energy level (HOMO) to the lowest occupied HOMO (n) 2 energy level (LUMO). Also, the length of the butadiene molecule is given by  single bond length  a =1 (single bond length) + 2 (double bond length) + 2   2    R C—C  a = RC—C + RC C + 2    2 

4.4 Particle in One Dimensional Box: Quantisation of Translational Energy

167

Whenever, length of molecule is to be determined, we add extra 1 R C—C to each end the molecule. This is called End effect length. 2 Here, in case of butadiene. 154  a = 1(154) + 2(135) + 2    2  = 578 pm = 578 × 10–12 m = 5.78 × 10–10 m m = mass of electron = 9.11 × 10–31 kg h = Planck’s constant = 6.6 × 10–34 Js ∆E =

(2n + 1)h 2 8ma 2

∆E =

[2(2) + 1] × (6.6 × 10−34 ) 2 8 × (9.11 × 10−31 ) × (5.78 × 10−10 ) 2

∆E = 8.945 × 10–19 Joules The wave number of spectral line in butadiene molecule is: ( 2n + 1)h v= 8ma 2c [(2 × 2) + 1] × (6.6 × 10−34 ) v = 8 × (9.11 × 10−31 ) × (5.78 × 10−10 ) 2 × (3 × 108 ) v = 4.52 × 106 m–1 v = 4.52 × 104 cm–1 The experimental value of wave number of spectral line in case of butadiene molecule (n2 → n3) is found to be 4.61 × 104 cm–1, which is agreed by calculated value, indicating that treating a linear polyene molecule like a particle in a box is justified.

Problems based on aPPlication of Particle in one dimensional box 1. Calculate the wavelength of the light absorbed when an electron in a linear molecule of length 10 Å, makes a transition from energy level, n = 1 to n = 2. Ans.

n+1

2 E

n

1

Here, h = 6.6 × 10–34 Js m = 9.11 × 10–31 kg a = 10 Å = 10 × 10–10 m = 10–9 m

168

Chapter 4

λ=

=

Particle in a Box: Quantisation of Translational Energy

8ma 2c (2n + 1)h 8 × (9.11 × 10−31 ) × (10−9 ) 2 × (3 × 108 ) [(2 × 1) + 1] × (6.6 × 10−34 )

= 1.1 × 10–6 m. 2. Calculate the wavelength in nm, corresponding to the transition from HOMO to LUMO in 1, 3, 5-Hexatriene molecule. The C—C and C C bond lengths are 154 pm and 135 pm respectively. 1

2

3

4

5

6

Ans. H2C HC—HC HC—HC CH2 The 6π electrons in 1, 3, 5-Hexatriene molecule are distributed in various energy levels by Aufbau rule as: E4

n4 (LUMO)

E3

n3 (HOMO)

E2

n2

E1

n1

n+1

4 (LUMO) E

n

3 (HOMO)

The length of hexatriene molecule is given by  single bond length  a = 2(single bond length) + 3(double bond length) + 2   2    154  a = 2(154) + 3(135) + 2    2  a= m= h= c=

867 pm = 867 × 10–12 m 9.11 × 10–31 kg 6.6 × 10–34 Js 3 × 108 ms–1

λ=

8ma 2c (2n + 1)h

8 × (9.11 × 10−31 ) × (867 × 10−12 ) 2 × (3 × 108 ) = [2(3) + 1] × (6.6 × 10−34 ) λ = 3.55 × 10–7 m = 3.55 × 10–2 nm.

4.4 Particle in One Dimensional Box: Quantisation of Translational Energy

169

3. Treat the π-electrons in butadiene as particle in a box problem of length 1 4RCC (as in this case an extra RCC is added to each end of molecule, take 2 RCC = 140 pm). Calculate the absorption frequency for the transition of an electron from the ground state (i.e., valence state) to the lowest unoccupied orbital. 1

2

3

4

Ans. CH2 CH—CH CH2 The 4π electrons in butadiene molecule are distributed in various energy levels by Aufbau rule as: E3

n3 (LUMO)

E2

n2 (HOMO)

E1

n1

n+1

3

LUMO

2

HOMO

E n

The length of the butadiene molecule is given by  1 a = 4R CC + 2  R CC   2 1  = 4(140) + 2  × 140  2  = 700 pm = 700 × 10–12 m m = 9.11 × 10–31 kg, h = 6.6 × 10–34 Js, c = 3 × 108 ms–1 v=

(2n + 1)h 8ma 2

v=

[2(2) + 1] × (6.6 × 10−34 ) 8 × (9.11 × 10−31 ) × (700 × 10−12 ) 2

=

5 × (6.6 × 10−34 ) 8 × (9.11 × 10−31 ) × (700 × 10−12 ) 2

v = 9.24 × 1014 ms–1. 4. Show that the model of particle in a one dimensional box can be applied to calculate the π electron transition energy, when such an electron is excited from HOMO → LUMO of a butadiene molecule.

