Advanced Engineering Physics (WBUT–2014) [4 ed.] 9789339214227, 9339214226

Table of contents :
Title
Contents
1 Vector Analysis
2 Electrostatics
3 Dielectrics
4 Magnetostatics
5 Electromagnetic Field Theory
6 Classical Mechanics
7 Quantum Mechanics
8 Statistical Mechanics
Appendix-A
Appendix-B
Appendix-C
Appendix-D
Appendix-E
Model Question Papers
Solved Question Papes

Citation preview

Advanced Engineering Physics Fourth Edition

WBUT–2014

About the Authors Sujay Kumar Bhattacharya is Former Head of Department of Sciences and Humanities at West Bengal Survey Institute (Govt. of West Bengal), Bandel, Hooghly. He did his MSc from Calcutta University and PhD from Jadavpur University. In the past, he has taught physics at Meghnad Saha Institute of Technology, Kolkata, for seven years. Here he served as Assistant Professor of Physics and Head of Department of Physics. He has also taught physics at the BSc (Honours) level at R K Mission Vidyamandir, Belur Math, Howrah. Prof. Bhattacharys’s research area of interest is plasma physics. He has contributed more than 32 research papers in reputed national and international journals and symposia. He is also a life member of Plasma Science Society of India (PSSI) and Indian Association for Cultivation of Science (IACS). Saumen Pal, a doctorate in physics, is currently Assistant Professor of Physics in the Department of Basic Sciences and Humanities, Meghnad Saha Institute of Technology, Kolkata. He obtained his BSc (Honours) and MSc from Dhaka University, and MPhil and PhD from Pune University. He has published and presented eight research papers in various journals and conferences. Prof. Pal is also a poet and a writer in Bangla. His scientific-mindedness is reflected in his poems and articles. Some of his Bangla poems and articles have been published in local magazines in Kolkata. He has also published a book of poetry titled Bilambita Prayas. He has nearly one year of experience as Scientific Officer under Bangladesh Atomic Energy Commission, Dhaka and three and half years’ of experience of teaching Physics at BSc (Hons) level in Basanti Devi College, Kolkata. Additionally, he has taught Physics at the BTech level for thirteen years and has been involved in research in the subject for eight years. His areas of research interest include nuclear physics and biophysics. Sujay Kumar Bhattacharya and Saumen Pal have collectively authored two other books, Basic Engineering Physics and Physics-II (PH-EE-401), both published by McGraw Hill Education (India).

Advanced Engineering Physics Fourth Edition

WBUT–2014

Sujay Kumar Bhattacharya Former Head Department of Sciences and Humanities West Bengal Survey Institute Bandel, Hooghly

Saumen Pal Assistant Professor (Physics) Department of Basic Sciences and Humanities Meghnad Saha Institute of Technology Kolkata

McGraw Hill Education (India) Private Limited NEW DELHI McGraw Hill Education Offices New Delhi New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto

McGraw Hill Education (India) Private Limited Published by the McGraw Hill Education (India) Private Limited P-24, Green Park Extension, New Delhi 110 016 Advanced Engineering Physics, 4/e (WBUT–2014) Copyright © 2015, 2013, 2012, 2011 by McGraw Hill Education (India) Private Limited. No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the author. The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, McGraw Hill Education (India) Private Limited ISBN (13): 978-93-3921-422-7 ISBN (10): 93-3921-422-6 Managing Director: Kaushik Bellani Head—Higher Education (Publishing and Marketing): Vibha Mahajan Senior Publishing Manager (SEM & Tech Ed.): Shalini Jha Assistant Sponsoring Editor: Koyel Ghosh Editorial Executive: Piyali Chatterjee Manager—Production Systems: Satinder S Baveja Assistant Manager—Editorial Services: Sohini Mukherjee Senior Manager—Production: P L Pandita Assistant General Manager (Marketing)—Higher Education: Vijay Sarathi Assistant Product Manager (SEM & Tech Ed.): Tina Jajoriya Senior Graphic Designer—Cover: Meenu Raghav General Manager—Production: Rajender P Ghansela Manager—Production: Reji Kumar

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Dedicated to science and technology students, wishing for the development of their scientific-mindedness.

Contents Preface

xi

Roadmap to the Syllabus

xv

1. Vector Analysis 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 1.14

1.1–1.39

Introduction 1.1 Representation of a Vector 1.1 Some Important Definitions About Vectors 1.1 Resolution of a Vector into Components 1.2 Product of Two Vectors 1.3 Triple Product 1.5 Scalar and Vector Fields 1.6 Gradient of Scalar Field 1.6 Divergence of Vector Field 1.7 Curl of a Vector Field 1.9 Curl in the Context of Rotational Motion 1.9 Vector Integration 1.10 Integral Theorems 1.12 Coordinate System 1.13 Worked Out Problems 1.17 Review Exercises 1.35

2. Electrostatics 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8

Introduction 2.1 Quantization of Charge 2.1 Conservation of Charge 2.1 Coulomb’s Law 2.1 Principle of Superposition 2.2 Electric Field 2.3 Continuous Charge Distributions Electric Potential 2.4

2.1–2.33

2.3

viii

Contents

2.9 2.10 2.11 2.12 2.13

Electric Potential Energy 2.4 Electric Flux 2.5 Solid Angle 2.5 Gauss’ Law 2.6 Poisson’s and Laplace’s Equations 2.11 Worked Out Problems 2.14 Review Exercises 2.28

3. Dielectrics 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8

Introduction 3.1 Polarization 3.1 Types of Dielectrics 3.2 Relation Between Dielectric Constant and Electrical Susceptibility 3.4 Polarizability 3.5 Types of Polarization 3.5 Polarization in Monoatomic Gases 3.7 Polarization in Polyatomic Gases 3.9 Worked Out Problems 3.10 Review Exercises 3.16

4. Magnetostatics 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13 4.14

3.1–3.18

4.1–4.26

Introduction 4.1 Electric Current 4.1 Current Density 4.1 Equation of Continuity for Current 4.2 Force on a Moving Charge in a Static Magnetic Field 4.3 Force on Current Element Placed in a Static Magnetic Field 4.4 Biot–Savart Law 4.4 Applications of Biot–Savart Law 4.5 Force Between Two Straight Parallel Wires 4.8 Magnetic Force Between Two Finite Elements of Current 4.8 Divergence of Magnetic Field 4.9 Ampere’s Circuital Law 4.10 Applications of Ampere’s Circuital Law 4.11 Magnetic Potentials 4.13 Worked Out Problems 4.15 Review Exercises 4.21

5. Electromagnetic Field Theory 5.1 Introduction 5.1 5.2 Magnetic Flux 5.1

5.1–5.33

Contents

5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 5.12 5.13 5.14 5.15 5.16

ix

Faraday’s Law of Electromagnetic Induction 5.2 Electromotive Force 5.2 Integral Form of Faraday’s Law 5.3 Displacement Current 5.4 Modified Ampere’s Law 5.5 Continuity Property of Current 5.6 Maxwell’s Equations 5.7 Wave Equations in Free Space 5.10 Transverse Nature of Electromagnetic Wave 5.11 Potentials of Electromagnetic Field 5.13 Electromagnetic Waves in a Charge-Free Conducting Media and Skin Depth 5.13 Skin Depth or Depth of Penetration (d) 5.15 Electromagnetic Energy Flow and Poynting Vector 5.16 Average Power Calculation using Poynting Vector 5.18 Worked Out Problems 5.19 Review Exercises 5.27

6. Classical Mechanics

6.1–6.44

6.1 Introduction 6.1 6.2 Lagrangian Formulation 6.13 6.3 Hamiltonian Formulation 6.20 Worked Out Problems 6.23 Review Exercises 6.40 7. Quantum Mechanics 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8

Introduction 7.1 Wave Function, Probability and Probability Density 7.2 Normalization of Wave Function and Orthogonality of Wave Functions 7.4 Physical Significance and Interpretation of Wave Function 7.5 Operators in Quantum Mechanics 7.7 Fundamental Postulates of Quantum Mechanics 7.15 Basic Considerations for Developing Schrödinger’s Equation 7.17 Applications of Schrödinger’s Equation 7.20 Worked Out Problems 7.32 Review Exercises 7.50

8. Statistical Mechanics 8.1 8.2 8.3 8.4

7.1–7.58

Introduction 8.1 Concept of Phase Space 8.2 Concept of Energy Levels and Energy States 8.3 Macrostate and Microstate 8.3

8.1–8.28

x

Contents

8.5 8.6 8.7 8.8 8.9 8.10 8.11 8.12 8.13 8.14 8.15

Thermodynamic Probability and Entropy 8.4 Equilibrium Macrostate 8.4 MB, BE and FD Statistics 8.5 Maxwell – Boltzmann (MB) Statistics 8.5 Bose–Einstein (BE) Statistics 8.7 Fermi–Dirac (FD) Statistics 8.9 Classical Statistics as a Special Case of Quantum Statistics 8.10 Density of States or Quantum States in Energy Range Between e and e + d e Fermi Distribution at Zero and Non-zero Temperature 8.12 Derivation of Planck’s Law of Radiation from BE Statistics 8.17 Comparative Study of Three Statistical Distribution Functions 8.19 Worked Out Problems 8.19 Review Exercises 8.25

8.11

Appendix A: Surface Integrals

A.1

Appendix B: Angular Concept

B.1–B.2

Appendix C: Useful Physical Constants

C.1

Appendix D: Derivation of Gradient, Divergence, Curl and Laplacian

D.1–D.10

Appendix E: Set of Experiments (For Practical Classes)

E.1–E.39

Model Question Paper 1

M1.1–M.3

Model Question Paper 2

M2.1–M2.3

Model Question Paper 3

M3.1–M3.3

Model Question Paper 4

M4.1–M4.3

Model Question Paper 5

M5.1–M5.3

Solved WBUT Question Paper (2010)

S1.1–S1.10

Solved WBUT Question Paper (2011)

S2.1–S2.17

Solved WBUT Question Paper (2012)

S3.1–S3.22

Solved WBUT Question Paper (Dec. 2012)

S4.1–S4.10

Solved WBUT Question Paper (June 2013)

S5.1–S5.7

Solved WBUT Question Paper (Dec. 2013)

S6.1–S6.8

Solved WBUT Question Paper (2014)

S7.1–S7.14

Preface Engineering and technology, in essence, are applied aspects of science. Science provides concepts, theories as well as formulae to engineering and technology and helps in their progress. Any technological development depends on better scientific understanding of the working of nature. New inventions are the results of varied scientific research and experimentation. Physical science and technology are very closely interlinked. Some modern technologies like biotechnology, food technology, etc., are very much dependent on bioscience along with physics. Technologists, today, require sound knowledge of physics to be able to face current and new challenges in this field at present and in the future.

Aim of this Book The West Bengal University of Technology recently reorganized the physics syllabus for its BTech program. It has divided the entire syllabus into two parts. Topics to be studied by the students of both physics-based and chemistry-based streams of engineering have been compiled as the first part and have been made compulsory for all first-year students. Some other topics have been put under the second part, which are to be studied as part of the compulsory curriculum by the second-year students of physics-based streams only. We have compiled the following topics of the second part of the BTech syllabus in this book: (a) vector analysis, (b) electrostatics, (c) dielectrics, (d) magnetostatics, (e) electromagnetic field theory, (f) classical mechanics, (g) quantum mechanics and (h) statistical mechanics. The book is, hence, appropriately titled, Advanced Engineering Physics.

Scope of this Book This book is designed as a first reader for second-year students of Mechanical Engineering, Civil Engineering, Production Engineering, Automobile Engineering, Computer Science Engineering, Information Technology, and Electronics and Communication Engineering studying the course, Physics-II (PH301, PH401) at West Bengal University of Technology. It will also be useful for students of other universities pursuing BSc (Honours) and BSc (General) courses.

Salient Features of the Book Though many books covering the aforesaid topics are available in the market, none of them entirely cover the new syllabus and a majority of them discuss topics either very briefly or in a highly exaggerated manner. For this reason, these books are not completely suitable for students pursuing the second year of BTech at WBUT, for understanding their physics course. Keeping all these points in mind, we decided to pen this physics textbook.

xii

Preface

This book has been written on the basis of our class lecture notes; prepared and taught during the last ten and half years. It has been written in the most coherent and exhaustive manner so that the students can grasp the subject with minimum labor and time. Much effort has been given to elucidate the fundamental concepts of the subject in an easy-to-understand manner. The book is enriched with a large number of illustrations and solved exercises, so that students can develop a comprehensive and thorough understanding of the topics.

Salient Features of the Book ✦ Complete coverage of the WBUT Physics-II (PH301, PH401) syllabus ✦ Excellent coverage of topics such as quantum mechanical operators, commutation relations, ✦ ✦ ✦ ✦ ✦

✦

Schrodinger’s wave equation, free particles in 1D and 3D box Appendices on surface integrals, angular concept, list of useful physical constants, and derivation of gradient, divergence, curl and Laplacian for easy reference 11 preparatory experiments for practical classes with detailed step-wise procedures with illustrations and questions and answers for viva voce 5 model question papers according to WBUT pattern Latest solved WBUT examination papers for students’ practice Comes with an APP! ■ Study advanced engineering physics on the go ■ Revise key topics before an examination ■ Test your understanding ■ All this and much more Enhanced pedagogy includes ■ Illustrations: 150 ■ Worked Out Problems: 190 ■ Multiple Choice Questions: 295 ■ Short Questions with Answers: 150 ■ Descriptive Questions: 235 ■ Numerical Problems: 170

Organization of the Book This book consists of eight chapters. Chapter 1 explains vectors and discusses algebra and calculus along with gradient of scalar field, divergence of vector field and curl of a vector. Chapter 2 elucidates electrostatics. Under electrostatics, quantisation of charge, conservation of charge, Coulomb’s law, electric field, electric potential, electric flux, Gauss’ law and Poisson’s and Laplace’s equations have been discussed. Chapter 3 presents discussions of dielectrics. Under this topic, electric polarizations, polarizability and polarization in monatomic and polyatomic gases are also described. Chapter 4 elucidates magnetostatics. Biot–Savart Law, Ampere’s circuital law and magnetic scalar and vector potentials. Chapter 5 discusses electromagnetic field theory. Maxwell’s equations in both differential and integral forms are included here. Chapter 6 discusses topics on classical mechanics elaborate such as constraints of motion, generalized coordinates, degrees of freedom, D’ Alembert’s principle, lagrangian formulation, hamiltonian formulation, etc. Chapter 7 discusses quantum mechanics, especially wave function representing a particle in space, operators in quantum mechanics and development of Schrodinger’s wave equation. Chapter 8 elucidates statistical mechanics. Bose–Einstein and Fermi–Dirac statistics are also discussed along with a comparative study of the three statistical distribution functions.

xiii

Preface

Appendix A and Appendix B explain surface integrals and angular concept, respectively. Appendix C provides a list of useful physical constants. Appendix D elucidates curl of gradient, divergence, curl and Laplacian. A set of 11 experiments in proper format, with required aims, theories, tables and procedures are included in the book as Appendix E. This section covers the syllabus of the practical classes for this course. Hence, the students will no longer require a separate book for their practical classes. Review exercises are divided into three parts. Part 1 comprises multiple choice questions and short questions with answers. Part 2 consists of descriptive questions and Part 3 comprises unsolved problems under the heading numerical problems. For solving such problems, students will have to apply critically, concepts which they have already learnt in the text.

Additional Features of this Book Five model question papers, set as per the pattern of WBUT examinations and solutions to questions of WBUT examinations from 2010 to 2014 are provided toward the end of the book for self-practice. We were very surprised to observe that in our country the study of science fails to have an impact in the development of scientific-mindedness in most science and engineering students. It may be due to the faulty method of teaching of science. We feel that practical use of scientific knowledge in one’s daily life helps in eradicating superstition. With this view in mind, the uses of scientific theories have been discussed at the end of each chapter. We hope that it will help develop a scientific attitude in the students.

Acknowledgements The following reviewers deserve special mention for taking out time to give their valuable comments on various chapters of the book: Apurba Kumar Roy

Dr. Sudhir Chandra Sur Degree Engineering College Kolkata

Gouri Shankar Pal

Gargi Memorial Institute of Technology

Mahadev Das

Abacus Institute of Engineering

Mrs Subhra Sarkar

Techno India, Kolkata

We acknowledge with gratitude our colleagues for their encouragement and constructive suggestions during the preparation of the manuscript. We are very thankful to our family members who fully cooperated with us throughout the writing of this book. Suggestions to improve the quality of this book are always welcome. These can be sent to us at [email protected] and [email protected]. We would also like to thank the entire team at McGraw Hill Education (India) for bringing out this book in a very short span of time. Sujay Kumar Bhattacharya Saumen Pal

Publisher’s Note Remember to write to us. We look forward to receiving your feedback, comments and ideas to enhance the quality of this book. You can reach us at [email protected]. Please mention the title and authors’ name as the subject. In case you spot piracy of this book, please do let us know.

ROADMAP TO THE SYLLABUS This text is useful for subject codes: PH301—ME, Civil, CSE, IT PH401—ECE

Module 1: Vector Calculus Physical significances of grad, div, curl. Line integral, surface integral, volume integral - Physical examples in the context of electricity and magnetism and statements of Stokes theorem and Gauss’s theorem [No Proof]. Expression of grad, div, curl and Laplacian in spherical and cylindrical coordinates. GO TO: CHAPTER 1. Vector Analysis

Module 2: Electricity Coulomb’s law in vector form. Electrostatic field and its curl. Gauss’s law in integral form and conversion to differential form. Electrostatic potential and field, Poisson’s equation. Laplace’s equation (Application to Cartesian, Spherically and Cylindrically symmetric systems – effective 1D problems). Electric current, drift velocity, current density, continuity equation, steady current. Dielectrics – Concept of polarization, the relation D = e0E + P. Polarizability. Electronic polarization and polarization in monoatomic and polyatomic gases. GO TO: CHAPTER 2. Electrostatics CHAPTER 3. Dielectrics CHAPTER 4. Magnetostatics

Module 3: Magnetostatics and Time Varying Field Lorentz force, force on a small current element placed in a magnetic field. Biot–Savart law and its applications, divergence of magnetic field, vector potential, Ampere’s law in integral form and conversion to differential form. Faraday’s law of electromagnetic induction in integral form and conversion to differential form. GO TO: CHAPTER 4. Magnetostatics CHAPTER 5. Electromagnetic Field Theory

xvi

Roadmap to the Syllabus

Module 4: Electromagnetic Theory Concept of displacement current Maxwell’s field equations, Maxwell’s wave equation and its solution for free space. E.M. wave in a charge free conducting media, skin depth, physical significance of skin depth, E.M. energy flow, Poynting vector. GO TO: CHAPTER 5. Electromagnetic Field Theory

Module 5: Quantum Mechanics Generalised coordinates, Lagrange’s Equation of motion and Lagrangian, generalised force potential, momenta and energy. Hamilton’s Equation of motion and Hamiltonian. Properties of Hamilton and Hamilton’s equation of motion. Concept of probability and probability density, operators, commutator. Formulation of quantum mechanics and basic postulates. Operator correspondence. Time-dependent Schrödinger’s equation, formulation of time independent Schrödinger’s equation by method of separation of variables. Physical interpretation of wave function y (normalization and probability interpretation). Expectation values. Application of Schrödinger equation – Particle in an infinite square well potential (1-D and 3-D potential well). Discussion on degenerate levels. GO TO: CHAPTER 6. Classical Mechanics CHAPTER 7. Quantum Mechanics

Module 6: Statistical Mechanics Concept of energy levels and energy states. Microstates, macrostates and thermodynamic probability, equilibrium macrostate. MB, FD, BE statistics (No deduction necessary), fermions, bosons (definitions in terms of spin, examples), physical significance and application, classical limits of quantum statistics. Fermi distribution at zero & non-zero temperature, calculation of Fermi level in metals, also total energy at absolute zero of temperature and total number of particles. Bose-Einstein statistics – Planck’s law of blackbody radiation. GO TO: CHAPTER 8. Statistical Mechanics

Chapter

1 Vector Analysis 1.1

IntroduCtIon

The physical quantities which we usually come across in physics and also our daily lives are broadly classified into two types: (i) scalars, and (ii) vectors. A scalar is a quantity which is completely specified by its magnitude and has no reference to the idea of direction. For example—mass, length, time, temperature, work, speed, electric charge, etc., are scalars and are added according to the ordinary rules of algebra. The physical quantities which have both magnitude and direction and which can be added according to the triangle rule are called vector quantities. Acceleration, momentum, velocity, weight, electric field intensity, magnetic field intensity etc., are vector quantities as each involve magnitude and direction and follow the triangle rule. If any physical quantity has both magnitude and direction but does not add up according to the triangle rule, it will not be called a vector quantity. Electric current in a wire has magnitude as well as direction but there is no meaning of triangle rule there. Thus, electric current is not a vector quantity.

1.2

representatIon of a veCtor Q

A vector is represented analytically by _› bold-faced letter such as A or putting an arrow on the top of it such as A in Fig. 1.1.

A P

Fig. 1.1

1.3

Vector representation.

some Important defInItIons about veCtors _›

_›

Equality of vectors Two vectors A and B are said to be equal if they have the same magnitude and direction regardless of their initial points [Fig 1.2]. Negative vector A vector having the same magnitude but direction opposite to that of the given vector is called a negative vector relative to _›

_›

that vector. In Fig. 1.3, A and B are opposite vectors.

A B

Fig. 1.2

Equality of two vectors.

1.2

Advanced Engineering Physics

Unit vector A vector of unit magnitude is called a unit vector. _› If any vec_› _› A tor A is of magnitude |A| then the unit vector is represented by ___ _› . |A| Null or zero vector It is a vector having zero magnitude. It has no definite __›

A

B

Fig. 1.3

direction. It is denoted by O.

Opposite vectors.

Coplanar vectors If a system of vectors _› _› lie in__›the same plane then they are called coplanar vectors. In Fig. 1.4, A, B and C are coplanar vectors. Collinear or parallel vectors Vectors which have the same line of action or having lines of action parallel to the same line are called collinear vectors.

A B

_›

–1

Reciprocal_vectors A vector A having the same direction as that of a _› C › given vector A but magnitude as the reciprocal of A is known as the recipro_› _› Fig. 1.4 Coplanar vectors. 1 cal vector of A. Thus A–1 = ___ _› Â. |A| Polar vectors A vector which has a linear motion in a particular direction and changes its sign under inversion or reflection is called a polar vector. Linear velocity, linear momentum, force, etc., are examples of polar vectors. Axial vector A vector corresponding to the rotation about a certain axis is called an axial vector. Angular velocity, angular momentum, torque, etc., are axial vectors.

1.4

resolutIon of a veCtor Into Components

Any vector can be resolved into component vectors along three axes of the cartesian coordinate system. Here, the three axes OX, OY and OZ are mutually perpendicular to each other. If , and be the unit vectors along the x, y and z axes and Ax, Ay and

z R

Az be the vector intercepts of A along the x, y and z axes respectively, then _›

we may write A = Ax + Ay + Az .

C k

_›

By geometry, the magnitude of the vector A is _›

O

_______________

A = |A| = ÷OP2 + OQ2 + OR2

i

___________

or,

A=÷

A2x + _›

A2y

+

A2z

g

…(1.1)

b a

A j

Q

y

P

The unit vector along A is given by _› x Ay Az A Ax  = __ = ___ + ___ + __ …(1.2) Fig. 1.5 Resolution of a vector A A A A into components. A A A y ___x = cos a, ___ = cos b, __z = cos g Here, …(1.3) A A A _ › are called the direction cosines of vector A. Generally they are represented by the letters l, m and n respectively and a, b, g are the angles as shown in Fig. 1.5. If we put the values of Ax, Ay and Az from Eq. (1.3) in Eq. (1.1), we get

Vector Analysis

1.3

cos2 a + cos2 b + cos2 g = 1 or,

2

2

…(1.4)

2

l +m +n =1 Thus, the sum of squares of the direction cosines is equal to unity.

1.5

…(1.5)

produCt of two veCtors

The product of two vectors is classified as (i) scalar product or dot product, and (ii) vector product or cross product.

1.5.1

scalar or dot product _›

A

_›

The scalar product of two vectors A and B is equal to the product of the magnitudes of these vectors multiplied by the cosine of the angle between them as shown in Fig. 1.6. _› _›

_› _›

B

A ◊ B = |A| |B| cos q

Hence,

Fig. 1.6 Scalar__product __› of two › vectors A and B.

where q is the_ angle between directions of A and B. › _› Note that A ◊ B is a scalar and not a vector.

1.5.2

q

properties and other results of scalar product

(i) Scalar product_ of_two_vectors obeys commutative law › › › _› i.e., A◊B=B◊A _›

_›

_› _›

(ii) If two vectors A and B are mutually perpendicular, cos q = 0, A ◊ B = 0. This condition is known as the orthogonality condition of two vectors. If this property is applied to a unit vector, then ◊ = ◊ = ◊ =0 (iii) Scalar product _› _› of__›two_vectors _›is distributive __› › _› i.e., A ◊ (B + C) = A ◊ B + A ◊ C (iv) Scalar multiplication If_ m, n are _› _› _› scalars _› _› is associative. _› and A, B are two vectors, then _› two › mA ◊ nB = mn (A ◊ B) = (nA) ◊ (mB) = A ◊ mn B

1.5.3

F

physical applications of scalar product _›

_›

(i) Work done Work done by a force F causing a displacement d is given by Fig. 1.7.

q

d F cos q

_› _›

W = F cos q d = F ◊ d

Fig. 1.7 Work done by a force.

(ii) Power The rate of doing work is power. So, power is dW d _› _› _› _› p = ___ = __ (F ◊ d ) = F ◊ v dt dt _› where v is the velocity of the body.

n

E

__›

(iii) Electric flux _› Let us consider an elementary area ds in an electric field E _› [Fig. 1.8]. The normal electric flux coming out of the area = E. ds where is the outward unit normal to the surface.

ds

Fig. 1.8 Normal component of electric flux.

1.4

Advanced Engineering Physics _›

(iv) Magnetic flux (j) The magnetic flux of a magnetic field of flux density B passing normally _›

through an area S is

_› _›

j = B. S

1.5.4 vector or Cross product _›

n

_›

The vector product of two vectors A and B is defined as a vector whose magnitude is equal to the product of the magnitudes of the vectors and the sine of the angle between their _directions _› and its › direction is perpendicular to the plane containing A and B. The vector product is represented as __›

_›

_›

q

_› _›

C = A × B = |A| |B| sin q The direction of C is determined by the screw rule [Fig. 1.9] or any right-hand rule. __›

1.5.5

B

A

Fig. 1.9 Representation__›of cross __› product of two vectors A and B.

properties and other results of vector product

(i) Vector product does not obey commutative law. _›

_›

_›

_›

A×B=–B×A (ii) The distributive law for cross products holds good. _›

_›

__›

_›

_›

_›

__›

A × (B + C) = A × B + A × C (iii) The vector product is associative. _›

_›

_›

_›

A × (mB) = (mA) × B (iv) For the orthonormal vector triad , , , × =– × = ; and

_›

× = – × = and × = – × =

× = × = × =0 _›

(v) If A = Ax + Ay + Az and B = Bx + By

|

A × B = Ax Ay Az Bx By Bz

1.5.6

+ Bz , then

|

physical application of vector product _›

(i) Moment of a _force or torque ( t ) The moment of a force (or torque) about a fixed point is the _› _› _› _› › vector t = r × F, where r is the position vector of the particle and F is the applied force. _›

(ii) Angular momentum _› _ (_L) Angular momentum of a particle is defined as the moment of linear › › momentum, i.e., L = r × p where p is the momentum of the particle. (iii) Force on a moving charge in a magnetic field When a charged particle moves in a_ magnetic _› › field, a force acts on it. If a charge q moves with a velocity v in a uniform magnetic field B, then the force experienced_by the charge is _› _› › F = q ( v × B) …(1.6)

Vector Analysis

1.6

1.5

trIple produCt _› _›

__›

__›

_›

Suppose we have three vectors A, B and C. If the vector product of two vectors B and C is a vector, this may _› be multiplied scalarly or vectorially with the first vector A. This is known as triple product. There are two types of triple product _› _›

__›

(i) A. (B × C) is known as scalar triple product. _›

_›

__›

(ii) A × (B × C) is known as vector triple product.

1.6.1

n

scalar triple product

The scalar triple product of three vectors is a scalar and is represented as _› _› __›

_› _›

__›

__›

_›

[A B C] = A. (B × C) _›

__› _›

q C C cos q

_›

= B. (C × A) = C. (A × B)

B

Properties of _scalar triple product The scalar triple product __› › _› of three vectors A, B and C represents _› _› the __› volume of a parallelepiped whose three adjacent sides are A, B and C. _› _› __› The volume of the parallelepiped with sides A, B and C [Fig 1.10] = (Height of the parallelepiped) × Area of the base _›

_›

Fig. 1.10 Volume of a parallelepiped.

_›

= C cos q |(A × B)| __› _›

A

_›

_› __›

= C. (A × B) = (A × B). C

__› _›

__› _›

_›

_›

The sign of C. (A × B) can be either _› positive _› __› or negative according as C, A and B do or do not form a righthanded system. The scalar quantity (A × B). C denotes the volume of the parallelepiped. Any face of the parallelepiped may be taken as base. Hence, its volume can be represented by any of the three expressions _› _›

__›

_› __›

_› __› _›

_›

A. (B × C), B. (C × A), C. (A × B)

A1 Scalar triple product of A, B and C is given by A. (B × C) = B1 C1 _› _›

__›

_› _›

__›

|

A2 A3 B2 B3 C2 C3

|

1.6.2 vector triple product The vector triple product of three vectors is a vector product vector product of the other _› _› of one _› with __› vector _› __the › two vectors. The vector triple product of three vectors A, B and C is A × (B × C). Vector triple product as __the_sum of two terms _› __› be_written _› can › › _› › _› __› A × (B × C) = B (A. C) – C (A. B) …(1.7) In general, _› _› _› __› _› __› A × (B × C) π (A × B) × C _›

_›

__›

_›

__›

_›

_›

__›

The vector _A × (B_ × C) represents a vector coplanar with B and C, but (A × B) × C represents a vector › › coplanar with A and B. So, the product does not represent the same vector.

1.6

1.7

Advanced Engineering Physics

sCalar and veCtor fIelds

A mathematical function, or a graphical sketch, constructed so as to describe the variation of a quantity in a given region is said to represent the field of that quantity associated with that region. According to the nature of the physical quantity, there are two main kinds of fields. (a) Scalar fields (b) Vector fields A field is a spatial distribution of quantity, which may or may not be a function of time. The scalar field is for a scalar quantity and vector field for a vector quantity. The distribution of temperature in a medium, distribution of electrostatics or gravitational potential, etc., are examples of scalar fields. A scalar field is represented by a function in space as y (x, y, z). If the scalar quantity is also varying with time then the function is y (x, y, z, t). The distribution of magnetic and electric field intensity, the distribution of velocity in a _› fluid, the force _› field are examples of vector fields. The vector field F in the cartesian coordinate system can be written as F (x, y, z, t) = Fx (x, y, z, t) + Fy (x, y, z, t) + Fz (x, y, z, t)

1.8

GradIent of sCalar fIeld

If y (x, y, z) be a scalar function of position of coordinates (x, y, z) in space, then y can be differentiated with respect to x keeping the other two coordinates y and z constant. This type of differentiation is known as partial ∂y ∂y ∂y derivatives. The partial derivatives along the three orthogonal axes are ___, ___ and ___. ∂z __› ∂x ∂y The gradient of a scalar point function y (x, y, z) is defined as —y and given by __› ∂y ∂y ∂y …(1.8) grad y = —y = ___ + ___ + ___ ∂x ∂y ∂t where vector differential operator — (del) is defined as __› ∂ ∂ ∂ — = ___ + ___ + __ ∂x ∂y ∂z The ‘del’ operator is a vector operator, __› when it operates on a scalar point function; it converts the scalar function into a vector function, grad y (—y) is a vector quantity. According to the theory of partial derivatives, ∂y ∂y ∂y …(1.9) dy = ___ dx + ___ dy + ___ dz ∂x ∂y ∂z This shows how y varies as we go a small distance (dx, dy, dz), away from the point (x, y, z). The above relation can be written as ∂y ∂y ∂y dy = ___ + ___ + ___ ◊ ( dx + dy + dz) ∂x ∂y ∂z

(

__›

)

_›

= —y. d r where, d r = dx + dy + dz is the infinitesimal displacement vector. _›

1.8.1

…(1.10)

Geometrical Interpretation of the Gradient

Like an ordinary vector, a gradient has both magnitude and direction. From Eq. (1.10), we have __›

_›

__›

_›

dy = —y. d r = | —y | d r | cos q

…( 1.11)

Vector Analysis __›

_›

1.7 _›

where q is the angle between the vector —y and d r . For a fixed value of | d r |, the maximum charge __ of y › _› occurs when q = 0. So, for fixed distance |d r |, dy is greatest when one moves in the same direction as —y. Putting cos q = 1 in Eq. (1.11) we get maximum rate of increase of the function, _› dy ___ = |— y| …(1.12) dr max The gradient of a scalar field is a vector whose magnitude is equal to the maximum rate of change of scalar field and direction is along that change. _ __› › Example: The electric field intensity E = – —V where V is the electric potential.

|

1.8.2

directional derivative

__›

__›

Suppose ê represents a unit vector along a specific direction. The component of —y along ê is —y. ê. This is known as the directional derivative of y along ê. The directional derivative of y in the direction ê is the __›

__›

_›

__›

component of —y along ê. Since —y. ê £ |— y |, —y is equal to the largest directional derivative of y.

1.9

dIverGenCe of veCtor fIeld

The divergence of a vector field at any point is defined as the net outflow or flux of that field per unit volume. _› _› The divergence of a vector point function A is denoted by div A and can be written as _› __› _› ∂ ∂ ∂ div A = —. A = ___ + ___ + __ . ( Ax + Ay + Az) ∂x ∂y ∂z ∂A ∂A ∂A …(1.13) = ____x + ____y + ____z ∂x ∂y ∂z __› _› _› Physical meaning —. A is the measure of how much the vector A spreads out (i.e., diverges) from the point in question.

(

A

)

A

(a) Positive divergence.

Fig. 1.11

(b) Negative divergence.

A

(c) Zero divergence.

Divergence of a vector field.

For example, the vector function in Fig. 1.11(a) has a large positive divergence at the point A. It indicates a net outflow, while a negative value of divergence [Fig. 1.11(b)] represents a net__ inflow and the function in › _› Fig. 1.11(c) has zero divergence at A; it is not spreading out at all. In this case, —. A = 0 and it implies that there is no inflow or outflow. _› _› A vector _A› , which satisfies the condition div A = 0 is called a solenoidal vector. For example, the magnetic field vector B is a solenoidal vector.

1.8

Advanced Engineering Physics _›

The points _at which the divergence of A is greater than zero are called sources and the points at which › divergence of A is less_than zero are called sinks. The divergence of a vector field is a scalar quantity. The › divergence of a vector A may be written as y _› __›

H

ÚÚ A.dS

__› _›

—. A = Lt

DV Æ 0

S ______ DV

…(1.14)

E vx – 1 ∂vx Dx 2 ∂x

where dS is the surface enclosing the volume DV.

1.9.1

Q

vx + 1 ∂vx Dx 2 ∂x

F

D A

divergence of a fluid

C B x

Let us consider an elementary parallelepiped of volume Dx Dy Dz within a fluid as shown in Fig. 1.12, where Dx, Dy and Dz are length, breadth _› and height of the parallelepiped. Suppose v represents a vector point at the centre point Q of the parallelepiped. _›

Here

G

z _

Fig. 1.12 Representation of the divergence of v› in cartesian coordinates.

v = vx + vy + vz

1 ∂vx The x component of v at the face ADHE = vx – __ ___ × Dx 2 ∂x _› 1 ∂vx The x component of v at the face BCGF = vx + __ ___ Dx 2 ∂x _›

Per unit time, the volume of the fluid crossing ADHE 1 ∂vx = vx – __ ___ Dx Dy Dz 2 ∂x

(

)

…(1.15)

Per unit time, the volume of the fluid crossing BCGF 1 ∂vx = vx + __ ___ Dx Dy Dz 2 ∂x

(

)

…(1.16)

Hence in the x direction, the gain in volume of the fluid per unit time 1 ∂vx 1 ∂vx = vx + __ ___ Dx – vx – __ ___ Dx Dy Dz 2 ∂x 2 ∂x ∂vx = ___ Dx Dy Dz ∂x Similarly, the gain in volume of the fluid per unit time along the y direction ∂vy = ___ Dx Dy Dz ∂y and along the z direction ∂vz = ___ Dx Dy Dz ∂z So, the total flux or gain of fluid per unit volume per unit time ∂vy ∂vz ∂vx ___ ___ + + ___ Dx Dy Dz ∂x ∂y ∂z _______________________ = Dx Dy Dz

[(

(

)(

)

)]

…(1.17)

…(1.18)

…(1.19)

Vector Analysis

∂vx ∂vy ∂vz = ___ + ___ + ___ = ∂x ∂y ∂z

(

__› _ ›

= —. v

1.9

∂ ∂ ∂ ___ + ___ + __ . ( vx + vy + vz) ∂x ∂y ∂z

)

…(1.20)

__› _ ›

If there is no gain of fluid anywhere, then —. v = 0. This is called the continuity equation for an incompressible fluid.

1.10 Curl of a veCtor fIeld The curl of a vector field at any point measures the rate of rotation of that vector. The curl of a vector _› field is also known as circulation or rotation. The curl of a continuously differentiable vector point function A (x, y, z) is defined by the equation _› __› _› ∂ ∂ ∂ Curl A = — × A = ___ + ___ + __ × ( Ax + Ay + Az) ∂x ∂y ∂z j k i ∂ ∂ ∂ = ___ ___ __ ∂x ∂y ∂z Ax Ay Az

(

)

|

=

1.10.1

|

(

∂Ay ∂Az ____ ___ – – ∂y ∂z

) (

∂Az ____ ∂Ax ___ – + ∂x ∂z

) (

∂A ∂Ax ____y – ____ ∂x ∂y

)

…(1.21)

physical meaning

Curl is_a measure of how much the vector ‘curl around’ the point in question. For example, the existence of › curl of _v , the velocity at a point in a space indicates circulation or vorticity at that point _› of the liquid flow. › If curl v = 0, it means that if a wheel is placed in the liquid, it will not rotate. But if curl v π 0, the wheel will rotate. If a free magnetic __› __› pole is placed __› near a current-carrying conductor, the pole rotates _› __›around the conductor, _› which means cH. dl π 0, so curl H π 0. But in the case of an electrostatic field, c E. dl = 0, so curl E = 0. The curl of a vector field at a point is defined as the amount of maximum line integral at any point in a vector field per unit area around a closed curve and is directed along the normal to the plane of the area. Thus,

_›

__›

_› __›

_›

curl A = — × A = Lt

DsÆ0

[ A◊dl ] __________ c

max

Ds

_›

…(1.22) _›

Irrotational vector If the curl of a vector field A is zero then the vector field A is called an irrotational vector. Gravitational field, electrostatic fields, etc., are irrotational fields.

1.11 Curl In the Context of rotatIonal motIon __›

_›

Consider a rigid body R rotating about an axis passing through O [Fig 1.13] with an angular velocity w. If r be the position vector of a point P on the rigid body, then its linear velocity _› __› _› v =w× r ...(1.23) = ( w1 + w2 + w3) × ( x + y + z)

1.10

Advanced Engineering Physics

|

= w1 x

w3 z

w2 y

w

|

v

= (w2z – w3y) – (w1z – w3x) + (w1y – w2x)

P r

__›

_›

_›

curl v = — × v =

So,

|

∂ ∂ ∂ ___ __ ___ ∂y ∂z ∂x (w2z – w3y) (w3x – w1z) (w1y – w2x)

|

O R

Fig. 1.13 Rotational motion of a rigid body.

= (w1 + w1) + (w2 + w2) + (w3 + w3) __›

=2w

…(1.24) _›

Thus, we see that the curl of a vector field v is associated with the rotational properties of the vector field and shows that the angular velocity of a uniformly rotating body is one half the curl of the linear velocity.

1.11.1

some Important rules on Gradient, divergence and Curl

__›

__›

__›

(i) — (j + y) = —j + —y __› _›

__› _›

__›

__› __›

(ii) — ◊ (P + Q) = — ◊ P + — ◊ Q __›

_›

__›

_›

__›

__›

__›

_›

__›

(iii) — × (P + Q) = — × P + — × Q __›

__› _›

_›

(iv) — ◊ (y P) = —y ◊ P + y — ◊ P __›

_›

__›

__› _›

__›

__› __›

__›

_›

_›

(v) — × (y P) = —y × P + y — × P _›

_› __›

__›

(vi) — ◊ (P × Q) = Q ◊ (— × P) – P ◊ (— × Q) __›

__› __› _›

__›

_›

__› __› _›

_› __› __›

_› __› __›

(vii) — × (P × Q) = (Q ◊ —)P – Q(— ◊ P) – (P ◊ —) Q + P (— ◊ Q) __› _› __›

__› __› _›

_› __› __›

__›

__›

_›

_›

__›

__›

(viii) — (P ◊ Q) = (Q . —)P + (P ◊ —)Q + Q × (— × P) + P × (— × Q) __› __› ∂2y ∂2y ∂2y (ix) — ◊ (—y) = —2y = ____2 + ____2 + ____ ∂x ∂y ∂z2 __› __› (x) — × (—y) = 0 __› __›

_›

(xi) — ◊ (— × P) = 0 __›

__›

_›

__› __› _›

_›

(xii) — × (— × P) = — (— ◊ P) – —2 P _›

__›

Here P and Q are differentiable vector functions and j and y are differentiable scalar functions.

1.12 veCtor InteGratIon In vector analysis, the integrals that generally come are the line integral, the surface integral and the volume integral.

Vector Analysis

1.12.1 __›

line Integration

1.11

_›

Let dl be an_ element of length on a__smooth curve PQ and A be continuous _vector point function. The scalar › › › product of A with the line element dl is called the line integral of the vector A and for an extended path it will be equal to the integral Q

_› __›

Ú

Q

A.dl = Ú A dl cos q

P

…(1.25)

P

_›

It is defined as the line integral of the vector A along the curve PQ __› _› [Fig. 1.14], where q is the angle _between A and elementary length dl. › In terms of the components of A along three cartesian coordinates, we have Q

Ú

P

Q

Q

_› __›

dl

A.dl = Ú ( Ax + Ay + Az). ( dx + dy + dz) P Q

P

Ú (Ax dx + Ay dy + Az dz)

=

…(1.26)

P

Fig. 1.14 Elementary length dl along the curve PQ.

If the path of integration is a closed curve, then we write instead of Ú. If the value of line integral depends only on the initial and final points in the vector field and independence of the path then the vector field is called conservative field. All central force fields such as gravitational field and electrostatic field are conservative fields. For a conservative force field, _› __›

A ◊ dl = 0 …(1.27) _› _› c If the_line__integral over a closed path in a vector field A_ is zero, then A will be__the _gradient of a scalar func› › › › › tion i.e, A = —y, where y is the scalar point function. If A is conservative then — × A will be zero.

1.12.2

surface Integral

_›

The surface integral of a vector _› field F over a piecewise smooth surface S in space is defined as the integral of the normal component of F across the surface and can be written as _› __›

_›

ÚÚ F. dS or ÚÚ F ◊ dS S

S

where ds is the elementary surface on S and is a unit vector along the outward drawn normal to the surface. __› _› If F = Fx + Fy + Fz and ds = dy dz + dx dz + dx dy _› __›

ÚÚ F. dS = ÚÚ (Fx dy dz + Fy dx dz + Fz dx dy)

Then

S

S

The surface integral can also be written as _› __›

_›

Ú F. dS = Ú F ◊ dS

S

S

The notation ÚÚ is used for a closed surface S. Sometimes s may also be used. S

1.12

Advanced Engineering Physics

1.12.3 _›

volume Integral

Let F be a single-valued continuous vector function in volume V. _› Suppose the volume is divided into a large number of small volume elements (dV). The volume integral of F is the sum of the products of values of vector fields and the volume for all elements and for infinite large volume elements, volume integral _›

= ÚÚÚ F dV

_›

V

If F = Fx + Fy + Fz then volume integral = ÚÚÚ ( Fx + Fy + Fz) dV V

=

ÚÚÚ Fx dV + ÚÚÚ Fy dV + ÚÚÚ Fz dV V

V

V

For scalar point function y (x, y, z), volume integration will be = ÚÚÚ y dV = ÚÚÚ y dxdydz V

It is a scalar quantity.

V

_›

The notations Ú FdV or Ú y dV are also used to indicate volume integration for the respective vector and V V scalar fields.

1.13 InteGral theorems There are mainly three fundament integral theorems in relation to the integration of vector fields: (i) Gauss’ Divergence Theorem (ii) Stoke’s Theorem, and (iii) Theorem for Gradient. The first one is a correlation between a closed surface and its enclosed volume. The second one is the theorem for curls, which correlates a closed line to its enclosed surface. The third one is the theorem for gradient, which correlates closed points to a line.

1.13.1

Gauss’ divergence theorem

Statement The theorem states that the surface integral of the normal component of a vector field taken around a closed surface is equal to the integral of the divergence of the vector taken over the volume enclosed by the surface. _› Let us consider a volume V enclosed by a closed surface S. If a vector function F is continuously differentiable throughout V then from Gauss’ divergence theorem. __› _›

_› __›

ÚÚ F. dS = ÚÚÚ (—◊ F) dV or,

S_ ›

V

__› _›

ÚÚ F. dS = ÚÚÚ (—◊ F) dV S

…(1.28)

V

where is the outward __› _› drawn unit normal to the surface. Significance If F represents the velocity of a fluid, then the flux of F is the total amount of fluid passing through the surface per unit time. A point of positive divergence acts as a surface of the fluid measuring the amount of fluid it is producing but a point of negative divergence measures the amount of fluid it is absorbing. The divergence theorem is basically a statement of incompressibility of fluid. For incompressible fluid, the volume integral of the _divergence measures the net amount of fluid that is produced inside the region, › and the surface integral of F (velocity flux) over the closed surface S gives the net amount of fluid flowing

Vector Analysis

1.13

out through the surface per unit time. Hence, both are equal from the law of conservation of mass. So, from divergence theorem Amount of fluid produced inside the volume = net amount of fluid flowing out through the surface

1.13.2

stoke’s theorem

Statement The line integral of the tangential component of a vector taken around a closed path is equal to the surface integral of the normal component of the curl of the vector taken over any surface having the path as the boundary. _› Mathematically, the theorem states that if F is a continuously differentiable vector point function in a region of space and S is an open two-sided surface bounded by a closed, non-intersecting curve C, then _› __› c

c

_› __›

__›

_›

s

_› __›

or,

__›

F. dr = ÚÚ (— × F) ◊ dS F. dr = ÚÚ (— × F) ◊ dS

…(1.29)

s

where is the outward drawn unit normal to the surface S. Stoke’s theorem relates surface integral with line integral. _›

Significance Curl measures the amount of twist or rotation of the vector F at each point, the integral of the curl through a surface or flux of the curl can be determined if we go all around the edge and find how much of the fluid is flowing out of the boundary. There is no restriction on the shape of the surface S. The only condition is that the boundary of the surface S must coincide with the curve C.

1.14 CoordInate system For solving three-dimensional problems, we require a coordinate system. There are mainly three types of coordinate systems (i) cartesian coordinate system, (ii) cylindrical coordinate system, and (iii) spherical coordinate system.

1.14.1 Cartesian Coordinate system (x, y, z)

Z

In the cartesian coordinate system, let P be the position _› vector with respect to origin O given by [Fig 1.15] by r = x + y + z, where , , , are unit vectors along the X, Y and Z respectively. Here unit vectors are not the functions of the coordinates. Now elementary displacement _›

d r = dx + dy + dz

dz r

P dy

dx

z O

Y

...(1.30)

x

Elementary surface area in XY plane is y

dS = dx dy and the corresponding volume element dV = dx dy dz

X

Fig. 1.15 The elementary volume element dv in the cartesian coordinate system.

1.14

Advanced Engineering Physics

Cylindrical polar Coordinate system (r, j, z)

1.14.2

In the cylindrical coordinate system, the position of the point P is P(r, j, z) [Fig. 1.16(a,b)]. The unit vector at P is directed radially outward from the z axis. The unit vector is directed along the direction of increasing of j and the unit vector is along the direction of the z axis. z rdj

dj

z K

r

dz

dr dz

f

P

r

z

O

y

y x

j

f

dj

y

dr

x

x (a)

Fig. 1.16

(b)

(a) A point in cylindrical coordinate system. (b) The elementary volume element dv in cylindrical coordinate system.

From Fig. 1.16(b) x = r cos j, y = r sin j and z = z _› Again in the cartesian coordinates, the position vector r can be written as _› r = x+ y+ z = r cos j + r sin j + z 2

2

2

2

2

2

…(1.31) 2

The elementary curve length (dr) = (dx) + (dy) + (dz) = (dr) + r (dj) + (dz) and

2

_›

d r = dr + j (rdj) + dz

The volume element in cylindrical coordinate can be expressed as [from Fig 1.16(b)] dV = dr × rdj × dz = rdr dj dz The unit vectors in the cylindrical coordinates system are: _›

…(1.32) …(1.33) …(1.34)

_›

∂r ∂r = ___ ___ = cos j + sin j ∂r ∂r _›

/| /|

| | /| | _›

∂r ∂r j = ___ ___ = – sin j + cos j ∂j ∂j _›

_›

∂r ∂r = ___ ___ = ∂z ∂z

1.14.3

spherical polar Coordinate system (r, q, j)

In the spherical coordinate system, the position of the point A is A (r, q, j) [Fig 1.17]. The unit vector is directed radially outward. The unit vector is normal to the conical surface and it is directed to the directional in which q increases.

Vector Analysis

1.15

q dq

The unit vector j is along the direction in which j increases. The relation between cartesian and spherical polar coordinates is x = r sin q cos j, y = r sin q sin j and z = r cos q z So, the position vector _› r = x+ y+ z E = r sin q cos j + r sin q sin j B A + r cos q …(1.35) r If dl is the elementary length then in spherical D C coordinates __›

dl = dr + rdq + j r sin q d j 2

2

2

2

2

…(1.36)

2

y¢

dj

O

2

y f

H …(1.37) and dl = (dr) + r (dq) + r sin q (dj) G The surface element ds in spherical coordinates is obtained from Fig. 1.17. dS = AB × AC = r sin qdj × rdq = r2 sin q dq dj …(1.38) x F and volume element dV = dS × height (dr) Fig. 1.17 Surface element in spherical polar = r2 sin q dq dj dr …(1.39) coordinate system. The unit vectors in the spherical polar coordinate system are _› _› ∂r ∂r = ___ ___ = sin q cos j + sin q sin j + cos q ∂r ∂r

/| /|

|

_›

_›

_›

_›

∂r ∂r = ___ ___ = cos q cos j + cos q sin j – sin q ∂q ∂q

|

∂r ∂r j = ___ ___ = – sin j + cos j ∂j ∂j

and

1.14.4

/|

|

Gradient, divergence, Curl and laplacian in Cartesian, Cylindrical and spherical Coordinate system

(i) In cartesian coordinate system (a) Gradient of a scalar function y(x, y, z) __› ∂y ∂y ∂y —y = ___ + ___ + ___ ∂x ∂y ∂z _› (b) Divergence of vector field A (x, y, z) ∂Ax ∂Ay ∂Az — ◊ A = ____ + ____ + ___ ∂x ∂y ∂z _› (c) Curl of vector field A(x, y, z) __› _›

__›

| |

_› ∂ — × A = ___ ∂x Ax

∂ __ ∂ ___ ∂y ∂z Ay Az

1.16

Advanced Engineering Physics

(

=

∂Ay ∂Az ____ ___ – + ∂y ∂z

) (

∂Az ∂Ax ___ ____ – + ∂z ∂x

) (

∂A ∂Ax ____y – ____ ∂x ∂y

)

(d) Laplacian of y (x, y, z) ∂2y ____ ∂2y ____ ∂2y —2y = ____ + + ∂x2 ∂y2 ∂z2 (ii) In cylindrical coordinate system (a) Gradient of scalar function y (r, j, z) __› ∂y ∂y 1 ∂y —y = ___ + j __ ___ + ___ r ∂r ∂j ∂z _›

(b) Divergence of vector field A (r, j, z) __› _› ∂ ∂ 1 ∂ — ◊ A = ___ + j __ ___ + __ ◊ ( Ar + j Aj + Az) r ∂r ∂j ∂z ∂A 1 ∂ 1 j ∂Az = __ ___ (rAr) + __ ____ + ___ r ∂r r ∂j ∂z

(

)

_›

(c) Curl of vector field A(r, j, z) __›

rj ∂ ___ ∂j Aj

| |

_›

1 ∂ — × A = __ ___ r ∂r Ar (d) Laplacian of y (r, j, z)

∂ __ ∂z Az

2 2 ∂y 1 ∂ 1 ∂y ∂y —2y = __ ___ r ___ + __2 ____2 + ____ . r ∂r ∂r r ∂j ∂z2 (iii) In spherical coordinate system

( )

(a) Gradient of a scalar function y (r, q, j) __›

∂y 1 ∂y 1 ∂y —y = rˆ ___ + __r ___ + ______ ___ r sin q ∂j ∂r ∂q _›

(b) Divergence of vector field A (r, q, j) __› _› ∂ 1 ∂ 1 1 ∂Aj — ◊ A = __2 __ (r2 Ar) + ______ ___ (sin q Aq) + _____ ____ r sin q ∂q r sin q ∂j r ∂r _› (c) Curl of vector field A(r, q, j) r sin q j r ∂ ∂ ∂ 1 ___ __ ___ — × A = _______ 2 ∂j ∂r ∂q r sin q Ar rAq r sin q Aj (d) Laplacian of y (r, q, j) ∂y ∂y ∂2y ∂ 1 ∂ 1 1 ___ ___ _______ ____ —2y = __2 __ r2 ___ + _______ sin q + . ∂r ∂q r ∂r r2 sin q ∂q r2 sin2 q ∂j2 __›

_›

|

( )

|

(

)

Vector Analysis

1.17

Worked Out Problems __›

__›

__›

__›

If A = 4 + 3 + __, B = 2__ – + 2 find a unit vector perpendicular to vectors A and B. Find › › also the angle between the vector A and B. Example 1.1

Sol.

_›

_›

|

A×B= 4 2 _›

|

1 = 7 – 6 – 10 2

3 –1

_________________

_›

____

and |A × B| = ÷72 + (– 6)2 + (– 10)2 = ÷185 _›

_›

– 10 A × B 7 – 6____ The unit normal = ______ _› _› = ___________ 185 ÷ |A × B| _›

_›

If q is the angle between A and B, then__________ _›

_›

_› _›

_›

___

|A × B| = | A| | B | sin q, here | _A| = ÷__________ 42 + 32 + 12 = ÷26 › |B| = ÷22 + 12 + 22 = 3 _›

_›

____

÷185 |A × B| ___ \ sin q = ______ _› _› = _____ |A| ×____ |B| 3÷26 ÷185 ___ = 62.77° \ q = sin–1 _____ 3÷26

( )

_›

__›

_›

Example 1.2 Show that the vectors A = 3 – 2 + , B = – 3 + 5 and C = 2 + – 4 from the sides of a right-angled triangle. Sol. In our problem __› find that _› we B + C = ( – 3 + 5 ) +_(2 + – 4 ) › =3 –2 + =A \ The given vectors form a triangle. _›

________

_›

_________

___

__›

_________

___

___

|A| = ÷9 + 4 + 1 = ÷14

Again

|B| = ÷1 + 9 + 25 = ÷35 |C| = ÷4 + 1 + 16 = ÷21

_›

__›

_›

|A|2 + |C|2 = 14 + 21 = 35 = |B|2.

\

_› _› __›

A, B, C form a right-angled triangle.

Example 1.3 Sol.

_›

_›

_›

_›

_›

_›

_›

_›

_›

_›

_›

_›

If |A + B| = |A – B|, then show that A and B are perpendicular. |A + B| = |A – B|

or,

_›

_›

|A + B|2 = |A – B|2 _› _›

_› _›

or,

2 A ◊ _B = – 2 A ◊ B

or,

4 A ◊B = 0, So A and B are perpendicular to each other.

_›

›

_›

_›

1.18

Advanced Engineering Physics

Example 1.4 The three adjacent sides of a parallelepiped are represented by + 2 , 4 and + 3 . Calculate its volume. _ __› _› › Sol. Let A = + 2 , B = 4 and C = + 3 _› _›

__›

The volume of the parallelepiped is V = A ◊ (B × C) _›

__›

| |

B×C= 0 0

Here

_› _›

__›

4 0 = 12 1 3

\ A ◊ (B × C) = ( + 2 ) ◊ 12 = 12 So, volume V = 12 units For what_ value of m are the coplanar? __› _› following three vectors › A = 3 + 2 + , B = 3 + 4 + 5 and C = + – m

Example 1.5 Sol.

_› _›

__›

|

__›

|

_›

_›

Find the area of the parallelogram determined by the vectors A = 3 + 2 and B = 2 – 4 . The area of the parallelogram is

Example 1.6 Sol.

_› _›

Three vectors A, B and C will be coplanar if their scalar triple product is zero, i.e., A ◊ (B × C) = 0 3 2 1 So, 3 4 5 =0 1 1 –m or, 3(– 4m – 5) + 2 (5 + 3 m) + 1 (3 – 4) = 0 or, m = –1

_›

_›

|

|

A × B = 3 2 0 = –8 + 12 + 6 0 2 –4 ____________ ___ The magnitude of the area of the parallelogram is ÷64 + 144 + 36 = 2÷61 units. Example 1.7 Sol.

\

q If â and are unit vectors and q is the angle between them, show that 2 sin __ = |â – |. 2 |â – |2 = (â – ) ◊ (â – ) [WBUT 2012] = |â|2 + | |2 – 2â ◊ = 1 + 1 – |â| | | cos q = 2 (1 – cos q) q = 4 sin2 __ 2 q |â – | = 2 sin __ 2

|â| = 1 | |=1

_›

Example 1.8 Find the torque about the point 0 (2, – 1, 3) of a force F( 3, 2, – 4) passing through the point P (1, – 1, 2). Sol. The position vector of P relative to O is _› r = ( – + 2 ) – (2 – + 3 ) =– –

Vector Analysis

1.19

_›

Again force F=3 +2 –4 _› _› _› The required moment or torque is t = r × F = –1 3

|

_›

_›

_›

_›

_›

_›

_›

_›

_›

_›

_›

_›

_›

_›

_›

_›

Show that a × ( b + c ) + b × ( c + a ) + c × ( a + b ) = 0 _› _› _› _› _› _› _› _› _› _› _› _› We have a × b + a × c + _b × c + b × _a + c × a + _c × b _› _› _› _› _› _› _› _› › › › _› =a×b+a×c+b×c–a×b–a×c–b×c =0

Example 1.10 Sol.

_›

_›

|

–1 = 2 – 7 – 2 –4

If a , b , _c satisfy the relation a + b + c = 0 show that a × b = b × c = c × a . _› _› › We have a +_b = – _c _› _› › › \ _ b × ( a_ + b_) = – _b × c _› _› › › › › or, b × a + _b × b = – _b × c _› _› _› _› › › or, b × _a = –_ b × c = c × b _› _› › › or, _a› × _b = _b × _c › › › Similarly, b × _c = _c × a _› _› _› _› › › \ a×b=b×c=c×a

Example 1.9 Sol.

_› _› _›

0 2

Example 1.11 A particle is acted upon by constant forces ( 5 + 2 + ) and (2 – – 3 ) and is displaced from the origin to the point (4 + – 3 ). Show that the total work done by the forces is 35 units. Sol. The displacement produced in moving from origin to the point (4 + – 3 ) is (4 + – 3 ). So, the total work done = (5 + 2 + ) ◊ (4 + – 3 ) + (2 – – 3 ) ◊ (4 + – 3 ) = 20 + 2 – 3 + 8 – 1 + 9 = 35 units Example 1.12 A charge of 2.0 mC moves with a speed of 2.0 × 106 m/s along the positive x axis. A magnetic field B of strength (0.20 + 0.40 ) _tesla exists in space. Find the magnetic force acting on the charge. _› › Sol. The force on the charge = q( v × B) = 2 × 10– 6 × [2 × 106 × (0.2 + 0.4 )] = (0.8 – 1.6 ) Newton _›

_› _› dA If A(t) has a constant magnitude then show that ___ is perpendicular to A. dt _› _› We know that A ◊ A = | A |2 = constant

Example 1.13 Sol.

_›

_›

_›

_›

_› dA dA _› dA \ _ A ◊ ___ + ___ ◊ A = 0 or, 2A ◊ ___ = 0 dt _ dt dt › › dA So, ___ is perpendicular to A. dt

Example 1.14 Sol.

Here

__›

If j (x, y, z) = 3x2y – y3z2, find —j at the point (1, – 2, – 1). j (x, y, z) = 3x2y – y3 z2

[WBUT 2004, 2006]

1.20

Advanced Engineering Physics __›

∂j ∂j ∂j —j = ___ + ___ + ___ ∂x ∂y ∂z ∂j ___ ∂ 2 3 2 ___ = (3x y – y z ) = 6xy ∂x ∂x ∂j ___ ∂ ___ = (3x2y – y3z2) = 3x2 – 3y2z2 ∂y ∂y ∂j __ ∂ ___ = (3x2y – y3z2) = – 2y3z ∂z ∂z

Again

__›

\

—j = 6xy + (3x2 – 3y2z2) – 2y3z __›

Now at (1, – 2, – 1), —j |1, – 2, – 1 = [(6 × (1) (– 2)] + [3 – 3 × (– 2)2 × (– 1)2] – [2 × (– 2)3 (– 1)] = – 12 – 9 – 16 Example 1.15

Find the directional derivative of y (x, y, z) = xy2z + 4 x2z at (– 1, 1, 2) in the direction

(2 + – 2 ). Sol.

__›

—y =

Here

∂ ∂ ∂ ___ + ___ + __ (xy2z + 4x2z) ∂x ∂y ∂z

(

)

= (y2z + 8xz) + 2xyz + (xy2 + 4x2) at (–1, 1, 2)

__›

—y = – 14 – 4 + 3

The unit vector in the direction of 2 + – 2 is 2 + –2 2 __________ = __ = ___________ 2 2 2 ÷2 + 1 + 2 3

1 2 + __ – __ 3 3

__

› 2 1 2 Hence the directional derivative is —y ◊ = (– 14 – 4 + 3 ) ◊ __ + __ – __ = – 38/3 3 3 3 __› Example 1.16 Show that —y is a vector perpendicular to the surface y (x, z, y) = c, where c is a constant.

Sol.

(

)

_›

_›

Let the position vector r = x + y + z belong to the surface y (x, z, y) = constant. Then d r = dx + dy + dz, represents a vector which is tangent to the surface y. ∂y ∂y ∂y Again dy = ___ dx + ___ dy + ___ dz = 0 ∂x ∂y ∂z ∂y ∂y ∂y = ___ + ___ + ___ ◊ ( dx + dy + dz) ∂x ∂y ∂z

(

__›

)

_›

= —y ◊ d r = 0 _› So, —y is perpendicular to d r and therefore perpendicular to the surface. __›

Example 1.17 Sol.

Let \

Find a unit normal to the surface z2 = x2 – y2 at the point (1, 0, – 1). y (x, y, z) = x2 – y2 – z2 __› ∂ ∂ ∂ —y = ___ + ___ + __ (x2 – y2 – z2) ∂x ∂y ∂z

(

)

Vector Analysis

1.21

= 2x – 2y – 2z —y |1, 0, –1 = 2__ + 2 = 2( + ) › —y ( + ) 2( + ) 1__ ____ __ __ + ___ __ = ± ___ ______ = ± _______ Now unit normal = __› = ± 2 _______ = ± ___ ( + ) 2 2 2 2 2 2 2 ÷ ÷ ÷ ÷ ÷2 + 2 |—y| 1, 0, – 1 Positive and negative signs imply that the unit normal is either outward or inward. __›

[

|

] [

]

Example 1.18 In what direction from the point (1, 3, 2) is the directional derivative of y = 2xz – y2 a maximum? What is the magnitude of this maximum? __› ∂ ∂ ∂ Sol. Here —y = ___ + ___ + __ (2xz – y2) ∂x ∂y ∂z

(

__›

)

= 2z – 2y + 2x

—y |1, 3, 2 = 4 – 6 + 2 __› The directional derivative is maximum in the direction —y = 4 – 6 + 2 ________________

__›

___

___

The magnitude of this maximum is |—y | = ÷(4)2 + ( – 6)2 + (2)2 = ÷56 = 2÷14

Example 1.19 Find the values of the constants a, b, c so that the directional derivative of y = axy2 + byz + cz2 x3 at (1, 2, – 1) has a maximum of magnitude 64 in a direction parallel to the z axis. __› ∂ ∂ ∂ Sol. Here —y = ___ + ___ + __ (axy2 + byz + cz2 x3) ∂x ∂y ∂z = (ay2 + 3x2 cz2) + (2 axy + bz) + (by + 2cz x3)

(

)

__›

—y|1, 2, – 1 = (4a + 3c) + (4a – b) + (2b – 2c)

Now

Since y has a maximum magnitude along the z axis, so __›

and

—y|1, 2, – 1 ◊ = 64 4a + 3c = 0 4a – b = 0

__›

_›

Show that —rn = nr n – 2 r __› ∂ ∂ ∂ —rn = ___ + ___ + __ (x2 + y2 + z2)n/2 ∂x ∂y ∂z n n = __ (x2 + y2 + z2)n/2 – 1 ◊ 2x + __ (x2 + y2 + z2)n/2 – 1 ◊ 2y + 2 2 = n(x2 + y2 + z2 )n/2 – 1 ( x + y + z) _› _› = n(r2)n/2 – 1 r = nrn – 2 r

Example 1.20 Sol.

or, (2b – 2c) = 64 Now solving these equations, we have a = 6 b = 24 c=–8

(

{

)

} {

} { __n2 ( x + y + z ) 2

2

2 n/2 – 1

◊ 2z

}

Example 1.21 Find the angle between the surfaces x2 + y2 = 9 – z2 and x2 + y2 = (z + 3) at the point (2, – 1, 2). Sol. Let y1 = x2 + y2 + z2 – 9 and y2 = x2 + y2 – z –3 These two surfaces are constant.

1.22

Advanced Engineering Physics __› __›

—y1

and

| __

=4 –2 +4

| __

=4 –2 –

2, –1, 2 ›

__›

—y2 = —(x2 + y2 – z – 3) = 2x + 2y –

Again __› __›

__›

—y1 = —(x2 + y2 + z2 – 9) = 2x + 2y + 2z

Now

—y2

2, –1, 2 ›

Since —y1 and —y2 are normal to the surfaces y1 and y2, the angle between the surfaces is the angle between the normal to the surfaces. Let q be the angle between the surfaces. So,

__›

__›

__›

__›

__›

__›

—y1 ◊ —y2 = |—y1| |—y2| cos q —y1 ◊ —y2 _______________________________ (4 – 2 + 4 ) ◊ (4 – 2 – ) 16 ___ cos q = __________ = _____________ ________________ = _____ __› __› 2 2 2 2 2 2 6 21 ÷ |—y1| |—y2| ÷4 + (– 2) + 4 ÷4 + (– 2) + ( – 1)

or,

( )

16 ___ = 54.41° q = cos–1 _____ 6÷21

or,

Example 1.22 Find the unit vector perpendicular to x2 + y2 – z2 = 100 at the point (1, 2, 3). [WBUT 2007] Sol. Here x2 + y2 – z2 – 100 __›y = __ › or, —y = —(x2 + y2 – z2 – 100) = 2x + 2y – 2z or

__›

—y |1, 2, 3 = 2 + 4 – 6 __›

—y + 2___– 3 2 +4 –6 _____________ = ± _________ Now unit normal = ____ __› = _______________ 2 2 2 ÷14 |—y | ÷2 + 4 + (– 6 ) Example 1.23 Sol.

Here

Example 1.24 Sol.

__› _ ›

Show that — ◊ r = 3 __› _ ∂ ∂ ∂ › — ◊ r = ___ + ___ + __ ◊ ( x + y + z) ∂x ∂y ∂z =1+1+1=3

(

)

__› _ _ › ›

_›

_›

Show that —( a ◊ r ) = a , where a is a constant vector. _›

_› _›

r = x + y + z, a ◊ r = ( a1 + a2 + a3) ◊ ( x + y + z) = a1x + a2y + a3z __› _ _ ∂ ∂ ∂ › › —( a ◊ r ) = ___ + ___ + __ (a1x + a2y + a3z) ∂x ∂y ∂z

Here

(

Now

)

_›

= a1 + a2 + a3 = a __› _›

Sol.

(

Example 1.26 Sol.

_›

Find — ◊ F if F = 2x2z – xy2z + 3yz2 __› _› ∂ ∂ ∂ We have — ◊ F = ___ + ___ + __ ◊ (2x2z – xy2z + 3yz2 ) ∂x ∂y ∂z = 4xz – 2xyz + 6yz

Example 1.25

)

Determine _› the constant a so that the vector field F = (2x + 3y) + (3y –2z) + (y + az) is solenoidal.

_›

__› _›

If vector F is solenoidal then — ◊ F = 0

Vector Analysis

1.23

__› _›

∂ ∂ ∂ — ◊ F = ___ (2x + 3y) + ___ (3y – 2z) + ___ (y + az) = 0 ∂x ∂y ∂z =2+3+a=0 or, a=–5 _› _› _› i x + jy Example 1.27 Show that the vector field F = ________ ______ is a “source” field. 2 __› _› __› _› ÷x + y2 Sol. For a source field, — ◊ F should be positive and for a sink field, — ◊ F should be negative. _› x+ y ______ Here F = ________ ÷x2 + y2 Here

__› _›

)(

x+ y ∂ ∂ ∂ ___ ______ + ___ + __ ◊ _______ ∂x ∂y ∂z ÷x2 + y2 y ∂ ______ ∂ ______ x = ___ _______ + ___ _______ ∂x x2 + y2 ∂y x2 + y2

—◊F =

\

(

) (÷

(÷

)

)

y2 x2 1 1 _________ _________ = – _________ + – + _________ 2 2 3/2 2 2 1/2 2 2 3/2 2 (x + y ) (x + y ) (x + y ) (x + y2)1/2 (x2 + y2) 2 = – _________ + _________ 2 2 3/2 2 (x + y ) (x + y3)1/2 1 2 1 ______ = – _________ + _________ = _______ 2 2 1/2 2 2 1/2 2 (x + y ) (x + y ) ÷x + y2

__› _›

_›

\ — ◊ F is + ve, so F is a source field. Example 1.28 Sol.

_›

__›

()

r Evaluate — ◊ __3 r __›

_›

__›

_›

__› _›

We know that — ◊ (y A) = —y ◊ A + y — ◊ A _› _ › 1 Here we consider y = __3 and A = r r __› _› __› _ __ 1 › 1 › _› r __ So, — ◊ 3 = —__3 ◊ r + __3 — ◊ r r r r

()

__

__

› › _› _› 3 _› = – 3r–5 r ◊ r + __3 [ —rn = nrn – 2 r and — ◊ r = 3] r 3 3 3 = – 3r–3 + __3 = – __3 + __3 = 0 r r r

Example 1.29 Sol.

_›

__› _›

__› _›

__›

_›

Find — ◊ F and — × F where F = —(x3 + y3 + z3 – 3xyz). __›

[WBUT 2001]

Here, F = —(x + y + z – 3xyz) ∂ ∂ ∂ = ___ (x3 + y3 + z3 – 3xyz) + ___ (x3 + y3 + z3 – 3xyz) + ___ (x3 + y3 + z3 – 3xyz) ∂x ∂y ∂z = (3x2 – 3yz) + (3y2 – 3xz) + (3z2 – 3xy) __› _› ∂ ∂ ∂ Now — ◊ F = ___ + ___ + __ ◊ [(3x2 – 3yz) + (3y2 – 3xz) + (3z2 – 3xy) ] ∂x ∂y ∂z 3

3

3

(

)

1.24

Advanced Engineering Physics

= 6x + 6y + 6z = 6 (x + y + z) __›

_›

—×F=

Again

|

+

|

∂ ___ ∂y

∂ ___ ∂x 3x2–3yz

∂ __ = ∂z 3z2–3xy

2

3y –3xz

[

[

∂ ∂ ___ (3z2 – 3xy) – ____(3y2 – 3xz) ∂y ∂z

∂ ∂ __ (3x2 – 3yz) – ___ (3z2 – 3xy) + ∂z ∂x

] [

]

∂ ∂ ___ (3y2 – 3xz) – ___ (3x2 – 3yz) ∂x ∂y

]

= (–3x + 3x) + (–3y + 3y) + (–3z + 3z) = 0 Example 1.30 Sol.

Here

1 Prove that —2 ln r = __2 r 1 ln r = ln (x2 + y2 + z2)1/2 = __ ln (x2 + y2 + z2) 2 __› __› __› __› 1 2 __ — ln r = — ◊ — ln r = — ◊ — ln (x2 + y2 + z2) 2 __› ∂ ∂ ∂ 1 ___ __ ___ + + __ ln (x2 + y2 + z2) = —◊ 2 ∂x ∂y ∂z

[{

}

(

[WBUT 2007]

]

__› _›

)

__ x+ y+ z r 1 › = __ — ◊ 2 __________ = — ◊ __2 2 2 2 2 r x +y +z __› _› r 2 __ So, — ln r = — ◊ 2 r__ __› __› _› _› _› › We know that — ◊ (y A) = —y ◊ A + y — ◊ A __

So,

Example 1.31

__›

Sol.

|

_› ∂ — × r = ___ ∂x x

Sol.

Example 1.32 _›

__

› 1 _› 1 › _› —2 ln r = — __2 ◊ r + __2 — ◊ r r r _ _ 3 2 3 1 –4 › › = – 2r ( r ◊ r ) + __2 = – __2 + __2 = __2 r r r r __› _ › Show that — × r = 0

|

∂ ___ ∂y y

∂ __ = ∂z z

(

∂ ∂ ___ z – __y + ∂y ∂z

) (

∂ ∂ __ x – ___z + ∂z ∂x

) (

∂ ∂ ___ y – ___ x = 0 ∂x ∂y

)

_›

Show that the vector F = (4 xy – z3) + 2x2 – 3xz2 is irrotational.

If F is an irrotational vector then

__›

_›

__›

_›

—×F=0 —×F=

Now

=

[

∂ ∂ ___ (– 3xz2) – __ (2x2) + ∂y ∂z

] [

|

∂ ___ ∂x 4xy–z3

∂ ___ ∂y 2x2

∂ __ ∂z –3xz2

|

∂ ∂ __ (4xy – z3) – ___ (–3xz2) + ∂z ∂x

] [

∂ ∂ ___ (2x2) – ___(4xy – z3) ∂x ∂y

]

Vector Analysis

1.25

= 0 + (– 3z2 + 3z2) + (4x – 4x) = 0 Example 1.33 Sol.

Let

__› __›

__›

_›

__› __›

_›

(

—×F=

Now Again

_›

Show that _› — ◊ — × F = 0 F = F1 + F2 + F3

—◊— × F =

(

∂F3 ∂F1 ____ ____ – + ∂z ∂x ∂F3 ∂F2 ∂ ∂ ∂ ___ + ___ + __ ◊ ____ – ____ ∂x ∂y ∂z ∂y ∂z

∂F3 ____ ∂F2 ____ – + ∂y ∂z

) (

) (

∂F1 ∂F2 ____ ____ – ∂x ∂y ∂F1 ∂F3 + ____ – ____ + ∂z ∂x

)

)[( ) (

) ( ) (

) ( )

)(

)(

)

∂ ∂F3 ∂F2 ∂ ∂F1 ∂F3 ∂ ∂F2 ∂F1 = ___ ____ – ____ + ___ ____ – ____ + __ ____ – ____ ∂x ∂y ∂z ∂y ∂z ∂x ∂z ∂x ∂y

(

(

∂F1 ∂F2 ____ ____ – ∂x ∂y

)]

∂2 F3 ∂2F2 ∂2F1 ∂2F3 ∂2F2 ∂2F1 = _____ – _____ + _____ – _____ + _____ – _____ = 0 ∂x ∂y ∂x ∂z ∂y ∂z ∂y ∂x ∂z ∂x ∂z ∂y Example 1.34 Sol.

__›

_›

Show that F = (2xy + z3) + x2 + 3xz2 is a conservative force field. [WBUT 2004, 2006]

|

_›

Here — × F =

∂ ___ ∂y

∂ ___ ∂x 2xy + z3

x

2

∂ __ ∂z 3xz2

__›

|

=

(

)

∂ ∂ ___ 3xz2 – __x2 + ∂y ∂z

[

∂ ∂ __ (2xy + z3) – ___ 3xz2 ∂z ∂x

]

∂ 2 ___ ∂ ___ x – (2xy + z3) ∂x ∂y

+

[

_›

_›

]

2 2 _› = 0 + (3z – 3z ) + (2x – 2x) = 0

For a conservative force field, — × F = 0 _›

Sol.

_›

If the vectors A and B are irrotational then show that the vector A × B is solenoidal. [WBUT 2006] _› _› If A and B are then __› irrotational __› _› _› — × A = 0 and — × B = 0

Example 1.35

__› _›

__›

_›

_› _›

__›

_›

_›

Now _— ◊ (_A × B) = (— × A) ◊ B – (— × B ) ◊ A = 0 – 0 = 0 › › Hence (A × B) is solenoidal. Example 1.36 Sol.

Let So that Again So, or,

Example 1.37 Sol.

__›

__›

_›

_›

__› __› _›

2 Show that — ×__(— × _› A) = – — A + — (— ◊ A) › = _P _ _ __› __› _— › › › › — × (— × A) = P × (P × A)

_› _ _› _ _ _› _› _› _› › › › a × ( b × c ) = b ( a ◊ c ) – ( _› _› _› _› _› _› _a› ◊ _b›) _c› P__× (P__ × A_) = __ P (P ◊ A ) – ( P ◊_P) A › › › __› _› › ›

— × — × A = — (— ◊ A) – —2 A

__›

_›

_›

__›

_›

__› _ ›

If w is a constant vector, r is the position vector and v = w × r . Prove that — ◊ v = 0 __› _ ›

__›

__›

_›

— . v = — ◊ (w × r )

1.26

Advanced Engineering Physics

)| )

|

∂ ∂ ∂ ___ + ___ + __ ◊ w1 w2 w3 ∂x ∂y ∂z z x y ∂ ∂ ∂ = ___ + ___ + __ ◊ [ (w2z – w3y) + (w3x –w1z) + (w1y – w2 x) ∂x ∂y ∂z =0+0+0=0 =

__›

__›

Show that — × —y = 0 __› __› ∂ ∂ ∂ — × —y = ___ + ___ + __ × ∂x ∂y ∂z

Example 1.38 Sol.

( (

)(

(

i ∂ ___ = ∂x ∂y ___

j k ∂ ∂ ___ __ = ∂y ∂z ∂y ∂y ___ ___ ∂y ∂z

| | ∂x

Example 1.39 ∂2y —2y = ____ . ∂t2 Sol.

[

∂y ∂y ∂y ___ + ___ + ___ ∂x ∂y ∂z

] [

∂2y ∂2y _____ _____ – + ∂y ∂z ∂z ∂y

)

] [

∂2y ∂2y _____ _____ – + ∂z ∂x ∂x ∂z

_

]

∂2y ∂2y _____ _____ – =0 ∂x ∂y ∂y ∂x

_

› __ › _› _› ∂B › _› ___ ∂E , then show that E and B satisfy If — ◊ E = 0, — ◊ B = 0, — × E = – ___ _› , — × B = ∂t ∂t

__›

_›

__›

__›

_›

_›

_›

__›

__›

_›

__›

_›

__› __› _›

_›

__

( )

( )

_›

∂B ∂ › _› ∂ ∂E ∂2 E We have — × — × E = — × – ___ = – __ (— × B) = – __ ___ = – ____ ∂t ∂t ∂t ∂t ∂t2 __›

_›

_›

But — × — × E = —(— ◊ E) – —2E = – —2E _›

_ 2 ›

∂2 E – — E = – ____ ∂t2

_›

_ 2 ›

∂2E So, \ — E = ____ ∂t2 _› _› _› __› __› _› __› __› _› ∂E ∂ ∂ ___ ∂B ∂2 B ___ __ __ ____ Similarly, — × — × B = — × = (— × E) = – =– 2 ∂t ∂t ∂t ∂t ∂t __›

__›

_›

__› __› _›

( )

( )

_›

_›

But — × — × B = — (— ◊ B) – —2B = – —2B _ 2 ›

_›

∂2 B — B = ____ ∂t2

So,

_› _› ∂2y i.e., E and B satisfy the equation —2y = ____ ∂t2

()

1 Show that —2 __r = 0 ∂2 ___ ∂2 ___ ∂2 ___________ 1 1 __ ___ __________ r = ∂x2 + ∂y2 + ∂z2 2 ÷x + y2 + z2

Example 1.40 Sol.

—2

()

(

[WBUT 2011]

)

∂ ∂2 1 Now ___2 (x2 + y2 + z2)– ½ = ___ – __ (x2 + y2 + z2)–3/2 (2x) 2 ∂x ∂x ∂ = – ___ [x (x2 + y2 + z2) – 3/2 ] ∂x

[

]

Vector Analysis

1.27

= – (x2 + y2 + z2)– 3/2 + 3x2(x2 + y2 + z2)–5/2 2x2 – y2 – z2 = _____________ (x2 + y2 + z2)5/2 2y2 – z2 –x2 ∂2 Similarly, ___2 (x2 + y2 + z2)–1/2 = _____________ ∂y (x2 + y2 + z2)5/2 2z2 – x2 – y2 ∂2 2 2 2 –½ ___ _____________ (x + y + z ) = ∂z2 (x2 + y2 + z2)5/2

and

(

)

∂2 ∂2 ∂2 1 —2 __r = ___2 + ___2 + ___2 (x2 + y2 + z2) – ½ ∂x ∂y ∂z

()

So,

2x2 – y2 – z2 + 2y2 – z2 –x2 + 2z2 – x2 – y2 = _________________________________ =0 (x2 + y2 + z2)5/2 Example 1.41

_›

_›

y. Sol.

__›

Show that A = (6xy + z3) + (3x2 – z) + (3xz2 – y) is irrotational. Find y such that A = — [WBUT 2012] __›

_›

_›

A vector A is called irrotational if — × A = 0

Here

__›

_›

—×A=

|

∂ ___ ∂x 6xy + z3

∂ ___ ∂y 2

3x –z

|

∂ __ = ∂z 3xz2–y

[

∂ ∂ ___ (3xz2 – y) – __ (3x2 – z) ∂y ∂z

[

+

+

]

∂ ∂ __ (6xy + z3 ) – ___ (3xz2 – y) ∂z ∂x

]

∂ ∂ ___ (3x2 – z) – ___ (6xy + z3) ∂x ∂y

]

[

= (– 1 + 1) + (3z2 – 3z2) + (6x – 6x) =0 _› So, A is irrotational. Here Then

__› ∂y ∂y ∂y A = —y = ___ + ___ + ___ = (6xy + z3) + (3x2 – z) + (3xz2 – ) ∂x ∂y ∂z ∂y ∂y ∂y ___ = 6xy + z3, ___ = 3x2 – z, ___ = 3xz2 – y ∂x ∂y ∂z _›

Integrating first with respect to x, keeping y and z constant, x2 y = 6 __y + xz3 + C1(y, z) = 3x2y + xz3 + C1 (y, z) 2 Integrating second with respect to y, keeping x, and z constant,

1.28

Advanced Engineering Physics

y = 3x2y – yz + C2 (x, z) Integrating third with respect to z, keeping x and y constant, y = xz3 – yz + C3 (x, y) Comparison of all equations in y, shows that there will be a common value of y if we choose C1 (x, z) = – yz, C2 (x, z) = xz3, C3 (x, y) = 3x2y y = 3x2y + xz3 – yz + C [where C is pure constant]

So that Example 1.42 a > 0. Sol.

_›

_›

__

_› › r Show that E = __2 is irrotational. Find y such that E = – —y and such that y (a) = 0 where r __›

__

_›

__

__

_› › r 1 › _› › 1 _› — × E = — × __2 = __2 — × r + —__2 × r r r r

2 _› _› = 0 + – __4 r × r = 0 r

(

)

So E is irrotational.

_

› __› _› ∂y r Since E is irrotational, so E = – — y = __2 = _r , or, – ___ = _r ∂r r ∂y 1 ___ __ =– r or, ∂r Now integrating on both sides, y = – ln r + C where C is constant Applying boundary condition, y (a) = 0, so C = ln a a \ y = – ln r + ln a = ln __r d2f (r) 2 _____ df (r) Example 1.43 Show that —2 f (r) = ______ + __ r 2 dr dr

( )

Sol.

We know that

__› __›

—2 f (r) = — . — f (r) _ __› df (r) __› r› _____ df (r) = — . _____ = — . __ r dr dr __› _ __› df (r) › 1 df (r) _› 1 = __r _____ (— . r ) + — __r _____ . r dr __ __dr _ __ _ _

(

›

›

›

›

›

)

›

[Here we have used —.(y A) = y —. A + —y . A] 3 df (r) d 1 df (r) _› = __r _____ + __ __r _____ . r dr dr dr

(

3 df (r) = __r _____ + dr

[

)

]

d2f (r) __ df (r) _› 1 ______ 1 _____ __ r dr2 – r2 dr . r

2

3 df (r) d f (r) __ 1 df (r) = __r _____ + ______ – r _____ [ 2 dr dr dr d2f (r) __ 2 df (r) = ______ + r _____ 2 dr dr

_›

. r = r]

Vector Analysis

Example 1.44 Sol.

1.29

Find f (r) such that —2 f (r) = 0

In the previous problem, we have seen that d2f (r) __ 2 df (r) —2f(r) = _____ + r _____ 2 dr dr 2 d f (r) 2 df (r) + __r _____ = 0 But, here —2 f(r) = 0 so ______ 2 dr dr df (r) d 1 2 __ __ r _____ = 0 or, dr r2 dr

(

)

df (r) r2 _____ = A (constant) dr

Now integrating

df (r) __ A _____ = 2 dr r A f(r) = – __ r + B (where B is constant) A A¢ __ f(r) = B – __ r = B + r (where A = – A¢)

or, Again integrating So,

_›

Calculate the amount of work done in moving a particle in a force field F = xy + yz + xz _› along the curve r = t + t2 + t3, where t is varying from – 1 to 1. Sol. We know that position vector r= x+ y+ z Example 1.45

= t + t2 + t3 _› So x = t, y = t and z = t , the parametric equation of the curve. Here, F = xy + yz + xz = t (t2) + t2 (t3) + t (t3) = t3 + t5 + t4 2

and

3

dr __ = dt

+ 2t + 3t2 _›

+1

_›

Now work done W = Ú F. d r = Ú (t3 + t5 + t4 ) . ( + 2t + 3t2 ) dt c +1

–1 +1

= Ú (t3 + 2t6 + 3t6) dt = Ú (t3 + 5t6) dt –1

|

–1

|

7 +1

4

t t = __ + 5 __ 7 4

–1

10 = ___ 7 _›

Example 1.46 Find the work done in moving a particle in the force field F = 3x2 + (2xz – y) + z along the straight line from (0, 0, 0) to (2, 1, 3). Sol.

Work done

_› __›

W = Ú F.dr = Ú [3x2 + (2xz – y) + z ] . ( dx + dy + dz) c

c

= Ú [3x dx + (2xz – y) dy + zdz] 2

c

We know that the equation of a straight line x – x1 ______ y – y1 _____ z – z1 ______ x2 – x1 = y2 – y1 = z2 – z1 = t (say)

1.30

Advanced Engineering Physics

or,

y – 0 _____ z–0 x – 0 _____ _____ = = =t 2–0 1–0 3–0

So,

x = 2t,

y=t

and z = 3t

The points (0, 0, 0) and (2, 1, 3) correspond to t = 0 and t =1 So, work done W = Ú [3x2 dx + (2xz – y) dy + zdz] c

1

= Ú [12t2 (2dt) + (12t2 – t) dt + 9 t dt] 0 1

= Ú (36 t2 + 8t) dt = 16. 0

_›

Example 1.47 Show that the force F = (2xy + z3) + x2 + 3xz2 is conservative. If this force acting on a particle displaces it from the point (0, 1, 2) to (4, 2, 3), calculate the work done. Sol. For the first part, see Example 1.34. B_ ›

B

__›

For the second part, W = Ú F.dr = Ú Fx dx + Fy dy + Fz dz) A

A

= Ú (2xy + z ) dx + x2dy + 3xz2 dz 3

B

= Ú (2xy dx + x2 dy) + (z3 dx + 3xz2 dz) A

(4, 2, 3)

=Ú

(4, 2, 3)

d (x2y) + d(xz3) = Ú

(0, 1, 2)

d (x2y + xz3)

(0, 1, 2)

2

=x y+

2, 3 xz3 |4, 0, 1, 2

= 16 × 2 + 4 × 27 – 0

= 140 Example 1.48 Show that for a conservative force field, work done in moving a particle from one point P1 (x1, y1, z1) in this field to another point P2 (x2, y2, z2) is independent of the path joining the two points. Sol.

__›

_›

For a conservative force field, _› F_ = —y › Now work done dW = F. d r Total work done

P2 _ ›

_›

P2

__›

_›

W = Ú dW = Ú F. d r = Ú —y . d r P1

P2

( Ú(

=Ú

P1

P2

=

P1

P2

P1

∂y ∂y ∂y ___ + ___ + ___ . ( dx + dy + dz) ∂x ∂y ∂z

)

∂y ∂y ∂y ___ dx + ___ dy + ___ dz ∂ ∂y ∂z

= Ú dy = y (P2) – y (P1) P1

)

Vector Analysis

1.31

Thus we see that work done depends only on the initial and final points but is independent of the path joining them. Example 1.49 Show that the work done on a particle in moving it from A to B equals its change in kinetic energies at these points whether the force field is conservative or not. Sol. Let a force acting on a particle displace it from the point A to another point B. Then work done B_ ›

__›

W = Ú F. dr A

_›

_›

_› dv We know that F = m ___ where v is the velocity of the body. dt _›

B

B

_› dv _› dv W = Ú m ___ . d r = Ú m ___ . v dt dt dt A A

So,

B

B

_› _› 1 = m Ú d v . v = m Ú v dv = __ m (vB2 – vA2) 2 A A

So, work done on a particle in moving it from A to B equals its change in kinetic energies at these points and does not depend on the nature of the force. _› __›

_›

Evaluate F. dr where F = (2x – y + z) + (x + y – z2) + (3x – 2y + 4z) and C is the closed

Example 1.50

c

curve in the xy plane, x = 3 cos t, y = 3 sin t from t = 0 to 2p. Sol.

__›

_›

In the plane xy, z = 0, F = (2x – y) + (x + y) + (3x – 2y) and dr = dx + dy _›

_›

F. d r = (2x – y) dx + (x + y) dy

So,

Again x = 3 cos t or, dx = – 3 sin t dt So, the total work done _›

_›

and and

y = 3 sin t dy = 3 cos t dt

2p

W = F. d r = Ú (6 cos t – 3 sin t) (– 3 sin t dt) + (3 cos t + 3 sin t) (3 cos t dt) c

0

2p

2p

)

(

1 = Ú (9 – 9 sin t cos t) dt = 9 Ú 1 – __ sin 2t dt 2 0 0

[

1 = 9 t + __ cos 2 t 4

]

2p 0

[

]

1 1 = 9 2p + __ – __ = 18 p 4 4

_›

Evaluate ÚÚ r . dS over the surface S of the unit cube bounded by the coordinate planes and

Example 1.51

S

the planes x = 1, y = 1, z = 1 Sol.

_›

For any surface S, the projection of that surface on the plane xy (See Appendix A) will be ÚÚ A . S

dS = ÚÚ R

_›

dxdy A . ______ where R is the region on the plane xy. Similarly, on the other planes, yz and zx | . |

1.32

Advanced Engineering Physics _› _› dzdx dydz will be ÚÚ A . _____ and ÚÚ A . _____ From Fig. 1.1W, over the surface ABEF, unit normal = and R R | . | | . | x=1 _› So, r . = ( x + y + z) . = x = 1 11

_

dydz

11

| . |

00

ÚÚ r›. dS = Ú Ú _____ = Ú Ú dydz = 1

\

ABEF

00

z

Over the surface OCDG, unit normal =– and x = 0 _›

r.

So,

=–x=0

F

E

12

_›

ÚÚ r . dS = Ú Ú 0 dydz = 0

\

OCDG

00

Over the surface BCDE, unit normal = and y = 1 = ( x + y + z) . = y = 1

_›

r.

So,

11

_

dxdz

11

ÚÚ r›. dS = Ú Ú ____ = Ú Ú dxdz = 1

\

BCDE

. | Over the surface AOGF, unit normal =– and y = 0

00 |

_›

r.

So, \

00

O A

_

=–y=0

00

Over the surface EDGF, unit normal = and z = 1 _› So, r . = ( x + y + z) . = z = 1 11

dxdy

11

| . |

00

› ÚÚ r . dS = Ú Ú 1 ____ = Ú Ú dxdy = 1

EDGF

00

Over the surface OABC, unit normal =– and z = 0 _› So, r . = ( x + y + z) . (– ) = – z = 0 _

11

› ÚÚ r . dS = Ú Ú 0 dxdy = 0

OABC

So,

y

B

Fig. 1.1W Calculation of surface area of a unit cube bounded by coordinate planes and planes x = 1, y = 1 and z = 1.

› ÚÚ r . dS = Ú Ú 0 dxdz = 0

_

C

x

11

AOGF

\

D

G

00

_›

ÚÚ r . dS = 1 + 0 + 1 + 0 + 1 + 0 = 3 S

[Note: By applying divergence theorem it can be easily shown that

Vector Analysis 111

__› _

_

1.33

› › =3 ÚÚ r . dS = Ú (—. r ) dV = Ú Ú Ú 3 dxdydz = | 3xyz |1,1,1 0,0,0 S

V

Example 1.52

000

Evaluate ÚÚÚ (2x + y) dV, where V is the enclosed volume bounded by the cylinder z = 4 – x2 V

and the planes x = 0, y = 0, y = 2 and z = 0 Sol.

2

2

Here ÚÚÚ (2x + y) dV = Ú

Ú

V

4 – x2

Ú

(2x + y) dx dy dz

x=0 y=0 z=0 2

2

=Ú

Ú

x=0 y=0 2

=Ú

x=0

[

[

4 – x2

Ú

z=0

]

__›

|

__› If__›H =_› — × A, prove that ÚÚ H.

Example 1.53 Sol.

__›

2

Ú

[2x (4 – x2) + y (4 – x2)] dxdy

x=0 y=0

y2 2x (4 – x2) y + __ (4 – x2) 2

2x3 = 8x2 – x4 + 8x – ___ 3 __›

2

(2x + y) dz dxdy = Ú

]

2

2 y=0

dx = Ú

(16x – 4x3 + 8 – 2x2) dx

x=0

2

80 = ___ 3 0 dS = 0 for any closed surface S.

S

Here, H = — × A

__› __›

__›

Now from divergence theorem, ÚÚH. dS = ÚÚÚ (— ◊ H)dV __› __›

__› __›

_›

V

S

But we have — ◊ H = — ◊ — × A = 0 [since divergence of any curl is zero] __›

ÚÚ H ◊ dS = 0

\

S

Example 1.54

is the unit outward drawn normal to any closed surface of area S, show that

If

ÚÚÚ div

dV = S. Sol.