170

Chapter 4

1

2

3

Particle in a Box: Quantisation of Translational Energy

4

Ans. CH2 CH—CH CH2 The 4π electrons in butadiene molecule are distributed in various energy levels by Aufbau rule as E3

n3 (LUMO)

E2

n2 (HOMO)

E1

n1

n+1

3 (LUMO)

n

2 (HOMO)

The length of butadiene molecule is given by single bond length  a = 1(single bond length) + 2(double bond length) + 2   2    154  a = 1(154) + 2(135) + 2    2  = 578 pm = 578 × 10–12 m DE =

(2n + 1)h 2 8ma 2

[2(2) + 1] × (6.6 × 10−34 ) 2 8 × (9.11 × 10−31 ) × (578 × 10−12 ) 2 DE = 8.945 × 10–19 J The wave number of spectral line is (2n + 1)h v= 8ma 2c =

=

[2(2) + 1] × (6.6 × 10−34 ) 8 × (9.11 × 10−31 ) × (5.78 × 10−10 ) 2 × (3 × 108 )

= 4.52 × 106 m–1 = 4.52 × 104 cm–1 The experimental value of wave number of spectral line in case of butadiene molecule (n2 → n3) is found to be 4.61 × 104 cm–1, which is agreed by calculated value, indicating that treating a linear polyene molecule (butadiene) like a particle in a box is justified. 5. The π-electrons of a conjugated molecule can be regarded as moving in a particle in a box, where the box length is somewhat more than the length of the conjugated chain. For butadiene, take this length as 7.0 Å. Estimate the wavelength of the light absorbed when a π-electron is excited from a highest occupied to lowest vacant level. The experimental value is 217 nm.

4.5 Particle in Two-dimensional Box

1

2

3

171

4

Ans. CH2 CH—CH CH2 The 4π electrons in butadiene molecule are distributed in various energy levels by Aufbau rule as E3

n3 (LUMO)

E2

n2 (HOMO)

E1

n1

n+1

3 (LUMO)

n

2 (HOMO)

The length of the butadiene molecule is given as a = 7 Å = 7 × 10–10 m = 7 × 10–10 × 1012 pm = 700 pm λ=

8ma 2c (2n + 1)h

λ=

8 × (9.11 × 10−31 ) × (700 × 10−12 ) 2 × (3 × 108 ) [2(2) + 1] × (6.6 × 10−34 )

λ = 3,246,472.7 × 10–13 m λ = 324.6 × 10–9 m λ = 324.6 nm.

4.5

PARTICLE IN TWO-DIMENSIONAL BOX

Particle in a one-dimensional box serve as the simplest case for the treatment of bound electrons in atoms and molecules. Thus, it is a very convenient model for several types of atomic and molecular systems. The calculated values for various observables for this system are approximate but they agree fairly well with the experimental results. Here we will discuss the quantum mechanical treatment of a single particle moving in a two dimensional box. We have two cases: Case I : Particle moving in two dimensional rectangular box (a ≠ b) Case II : Particle moving in two dimensional square box (a = b).

4.5.1 Two Dimensional Rectangular Box Suppose there is a particle moving in a two dimensional box whose dimensions are a and b along the two coordinates x and y.

172

Chapter 4

Particle in a Box: Quantisation of Translational Energy

To calculate energy for particle in a two dimensional box, we have to first write the Hamiltonian (Ĥ) part and thus the general Schrodinger equation would be: Ĥ= −

∂2  2  ∂ 2  2 + 2  + V ( x, y ) 2m  ∂x ∂y  (in two dimension)

Ĥψ = Eψ   2  ∂ 2 ∂2  −  2 + 2  + V ( x, y )  ψ = Eψ ∂y    2m  ∂x (i) Schrodinger equation when particle is outside the box (V = ∞): −

2  ∂ 2 ∂2   2 + 2  ψ + ∞ψ = Eψ 2m  ∂x ∂y 

∂2  2  ∂ 2  2 + 2  ψ + ( E − ∞)ψ = 0 2m  ∂x ∂y 

As E