From divergence theorem ÚÚÚ div dV = ÚÚ ◊ dS = V

Example 1.55

S

ÚÚ dS = S S

_›

S

= (a + b + c) V. Sol. Applying divergence theorem __› _›

_›

ÚÚ A ◊ dS = ÚÚÚ (— ◊ A) dV S

So

__› _›

V

∂ ∂ ∂ ___ + ___ + __ ◊ (ax + by + cz ) ∂x ∂y ∂z =a+b+c

—◊A =

Here

(

__› _›

)

ÚÚÚ (— ◊ A) dV = ÚÚÚ(a + b + c) dv = (a + b + c) V V

_›

If S is any closed surface enclosing a volume V and A = ax + by + cz , prove that ÚÚ A ◊ dS

V

1.34

Advanced Engineering Physics

Example 1.56

__›

_›

_›

Evaluate ÚÚ (— × A) ◊ dS, where A = (y – z + 2) + (yz + 4) – xz and S is the surface of S

the cube x = 0, y = 0, z = 0, x = 2, y = 2, z = 2 above the xy plane.

Sol.

__›

_›

Here — × A =

|

|

∂ ∂ ∂ ___ __ ___ = (0 – y) + (– 1 + z) + (0 – 1) = – y + (z – 1) – ∂y ∂z ∂x y – z + 2 yz+4 – xz

__› _› __› _› dxdy Now ÚÚ (— × A) ◊ dS = ÚÚ (— × A) ◊ ____ R S | ◊ | __› _› __› _› dxdy above the xy plane = , so ÚÚ ((— × A) ◊ dS = ÚÚ (— × A) ◊ ____ R R ◊ | __›

__›

_›

2 2

_›

Here (— × A) ◊ = – 1, so ÚÚ (— × A) ◊ dx dy = – Ú R

Ú dxdy = – 4

0 0

_›

_›

Example 1.57 Evaluate Ú A ◊ dS where A = 4xz – y2 + yz and S is the surface of the cube bounded by x = 0, x = 1, y = 0, y = 1, z = 0, z = 1 Sol. By applying divergence theorem __› _› _› ∂ ∂ ∂ ÚÚ A. dS = ÚÚÚ (—◊ A) dV = ÚÚÚ ___ (4xz) + ___ (– y2) + __ (yz) dV V

S

V

[

∂x

∂y

1

= ÚÚÚ (4z – y) dx dy dz = Ú V

1

=Ú

]

∂z

1

Ú Ú

(4z – y) dx dy dz

x = 0y = 0 z = 0

1

1

Ú

2z2 – yz|10 dxdy = Ú

x=0 y = 0 1

1

( )

1

1

Ú

(2 – y) dxdy = Ú

x = 0y=0

x=0

(

y2 2y – __ 2

)|

1

dx. 0

1

3 3 1 = Ú 2 – __ dx = Ú __ dx = __ 2 2 0 0 2 Example 1.58

Evaluate ÚÚ r ◊ dS over the surface of a sphere of radius a with centre at (0, 0, 0). S

Sol.

z

By applying divergence theorem __› _

_

D

› › ÚÚ r ◊ dS = ÚÚÚ(— ◊ r ) dV = ÚÚÚ 3 dV = 3V S

V

V

The volume of the surface of a sphere of radius a is _› 4 4 V = __ p a3 So ÚÚ r ◊ dS = 3 × __ p a3 = 4p a3 3 3 S _›

Verify Stoke’s theorem for A = (x + 3yz) + xy for the square surface as shown in Fig. 1.2W. Example 1.59

C

A

B

x

Fig. 1.2W Verification of Stoke’s theorem.

y

Vector Analysis

Sol.

|

__›

_› ∂ — × A = ___ ∂x 0

Here

∂ ___ ∂y x + 3yz

∂ __ ∂z xy

1.35

|

= (x – 3y) + (– y) + (1) Since x = 0, for this surface, __›

1 1

_›

dydz Ú (— × A) ◊ dS = Ú Ú [ (x – 3y) – y + ] ◊ ____

S

| ◊ |

0 0 1 1

=Ú

Ú – 3y dy dz

[Here, x = 0]

0 0

3 = – __. 2 _›

_

_›

_›

_›

_›

_

_›

_

_

_› __›

› › › › Ú A ◊ d r = Ú A ◊ d r + Ú A ◊ d r + Ú A ◊ d r + Ú A ◊ dr

Now For the first part

AB

BC

_›

CD

DA

Ú A ◊ d r = Ú A ◊ dy = Ú 0 dy = 0

AB

_›

[Here, z = 0 x = 0]

_›

_›

For the second part Ú A ◊ d r = Ú A ◊ dz = Ú 0 dz = 0 BC _›

For the third part

_

› Ú A ◊ d r = Ú A ◊ dy = Ú 3y dy

CD

For the fourth part

[Here, x = 0]

0

_›

[Here, z = 1]

1

_›

3 = – __ 2 _

_›

› Ú A ◊ d r = Ú A ◊ dz = Ú0 dz = 0

DA

__› __›

3

3

Ú A ◊ dr = 0 + 0 – __2 + 0 = – __2

So, total

So, Stoke’s theorem is verified.

Review Exercises

Part 1: Multiple choice questions

_›

1. If is the unit _› vector in the direction A then _› _› A (a) = ___ (b) = A|A| _› |A| _ _› › 2. Two vectors A and B are parallel when _›

_›

_› _›

(a) A × B = 0 (b) A ◊ B = 0 3. The angle between and (2 + ) is 2 2__ (a) cos–1 __ (b) cos–1 ___ 5 ÷5

_›

(c)

|A| = ___ _› A _› _›

(c) A ◊ B = 1 2 (c) cos–1 __ 3

[WBUT 2004] (d) None of these [WBUT 2004] (d) None of these [WBUT 2005] (d) None of these

1.36

Advanced Engineering Physics _› _› _›

4. Condition of coplanarity of three vectors (a, b , g ) is _› _›

_› _›

_›

_›

(a) a ◊ ( b + g ) = 0 (b) a ◊ ( b × g ) (c) 5. The unit vector along the direction 2 + 3 + 4 is 1 ___ (2 + 3 + 4 ) (a) ____ (b) ÷19 1 ___ (2 – 3 + 2 ) (c) ____ (d) ÷29 _›

_›

_›

_› _›

__›

(d) a ◊ ( b – g ) = 0

a × (b × g )

1 ____ ___ (2 + 3 + 4 ) ÷29 1 ____ ___ (2 – 3 + 4 ) ÷19

6. When the magnitude of A is constant which one of the following is true? _›

_›

_›

dA (a) ___ = 0 dt

dA (b) A ◊ ___ = 0 dt

__›

[WBUT 2006] __›

_›

_›

dA (c) A × ___ = 0 dt

_›

[WBUT 2008]

| |

dA (d) ___ = 0 dt

7. The angle between —j and the surface j = constant is [WBUT 2006] p p __ __ (a) (b) (c) p (d) zero _› 2 4 8. For arbitrary scalar and vector fields j and A, which of the following is always correct? [WBUT 2008] __›

__›

__›

_›

__›

__› __› _›

__› __›

(a) — × (— × A) = 0 and — × (—j) = 0

(b) — (— ◊ A) = 0 and — ◊ (—j) = 0

(c) — ◊ (— × A) = 0 and — × (—j) = 0

(d) — ◊ (— × A) = 0 and — ◊ (—j) = 0

__› __›

_›

__›

_›

__›

_›

_›

_›

9. If A and B are irrotational then A × B is (a) solenoidal (b) irrotational __› _ ›

__› __›

(c) rotational

10. — ◊ r is equal to (a) 2 (b) zero (c) 1 _› 11. A fluid with velocity v is said to be incompressible when __› _ ›

__›

(a) — ◊ v = 0

_›

(b) — × v = 0

_›

__›

__› __›

(d) None of these [WBUT 2004] (d) 3

__›

_›

(c) — × (— × v ) = 0

(d) None of these

_›

12. The value of r ◊ dV is S (a) V (b) 2 V (c) zero (d) 3 V 13. A solenoidal _› field is one for which _› _› (a) grad A = 0 (b) (c) curl A = 0 (d) None of these _› div A = 0 14. The value of a for which A = 2ax + 2y + 4z is solenoidal is equal to (a) 2 (b) 3 (c) – 3 (d) 1 _› 15. For conservative field, a vector field A can be written as _›

_›

__› __›

(a) A = —2y (b)_A = — ◊ — y › 16. For irrotational vector field A __›

__› _›

_›

(a) — × A = 0

__› __›

_›

(b) — ◊ A = 0

17. The values of — ◊ — × A is (a) 1 (b) 2

_›

__›

(c) A = – — y _›

(d) None of these

(c) grad A = 0

(d) None of these

(c) zero

(d) – 1

Vector Analysis

1.37

18. Gradient of y represents the rate of change which is (a) minimum (b) maximum (c) neither maximum nor minimum (d) None of these [Ans. 1 (a), 2 (a), 3(b), 4 (b), 5 (b), 6(b), 7 (a), 8 (c), 9 (a), 10 (d), 11 (a), 12 (d), 13 (b), 14 (c), 15 (c), 16 (a), 17 (c), 18 (b)]

Short questions with Answers 1. Ans. 2. Ans.

Define scalar field and vector field. See section 1.7 What is solenoidal vector? A vector field_› is said to be solenoidal if the divergence of that vector field is zero. So for solenoidal vector field (A) __› _›

—◊A = 0 3. What is irrotational or lamellar vector field? Ans. A vector field is said _to be irrotational if the curl of that vector field is zero. So for irrotational or › lamellar vector field ( A ) __› _› —×A=0 4. Define conservative field. Ans. If the line integral of the vector field depends on the initial and final points but is independent of the path of the integral then the vector field is called a conservative field. _› __›

_›

_›

__›

A ◊ dl = 0 then A will be the gradient of a scalar function, i.e., A = —y.

For conservative field, __›

5. Prove that Ú —y dV = Ú y dS V

_›

_›

S

_›

Ans. Let P = y c where c is a constant vector. __›

_›

_›

Then from divergence theorem Ú — ◊ (y c ) dV = Ú y c ◊ dS V

__›

_›

__›

_›

_› __›

S

_›

_›

Again we know that — ◊ (y c ) = —y ◊ c = c ◊ —y and y c ◊ = c ◊ (y ) _ __›

_

› › Ú c ◊ — y dV = Ú c ◊ (y ) dS

So,

V

_›

S

Now taking c outside of the integration __›

_›

_›

c ◊ Ú — y dV = c ◊ Ú y dS V

S

__›

Ú — y dV = Ú y dS

So,

V

S

6. Define line integral. Ans. The integral of a point function along a curve is called line _› integral. _› _› Let r = r (x, y, z) be the equation of a curve. If y and A are scalar and vector fields respectively, __›

and dr is the displacement then the integral may be

_› __›

_›

__›

Ú y dr, Ú A ◊ dr and Ú A × dr

C

1.38

Advanced Engineering Physics

Each integral is called the line integral along the curve C. 7. What is gradient? Ans. Gradient is the maximum variation of a scalar considering all directions. It is the maximum rate of growth of scalar y. Gradient is a vector that represents both the magnitude and the direction of the maximum space rate of increase of a scalar. 8. Define source and sink. Ans. If the divergence of a vector field is positive, it indicates a net outward flow from the point. The given point is a ‘source point’. But if divergence of a vector field is negative, it indicates a net flow towards the point. The given point is a ‘sink point’. 9. State Stoke’s theorem. Ans. See Section 1.13.2 10. What is the physical meaning of divergence of a vector? Ans. The physical meaning of the divergence of a vector is the limiting value of the net outward flow to some physical quantity like a fluid or electric flux through the surface area of unit volume as the volume tends to approach zero.

Part 2: Descriptive questions 1. What are scalar and vector fields? Give examples. 2. (a) Define vector field. Give an example. __› (b) What do you mean by a scalar field? If y (x, y, z) = 3x2y – y3z2, find —y at the point (1, – 2, 1). [WBUT 2004] 3. Define divergence of a vector point function. What is its physical significance? 4. Explain the terms surface integral and volume integral. _› 5. Show that (i) curl grad v = 0 (ii) div curl A = 0. 6. Define curl of a vector point function. What is its physical significance? 7. State Gauss’ __› divergence theorem. Explain the physical significance of this theorem. 8. Show that —y is perpendicular to the surface over which y is constant. 9. State Stoke’s theorem. Define line integral of a vector. _›

__›

_›

_›

10. If A = — × B, show that ÚÚ A ◊ dS = 0 for any closed surface ‘S’. __›

__›

S

__›

__›

11. Show that Ú j —y ◊ dr + Ú y —j ◊ dr = 0; j and y are scalar fields. C

C

__›

12. If a rigid body rotates about an axis passing through the origin with angular velocity w and with __› 1 __› _› _› __› _› linear velocity v = w × r , then prove that w = __ (— × v ). 2 d 2 f (r) __ df (r) 2 _____ 2 ______ 13. Prove that — f (r) = +r . dr dr2 14. 15. 16. 17.

__›

__›

_›

__› __› _›

_›

Prove that — × — × A =_—(— ◊__A) – —2 A. _› __› › › If the vector functions F and G are irrotational, show that F × G is solenoidal. The gradient of a scalar quantity is a vector quantity. Explain. 1 Show that —2 (loger) = __2 . r

[WBUT 2002] [WBUT 2006] [WBUT 2002] [WBUT 2007]

Vector Analysis _›

__›

_›

1.39

curve.

_› __›

_› __›

_›

18. Given F = f (r) r , show that — × F = 0 and hence show that

c

F ◊ dr = 0 where C is a simple closed [WBUT 2008]

_›

19. Show that S B ◊ dS = 0 where B is the magnetic field and S is a closed surface. State the theorem that you have used. [WBUT 2006]

Part 3: Numerical Problems 1. Find the directional derivative of y (x, y, z) = xy2 z + 4x2 z at (– 1, 1, 2) in the direction (2 + – ). 38 Ans. – ___ 3 _› 2. Find the torque about the point O (3, –1, 3) of a force F (4, 2, 1) passing through the point A (5, 2, 4). [Ans. + 2 – 8 ] _› 3. A fluid motion is given by _V = (y + z) + (z + x) + (x + y) . Show that the motion is irrotational. › 4. Find the constant a so that V is a conservative vector field where __ 2 V = (axy – z2) + (a – z) x2 + (1 – a) [Ans. a = 4] _› az . 3 2 2 5. Find the work done in moving an object in the field F = (2xy + z ) + x + 3xz from the point (1, – 2, 1) to (3, 1, 4) independent of the path. [WBUT 2007] _› 6. Find the work done to move an object along a vector r = 3 + 2 – 5 if the applied force _› is F = 2 – – . [WBUT 2005]

[

7. Find the unit vector perpendicular to x2 + y2 – z2 = 100 at the point (1, 2, 3). _›

_›

8. Evaluate ÚÚ F ◊ dS, where F = 4xz – y2 + yz S

[

]

± + 2___– 3 Ans. ___________ ÷14 [WBUT 2007]

and S is the surface of the cube bounded by x = 0,

x = 1, y = 0, y = 1, z = 0, and z = 1.

[Ans. – 2] [WBUT 2002]

______

9. Evaluate ÚÚ ÷x2 + y2 dx dy over the region R in the xy plane bounded by x2 + y2 = 36. R

_›

]

_›

2 3

2

3

[Ans. 144 p]

2 2

10. Show that the force field F defined by F = (y z – 6 xz ) + 2xyz + (3xy z – 6x2z) is conservative. _› 11. For what value of a is the vector A = (x + 3y) + (y – 2z) + (x + az) solenoidal? [Ans. a = – 2] _›

12. Evaluate ÚÚ r ◊ dS. S 13. Find the unit vector perpendicular to x2 + y2 – z2 = 100 at the point (1, 2, 3). 3 1 2 ___ + ____ ___ – ____ ___ Ans. ____ ÷14 ÷14 ÷14

[

14. If y (x, y, z) = 3xy2 – 5x2z + 2z2 find —2y. _›

_›

15. Evaluate ÚÚ A ◊ dS, where A = (2x + 3z) – (xz + y) + (y2 + 2z) S

having centre at (3, – 1, 2) and radius 3.

[Ans. 3V]

]

[WBUT 2007]

[Ans. 6x – 10z + 4] and S is the surface of the sphere [Ans. 108 p]

CHAPTER

2 Electrostatics 2.1

INTRODUCTION

Electrostatics is an important branch of physics which deals with electric charge at rest. The electric force between two electrons is the same as the electric force between two protons placed at the same distance. The amount of charge on an electron is the same as that on a proton. The charge on a proton is positive and that on an electron is negative. The net charge on the electron–proton system is zero. Stationary charges produce an electric field that is constant with time, hence the term electrostatics. Both electrostatics and magnetostatics can be explained by using vector calculus.

2.2

QUANTIZATION OF CHARGE

In 1911, Millikan successfully showed that charges in tiny oil drops are exact multiples of elementary charges. The magnitude of charge on a proton or an electron (e = 1.6 × 10–19 c) is called elementary charge. Quantization of charge means that all observable charges are integral multiple of elementary charge e(= 1.6 × 10–19c).

2.3

CONSERVATION OF CHARGE

The law of conservation of charge states that for an isolated system, the net charge always remains constant. In b-decay a neutron converts itself into a proton and creates an electron. The net charge remains zero before and after the decay. In nuclear fission, the total charge is always conserved. 1 235 Æ 56 Ba 141 + 36 K 92 + 30 n1 0n + 92 U Before collision total charge = + 92 e and total charge after collision = (56 + 36) e = 92 e. So total charge is conserved.

2.4

COULOMB’S LAW

Statement The force between two small charged bodies separated by a distance in air is (a) directly proportional to the magnitude of each charge

2.2

Advanced Engineering Physics

(b) inversely proportional to the square of the distance between them F12 q1 q2 F21 (c) directed along the line joining the charges A B The distance between charges must be large compared to their linear r dimension. Fig. 2.1 Force between two In mathematical form, if q1 and q2 be two like charges and r is the discharges. tance between them [Fig. 2.1] then the force exerted on q1 due to the charge q2 is ___› 1 q1 q2 F12 = ____ _____ ...(2.1) 4pe0 r2 21 Here, 21 is unit vector pointing from q2 to q1 and e0 is the permittivity of free space. Experimentally, mea1 sured value of ____ = 9 × 109 Nm2/C2 4pe0 Similarly, the force exerted on q2 due to the charge q1 is ___› 1 q1 q2 F21 = ____ _____ ...(2.2) 4pe0 r2 12 _› _› _› 1 q1 q2 Here, 12 is a unit vector pointing from q1 to q2. So F12 = – F21. For two unlike charges, F12 = ____ _____ 12 2 4pe 0 r ___› q q 1 1 2 and F21 = ____ _____ are attractive 4pe0 r2 21 Relative permittivity (er ) The relative permittivity or dielectric constant of a medium is defined as the ratio of the force between two charges placed at a distance in vacuum (or air) to the force between the same charges placed at the same separation in that medium. e 1 q1 q2 1 q1 q2 er = __ Fvacuum = ____ _____ and Fmedium = ____ _____ ...(2.3) 2 e0 4pe0 r 4pe r2

[

2.5

]

PRINCIPLE OF SUPERPOSITION

The principle states that when a number of charges are interacting, the total force on a given charge is the vector sum of the individual forces exerted by all other charges on the given charge. If q1, q2, q3, q4, ... are the charges situated at A, B, C, D, ..., as shown in Fig. 2.2. The total force on q1 due to all other charges is __› q1 q3 1 q1 q2 _____ F1 = ____ _____ 21 + 31 + ... + ... 2 4pe0 r r231 21 _›

[

]

_›

= F12 + F13 + .... The total force on q2 due to all other charges is _› q2 q3 1 q2 q1 _____ F2 = ____ _____ 12 + 2 4pe0 r r2

[

12

32

32

+ ...

r41

q3 C r31 r 21

r51

q1

D q1

E q5

_›

_›

N

__›

F = F1 + F2 + ... = S Fi i =1

A r61 q6 F

Fig. 2.2 Principle of superposition.

]

If there is a test charge q0, then total force on the test charge q0 due to all other charges is _›

B q2

Electrostatics

1 = ____ 4pe0

2.6

N

S

i=1

q0 qi ____ r2i

i

2.3

...(2.4)

ELECTRIC FIELD

The electric field due to a charge is the space around the charge in which any other charge is acted upon by an electrostatic force. If we have many charges q1, q2, ... qn at distances r1 r2 ... rn respectively from a test charge q0 then from the principle of superposition, the total force on q0 is _› 1 N q0 qi F = ____ S _____ 4pe0 i = 1 r2 i i

q0 = ____ 4pe0

qi S __2 i=1 r i The electric field intensity at the point is _›

N

_›

i

qi __ ...(2.5) 2 i i=1 r i Thus, the electric field intensity at a point in the electric field is the force on a unit test charge placed at the point concerned. F ____ 1 E = __ q0 = 4pe0

2.7

N

S

CONTINUOUS CHARGE DISTRIBUTIONS

On a uniform charge body, there are three types of distribution of charge: (i) Line charge distribution If q is the total charge over a conducting wire of length l and infinitesimally small thickness, then charge per unit length l (line charge density) is q l = __ coulomb/m l For non-uniform distribution of charge Dq dq l = Lt ___ = ___ D l Æ 0 Dl dl where Dl is a small element of length which carries a charge Dq. So, total charge over the whole length q = Ú l dl

...(2.6)

l

(ii) Surface charge distribution If q is the charge uniformly distributed over the conducting surface S, then surface density of charge (charge per unit area) s is q s = __ coulomb/m2 S If Dq be the charge contained by a small element DS then surface charge density, Dq dq s = Lt ___ = ___ DS Æ 0 DS dS So, total charge on the surface q = Ú s dS S

...(2.7)

2.4

Advanced Engineering Physics

(iii) Volume charge distribution If q is the charge uniformly distributed over the volume V then the volume density of charge (charge per unit volume) q r = __ coulomb/m3 V If charge distribution is not uniform then Dq dq r = Lt ___ = ___ DV Æ 0 DV dV where DV is a small volume element which carries a charge Dq. So, total charge over the whole volume q = Ú r dV

...(2.8)

V

2.8

ELECTRIC POTENTIAL

The concept of potential is based on energy consideration. The electric potential at a point in an electric field near a charged conductor is defined as the amount of work done in bringing a unit positive charge from infinity to that point against the electrostatic O P force. A positively charged body always tends to move from higher potential +q to lower potential. r

Potential at a point due to a point charge The potential at P [Fig. 2.3] is given by r

Fig. 2.3 Potential at point P due to charge +q at O.

_› __›

V = – Ú E ◊ dr r

q 1 q = – Ú ____ __2 dr = _____ 4pe0 r 4pe0r

...(2.9)

Electric field Intensity as a gradient of potential _› _› Let the electric field at a point r due to a charge distribution be E_ and electric potential at the same point be _› _› › V. Suppose a test charge q0 is displaced slightly from r to r + d r . Then the force on the test charge q0 is _›

__›

_›

_›

_›

_›

_›

F = q0E and work done for small displacement dr is dw = F ◊ d r = q0 E ◊ d r _›

_›

The change in potential energy = – dw = – q0E ◊ d r _›

_›

So the change in potential dV __= – E ◊ d r _ › _› _› › Again dV = —V ◊ d r = – E ◊ d r _›

__›

E=–—V ...(2.10) Hence, the electric field at a point is equal to the negative gradient of the electrostatic potential at the point. or,

2.9

ELECTRIC POTENTIAL ENERGY

We define electric potential energy of a system of point charges as the work required to assemble this system of charges by bringing them from infinite distances.

Electrostatics

2.5

If two point charges q1 and q2 are separated by a distance r12 then potential energy of the system q1 and q2 is 1 q1 q2 U = ____ _____ 4pe0 r12 If another charge q3 is at a distance r13 from q1 and distance r23 from q2 then potential energy of the system (q1 + q2 + q3) is q1 q3 q2 q3 1 q1 q2 ____ U = ____ ____ + r + ____ r r23 4pe0 12 13

(

)

Generally, the potential energy for a system of n point charges is n n q q i j 1 1 U = __ × ____ S S ____ rij 2 4pe0 i=1

...(2.11)

j=1

In summation, each term is counted twice, so half factor is included here to avoid double counting each term for calculation of potential energy.

2.10 ELECTRIC FLUX __›

We know that any area element __› dS is a vector dS. If is the unit normal to the area element then dS = dS E The total number of lines of force passing through a surface placed in an electric field is known as electric flux (jE). _› q Consider a surface of area S inside an electric field__E› [Fig. 2.4]. The dS E cos q surface S is divided into a number of elementary areas dS (known as__ area › vector). The component of the electric field along the area vector dS is S given by Fig. 2.4 Component of electric En = E cos q ___ __› › field E along dS. So, electric flux djE = En dS = E cos q dS _› __›

_›

= E ◊ dS = E ◊ dS The total electric flux through S is _›

jE = Ú E ◊ dS

...(2.12) ...(2.13)

s

2.11 SOLID ANGLE dS In Fig. 2.5, the solid angle subtended by any surface dS at a point O, disdS ¢ tance r away, is given by dw dS¢ _______ dS cos q ___ O r dw = 2 = q r r2 n where dS¢ = dS cos q is the projection of the surface dS. dS cosq For sphere dS¢ = 4pr2 and q = 0, so solid angle subtended by the sphere r at its center is dS cos q Fig. 2.5 Solid angle at point O. = 4p w = Ú _______ s r2

2.6

Advanced Engineering Physics

2.12 GAUSS’ LAW 1 Gauss’ law states that the total electric flux through a closed surface in an electric field is equal to __ times e0 the total charge enclosed by the surface, where e0 is the permittivity of free space. _› q _› Mathematically, E ◊ d S = __ when S encloses q e0 s =0 when S does not enclose q ...(2.14) Proof of Gauss’ law E We consider a spherically symmetric closed surface. Suppose a charge q is _› R dS placed at the center of a sphere a radius r and E is the electric field intensity r at a point R on the surface [Fig. 2.6]. O r +q From Coulomb’s law, _› q 1 E = ____ __2 4pe0 r __› Fig. 2.6 Proof of Gauss’ law. The flux of electric field through dS is __› _› __› q q djE = E ◊ dS = ______2 ◊ dS = ______2 dS 4pe0 r 4pe0 r The total flux of the electric field due to the internal charge q through the closed surface q q ____ q 4pr2 __ dS ____ jE = Ú djE = ____ Ú ___ = = e0 4pe0 s r2 4pe0 r2 which is Gauss’ law in electrostatics.

2.12.1

Differential Form of Gauss’ Law

The integral form of Gauss’ law is _› __› q E ◊ dS = __ ...(2.15) e0 S If r be the volume charge density over a small volume element dV within the closed surface S, then q = Ú r dV

...(2.16)

V

Again from Gauss’ divergence theorem __› _›

_› __› S

E ◊ dS = Ú (— ◊ E) dV

Now from Eqs (2.17) and (2.16) into Eq. (2.15) __› _› 1 r dV Ú (— ◊ E) dV = __ e0 VÚ V __› _› r or dV = 0 Ú — ◊ E – __ e0 V This is true for any arbitrary volume V. __› _› r — ◊ E = __ \ e0

(

...(2.17)

V

...(2.18)

)

...(2.19)

2.7

Electrostatics

This is the differential form of Gauss’ law. __›

_›

In vacuum, electric displacement vector D = e0 E __› __›

So —◊D=r This is also the differential form of Gauss’ law in terms of electric displacement vector.

2.12.2

...(2.20)

Coulomb’s Law from Gauss’ Law

Here we would like to deduce Coulomb’s _law from __› Gauss’ law. Here, S is › the spherical gaussian surface. In _Fig.__2.7, E and dS on the gaussian surface › › are directed radially outward. So E ◊ dS = E dS. Now from Gauss’ law

E ds r

q

r

O

_› __› s

S

q E ◊ dS = __ e0

q Fig. 2.7 Derivation of E dS = E × 4pr2 = __ [E is constant] e0 Coulomb’s law from s q Gauss’ law. ______ or, E= 2 4pe0 r _› If another point charge q0 be placed at the point at which E is calculated then the force on q0 due to q is _› _› 1 q0 q F = q0 E = ____ ____ ...(2.21) 4pe0 r2 where is the unit vector. Equation (2.21) is Coulomb’s law. or,

2.12.3

Application of Gauss’ Law

For calculating electric field due to a charge distribution, Gauss’ law provides the easiest way. Here, we consider some important applications of Gauss’ law. (i) Electric field due to uniformly charged sphere Case I Field at a point outside the charged sphere Let P be a point situated outside the charged sphere having charge q uniformly distributed throughout the volume of the sphere of radius R [Fig. 2.8]. In order to find the electric field intensity at P, a concentric sphere of radius r is drawn as gaussian surface, over which the electric_›field intensity is directed normal to every point of this surface. If E is the electric field intensity at the point P then the total flux through the gaussian surface is _› __› q jE = E ◊ dS = E (4p r2) = __ e0 s q or, E = ______2 4pe0 r For continuous charge distribution of density r within the sphere

E

E

E

E R E

P

r

E

E E

Fig. 2.8

E

E

Electric field outside of a charged sphere.

2.8

Advanced Engineering Physics

So,

4 q = __ p R3 r 3 4 3 __ p R r R3r 3 1 ______ ____ E= = _____2 2 4pe0 r 3e0r

...(2.22)

Case II Field at a point on the surface of the sphere For any point on the surface of the sphere r = R, from Eq. (2.22), we have R3r Rr E = _____2 = ___ 3e0 3e0R

...(2.23)

P r

O

r

E

Case III Field at a point inside the charged sphere We want to find the electric field at a point P inside the sphere at a distance r from the center [Fig. 2.9]. The total charge inside the gaussian surface of R radius r 4 Fig. 2.9 Electric field inside q = __ p r3r 3 a charged sphere. The total electric flux over the gaussian surface is given by _› _› q 4pr3r jE = E ◊ d S = E × 4pr2 = __ = _____ E e0 s 3e0 R rr or, E = ___ ...(2.24) 3e0 The variation of electric field intensity in different cases as discussed is shown in Fig. 2.10. E

μ

1 2 μ r

(ii)

Electric field due to a charged spherical shell

Case I At a point outside the charged shell Let P be a point outside the shell at a distance r [Fig. 2.11]. Here r > R The total flux over the gaussian surface _› q _› jE = E ◊ d S = __ e0 s q 2 __ or, E (4pr ) = e0 q or, E = ______2 ...(2.25) 4p e0r Case II At a point on the surface of the charged shell Here, r = R and the sphere itself behaves as gaussian surface. The electric field intensity on the surface is then q E = _______2 4p e0 R Again for spherical shell the surface density of charge q s = _____2 4pR

R

O

r=R

r

__›

Fig. 2.10 Variation E with distance from the center of a charged sphere. E E

E

R

E

r

P

E E

Fig. 2.11 Electric field due to a charged spherical shell.

2.9

Electrostatics

s E = __ e0

\

...(2.26) E

R

Case III At a point inside the shell Here r < R, the total charge is situated on the surface of the shell of radius R and no charge is therefore enclosed by the gaussian surface, therefore, _›

_›

E ◊ dS = 0 s \ E=0 ...(2.27) The variation of the electric field intensity with distance from the centre is shown in Fig. 2.12 with the help of Eqs (2.25), (2.26) and (2.27). (iii)

R

r

Electric field intensity due to long uniformly charged cylinder

where l is the line charge density.

\

r=R

Fig. 2.12 Electric field at a point inside a spherical charged shell.

Case I At a point outside of the cylinder Let P1 be the point at a distance r from the axis of the cylinder of radius R [Fig. 2.13]. Imagine a cylindrical gaussian surface of length l through P1. The field will have a cylindrical symmetry and the total flux over the gaussian surface _› __› q ll jE = E ◊ dS = __ = ___, e0 e0 s

or,

O

ll E × 2prl = ___ e0 l E = ______ 2p re0

...(2.28)

R E

E

l

E

r

l

r

P2

P1

E

E

Fig. 2.13 Electric field at a point from the axis of a uniformly charged cylinder.

Case II At a point on the surface of the cylinder Here r = R, the cylinder itself is the gaussian surface. The electric filed at the surface is obtained by replacing r by R in Eq. (2.28) l E = ______ ...(2.29) 2p R e0 Case III At a point inside the cylinder Let P2 be the point at a distance r (r < R) from the axis of the cylinder [Fig. 2.13]. A coaxial cylinder of radius r and length l is constructed as gaussian surface such that the point P2 lies on the curved surface of the cylinder. The total charge enclosed by the gaussian surface q¢ = p r2lr [where, r is the volume charge density] Again(p R2l)r = l l

2.10

or,

Advanced Engineering Physics

l r = ____2 , pR From Gauss’ law, the electric flux

E

_› __› q¢ pr2l r jE = E ◊ dS = __ = _____ e0 e0 s

or,

pr2lr E × (2prl) = _____ e0

Eμr Eμ

1 r

r=R r rr l lr r __› E = ___ = ___ ____2 = _______ ...(2.30) 2e0 2e0 pR Fig. 2.14 Variation of E with 2p R2e0 distance r from the axis of The variation of electric field intensity with distance r from the a charged cylinder. axis of a charged cylinder is shown in Fig. 2.14. [Note: For a hollow charged cylinder, the charge inside the cylinder is zero and so the electric flux inside the cylinder, jE = 0 which gives E = 0.]

( )

or,

(iv) Electric field due to an infinite plane charge sheet Here we consider an infinite thin plane charge Infinite plane sheet of positive charge having surface densheet sity of charge s [Fig. 2.15]. To find the electric field at a point P1, let us consider another P1 point P2 on the other side of the sheet so that P2 E two points P1 and P2 are equidistant from the Ds sheet. We construct a cylindrical gaussian sur- E DS n –n face normally through the plane which extends DS equally on two sides of the plane. The flux of the electric field crossing through the gaussian surface jE = E Ds + E DS Fig. 2.15 Electric field due to an infinite charged sheet. = 2E DS [where DS is the area of cross section of each end face] s DS = _____ e0

s

E = ___ …(2.31) 2e0 We see that electric field is uniform and does not depend on the distance from the charge sheet. or,

(v) Electric field near a charged conducting surface Here we consider a plane conducting sheet. All the charges of the conductor lie on the surface, so the electric field inside the conductor is zero. We have to find the electric field at a point P which is near but outside the conductor. To find the electric field, we construct a gaussian surface as follows [Fig. 2.16]. The total flux through the gaussian surface is

DS

P

E

Fig. 2.16 Electric field at a point near a charged conducting surface.

2.11

Electrostatics

jE = E DS and charge enclosed inside the closed surface is s DS. So from Gauss’ law s DS E DS = _____ e0 s E = __ e0

or,

…(2.32)

The electric field near a plane charged conductor is twice the electric field due to a non-conducting plane charge sheet. This is also known as Coulomb’s theorem. The theorem states that the electric field at any point very close to the surface of a charged conductor is equal to charge density of the surface divided by free space permittivity.

2.13 POISSON’S AND LAPLACE’S EQUATIONS The differential form of Gauss’ law is __› _› r — ◊ E = __ [From Eq. (2.19)] e0 _›

Again, the electric field (E) at any point is equal to the negative gradient of the potential V i.e.,

__›

E=–—V

[From Eq. (2.10)]

Now combining these two equations, we have __› __› r — ◊ (– — V) = – —2 V = __ e0 r 2 __ or, — V=– e0

…(2.33)

This is known as Poisson’s equation. Now in a charge-free region (r = 0), the Poisson’s equation becomes —2 V = 0

…(2.34)

This equation is known as Laplace’s equation and is valid only in the charge-free region. The laplacian operator —2 is a scalar operator and its form in three coordinate systems are: ∂2 ∂ 2 ∂ 2 (a) Cartesian system (x, y, z): —2 ∫ ___2 + ___2 + ___2 ∂x ∂y ∂z (b) Cylindrical coordinate system (r, f, z): ∂ ∂2 1 ∂ 1 ∂2 —2 ∫ __ ___ r ___ + __2 ____2 + ___2 r ∂r ∂r r ∂j ∂z

( )

(c) Spherical polar coordinate system (r, q, j) ∂ ∂ ∂ ∂2 1 ∂ 1 1 ___ ___ _______ ____ —2 ∫ __2 __ r2 __ + _______ sin q + ∂r ∂q r ∂r r2 sin q ∂q r2 sin2q ∂j2

( )

(

)

2.12

2.13.1

Advanced Engineering Physics

Application of Laplace’s Equation (Effective 1D Problem)

(i) Potential between the plates of a parallel-plate capacitor Let us consider a parallel-plate condenser (capacitor) having two plates, one at z = 0 and other at z = d [Fig. 2.17]. The potential at the upper plate is VA and potential at the lower plate is zero. So, the potential exists only along the z direction. The Laplace’s equation in the cartesian system is 2

2

z VA d

2

∂ V ____ ∂ V ____ ∂ V ____ + + =0 ∂x2 ∂y2 ∂z2

V=0

∂ 2V ____ ∂ 2V ____ = =0 ∂x2 ∂y2 Since the potential exists only along the z direction

y

Here

∂ 2V ∂V ____ So, = 0 or ___ = C (constant) 2 ∂z ∂z or, V = Cz + D (D is another constant) Now applying boundary conditions, i.e., z = 0, V = 0 and z = d, V = VA The first condition gives D = 0 From the second condition at z = d, V = VA VA = Cd VA or C = ___ d VAz So, from Eq. (2.35), V = ____ d

x

Fig. 2.17 Potential difference between the plates of a parallel-plate capacitor.

…(2.35)

…(2.36)

Equation (2.36) is the solution of Laplace’s equation, which gives the potential between the plates. (ii) Potential of a coaxial cylindrical capacitor Here, we consider a cylindrical capacitor of inner and outer radii a and b (b > a) as shown in Fig. 2.18. The potential of the inner cylinder is VA and the potential of the outer cylinder is zero. Since the variation of potential exists only along the radial direction.

[

( )

rp b

]

∂ 2V ____ ∂ 2V ____ then = =0 …(2.37) ∂j2 ∂z2 Integrating Eq. (2.37) twice with respect to r, the potential at an arbitrary distance r is V = C ln r + D …(2.38) [where C and D are constants of integration] ∂ ∂V 1 __ __ ___ r ∂r r ∂r = 0

z

a

V=O VA y

x

Fig. 2.18 Potential of a cylindrical capacitor.

2.13

Electrostatics

Now applying boundary conditions, i.e., r = b, V = 0 and r = a, V = VA a we have, from Eq. (2.38), D = – C ln b and VA = C ln __ b V ln b A ____ so, C = ____ a and D = – VA __ a __ ln ln b b Now from Eq. (2.38), we have VA ln b ____ V = ____ a ln r – VA __ a __ ln ln b b r __ ln b = VA ____ a __ ln b

….(2.39)

Equation (2.39) is the solution of Laplace’s equation in cylindrical coordinate, which gives the potential inside a cylindrical capacitor. z (iii) Potential of a concentric spherical capacitor Let us consider a spherical capacitor of inner and outer radii a and b (b > a) as shown in Fig. 2.19. The potential of the inner sphere is VA and the outer sphere is zero. Since the variation of potential exists only along the radial direction then from Laplace’s equation

[

b

r P

∂ 2___ ∂V ∂ V 1 __ __ r ∂V = 0 Here, ___ = ____ = 0 …(2.40) 2 ∂r ∂q ∂j ∂r r Integrating Eq. (2.40) twice with respect to r, the potential at P will be C V = – __ …(2.41) r +D

( )

y

VA

x

]

2

a

Fig. 2.19 Potential of concentric spherical capacitor.

[where C and D are integrating constants] Now applying boundary conditions, i.e., r = b, V = 0 and r = a, V = VA C we have from Eq. (2.41) D = __ b So,

(

C __ C 1 __ 1 __ V = – __ r + b =C b– r

)

…(2.42)

(

1 1 and from second boundary condition (r = a, V = VA), Eq. (2.41) gives VA = C __ – __ b a V A or, C = _______ 1 __ 1 __ – b a

(

)

)

2.14

Advanced Engineering Physics

Now putting the values of c in Eq. (2.42) we have VA 1 1 V = _______ __ – __r …(2.43) b 1 __ 1 __ –a b Equation (2.43) is the solution of Laplace’s equation in spherical polar coordinate, which gives the potential inside a spherical capacitor.

(

)

(

)

Worked Out Problems Example 2.1

Compare the electrostatics force and gravitational force between a proton and electron in a hydrogen atom. Given e = 1.6 × 10–19 C, me = 9.1 × 10–31 kg, mp = 1.7 × 10–27 kg and G = 6.67 × 10–11 Nm2 kg–2 Sol. The electrostatic force between a proton and electron is (1.6 × 10–19)2 _____ 1 e2 1 Fe = _____ __2 = 9 × 109 × ___________ = 9 × 109 Nm2/C2 2 4p e0 r 4p e r 0

]

[

The gravitational force between a proton and electron mp me 1.7 × 10–27 × 9.1 × 10–31 Fg = G _____ = 6.67 × 10–11 ____________________ 2 r r2 –19 2 Fe __________ (1.6 × 10 ) 9 × 109 ___ or = × ______________ = 2.2 × 1034 –11 –58 Fg 6.67× 10 9.1 × 1.7 × 10 So, electrostatics force between electron and proton is much greater than gravitational force. Example 2.2 Two particles P and Q having charges 8.0 × 10 – 6 C and – 2.0 × 10– 6 C respectively are held fixed with a separation of 20 cm. Where should a third charged particle be placed so that it does not experience a net electric force? Sol. Since the particles are charged with opposite sign, the point R where net electric force is zero, can’t be between P and R. Suppose QR = x metre and charge on R is q. –6

8.0 × 10 × q The force at R due to P is = _____________2 4p e0 (x + 0.2) 2.0 × 10– 6 × q The force at R due to Q is = ____________ 4p e0 x2 They are oppositely directed and the resultant is zero. So, or, or,

8 × 10–6 _______ 2 × 10–6 ________ = 2 (0.2 + x) x2 0.2 + x = 2x x = 0.2 m

Q

P 20 cm

Fig. 2.1W

R x

2.15

Electrostatics

Example 2.3 Four equal charges + q are placed at the corner of a square. Find the point charge at the center of the square so that the system will remain in equilibrium. Sol. The charge Q at the center (O) must be negative [Fig. 2.2W]. If the net force on a charge at D is zero, then by symmetry it follows that the net force experienced by charges at other points will also be zero. The resultant force (FR) at D due to all other charges at different corner will be FR = FB + FC cos 45° + FA cos 45°

[

q2 q2 1 _____ q2 1 __ + = _____ ______2 + __2 ___ __ 4p eo (DO) a ÷2 a2÷2 q2 1 2 = _____ ______ + _____ __ __ 4p e0 (÷2 a)2 ÷2 a2

[

]

FA FR FC

]

+q C Q O

Qq That force must be equal to ___________2 1__ 4p e0 ___ a ÷2 Qq × 2 _______ q2 1 ___ 2 _______ __ So = + __ 4p e0 a2 4p e0 a2 2 ÷2

( ) ( )

+q

+q B

A

Fig. 2.2W Equilibrium of four equal charges at corners of a square when a charge is placed at the center.

__

or,

+q D

(1 + 2÷2 ) Q = q ________ 2

__

q(1 + 2÷2 ) So, the charge at the center will be – _________. 2 Example 2.4 Two similar balls of mass m are hung from silk threads of length l and carry same charges. Prove that for a small angle j, the separation of the charges, will be

(

q 2l x = ________ 2p eo mg Sol.

)

1 __ 3

j In Fig. 2.3W, each ball of charge + q are suspended from O by silk threads. Here q = __ 2 qq The restoring force = mg sin q and electrostatic repulsive force between the balls is = _______2 4p e0 x q2 O _______ In equilibrium, mg sin q = where x is the separa4p e0 x2 tion of the balls. j

x/2 x From the figure for small q, sin q = ___ = __ l 2l 2 q x \ mg __ = _______2 2l 4p e0 x or, Hence proved.

q2l x3 = ________ 2p e0 mg

or,

(

q2l x = _______ 2p e0 mg

q

l

q

mg sin q q

F

x

)

1 __ 3

mg cos q mg

Fig. 2.3W Separation of two like and equal charges suspended from a point.

2.16

Advanced Engineering Physics

Example 2.5 An amount of charge Q is divided into two particles. Find the charge on each particle so that the effective force between them will be maximum. Sol. Suppose the charge on one particle be q, then charge on the other is (Q – q). If they are separated by a distance r then force 1 q(Q – q) F = _____ _______ 4p e0 r2 dF ___ For maximum F, =0 dq

[

]

So,

q(Q – q) d _____ 1 _______ ___ =0 dq 4p e0 r2

or,

d ___ (qQ – q2) = 0 dq

or,

Q – 2q = 0 or,

Q q = __ 2

So, for maximum F, Q is to be equally divided in the particles. Example 2.6 Find out electric field intensity at any point on the axis of the uniformly charged rod. Sol. In Fig. 2.4W, let L is the length of the rod AB uniformly charged (q). If l be the linear charge denq sity then l = __ and charge on an elementary length L dx is dq = l dx. The electric field at P due to dx is ___› _› dq dE = _______2 x along BP 4p e0 x For the entire charged rod, electric field acts in the same direction and total field at P is a+L

If a >> L,

A

dx

B

P E

L x a

Fig. 2.4W Electric field at a point P on the axis of a uniformly charged rod AB.

l dx l lL 1 _____ 1 E = Ú _____ ___2 = _____ __ – = ____________ 4p e0 a a + L 4p e0 a (a + L) a 4p e0 x 1 q E = _____ __2 . 4p eo a

[

]

So under this condition, a charged rod behaves as a point charge. Example 2.7 Find out the electric field intensity at any point on the perpendicular bisector of a uniformly charged rod. Sol. Consider an elementary length dx at a distance x from the center of the rod of length L [Fig. 2.5W]. q q The charge on this element dx is dq = __ dx = l dx, where l = __ the linear charge density. L L The electric field at P due to this element is dq dq dE = __________2 = ____________ 4p e0 (AP) 4p e0 (a2 + x2)

2.17

Electrostatics

Again, the component of dE along OP is dE cos q. The resultant field at P due to the whole charged rod is 1 E = Ú dE cos q = _____ 4p e0

L /2

Ú – L /2

P

q

a A

a ______ cos q = _______ 2 ÷x + a2

L /2

dx Ú _________ 2 2 3/2 – L /2 (a + x ) We have x = a tan q or dx = a sec2q dq qa qa2 a sec2q dq ________ or, E = ______ Ú _________ = Ú cos q dq 4p e0 L a3 sec3q 4p e0 a3L q q L /2 x = _______ sin q = _______ _________ 4p e0 L a 4p e0 L a (a2 + x2)1/2 – L /2 2q q 1 _______ = ______________ _______ = _____. __________ 4p e0 a÷L2 + 4a2 2p e a÷L2 + 4a2

[

Y q

q a dx __________ L(a2 + x2)3/2

[ qa = ______ 4p e0 L

dE

]

x O

dx

L

Fig. 2.5W Electric field at a point on the perpendicular bisector of a uniformly charged rod.

]

0

Example 2.8 Sol.

Find out the electric field intensity at a point on the axis of a uniformly charged ring. Consider an elementary length dl of the ring (Fig. 2.6W). The charge of the ring dq = l dl. The field at P due to dq of dl is 1 dq A dl dE = _____ ___ along AP r 2 4p e0 r a P q dEx The component of dE along the x axis is dEx = dE cos q c q dE x l dl 1 \ dEx = _____ ____ cos q 4p e0 r2 Fig. 2.6W Electric field intensity at a Total field intensity at P due to the whole ring point on the axis of a charged ring. l l dl x E = _____ Ú __2 cos q = _______2 Ú __r dl 4p e0 r 4p e0 r l __ x _____ = × 2pa 4p e0 r3 qx 2p a x l = _____ _________ = ______________ along the x axis 4p e0 (a2 + x2)3/2 4p e0 (a2 + x2)3/2 q When x >> a, E = _______2 4p e0 x So, at large distance the ring behaves like a point charge.

Example 2.9 Sol.

Find out the electric field intensity at any point on the axis of a uniformly charged disc. In Fig. 2.7W, Let R be the radius of the disc and x be the distance of the field point P from the center

O. We consider a concentric ring within radii r and r + dr. If s is the surface charge density then total _› s dS charge on a surface element dS is s dS. The field at P due to ds is given by d E = _______2 â, here 4p e0 a a2 = x2 + r2.

2.18

Advanced Engineering Physics

The component along the axis of the disc

dS

s dS 1 dE1 = _____ _______ cos q 4p eo (r2 + x2) s dS ________ x 1 = _____ _______ 4p e0 (r2 + x2) (x2 +r2)1/2 s dS x = _____________ 4p e0 (r2 + x2)3/2

a P

q

R

O r

E1

q

x E

Fig. 2.7W Electric field at a point P on the axis of a charged disc.

Again ds = 2 p r dr or,

s × 2 p r dr x dE1 = ______________ 4 p e0 (r2 + x2)3/2

E s 2e0

R

2p x s r dr Total intensity at P is E = Ú dE1 = _____ Ú _________ 4 p e0 0 (r2 + x2)3/2 x s 1 _________ 1 E = ___ __ – 2e0 x (R2 + x2)1/2 s x = ___ 1 – _________ 2 2e0 (R + x2)1/2

[ [

–X

+X s 2e0

] ]

Fig. 2.8W Variation of field intensity with distance from the center of a charged disc.

For infinite charge sheet R Æ a sx \ E = ______ 2E0 |x| For positive s, E is positive when x is positive and it is negative when x is negative. The magnitude s of the field is independent of the distance x and is given by ___. The variation of field intensity with 2e0 distance from the center of a uniformly charged disc is shown in Fig. 2.8W. Example 2.10 Five equal charges of 40 nC each are placed at five vertices of a regular hexagon of 6 cm side. The sixth vertex is free. Determine the electric field at the center of the hexagon due to the distribution. Sol. The field at the center due to the charges located at two opposite vertices is zero. Since there is no charge at F (free vertex) [Fig. 2.9W] so the resultant field will be due to charge q located at C. The field is directed from the center to the vacant

A

B

F

C E

D

Fig. 2.9W Electric field at center due to five equal charges placed at five corners of a regular hexagon.

q corner and its magnitude is _______2 , where a is the distance 4p e0 a 9 × 109 × 40 × 10–9 of the center from each of the vertices. So, the field is ________________ N/C or, 105 N/C –2 2 (6 × 10 )

2.19

Electrostatics

Example 2.11 Infinite number of positive charges, each of magnitude q is placed on the x axis at the point x = 1, 2, 4, 8, … What will be intensity of electric field at x = 0? Also calculate the electric field if alternative charges are of opposite signs. q q q O Sol. The resultant intensity at x = 0 [Fig. 2.10W] is ... x = 0 1 2 4 q q q q 1 E = _____ __2 + __2 + __2 + __2 + … 4p e0 1 Fig. 2.10W Electric field at x = 0 due 2 4 8 to equal charge at x = 1, 2, 4, 8, ... q 1 1 1 = _____ 1 + __ + ___ + ___ + … 4p e0 4 16 64 q 1 = _____ _______ 4p e0 1 1 – __ 4 q __ 4 _____ = units 4p e0 3

[

]

[

]

( )

If alternate charges are of opposite signs then electric intensity at x = 0 is q q q q 1 E1 = _____ __2 – __2 + __2 – __2 + … 4p e0 1 2 4 8 q q 4 1 = _____ _______ = _____ __ units 4p e0 4p e0 5 1 1 + __ 4

[

(

]

)

Example 2.12 Three charges q1 q2 and q3 are at the vertice of an equilateral triangle of 1 m side. The charges are q1 = – 2 m C, q2 = 6 m C and q3 = 4.5 m C. Find the total potential energy of this charge distribution. q2 q3 q1 q3 q1 q2 _____ 1 Sol. The total potential energy, U = _____ _____ + r + _____ r13 4p e0 r12 23

[

]

1 = 9 × 109 × __ a (q1 q2 + q2 q3 + q1 q3)

[r12 = r23 = r31 = a (Say)]

9

= 9 × 10 × [– 2 × 6 + 6 × 4.5 – 2 × 4.5] × 10–12 = 0.054 Joule Example 2.13 Three charges q, 2q, and 4q are placed along a straight line of 6 cm length. Where should the charges be placed so that potential energy of the system is minimum. Find out the distance of the charges. Sol. Let 2q be placed between the other two charges [Fig. 2.11W]. Suppose, the distance between q and 2q is x m. so, the distance between 2q and 4q is (0.06 – x) m. The potential energy of the whole system, q × 2q _______ 2q × 4q ______ q × 4q 1 U = _____ ______ x + 0.06 – x + 0.06 4p e0

[

For minimum value of U or, or,

]

dU ___ =0 dx

1 4 – __2 + _________2 = 0 x (0.06 – x) x = 0.02 m

x q

(0.06 – x) 2q

4q

Fig. 2.11W Minimum PE due to charges q, 2q, 4q placed on a line of 6 cm length.

2.20

Advanced Engineering Physics

So, the distance between q and 2q is 0.02 m and the distance between 2q and 4q is 0.04 m. _›

Example 2.14 If the electric field is given by E = 6 + 3 + 4 , calculate the electric flux through a surface of area of 20 units laying in the yz plane. _› _› Sol. Since the surface area lies in the yz plane, the area vector S is directed along the x direction. So S = 20 . _› Here E=6 +3 +4 _› _› \ electric flux through the surface is jE = E . S = (6 +3 + 4 ) . 20 = 120 units Example 2.15 Find out electric field intensity at regions I, II , III due to two infinite plane parallel sheets of charge [Fig. 2.12W].

E2

E2

s1

E1

E1

E2

s21

E1

II I

III A

Fig. 2.12W

Sol.

B

Electric field at three points I, II and III due to two infinite plate charge sheets.

Let A and B be two infinite plane parallel charge sheets and s1, s2 be uniform surface densities of charge on A and B respectively. Here s1 > s2 (say). In region I, the net electric field – s1 – s2 1 EI = E1 + E2 = ____ + ____ = – ___ (s1 + s2) 2e0 2e0 2e0 s1 s2 1 In region II, EII = E1 – E2 = ___ – ___ = ___ (s1 – s2) 2e0 2e0 2e0 s1 s2 1 In region III, EIII = E1 + E2 = ___ + ___ = ___ (s1 + s2) 2e0 2e0 2e0 s Now if s1 = s and s2 = – s then EI = 0 EIII = 0, and EII = __ e0

( )( )

Example 2.16

The electric field components in Fig. 2.13W are Ex = ax1/2, Ey = Ez = 0 in which

a = 800 NC –1 m–1/2. Calculate electric flux through the cube and the charge within the cube. Assume that a = 0.1 m. Sol. The electric flux is zero for each face of the cube except the two faces ABCD and EFGH. The magnitude of electric field at the face ABCD, E1 = a x1/2 = a a 1/2 and at the face EFGH, E2 = a x 1/2 = a (2a)1/2.

2.21

Electrostatics _› __›

y

j1 = E. dS = E1S cos 180°

So, flux

B

= a a1/2 × a2 (–1) = – a5/2 a _› __›

j2 = E. dS = E2S cos 0°

and flux

1/2

= a (2a) So, net flux

1/2

a

1/2

5/2

a – a5/2 a

a =2

j = (E2 – E1) = 2 5/2

a

E

A

2

5/2

F

a C

1/2

= a a (2 –1) Now, putting the value a = 0.1 m and a = 800 NC –1 m–1/2, we have net flux = 1.05 Nm2 C –1 and charge q = e0j So, q = 8.85 10–12 × 1.05 = 9.3 × 10–12 C

G

a D

z

x

H a

Fig. 2.13W Electric flux through a cube and charge inside it.

Example 2.17 Using Gauss’ law in integral form, obtain the electric field due to the following charge distribution in spherical coordinates. [WBUT 2012] r2 r (r, q, j) = r0 1 – __2 0 a) _› q E ◊ d S = __ e0

_› S

a

a

a

q = Ú 2p rl dr r = 2p l Ú r dr l r = 2p l l Ú r2 dr

But

0

_› __›

Now S

0 0

2p l l E ◊ dS = _____ Ú r2 dr e0 a

E

Fig. 2.15W Electric field at points where r < a and r > a due to long cylindrical charged body.

l r2 E = ____ 3e0

or,

E A1

_› __›

Now, from Gauss’ law

r

l

0

2 = __ p l l r3 3

0

2.24

Advanced Engineering Physics

2 pa3 E × 2p r l = __ ___ l l 3 e0

or,

1 a3 l E = __ ____ 3 e0 r

or,

Example 2.23 The potential field at any point in free space is given by V = 5x2y + 3yz2 + 6xz volt, where x, y, z are in meters. Calculate the volume charge density at point (2, 5, 3) m. Sol. From Poisson’s equation r —2 V = – __ e0 Here

∂ 2V ____ ∂ 2V ∂ 2V —2 V = ____ + 2 + ____ 2 ∂x ∂y ∂z2

So,

∂ 2V ___ ∂2 ∂ ____ = (5x2y + 3yz2 + 6xz) = ___ (10xy + 6z) = 10y 2 2 2 ∂x ∂x ∂x ∂ 2V ___ ∂2 ∂ ____ = 2 (5x2y + 3yz2 + 6xz) = ___ (5x2 + 3z2) = 0 2 ∂y ∂y ∂y ∂ 2V ___ ∂2 ∂ ____ = 2 (5x2y + 3yz2 + 6xz) = __ (6yz + 6x) = 6y 2 ∂z ∂z ∂z

∂ 2V ____ ∂ 2V ____ ∂ 2V or, ____ + + = 10y + 6y = 16y ∂x2 ∂y2 ∂z2 ∂ 2V ____ ∂ 2V ____ ∂ 2V Now at point (2, 5, 3), ____ + + = 80 ∂x2 ∂y2 ∂z2 r —2 V = 80 = – __ e0

or,

r = – 80 e0 = – 80 × 8.854 × 10–12 C/m3

or, _›

Example 2.24 In a field E = – 50y – 50x + 30 V/m, calculate the differential amount of work done in moving 2m C charge a distance 5m m from _P1 __(1, 2, 3,) to P2 (2, 4, 1). › › Sol. We know that work done dw = – qE. dl Here

__›

dl = 5êP P

1 2

(2 – 1) + (4 – 2) + (1 – 3) 1 _____________ êP P = _________________________ = __ ( + 2 – 2 ) 1 2 2 2 2 3 ÷1 + 2 + (– 2) or, or,

__›

5 dl = __ ( + 2 – 2 ) m . 3 _› __› 5 dw = – qE ◊ dl = – 2 × 10–6 (– 50y – 50x + 30 ). __ ( + 2 – 2 ) × 10–6 3

At initial point (1, 2, 3)

Electrostatics

Example 2.25 Sol.

2.25

5 dw = – 2 × 10–6(– 100 – 50 + 30 ). __ × 10–6 ( + 2 – 2 ) 3 10 = – ___ × 10–12 (– 100 – 100 – 60) 3 10 = – ___ × 260 × 10–12 Joule = 8.66 × 10–10 Joule 3 1 Show that V = __r satisfies Laplace’s equation. _›

The position vector r = x + y + z | r | = ÷x2 + y2 + z2 = (x2 + y2 + z2)1/2

(

)

∂2 ∂2 ∂2 1 —2 __r = ___2 + ___2 + ___2 (x2 + y2 + z2)– 1/2 ∂x ∂y ∂z

()

\ Now,

__________

_›

_›

Magnitude of r ,

∂2 2 ∂ ___ (x + y2 + z2)–1/2 = ___ {–x(x2 + y2 + z2)–3/2} 2 ∂x ∂x 2x2 – y2 – z2 = _____________ (x2 + y2 + z2)5/2

2y2 – x2 – z2 ∂2 Similarly, ___2 (x2 + y2 + z2)–1/2 = _____________ ∂y (x2 + y2 + z2)5/2 and

2z2 – y2 – x2 ∂2 2 2 2 –1/2 ___ _____________ (x + y + z ) = ∂z2 (x2 + y2 + z2)5/2 2x2 – y2 – z2 + 2y2 – x2 – z2 – 2z2 – y2 – x2 1 = __________________________________ —2 __ =0 r (x2 + y2 + z2)5/2

()

\

1 So, V = __r satisfies Laplace’s equation. Example 2.26

Sol.

qe– r/l The potential in a medium is given by j (r) = ________ 4p e0r

(i) Obtain the corresponding electric field. (ii) Find the charge density that may produce the potential mentioned above. Here j (r) is purely a function of r. So, electric field __› _› ∂j q ∂ e–r/l E = – — j = – ___ = – _____ __ ____ (i) 4p e0 ∂r r ∂r

( )

q 1 1 = – _____ – ___ e– r/l – __2 e– r/l 4p e0 lr r

[

q = _____ 4p e0

(

(

)

1 __ 1 – r/l ___ e + l r r2

_›

]

_›

)

q r r = _____ ___2 + __3 e–r/l 4p e0 lr r

[

_›

r where = __ r

]

[WBUT 2008]

2.26

Advanced Engineering Physics __› _›

r — ◊ E = __ e0

(ii) We know that

_

_

› › __› _› __› q r r r = e0 — ◊ E = e0 — ◊ _____ ___2 + __3 e– r/l 4 p e0 lr r

[ (

or,

__›

_›

_› __› q __› ____ r – r/l ___ = —◊ e + —◊ 4p l r2 __›

[ ( ) _›

__› _›

)

_›

]

() ] r –r/l __ e r3

— ◊ (j A ) = —j ◊ A + j— ◊ A

By applying

_› _› __ __ _› __› __› _› q __ 1 › – __lr __ r __ 1 –r/l › __ r r r –r/l __ –r/l ___ r= — e ◊ 2+ e — ◊ 2 + —e ◊ 3 + e — ◊ __3 4p l l r r r r __› r __ – 1 –r/l —(e ) = – __ e l l

[

We have Again

__›

_›

( )

_›

__

()

()

› r r 1 — ◊ __3 = 0 and — ◊ __2 = __2 r r r r __ q – 1 1 1 – __r 1 1 – __r 1 r = ___ – __2 e l __r + __ e l __2 – __ e l __2 4p l l r l r q = – ______ e– r/l. 4p l2r

[

So,

( )]

( )

]

_›

Example 2.27 Show that F = (2xy + z3) + x2 + 3xz2 is a conservative field. Find also the scalar potential. [WBUT 2003] __› _› Sol. We know that for conservative force field — × F = 0 _›

F = (2xy + z3) + x2 + 3xz2

Here __›

\

_›

—×F= _›

|

∂ ___ ∂x (2xy + z3)

∂ ___ ∂y x2

|

∂ __ = (0 – 0 ) + (3z2 – 3z2) + (2x – 2x) = 0 ∂z 3xz2

So, F is a conservative field. _› __› Again, we know that for a conservative field F = — V, where V is the scalar potential. \ \ here So,

\

_› __›

__

__› › ∂V ∂V ∂V F ◊ dr = —V. dr = ___ dx + ___ dy + ___ dz = dV ∂x ∂y ∂z

_› __›

F ◊ dr = (2xy + z3)dx + x2dy + 3xz2 dz _› __›

dV = F. dr = (2xy +z2) dx + x2dy + 3xz2 dz = 2xy dx + x2dy + z3 dx + 3xz2 dz = d (x2y) + d (xz3) = d (x2y + xz3) V = Ú dV = Ú d (xy2 + xz3) = x2y + xz3 + constant.

2.27

Electrostatics

Example 2.28

Sol.

Find the electric field due to the following electric potential. sin q cos j V = _________ r2 __› _› We know E = – — V. Here, V is the function of r, q, j. So in spherical polar coordinates __

_›

[

› ∂V 1 ∂V 1 ∂V E = – — V = – ___ êr + __r ___ êq + ______ ___ êj r sin q ∂j ∂r ∂q

]

2 sin q cos j ___ cos q cos j ___ sin q sin j ∂V ∂V __________ ∂V ___ = – ___________ = = – _________ , , 3 2 ∂r ∂q ∂j r r r2 So,

Example 2.29

_›

2 sin q cos j 1 1 sin q sin j E = ___________ êr – __3 cos q cos j êq + ______ _________ êj 3 r sin q r r r2 1 = __3 [2 sin q cos j êr – cos q cos j êq + sin q êj] units r

Is it possible for the electric potential in a charge-free space to be given by (a) V = x2 + y2

– y2 (b) x2 + y2 – 2z2. If not, find the charge density. Sol. (a) V = x2 + y2 – z2 or,

__›

—V =

(

)

∂ ∂ ∂ ___ + ___ + __ (x2 + y2 – z2) ∂x ∂y ∂z

= 2x + 2y + – 2z and

__› __›

—2V = — ◊ —V =

(

)

∂ ∂ ∂ ___ + ___ + __ ◊ (2x + 2y – 2z ) ∂x ∂y ∂z

= 2 + 2 – 2 = 2. r Now by using Poisson’s equation, —2V = __ e0 \ r = – 2 e0 The space is not charge-free. 2 2 2 (b) __›V = x + y – 2z or, —V = 2x + 2y – 4z —2V = 2 + 2 – 4 = 0 r By Poisson’s equation, —2V = – __ = 0 or, r = 0 e0 The space is a charge-free region. Example 2.30 Region between the two coaxial cones is shown in Fig. 2.16W. A potential Va exists at q1 and V = 0 at q2. The cone vertices are insulated at r = 0. Solve Laplace’s equation to get potential at a cone at any angle q. Sol. The potential is constant with respect to r and j. So, in spherical polar coordinates, Laplace’s equation takes from

2.28

Advanced Engineering Physics

(

)

∂V d 1 _____ 1 ___ __ sin q ___ = 0 2 sin q dq ∂q r Now after integration with respect to q

Again integrating

q2 q1

( )

Here, boundary constants are q = q1, V = Va q = q2 , V = 0 Now using boundary conditions q1 Va = C1 ln tan __ + C2 2

( (

) )

q2 0 = C1 ln tan __ + C2 2 After simplification

Va

∂V sin q ___ = constant = C1 (say) ∂q q V = C1 ln tan __ + C2 (constant) 2

V=0

Fig. 2.16W Potential at a cone of angle q due to two other equipotential cones at angles q1 and q2.

q2 q ln tan __ – ln tan __ 2 2 V = Va ____________________ q q 1 2 ln tan __ – ln tan __ 2 2

( ) (

) ) ( )

(

Review Exercises

Part 1: Multiple Choice Questions 1. Which of the following statements is not correct regarding electrostatic field vector E ? [WBUT 2008] _› __›

E ◊ dr = 0 , where c is a simple closed curve.

(a) C b

(b)

_› __›

Ú E ◊ dr is independent of the path for given end points a and b. a

_›

__›

_›

__›

_›

_›

(c) E = — × A for some vector potential A. (d) E = —j, for some scalar field j. 2. Flux of the electric field for a point charge (q) at origin through a spherical surface centered at the origin is [WBUT 2006] 2q q q (a) ___ (b) __ (c) _____ (d) zero e0 e0 4p e0 3. Two concentric spheres of radii r1 and r2 carry charges q1 and q2 respectively. If the surface charge density (s ) is the same for both spheres, the electric potential at the common center will be s r1 s r2 s s (a) __ __ (b) __ __ (c) __ (r1 – r2) (d) __ (r1 + r2) e0 r2 e0 r1 e0 e0

2.29

Electrostatics

4. Six charges, each equal to + q, are placed at the corners of a regular hexagon of side a. The electric potential at the point where the diagonals of the hexagon intersect will be given by ___ 3q q 6q ÷ 1 __ 1 ___ 1 ____ _____ _____ _____ (a) zero (b) (c) (d) 4p e0 a 4p e0 a 4p e0 2a 5. In free space Poisson’s equation is

[WBUT 2005] e 0 (a) —2 V = 0 (b) —2V = __ (c) —2V = a (d) None of these r 6. The electric flux through each of the faces of a cube of 1 m side if a charge q coulomb is placed at its centre is [WBUT 2007] q (a) ___ 4e0

(b) 4e0 q

q (c) ___ 6e0

e0 (d) ___ 6q

7. Let (r, q, j) represent the spherical polar coordinates of a point in a region where the electrostatics potential V is given by V = Kj2. Then the charge density in that region [WBUT 2007] (a) is also a function of j only (b) is constant in that region (c) is a function of all the coordinates (r, q, j) (d) is a function of (r, q) only 8. The electrostatic potential energy of a system of two charges q1 and q2 separated by a distance r is 1 q1 q2 (a) _____ _____ 4p e0 r2

e0 q1 q2 (b) ___ _____ 4p r _›

1 q1 q2 (c) _____ _____ 4p e0 r

q21 q2 1 _____ _____ (d) 4p e0 r2

9. The magnitude of electric field E in the annular region of a charged cylindrical capacitor (a) same anywhere

1 (b) varies as __r

1 (c) varies as __2 (d) None of these r 10. Electric field and potential inside a hollow charged conducting sphere are respectively q (a) 0, 4p e0 __r

1 q (b) _____ __2 , 0 4p e0 r

q q (d) ______2 , ______ 4p eo r 4p e0r 11. For a closed surface which does not include any charge, the Gauss’s law will be _› __›

(a)

E. ds = 0

S

_› __›

(b) S

q E. ds = __ eo

1 q (c) 0, _____ __r 4p e0 _› __›

(c) S

q E. ds = _____ 4p e0

(d) None of these

12. Electrostatic field is (a) conservative (b) non-conservative 13. Laplace’s equation for an electrostatic field is r (a) —2V = __ e0

(b) —2V = 0

(c) rotational

(d) None of these

r (c) —2V = – __ e0

__› r (d) —V = __ e0

14. Electric field intensity at any point distant r from a plane charged conducting sheet varies as (a) r–1 (b) r0 (c) r–2 (d) r 15. In a region of space, if the electrostatic potential is constant, then the electric field at that region is (a) zero (b) infinite (c) constant (d) None of these

2.30

Advanced Engineering Physics

16. If the flux of the electric field through a closed surface is zero, (a) the charge inside the surface must be zero (b) the electric field must be zero everywhere on the surface (c) the charge in the vicinity of the surface must be zero (d) None of these [Ans. 1 (c), 2 (b), 3 (d), 4 (c), 5 (a), 6 (c), 7 (d), 8 (c), 9 (b), 10 (c), 11 (a), 12 (a), 13 (b), 14 (b), 15 (a), 16 (a)]

Short Questions with Answers 1. What is an electric line of force? What is its importance? Ans. An electric line of force is an imaginary straight or curved path along which a positive test charge is supposed to move. The lines of force originate from a single positive charge and converge at an isolated negative charge. The relative closeness of electric line of force in a certain region provides an estimate of the electric field strength in that region. 2. Sketch the electric lines of force due to point charges (i) q > 0 (ii) q < 0. Ans.

q>0

q> 1, P0 Æ 1 complete alignment, but this does not occur in gases. For most practical cases (x is small, i.e., at high temperature) x x R is where

_› __›

�Ú B ◊ dl = m0I or, or,

c

B × 2pr¢ = m0I m0I B = ____ 2pr¢

m 0I At the surface of the wire, r ¢ = R, B = ____ 2pR

…(4.43) …(4.44)

Advanced Engineering Physics

4.12

B

R=r

Fig. 4.15

r

Variation of magnetic field with distance of a current carrying cylindrical wire.

The variation of the magnetic field with distance from the axis of the cylinder is shown in Fig. 4.15. Note: (i) For long straight conducting wire of infinite length carrying current I, the magnetic field at any point m0I at a perpendicular distance r from the wire will be B = ____. 2pr (ii) For a hollow cylinder, since the current I exists only on the surface of the cylinder and inside the cylinder current is zero, so magnetic field inside the cylinder will be zero.

4.13.2

magnetic field Inside a long solenoid

When a current (I) carrying wire is wound tightly on the surface of a cylindrical tube, we get a solenoid. Generally, the length (L) of the solenoid is large as compared to the transverse dimension. If N is the total N number of turns over a length L, we get __ = n as the L number of turns per unit length of the solenoid. Keeping

l S

R

P

Q

the product nI fixed, if we make n very large and corresponding I very small, then we obtain a surface current Fig. 4.16 Magnetic field inside a solenoid. of value nI over the curved surface of the cylinder. It turns out that the magnetic field inside a closely wound solenoid is almost uniform over its cross section and can be taken to be negligible outside the volume of the _›

solenoid. Ampere’s law thus can easily applied to find out the value of B inside the solenoid. In Fig. 4.16, we draw a rectangle PQRS of length l. The line PQ is parallel to the solenoid axis and hence _› parallel to the field B inside the solenoid. Thus, Q

Ú P

_› __›

B ◊ dl = Bl _› __›

_›

Along QR, RS and SP, B. dl is zero everywhere as B is either zero (outside the solenoid) or perpendicular

__›

to dl (inside the solenoid). Thus, from PQRSP,

Magnetostatics

4.13

_› __›

B ◊ dl = Bl …(4.45) If n be the number of turns per unit length, then total of nl turns cross the rectangle PQRS. Each turn carries a current I. Hence net current passes through the area PQRS is nlI. Now, from Eq. (4.45) we have from Ampere’s law _› __›

B ◊ dl = m0 nlI Bl = m0 nlI B = m0 nI …(4.46) The above equations gives the magnetic field inside a long closely wound solenoid. The relation does not depend on the diameter or the length of the solenoid and magnetic field B is constant over the solenoid cross section. or, or,

4.13.3 magnetic field due to toroid An endless solenoid in the form of circular shape is called toroid. The magnetic field in such a toroid can be obtained by using Ampere’s law. _› Let P be a point on the concentric circular path at which magnetic field B is to be calculated. By symmetry, the field will have equal magnitude at all points of this circle [Fig. 4.17]. Let the distance of P from the center be r. The field B is tangential at every point of the circle. Hence, _› __›

B ◊ dl = B Ú dl = B × 2pr = m0 NI where N is the total number of turns and the current crossing the area bounded by the circle is NI. m0 NI So, B = _____ …(4.48) 2pr Thus B is inversely proportional to r. If the cross section of the toroid is N very small, the variation in r can be neglected and ____ can be written as n, 2pr the number of turns per unit length. So B = m0nI …(4.49) The field at an external point (P¢) of the toroid, from Ampere’s law B ◊ dl = 0 or, B = 0 …(4.50) Thus the field outside the toroid is zero.

…(4.47)

P r o

P¢

Fig. 4.17

Magnetic field due to a toroid.

4.14 magnetIC potentIals 4.14.1 magnetic scalar potential

_›

__›

_›

_›

In electrostatics, scalar potential V plays an important role to find electric field intensity E. Since — × E = 0, E can be expressed as the gradient of a scalar quantity V. The two powerful relations are: _›

__›

E = – — V and

—2V = 0

…(4.51) …(4.52)

Advanced Engineering Physics

4.14

If__ in _some region of space, current density J = 0, then from Ampere’s circuital law in magnetostat_› › › ics, — × B = 0. We may therefore express B as a gradient of scalar quantity Vm __›

_›

i.e. B = – — Vm where Vm is called the magnetic scalar potential. __› _›

__›

…(4.53)

__›

Since — ◊ B = 0 so — ◊ (– —Vm) = 0 or, —2Vm = 0 …(4.54) We see that Vm satisfies Laplace’s equation in homogeneous magnetic materials, it is not defined in any region where the current density exists. The magnetic scalar potential may be defined as a scalar whose negative gradient at any point gives the magnetic induction at that point due to a close loop of carrying current. The magnetic scalar potential is useful in describing the magnetic field around a current source

4.14.2

magnetic vector potential

__› _›

Gauss’ law in magnetostatics state that always — ◊ B = 0. Again we know that divergence of any curl = 0, __› __›

_›

_›

_›

__›

_›

_›

i.e., —. — × A = 0. Since the curl of B is not necessarily zero (only if J = 0, — × B = 0), so B can’t be the gradient of a scalar potential in general but as the curl of a vector field, in the form _›

_›

__›

_›

B=—×A

… (4.55) _›

The vector function A which satisfies Eq. (4.54) is known as vector potential. The vector potential _A is as › important in magnetostatics as the scalar potential function_V in electrostatics. The _vector potential A does › › not_have any physical significance. It can help to determine B at a given point, since B is the space derivative › of A. The magnetic vector potential may be defined as a vector, the curl of which gives the magnetic induction produced at any point by a closed-loop carrying current. __›

Again So, but,

_›

_›

— × B = m0 J

We know that

__›

_›

__›

_›

__›

_›

__›

__›

__›

_›

__› __› _›

B=—×A

…(4.56) _›

P

_›

— × B = — × — × A = m0 J

...(4.57)

_ 2 ›

J

— × — × A = —(—. A) – — A __› _›

__›

__›

_›

r

dv

For steady current, we take —. A = 0 So,

r

_›

— × — × A = – —2 A

_›

_›

Now from Eq. (4.57) for dc current only, —2 A = – m0 J …(4.58) Equation (4.58) is the same as Poisson’s equation in electrostatics, r —2V = – __ …(4.59) e0 where V is the electrostatics potential and satisfies r dV 1 V = _____ Ú ____ 4p e0 V r Similarly for Eq. (4.58) we have the general solution _› _› m 0 __ ___ A= Ú J dV 4p V r

Fig. 4.18 Magnetic vector potential at a distance r from a current element.

...(4.60)

…(4.61)

The magnetic vector potential is useful for studying radiation in transmission lines, wave guides, antennas.

Magnetostatics

4.15

Here r is the distance from the current _› being _› element to the point at which the magnetic vector potential is calculated [Fig. 4.18]. Thus the field B produced by a current can be calculated by first determining A using Eq. (4.61) and substituting this in Eq. (4.55).

Worked Out Problems Example 4.1 is 200 mA? Sol.

How many electrons pass through a wire in 1 minute if the current passing through the wire q ne I = __ = ___ t t 200 × 10–3 × 60 It _____________ n = __ = = 7.5 × 1019. e 1.6 × 10–19 What is the drift velocity of electrons in a Cu conductor having a cross-sectional area of

We know or,

Example 4.2

5 × 10–6 m2 if the current is 10 A? Assume that there are 8 × 1028 electrons/m3. Sol. Here, area of cross section A = 5 × 10–6 m2 Current I = 10 A Number density of free electrons, n = 8 × 1028 electrons/m3. 10 I We know vd = ____ = ___________________________ = 1.56 × 10–4 ms–1. neA 8 × 1028 × 1.6 × 10–19 × 5 × 10–6 Example 4.3

Calculate the magnetic field at the center of a regular hexagon [Fig. 4.1W] of side a meter

and carrying a current I A. Sol. The magnetic field at O due to part AB of the hexagon is m0 I B ¢ = ___ _r (sin q1 + sin q2) 4p __ m0 ÷3 Here r = ___a and q1 = q2 = 30°, ___ = 10–7 2 4p So

I __ (sin 30° + sin 30°) B ¢ = 10 ____ 3 ÷ ___a 2 2I 1 __ 1 –7 __ __ = 10 × ____ + ÷3 a 2 2

C B

D

r q1 I q2

O

A

E

–7

(

F

Fig. 4.1W Magnetic field at the center of a hexagon.

)

2I __ = 10–7 × ____ ÷3 a So, total magnetic field at O of the hexagon is

__

4÷3 2I __ = ____ × 10–7 Tesla B = 6B ¢ = 6 × 10 × ____ a ÷3 a –7

Example 4.4

A circular segment QR of a wire PQRS [Fig. 4.2W] of 0.1 m radius subtends an angle of 60°

at its center. A current of 6 amperes is flowing through it. Find the magnitude and direction of the magnetic field at the center of the segment.

Advanced Engineering Physics

4.16

10

Magnetic field due to current in a circular segment making an angle q at the center is m0 I dl 6A S P 6A dB = ___ ___ 4p r2 6A m0 I rdq ___ m0 ____ I dq R Here, dl = r dq So, dB = ___ _____ = Q 4p r 4p r 60° mo Iq So B = ___ __ 4p r p 1 = 10–7 × 6 × __ × ___ Tesla Fig. 4.2W Magnetic field at the 3 0.1 cm

center of a circular element of a

= 6.28 × 10–6 Tesla current-carrying wire. = 6.28 m Tesla Example 4.5 A magnetic field 4 × 10–3 tesla exerts a force of (4 + 3 ) × 10–10 N or a particle having change of 1 × 10–9 C and moving in the xy plane. Calculate the velocity of the particle. Sol.

_›

Lorentz force

_›

_›

F = q ( v × B)

Here

(4 + 3 ) × 10–10 = 1 × 10–9 [(vx + vy )] × 4 × 10–3

or,

(4 + 3 ) × 10–10 = 4 × 10–12 [(vx (– ) + vy ( )] 3 × 10–10 3 vx = – ________ = – __ × 102 = – 75 –12 4 4 × 10

or,

4 × 10–10 vy = ________ = 100 4 × 10–12 v = – 75 + 100 ms–1

So,

A test charge having a charge of 0.4 C is moving with a velocity of 4 – + 2 m/s through an electric field of intensity 10 + 10 and a magnetic field 2 – 6 – 6 . Determine the magnitude and direction of the Lorentz force acting on the test charge. [WBUT 2007] Example 4.6

Sol.

__›

_›

_›

Total Lorentz force = q E + q ( v × B) _›

Here electric force = q E = 0.4 (10 + 10 ) j i _› _› and magnetic force = q ( v × B) = 0.4 4 –1 2 –6

|

k 2 –6

|

= 0.4 [(6 + 12) + (4 + 24) + (– 24 + 2) ] _›

Now total Lorentz force,

= 0.4 (18 + 28 – 22 ) _›

_›

_›

F = q E + q ( v × B) = 0.4 (10 + 10 ) + 0.4 (18 + 28 – 22 ) = 0.4 (28 + 28 – 12 )

___________________

= 0.4 ÷(28)2 + (28)2 + (– 12)2 = 16.6 N

The magnitude of the force

Suppose, the total force makes an q with the x axis, then _›

F. = 0.4 (28 + 28 – 12 ). = 0.4 × 28

Magnetostatics

4.17

___________________

or,

(0.4) ÷(28)2 + (28)2 + (– 12)2 cos q = 0.4 × 28 28 __________________ cos q = ____________________ 2 ÷(28) + (28)2 + (– 12)2 7 ____ cos q = _____ ÷107 7 ____ = 47.41° q = cos–1 _____ ÷107

or, or,

( )

\

Example 4.7 A straight wire carrying a current of 10 A is bent into a semicircular arc of p cm radius as shown in Fig. 4.3W. What is the magnetic field and direction of the magnetic field at center O of the arc? Sol. The magnetic field at center O due to each straight portion of the wire is O. The magnetic field at center O is only due to half the circular loop. The magnitude of magnetic field O m0I ____________ m0 p I 4p × 10–7 × 10 –4 ___ B = ___ ___ = = = 10 T 4p a 4a Fig. 4.3W Magnetic field at 4 × p × 10–2

( )

The current in the loop is anticlockwise and the direction, of the field is perpendicular to the paper.

the center of a semicircular wire carrying current.

Example 4.8 The wire loop ABCDA formed by joining two semicircular wires of radii R1 and R2 carries a current I as shown in Fig. 4.4W. Find out the magnetic field at center O. Sol. The magnetic field due to a semicircular loop of radius R1 is mo p I B1 = ___ ___ 4p R1

( )

[Here, direction of current is anticlockwise] and direction of the field is normal to the plane of the loop, directed upward. For a bigger loop, direction of the current is clockwise. The value of the magnetic field due to semicircular loop of radius R2 is m0 p I B2 = ___ ___ 4p R2

( )

Here direction of the magnetic field is into the plane of the paper. So, net magnetic field m0 1 1 B = B1 – B2 = ___ p I ___ – ___ R1 R2 4p m 0I 1 1 = ___ ___ – ___ 4 R1 R2

(

)

(

R2

R1 D

C

O

B

A

Fig. 4.4W Magnetic field at the center of two concentric semicircular wires carrying current.

)

and direction is perpendicular to the plane of the paper, hence directed upward. Example 4.9 A current-carrying straight wire cannot move but a current-carying square loop adjacent to it can move under the influence of a magnetic force. Show that the square loop in Fig. 4.5W will move towards the wire.

Advanced Engineering Physics

4.18

Sol.

In Fig. 4.5W, we see that the force acting on arms AB and DC are equal and opposite. But the force on arm AD is given by m0 2Ii F1 = ___ ___ 4p a

( )

which is directed towards the wire. The force on arm BC is given by m0 2Ii F2 = ___ ___ 4p b

( )

which is directed away from the wire. Here, b > a hence F1 > F2. So the loop will move towards the wire.

D

i

C

a I

A

B b

Fig. 4.5W Current-carrying square loop moves under the influence of magnetic force due to the current-carrying wire.

Example 4.10 A long solenoid of 40 cm length has 300 turns. If the solenoid carries a current of 3.5 A, calculate (i) magnetic field at the center of the solenoid, and (ii) magnetic field of the axis at one end of the solenoid. Sol. (i) The magnetic field at the center of the solenoid is N 300 B = m0 n I = m0 __ I = 4p × 10–7 × ____ × 3.5 = 3.3 × 10–3 Tesla 0.4 l (ii) The magnetic field at one end of the solenoid N 1 1 1 300 B = __ m0 n I = __ m0 __ I = __ × ____ × 3.5 2 2 2 0.4 l = 1.65 × 10–3 Tesla. Example 4.11 In Fig. 4.6W, in between two rails, a metal wire of mass m slides without friction. The distance of separation between the two rails is l. The break lies in a vertical uniform magnetic field B (perpendicular to the paper). Find the velocity of the wire as a function of time t, if a constant current I flows along one rail and then through the wire and back down the other rail. Sol. The force exerted on the wire of length l is F = B I l sin 90° = B I l The direction of the force will be to the left according to Fleming’s left-hand rule. The acceleration of the wire BIl F ____ a = __ m= m

Rail I

l

I

I

Fig. 4.6W A metal wire slides without friction between two rails carrying current in an opposite direction.

BIl Let initial velocity of wire u = 0, then velocity at any time t is v = at = ____ m t. Example 4.12 Two straight wires, each 2 m long, are parallel to one another and are separated by a distance of 2 cm. If each carries a current of 8 A, calculate the force experienced by either of the wires. Sol. The force per unit length experienced by each wire carrying currents I1, I2 separated by a distance d is given by

Magnetostatics

4.19

m0 I1 I2 F = ______ 2p a Here

I1 = I2 = 8 A d = 2 cm = 0.02 m

so,

4p × 10–7 × 8 × 8 F = _______________ = 64 × 10–5 N/m 2p × 0.02

The total force on either of the wires is F = 2 × 64 × 10–5 N = 128 × 10–5 N In the Bohr model of a hydrogen atom, an electron is revolving in a circular path of 0.4 Å radius with a speed of 106 m/s. What is the value of magnetic field at the center of the orbit? Sol. Here r = 0.4 Å = 0.4 × 10–10 m, v = 106 m/s. Example 4.13

2p r ______________ 2p × 0.4 × 10–10 We know that time period T = ____ = 8p × 10–17 s 6 v = 10 1.6 × 10–19 __ e _________ 2 –3 __ Again, current I = = = × 10 A T 8p × 10–17 p The magnetic field at the center of the orbit mo 2pI 2×p 2 B = ___ ___ = 10–7 × _________ × __ = 10 Tesla. –10 p 4p r 0.4 × 10

( )

Example 4.14

The volume current density distribution in cylindrical coordinates is J (r, j, z) = 0 0 1; a sample of silicon will act like a conductor at frequency £ 109 Hz. we The penetration depth for good conductor _____

÷

2 d = ____ wms _______________________

÷

2 = 3.6 × 10–2 m = ______________________ 2p × 106 × 4p × 10–7 × 200 = 3.6 cm Example 5.18

Calculate the skin depth for a frequency 1010 Hz for silver.

Given

s = 2 × 107 Sm–1

and m = 4p × 10–7 Hm–1

Here w = 2pf = 2p × 1010, s = 2 × 107 Sm–1 and m = 4p × 10–7 Hm–1

Sol.

Now skin depth

( )

2 d = ____ wms

1/2

(

2 = __________________________ 2p × 1010 × 4p × 10–7 × 2 × 107

)

1/2

= 1.12 × 10–6 m Discuss the behavior of copper to electromagnetic waves of frequency 0.5 × 1016 Hz and 7.0 × 10 Hz. Given the conductivity of copper s = 5.8 × 107 mho m–1 and permittivity e = 9 × 10–12 C2/Nm2 Example 5.19 20

Sol.

Here

w = 2p × 0.5 × 1016 s ______________________ 5.8 × 107 ___ = 205 = we 2p × 0.5 × 1016 × 9 × 10–12

s Since ___ > 100, the conduction current dominates. Hence for frequency 0.5 × 1016 Hz copper is a we conductor. Now for frequency f = 7 × 1020 Hz, w = 2p × 7 × 1020 \

s _____________________ 5.8 × 107 ___ = 1.47 × 10–3 = we 2p × 7 × 1020 × 9 × 10–12

s Since ___ < 100, the displacement current dominates. Hence for frequency 7 × 1020 Hz, copper is a we dielectric.

5.27

Electromagnetic Field Theory

Example 5.20

Calculate the value of Poynting vector for a 60 W lamp at a distance of 0.5 m from it.

Sol. Total average power emitted by the lamp = 60 W The light emitted by the lamp will spread out in the form of a sphere, the radius of which is equal to the distance from it. \ radius of the sphere R = 0.5 m Let P be the average Poynting vector over the surface of the sphere, then Power 60 P = ______ = ________2 = 19.1 W/m2. 4pR 4p (0.5)

Review Exercises

Part 1: Multiple Choice Questions 1. The magnetic flux linked with a coil at any instant ‘t’ is given by jt = 5t3 – 100 t + 200, the emf induced in the coil at t = 2 seconds is (a) 200 V (b) 40 V (c) 20 V (d) –20 V 2. A cylindrical conducting rod is kept with its axis along the positive z axis, where a uniform magnetic field exists parallel to the z axis. The current induced in the cylinder is (a) clockwise as seen from the + z axis (b) zero (c) anticlockwise as seen from the – z axis (d) None of these 3. In an electromagnetic wave in a free space, the electric and magnetic fields are [WBUT 2008] (a) parallel to each other (b) perpendicular to each other (c) at an acute angle (d) inclined at an obtuse angle _› inclined _› 4. If E and B are the electric field and the magnetic field of electromagnetic waves traveling in vacuum with propagation vector then [WBUT 2006] _› _›

_›

_›

_›

_›

_›

_›

_›

(a) k ◊ E = 0 (b) k × E = 0 (c) B × E = 0 5. The velocity of a plane electromagnetic wave is given by

(d) k × E = – B

1 _____ (a) c = ______ m ÷ 0 e0 __› _› r 6. — ◊ E = __ represent e0

e0 (d) ___ m0

1 (b) c = _____ m0 e0

(c) c = m0 e0

(a) Ampere’s law (b) Laplace’s equation (c) Gauss’ law in electrostatics (d) Faraday’s law of electromagnetic induction 7. The differential form of Faraday’s law of electromagnetic induction is __›

_›

__›

_›

_›

∂B (a) — × E = – ___ ∂t _›

∂E (c) — × B = ___ ∂t

__›

_›

__›

_›

_›

∂B (b) — × E = ___ ∂t

_›

∂E (d) — × B = – ___ ∂t

5.28

Advanced Engineering Physics _›

8. Maxwell’s electromagnetic wave equations in terms of an electric field vector E in free space is [WBUT 2005] _ _ _› _› 2 › 2 › ∂ ∂ E E (a) —2 E = – m0 e0 ____ (b) —2 E = m0 e0 ____ ∂t2 ∂t2 _ 2 ›

_›

_›

__

› ∂2 E ∂E (d) — E = m0 e0 ____ (c) — E = m0 e0 ___ ∂t ∂t2 9. Displacement current arises due to [WBUT 2004, 2007] (a) positive charge only (b) negative charge only (c) time-varying electric field (d) None of these 10. In electromagnetic induction (a) mechanical energy is converted into magnetic energy (b) mechanical energy is converted into electrical energy (c) magnetic energy is converted into mechanical energy (d) magnetic energy is converted into electrical energy 11. Waves originating from a point source and traveling in an isotropic medium are described as [WBUT 2007]

(a) j = j0 exp i (kr – w t)

(b) j = j0 exp i (kr – w t)/r 2

(c) j = j0 exp i (kr – w t)/r (d) j = j0 exp i (kr + w t)/r 12. Electromagnetic wave is propagated through_ a region of vacuum, which does not contain any charge __› › or current. If the electric vector is given by E = Eo exp i (kx – w t) then the magnetic vector is [WBUT 2007] (a) in the x direction (b) in the y direction (c) in the z direction (d) rotating uniformly in the xy plane 13. Steady current produces (a) magnetostatic field (b) electrostatic field (c) time varying electric field (d) time-varying magnetic field 14. A conducting rod is moved with a constant velocity v in a magnetic field. A potential difference appears across the two ends _› _› _› _› _› _› (a) if v || l (b) if v || B (c) if l || B (d) None of these 15. A bar magnet is released from rest along the axis of a very long vertical copper tube. After some time, the magnet (a) will stop in the tube (b) will move with almost constant speed (c) will move with an acceleration g (d) will oscillate 16. The dimension of m0 e0 is (a) L–2 T –2

(b) L–2 T 2

(c) LT –1

17. The modified Ampere’s circuital law is _› __›

_› __›

(a)

B ◊ dl = m0 (I + Id)

(b)

B ◊ dl = m0 I

(c)

B ◊ dl = e0 I

(d)

B ◊ dl = e0 (I + Id)

_› __›

_› __›

(d) L–1 T –1

5.29

Electromagnetic Field Theory

18. The electromagnetic wave is called transverse wave because [WBUT 2008] (a) the electric field and magnetic field are perpendicular to each other (b) the electric field is perpendicular to the direction of propagation (c) the magnetic field is perpendicular to the direction of propagation (d) both the electric field and magnetic field are perpendicular to the direction of propagation _›

_ 2 ›

∂2 B 19. The solution of a plane electromagnetic wave — B = m0 e0 ____ is ∂t2 _›

_›

_› _›

_›

k◊ r )

_›

_› _›

(b) B = B0 ei(w t + k_◊ r ) › _› › _› _› _› _› (c) B = B0 ei( k ◊ r – w t) (d) B = B0 e–i(w t + k ◊ r ) 20. When a magnet is being moved towards a coil, the induced emf does not depend upon (a) the number of turns of the coil (b) the motion of the magnet (c) the magnetic moment of the magnet (d) the resistance of the coil 21. The variation of induced emf (e) with time (t) in a coil if a short bar magnet is moved along its axis with a constant velocity is best represented as (a) B = B0 ei(w_ t –

e e

(a)

(b)

t

t e

e

(c)

(d) t

t

22. The SI unit of Poynting vector is (a) Wm (b) Wm–1 (c) Wm2 23. The Poynting vector is given by the expression _›

__›

(a) E × H

__›

_›

(b) H × E

_› __›

(c) E ◊ H

(d) Wm–2 (d) None of these

24. Skin depth for a conductor in reference to electromagnetic wave varies (a) inversely as frequency (b) directly as frequency (c) inversely as square root of frequency (d) directly as square of frequency 25. The ratio of the phase velocity and velocity of light is (a) one (b) less than one (c) greater than one (d) zero 26. The value of skin depth (d) in a conducting medium is ___

÷

2s (a) d = ___ mw

_____

÷

2 (b) d = ____ mws

_____

÷

1 (c) d = ____ mws

___

÷

2m (d) d = ___ ws

5.30

Advanced Engineering Physics

27. Skin depth is proportional to (a) w

(b) m

1__ (c) ___ ÷s

__

(d) ÷s

[Ans. 1 (b), 2 (b), 3 (b), 4 (a), 5 (a), 6 (c), 7 (a), 8 (b), 9 (c), 10 (d), 11 (a), 12 (a), 13 (a), 14 (d), 15 (b), 16 (b), 17(a), 18 (d), 19 (c), 20 (d), 21 (b), 22 (d), 23 (a), 24 (c), 25 (b), 26 (b), 27 (c)]

Short Questions with Answers 1 State Faraday’s laws of electromagnetic induction. Ans. (i) Whenever there is a change in the magnetic flux linked with a coil an emf is set up in it and stays as long as the magnetic flux linked with it is changing. (ii) The magnitude of the induced emf is proportional to the rate of change of magnetic flux linked with the coil, i.e., dj e a ___ dt 2 What is the difference between conduction current and the displacement current? Ans. Conduction Current Displacement Current (i) It does obey Ohm’s law. (i) It does not obey Ohm’s law. (ii) Conduction current is due to the actual (ii) Displacement current is due to the time-varyflow of charge in a conductor. ing electric field in a dielectric. _› _› _› _› ∂E ___ (iii) Conduction current density J = s E. (iii) Displacement current density J D = e0 ∂t 3. An electron moves along the line AB which lies in the same plane as a circular loop of conduction wire, as Loop shown in Fig. 5.2W. Find out the direction of the induced current in the loop. A B Ans. The electron moves from A to B so the direction of the curFig. 5.2W rent will be from B to A. The magnetic field generated in the loop due to the motion of the current will be directed into the plane of the paper. To oppose this, the current in the coil must be anticlockwise, in accordance with Lenz’s law. 4. Why is electromagnetic wave called transverse wave? Ans. An electromagnetic wave is called a transverse wave because by the direction of the propagation, the electric field and magnetic field are mutually perpendicular to each other. 5. What is displacement current? Ans. See Section 5.6. 6. Why do birds fly off a high-tension wire when current is switched on? Ans. When current begins to increase from zero to maximum value, a current is induced in the body of the bird. This produces a repulsive force and the bird flies off. 7. Two similar circular coaxial loops carry equal currents in the same direction. If the loops be brought nearer, what will happen to the currents in them?

5.31

Electromagnetic Field Theory

Ans. When the loops are brought closer, there is an increase of magnetic flux. An induced emf is produced. According to Lenz’s law, the induced emf has to oppose the change of magnetic flux. So, the current in each loop will decrease. 8. Two coils are being moved out of a magnetic field. One coil is moved rapidly and the other slowly. In which case is more work done and why? Ans. More work will be done in the case of a rapidly moving coil. This is because the induced emf will be more in this coil as compared to slow moving coil. 9. Define skin depth. Ans. Skin depth is defined as the distance in the conductor over which the electric field vector of the wave propagating in the medium decays to 1/e times its value at the surface. 10. What is Poynting vector? _› __› Ans. The cross product of the electric vector E and the magnetic field vector H is known as a Poynting _›

_›

__›

vector. Mathematically Poynting vector P = E × H 11. What is the effect of frequency on skin depth? Ans. We know that skin depth

( )

2 1/2 d = ____ wms Thus skin depth is inversely proportional to the square root of the frequency. So, skin depth decreases with increase in frequency. 12. How is skin depth useful in practical situation? Ans. A poor conductor can be made a good conductor with a thin coating of copper or silver. A silver coating on a piece of glass may be an excellent conductor at microwave frequency. The performance of gold and gold-coated brass wave guides will be the same if the coating thickness is equal to skin depth. So in a practical situation, skin depth method is useful to reduce the cost of the material.

Part 2: Descriptive Questions 1. (a) Write down Faraday’s law of electromagnetic induction. [WBUT 2004] (b) Express it in differential form. [WBUT 2006] 2. (a) Write down Maxwell’s equations in differential form and explain the physical significance of each equation. [WBUT 2002, 2004] _›

_ 2 ›

_›

∂ 2E (b) Show that the wave equation in free space for electric field E is given by — E = m0 e0 ____ ∂t2 [WBUT 2004] 3. (a) State Maxwell’s equations. From these equations, derive the wave equations for an electromagnetic wave. What is the velocity of this wave? _› (b) Assuming_a plane wave solution, _establish the relation between the propagation vector ( k ), elec› › tric field (E) and magnetic field (B). [WBUT 2008] 4. (a) Distinguish between the conduction current and displacement current. (b) Write down Faraday’s law of electromagnetic induction.

5.32

Advanced Engineering Physics

5. (a) Write down Maxwell’s field equations, explaining the term used. Show that in vacuum, both electric and magnetic vectors obey wave equation. Assuming a plane wave solution show that magnetic field is always orthogonal to the electric field. (b) Find the displacement current within a parallel-plate capacitor in series with a resistor which carries current I. Area of the capacitor plates are A and the dielectric is vacuum. [WBUT 2006] 6. (a) Write and explain differential and integral forms of Maxwell’s equations. (b) Explain the significance of displacement current. 7. (a) Write down Maxwell’s field equations. (b) From those equations identify Gauss’ law, Ampere’s law and Faraday’s law. (c) How does velocity of light depend on the properties of vacuum? _› [WBUT 2005] 8. Use Faraday’s law of electromagnetic induction and the fact that magnetic induction B can be derived _› from a vector potential A. Show that the electric field can be expressed as __›

_›

_›

∂A [WBUT 2007] E = – — j – ___ where j is the scalar potential ∂t 9. (a) What is displacement current? Distinguish between conduction and displacement current. _›

_›

_› _› i( k . r – w t)

_›

_›

_› _› i( k . r – w t)

(b) Show that E = E0 e and B = B0 e are possible solutions of Maxwell’s electromagnetic wave equations in terms of electric and magnetic field. _› 10. Starting _› from Maxwell’s equations in free space, show that the magnetic field B and the electric field E in an electromagnetic wave travel with the same speed. 11. Write down Maxwell’s equations in integral form and explain the physical significance of each equation. _

_

› › __› _› _› ∂B ∂E ___ ___ and — × B = m0 J + eo 12. Show that — × E = – ∂t ∂t

__›

_›

[

_›

]

_›

_›

13. Show that in free space, the electric field E, magnetic field B and propagation vector k are perpendicular to each other. _›

14. The electric field associated with an electromagnetic wave is E = E0 cos (kz – w t) + E0 sin (kz – w t), _›

where E0 is a constant. Find the corresponding magnetic field B. 15. (a) State Lenz’s law. Explain it from the principle of conservation of energy. (b) A wire is rotated about one of its ends at right angles to a magnetic field. Deduce the expression for induced emf. 16. Define skin depth. Show that in case of a semi-infinite solid conductor, the skin depth d is given by 1 ____ d = ______ ÷wms where symbols have their usual meanings. 17. What is Poynting vector? Show that Poynting vector measures the flow of energy per unit area per second in an electromagnetic wave. 18. State and prove Poynting theorem. 1 19. Show that average power

= __ Re (E × H*) 2

5.33

Electromagnetic Field Theory

20. What is Poynting vector? Find the expression of Poynting vector. What is the physical interpretation of this vector? 21. A plane electromagnetic wave is incident normally on a metal of electrical conductivity s. Show that the electromagnetic wave is damped inside the conductor and find the skin depth.

Part 3: Numerical Problems 1. A parallel-plate capacitor with plate area A and separation d between the plates is charged by a A constant current I. Consider a plane surface of area __ parallel to the plates and drawn symmetrically 4 I between the plates. Calculate the displacement current through this area. Ans. ID = __ 4 2. A parallel-plate capacitor with circular plates of radius a = 5.5 cm is being charged at a uniform rate dE so the electric field between the plates changes at a constant rate ___ = 1.5 × 1012 V/ms. Find the dt displacement current for the capacitor. [Ans. ID = 0.13A]

[

]

3. The magnetic field in a plane electromagnetic wave is given by By = 2 × 10–7 sin (0.5 × 103 x + 1.5 × 1011 t) Tesla. (a) What is the wavelength of the wave? (b) Write an expression for the electric 4.

5.

field. [Ans. l = 1.26 cm, E0 = 60 Vm–1] Capacitance of a parallel-plate capacitor is 2mF. Calculate the rate at which the potential difference dV between the two plates must charge to get a displacement current of 0.4 A. [Ans. ___ = 2 × 105 V/s] dt A current of 5 A is passed through a solenoid of 50 cm length, 3.0 cm radius and having 200 turns. When the switch is open, the current becomes zero within 10–3 s. Calculate the emf induced across dj N the switch. [Ans. e = 1.42 volt] Hints: e = N ___, j = BA = m0 ni A, n = __ dt l A rectangular loop of 8 cm side and 2 cm having a resistance of 1.6 W is placed P in a magnetic field and gradually reduced at the rate of 0.02 Tesla/s. Find out the –5 induced current. [Ans. 2 × 10 A] –1 A 50 cm long bar PQ is moved with a speed of 4 ms in a magnetic field B = 0.01 Tesla as shown in Fig. 5.3W. Find out the induced emf. B [Ans. 0.02 V] Q Calculate the skin depth for a frequency 1010 Hz for silver. Given s = 2 × 107 –1 –7 –1 –6 Sm and m = 4p × 10 Hm . [Ans. d = 1.12 × 10 m] Fig. 5.3W Find the skin depth d at a frequency of 1.6 MHz in aluminium, where s = 38.2 mS/m and mr = 1. Also find wave velocity. [Ans. d = 6.4 × 10–5 m, v = 6.47 × 102 m/s] A laser beam has a diameter of 2 mm, what is the amplitude of the electric and magnetic field in the beam in vacuum if the power of the laser is 1.5 mW? [Given m0 = 4p × 10–7 H/m, e0 = 8.85 × 10–12 F/m]. [Ans. E0 = 600 V/m H0 = 1.59 amp/m] Find the depth of penetration of a megacycle wave into copper which has conductivity of s = 5.8 × 107 mho/m and a permeability equal to that of free space. [Ans. d = 6.6 × 10–5 m]

[

6.

7.

8. 9. 10.

11.

]

Chapter

6 Classical Mechanics 6.1

IntroduCtIon

In its literal meaning, the very phrase ‘classical mechanics’ contains the topic of newtonian mechanics. But for some practical reasons, newtonian mechanics is discussed separately. The newtonian mechanics for its discussion relies on the three laws of newton. On the other hand, classical mechanics goes a step further. Some luminaries (like Euler, D’Alembert, Lagrange, Hamilton and others) made remarkable contributions to the development of the subject of classical mechanics by introducing some superior methods for solving real physical problems. These methods are basically discussed in the topic of classical mechanics. The entire subject of classical mechanics, which deals with all kinds of motion of material bodies at the macroscopic level throughout the universe, is basically based on newton’s laws of motion. But the aforesaid superior methods introduced by the said luminaries avoid explicit mention of newton’s laws of motion. So, under the topic of classical mechanics we will discuss the D’Alembert’s principle along with lagrangian and hamiltonian formulations. The generalized version of classical mechanics which is known as quantum mechanics was formulated in the early twentieth century to account for these phenomena at the microscopic level. In this chapter we will try to explore the superiority of lagrangian and hamiltonian formulations over the newtonian formulation in solving mechanical problems.

6.1.1 Limitations of newtonian Mechanics Though the subject of classical mechanics is entirely based on newton’s laws of motion, it often becomes very difficult to solve problems of mechanics by considering the forces acting on a system and the accelerations of the system under these forces. For analysis of the nature of motion of any system one has to proceed through two essential steps: at first one has to write the differential equation of motion of the system by considering the forces which are acting on the system and then one has to solve the said differential equation. But in newtonian mechanics, while using newtonian formulations, one comes across many limitations which are noted below: (a) newtonian mechanics is valid only in inertial frames of reference which are by definition cartesian-like and move with constant velocities. newtonian mechanics cannot be applied in case of non-inertial frames of reference. (b) For writing the differential equation of motion, a particular coordinate system has to be selected by considering the symmetry of the system.

6.2

Advanced Engineering Physics

(c) The newtonian formulation is not invariant under all coordinate systems. While coordinates are transformed from one system to another, the equations of motion change their forms. As an example, when a particle of mass m moves under the influence of an external force Fx, one can write its differential equation in cartesian coordinates as d2x m ___2 = Fx ...(6.1) dt But in the polar coordinates (r, q), the equation of motion of the particle is given by dq 2 d2r m ___2 = mr ___ ...(6.2) dt dt [i.e., mr = mr q 2] d2r which is not in the form m ___ = Fr. Here, the force depends on both the parameters r and q. Such a dt coupled dependency of coordinates is very much inconvenient. This inconvenience can be avoided if one selects a generalized coordinate system without making any reference to a specific coordinate system. (d) The newtonian formulation (while using newton’s laws) require complete specification of all the forces which act on the body whose motion is under consideration at all instants of time. But it has been observed that sometimes the independent form of some forces cannot be known though the effects of them on the concerned system are well known. There is no scope to substitute these effects of the forces in newtonian formulation. Consequently, it becomes difficult to find the solution of the equation of motion in order to know the nature of motion of the particles of the system. (e) Because of the aforesaid reasons, one cannot deal with the constrained motion by making use of the newtonian formulation. (f) In newtonian formulation, the equations of motion involve vector quantities like force, momentum, etc., which introduce complexity in solving the problems. (g) The greatest limitation faced in this formulation is that the mechanical problems are always tried to resolve geometrically rather than analytically. And in case of constraint motion, determination of all the reaction forces is difficult in newtonian formulation. One can easily avoid all these aforementioned difficulties if one uses a new formulation called lagrangian formulation.

( )

6.1.2

Constraints of Motion

The constraints are the obstacles or geometrical restrictions on the motion of a particle or a system of particles. The forces which are responsible for the constraints are called the forces of constraints. When there is a constraint, the particle motion is restricted to occur only along some specified path, or on a surface (plane or curved) arbitrarily oriented in space. Let us mention some examples of constraint motion as given below: (a) The motion of a bead along a horizontally stretched wire. It is restricted to move on a straight path. (b) The motion of the bob of a simple pendulum. It is restricted to oscillate in the vertical plane. (c) The motion of a train on the rails. It is confined along the rails only. (d) The motion of the gas molecules which are confined in a container. The motion is restricted by the walls of the container. (e) The motion of a rigid body is such that the distance between only two particles of the body remains always constant.

6.3

Classical Mechanics

To describe the motion of a body (e.g., a bead) in space we O require three coordinates but to describe its motion on a stretched q string we require only one coordinate. Thus, if we impose constraints l on a mechanical system we can reduce the number of coordinates required to describe its motion. And in this way we can simplify the mathematical description of its motion. The conditions imposed on P the system by the constraints can be written down mathematically as a relation satisfied by the coordinates of the particles of the system Fig. 6.1 Motion of a simple at any time. This is the way in which the constraints imposed on a pendulum on a vertical plane. system reduce the number of coordinates needed to simplify the configuration of a system. Let us consider, as an example, the motion of a simple pendulum which is confined to move in a vertical plane as shown in Fig. 6.1. We need only two coordinates (in cartesian coordinate system x and y; and in polar coordinates system, r and q), to locate the position of the bob at any time with respect to the point of suspension. The constraint of motion of the pendulum is that the distance (l) of the bob P from the point of suspension O remains always constant. The mathematical expression of the constraint when expressed in cartesian coordinates is given by x2 + y2 = l2 ...(6.3) If it is expressed in plane polar coordinates, it is given by r=l ...(6.4) It is seen that the mathematical expression describing the constraint is simpler in case of plane polar coordinates. One coordinate q, in plane polar coordinates, would suffice to describe the constraint motion, whereas in cartesian coordinates, we require two coordinates (namely x and y). So, choice of a suitable coordinate system is also very much important in describing constrained motion. Proper choice of coordinates helps us reduce the number of parameters and make expressions (describing the motion of the restricted system) simpler. If a particle is restricted to move on the surface of a sphere, then we can write the mathematical expression of the restriction in cartesian coordinates as follows: x2 + y2 + z2 = a2 ...(6.5) And in spherical polar coordinates, it can be expressed as r= ...(6.6) where a is the radius of the sphere. In this examO ple we can see that the spherical polar coordinates can be used with a greater advantage. q1 l 1 In the above two examples, we have so far considered a single particle (or body) system. m1 Let us now consider a two-body system. A doul2 q ble pendulum is shown in Fig. 6.2. 2 In this case, we would require four coordinates (two for each pendulum) if we had chom2 sen cartesian coordinates to describe the system completely. But if we choose plane polar coordinates, we require only two coordinates, namely, Fig. 6.2 A double pendulum having two bobs of masses q1 and q2. m and m suspended from a rigid support O. 1

2

6.4

Advanced Engineering Physics

6.1.3

Classification of Constraints

The constraints of motion can be classified on the basis of any parameter of interest. We will classify the constraints in this case on the basis of only two parameters, namely, time and velocity. Depending on time, constraints are classified as scleronomic constraints and rheonomic constraints. (i) Scleronomic constraints These are the constraints which do not explicitly depend on time. (ii) Rheonomic constraints These are the constraints which explicitly depend on time. We can represent scleronomic constraints in an equational form as follows: f (x1, y1, z1; x2, y2, z2 : xn, yn, zn) = 0 Similarly, a rheonomic constraint can be expressed as follows: f (x1, y1, z1; x2, y2, z2 : xn, yn, zn; t) = 0 Depending on velocity the constraints can be classified as holonomic and non-holonomic constraints. (i) Holonomic constraints If the relation of the constraints can be expressed as an equation and if they are independent of velocity then such constraints are called holonomic constraints. When expressed in equational form we get. f (x1, y1, z1; x2, y2, z2 : xn, yn, zn; t) = 0 (ii) Non-holonomic constraints If the constraints cannot be expressed in equational form and they are dependent on velocity then such constraints are called non-holonomic constraints. Examples of various constraints: (a) Scleronomic constraints The motion of a point mass in a simple pendulum with a rigid support. The equation of constraint x2 + y2 = l2 is independent of time. (b) Rheonomic constraints A bead sliding on a moving wire is an example of a rheonomic constraint. If the effective length ‘l’ of a simple pendulum varies with time because of change of temperature, then it becomes an example of rheonomic constraint as l = f (t). (c) Holonomic constraints The constraints involved in the motion of a rigid body is an example of holonomic constraint because in this case (ri – rj)2 – cij2 = 0 where ri and rj are respectively the position vector of the ith and jth particles and cij is the distance between them. (d) Non-holonomic constraints If a gas is kept in a spherical container with radius a and r is the position vector of a gas molecule, then the condition of constraint for motion of a molecule is given by |r| £ a fi r – a £ 0 It is thus an example of a non-holonomic constraint.

6.1.4

degrees of Freedom

The number of independent ways in which a mechanical system can execute its motion without violating any constraint imposed on it is called degrees of freedom of the said system. In other words, the degrees of freedom is the least possible number of coordinates to describe the motion of the system completely. A bead constrained to move along a stretched string has one degree of freedom as it requires only one coordinate for describing its motion.

Classical Mechanics

6.5

Let us consider again the simple pendulum constrained to move on a vertical plane. If we choose the cartesian coordinate system, we require two coordinates (x and y) to describe its motion completely but if we choose the plane polar coordinate system instead, we require only one coordinate (q) and as it is the least number of coordinates for describing its motion completely its degree of freedom is one. We can look at the degrees of freedom from another point of view as stated below: The kinetic energy of a particle when it moves in one-dimensional, two-dimensional and three-dimensional 1 1 1 spaces (i.e., along the x axis, in the xy plane and in the xyz space) is given by __ mx2, __ m (x2 + y2) and __ m (x2 2 2 2 + y2 + z2) respectively. So, the degrees of freedom may also be defined as the number of squared terms appearing in the expression of the kinetic energy of the system. If we impose more constraints on the motion of a body, we get the number of coordinates required to describe its motion reduced. So, imposing constraints is a way of simplifying the problems mathematically in the sense that the number of equations of motion are reduced to the same number as the degrees of freedom. In a system of n particles subjected to k constraints which are expressible in k equations of the form g1 (x1, y1, z1; x2, y2, z2 : xn, yn, zn, t) = a1 g2 (x1, y1, z1; x2, y2, z2 : xn, yn, zn, t) = a2 ... ... ... gk (x1, y1, z1, ; x2, y2, z2, : xn, yn, zn, t) = ak The degrees of freedom nf is given by 3n – k; i.e., nf = 3n – k.

6.1.5 Generalized Coordinates A set of independent coordinates which is sufficient in number to specify unambiguously the configuration of the system is known as the set of generalized coordinates. In other words, the generalized coordinates are such coordinates which are capable of replacing coordinates of any coordinate system like cartesian, spherical polar and cylindrical polar coordinate systems. We sometimes wish to introduce not all coordinates with respect to a fixed coordinate system but some of them may be selected with respect to a new origin or a moving coordinate system. As for example, while dealing with rigid body motion, we specify three cartesian coordinates to locate the center of mass of the rigid body with respect to an external origin and three angular coordinates relative to the origin at the center of mass of the rigid body. Thus, the generalized coordinates should be all chosen as the conventional orthogonal position coordinates or all may be angular coordinates.

6.1.6 Choice of Generalized Coordinates While one chooses generalized coordinates for a system, one must be guided by the following three principles: (a) Their values should be capable of determining the configuration of the concerned system. (b) They should be in a position to vary arbitrarily and independently of each other, without violating the constraints of the system. (c) There is no uniqueness in the choice of generalized coordinates. The choice should fall on a set of coordinates that will be able to give us a reasonable mathematical simplification of the concerned problem.

6.6

6.1.7

Advanced Engineering Physics

notation for Generalized Coordinates

The generalized coordinates are denoted by the letter q of Roman alphabet with a numerical subscript, i.e., a set of n generalized coordinates can be expressed as q1, q2, q3, ....., qn respectively. Alternatively, the set of n generalized coordinates can be expressed by q with a letter subscript to it and specifying within a bracket the numerical values that the letter subscript is allowed to take. e.g., qj (j = 1, 2, 3, ...., n). Thus, when a particle moves in a plane, it may be described by cartesian coordiantes (x, y) or the polar coordinates (r, q), and so on. And we can have the following mapping between the set (q1, q2) of generalized coordinates and any one of the sets (x, y) or (r, q), as follows: q1 = x q1 = r or q2 = y q2 = q ______ y where r = ÷x2 + y2 and q = tan– 1 __x

()

When the problem involves some spherical symmetry, it is convenient to use spherical polar coordinates while transforming generalized coordinates to any specific coordinates. So, we can write, 1 __

and or,

q1 = r = (x2 + y2 + z2 ) 2 z _______ q2 = q = cot– 1 _________ 2 ÷(x + y2) y q3 = f = tan– 1 __x z _______ q1 = q = cot– 1 _________ 2 ÷(x + y2) y q2 = f = tan– 1 __x

() ()

1 __

and q3 = r = (x2 + y2 + z2) 2 If one prefers to accept a coordinate system which is moving uniformly with a velocity v in the x direction, the generalized coordinates are given by (q1, q2, q3) as follows: q1 = x – xt q2 = y where x = v = constant. and q3 = z So, we can see that one of our generalized coordinates (q1, q2, q3) is Y A(x, y) or A(r, q) velocity dependent, i.e., q1 = q1 (v). It is also dependent on time, i.e., q1 = q1 (t) r y \ we can write q1 = q1 (x, v, t) = q1 (x, x, t) q With a bit of analysis we can show that each coordinate of a generX O alized coordinate system can be expressed as a function of all the (or x some of them) coordinates of another system including time and/or velocities. Fig. 6.3 A single particle system Let us now consider a single particle system in a plane. If A be a in XY plane. point in the XY plane as shown in Fig. 6.3,

Classical Mechanics

6.7

then the point A can be represented by A (x, y) in cartesian coordinates and by A (r, q) in polar coordinates. And from the diagram we can write, x = r ______ cos q and y = r sin q So, r = ÷x2 + y2 and q = tan–1(y/x) Hence, x = x (r, q) and y = y (r, q) and r = r (x, y) and q = q (x, y) So, we can show x and y as functions of r and q and vice versa. now, we can, in general, represent the point A in the XY plane by two generalized coordinates q1 and q2 as A(q1, q2). And (q1, q2) may represent either (x, y) or (r, q) or any other coordinate pair belonging to any other coordinate system. Let us now extend our idea by considering a system of two Y particles A and B in the XY plane as shown in Fig. 6.4. If the disB(x2, y2) tance AB between the points A and B be constant (i.e., the points r r2 A and B are constrained) then we can write, A(x1, y1) y2 AB = {(x2 – x1)2 + (y2 – y1)2}1/2 y1 r1 q 2q = constant = r (say) 1 X 0 where (x1, y1) and (x2, y2) are the cartesian coordinates of the x2 points A and B respectively. x1 Similarly, the system of points A and B can be denoted by Fig. 6.4 A system of two particles A and (r1, q1) and (r2, q2) in the polar coordinate system. So, in genB in the XY plane. eral, the two points can be represented by the ordered pairs (q1, q2) and (q3, q4) in the generalized coordinate system. now, we can write the following expressions from Fig. 6.4: r2 = r1 + r fi r = r2 – r1 _

where

r1 = f1 (x1, y1, q1), _

r2 = f2 (x2, y2, q2) _

_

and r = f3 (r1, r2) = f3 (x1, y1, x2, y2, r1, r2, q1, q2) So, the above four equations indicate that any coordinate of one coordinate system can be expressed as a function of the coordinates of any other coordinate system. Hence, by choosing a generalized coordinate system, we can write q1 = q1 (x1, y1, x2, y2; t) and q2 = q2 (x1, y1, x2, y2; t) Also, q1 = q1 (r1, q1, r2, q2; t) and q2 = q2 (r1, q1; r2, q2; t) now, if we consider a system of n particles, then we can write, in general q1 = q1 (x1, y1, z1; x2, y2, z2; ..., xn, yn, zn; t) q2 = q2 (x1, y1, z1; x2, y2, z2; ..., xn, yn, zn; t) ...(6.7) ... q3n = q3n (x1, y1, z1; x2, y2, z2; ..., xn, yn, zn; t) As the system contains n particles, it requires 3n space coordinates as well as one time coordinate. The set of above 3n equations forms a set of 3n transformation equations which transform the coordinates of the

6.8

Advanced Engineering Physics

cartesian coordinate system to generalized coordinates qj (where j = 1, 2, ..., 3n) of the generalized coordinate system. Similarly, the generalized coordinates can be transformed into cartesian coordinates as given below: x1 = x1 (q1, q2, q3, ..., q3n, t) y2 = y2 (q1, q2, q3, ..., q3n, t) ...(6.8) ... ... ... zn = zn (q1, q2, q3, ..., q3n, t) The necessary and sufficient condition for the transformation from a set of coordinates qj (j = 1, 2, 3, ..., 3n) to another set (x1, y1, z1; x2, y2, z2, ..., xn, yn, zn) to be effective is that the value of the Jacobian determinant J will not be equal to zero. q1, q2, ..., q3n ∂ (q1, q2, ..., q3n,) ______________ i.e., J ___________ x1, y1, ..., zn ∫ ∂ (x , y , ..., z )

(

(

)

=

1

∂q1 ___ ∂x2 ∂q ___1 ∂x2 ... ∂q ___1 ∂zn

1

n

)

∂q3n ∂q2 ____ ... ____ ∂x1 ∂x1 ∂q3n ∂q2 ___ ... ____ π0 ∂x2 ∂x2 ... ... ... ∂q3n ∂q ___2 ... ____ ∂zn ∂zn

If this condition fails to hold then Eq. (6.7) does not define a consistent set of generalized coordinates.

6.1.8

notations for Generalized Variables

Let us now develop notations for generalized variables like generalized displacement, generalized velocity, generalized acceleration, generalized momentum, generalized force, etc., in terms of generalized coordinates qj (j = 1, 2, 3......, 3n). Generalized displacement Let d ri represent a small displacement of the i th particle of the system of n particles. ri represents the cartesian coordinates of the i th particle. \ ri = ri (q1, q2, ..., q3n, t) For scleronomic system, we get ri = ri (q1, q2, ..., q3n) Hence,

3n ∂r _ i d ri = S ___ dqj j = 1 ∂qj

...(6.9)

where dqi are called the generalized displacements or virtual arbitrary displacements and d t = 0. Generalized velocity . The generalized velocity associated with the generalized position coordinate qj is denoted by qj. For an unconstrained system of n particles, we get ri = ri (q1, q2 ,..., q3n , t) \

. 3n ∂ri . ∂ri ri = S ___ qj + ___ ∂t j = 1 ∂qj

...(6.10)

6.9

Classical Mechanics

If the n-particle system contains k constraints, the number of generalized coordinates is 3n – k = f and in that case f ∂r . ∂ ri . i ri = S ___ qj + ___ ∂t j = 1 ∂qj

...(6.11)

If the dimensions of a generalized coordinate is that of momentum then the generalized velocity will have ∂ ri the dimension of force, and so on. The term ___ appears in the expression only when there are coordinates ∂t depending on time or in some cases where it is convenient to introduce moving coordinate axes. Generalized accelerations Let us again consider Eq. (6.10) . 3n ∂ ri . ∂ ri ri = S ___ qj + ___ ∂t j = 1 ∂qj now, differentiating it with respect to time, we get d 3n ∂ ri . ∂ ri ri = __ S ___ qj + ___ dt j = 1 ∂qj ∂t

(

)

. ∂ ri . 3n ___ ∂ ri d ___ d ∂ri __ or, ri = S qj + S qj + __ ___ dt dt ∂q ∂q ∂t j=1 j=1 j j . . 3n ∂ r . 3n ∂r ∂ri i i or, ri = S ___ qj + S ___ qj + ___ ∂q ∂q ∂t j=1 j . j=1 j now, putting the value of ri from Eq. (6.10) and changing the index j to k, we get 3n

( )

3n ∂ ri = S ___ j = 1 ∂qj

(S

( )

)

∂ ri . ___ ∂ ri . 3n ___ ∂ ri ∂ ___ qk + qj + S qj + __ ∂t ∂t k = 1 ∂qk j = 1 ∂qj 3n

(S

∂ ri . ___ ∂ ri ___ qk + ∂t k = 1 ∂qk 3n

)

or

3n ∂2r ∂2ri . . 3n _____ ∂2ri . 3n ___ ∂ ri ∂2ri i . ______ _____ ____ q q ri = S S qk qj + S qj + S j+S k+ ∂ti2 k = 1 ∂qk ∂t j = 1k = 1 ∂qj dqk j = 1 ∂qj ∂t j = 1 ∂qj

or,

3n ∂ r 3n 3n ∂ 2r 3n ∂2r ∂2ri . . i i i . ri = S ___ qj + S S ______ qj qk + 2 S _____ qj + ____ ∂t2 j = 1 ∂qj j = 1k = 1 ∂qj ∂qk j = 1 ∂qj ∂t

3n 3n

...(6.12)

now, if we look closely into Eq. (6.12) from the dimensional point of view, we get the generalized accelerations as qj . So, qj are the generalized accelerations. Generalized momentum The kinetic energy T of a system of n free particles in terms of cartesian coordinates is given by n n . . . 1 1 T = S __ mi ri2 = S __ mi (ri . ri) 2 2 i=1 i=1 . now substituting for ri from Eq. (6.10), we get n 3n ∂ r . 3n ∂ r . ∂ ri ∂ ri i i 1 T = S __ mi S ___ qj + ___ . S ___ qk + ___ 2 ∂t ∂t i=1 k = 1 ∂qk j = 1 ∂qj

[

or,

][

n 3n 3n ∂ ri ∂ ri . . 1 n 1 T = __ S S S mi ___ . ___ qj qk + __ S mi 2 i = 1 j = 1 k = 1 ∂qj ∂qk 2 i=1

...(6.13)

]

[

3n

S

j=1

]

( )

n ∂ ri . 3n ___ ∂ ri . ___ ∂ ri __ ∂ ri 1 ___ + S mi ___ qj + S qk . ∂qj ∂t 2 i = 1 ∂t k = 1 ∂qk

2

6.10

Advanced Engineering Physics

The second term on the right-hand side consists of two sums which are identical so that n 3n 3n n 3n n ∂ ri ∂ ri . . ∂ ri . ∂ ri ∂ ri 1 1 T = __ S S S mi ___ . ___ qj qk + S S mi ___ qk . ___ + __ S mi ___ 2 i = 1 j = 1 k = 1 ∂qj ∂qk 2 ∂qk ∂t ∂t i=1 k=1 i=1

( )( )

( )

2

...(6.14)

Thus, the general kinetic energy in terms of generalized velocities comprises three distinct terms—the first term contains quadratic terms of velocities, the second term contains linear terms of velocities and the third term is independent of generalized velocities. If we denote these terms respectively by T (2), T (1) and T (0), we can represent Eq. (6.14) as T = T (2) + T (1) + T (0) ...(6.15) ∂ ri T (1) and T (0) will vanish when ___ = 0, i.e., when our defining Eq. (6.8) of generalized coordinates do ∂t not have explicit dependence on time (i.e., when no moving coordinates are involved). And in such a case the term T(2) survives and contains cross-terms also (i.e., terms qj qk where j π k) unlike the kinetic energy Eq. (6.13) in cartesian coordinates which is purely a quadratic function of linear velocities containing only the squared terms. Such a form is called homogeneous quadratic form or canonical form. If the term T(2) is ∂ ri ∂ ri free from cross-terms which will happen when ___ . ___ = 0 for j π k then the generalized coordinate system ∂qj ∂qk in qj’s is referred as an orthogonal system. n . 1 now, if we consider T = S __ mi ri2 and rewrite T in the following form: i=1 2 n . . . 1 T = S __ mi (x2i + y2i + 2:i) 2 i=1 . . then we can show that the linear momentum (mi xi) associated with the linear velocity xi is given by . ∂T pxi = ___ = mxi ∂xi Then, we can similarly define the generalized momentum associated with generalized coordinate qk as ∂T pk = ___ ...(6.16) ∂ k The generalized momentum pk need not always have the dimension (MLT –1) of linear momentum, for if qk is an angular coordinate, pk is the corresponding angular momentum with dimensions (ML2T –1). . now, differentiating Eq. (6.14) with respect to qk, we obtain n

pk = S

3n

S

i=1 j=1

∂ri ___ ∂ ri 1 ___ __ m . 2 i ∂qj ∂qk

j

n ∂ ri ∂ ri + S mi ___ ◊ ___ ∂qk ∂t i=1

...(6.17)

If the generalized system is stationary then the last term will again be absent. It is again a linear function of generalized velocities. Let us now compute generalized momenta associated with the plane polar coordinates (r, q) as follows: For a single particle system in two dimensions, Eq. (6.13) reduces to . . 1 T = __ m (r. r) 2 . . 1 or, T = __ m (x2 + y2) 2

6.11

Classical Mechanics

and Eq. (6.14) reduces to ∂r ∂r . . 1 2 2 T = __ S S m ___ . ___ qj qk [as the generalized coordinate system is stationary] 2 j = 1 k = 1 ∂qj ∂qk or,

2 ∂r ∂r . . ∂r ∂r . . 1 T = __ m S ___ . ___ qj q1 + ___ . ___ qj q2 2 j = 1 ∂qj ∂q1 ∂qj ∂q2

(

)

∂r ∂r . ∂r ∂r . . ∂r ∂r . . ∂r ∂r . 1 T = __ m ___ ___ q12 + ___ . ___ q1 q2 + ___ ___ q1 q2 + ___. ___ q22 2 ∂q1 ∂q1 ∂q1 ∂q2 ∂q1 ∂q2 ∂q2 ∂q2 We are considering a single particle system in two dimensions in the polar coordinate system. So we have two polar coordinates r and q. We can assume either q1 = r and q2 = q, or q1 = q and q2 = r. In either of the cases, we get ∂r ∂r . ∂r ∂r . . ∂r ∂r . 1 T = __ m ___ . ___ r2 + 2 ___ ___ rq + ___ . ___ q 2 2 ∂r ∂r ∂r ∂q ∂q ∂q ∂y ∂y . ∂y ∂y . . ∂x ∂x ∂x ∂x 1 or, T = __ m ___ + ___ . ___ + ___ r2 + 2 ___ + ___ . ___ + ___ r q 2 ∂r ∂r ∂r ∂r ∂r ∂r ∂q ∂q . ∂y ∂y ∂x ∂x + ___ + ___ . ___ + ___ q 2 ∂q ∂q ∂q ∂q From the relation x = r cos q and y = r sin q, we get ∂y ∂y ∂x ∂x ___ = cos q, ___ = – r sin q, ___ = sin q and ___ = r cos q ∂r ∂q ∂r ∂q Putting the values of these derivatives in the above equation we get the kinetic energy in polar coordinates (r, q) as follows: . . . 1 1 T = __ m (cos2 q + sin2 q) r2 + 2 × __ m (– r cos q sin q + r cos q sin q) r q 2 2 . 1 + __ m (r2 sin2q + r2 cos2q ) q 2 2 . . 1 or T = __ m (r2 + r2q 2) 2 which is a familiar expression. \ the associated generalized momenta are given by Eq. (6.16) as follows: . ∂T pr = ___ = mr (Linear momentum) ∂r . ∂T and pq = ___ = mr2q (Angular momentum) ∂q However, the product of any generalized momentum and the associated (also known as conjugate) coordinate must have a dimension of angular momentum.

(

or,

)

(

(

(

)

)( )(

) (

)(

)

)

Generalized force The definition of a generalized force associated with generalized displacement is given as follows: Let us consider the amount of work done on the system by all forces Fi while causing all arbitrary displacements d ri of the system. We can write

6.12

Advanced Engineering Physics n

d w = S Fi ◊ d ri i=1

or,

n 3n ∂ r i d w = S Fi . S ___ d qj i=1 j = 1 ∂qj

or,

dw = S

n

3n

S

i=1 j=1

( )

∂ ri Fi. ___ d qj ∂qj

3n

d w = S Qj dqj

or,

...(6.18)

j=1 n

∂ ri Qj = S Fi . ___ ...(6.19) ∂qj i=1 Let us note that Qj is dependent on the force acting on the particles and the coordinate qj and possibly on the time t. As Qj dqj gives the dimension of work, it is natural to call Qj as the generalized force associated with a generalized coordinate qj . So, the product of Qj with the arbitrary displacement dqj is equal to the work done corresponding to that displacement. And the generalized force need not always have the dimension of force as d qj will not always have the dimension of distance. If we consider again a particle having coordinates (r, q), then the components of the force acting on the particle is given by __ F = Fr + Fq where and are the unit vectors acting along the directions of increasing r and q respectively. _ _ We can express the vector r as r = r . The generalized forces associated with r– and q –coordinates are respectively given by __ ∂ r ∂r Qr = F . ___ = (Fr + Fq ) . __ ∂r ∂r \ Qr = Fr ...(6.20) where

_

Again, or,

__ ∂ r ∂ Qq = F ◊ ___ = (Fr + Fq ) . r ___ ∂q ∂q Qq = r Fq

( )

[ We have from Fig. 6.5, = cosq + sin q and = – sinq + cosq ∂ ___ = – sinq + cosq = \ ∂q From Eqs (6.20) and (6.21), we find that Qr is the component of the force in the r direction and Qq is the torque acting on the particle to the direction of increasing q. Qr has the dimension of (MLT –2) and Qq has the dimension of (ML2T – 2 ) and rQr and qQq have the dimension of work.

y

q

...(6.21) ∂ ___ = ∂q

]

r

r

j

q

x i z

q ¢ = 90°– q

Fig. 6.5 Unit vector and in polar coordinate system.

6.13

Classical Mechanics

Generalized potential If the forces acting on the system under consideration be derivable from a scalar potential V depending on the position coordinates only then V is the potential energy associated with the system and in such a case, we can write dw = – dV n

dw = – S

or,

i=1

(

∂V ∂V ∂V ___ d xi + ___ dyi + ___ d zi ∂xi ∂yi ∂zi

)

...(6.22)

And in terms of generalized coordinates qj we can write 3n ∂V d w = – d V = – S ___ d qk k = 1 ∂qk 3n

d w = – S Qk d qk

or,

...(6.23)

k=1

∂V Qk = – ____ ...(6.24) ∂ qk In this sense, the definition of Qk as the generalized force is justified. When the system is not conservative and the potential depends on the generalized velocities qj, the generalized force in such a case is defined as Thus, we have

∂U d ∂ U Qj = – ___ + __ ___ ...(6.25) ∂qj dt ∂ qj where U may be called velocity dependent potential or simply generalized potential as the generalized force Qj can be derived from it.

( )

6.2

LaGranGIan ForMuLatIon

In Section 6.1.1 we have discussed in detail the limitations of newtonian mechanics. We observed there that newton’s laws of motion are dependent on the type of reference frames. In non-inertial frames of reference, the newton’s laws of motion are not valid. The equations involved in newtonian mechanics are usually in vector form, so they are inconvenient to handle. To overcome all these difficulties, a set of equations called Lagrange’s equations of motion are used. This set of equations can be written in any system of coordinates.

6.2.1

System point and Configuration Space

The description of the motion of a single particle in three-dimensional space is a simple, mechanical problem. Any problem involving two particles (where each particle is described by a set of three coordinates) can be reduced to a single particle problem simply by regarding that the said single particle moves in a six-dimensional space. Though it is not possible to physically visualize a space having more than three dimensions, one can logically do so and this way one can convert a many-body problem to a single-body problem. Thus, in general, a problem involving n particles can be treated as a problem of a single particle moving along a trajectory of 3n dimensional space. So, we have constructed a 3n dimensional space to give meaning to the concept of the system point. When an n particle problem is thus converted to a single particle problem, the 3n dimensional space is regarded as configuration space and the associated single particle is regarded as the system point. The motion of the system point in the ‘configuration space’ is called the motion of the system between any two given instants of time. As stated above, the configuration space has no necessary connection with a real three-dimensional space.

6.14

6.2.2

Advanced Engineering Physics

principle of Virtual Work

When the sum of all the forces acting on a mechanical system is zero, the system is regarded to be in an equilibrium state. The system can have several such equilibrium states at a given point of time without violating the constraints acting on it. The virtual displacement is such a displacement which takes the system from one equilibrium state to another infinitely close equilibrium state at the same instant of time, i.e., _ it is an infinitesimal displacement when the change of time (dt = 0). The virtual displacement d ri differs _ from the real displacement d ri in the sense that the virtual displacement does not involve any change in time while the real displacement involves a change in time dt. And during this time interval dt the forces and the constraints may change. Let us consider a simple pendulum. The tension in the string is the force of constraint. Let l be the length of the pendulum. When this length (l) remains constant, the bob moves in the arc of a vertical circle having a constant radius l. If a very small displacement of the bob takes place, the force of constraint remains perpendicular to it. And the work done by the force of constraint is zero. On the other hand, for a pendulum whose length changes with time, the bob does not move in the arc of a circle and hence, the displacement is no more perpendicular to the force of constraint. And the work done by the force of constraint in any real displacement is not zero. The virtual displacement at an instant t is a displacement of the bob along the arc of a circle whose radius is equal to the length of the pendulum at the same instant (t). And when the bob makes a virtual displacement, the length l of the pendulum remains constant. This is because the virtual displacement in any instant of time must be consistent with the constraint at that instant. no passage of time takes place during any virtual displacement. It is an infinitesimal imaginary displacement which is normal to the force of constraint. Hence, the virtual work done by the force of constraint is zero. This is the principle of virtual work. This principle can be stated as follows: A system of particles, will be in a state of equilibrium if and only if the total virtual work done by the forces of constraint in any arbitrary infinitesimal virtual displacement is zero.

6.2.3

d’alembert’s principle

This principle is based on the principle of virtual work. The system is subjected to an infinitesimal displacement consistent with the forces of constraints imposed on the system of particles at a given instant of time t. This change in the configuration of the system is not associated with a change in time, i.e., there is no actual displacement during which the forces of constraint may change. __ now, let us suppose that the system is in an equilibrium state, i.e., the total force Fi acting on each particle is zero; then the work done by this force in a small virtual displacement d ri will be zero. That is, for the whole system of n particles, we can have n

S i=1

__

__

Fi .d ri = 0 __

Let this total force Fi acting on the ith particle be expressed as the sum of the applied force Fia and the force __ of constraint Fic. Then the above equation takes the following form: n

S i=1

__

n

__

(Fia . d ri) + S (Fic . d ri) = 0 i=1

now, let us consider a system of n particles for which the virtual work done by the forces of constraints becomes zero. An example of such a system can be that of a group of n particles which are constraint to move on a smooth horizontal surface so that the forces of constraints on them are normal to the surface while the virtual displacement will be tangential to the surface. In such a situation, the virtual work done by the forces of constraints will be equal to zero.

6.15

Classical Mechanics

Thus, the above equation becomes, n

S i=1

__

Fia . d ri = 0

...(6.26)

This equation gives us the mathematical expression for the principle of virtual work. In order to interpret the equilibrium of the system, D’ Alembert adopted the idea of a ‘reversed force’. He assumed that a system will be in equilibrium under the action of a force which is equal to the sum of the actual __ . force Fi and the reversed effective force pi. Thus, we can have __ __ . . . Fi + (– p) = 0 [ F = ma = mv = p] __ . or, Fi – pi = 0 . where – pi appears as an effective force called the reversed effective force of inertia on the ith particle. And for the whole system of n particles we can write n __ . ...(6.27) S (Fi – p) . d ri = 0 i=1

__

__

__

Again writing Fi as the sum of applied force Fia and force of constraint Fic, we get __ __ __ Fi = Fia + Fic __ Putting this expression of Fi in Eq. (6.27), we get n n __ __ . S (Fia – pi) . d ri + S Fc. d ri = 0 i=1

...(6.28)

i=1

Since we are considering a system for which the virtual work done by the forces of constraint in zero, Eq. (6.28) reduces to n __ . ...(6.29) S (Fia – pi) . d ri = 0 i=1

Equation (6.29) is the principle of virtual work in this case and it is known as D’Alembert’s principle. The key point of this equation (i.e., Eq. (6.29)) is that we have got rid of the forces of constraints. In order to satisfy Eq. (6.29), we cannot equate separately the coefficient of d ri to zero since d ri’s are not independent of each other. And hence it becomes necessary to transform d ri’s into corresponding set of displacements of generalized coordinates, d qj which are independent of each other. The coefficient of each d qj will be equated to zero. The D’Alembert’s principle does not involve with it the forces of constraints. Anyway it is now sufficient to specify all the applied forces only. Further, it is valid for all scleronomic and . rheonomic systems that are either holonomic or homogenous non-holonomic. The inertial force – p is introduced in order to reduce the dynamical problem to an equivalent statical problem. This inertial force can be considered __ .as an inertial force arising in an accelerated frame of reference, e.g., in a freely falling accelerated frame Fi – pi = 0 and hence gravity is nullified by the action of the inertial force even though the whole system is in motion and not in either static or dynamic equilibrium.

6.2.4

Lagrange’s equations from d’alembert’s principle

We have discussed about coordinate transformation in Section 6.1.7. The coordinate transformation equations in vector form can be written as ri = ri (q1, q2, q3,..., qf , t) f being degree of freedom and f = 3n – k, k being the number of constraints and all q’s are independent of each other. Taking the derivative of ri with respect to time, we get

6.16

Advanced Engineering Physics

d ri ___ ∂ ri ___ ∂ ri ___ ∂ ri dqf ∂ ri dt dq1 ___ ∂q2 ___ = + + º +___ ___ + ___ __ dt ∂q1 dt ∂q2 dt ∂qf dt ∂t dt f ∂ ri ∂ ri ...(6.30) vi = S ___ q◊ j + ___ ∂t j = 1 ∂qj Again, we can connect the infinitesimal displacement d ri with d qj as follows: f ∂r ∂ ri i d ri = S ___ d qi + ___ d t ∂t j = 1 ∂qj But in this equation the last term on the right-hand side is zero since in virtual displacement, only the coordinate displacement is considered but not that of time. So the above equation of d ri will reduce to f ∂r i d ri = S ___ d qj ...(6.31) ∂q j=1 j We can now write the D’Alembert’s principle (i.e., Eq. (6.29) ) as follows and modify it with the help of Eq. (6.31): n __ . S (Fia – pi). d ri = 0 ...(6.29)

or,

i=1

n

S

or,

i=1

f ∂r . i (Fi – pi) ◊ S ___ ∂qj = 0 ∂q j=1 j

__

a

f n ∂ ri . ∂ ri Fia ◊ ___ d qi – S S pi ◊ ___ d qj = 0 ...(6.32) ∂q ∂qj i=1 j=1 i=1 j=1 j n __ ∂ ri Let us now write = Qj S Fia ◊ ___ ∂qj i=1 where Qj’s are the components of the generalized force as we have defined it in Section 6.1.8. Qj need not have the dimensions of force but the product (Qj d qj) must have the dimensions of force. Thus, we can now write Eq. (6.32) as given below: n f f . ∂ ri dq = 0 ...(6.33) S Qj dqj – S S pi . ___ ∂qj i i=1 j=1 j=1 The second term of Eq. (6.33) can be evaluated as follows: n

f

S S

or,

n

f

S S i=1 j=1 n

f

S S

or,

i=1 j=1 n

f

S S

or,

i=1 j=1

Again

__

( ) S S { ( ) S S { ( )

n f . ∂ ri ∂ ri pi ◊ ___ dqj = S S mi ri . ___ d qj ∂ qj ∂qj i=1 j=1

( )} ( )}

f . ∂ ri . ∂ ri . d ∂ ri d __ pi ◊ ___ dqj = mi ri . ___ – mi ri . __ ___ dt ∂ qj ∂qj ∂ qj i = 1 j = 1 dt n

n f . ∂ ri ∂ ri d ∂ ri d __ pi ◊ ___ dqj = mvi . ___ – mi vi ◊ __ ___ dt ∂qj ∂qj ∂qj i = 1 j = 1 dt

( ) ( )

∂ ri ∂ dri d ___ __ = ___ ___ dt ∂qj ∂qj dt

d qj

d qj

...(6.34)

6.17

Classical Mechanics

( )

∂ ri ∂ d ___ __ = ___ (vi) dt ∂qj ∂qj

or,

...(6.35)

Let us now differentiate Eq. (6.30) with respect to ∂ ∂ f ∂ ri . ∂ ri ___ (vi) = ___ S ___ qj + ___ ∂ j ∂ j j = 1 ∂qj ∂t

( (

j

to obtain,

)

_

∂ ri . ∂ ri . ∂ ri ∂ ∂ ∂ ri . ___ (v ) = ___ ___ q + ___ q + ... + ___ qj + ... + ___ ∂ i i ∂ j ∂q1 1 ∂q2 2 ∂qj ∂ qf

or,

f

∂___ vi ___ ∂ ri = ∂ i ∂ qi

or,

∂ ri + ___ ∂t

) ...(6.36)

Let us now modify Eq. (6.33) with the help of Eqs. (6.34), (6.35) and (6.36) n f f . ∂ ri d q = 0 ...(6.33) S Qj d qj – S S pi . ___ ∂qj j i=1 j=1 j=1 n

f . ∂ ri pi . ___ d qj = S Qj d qj ∂qj j=1

f

S S

or,

i=1 j=1 f

n

S S

or,

i=1 j=1

f

n

S S

or,

i=1 j=1

n

{ ( ) ( )} { ( ) ( )} {( ) } ∂ ri d d ∂ ri __ mi vi . ___ – mi vi . __ ___ dt dt ∂qj ∂qj

∂ vi d d ∂ ri __ m v . ___ – mi vi . __ ___ dt i i ∂ j dt ∂qj f

i=1 j=1

f

or,

[

n

S S j=1

i=1 f

j=1

(

[{

)

∂ d ___ __ dt ∂ j f

S

or,

j=1

1 where Ti = __ mi vi2 2

n

S i=1

i=1

(

}

d qi = S Qj d qj j=1

[by using Eq. (6.36)]

]

f

[by using Eq. (6.35)] f

) dq = S Q dq

n

j

j

j

j=1

]

n

n

[ ( ) ( )] ∂ d ___ __ dt ∂ j f

j=1

where T = S Ti

[by using Eq. (6.34)] f

f ∂ 1 1 __ mi vi2 – ___ S __ mi vi2 dqj = S Qj d qj 2 ∂qj i = 1 2 j=1

S

or, n

n

∂ __ ∂ 1 d ___ 1 __ m v 2 – S ___ __ mi vi2 dt ∂ j 2 i i ∂q i=1 j 2

S

or,

j=1

∂ vi ∂ vi d __ m v ◊ ___ – mi vi ◊ ___ dqj = S Qj d qj dt i i ∂ i ∂qi j=1

S S

or,

f

dqj = S Qj dqj

S

i=1

[

∂ Ti – ___ ∂qj

S

i=1

Ti

]

f

d qj = S Qj d qj j=1

∂ ∂ d ___ __ (T) – ___ (T) d qj = S Qj d qj dt ∂ j ∂qj j

6.18

Advanced Engineering Physics f

S

or,

j=1

[( )

]

∂T ∂T d ___ __ – ___ – Qj d qj = 0 dt ∂ j ∂ j

...(6.37)

Since the constraints are holonomic, qj are independent of one another and hence to satisfy Eq. (6.37), the coefficient of each d qj should separately be zero, i.e., ∂T ∂ T ___ d ___ __ – – Qj = 0 dt ∂ j ∂qj ∂ T ___ ∂T d ___ __ – = Qj dt ∂ j ∂ qj

or,

...(6.38)

As j ranges from 1 to f (where f represents the degrees of freedom), there will be f such second-order differential equations of motion of the system of n particles. Let us now find the values of Qj in case of conservative and non-conservative force fields. (i) Conservative system now, if the system is a conservative one, Fi or Qj will be derivative of the potential function V, which is again a function of the coordinates only. That is, in this case V = V(r). In such a case, we can write __ ∂V Fi = –—i V = – ___ ∂ri where is a unit vector normal to the equipotential surface. The generalized force can be expressed as follows: n n __ ∂ri ∂ ri Qj = S Fi ◊ ___ = – S —i V ◊ ___ ∂qj ∂qj i=1 i=1 n

Qj = – S

or,

i=1

or,

( )( ) ∂V ___ ∂ri

∂ ri ___ . ∂qj

∂V Qj = – ___ ∂qj now, from Eqs (6.38) and (6.39), we can have

...(6.39)

( )

∂T ∂T ∂V d ___ __ – ___ = – ___ dt ∂ j ∂qj ∂qj or,

∂ T ___ ∂ d ___ __ – (T– V) = 0 dt ∂ j ∂qj

or,

∂ ∂ d ___ __ (T – V) – ___ (T – V) = 0 dt ∂ j ∂qj

(

)

[since V is independent of j]

now writing T – V = L, where L is called the lagrangian for the conservative system, the set of above equations becomes ∂L ___ ∂L d ___ __ – = 0, j = 1, 2, 3 ... f ...(6.40) dt ∂ j ∂qj Each equation of the set of Eq. (6.40) is known as Lagrange’s equation of motion for a conservative system.

Classical Mechanics

6.19

(ii) Non-conservative system If the potentials are velocity dependent (called generalized potential) then though the system is not a conservative one, yet the above mentioned form of Lagrange’s equations can be retained provided Qj , the generalized forces, are derived from a function U (qj, j), called the generalized or velocity-dependent potential, such that we have, ∂U d ∂U Qj = – ___ + __ ___ ...(6.41) ∂qj dt ∂ j

( )

now, substituting Qj by this value of it in Eq. (6.38), we can write

( )

( )

∂T ∂T ∂U d ∂U d ___ __ – ___ = – ___ + __ ___ dt ∂ j ∂ qj dt ∂ j ∂ qj

(

)

∂ ∂ d ___ __ (T – U) – ___ (T – U) = 0 dt ∂ j ∂qj

or,

now, if we put lagrangian L = T – U then the above equation becomes ∂L ___ ∂L d ___ __ – = 0; j = 1, 2, 3, ..., f dt ∂ j ∂qj which is exactly of the same form of Eq. (6.40). Some characteristics of Lagrange’s equations Some characteristics of Lagrange’s equations are noted below: (a) Lagrange’s equations of motion for a system of particles are derived by considering the energy of the system instead of considering the forces. (b) While deriving Lagrange’s equations the effects of the forces of constraint are eliminated. (c) As the lagrangian L (= T – V) is a scalar quantity, Lagrange’s equations are in scalar form. But newton’s equations can appear in vectorial form which are difficult to handle. (d) Lagrange’s equations are independent of coordinate systems. (e) Since all Lagrange’s equations are second-order differential equations, their solutions contain two arbitrary constants and all equations are of the same form.

6.2.5 Cyclic or Ignorable Coordinates The lagrangian as described in the previous section is a function of the generalized coordinates qj ; generalized velocities j and time t. now, if the lagrangian of a system of particles does not contain a particular coor∂L dinate qk , then obviously in case of such a system of particles; ___ = 0. Such a coordinate (qk) which does not ∂qk appear in the expression of the lagrangian is called an ignorable (or a cyclic) coordinate.

6.2.6

Conservation of the Linear Momentum in Case of a Cyclic Coordinate

For a conservative system of particles, the equations of Lagrange are ∂L ∂L d ___ __ – ___ = 0, j = 1, 2, ..., f dt ∂ j ∂qj

( )

6.20

Advanced Engineering Physics

where the lagrangian L = T – V. Let us now assume that the qk coordinate is cyclic, i.e., it is missing in the expression of L. Then for this d ∂L coordinate, the Lagrange’s equation reduces to __ ___ = 0 dt ∂ j ∂L ___ or, = constant ...(6.42) ∂ j

( )

now, we can write ∂L ___ ∂ ∂T ___ = (T – V) = ___ ∂ j ∂ j ∂ j

[ V π V ( j)]

∂L ___ = pj ∂ j

or,

...(6.43)

now, considering Eqs (6.42) and (6.43), we can write pj = constant i.e., the linear momentum corresponding to a cyclic or ignorable coordinate qk is conserved.

6.3

haMILtonIan ForMuLatIon

In the previous section, we have derived Lagrange’s equations of motion for a system of particles. These equations are a set of second-order, first-degree differential equations. In this section, we shall discuss an alternative set of 2f first-order, first-degree differential equations known as Hamilton’s equations of motion. Lagrange’s equations are derived on the basis of the lagrangian L. Similarly, the Hamilton’s equations of motion will be derived on the basis of another function H which is known as the hamiltonian. In case of lagrangian formulations, the independent variables are the generalized coordinates as well as time. Though the generalized velocities also appear explicitly in this formulation, but they are simply time derivatives of the generalized coordinates. So they are treated ultimately as some dependent variables. This fact is reflected in the definition of configuration space where a single particle supplies only three initial values namely the three position coordinates. In the hamiltonian approach, the aforesaid dependency of velocities are eliminated by introducing a new independent variable—known as the generalized momentum pj. In this approach, we shall use the hamiltonian H instead of the lagrangian. This hamiltonian is a function of generalized coordinates, generalized momenta as well as time, i.e., H (qj, pj, t) while the lagrangian is L(qj, j, t). So there is a change of basis from the set (qj, j, t) to the set (qj, pj, t).

6.3.1

phase Space

While defining configuration space, we considered three space coordinates of real three-dimensional space for each particle along with time t. So in a system of n particles without any constraint, we had 3n space coordinates in the corresponding configuration space. But having provided equal status to ‘coordinates’ and ‘momenta’, the configuration space no more remains sufficient to represent the history of the system of particles. So, the path adopted by the system of particles during its motion must now be represented by a space of 6n coordinates in which three coordinates will represent the three position coordinates of real three-dimensional space and three momenta of the same space for a single particle. So, the new space will have 6n dimensional space called phase space. While there will be k constraints the configuration space will have f dimensions and phase space will have 2f dimensions where f is the degree of freedom and is given by f = 3n – k.

Classical Mechanics

6.3.2

6.21

hamilton’s equations of Motion

The hamiltonian H is a function of generalized coordinates, generalized momenta and time, i.e., H = H (qj, pj, t). And the generalized mometa pj are given by ∂L pj = ___ ∂ j which show that for each generalized coordinate qj there is one corresponding generalized momentum pj. The mechanical state of the system under consideration can thus be described completely, provided the variables qj and pj are given as functions of time. The hamiltonian H is defined by the following equation: 2f

H = S pj

j

–L

...(6.44)

j=1

H is a constant of motion with the condition that the lagrangian L does not depend on time explicitly. We can express the hamiltonian H as H = S pj j – L(qj, j) j

\

H = H(qj, pj) As in hamiltonian formulation we provide generalized momenta an independent status, having them placed on equal footing along with the generalized coordinates, the hamiltonian is regarded, in general, as a function of the position coordinates qj, the momenta pj and time t, i.e., H = H (qj, pj, t) Hence, in differential form we can write 2f 2f ∂H ∂H ∂H ...(6.45) dH = S ___ dqj + S ___ dpj + ___ dt ∂q ∂p ∂t j=1 j=1 j j Again, H is defined as 2f

H = S pj j=1 2f

Hence,

j

–L 2f

dH = S

dpj + S pj d

j

i=1

j=1

j

– dL

...(6.46)

But the lagrangian is given by L = L (qj, j, t) Hence,

2f 2f ∂L ∂L ∂L dL = S ___ dqj + S ___ d j + ___ dt ∂q ∂t j=1 j=1 ∂ j j

...(6.47)

now putting the value of dL from the Eq. (6.47) in Eq. (6.46), we have 2f

dH = S

2f 2f 2f ∂L ∂L ∂L dpj + S pj d j – S ___ dqj – S ___ d j – ___ dt ∂q ∂t j=1 j=1 j=1 ∂ j j

j

j=1

∂L ∂L now recognizing, ___ = pj and ___ = ∂qj ∂ j we have 2f

dH = S j=1

j and

putting these in Eq. (6.48)

2f

j

2f

dpj + S pj d j – S j=1

j=1

j

2f ∂L dqj – S pj d j – ___ dt ∂t j=1

...(6.48)

6.22

Advanced Engineering Physics 2f

or,

2f

∂L ...(6.49) dqj – ___ dt ∂t now, comparing the coefficients in Eqs (6.49) and (6.45), we arrive at the following set of equations: dH = S

j

j=1

j

∂H = ___ ∂pj

j

∂H = – ___ ∂qj

dpj – S

j

j=1

...(6.50)

∂L ∂H ...(6.51) – ___ = ___ ∂t ∂t The set of Eqs. (6.50) are known as the Hamilton’s canonical equations of motion or simply Hamilton’s equations of motion. They make a set of 2f first-order, first-degree differential equations of motion. On integration of these 2f differential equations, we get 2f constants of integration which can be evaluated in terms of the initial conditions of the system of n particles.

6.3.3

a Coordinate Cyclic in hamiltonian

A coordinate which is cyclic in the lagrangian is also cyclic in the hamiltonian. Let us assume that qj does not appear in the lagrangian L. Then we can write ∂H ∂L ___ =0 L π L (qi) and pj = ___ j =– ∂qj ∂qj

[

]

So, we have pj = constant. But if pj is constant, then from Eq. (6.50), we can have ∂H ___ =0 j =– ∂qj ∂H ___ or, = 0 fi H π H (qj) ∂qj i.e., qj does not appear in H. So, qj is cyclic in the hamiltonian. If a position coordinate is cyclic in H, it would reduce the number of variables by two in the hamiltonian formulation.

6.3.4

physical Significance of the hamiltonian

(a) The lagrangian L possesses the dimensions of energy. Similarly, the hamiltonian H also possesses the dimensions of energy. (b) In all circumstances, it is not equal to the total energy E of the system of particles. Only in a special case the hamiltonian is equal to the total energy, i.e., when the system is conservative and the coordinate transformation equations are independent of time; then H = E. (c) In all other cases except the case (b), H is a constant of motion but not equal to the total energy of the system of particles. (d) When L is not an explicit function of t, H is also not an explicit function of t. Let us first write, H = H (qj, pj, t) Hence, by differentiating H with respect to t, we get

Classical Mechanics

∂H dH 2f ___ ___ =S dt j = 1 ∂qj

j

2f ∂H + S ___ j = 1 ∂pj

From Eq. (6.50), we have ∂H ___ = – j and ∂qj Therefore, we can write 2f dH ___ =–S dt j=1

j

∂H ___ = ∂pj

j

+S j=1

j

∂H + ___ ∂t

j

2f

j

6.23

j

∂H + ___ ∂t

∂H dH ___ ___ = dt ∂t ∂L dH ___ = – ___ or, [by Eq. (6.51)] dt ∂t now, if L is not a function of t, then ∂L dH ___ = 0, so we get ___ = 0 dt ∂t This means that t will also not appear in the expression of the hamiltonian H, i.e, t must be cyclic in the hamiltonian for it to be a constant of motion. Our interest is more in those systems for which the hamiltonian H is both a constant of motion and equal to the total energy E of the system of particles. (e) The lagrangian formulation can be suitably used in the macroscopic world only (i.e., in classical mechanics) but the hamiltonian formulation can be used in both the microscopic as well as the macroscopic world, i.e., in quantum as well as classical mechanics). or,

6.3.5

use of Lagrangian and hamiltonian Methods

In order to solve any mechanical problem by either lagrangian or hamiltonian formulation, one has to construct the lagrangian function L first. It is expressed in terms of the generalized coordinates and generalized velocities. In order to construct the lagrangian function one needs to express the kinetic energy and the potential energy as functions of these generalized quantities. Then the hamiltonian function can be derived from the expression of the lagrangian function and then the hamiltonian is constructed in terms of generalized coordinates and generalized momenta. In the following section named Worked Out Problems we shall show how to solve mechanical problems by using lagrangian and hamiltonian formulations.

Worked Out Problems Example 6.1

Sol.

(a) (b) (c) (d) (a)

Find the degrees of freedom of the following systems: The bob of a simple pendulum oscillating in a vertical plane. Two particles, connected by a rigid rod moving freely in a plane. Three particles connected by a rigid rod moving freely in a plane. A particle moving on the circumference of a circle. [WBUT 2006] A simple pendulum is shown in Fig. 6.1W. The motion of bob of this simple pendulum can be described by the angular coordinate q. The radial coordinate r = l (a constant). This is the equa-

6.24

Advanced Engineering Physics

tion of constraint and l is the effective length of the pendulum. Hence, the simple pendulum (i.e., the bob of it) has one degree q l of freedom. l (b) In this case, the motion is confined in a plane. So the two particles require (x1 y1) and (x2, y2), i.e., 4 coordinates. There is only one constraint in the present case which is (x2 – x1)2 + (y2 – y1)2 Fig. 6.1W Simple = l2 pendulum. where l is the distance between the particles. So the number of degrees of freedom is given by f =2×2–1=3 (c) The given three particles may be connected in two different ways: (i) Linear arrangement of particles In this case, the distance between particles 1 and 2, and 2 and 3 remain constant. Hence the number of constraints is given by k = 2. \ the degrees of freedom f is given by f=3×2–2=4 (ii) Triangular arrangement of particles In this case, the number of constraints k = 3 as there are three equations representing three sides of the triangle. So, the required number of degrees of freedom is given by f =3×2–3=3 (d) The locus of the particle is a circle. So, the equation of the constraint is given by x2 + y2 = a2 or r = a where a is the radius of the circle. In cartesian coordinates, one of the two variables i.e., either x or y is sufficient to find the position of the particle. And in polar coordinates, one variable q is sufficient to locate the position of the particle. \ the degrees of freedom is given by f =2–1=1 Example 6.2 A mass m is tied with a horizontal spring as shown in Fig. 6.2W. It is able to move to and fro on a frictionless plane in one line. If the spring is pulled and released, the mass executes simple harmonic motion. Find its time period by using lagrangian method. Sol. In this case, force F = – kx, where k is the spring constant. So, the kinetic energy (T) and potential energy (V) of the mass m is given by x

1 1 T = __ mx2 and V = – Ú F dx = __ kx2 2 2 0 \ the lagrangian

L=T–V

or,

1 1 L = __ mx2 – __ kx2 2 2

m k

x

Fig. 6.2W

Spring and mass system.

…(1)

6.25

Classical Mechanics

now, the Lagrange’s equation of motion for a simple harmonic oscillator is given by ∂L ___ ∂L d ___ __ =0 – dt ∂x ∂x now, differentiating Eq. (1) with respect to x and x, we get

( )

…(2)

∂L ∂L ___ = mx and ___ = – kx ∂x ∂x Putting these, in Eq. (2) we get d __ (mx) – (– kx) = 0 dt or,

mx + kx = 0

or,

k x + __ mx=0

or,

x + w2 x = 0

__

÷

k where w = __ m

\ the time period of oscillation of the spring-mass system is given by __

2p m Tp = ___ = 2p __ w k

÷

Example 6.3

Obtain the equation of motion of a simple pendulum form its lagrangian representation. Hence deduce the expression for its time period for small amplitude motion. [WBUT 2006] Sol. The simple pendulum in oscillation has been shown in Fig. 6.3W. Let at any instant of time during its oscillation, the string OB make an angle q with its equilibrium position OA. The kinetic energy of the bob is given by O 1 1 1 T = __ mv2 = __ m (lq)2 = __ ml2 q 2 2 2 2

q

l

where l is the length of the string and v, the velocity of the bob is given by v = lq. C B While passing from B to A the bob falls freely through a vertical distance A CA. The potential energy of the bob of the simple pendulum is thus, given Fig. 6.3W A simple by pendulum. V = mg (OA – OC) = mg (l – l cosq) or, V = mgl (l – cosq) We have considered the horizontal level passing through A as reference level. \ the lagrangian L is given by 1 L = T – V = __ ml2 q2 – mgl (1 – cos q) 2 The Lagrange’s equation of motion is

...(1)

6.26

Advanced Engineering Physics

∂L ___ ∂L d ___ __ = 0 [ q is the only generalized coordinate] – dt ∂q ∂q From Eq. (1), we get

( )

...(2)

∂L ∂L ___ = ml2q and____ = – mgl sin q ∂q ∂q ∂L ∂L Putting these values of ___ and ___ in Eq. (2), we get ∂q ∂q d 2 __ (ml q) + mgl sin q = 0 dt g or, q + __ sin q = 0 l

...(3)

This is the required differential equation of motion of the simple pendulum. now, for small oscillation of the bob, we can write, sin q ª q \ Eq. (3) reduces to g q + __ q = 0 l q + w2 q = 0

or,

...(4)

__

where

g w = __ l

÷

So, Eq. (4) represents simple harmonic motion whose time period Tp is given by __

÷

2p l Tp = ___ = 2p __ g w Example 6.4 Derive the equations of motion of a bead which is sliding on a uniformly rotating wire in a horizontal space. Sol. Let P be the position of a bead [Fig. 6.4W] which is sliding along the wire OQ. The wire OQ is rotating with a uniform angular velocity w. Let the angular displacement q = 0 at t = 0, so, we can write q = wt This equation helps us to determine the angular position q at any time t. To locate the bead in the XY plane we require the variable r (radius vector). The transformation equations relating the cartesian and plane polar coordinates are given by

y

P (x, y) r q

O

x = r cos q = r cos w t and y = r sin q = r sin w t Differentiating these equations with respect to t, we get . x = r cos (w t) – r w sin (w t)

Q w

x

Fig. 6.4W A bead is sliding on a wire which is rotating on a horizontal plane.

6.27

Classical Mechanics

. and = r sin (w t) + rw cos (w t) Hence, the kinetic energy of the bead is given by

or,

1 T = __ m (x2 + 2) 2 . 1 __ T = m (r2 + r2w2) 2

As the plane of rotation is horizontal, the potential energy is constant anywhere in the plane. Let it be V (= constant). Then, the lagrangian L is given by

or,

. 1 L = T – V = __ m (r2 + r2 w2) – V 2 . 1 L = __ m (r2 + r2 q2) – V [ w = q] 2

...(1)

The Lagrange’s equation for r and q are given by ∂L ___ ∂L d ___ __ =0 – dt ∂r ∂r

( )

...(2)

∂L ___ ∂L d ___ __ =0 – dt ∂q ∂q now, differentiating Eq. (1), we get and

( )

...(3)

∂L ∂L ___ = mr, ___ = mr2q, ∂r ∂q ∂L ∂L 2 ___ = mrq and ___ = 0 ∂r ∂q Putting these value in Eqs. (2) and (3), we get mr – mrq2 = 0 mr2q = 0 Equation (4) can be rewritten as r = rq2 = rw2 Equation (6) is an expression which indicates that the bead moves under the action of the centrifugal force mw2r directed outwards along the wire. Example 6.5 Derive the equation of motion of a system of two masses connected by an inextensible string passing over a small smooth pulley (called Atwood’s machine). Sol. Figure 6.5W shows the principle of Atwood’s machine consisting of two masses m1 and m2 suspended over a frictionless pulley P of radius ‘a’ and connected by an inextensible string of length l. Let Q1A = x, and Q2B = x1 From the diagram, we can write l = x + x1 + pa [ a is radius]

...(4) ...(5) ...(6)

P a

Q1

Q2

Q x1

x

m2 B

A m1

Fig. 6.5W Atwood’s machine

6.28

Advanced Engineering Physics

\ x1 = l – pa – x \ x1 = – x and x 1 = – x now, the kinetic energy of the mass m1 is 1 T1 = __ m1x2 2 and the kinetic energy of m2 is 1 1 T2 = __ m2x21 = __ m2x2 2 2 \ the total kinetic energy of the system is given by

or,

1 1 T = T1 + T2 = __ m1x2 + __ m2x2 2 2 1 2 __ T = (m1 + m2) x 2

With reference to the horizontal line passing through Q, the potential energy of m1 is given by V1 = – m1gx and that of m2 is V2 = – m2g (l – pa – x) \ total potential energy is given by V = V1 + V2 = – m1gx – m2g (l – pa – x) The lagrangian of the system is given by 1 L = T – V = __ (m1 + m2) x2 + m1 gx + m2g (l – pa – x) 2 ∂L ___ = (m + m ) x \ 1 2 ∂x ∂L ___ = (m1 – m2) g and ∂x Then Lagrange’s equation of motion is given by

...(1) ...(2) ...(3)

∂L ___ ∂L d ___ __ =0 – dt ∂x ∂x (m1 + m2) x – (m1 – m2) g = 0

( )

or, or,

m1 – m2 x = _______ g m1 + m2

(

)

...(4)

Equation (4) is the required equation of motion of the system. Example 6.6 Sol.

Find the equation of motion of a free particle. If a particle is not acted upon by any external force, the particle is called a free particle. So the potential energy of this particle V = 0. Hence, the lagrangian is given by 1 L = T – V = __ m (x2 + 2 + z2) 2 where x, y and z are cartesian coordinates.

6.29

Classical Mechanics

The coordinates (x, y, z) are cyclic, ∂L ___ ∂L ∂L ___ ___ = = =0 ∂x ∂y ∂z So, the Lagrange’s equations of motion are:

\

∂L ___ ∂L d ___ __ = 0 fi mx = 0 – dt ∂x ∂x ∂L ___ ∂L d ___ __ = 0 fi my = 0 – dt ∂ ∂y

( )

( )

∂L ___ ∂L d ___ __ = 0 fi mz = 0 – dt ∂z ∂z

( )

and Example 6.7 Sol.

Derive the lagrangian and Lagrange’s equation of a freely falling body. Let us assume that a body of mass m is falling freely under gravitational force. The point P is at a height h. A body of mass m is falling from the point P. Let it reach the point Q at time t [Fig. 6.6W]. \ the potential energy of the body is given by V = mg (h – x) P and kinetic energy is given by

( )

dx 1 T = __ m ___ 2 dt

2

x

1 = __ mx2 2

h

Q

\ its lagrangian is given by

h–x

1 L = T – V = __ mx2 – mg (h – x) 2 As there is only one variable x, the Lagrange’s equation is given by ∂L ___ ∂L d ___ __ =0 – dt ∂x ∂x now, differentiating the lagrangian, we get

( )

∂L ∂L ___ = mx and ___ = mg ∂x ∂x Putting these values in Eq. (1), we get d __ (mx) – mg = 0 dt or, x–g=0 \ Eq. (2) is the required equation.

R

Fig. 6.6W

A freely falling body.

...(1)

...(2)

Example 6.8 Find equation of motion of a bead which is sliding on a wire which is rotating on a vertical plane with uniform angular velocity. Sol. Let us assume that XY is a vertical plane where the y axis is along the vertical direction. The gravitational force acting on the bead is acting along the negative y axis [Fig. 6.7W]. The potential energy of the bead is given by

6.30

Advanced Engineering Physics

V = mgy = mgr sin q Let the angular displacement q = 0 at t = 0. So, we can write, q = wt ...(1) This equation helps us to find q at any instant of time. To locate the bead in the XY vertical plane, we require the variable r (the radius vector). The transformation equations are: w x = r cosq = r cos w t y A and y = r sin q = r sin w t now, differentiating these two equations with respect P (x, y) to t, we get . x = r cos w t – rw sin w t r . and = r sin w t + rw cos w t q x O Hence, the kinetic energy of the bead is given by

or,

1 T = __ m (x2 + 2) 2 . 1 T = __ m (r2 + r2 w2) 2

Fig. 6.7W A bead sliding over a wire which is rotating uniformly on a vertical plane.

So, the lagrangian is given by

or,

. 1 L = T – V = __ m (r2 + r2 w2) – mg r sinq 2 . 1 L = __ m (r2 + r2w2) – mgr sin w t 2

\ the Lagrange’s equation for r is given by

or,

∂L ___ ∂L d ___ __ =0 – dt ∂r ∂r . d __ (mr) – mr w2 + mg sin w t = 0 dt

or, r – rw2 + g sin w t = 0 The Eqs (1) and (2) are the required equations of motion of the bead in a vertical plane.

...(2)

Example 6.9 Find the equation of motion of a simple pendulum whose point of suspension is moving along the x axis with an acceleration f. Sol. The system is shown in Fig. 6.8W. O O¢ X, X¢ The point of suspension O ¢ moves along OX line with an acceleration of f. Hence X¢ Y ¢ (with origin at O ¢) q l is a non-inertial frame and XY is an inertial frame of reference. The equation of constraints are given by m Y¢ Y x¢ = l sin q y¢ = l cos q Fig. 6.8W A simple pendulum with sliding point of suspension.

6.31

Classical Mechanics

The motion is, here, described by the generalized coordinate q. These equations of constraints can be transferred from the non-inertial frame to inertial frame of reference as follows: 1 x = x¢ + __ ft2 and y = y ¢ 2 1 or, x = l sin q + __ ft2 and y = l cos q 2 Again, x = l cos q q + ft and = – l sin q q The kinetic energy of the bob is given by

or, or,

1 1 T = __ m (x2 + 2) = __ m [(l cos q q + ft)2 + (– l sin q q)2] 2 2 1 T = __ m [l2 cos2 q q 2 + f 2t2 + 2 lft cos q q + l2 sin2 q q 2] 2 1 T = __ m [l2q 2 + f 2t2 + 2 lft cos q q] 2

The potential energy of the bob is given by V = – mgy = – mgl cos q [Assuming V = 0 at y = 0] \ the lagrangian of the system is given by 1 L = T – V = __ m [l2q2 + f 2t2 + 2lft cos q q] + mgl cos q 2 now, Lagrange’s equation of motion is given by

( )

or, or, or, or, or,

∂L ___ ∂L d ___ __ =0 – dt ∂q ∂q d __ (ml2q + mlft cos q) – (– mlft sin q q) + mgl sin q = 0 dt ml2 q + mlf cos q – mlft sin q q + mlft sin q q + mgl sin q = 0 ml2 q + lfm cos q + mgl sin q = 0 lq + f cos q + g sin q = 0 f g q + _ cos q + __ sin q = 0 l l

...(1)

Equation (1) represents the equation of motion of a simple pendulum whose point of suspension is moving along the x axis with an acceleration f. Example 6.10 Obtain the Lagrange’s equation of motion for a particle in central force field in the space. Sol. Let P (r, q) be the position of the particle of mass m moving under a central force field in space [Fig. 6.9W]. Then the lagrangian of the system is given by 1 L = T – V(r) = __ m (r2 + r2 q2) – V(r) ...(1) 2

Y P(r, q)

r

O

X

Fig. 6.9W Motion of a particle under central force.

6.32

Advanced Engineering Physics

where the velocity of the particle in plane polar coordinate is given by v2 = r2 + r2 q2 \ kinetic energy of the particle is given by 1 1 T = __ mv2 = __ m (r2 + r2 q2) 2 2 And the potential energy is given by r

r

k k V = – Ú F dr = Ú – __2 dr = – __r – – r Since the force F is attractive in nature and varies inversely as the square of the distance from the origin, k F = – __2 where k is a constant of proportionality. r \ the lagrangian is given by L=T–V i.e.,

or,

k 1 L = __ m (r2 + r2 q2) + __r 2

Differentiating L, we get ∂L ∂L ___ ___ = mr, = mr2q ∂r ∂q ∂L ∂L k ___ ___ = mr q2 – __2 , =0 ∂r ∂q r The equations of motion for this case are:

( )

∂L ___ ∂L d ___ __ =0 – dt ∂r ∂r

...(1)

...(2)

( )

∂L ___ ∂L d ___ __ =0 – dt ∂q ∂q From Eqs (1) and (2), we get

and

d k __ (mr) – mr q2 + __2 = 0 dt r d __ (mr2 q) = 0 dt

Rearranging again, we get k mr – mr q2 + __2 = 0 ...(3) r 2 2 mr rq + mr q = 0 The two equations of the equation set (3) are the required equations of motion for the particle that moves under a central attractive force.

Classical Mechanics

6.33

z Example 6.11 Find Lagrange’s equations of motion of a spherical pendulum. Sol. In case of a spherical pendulum, the bob is constrained to move on the surface of q the sphere. The position of the bob can be y located by spherical polar coordinates r, q and f as shown in Fig. 6.10W. The disf r tance r of the bob from the origin is the radius of the sphere on which the particle m x (bob) moves. The force in this case is not central, but is constant in the vertical direcFig. 6.10W A spherical pendulum of a bob of mass m. tion. Among the three coordinates r, q and The bob traces on the lower hemispherical surface. f, here r is a constant. So, the bob can be The greatest circle lies on the horizontal XY plane. located by q and f only and they are the generalized coordinates. The transformation equations which relate the spherical polar coordinates to cartesian coordinates are given by x = r sin q cos f, y = sin q sin f and z = r cos q Using these transformation equations and keeping in mind the fact that r = 0, the kinetic and potential energies are expressed as follows:

1 T = __ m (x2 + 2

2

1 + z2) = __ mr2 (q2 + sin2 q f2) and 2

V = mgz = mgr cos q So, the lagrangian of the system is given by 1 L = T – V = __ mr2 (q2 + sin2 q f2) – mgr cos q 2

...(1)

now, differentiating L partially with respect to q, q, f and f, we get ∂L ___ = mr2 sin q cos q f2 + mgr sin q, ∂q ∂L ___ = mr2 q, ∂q ∂L ___ =0 ∂f

...(2)

∂L ___ = mr2 sin2 q f ∂f The Lagrange’s equations for such a system can be written as

and

and

( ) ( )

∂L ___ ∂L d ___ __ =0 – dt ∂q ∂q ∂L ___ ∂L d ___ __ =0 – dt ∂f ∂f

...(3)

6.34

Advanced Engineering Physics

By using Eq. (2), Eq. (3) can be written as follows: d __ (mr2 q) – mr2 sin q cos q f2 – mgr sin q = 0 dt d __ and (mr2 sin2 q f) = 0 dt On solving these two equations further, we get mr2 q – mr2 sin q cos q f2 – mgr sin q = 0 and mr2 sin2 q f = 0 Equations of set (4) are the required equations of motion for a spherical pendulum.

...(4)

Example 6.12 Use Hamilton’s canonical equations of motion to obtain the equation of motion of simple pendulum. [WBUT 2006] Sol. A simple pendulum is shown in Fig. 6.11W. Referring to the figure and assuming m to be the mass of the bob, l, the effective length and q, angular displacement for small oscillations, we can q l express the kinetic energy as 1 T = __ ml2 q 2 C B 2 A

and the potential energy as V = mgl (1 – cos q) So, the lagrangian is given by 1 L = T – V = __ ml2 q 2 – mgl (1 – cos q) 2 now, the generalized momentum pq can be written as

Fig. 6.11W A simple pendulum.

...(1)

∂L pq = ___ = ml2 q ∂q \ the hamiltonian is given by H = S pi i

or, or,

i

...(2)

– L = pq q – L

[

1 H = ml2 q 2 – __ ml2 q2 – mgl (1 – cos q) 2 1 2 2 H = __ ml q + mgl (1 – cos q) 2

Hence H=T+V=E where E is the total energy From Eq. (2), we can write pq q = ___2 ml pq 1 \ H = __ ml2 ___2 2 ml

( )

2

+ mgl (1 – cos q)

]

[from Eq. (3)]

...(3)

Classical Mechanics

pq2 H = ____2 + mgl (1 – cos q) 2ml pq ∂H ∂H ___ ___ \ = 2 , ___ = mgl sin q ∂pq ml ∂q now, Hamilton’s equations for q and q are or,

∂H ∂H ___ q = ___ and q=– ∂pq ∂q In this case, we have pq q = ___2 ml and = – mgl sin q q Equations (6) and (7) are Hamilton’s equations for simple pendulum. From Eq. (2), we have pq = ml2 q 2 \ q = ml q Putting this value of q in Eq. (7), we get ml2 q = – mgl sin q g or, q = – __ sin q l g or, q + __ sin q = 0 l

Equation (8) is the equation of a simple pendulum which represents SHM for small q. g For small q, it is q + __ q = 0 l

6.35

...(4) ...(5)

...(6) ...(7)

...(8)

...(9)

Example 6.13 Derive the Hamilton’s equations of motion for a spring-mass system which oscillates on a frictionless horizontal plane. (The spring is unstretched at x = 0.) Sol. The spring-mass system is shown in Fig. 6.12W. The kinetic energy of the mass m is given by M 1 2 __ T = mx K 2 where x is its displacement from its mean position at x any time t. x=o The potential energy is given by Fig. 6.12W A spring-mass system. If the 1 V = __ kx2 mass m is pulled or pushed and 2 then released, it executes SHM where k is the spring constant. along the x axis. \ the lagrangian L of the system is given by 1 1 L = T – V = __ mx2 – __ kx2 2 2

6.36

Advanced Engineering Physics

∂L px = ___ = mx ∂x px __ x=m

\ or,

...(1)

now, the hamiltonian is given by H = S pi i

i

1 1 – L = px x – __ mx2 + __ kx2 2 2

or,

px __ px 1 __ H = px. __ m – 2 m. m

or,

px2 1 H = ___ + __ kx2 2m 2

( ) + __12 kx 2

2

[by Eq. (1)]

px ∂H ∂H __ ___ = , ___ = kx ∂px m ∂x now, the Hamilton’s equations are given by px ∂H x = ___ or, x = __ m ∂p or,

x

∂H ___ or, and x =– x = – kx ∂x These are the desired Hamilton’s equations of motion. Example 6.14 Derive the Hamilton’s equations of motion for a particle which is acted upon by a central force. Sol. The particle which is being acted upon by a central force is shown in Fig. 6.13W. The lagrangian of the system is given by 1 L = T – V(r) = __ m (r2 + r2q2) – V(r) ...(1) 2 For an attractive system the potential V(r) is given by k V(r) = – __r \

Y P (r, q) m r q

O

X

Fig. 6.13W The particle P with mass m is acted upon by a central force.

...(2)

k 1 L = __ m (r2 + r2q2) + __r 2

...(3)

now, from Eq. (3), we get ∂L pr = ___ = mr ∂r ∂L and pq = ___ = mr2 q ∂q \ the hamiltonian H is given by H = S pi i

i

– L = pr r + pq q – L

...(4) ...(5)

6.37

Classical Mechanics

now, using Eqs (4) and (5), we get H = (mr) r + (mr2q) q – L or, or,

k 1 H = mr2 + mr2q2 – __ m (r2 + r2q2) – __r 2 k 1 1 H = __ mr2 + __ mr2 q2 – __r 2 2

or,

k 1 1 H = ___ (mr)2 + _____2 (mr2 q)2 – __r 2m 2mr

or,

pq2 k pr2 H = ___ + _____2 – __r 2m 2mr

...(6)

The hamiltonian equations for the generalized coordinates qj and pj are: ∂H ∂H ___ = ___ and j=– ∂pj ∂qj In the present case, the generalized coordinates are r, q, pr and pq. \ the hamiltonian equations for r, q, pr and pq are given by j

∂H r = ___, ∂pr

∂H q = ___, ∂pq

r

∂H = – ___ ∂r

and

∂H pq = – ___ ∂q

now, using Eq. (6), we get pr r = __ m, pq q = ____2 , mr pq __ k ____ – 2, r = 3 mr r and = 0 q These are the required hamiltonian equations. y Example 6.15 Find the lagrangian as well as hamiltonian equations of motion for a compound pendulum. O Sol. The compound pendulum has been shown in l Fig. 6.14W. It is pivoted at the point O and capable q of oscillating about this point. When the compound c pendulum oscillates, its center of mass gets dismg placed. Let at time t its angular displacement be q. 1 \ its kinetic energy, T = __ Iq2 x 2 and its potential energy, V = – mgl cos q Fig. 6.14W Motion of a compound pendulum. [Here, O is the reference point] where OC = l and I is the moment of inertia of the body about the axis of rotation.

6.38

Advanced Engineering Physics

\ the lagrangian L is given by 1 L = T – V = __ I q2 + mgl cos q 2 The Lagrange’s equation of motion is given by

...(1)

( )

∂L ___ ∂L d ___ __ =0 – dt ∂q ∂q d __ (Iq) + mgl sin q = 0 dt mgl q + ____ sin q = 0 I

or, or,

If q is very small then we can write sin q ª q. So the equation reduces to mgl q + ____ q = 0 I

...(2)

This equation (i.e., Eq. (2)) is the Lagrange’s equation of motion. It is a simple harmonic motion and its time period is given by ____

÷

I T = 2p ____ mgl Differentiating Eq. (1) with respect to q, we get

or,

∂L pq = ___ = Iq ∂q p q q = __ I

\ the hamiltonian H is given by H = S pj

j

– L = pq q – L

j

or, or,

1 H = pq q – __ Iq2 – mgl cos q [by Eq. (1)] 2 2 p 1 pq H = pq __q – __ I. ___ – mgl cos q I 2 I2 2 pq H = __ – mgl cos q 2I

\ the hamitonian equations of motion are given by ∂H pq q = ___ = __ or pq = Iq ∂pq I ∂H = – ___ = – mgl sin q ∂q Equations (3) and (4) are the desired hamiltonian equations of motion. Differentiating Eq. (3) with respect to time and combining with Eq. (4), we get Iq = – mgl sin q and

q

...(3) ...(4)

6.39

Classical Mechanics

or,

mgl q + ____ sin q = 0 I

If q is very small then sin q ª q mgl q + ____ q = 0 \ I

____

÷

I fi T = 2p ____ mgl

Example 6.16 Derive the hamiltonian and hamiltonian equation of motion for a particle falling freely under the influence of gravity. P Sol. Let us consider a body of mass m which is falling freely under the influence of gravitational x force. The body initially remains at point P. Then it falls freely and at time t it reaches the point Q h Q [Fig. 6.15W]. Let h–x PQ = x, \ QR = h – x The potential energy at Q is given by R V = mg (h – x) Fig. 6.15W A freely falling body under gravity. and kinetic energy at Q is given by

( ) = __12 mx

dx 1 T = __ m ___ 2 dt

2

2

\ the lagrangian of the system is given by 1 L = T – V = __ mx2 – mg (h – x) 2 px ∂L \ px = ___ = mx or x = __ m ∂x So, the hamiltonian H is given by px __ px2 1 ___ H = px x – L = px __ – m – mg (h – x) m 2 m2

[

or,

px2 ___ px2 H = ___ – m 2m + mg (h – x)

or,

px2 H = ___ + mg (h – x) 2m

]

\ the hamiltonian equations of motion are given by ∂H ∂H px ___ = mg and x = ___ = __ x =– m ∂x ∂p x

now,

px x = __ m fi px = mx

\ mg = mx [ x = mg] \ x–g=0 This is the expected equation of motion for a freely falling body.

6.40

Advanced Engineering Physics

Review Exercises

Part 1: Multiple Choice Questions 1. If the constraint relations are independent of velocity, then the constraints are called (a) scleronomic (b) conservative (c) bilateral (d) holonomic 2. A system is called scleronomous, if the constraint equations do not contain (a) velocity (b) time (c) velocity and time (d) displacement 3. When a particle placed on the surface of a sphere moves on the surface, the constraint is (a) holonomic (b) non-holonomic (c) scleronomic (d) rheonomic 4. The constraint in case of rigid body motion is (a) holonomic (b) non-holonomic (c) rheonomic (d) none of these 5. The constraint associated with the motion of a pendulum is (a) dissipative (b) holonomic (c) rheonomic (d) All of these 6. The constraint involving rolling without sliding is (a) holonomic (b) non-holonomic (c) rheonomic (d) conservative 7. The characteristics of generalized coordinates are that (a) they are independent of each other (b) they are necessarily spherical coordinates (c) they may be cartesian coordinates (d) Both (a) and (c) 8. The dimensions of generalized momentum (a) are those of linear momentum (b) are those of angular momentum (c) may be either those of linear or angular momentum (d) none of these 9. If the lagrangian is free of a particular generalized coordinate qj , then qj is called (a) superfluous (b) cyclic (c) cartesian (d) spherical 10. The number of generalized coordinates necessary to describe the motion of a particle in a circular wire of radius a is/are (a) two (b) three (c) one (d) none 11. The virtual displacement d ri (a) does not involve time (b) may involve time (c) may or may not involve time (d) none of these 12. The double pendulum requires n generalized coordinates to be described where n is (a) one (b) three (c) two (d) none of these 13. A generalized coordinate qj is absent in the lagrangian of a system. The corresponding conserved quantity is (a) energy (b) velocity (c) momentum (d) force 14. When a system is conservative, the hamiltonian is given by (a) H = T – V (b) H = T + V (c) H = V – T (d) H = TV

Classical Mechanics

6.41

15. If a generalized coordinate qj is cyclic in the lagrangian then in the hamiltonian it is (a) not cyclic (b) may or may not be cyclic (c) cyclic (d) unpredictable 16. The hamiltonian of a system is a function of (a) qj , pj , t (b) pj , j , t (c) qj j, t (d) pj , j , t 17. The lagrangian of a system is a function of (a) qj , pj , t (b) qj , j , t (c) j , qj , t (d) pj , j , t 18. A system consists of three point masses and forms a triangle. If the mutual distances between the masses do not vary with time, then the degrees of freedom of the system are (a) zero (b) three (c) six (d) none of these 19. A rigid body has constraints classified in the following group: (a) Rheonomic and holonomic (b) Rheonomic and non-holonomic (c) Sclerononic and holonomic (d) Scleronomic and non-holonomic [WBUT 2007] 20. The generalized coordinate can be of any dimension. However, (a) the product of the generalized coordinate and the generalized momentum is always ml2t (b) the dimension of the work done depends on the dimension of the generalized coordinate (c) the generalized momentum is always of the dimension ml/t2 (d) the generalized force is always of the dimension ml/t2 [WBUT 2007] 21. If the constraints are time dependent, then the relation between the hamiltonian H and the total energy E is given by (a) H = E (b) H π E (c) H = E – V (d) H = E – T [Ans. 1(d), 2 (b), 3 (b), 4 (a), 5 (d), 6 (b), 7 (d), 8 (c), 9 (b), 10 (c), 11 (a), 12 (c), 13 (c), 14 (b), 15 (c), 16 (a), 17 (b), 18 (c), 19 (c), 20 (a), 21 (b)]

Short Questions with Answers 1. Discuss the nature of constraints and specify the force of constraint in case of a pendulum with variable length and a particle moving on an ellipsoidal under gravity. Ans. In case of a pendulum with variable length, the length can be expressed as l = l(t). And its equation of constraint can be expressed as | r1 – r2 |2 = l2(t) which is dependent on time. So, the constraint is rheonomic and holonomic. The force of constraint is the tension in the string. A particle which is moving on the surface of an ellipsoid under gravity will leave the ellipsoid after reaching a certain point on the ellipsoid. Hence, the constraint is non-holonomic. And the force of reaction of the ellipsoidal surface on the particle is the force of constraint. 2. What difficulties do the constraints introduce while solving mechanical problems? Ans. The difficulties introduced by the constraints while solving mechanical problems are of two kinds. They are as follows: (a) The equation representing the constraints for a system of ‘n’ particles with position vectors r1, r2, r3, ..., rn is f (r1, r2, ... rn, t) = 0

6.42

3. Ans.

4. Ans.

5. Ans. 6. Ans. 7. Ans. 8. Ans. 9. Ans.

Advanced Engineering Physics

Usually, these n coordinates are not all independent but they are connected by the aforementioned equation. So, the n equations which are connected by this constraint equation are not all independent. Hence, in order to have a solution of the problem, these equations are to be written again by considering the effect of the constraints. This is the first difficulty. (b) In many situations, the nature of the forces of constraint remain unknown and they belong to the unknown parameters of the problems. These unknown forces of constraints introduce some complications in the process of solution of the problem. This is the second difficulty. What is meant by degrees of freedom of a dynamical system? The term degrees of freedom of a dynamical system means the number of independent ways in which that dynamical system can execute motion without violating the constraint imposed on the said system. It is defined as the minimum number of independent coordinates which can specify the position of the said dynamical system completely. It can also be defined as the number of squared terms in the expression of the kinetic energy of the system. If a system has n particles and k constraints, then the minimum number of coordinates to specify its position is (3n – k). Hence, the degrees of freedom of the system of n particles is given by f = 3n – k What do you mean by generalized coordinates? State their advantages. If a rigid body consists of n particles and there are k constraints associated with the body then its degrees of freedom is f = 3n – k This suggests that the system can be described by a minimum number of f independent coordinates where the effects of the constraints are included. These f independent coordinates qj (j = 1, 2, 3, ..., f) are called the generalized coordinates. Use of generalized coordinates gives us the following advantages: (i) The generalized coordinates may not essentially be cartesian, spherical, polar, etc. Thus, the dependence on the description of the configuration of the dynamical system is eliminated. (ii) The individual variation (D qj) may be considered as the generalized coordinates qj are all independent. (iii) A dynamical system described in terms of generalized coordinates may be treated as a free system. What are the limitations of newtonian mechanics? See Section 6.1.1. What are system-point and configurations space? See Section 6.2.1. State characteristics of lagrangian equations. See Section 6.2.4.1. What is phase space? Describe. See Section 6.3.1. What is meant by cyclic coordinate? The lagrangian (L) of a system is, in general, a function of all the generalized coordinates (qj) as well as time t, i.e., L = L (q1, q2, q3, ... , qf ; t)

Classical Mechanics

6.43

If a generalized coordinate remains absent in the expression of the lagrangian (i.e., it does not appear explicitly) then that coordinate is called a cyclic coordinate. Hence, for a cyclic coordinate, we get ∂L ___ =0 ∂qj 10. What are the advantages of the hamiltonian formulation over the lagrangian formulation? Ans. (i) In lagrangian formulation, the two variables qj and j are not given equal status since j is simply the time derivative of qj. But in case of hamiltonian formulation, both the coordinate (qj) and the momentum (pj where pj = mj j) are given equal status. This provides a frequent freedom of choosing coordinates and momenta. (ii) The fact that, for a conservative system, the numerical value of the hamiltonian (H) is equal to the total energy is very important in case of energy changes in atoms and molecules. (iii) Assigning equal status to qj and pj in hamiltonian formulation provides a convenient basis for the development of quantum mechanical theories as well as statistical mechanics. (iv) While quantizing a dynamical system, the knowledge of the hamiltonian is extremely important. While we set up Schrödinger equation in wave mechanics, we simply replace generalized momenta by corresponding differential operators. 11. State the conservation theorem of generalized momentum. Ans. The conservation theorem of generalized momentum states that the generalized momentum corresponding to a cyclic coordinate is constant. ∂L If qj be a cyclic coordinate then one can find that ___ = 0 ∂qj And the lagrangian equations of motion for a conservative system

( )

∂L ∂L d ___ __ – ___ = 0 dt ∂ j ∂qj reduces to

( )

∂L d ___ __ =0 dt ∂ j ∂L ___ = constant ∂ j The pair (qj , pj) is called the canonical or conjugate variables. This imples

Part 2: Descriptive Questions 1. State the principle of virtual work. Derive the mathematical expression for the principle of virtual work. 2. State the principle of D’Alembert. Derive a mathematical expression to represent D’Alembert’s principle. 3. What is the advantages of using ‘generalized coordinates’? Obtain the expression of ‘generalized force’. 4. What are the advantages of Lagrange’s equation over newton’s equation of motion?

6.44

Advanced Engineering Physics

5. Obtain Lagrange’s equation of motion from D’Alembert’s principle. Discuss the effect of dissipative forces. 6. Obtain Hamilton’s canonical equations by using the relations between the hamiltonian and the lagrangian of a system. 7. What do you mean by cyclic coordinates and superfluous coordinates? 8. From the lagrangian obtain hamiltonian. Is the hamiltonian equal to total energy? Is this equality of hamiltonian and total energy valid in general? 9. What do you mean by configuration space and degrees of freedom of a dynamical system? What are generalized coordinates and what are their advantages? Write down the equation of transformation from generalized coordinates to cartesian coordinates. 10. Write down the hamiltonian and derive the hamiltonian equation of motion of a simple pendulum. 11. Write down the hamiltonian and derive the hamiltonian equations of motion for a compound pendulum. 12. Derive the expression of kinetic energy of a system of particles in terms of generalized velocities. 13. Show that the kinetic energy for a scleronomous system is a homogeneous quadratic function of the generalized velocities.

Part 3: Numerical Problems 1. Use Lagrange’s equations to find the equations of motion of a compound pendulum which oscillates in a vertical plane about a fixed horizontal axis. Hence find the period of oscillation of the compound pendulum. 2. A particle of mass m moves in a two-dimensional potential V(x, y) = –k1x2 + k2x2. Write down the Lagrange’s equation of motion. 3. The bob of a simple pendulum moves in a horizontal circle. Find the angular frequency of the circular motion in terms of angle q and the length l of the rod. 4. Find out the degrees of freedom of the CH4 molecule. 5. Use Lagrange’s equation of motion to determine the motion of a mass m which slides without friction down an inclined plane. 6. The kinetic energy of an electron of charge e moving in the field of the nucleus of charge + 2e in 1 spherical polar coordinates is given by T = __ m (r2 + r2q 2 + r2 sin q2 f2) and potential energy V by 2 2 Ze V = ___ r . Obtain the Lagrange’s equation of motion. What is the hamiltonian of the system? 1 1 7. The lagrangian of a system is given by L = __ a 2 – __ b q2, where a and b are constants. Obtain 2 2 lagrangian equation of motion and also find the hamiltonian of the system.

Chapter

7 Quantum Mechanics 7.1

IntroduCtIon

In the book Basic Engineering Physics, the old quantum theory has been discussed. This theory was a collection of the results of researches in this field between the years 1900 and 1925, which predated the modern quantum mechanics. The theory was never complete or self-consistent, but was a collection of heuristic prescriptions, which are now understood to be the first quantum corrections to classical mechanics. The motions of particles which are subjected to external forces are discussed in newtonian (or classical) mechanics. And in newtonian mechanics it is also taken for granted that one can measure correctly the properties of particles like mass, position, velocity, acceleration, etc. Of course, this idea is valid in our common sense which we acquire through the experiences of our daily life. Classical mechanics is able to explain correctly the dynamical behavior of the material bodies in terms of the values it predicts for observable quantities and the values of the same quantities which are measured experimentally. On the other hand, quantum mechanics also deals with the values of observable quantities related to dynamical systems but in this case, the term ‘observable quantity’ bears different significance in the light of the uncertainty principle in the atomic realm. The uncertainty principle plays an important role in quantum mechanics. According to quantum mechanics, simultaneous measurement of the position and momentum of a moving particle with precision is impossible while according to classical mechanics, both the position and momentum of a moving particle can be ascertained accurately at every instant of time. And for this reason, classical mechanics is considered to be deterministic, whereas quantum mechanics is considered to be probabilistic. For example, according to classical mechanics, the radius of the electronic orbit in the ground state of hydrogen atom is exactly 5.3 × 10–11 m while according to quantum mechanics, this is most probable value of the said radius. Classical mechanics is a special case of quantum mechanics; in other words, it is an approximate version of quantum mechanics. The certainties which one comes across in classical mechanics are only apparent. The agreement which is observed between the predicted value and the experimental value of an object is due to the fact that the macroscopic material bodies consist of a large number of individual atoms and the departure of the collective behavior of the atoms from their individual behavior is unnoticeable. Thus, unlike classical mechanics, quantum mechanics includes two sets of physical principles; one for the macroscopic world and the other one for the microscopic world.

7.2

Advanced Engineering Physics

Quantum mechanics was not developed uniquely through a single formulation like classical mechanics. Rather initially it was developed through two different formulations by two renowned physicists, namely, Werner Heisenberg and Erwin Schrödinger. In 1925, on one hand, Heisenberg formulated the quantum theory in terms of observable quantities alone (like intensity and frequencies of spectral lines) to replace position, momentum and other dynamical variables of classical mechanics by matrices, while keeping the form of the equations of motion superficially the same as in classical mechanics. This formulation of quantum mechanics is known as matrix mechanics. On the other hand, in 1926, Schrödinger formulated the quantum theory on the basis of de Broglie hypothesis of matter waves even before it was experimentally verified by Davison and Germer in 1927. Schrödinger’s quantum theory is known as wave mechanics. He proposed that the wave function Y describing the matter waves satisfies a partial differential equation, and gave the prescription to write down the equation for any particular system of particles. Despite the vastly different appearance of the two aforesaid theories, it was very soon recognised that they are completely equivalent. Schrödinger’s equation is applicable only to non-relativistic particles. So, a new wave equation, which meets the requirements of the theory of relativity, was formulated by Paul Adrien Maurice Dirac in 1928 which is known as relativistic wave mechanics. Out of the two alternative forms of quantum mechanics — wave mechanics and matrix mechanics — it is the former one which lends itself more easily to the solution of wide variety of practical problems. For this reason, we will be dealing with Schrödinger’s version of quantum mechanics, i.e., wave mechanics in this chapter. It was again Dirac who ultimately set up a general formalism of quantum mechanics in the year 1930 to unify the two formulations of Heisenberg and Schrödinger.

7.2

Wave FunCtIon, probabIlIty and probabIlIty densIty

We now know well that nature is strikingly symmetrical in many ways. Our observable universe is entirely composed of matter and energy. Light is one of the various forms of energy. We have learnt through Compton and photoelectric effects that light has dual, wave as well as particle nature. So, one can expect from the concept of symmetry of nature that matter also may have dual, wave as well as particle nature (i.e., character). The particle nature of matter is well known to every one. Louis de Broglie proposed the wave nature of matter in 1924 in his hypothesis which was later named after him and Davison and Germer confirmed this experimentally in case of electrons in the year 1927. When light propagates through free space, its motion in space and time can be represented by the electromagnetic wave equation as given below: _ 1 ∂2 _ —2 x = __2 ___2 x ...(7.1) c ∂t _ __ _ where c is the velocity of light in the vacuum and x is either the electric vector E (r, t) or the magnetic vector __ _ B (r, t) of the light wave. So, again from the concept of symmetry of nature one can expect a similar equation to represent the motion of a wave-particle in space and time. This is what was done by Erwin Schrödinger in 1926. The idea that the stationary states of an electron in an atom corresponding to the standing matter waves was taken up and used by him as the foundation of wave mechanics. He developed an equation, now known as Schrödinger’s equation, to represent the motion of the matter-wave (in space and time) related to a free particle which is given below: _ ∂Y _ h2 – ___ —2 Y (r, t) = i h ____ (r, t) 2m ∂t

...(7.2) _

where h is Planck’s constant and h = h/(2p) and Y(r, t) is a function of space and time which satisfies the _ Eq. (7.2). The function Y (r, t) is known as Schrödinger’s wave function (i.e., the function which represents Schrödinger’s matter wave in space and time).

Quantum Mechanics

7.2.1

7.3

probability and probability density

When the double-slit experiment is carried out with light, such as the Young’s double-slit experiment, one considers the superposition of the secondary waves arising from the two slits on the screen of recording of the interference pattern. The light intensity on this screen is determined by the square of the amplitude of the wave, which is formed as a result of superposition of the two secondary waves coming from the two slits. The intensity of light on the screen is also proportional to the number of photons reaching the screen per unit area. The D probability (or chance) of finding a photon at a given point on the screen depends on the intensity of light wave at the point. x S1 This intensity (as stated above) is proportional to the square of Sr the resultant amplitude of the waves (i.e., the electric field) at S2 the point. Hence, the probability of finding a photon on a given point is proportional to the square of the resultant electric field Electron at the point. Now comparing the duality (i.e., dual behavior) of source S photon with that of a material particle (e.g., an electron) one can arrange a similar double-slit experiment with monoenergetic Fig. 7.1 Electron double-slit set-up for material particles (i.e., electrons) as shown in Fig. 7.1. interference of electron waves. Sr: source By putting forward similar arguments in case of electron of electrons, S1, S2; double slits, S: the double-slit experiment as we have done in case of the doublescreen and D: the detector. slit experiment of photon (i.e., Young’s double-slit experiment), we can conclude that the probability of finding an electron per unit area of the screen S of Figure 7.1 is proportional to the square of the wave function (or state function) Y for the electron. If one compares _Eqs (7.1) and (7.2), the logic which __has been put forward above will be more clear. The wave function x (and hence the electric field vector E) represents wave behavior __ of the photon while Schrödinger’s wave function Y represents the wave behavior of the electron. So, if | E |2 gives the probability of finding the photon on the screen of Young’s double-slit experiment, then | Y |2 will give the probability of finding the electron on the screen of the electron double-slit experiment [Fig. 7.1]. _ Therefore in general, we can state that if Y (r, t) represents the wave function of a material particle, then _ _ 2 | Y (r, t) | will represent the probability of finding that particle at point (x, y, z) at time t (since r = x + y + z). _ The wave function Y (r, t) plays the role of probability amplitude. This function is generally a complex quantity. Since probabilities are real positive numbers (varying from + 0 to + 1), the probability of finding a particle, at a given position (x, y, z) at a given time (t), is expected to be proportional to | Y |2. And it will be equal to | Y |2 when the wave function Y is a normalized one. Probability density of a particle It is the probability of finding the concerned particle in unit volume of a given space at a given instant of time. It is usually expressed as the square of the absolute value of the wave function when it is normalized. If the normalized wave function Yn is complex, then | Yn |2 = (Yn) (Yn*) where Yn* is the complex conjugate of Yn. So, the probability density P can be expressed as follows: _ _ _ P = Yn* (r, t) Yn (r, t) = | Yn (r, t)|2 ...(7.3)

7.4

Advanced Engineering Physics

Therefore, the probability of finding a particle having normalized wave function Yn in a volume element dV at the point (x, y, z) at time t is given by PV = Yn* Yn dV Since at a given point of time, the particle must be somewhere in the space, the total probability Pt to find the particle in space must be equal to unity. +

Pt = Ú Yn* Yn dV = 1

...(7.4)

–

Now, if the particle always lies in a point of a straight line, then the probability of finding it within a distance dx is given by P dx = Yn* Yn dx Since the particle must be somewhere on the straight line (in this case the x axis), the total probability of finding it on the line (i.e., total probability for one-dimensional motion) is given by +

Ú

–

7.3

Yn* Yn dx = 1

...(7.5)

normalIzatIon oF Wave FunCtIon and orthogonalIty oF Wave FunCtIons

One should note that the wave function Y (x, t) which describes the complete space-time behavior of a particle in one-dimensional motion has an appreciable amplitude in a region where the particle is likely to be found with a greater probability. One can assume that the quantity | Y (x, t) |2 dx = Y *(x, t) Y (x, t) dx is proportional to the probability of finding the particle in the interval between x and x + dx at the time t where Y *(x, t) is the complex conjugate of Y(x, t). The total probability of finding the particle anywhere in the onedimensional space is given by +

Ú

Pt

| Y (x, t) |2 dx

–

Let us define the position probability density (or simply probability density) as +

/[

Ú

P(x, t) = | Y (x, t) |2 Hence, the total probability will be

| Y (x, t) |2 dx

–

]

...(7.6)

+

Pt = Ú P (x, t) dx –

+

=

Ú

–

2

| Y (x, t) | dx

/[

+

Ú

–

| Y (x, t) |2 dx

]

=1 ...(7.7) This is what should be, since the total probability Pt must be always equal to unity. Since | Y (x, y) |2 is necessarily positive, Eq. (7.6) indicates that the probability density P(x, t) is always positive. This is consistent with the expected behavior and definition of probability, If Y (x, t) is multiplied by one complex constant quantity N so that Yn (x, t) = N Y (x, t) where Yn (x, t) satisfies the following relation

Quantum Mechanics +

Ú

–

7.5

+

| Yn (x, t) |2 dx = | N |2 Ú | Y (x, t) |2 dx = 1

...(7.8)

–

Then Yn (x, t) is said to be the normalized wave function and N is called the norm of the unnormalized wave function Y (x, t) required to normalize it. From Eq. (7.8), one can have

/

2

|N| = 1

Ú

+

Ú

| Y (x, t) |2 dx

...(7.9)

–

N is also called the normalization constant. And obviously for a wave function to be normalizable | Y (x, t) |2 dx or in general Ú | Y (r–, t) |2 dV over all space must be finite. This condition is known as the

square-integrability of the wave function. Orthogonality Let us consider two wave functions Ym (x, t) and Yn (x, t) which satisfy Schrödinger’s Eq. (7.2) with y = z = 0. In case of this pair of wave functions one can show that +

Ú

–

Ym (x, t) Yn (x, t) dx = 0

.. (7.10)

where m π n. This property of two wave functions is called orthogonality of these functions. To be physically meaningful, both the wave functions must satisfy the condition of nomalization. +

i.e.,

| Ym (x, t) |2 dx = Ú Ym* (x, t) Ym (x, t) dx = 1

Ú

| Yn (x, t) |2 dx = Ú Y*n (x, t) Yn (x, t) dx = 1

– +

and

–

7.4

+

Ú

– + –

physICal sIgnIFICanCe and InterpretatIon oF Wave FunCtIon

As the wave function of a material particle represents the physical behavior of the particle while it is in motion, it must have some physical significance, otherwise it will not be able to represent the dynamical behavior of the particle. Some important physical significances of the wave function are listed below: (a) Wave function gives information regarding the space-time behavior of the concerned particle. (b) The square of the wave function gives the measure of the probability of finding the particle about a position. (c) The magnitude of the wave function is large in regions where the probability of finding the particle is high and small in the region where the probability of finding the particle is low. (d) The wave function representing a moving particle must be continuous and single valued at each point of space. Also its value at each point of space must be finite. Interpretation of wave function One can give a physical interpretation of the wave function as follows: During the propagation of any wave some physical quantities vary with respect to space and time. As for example, when a light wave propagates, the electric and the magnetic fields of it vary in space and time and when a sound wave propagates the pressure in the medium of propagation varies in space and time. Similarly, when a particle moves the matter wave related to it (i.e., the de Broglie wave) also propagates. And during

7.6

Advanced Engineering Physics

the time of propagation of this wave, some physical quantities also keep on varying in space and time. All the information about these varying physical quantities are carried by the wave function Y of the matter wave. Y is generally a complex quantity and is a function of position coordinates (x, y, z) as well as time (t). In case of a moving body, the value of Y is related to the probability (i.e., chance) of finding the particle at a particular point in space (x, y, z) at a given time t. The wave function Y can also give information regarding the future behavior of the particle. The wave function Y actually contains all the information which the uncertainty principle allows one to know about the related particle but this wave function cannot be measured directly. As stated earlier, the probability of any event must be a real positive quantity whose value lies within the range between 0 and 1. Since wave function is generally a complex quantity which can have both positive and negative values, the probability related to the existence of the particle at a point in space and time can be expressed by | Y |2. This is because for any allowed value of Y, | Y |2 will always be positive. In order to understand this, one can express Y as follows: Y = a + ib ___ where i represents the square root of (–1), i.e., i = ÷–1 and a and b are two real quantities which can be expressed as the function of position coordinates (x, y, z) and time t. Now, the complex conjugate of Y (denoted by Y *) is given by Y * = a – ib So, the square of the modulus of Y can be written as | Y |2 = (Y *)(Y) = (a – ib) (a + ib) = a2 + b2 The value of a2 + b2 is always real and positive. Thus, | Y |2 is always a positive real quantity. If Yn be normalized value of Y, then | Y |2 represents the probability density of the particle. For a particle that is represented by the wave function Y (or the normalized wave function Yn), the probability of experimentally observing it at the position (x, y, z) at time t is given by | Yn |2. At any point in space where | Yn | > 0, there is a definite probability of detecting it. It was Max Born who first gave this statistical interpretation of the wave function in 1926. Let us now consider an ensemble of a large number of identical particles where each of them is described by the same normalized wave function Yn. Then the actual density of particles at position (x, y, z) at a given time t is equal to | Yn |2. If one assumes that a particle is moving along the x axis, then its wave function is a function of x and t only. If this function is normalized and denoted by Yn (x, t), then the probability density P (x, t) is taken to be | Yn (x, t) |2. And the probability of finding the particle at a particular time t in a region between x and x + dx is equal to | Yn |2 dx. This can be expressed as follows: p (x, t) dx = Y *n (x, t) Yn (x, t) dx = | Yn (x, t) |2 dx ...(7.11) where the probability density p (x, t) is defined as the probability per unit length (along the x axis for finding the particle around the point (x, 0, 0) at the time t. The probability for finding the particle in the region between the points (x1, 0, 0) and (x2, 0, 0) is given by x2

p (x1, x2, t) = Ú p (x, t) dx x1

x2

= Ú Yn* (x, t) Yn (x, t) dx x1

Quantum Mechanics

7.7

x2

p (x1, x2, t) = Ú | Yn (x, t) |2 dx

or,

...(7.12)

x1

In a three-dimensional case, the probability of finding the particle in a small volume dV around the point (x, y, z) at time t is given by p (x, y, z, t) = Ú | Yn (x, y, z, t) |2 dV

...(7.13)

v

7.5

operators In Quantum meChanICs

The word ‘operation’ means action, so an operator is an actor which can change a physical quantity having taken some action on it. Basically an operator is a mathematical rule that can change a given function into a new function. If o be an operator, then it is mathematically expressed as . As an example, if be the differential operator d ___ , then when it operates on the function f (x) = 2x2 + 3x + 1 one gets the following result: dx d d ___ f (x) = ___ (2x2 + 3x + 1) = 4x + 3 = f1(x) (say) dx dx

( )

An operator can be either linear or complex in nature. If be a linear operator then it exhibits the following properties: (cf ) = c (f ) and ( f + g) = f + g where c is a constant and f and g are two functions. An operator can generate a new function by operating on a given function. In the example given d above, the operator = ___ has generated the new function f1(x) by operating on the function f (x). dx In quantum mechanics, each dynamical variable is represented by an operator and the former provides a link with the latter (i.e., the operator) through the correspondence principle. Usually if a be a dynamical variable then the corresponding operator is denoted by where the symbol (^) is known as caret. The operators used in quantum mechanics are linear operators. The linear operators obey distributive law. Thus, if f (x) and g(x) are two functions of x and is a linear operator, then one gets { f (x) + g(x)} = f (x) + g(x) The linear operators also obey associative laws. Sums and products of two linear operators are also linear operators. And the sum of two linear operators is commutative, i.e., ( + ) f (x) = ( + ) f (x). But the product of two linear operators may or may not be commutative. For example, the product of the two linear operators d d2 d = ___ and = ___2 is commutative while the product of the two linear operators 1 = x and 1 = ___ is not dx dx dx commutative. In order to verify the above statements let us consider the function f (x) and perform the following operations:

7.8

Advanced Engineering Physics

d d f (x) = ___ f (x) ( ) ( ___ ) dx dx d d d f (x) = ( ___ ) ( ___ ) f (x) = ___ f (x) dx dx dx 2

d f (x) = ___ dx

(a)

3

2

3

2

and

3

2

Hence (

–

3

) f (x)

(

)

d d2 d 2 d = ___ ___2 – ___2 ___ f (x) dx dx dx dx d d2 d 2 d f (x) = ___ ___2 f (x) – ___2 _____ dx dx dx dx d3 d3 = ___3 f (x) – ___3 f (x) = 0 dx dx

[

](

)

d d2 d d2 d2 d [ , ] = ___, ___2 = ___ ___2 – ___2 ___ = 0 dx dx dx dx dx dx

i.e.,

d d2 So, the operators ___ and ___2 commute. dx dx (b) Let

1

d = ___ and let f (x) be a function. Then one can write dx d f (x) d ___ f (x) = x ◊ _____ 1 f (x) = x dx dx

= x and 1

and

Hence (

1

1

1

1

–

1

d d f (x) = ___ (x) f (x) = ___ {x(f (x)} dx dx d f (x) = x _____ + f (x) dx 1

1)

(

)

d d f (x) = x ___ – ___ x f (x) dx dx

d d = x ___ f (x) – x ___ f (x) – f (x) dx dx = – 1. f (x) As f (x) is an arbitrary function, one can remove it from both sides and get the following expression:

[ x, ___dxd ] = x ___dxd – ___dxd x = –1 d So, the operators x and ___ are not commutative. In order to get a function operated upon by an dx operator, the function is placed at the right side of the operator.

Quantum Mechanics

7.9

7.5.1 Commutator If

and be two operators, the commutator of this pair of operators is denoted by [ , ] and it is defined as [ , ]=

–

.

If [ , ] = 0, i.e., = , then the pair of operators is known as commutative. If [ , ] π 0, i.e., π , then the pair of operators is known as non-commutative. As for example, the pair of operators and x is non-commutative, i.e., [ , x] π 0, actually [ , x] = i h. This implies that the simultaneous measurement of these variables (i.e., x and px) with absolute precision is not possible. One should note that this incompatibility is valid for position and momentum in the same direction. When the operators are in different direction, they commute.

7.5.2

some Quantum mechanical operators

Before discussing the quantum mechanical operators, let us consider two wave equations given below: y = a sin (kx – w t) and Y = Aei(kx – w t) The first equation represents a wave of a simple harmonic oscillator while the second one represents the de Broglie wave of a particle and both the waves propagate along the x axis. The second equation contains a complex expression because a wave representing a moving particle can be complex. Now, in order to understand the role of operators in wave mechanics, let us differentiate the wave function Y (x, t) for a free particle given by Y (x, t) = Aei(kx – w t) with respect to x and t. So, we get

∂ ∂ ___ Y (x, t) = A___ {Aei(kx – w t)} ∂x ∂x

∂ ∂ __ Y (x, t) = A__{Aei(kx – w t)} ∂t ∂t Now, multiplying the first equation by – i h ∂ – i h ___ Y (x, t) = A(– i h) (ik) Aei(kx – w t) ∂x or,

∂ – i h ___ Y (x, t) = (h k) Y (x, t) ∂x

or,

∂ – i h ___ Y (x, t) = px Y (x, t) ∂x

...(7.14) [ p = px = hk]

Again, multiplying the second equation by i h, we get ∂ i h __ Y (x, t) = A(i h) (– iw) Aei(kx – wt) ∂t or,

∂ i h __ Y (x, t) = hw Y (x, t) ∂t

or,

∂ i h __ Y (x, t) = E Y (x, t) ∂t

...(7.15) [ E = hw]

7.10

Advanced Engineering Physics

As the wave function Y (x, t) in Eqs (7.14) and (7.15) is arbitrary, by dropping Y (x, t) from both sides of the said equations, we get ∂ ∂ – i h ___ = px and i h __ = E ∂x ∂t So, the momentum and energy operators can be written as ∂ ___ ...(7.16) x ∫ –ih ∂x ∂ ...(7.17) ∫ + i h __ ∂t These two operators are mathematical operators for representing the dynamical variables of momentum and the energy of a particle. In addition, since the coordinate x is a multiplying operator, we can express it in operator form as follows: =x ...(7.18) The operator representation of Eqs. (7.16) and (7.18) is called Schrödinger representation or coordinate representation. The operators corresponding to the other dynamical variables that are functions of the coordinate x and momentum px are obtained by substituting them (i.e., x and px) in classical expressions by (= x) and x ∂ = – i h ___ respectively. ∂x Thus, if a ( , x) is a dynamical variable, and function of x and px, then the operator representation of a can be found as follows:

(

)

(

)

∂ = a , – i h ___ ...(7.19) ∂x As for example, by applying the above rule (i.e., Eq. 7.19) we can find operators for kinetic energy (Ek) and angular momentum about the x axis (Lx) as follows: ( , px) = a ( ,

x)

(i) Operator representation of kinetic energy px2 ___ Ek = Ek ( px) = 2m Now,

we get

x

∂ = – i h ___, putting this for px ∂x

(

)

∂ 2 – i h ___ ∂x _________ k = 2m

– h2 ∂2 = ____ ___2 2m ∂x (ii) Operator representation of Lx __ _ _ _ _ Lx = y pz – z py [ L = r × p, r = x + y + z and p = px + py + pz] \ x = z– y or,

k

...(7.20)

Quantum Mechanics

or,

x

) (

(

∂ ∂ = y – i h __ – z – i h ___ ∂z ∂y

)

(

)

...(7.21)

(

)

...(7.22)

(

)

...(7.23)

∂ ∂ = – i h y __ – z ___ ∂z ∂y Similarly we can find the operator representation of Ly and Lz as

or,

and

7.5.3

x

7.11

y

∂ ∂ = – i h z ___ – x __ ∂x ∂z

z

∂ ∂ = – i h x ___ – y ___ ∂y ∂x

operator representation of three-dimensional variables

One-dimensional operators like momentum operators as follows: _ =r

= x and

x

∂ = – i h ___ lead us to write the three-dimensional position and ∂x __

and

= –ih —

...(7.24)

_ ∂ ∂ ∂ where r = x + y + z and — = ___ + ___ + __ ∂x ∂y ∂z As we have seen in a one-dimensional case, the operator representation of a three-dimensional dynamical _ _ variable a (r, p) which is a function of position and momentum can be obtained by using the operator forms _ _ _ _ of r and p in the expression of a (r, p) by using the operators and as given in Eq. (7.24): __ _ _ (r, p) = a ( , ) = a ( , – i h —) ...(7.25) As an example, the hamiltonian operator for three-dimensional motion can be obtained by using the classical expression of hamiltonian function p2 H = ___ + V(r) 2m

Hence, or,

7.5.4

( )2 = ____ + V(r) 2m h2 = – ___ —2 + V(r) 2m

...(7.26)

measurement of observables

Observables are dynamical quantities like position, linear momentum, angular momentum, kinetic energy, total energy, etc., which can be measured by conducting experiments on a physical system. There is an operator corresponding to each observable in quantum mechanics which operates upon the wave function to produce a new function. And the wave function contains every information regarding the system it represents. By using the operators one can extract information regarding the observables from the wave function. And also one can theoretically predict the average value (or expectation value) of the experimental observations regarding a physical quantity by using these operators.

7.12

Advanced Engineering Physics

7.5.5

eigen Functions and eigen values: definition of eigen Function and eigen value

If an operator operates (i.e., acts) on a wave function and as a result of operation produces the same wave function multiplied by a constant factor, then the wave function is called an eigen function and the constant multiplier is called the eigen value of the operator. Let the operator act on the wave function Y and returns the same function Y multiplied by a, i.e., Y = aY ...(7.27) So, according to the definition given above Y is the eigen function of the operator and a the corresponding eigen value. If the operator represents an observable A, the measurement of this observable will yield the value a when the state of the particle is represented by the wave function Y which is an eigen function of the operator . And in this case, the wave function Y represents a state of the system which is called the eigen state of the observable A. One observable may have several eigen states. The system is found to be in any one of the eigen states of it with a characteristic eigen value whenever one performs an experiment to measure the observable. Equation (7.27), i.e., Y = a Y is called the eigen value equation. As for an example, the function d2 f (x) = sin (wx + f) is an eigen function of the operator = ___2 with an eigen value (– w2), because dx d2 ___ f (x) = 2 {sin (wx + f)} dx = (– w2) sin (wx + f) or, f (x) = (– w2) f (x) 2 So, (– w ) is the eigen value of . The time-independent Schrödinger’s equation, i.e., the equation u(x) = Eu (x) ...(7.28) with

[

]

_ h2 = – ___ —2 + V(r) is also an eigen value equation where u(x) is an eigen function of the operator 2m

(known as hamiltonian operator) corresponding to an energy eigen value E. Since the hamiltonian function H represents the total energy of the system, the function u(x) represents one energy eigen state of the system having the value of the energy equal to E. The hamiltonian operator may have many eigen functions representing energy eigen states with characteristic energy eigen values when there are several eigen states with the same eigen value, the states, in that case, are called the degenerate states. Let us consider the wave function Y = u(x) e–i (Et/h) and let operate on it. So,

Y=

[u(x) e– i (Et/h)]

or,

Y = e– i (Et/h)

or,

Y = e– i (Et/h) Eu (x) [by Eq. (7.28)]

or,

Y = E[u(x) e– i (Et/h)]

u(x)

Y = EY ...(7.29) The above equation [i.e., Eq. (7.29)] shows that a stationary state wave function Y is an eigen function of the hamiltonian operator corresponding to an eigen value E.

or,

7.13

Quantum Mechanics

7.5.6

expectation value (or expected average)

As stated earlier, wave function related to an observable is the solution of Schrödinger’s equation. An operator is used to extract information about the corresponding observable by operating on the wave function. In order to measure an observable one can conduct one single experiment on a large number of identical systems or one can measure the same (i.e., single) observable independently a large number of times. The arithemetic average of all the experimental results so obtained in either case can be theoretically predicted by making use of the wave function that is expected to contain all information about the identical systems in the former case or the system concerned in the earlier case. The average, thus, obtained is called the expected average or the expectation value or mathematical expectation of the concerned observable. If one wants to determine the expected value of the momentum ( px) of a particle (moving along the x axis) at a time t, one can perform an experiment with a large number of so moving identical particles. Let the number of so moving particles (which are described by the same wave function Y) be n. Let ni be the number of particles lying on the position denoted by x; at time t. So, the average momentum can be expressed as follows:

S ni pxi S ni pxi i i px(av) = ______ = ______ n S ni i

ni px(av) = S __ n pxi i

( )

...(7.30)

ni The average of Eq. (7.30) is the weighted average and the weight in this case is __ n . This weight (or ratio)

( )

can be regarded as the probability (Pi) of finding a particle with momentum pxi. Hence the expression for the average value can be written as follows: px(av) = S pi pxi

...(7.31)

i

The above expression of Eq. (7.31) is true for discrete distribution of particles. If the distribution of particles be continuous, then one can replace the summation symbol by the integration symbol in the expression of Eq. (7.31). And in such case, the average value px(av) of momentum can be calculated if the probability density function p(x, t) be such that the probability of finding a particle in the region between x and x + dx is p (x, t) dx at a time t. Then the average momentum of a particle can be expressed by the following equation: +

px(av) = Ú px p (x, t) dx

...(7.32)

–

Now, if the wave function is known one can easily calculate the probability density. The probability density is given by the following expression: p (x, t) = | Yn (x, t) |2 = Yn* (x, t) Yn (x, t) where Yn (x, t) is the normalized value of the Y (x, t). On substitution of this value of p (x, t) in Eq. (7.32), one can obtain +

px(av) = Ú px Yn* (x, t) Yn (x, t) dx –

7.14

Advanced Engineering Physics

This theoretically calculated average is called the expected average or expectation value of the observable which is denoted by the symbol < px >. Then the expression for the expected average can be written as +

< px > = Ú Yn* (x, t) px Yn (x, t) dx – +

< px > = Ú Yn* (x, t) px Yn (x, t) dx

or,

...(7.33)

–

Similarly one can calculate the expectation value of the position of a particle which is given by the following expression: +

< x > = Ú Yn* (x, t) Yn (x, t) dx

...(7.34)

–

In general, for any physical quantity f (x, px, t) which is a function of x, px and t, the expected value is given by +

< f (x, px, t) > = Ú Yn* (x, t) (x, px, t) Yn (x, t) dx – +

or,

< f (x, px, t) > = Ú Yn* (x, t) f( , – +

x,

( )

t) Yn (x, t) dx

)

∂ , t Y (x, t) dx < f (x, px, t) > = Ú Yn* (x, t) f x, – i h ___ ...(7.35) n – ∂x ∂ where (x, px, t) = f ( , x, t) = f x, – i h ___, t . ...(7.36) ∂x _ For the matter of a particle in three-dimensional space, the wave function Yn (r, t) of the particle is a func_ _ tion of the position vector r and time t and the physical quantities are generally functions of position vector r, _ momentum p and time t. _ _ The expected average of one such physical variable (quantity) g (r, p, t) is given by

or,

(

+

_

_ _

_

< g > = Ú Y*n (r, t) (r, p, t) Yn* (r, t) dV

...(7.37)

–

where dV is a differential volume element. For calculation of the mathematical expectation or expectation value or expected average one can follow the rule given below: In order to calculate the expected average, place the operator representing the dynamical variable _ _ to operate on the normalized wave function Yn (r) and multiply the result by Y n* (r) and then integrate over the spatial coordinates.

7.5.7

Correspondence principle or ehrenfest theorem-operator Correspondence

The notion of the expectation values (or mathematical expectations) of dynamical variables [as has been given in Eqs. (7.33), (7.34) and (7.35)] can be used to check the validity of the correspondence principle which is one of the basic postulates of quantum mechanics and known as postulate of operator-variable correspondence or simply operator correspondence. In fact this postulate states about the correspondence

Quantum Mechanics

7.15

Postulate 2 of Quantum Mechanics The classical dynamical variables relating to the motion of a particle are represented by mathematical operators in quantum mechanics. of dynamical variables of classical mechanics to the operators related to the expected averages of quantum mechanics. According to the principle of correspondence, one can expect that the relationships of classical mechanics between various dynamical variables (e.g., coordinate and momentum) of a particle will also hold good between the expected averages of these quantities for the quantum mechanical wave packet associated with the particle. This is known as Ehrenfest theorem. The measure of the positional uncertainty (Dx) of a particle is given by the width of the wave packet. Since the momentum of the particle is given by p = h k, the range of variation of the wave number (Dk) gives the measure of uncertainty in case of the momentum of the particle. For a classical particle (where large distance and momenta are involved) one can ignore the uncertainty principle in the measurement of position and momentum. It was shown by Ehrenfest that the motion of a wave packet agrees with the motion of a particle in classical mechanics if the position and momentum vectors are considered as the expected averages of these quantities. One can show that the time rate of change of the expected average of the position coordinate x of a particle corresponds to the expected average of the velocity of the same. It can be expressed as follows: d 1 __ < x > = __ ...(7.38) m < px > dt One can also show that the time rate of change of the expected average of px corresponds to the x component of the expected average of the force acting on the particle. It can be expressed as follows: ∂V d __ < px > = – ___ = < Fx > ...(7.39) dt ∂x Equation (7.39) confirms the wave packet description of a moving particle for which the time rate of change of the momentum equals to the negative of the expected average of the gradient of potential, that is equal to the impressed force acting on the particle. It can be seen from this equation the wave packet description reproduces Newton’s second law of motion. In fact, the Eqs. (7.38) and (7.39) together express the Ehrenfest theorem and it presents a correspondence between the wave packet formalism and the classical dynamics describing the motion of a particle.

7.6

Fundamental postulates oF Quantum meChanICs

In the approach of developing the theory of wave mechanics which has been done so far, it has been tried to make plausible various new ideas on the basis of underlying physical principles involved. However, there is another alternative approach in which the formal mathematical structure of wave mechanics is founded on a number of fundamental postulates. The theory can be developed more logically on the basis of these postulates and derives justification from the success of it, in accounting for a wide variety of atomic and subatomic phenomena. Let us enumerate and discuss briefly the fundamental postulates all of which have been introduced so far. Five fundamental postulates of wave mechanics, which are listed below, are applicable in general to any quantum mechanical system. But here we shall restrict ourselves to a single particle system for the sake of simplicity.

7.16

Advanced Engineering Physics _

Postulate 1 There is a wave function associated with a particle [denoted by Y (r, t) or Y (x, y, z, t)] which can completely describe the space-time behavior of the particle in consistence with the uncertainty principle. Postulate 2 In quantum mechanics, the dynamical variables related to the motion of a particle are represented by mathematical operators. Postulate 3 For a dynamical variable a, the possible results of measurement are given only by the eigen values of the operator satisfying the following eigen value equation Ye = ae Ye

where Ye is the eigen function of the operator belonging to the eigen value ae . The eigen functions Ye are single-valued in space and square-integrable. And they also form a complete set of orthogonal wave functions. Postulate 4 The probability p dV of finding a particle in the volume element dV is given by the equation p dV = Y *n Yn dV = | Yn |2 dV where Yn is the normalized wave function, p = | Yn |2 and it is called the probability density. In a finite volume V of space, the probability of finding the particle is given by

Ú p dV = Ú | Yn |2 dV V

V

And the probability of finding the particle in entire space is given by P=Ú

Y*n Yn dV = 1

all space

Postulate 5 The expected average of the results of a large number of measurements of a dynamical variable a of a particle is given by < a > = Ú Y*n v

Yn dV

where is the operator representing the variable a and Yn is normalized wave function. The operators corresponding to some observables or dynamical variables are listed in the following table: Table 7.1

Some observables and their corresponding operators.

Observable

Operator _

(a) Position

= r (or = x, = y, = z)

(b) Momentum

= – i h — or,

__

(

2 __

x

∂ = – i h ___, ∂x 2

(

2

y

∂ = – i h ___, ∂y

2

2

∂ ∂ –h –h ∂ = ____ —2 = ____ ___2 + ___2 + ___2 2m 2m ∂x ∂y ∂z _ _ (r) = V(r) or (x, y, z) = V(x, y, z)

(c) Kinetic Energy

k

(d) Potential Energy (e) Total Energy

∂ = i h __ ∂t

(f) Hamiltonian

_ – h2 __ = ____ —2 + V(r) or, 2m __

_

_

(g) Angular Momentum (L = r × p)

_

__

= – i h (r × —)

(

z

∂ = – i h __ ∂z

)

) )

∂2 ∂2 – h2 ∂2 = ____ ___2 + ___2 + ___2 + V(x, y, z) 2m ∂x ∂y ∂z

Quantum Mechanics

7.7

7.17

basIC ConsIderatIons For developIng sChrödInger’s eQuatIon

The laws of classical mechanics are not able to explain the origin of quantized motion in atomic or subatomic domain. The old quantum theory is also not adequate in this respect as the rules of quantization in this theory are introduced on ad hoc basis. According to Heisenberg, the failure of the old quantum theory is due to the formulation of this theory on the basis of some classical concepts which have no meaning in the context of quantized motion of particles. Thus, in order to overcome this discrepancy, the electronic orbits of atoms and the velocity of the rotating electrons in these orbits are considered. If, however, one tries to determine the position (or, the velocity) of the electron in the atomic orbits, one has to perform some experiment (e.g., gamma-ray microscopic experiment) which will disturb the system (i.e., the electron) and will introduce some uncertainties in the quantity to be measured as predicted by the principle of uncertainty. Thus, attempts should be made to develop atomic mechanics not on the basis of unobservable classical concepts, but on the basis of some new concepts which are in fact based on observable quantities. For example, when one performs experiments to investigate an atomic structure, one measures the energies of the atomic states through measurement of wavelengths of the spectral line, intensities, etc., and not through the positions or velocities of the electrons. So the new atomic mechanics should be built on the basis of quantities like the intensities of spectral lines and the energies of the atomic states. It was on the basis of such thinking that Schrödinger developed wave mechanics (in 1926) on the basis of de Broglie hypothesis of wave-particle duality. The concepts introduced by de Broglie to relate dynamical variables connected with the particle motion (e.g., momentum) with the characteristics of waves (e.g., wavelength) were extended by him in a more general way to a quantum system to develop a wave equation to describe the motion of atomic particles, like electrons. Schrödinger’s wave equation has become the foundation of non-relativistic wave mechanics. Later (in 1928) Dirac laid the foundation of relativistic wave mechanics. And he then generalized the quantum mechanics by unifying Heisenberg’s matrix mechanics and Schrödinger’s wave mechanics.

7.7.1 time-dependent schrödinger’s equation According to the second postulate of quantum mechanics, the dynamical variables of classical mechanics relating to the motion of a particle can be represented by mathematical operators in quantum mechanics. If one keeps this point in view, one can easily derive Schrödinger’s time-dependent equation. Let us consider a particle of mass m, moving along the x axis with a fixed momentum px and total energy E in a free region. As no force is acting on the particle, it is in a region of uniform potential energy which can be assumed to be zero. Hence, in this case, the total energy will be equal to the kinetic energy. And it is E = px2/(2m) for non-relativistic motion. Now, keeping in view, the aforesaid second postulate, one can write the corresponding quantum mechanical expression of the classical expression E = px2/(2m) as follows: 2

x = ___ 2m Now, multiplying this equation by the wave function of a free particle Y(x, t) from right side, i.e., allowing the operators to act upon Y (x, t), one gets the following equation: 2

x Y (x, t) Y (x, t) = ___ 2m

7.18

or, or,

Advanced Engineering Physics

(

)

∂ 2 – i h ___ ∂x ∂ i h __ Y (x, t) = _________ Y (x, t) 2m ∂t ∂2Y (x, t) ∂ – h2 ________ ____ = i h __ Y (x, t) 2 2m ∂ t ∂x

...(7.40)

This equation is Schrödinger’s one-dimensional time-dependent equation for a free particle having wave function Y (x, t) = Aei(kx – wt). The wave function Y (x, t) conveys the necessary information to locate the particle in time and space, so Y (x, t) is a function of x and t. Now, if the particle is not free one it will be acted upon by some external force which will be characterized by a non-zero potential energy of the particle. Let the potential energy be V(x, t). So, one can write the operator equation as 2

x + (x, t) = ___ 2m

So, the corresponding Schrödinger’s equation will be given by (– i h)2 ∂2 ∂ i h __ Y (x, t) = ______ ___2 Y (x, t) + V(x, t) Y (x, t) 2m ∂x ∂t ∂ – h2 ∂ 2 i h __ Y (x, t) = ____ ___2 Y (x, t) + V(x, t) Y (x, t) 2m ∂x ∂t

or, or,

∂2 – h2 ___ ____ ∂ Y (x, t) Y (x, t) + V(x, t) Y (x, t) = i h __ 2m ∂x2 ∂t

...(7.41)

This equation [Eq. (7.41)], is Schrödinger’s one-dimensional equation for one particle system. In a three-dimensional case, Schrödinger’s equation for one particle system is given by _ _ _ _ ∂ – h2 2 ____ — Y (r, t) + V(r, t) Y (r, t) = i h __ Y (r, t) 2m ∂t

...(7.42)

∂2 ∂2 ∂2 where —2 = ___2 + ___2 + ___2 is the laplacian operator named after the famous mathematician Laplace. By ∂x ∂y ∂z – h2 using the hamitonian operator = ____ —2 + V( , t), one can write the time-dependent Schrödinger’s equation 2m of three dimensions as _ _ ∂ Y (r, t) = i h __ Y (r, t) ∂t

...(7.43)

7.7.2 time-Independent schrödinger’s Wave equation In the last subsection we have seen that in case of time-dependent Schrödinger’s equation, the potential energy (V ) of a moving particle is a function of both time and space coordinates. But, in many physical situations the potential energy of a moving body does not depend explicitly on time. In such cases, the potential _ _ energy as well as the forces which act on the particle is a function of position (r) only. So V = V(r). In such cases, the time-dependent Schrödinger’s equation of three dimensions given in Eq. (7.42) can be rewritten as follows:

7.19

Quantum Mechanics _

∂ Y (r, t) _ _ _ – h2 2 ____ — Y (r, t) + V(r) Y (r, t) = i h _______ 2m ∂t

...(7.44) _

[since V = V(r) only] _ Equation (7.44) is a partial differential equation of variables which are functions of r and t. While the _ potential energy V is a function of position vector r only, the total energy E is a constant quantity. So the equation is separable into the time-dependent and time-independent parts. And because of this situation, the wave _ _ function Y (r, t) can be expressed as the product of two separate functions f (r) and f (t) where the former is _ _ a function of r only and the later is a function of t only. Hence Y (r, t) can be written as _ _ Y (r, t) = f (r) f (t) _ Now, putting this value of Y (r, t) in Eq. (7.44), one can get _ _ _ ∂ – h2 2 _ ____ — f (r) f (t) + V (r) f (r) f (t) = i h __ {f (r) f (t)} 2m ∂t _

Now, dividing both sides by f (r) f (t), one gets, _ _ h2 1 1 ∂ 2 – ___ ____ _ — f (r) + V (r) = i h ___ __ f (t) 2m f (r) f (t) ∂ t

...(7.45)

_

The left-hand side of Eq. (7.45) is a function r only while right-hand side is a function of t only. This is possible only when both of the sides are separately equal to a constant quantity. This constant is the total energy E of the particle. Thus, one can write two equations, when one equates the left side to E, the following equation is obtained: _ _ _ – h2 2 _ ____ — f (r) + V (r) f (r) = E f (r) 2m _ _ _ 2m or, —2 f (r) + ___ [E – V(r)] f (r) = 0 ...(7.46) 2 h This is the three-dimensional time-independent Schrödinger’s wave equation. From this equation, the onedimensional time-independent Schrödinger’s equation can be written as follows:

d2 2m ___ f (x) + ___ [E – V(x)] f (x) = 0 dx2 h2 Equation (7.46) can be again written as follows

...(7.47)

_ _ _ – h2 2 _ ____ — f (r) + V(r) f (r) = E f (r) 2m _

_

f (r) = E f (r)

or,

...(7.48)

2

where

_ –h = ____ —2 + V(r) 2m

Equation (7.48) is the eigen value equation where E and f (r) are the eigen value and eigen function respectively. Thus, when the hamiltonian operator of time-independent Schrödinger’s wave equation operates on the _ _ wave function f (r) it reproduces the same wave function f (r) multiplied by the total energy. The eigen value E of the hamiltonian operator which is only the possible value of total energy of a quantum mechanical system.

7.20

Advanced Engineering Physics

Now, equating the right side of Eq. (7.45) to E, one can get

[

d i h __ f (t) = E f (t) dt

∂ d there is only one independent variable __ = __ ∂ t dt

]

d iE __ f (t) = – ___ f (t) dt h df (t) ___ iE ____ = dt f (t) h

or, or,

Now, integrating both sides, one can get iE ln { f (t)} = – ___ t + ln c where c is a constant. h

{ }

f (t) iE ___ ln ___ c =– h t

or,

f (t) = ce– (i E/h)t _ _ Thus, from the equation Y (r, t) = f (r) f (t) one can write or,

_

...(7.49)

iEt – ____ h

_

Y (r, t) = f (r) ce

...(7.50)

This gives the solution of Schrödinger’s equation. _

iEt – ____

_

Here, f (r) and ce h are the amplitude and phase of the wave function Y (r, t) respectively. The timeindependent form of Schrödinger’s wave Eq. (7.46) is sometimes known as the amplitude equation. Every solution of the time-independent Schrödinger’s wave equation gives a definite energy value. If one _ writes, Y = Yn (r) as the solution for the energy value E = En, then the particular solution is given by _

_

Yn (r, t) = Yn (r) ce i En t/h Yn (r, t) belongs to the definite energy value En. One can find the probability of finding the particle with energy value En as follows: _ _ _ pn = | Yn (r, t) |2 = Yn* (r, t) Yn (r, t)

...(7.51)

_

_

i E nt ____ h

= Y n* (r) c*e _

_

_

pn = | Yn (r) |2 = [Yn (r)]2

or,

i Ent – ____ h

Yn (r) ce

_ Y n* (r)

[ Thus, the probability of finding the particle with energy En is independent of time t.

7.8

...(7.52) = Yn (r) and c*c = 1] _

applICatIons oF sChrödInger’s eQuatIon

Schrödinger’s equation is able to describe any system of quantum particles in space and time. As it is a differential equation, it is to be solved for finding all the information about the quantum system of particles. The solution of Schrödinger’s equation is known as Schrödinger’s wave function as this equation is basically a wave equation which represents the de Broglie wave of the system. In this section of the present chapter, attempts will be made to determine the wave function of a particle which remains trapped in a certain region

7.21

Quantum Mechanics

of space by applying the theory of Schrödinger. The possible energy values of the particles also will be calculated.

7.8.1

particle in a box with Infinitely deep potential Well

The simplest quantum-mechanical problem is that of a particle trapped in a box with infinitely hard walls. One may specify the motion of the particle along a straight line, then the problem will be one-dimensional. And when no such restriction in motion will be imposed then it will be a three-dimensional problem. The one-dimensional case will be discussed first and then it will be followed by the three-dimensional case. (i) One-dimensional case (1D box) Let us consider a particle of mass m capable of moving along the x axis only between two infinitely hard walls and face an infinitely deep potential as shown in Fig. 7.2.

L V

(b)

O

X (a)

Fig. 7.2

L

One-dimensional infinite potential well.

The variable x can assume any value between 0 and L (i.e., xmin = 0 and xmax = L). The two hard walls act as infinitely high potential barriers. The particle is said to be in a one-dimensional box. The potential energy is constant within this region of length L. But it is infinite on both sides of the box. Inside the box, the potential energy can be assumed to be zero. So, one can write V = 0 for 0 < x < L ...(7.53) V= for x ≥ L and x £ 0 ...(7.54) The particle will need infinite amount of energy to overcome these two barriers. Since a particle cannot have infinite energy, it remains confined within the box with 0 £ x £ L. It is not possible to find the particle outside the box and therefore, its wave function Y (x, t) is zero outside. So, the wave function Y (x, t) must satisfy the following condition: Y (x, t) = 0 for x ≥ L and x £ 0 ...(7.55) The wave function Y (x, t) represents a stationary wave as V is independent of time. So, it can be expressed as Y (x, t) = f (x) f (t) The time-dependent part of Y is given by the following equation: f (t) = e– i Et/h ...(7.56) Equation (7.55) implies that f (x) = 0 for x ≥ L and x £ 0 ...(7.57)

7.22

Advanced Engineering Physics

The function f (x) can be determined having solved Schrödinger’s time-independent equation. This equation is d2f – h2 ___ ____ + Vf = Ef 2m dx2

...(7.58)

As in this case, V = 0, one can write, 2 h2 d f – ___ ___2 = Ef 2m dx or,

d2f ____ 2mE ___ + 2 f=0 2 dx h The general solution of Eq. (7.59) is given by _____

(÷ )

...(7.59) _____

(÷ )

2mE 2mE f = A sin ____ x + B cos ____ x ...(7.60) 2 h h2 where A and B are two constants whose value can be determined from the boundary conditions given below: f = 0 at x = 0 and f = 0 at x = L Putting x = 0 and f = 0 in Eq. (7.60), one gets B=0 _____

\

f = A sin

(÷ ) 2mE ____ x h2

...(7.61)

Now, putting x = L and f = 0 and B = 0 in the Eq. (7.60), one can get _____

_____

÷

2mE 0 = A sin ____ L+0 h2

÷

2mE or, A sin ____ L=0 h2 In Eq. (7.62) A π 0, because in such case f (x) = 0 for all values of x. Then one can write

...(7.62)

_____

÷

2mE sin ____ L=0 h2 And hence, one can have _____

÷

2mE ____ ◊L = np h2

...(7.63)

where n = 1, 2, 3, ... Now, using Eqs. (7.63) and (7.61), one can write np f (x) = A sin ___ x L

(

)

As f (x) is dependent on n, one can write the above equation as np f n (x) = A sin ___ x L

(

)

...(7.64)

Quantum Mechanics

7.23

Squaring both sides of Eq. (7.63), one can write n2 p2 h2 E = _______ 2mL2 As E is dependent on n, one can write n2 p2 h2 En = _______ ...(7.65) 2mL2 Here, one can regard n as a quantum number which determines the possible energy values En corresponding to different states represented by the function fn (x). Now, multiplying fn (x) by the time-dependent part of the wave function, one can write Yn (x, t) = fn (x) e– i En t/h So, in general form, the wave function can be written as np Yn (x, t) = A sin ___ x e– i En t/h L

( )

...(7.66)

To find out the value of A, one has to apply the condition of normalization of the wave function as follows: +

Ú

–

Yn* Yn dx = 1

L

Ú Y *n Yn dx = 1

or,

0

Now using Eq. (7.66), one gets L

np x dx = 1 Ú A2 sin2( ___ L ) 0

or, or,

A2L ____ =1 2 __ 2 A = __ L

÷

...(7.67)

Putting this value of A in Eq. (7.66) __

÷

np 2 Yn (x, t) = __ sin ___ x e – i En t/h L L

( )

...(7.68)

This wave function represents a stationary state where the total energy of the particle is En, given by Eq. (7.65). n 2 p2 h2 Since En = _______ and n = 1, 2, 3, ..., the energy of the particle cannot be continuously distributed; the 2m L2 possible energy values (En) are discrete. The lowest energy corresponds to n = 1. \ the lowest energy is given by p2 h2 E1 = _____2 2mL

...(7.69)

7.24

Advanced Engineering Physics

One can express En in terms of E1 as n2 p2 h2 En = _______ = n2 E1 2mL2 i.e., En = n2 E1 and En + 1 = (n + 1)2 E1 Now, subtracting Eq. (7.70) from Eq. (7.71), one obtains En + 1 – En = (2n + 1) E1 2

...(7.70) ...(7.71) ...(7.72)

2

p h En + 1 – En = (2n + 1) ◊ _____2 ...(7.73) 2mL Now, in case of a macroscopic body, mL2 is very large and the difference between two successive energy levels becomes extremely small. Hence, the possible levels (in such case) seem to have a continuous distribution. So, in a macroscopic world, the energy distribution seems to be continuous. The probability density pn of the particle is given by the following equation: pn = | Yn (x, t) |2 = Y*n (x, t) Yn (x, t) or,

() ( )

np 2 pn = __ sin2 ___ x L L

or,

...(7.74) [by Eq. (7.68)]

npx npx p When sin2 ____ = 1, i.e., when ____ = (2q + 1) __ and q = 1, 2, 3, ..., the probability density pn becomes L L 2 maximum.

( )

L In other words, when x = (2q + 1) ___ for q = 1, 2, 3, 4, ..., the probability density pn becomes maximum. 2p Figure (7.3) shows the variation of Yn and | Yn |2 with respect to x for n = 1, n = 2 and n = 3. As the value of n increases, the probable positions increase, which means that as the energy of a particle increases the peak number in the probability density curve also increases. According to classical theory, the particle should be found everywhere in the box with the same probability. But according to the quantum theory, the probability of existence of the particle is maximum at the antinodes and it is minimum at the nodes. Expected average of the position The expected average of position of the particle is given by +

=Ú –

L

Y n*

Yn dx = Ú | Yn |2 x dx 0

L

or,

npx 2 < x > = __ Ú x sin2 ____ dx L0 L

or,

2 x2 x sin (2x p x/L) cos (2x p x/L) < x > = __ __ – ____________ – ___________ L 4 4n p/L 8(n p/L)2

( )

[

( ) 2

or,

2 L < x > = __ __ L 4

]

L 0

7.25

Quantum Mechanics

2

y3

E3

|y3|

y2

E2

|y2|

y1

E1

|y1|

x=0

x=L

(a)

2

2

x=0

(b)

x=L

Fig. 7.3 (a) The energy band corresponding to normalized wave functions in case of ground, first excited and second excited states in one-dimensional box. (b) Probability densities of | Y1 |2, | Y2 |2 and | Y3 |2.

L < x > = __ 2

or,

So, for any value of n, the expected average of the position (x) of the particle is always L/2 for a box of linear dimension or length L. It is clear from this fact that the average position of the particle is the middle point of the box in all quantum states of the particle. Expected average of the momentum The expected average or the mathematical expectation of the momentum of the particle is given by +

L

< px > = Ú Y*n px Yn dx = Ú Y *n px Yn dx –

0 L

or, __

(

)

∂ < px > = Ú Yn* – i h ___ Yn dx ∂x 0

i Ent n p x – ____ 2 where Yn (x, t) = __ sin ____ e h L L

÷

( )

Substituting the values of Yn and Yn* in Eq. (7.75), one obtains L

2 ihnp npx npx < px > = – _______ cos ____ dx Ú sin ____ 2 L L 0 L

( ) ( )

...(7.75)

7.26

npx ih < px > = – __ sin2 ____ L L

[

or, or,

Advanced Engineering Physics

( )]

L 0

< px > = 0 Thus, the expected average of the momentum is zero.

...(7.76)

Momentum eigen values The momentum (pxn) of a particle in a state of energy eigen value En is given by the following equation: n2 p2 h2 nph p2xn = 2mEn = 2m _______ = ____ 2 L 2mL nph pxn = ± ____ L

( )

2

( )

or,

...(7.77)

From Eq. (7.77) one can conclude that the momentum eigen values are also discrete like the corresponding energy eigen values. The (±) sign indicates that for any value of the quantum number n, the particle has momenta in two mutually opposite directions. And at any energy level En, the magnitude of the momentum nph is ____ but its direction can be either positive or negative, this means that the particle moves to and fro in L the box. The average momentum for any value of the quantum number n is given by the following equation:

( )

1 pxn (av) = __ 2 or,

[(

nph nph + ____ + – ____ L L

) (

)]

pxn (av) = 0 This is consistent with the expected average of px , i.e., < px > as can be seen in Eq. (7.76).

Eigen function of the momentum The eigen function Y n¢ for an operator is given by Y ¢n = a Y n¢ where a is the eigen value of . So for the momentum operator respect to the eigen function Y n¢ is given by x Y n¢ = pxn Y n¢

x

...(7.78)

the corresponding eigen value pxn with ...(7.79)

∂ = – i h ___. ∂x The energy eigen function is given by the following expression: Yn = fn (x) e– i Ent/h

where the momentum operator

n

is given by

x

__

or,

÷

np x 2 Yn = __ sin ____ e– i Ent/h L L

( )

This energy eigen function does not satisfy Eq. (7.79). So, the energy eigen function Yn is different from the momentum eigen function Y n¢ . eiq – e–iq One can write sin q = _______, so the wave function fn (x) can be written as 2i __

÷

npx 2 fn (x) = __ sin ____ L L

( )

7.27

Quantum Mechanics __

÷

(

inpx _____

–i n p x ______

2 e L –e L = __ ___________ L 2i __

÷

) __

÷

1 2 1 2 fn (x) = __ __ ei n p x/L – __ __ e– i n p x/L 2i L 2i L

or,

fn (x) = f+n (x) – f–n (x)

or,

...(7.80)

__

÷

1 2 where f+n (x) = __ __ ei n p x/L 2i L

...(7.81)

__

÷

1 2 and f–n (x) = __ __ e–i n p x/L 2i L

...(7.82)

Now, each of f+n (x) and f–n (x) of Eqs. (7.81) and (7.82), satisfies the eigen value Eq. (7.79). Hence f+n (x) and f–n (x) are the momentum eigen functions, i.e., eigen functions of nph respectively. ( – ____ L )

x,

nph with eigen values + ____ and L

(

)

y

(ii) Three-dimensional case (3D box) A B Let us suppose that a particle of mass m is confined in a rectany1 gular box ABCDHGFE with its edges parallel to the three axes D C (i.e., x axis, y axis and z axis) as has been shown in the diagram E x of Fig. 7.4. x1 F z 1 Let the three sides of the said box be x1, y1, and z1 respecH G tively. The particle can move freely inside the box having ranges z 0 < x < x1, 0 < y < y1 and 0 < z < z1 where the potential is zero _ (i.e., V = 0). The potential function V (r) = V (x, y, z) is having a Fig. 7.4 Particle in a three-dimensional constant value V = 0 in the regions given below: rectangular box ABCDHGFE with V (x, y, z) = 0 for 0 < x < x1, length of edges x1, y1, and z1 respectively. V (x, y, z) = 0 for 0 < y < y1 and V (x, y, z) = 0 for 0 < z < z1. And the potential outside the box (also on the walls of it) is always infinite, i.e., V (x, y, z) = for 0 ≥ x ≥ x1, V (x, y, z) = for 0 ≥ y ≥ y1 and V (x, y, z) = for 0 ≥ z ≥ z1 The time-independent Schrödinger’s wave equation for the particle inside the box is given by the following equation:

or,

2m E —2 Y + _____ Y=0 h2 ∂2 Y ____ ∂2 Y ∂ 2Y ______ 2m E Y ____ + 2 + ____ + =0 2 2 ∂x ∂y ∂z h2

[ V = 0] ... (7.83)

Equation (7.83) can be solved by using the method of separation of variables. Let us assume that the function Y can be expressed as the product of three variables X, Y and Z where X = X(x), Y = Y (y) and Z = Z(z). So, each of the three functions X, Y and Z depends only on one coordinate.

7.28

\

Advanced Engineering Physics

Y (x, y, z) = X(x) Y(y) Z(z) 2

Hence,

...(7.84)

2

∂ Y d ____ = YZ ___2 X, 2 ∂x dx

∂ 2Y ∂2 d2 d2 ____ = ZX ___2 Y and ___2 = XY ___2 Z 2 ∂y dy ∂z dz Now, substituting these in Eq. (7.82) and then dividing by (XYZ), one gets d2 2mE 1 ___ 1 d2 1 d2 __ X + __ ___2 Y + __ ___2 Z + ____ =0 2 X dx Y dy Z dz h2

...(7.85)

Now, it is to be noted that each of the terms of this equation [i.e., Eq. (7.85)] depends on a different variable and the three variables are independent of each other. And the last term of the Eq. (7.85) is a constant. The only way for this equation to be valid for all the variables of x, y and z in the intervals (0, x1), (0, y1) and (0, z1) respectively is to be a constant. That is each differential term in the equation will be separately constant. Thus, one can express them as d2X 1 ____ __ = – a2, X dx2

d2Y 1 ___ __ = – b2 Y dy2

1 d2Z and __ ___2 = – g 2 Z dz where, a, b and g are some constants. \

2mE a2 + b2 + g 2 = ____ h2 These three equations with dependent variables X, Y and Z can be rewritten as follows: d2X ____ + a2X = 0 dx2

...(7.86)

d2Y ___ + b2Y = 0 2 dy

...(7.87)

d2Z ___ + g 2Z = 0 ...(7.88) dz2 The general solutions of Eqs (7.86), (7.87) and (7.88) are given below: X = A1 sin a x + B1 cos a x, Y = A2 sin b y + B2 cos b y and Z = A3 sin g z + B3 cos g z The values of the constants A1, A2, A3, B1, B2 and B3 can be obtained by applying the boundary condition. The boundary conditions require that the wave function vanishes at the box walls where the potential is infinite (as stated earlier). So, Y (0, y, z) = Y (x, 0, z) = Y (x, y, 0) = 0 ...(7.89) and Y (x1, y, z) = Y (x, y1, z) = Y (x, y, z1) = 0 ...(7.90)

7.29

Quantum Mechanics

Now, by applying the boundary conditions of relations (6.89) to the above equations of X, Y and Z, one gets, B1 = B2 = B3 = 0 Again by applying the boundary conditions of relations (7.90) to the above equations of X, Y and Z, one gets, sin (a x1) = 0 fi a x1 = nx p or a = nx p/x1, sin (b y1) = 0 fi b y1 = ny p or b = ny p/y1 and sin (g z1) = 0 fi g z1 = nz p or g = nz p/z1 where nx, ny and nz are integers and none is equal to zero. nx p x Hence, X = A1 sin _____ ...(7.91) x1 ny p y Y = A2 sin _____ y

...(7.92)

nz p z Z = A3 sin _____ z

...(7.93)

1

and

1

Now, substituting these values of X, Y and Z in Eq. (7.84), one gets ny p y nz p z nx p x _____ _____ Y (x, y, z) = A1 A2 A3 sin _____ sin x1 y1 sin z1 ny p y nz p z nx p x _____ _____ or, Y (x, y, z) = A sin _____ sin x1 y1 sin z1 where A = A1 A2 A3 and A is the normalization constant. Now, A can be obtained by using the normalization condition as given below. x1 y1 z1

Ú Ú Ú Y * Y dx dy dz = 1 000 x1 y1 z1

or,

A2 Ú Ú Ú sin2 000

(

ny p y nx p x n2 p z 2 _____ 2 _____ _____ dx dy dz = 1 z1 x1 sin y1 sin

) (

) (

) x y z A ◊ ( __ ) ( __ ) ( __ ) = 1 2 2 2 2

or,

1

1

1

__

2÷2 ________ A = _________ ÷(x1 y1 z1)

or,

So, the normalized wave function Yn (x, y, z) is given by the following expression: __ ny p y nz p z nx p x 2÷2 _________ _____ Yn (x, y, z) = ________ sin _____ sin sin _____ z1 x y 1 1 (x y z )

÷

1 1 1

Now, one has

or,

2 mE a 2 + b 2 + g 2 = _____ h2 2 2 ny p n2z p 2 ____ n2x p 2 _____ 2mE _____ _____ + + = 2 2 2 2 x1 y1 z1 h

(

) (

) ( )

...(7.94)

7.30

Advanced Engineering Physics

[

ny2 nz2 nx2 ___ h2 p 2 __ _____ E= + + ___ 2m x12 y12 z12

or,

]

...(7.95)

Now, it is clear from the above equation [i.e., Eq. (7.95)] that E is dependent on nx, ny and nz, so one can make use of the suffixes with E and rewrite Eq. (7.93) as follows: ny 2 nz 2 h2 nx 2 __ Enx ny nz = ___ __ + y + __ ...(7.96) z1 x 8m 1 1

[( ) ( ) ( ) ]

It can be stated from the above discussion that Yn (x, y, z) represents the state of the particle with total energy Enx ny nz as given in Eq. (7.96). The three integers nx, ny and nz are called the quantum numbers and are required to specify each energy state completely. For each set of values of the constants nx, ny and nz, one gets a new wave function Yn (x, y, z) which can be represented by Yn(nx ny nz)(x, y, z). And since Yn (x, y, z) cannot be zero within the box, no quantum number of the set (nx, ny, nz) can be zero. So, the ground state energy is given by h2 1 1 1 Eg = E111 = ___ __2 + __2 + __2 8m x1 y1 z1

(

)

...(7.97)

The corresponding wave function of energy E111 is given by _______

Yn(111) =

py pz px 8 ______ ___ ___ ___ x1 y1 z1 sin x1 sin y1 sin z1

(÷

) ( ) ( ) ( )

...(7.98)

The momentum of the particle in a state is given by _________

÷

pnx ny nz = 2m Enx ny nz _______________

÷(

or,

2 2 2 h2 nx ny nz pnx ny nz = __ ___2 + __2 + __2 4 x1 y1 z1

or,

h pnx ny nz = __ 2

)

_________________

ny 2 nz nx 2 __ __ __ + x1 y1 + z1

÷( ) ( ) ( )

2

Degeneracy of the energy states If one considers a cubical box, then x1 = y1 = z1. Let x1 = y1 = z1 = L Then the wave function of the time-independent Schrödinger’s equation and its corresponding energy value are given by the following two equations respectively: ___ ny p y nz p z nx p x 8 Yn (nx ny nz) (x, y, z) = __3 sin _____ sin _____ sin _____ ...(7.99) L L L L

÷

and

(

) (

h2 Enx ny nz = _____2 (nx2 + ny2 + nz2) 8mL

) ( )

...(7.100)

As can be seen from Eq. (7.100), in a cubical box, the total energy Enx ny nz µ (nx2 + ny2 + nz2), i.e. the total energy is proportional to the sum of squares of the three quantum numbers nx, ny and nz. And the particle in such a cube can have the same value of energy for more than one set of values of the quantum numbers (nx, ny, nz). This means that more than one wave functions can have the same energy value. The state represented by such a set of wave functions, is known as denegenerate state.

7.31

Quantum Mechanics

Let us take help of an example to understand the concept of degeneration more clearly. Let us consider the following three sets of values of nx, ny and nz. Table 7.2

Some quantum numbers, energy levels and their wavefunctions.

nx

ny

nz

nx2 + ny2 + nz2

E

Yn

1 1 2

1 2 1

2 1 1

6 6 6

E112 E121 E211

Yn(112) Yn(121) Yn(211)

3h2 For the aforementioned three different states, the particle has the same energy _____2 . So, one can write 4mL 2 3h E112 = E121 = E211 = _____2 ...(7.101) 4mL For these three sets of quantum numbers (i.e., values of nx, ny and nz) the wave functions are quite different. And hence one can write the following inequation Yn(112) π Yn(121) π Yn(211) Such energy states like Yn(112), Yn(121) and Yn(211) related to the same value of energy are called degenerate states. Since in these cases the wave functions are different, the probability densities in these three states are 3h2 different from each other. The states having the energy equal to _____2 are considered to have a degeneracy 4mL of three fold. The number of the independent wave functions (and hence the number of independent states) corresponding to the same value of energy is called the degeneracy of the energy states. In case of a one-dimensional box, the degeneracy of energy states is not observed. For a particle in a cubical box, the degree of degeneracy for few energy states (i.e., energy levels) are presented in the Table 7.3 which follows: Table 7.3

Some energy levels with their degenerate quantum number sets.

Energy Levels

Energy Values

Quantum Numbers

Degree of Degeneracy

E111

3h2 _____ 8mL2

(1, 1, 1)

No degeneracy

E112, E121, E122

6h2 _____ 8mL2

(1, 1, 2), (1, 2, 1), (2, 1, 1)

Three-fold degeneracy

E221, E212, E122

9h2 _____ 8mL2

(2, 2, 1), (2, 1, 2), (1, 2, 2)

Three-fold degeneracy

E311, E131, E113

11h2 _____ 8mL2

(3, 1, 1), (1, 3, 1), (1, 1, 3)

Three-fold degeneracy

E222

12h2 _____ 8mL2

(2, 2, 2)

No degeneracy

E123, E132, E231,

14h2 _____ 8mL2

(1, 2, 3), (1, 3, 2), (2, 3, 1),

Six-fold degeneracy

E213, E312, E321

(2, 1, 3), (3, 1, 2), (3, 2, 1)

7.32

Advanced Engineering Physics

The stationary wave function (or states) Yn(112), Yn(121) and Yn(211) for the particle in a cubical box of dimension L3 are degenerate. The linear combination of them Yn = C1 Yn(112) + C2 Yn(121) + C3 Yn(211) is also 3h2 an eigen function related to the same energy value _____2 . 4mL

( )

Worked Out Problems

Wave Function and probability Example 7.1

A particle which is moving along the y axis has the wave function Y (y) as given below: Y (y) = cy for 0 £ y £ 1 Y (y) = 0 for 0 ≥ y ≥1

Find the probability that the particle lies between y = 0.35 and y = 0.75. Sol. The probability of finding a particle represented by the wave function Y (y) in the length dy between y and y + dy is given by Y *(y) Y (y) dy = | Y (y) |2 dy Hence, the probability within the range 0.35 £ y £ 0.75 is given by 0.75

P=

Ú

0.75

Ú

| Y (y) |2 dy = c2

0.35

y2 dy

0.35

2

2

c c = __ [y3]0.75 = __ [0.422 – 0.043] 0.35 3 3 \

P = 0.126 c2

Example 7.2 Consider the wave function Y = Aea x + i b t where A, a and b are real positive quantities. Could it represent the wave nature of a particle? Sol. Y to be an acceptable wave function, it must satisfy the normalization condition, i.e., we must get +

Ú

| Y (x, t) |2 dx = 1

...(1)

–

In the present case, we get +

Ú

–

+

+

| Y | dx = Ú Y Y dx = Ú A2 e2a x dx 2

*

–

–

2

+ A = ___ [e2a x]– = 2a

By comparing the result with Eq. (1), we realize that the function Y is not a normalizable wave func+

tion as Ú | Y |2 dx = . So, the given function cannot represent the wave nature of any particle. –

7.33

Quantum Mechanics

Example 7.3 (i) Show that Y = Ax + B where A and B are constants, is a solution to Schrödinger’s equation for E = 0 energy level of a particle in a box. (ii) Show, however, that the probability of finding a particle with this wave function is zero. Sol. (i) Schrödinger’s equation for a particle in a box is given by d2 Y ___ 2m ____ + 2 EY=0 2 dx h

...(1)

d2 Y Now, differentiating Y = Ax + B twice with respect to x gives ____ = 0 for the left-hand side dx2 of the Schrödinger’s equation. Also by putting E = 0 in Eq. (1), we get the same equation. So Y = Ax + B is a solution to this Schrödinger’s equation for E = 0. (ii) The wave function Y equals to zero outside the box (x < 0 and x > L). In order that Y = Ax + B may be continuous at x = 0, it must be true that Y = 0 at x = 0. So A ◊ 0 + B = 0 or B = 0. Similarly, in order that Y may be continuous at x = L, it must be true that Y = 0 for L = 0. So, A ◊ L + 0 = 0 or A = 0. With both A and B equal to zero, Y = Ax + B = 0. Thus, the wave function is equal to zero inside the box as well as outside the box and probability of finding the particle anywhere with this wave function is zero. Example 7.4 A particle is represented by the wave function Y (x) = e– | x | sin a x. What is the probability that its position to the right of the point x = 1 (i.e., the point (1, 0, 0))? Sol. Given Y (x) = e– | x | sin a x Let, Y1(x) = ex sin (a x) for x < 0 and Y2 (x) = e– x sin (a x) for x > 0 +

\

Ú

–

1 | Y (x) |2 dx = __2 , c

where c is the normalization constant.

0

+

Ú | Y1 (x)|2 dx + Ú

or,

– 0

or,

Ú

–

0 +

1 e2x sin2 a x dx + Ú e–2x sin2 a x dx = __2 0 c 0

\

1 | Y2(x)| dx = __2 c

[

]

+

[

]

1 – cos 2a x 1 – cos 2a x L.H.S = Ú e2x __________ dx + Ú e – 2x __________ dx 2 2 – 0 –

+

1 1 1 = __ – __ Ú e2x cos 2a x dx – __ Ú e – 2x cos 2a x dx 2 20 20 1 1 1 = __ – _________ – _________ 2 4 (1 + a2) 4 (1 + a2) \

2 1 _________ 1 __ – = __ 2 4 (1 + a2) c2

or,

1 + a2 – 1 __ 1 _________ = 2 2 2 (1 + a ) c

7.34

Advanced Engineering Physics ________

÷

2(1 + a2) c = ________ a2 So, the normalized wave function is given by \

_________

÷

2 (1 + a2) – | x | Y (x) = _________ e sin a x a2 The required probability is given by 1 P = Ú | Y |2 dx = __2 Ú e – 2x sin2 a x dx 1 c 1 e–2 = _________ [1 + a2 – cos 2a + a sin 2a] 4 (1 + a2)

normalization and orthogonality Example 7.5 A particle is in a cubic box with infinitely hard walls whose edges are L units long. The wave function of the particle is given by ny p y nz p z nx p x Y = A sin _____ sin _____ sin _____ L L 2

) (

(

) ( )

Find the value of the normalization constant A. Sol. The condition for normalization is that the value of the normalization constant should satisfy the following condition: +

Ú

Y (x, y, z) Y (x, y, z) dV = 1

–

i.e., the total probability in the entire space must be 1. +

Ú

Y (x, y, t) Y (x, y, z) dx dy dz = 1 where dV = dx dy dz

–

+

Ú

or,

– +

or,

A2

Ú

–

ny p y nz p z nx p x A2 sin2 _____ sin2 _____ sin2 _____ dx dy dz = 1 L L L

) (

(

) ( ) ( ) Ú ( )

+ + ny p y nz p z nx p x sin2 _____ dx Ú sin2 _____ dy sin2 _____ dz = 1 L L L – –

(

)

Since the wave function is zero outside the box, we can write L L L ny p y nz p z nx p x A2 Ú sin2 _____ dx Ú sin2 _____ dy Ú sin2 _____ dz = 1 L L L 0 0 0 or, or,

( ( ) L L L A ( __ ) ( __ ) ( __ ) = 1 2 2 2 2

8 A2 = __3 L

___

\

÷

8 A = __3 . L

)

( )

7.35

Quantum Mechanics

m px n px Two wave functions Ym (x) = A sin _____ and Yn (x) = B sin ____ represent the motion 2 2

(

Example 7.6

)

(

)

of a particle confined in a one-dimensional box with perfectly hard walls. A and B are normalization constants. Show that if m and n are two integers and m π n, then the two given wave functions are orthogonal to each other. +

Sol.

Let

I=

Ú

Ym* (x) Yn (x) dx

Ú

m px n px A sin _____ B sin ____ dx L 2

– +

or,

I=

–

(

)

(

)

L

m px n px I = AB Ú sin _____ sin ____ dx L L 0

(

or,

) (

)

[ the wave functions must be zero outside the box] L

or,

m px n px AB I = ___ Ú 2 sin _____ sin ____ dx L L 2 0

or,

AB I = ___ 2

(

L

Ú 0

) (

)

[ cos __Lp (m – n)x – cos __Lp (m + n)x ] dx

[

or,

(m + n) p AB L L I = ___ ________ sin __ (m – n)x – ________ sin p _______ x L L 2 p (m – n) p (m + n)

or,

AB I = ___ × 0 2

\

I=0 +

i.e.,

Ú

–

Y m* (x) Yn (x) dx = 0

Hence, Ym (x) and Yn (x) are orthogonal to each other.

Quantum mechanical operators Example 7.7 (a) Sol.

Determine the expression for the following two operators:

( ___dxd + x )

(

2

d (b) ___ (cos x) dx

Let Y be the corresponding wave function. (a)

( ___dxd + x ) Y = ( ___dxd + x ) ( ___dxd + x ) Y d dY = ( ___ + x ) ( ____ + x Y ) dx dx 2

d 2Y dY dY = ____ + x ____ + Y + x ____ + x2 Y dx dx dx2

)

]

L 0

7.36

Advanced Engineering Physics

d d + 2x ___ + (x + 1) }Y ( ___dxd + x ) Y = { ___ dx dx d d + 2x ___ + (x + 1) ( ___dxd + x ) = ___ dx dx ( ___dxd cos x ) Y(x) = ___dxd (cos x Y (x)) 2

2

or,

2

2

2

2

\

2

2

(b)

d d = cos x ___ Y + Y ___ (cos x) dx dx d = cos x ___ Y – Y sin x dx or, Hence Example 7.8

( ___dxd cos x ) Y = ( cos x ___dxd – sin x ) Y ( ___dxd cos x ) = cos x ___dxd – sin x

Obtain the expression for the eigen function of the momentum operator

responding to an eigen value px . Sol. The operator equation for calculation of the eigen function is given by xY = px Y. where Y is an wave function satisfying the corresponding Schrödinger’s equation. d \ – i h ___ Y = px Y dx or, or, or, or,

dY – i h ____ = px dx Y dY – i h Ú ____ = px x + c where c is the constant of integration dx px x c ln Y = – ____ – __ ih ih p c x ln Y = – __ x + ln a where ln a = – __ ih ih

[

]

or, Y = a e(ipx x)/h This is the eigen function of x. d If = cos x and ___, then show that dx Let Y be a wave function. d So, [ , ] Y = cos x, ___ Y dx d d = cos x ___ – ___ cos x Y dx dx

Example 7.9 Sol.

[ [

]

]

and

do not commute.

x

d = – i h ___ cordx

7.37

Quantum Mechanics

dY d = cos x ____ – ___ (Y cos x) dx dx

\ \ \

and

dY d cos x dY = cos x ____ – Y ______ – cos x ____ dx dx dx = + Y sin x [ , ] = sin x [ , ] π 0 (in general) do not commute.

Example 7.10 Sol.

Prove that [ , ] = – [ , ] where and are two operators. Let Y be a wave function corresponding to both of the operators. \ [ , ]Y=( – )Y= Y– Y Again [ , ]Y=( – )Y= Y– Y =–( – )Y = – [ , ] Y [by Eq. (1)] Hence [ , ]=–[ , ]

Example 7.11 Sol.

If [ , z] = i h, then prove that (a) [ x, ] = – i h and (b) [ x, z] = – i h We know that x = z– y Now, (a) [ x, ] = [ z – y, ] = [ z, ] – [ y, ] = z – z– z + y = z – z+ z– z– y + y = ( z – z) + ( – ) z – ( y – y) = [ z, ] + [ , ] z – [ y, ] = – [ , z] + 0 – 0 = – ih (b) x = z– y \ [ x, z] = [ z – y, z] = [ z, z] – [ y, z] = z z– z z– y z+ z y = ( z – y) z – y z + z y = [ , z] z – y z + z y – z y + z y or, [ x, z] = [ , z] z – ( y z – z y) – ( z – z ) y = [ , z] z – [ y, z] – [ . z] y = 0 + 0 – [ , z] y = – (i h) y \ [ x, z] = – i h y

...(1)

y.

7.38

Advanced Engineering Physics

Example 7.12 Sol.

(a)

Show that (a) [ x, ] = 0; (b) [ x, x] = 0; (c) [ x, x = z– y Now [ x, ] = x – x or, [ x, ] = ( z – y) – ( z – y) or, Now,

[ x, ] = – i h [ x, ] Y = – i h = –ih

or, Hence (b) Now or, or,

Now

y]

= –ih

[( [( [( [

∂ ∂ ∂ ∂ y __ – ___ x – x y __ – z ___ ∂z ∂y ∂z ∂y

[( [( [ [

∂2 ∂ ∂2 ∂ ∂ y _____ – z _____ – ___ y ___ – z ___ ∂z∂x ∂y∂x ∂x ∂z ∂y

) ( ) ( ) (

∂ ∂ ∂ ∂ y __ – z ___ x – x y __ – z ___ ∂z ∂y ∂z ∂y

)] )]

z.

Y

) ] ]

∂ ∂ ∂ ∂ y __ – z ___ x Y – x y __ – z ___ Y ∂z ∂y ∂z ∂y

∂Y ∂Y ∂Y ∂ = – i h xy __ Y – zx ____ – xy ____ + zx ____ ∂z ∂y ∂z ∂y =0 [ x, ] Y = 0 [ x, ] = 0. x = z– y [ x, x] = x x – x x [ x, x] = ( z – y) x – x ( z – y) x]

= – h2

n] Y

= – h2

[ x,

[ x,

)] ) ] ( ) ( )]

) ( ) (

∂2 ∂ ∂2 ∂ ∂ y _____ – z _____ Y – ___ y __ – z ___ Y ∂z∂x ∂ y ∂x ∂ x ∂z ∂y

∂2Y ∂ ∂Y ∂ ∂Y d2Y = – h2 y _____ – z _____ – ___ y ____ + ___ z ____ ∂z ∂x ∂y ∂x ∂x ∂z ∂x ∂y

or,

[ x, n] Y [ x, n] [ x, y] Y [ x, y] Y

or,

[ x,

or, Hence (c)

y] Y

∂2Y ∂ 2Y ∂2Y ∂2Y = – h2 y _____ – z _____ – y _____ + z _____ ∂z ∂x ∂y ∂x ∂x ∂z ∂x ∂y

]

=0 =0 = 0. = ( x y – y x)Y = ( z – y) ( x –

–

= – h2

[(

∂ ∂ y __ – z ___ ∂z ∂y

z)Y

)(

–(

x

–

z)

(

) (

z

y)Y

∂ ∂ ∂ ∂ z ___ – x __ Y – z ___ – x __ ∂x ∂z ∂x ∂z

)(

)]

∂ ∂ y __ – z ___ Y ∂z ∂y

7.39

Quantum Mechanics

or,

[ x,

y] Y

(

)

(

∂Y ∂Y ∂ ∂Y ∂ ∂Y = – h2 y __ z ____ – x ____ – z ___ z ____ – x ____ ∂z ∂x ∂z ∂y ∂x ∂z

(

)

(

∂Y ∂Y ∂ ∂Y ∂ ∂Y – z ___ y ____ – z ____ + x __ y ____ – z ____ ∂x ∂z ∂y ∂z ∂z ∂y or,

[ x,

y] Y

)

)

∂2Y ∂2Y ∂2Y ∂Y ∂2Y = – h2 y ____ + yz _____ – 0 – xy ____ – 0 – z2 _____ + 0 + zx _____ – 0 2 ∂x ∂z ∂x ∂y ∂x ∂y ∂z ∂z ∂2Y ∂2Y ∂2Y ∂2Y ∂Y – yz _____ + 0 + z2 _____ + 0 + xy ____ – x ____ – zx _____ 2 ∂x ∂z ∂x ∂y ∂y ∂y ∂z ∂z

]

[ x,

y] Y

∂Y ∂Y = – h2 y ____ – x ____ ∂x ∂y

or,

[ x,

y] Y

∂Y ∂Y = (– i h) y (– i h) ____ – x (– i h) ____ ∂x ∂y

or, or,

[ x, [ x,

y] Y

Example 7.13 (a)

Sol.

[

or,

[

]

= (– i h) [ x – y] ] Y = – i h z [ z = ypx – xpy] y

∂ ∂ Evaluate the following operators which are combinations of and ___, or __ and t: ∂x ∂t

[ ] ∂ , ___ ∂x

[

∂2 (b) ( )2, ___ ∂x

[

]

∂2 (c) ( )3, ___2 ∂x

)

[ ] (

]

∂ ∂ ___ ___ – Y ∂x ∂x ∂Y ∂Y ∂Y ∂ = x ____ – ___ (xY) = x ____ – x ____ – Y ∂x ∂x ∂x ∂x

∂ , ___ Y = ∂x

(a)

or, or, \

[ ] [ ] [ ]

∂ , ___ Y = – Y = (– 1)Y ∂x ∂ , ___ Y = ∂x

n

Y, where un = – 1

∂ , ___ = ∂x

n

(= unit negative operator)

[ ] ( ∂2 , ___2 Y = ∂x

(b)

∂2 ___ ∂2 ___ – ∂x2 ∂x2

)

Y

∂2 ∂2 = x ___2 Y – ___2 (xY) ∂x ∂x ∂2Y ∂2Y ∂2 ____ ___ = x ____ – x – Y (x) ∂x2 ∂x2 ∂x2 =0 \

[ ]

∂2 , ___2 = 0 ∂x

[ ]

∂ (d) t, __ ∂t

7.40

Advanced Engineering Physics

[

(c)

] {

}

∂2 ∂2 ∂2 ( )3, ___2 Y = ( )3 ___2 – ___2 ( )3 Y ∂x ∂x ∂x ∂2 ∂2 = x3 ___2 Y – ___2 (x3 Y) ∂x ∂x ∂2 ∂2Y ∂2 = x3 ___2 Y – x3 ____ – Y ___2 (x3) 2 ∂x ∂x ∂x = – 6xY

[

\

]

∂2 ( )3, ___2 = – 6 ∂x

[ ] (

)

∂ ∂ ∂ t, __ Y = t __ – __ t Y ∂t ∂t ∂t

(d)

∂Y ∂Y = t ____ – t ____ – Y ∂t ∂t or,

[ ]

∂ t, __ Y = – Y = ∂t

where un = –1 and

n

n

Y

is unit negative operator

[ ]

∂ t, __ = ∂t

\

eigen Functions and eigen values By the principle of eigen values and eigen functions, we get Y = a Y where numerical eigen value.

is the operator and a is its

Example 7.14 Sol.

Find the eigen function and eigen value of the momentum operator. The momentum operator is given by ∂ = – i h ___ ∂x Let px be the eigen value of x. x

\ or, or, or, or, or,

∂ – i h ___ Y = px Y(x) ∂x d Y (x) – i h ______ = px dx [Since there is no mention of other independent variables] Y(x) d Y (x) – i h Ú ______ = px Ú dx Y (x) – i h ln Y (x) = px x + c1,

where c1 is the constant of integration

i i ln Y (x) = __ px x + __ c1 h h i i ln Y (x) = __ px x + c2 where c2 = __ c1 h h

7.41

Quantum Mechanics

( )

ipx x Y (x) = ce ____ where c2 = ln c h which is the required eigen function. The eigen value px can be any number. On putting the boundary conditions such that Y (x) be a periodic function in some distance L (periodic potential). e(i px x/h) = eipx (x + L)/h Putting x = 0, we get 1 = eipx L/h px L px L or, 1 = cos ____ + i sin ____ h h or,

( ) ( )

( ) ( ) ( )

( )

px L px L or, cos ____ + i sin ____ = 1 + i 0 h h px L Hence cos ____ = 1 h px L cos ____ = cos (2n p) h

or,

where n = 1, 2, 3, .... 2n p h px = _____ L

or,

The eigen function is now given by Y (x) = ce(ipx x/h) = cei(2p n x/L) So, the eigen functions are discrete and real.

(

1 __

2

(

)

2

)

d Example 7.15 Show that the function Y (x) = cx e – 2 x is an eigen function of the operator x2 – ___ . 2 dx Find the eigen value. Sol. By the definition of eigen function and eigen value, one gets Y (x) = a Y (x) where a is eigen value of , i.e., the operator returns the same function multiplied by the eigen value.

(

d2 Now, x2 – ___2 dx

)

{ cx e } 1 – __x2 2

{

}

1 – __ x2 d2 – c ___2 xe 2 dx 1 2 1 1 __ – __x2 – __ x2 d 3 – 2x ___ = cx e –c e 2 – x2 e 2 dx 1 – __ x2 2

= cx3 e

{

1 – __ x2 2

= cx3 e

1 – __x2 2

+ cx e

1 – __ x2 2

+ 2 cx e

}

1 – __x2 2

= 3 cx e

= (3) Y (x) where Y (x) = cx e

1 – __ x 2

– cx3 e

1 – __x2 2

7.42

Advanced Engineering Physics

(

)

d2 or, x2 – ___2 Y (x) = (3) Y (x) dx 1 – __ x2 \ the eigen value of the eigen function Y (x) = cx e 2 is 3. Hence, cx e

1 – __x2 2

(

)

d2 is an eigen function of the operator x2 – ___2 . dx

expected averages Example 7.16

The wave function of a particle in a one-dimensional box of length L is given by __

÷

np x 2 Yn = __ sin ____ L L Find the expected average or expectation value of x and x2. Sol. The expected average of x is given by L

< x > = Ú Y n* (x) Yn (x) dx 0 L

npx 2 < x > = __ Ú x sin2 ____ dx L0 L L or, < x > = __ 2 The expected average of x2 is given by or,

L

< x > = Ú Y n* (x) ( )2 Y (x) dx 0 L

or,

npx 2 < x2 > = __ Ú x2 sin2 ____ dx L0 L L2 L2 = __ – _____ 3 2x2p2

\

(

1 1 < x2 > = L2 __ – _____ 3 2x2 p2

[

)

]

2(1 + a2) __12 – | x | Y (x) = ________ e sin a x is the wave function for a particle. Calculate the expected a2 average of position x and square of the momentum px. Sol. We have, Example 7.17

[ [

2 (1 + a2) Y (x) = _________ a2 and

2 (1 + a2) Y (x) = _________ a2

] ]

1 __ 2 ex

sin a x, for x < 0

1 __ 2 e–x

sin a x,

for x > 0

7.43

Quantum Mechanics +

Now

=

Ú

Y *(x) Y (x) dx

–

or, \

or, \

+

0

Ú x e2x sin2 a x dx + Ú

–

]

xe–2x sin2 a x dx = 0

0

=0 +

Also,

[

2 (1 + a2) < x > = _________ a2

2

2

< (px) > = – h

Ú

d2 Y *(x) ___2 Y (x) dx dx

[

d d (ex sin a x) dx + Ú e–x sin a x ___2 (e–x sin a x) dx Ú ex sin a x ___ 2

–

< (px)2 > = – h2

0

2

2

dx

–

0

dx

]

< (px)2 > = h2(1 + a2)

Example 7.18 For the one-dimensional motion of a particle with normalized wave function is Yn = A sin a x. Calculate the expected average of the kinetic energy for this particle which is confined in a region between x = 0 and x = 10. Sol. The kinetic energy operator is given by 2

or,

(

k

d 1 x = ___ – i h ___ = ___ 2m 2 m dx

k

– h2 d2 = ____ ___2 2m dx

) ( – i h ___dxd )

\ the expected average of the kinetic energy is 10

< Ek > = Ú Y n*

k

Y dx

0 10

or,

d2 – A2 h2 < Ek > = ______ Ú sin a x ___2 (sin a x) dx 2m 0 dx

or,

a2 A2 h2 < Ek > = _______ Ú sin2 a x dx 2m 0

or,

a2 A2 h2 1 – cos 2a x < Ek > = _______ Ú __________ dx 2m 0 2

or,

a2 A2 h2 sin 2a x < Ek > = _______ x – _______ 4m 2a

10

10

\

[ ] a A h sin 20a < E > = _______ [ 10 – _______ ] 4m 2a 2

k

2

2

10 0

7.44

Advanced Engineering Physics

miscellaneous problems Example 7.19 Compute the lowest energy of a neutron confined to the nucleus of an atom where the nucleus is considered a box with size of 10–14 m. (h = 6.26 × 10–34 Js, m = 1.6 × 10–24 g). Sol. Consider the nucleus as a cubical box of size 10–14 m. \ x = y = z = 10–14 m = l (say) For the neutron to be in the lowest energy state nx = ny = nz = 1 2 ny2 nz2 p 2 h2 nx __ Now, E = _____ __ + + ___ [ lx = ly = lz = l] 2m lx2 ly2 lz2

[

or, or, \

]

p 2 h2 3 E = _____ __2 2m l

[]

3 × (6.62 × 10–34)2 3h2 E = ____2 = ___________________ 8ml 8 × 1.6 × 10–27 × 10–28 E = 10.29 × 10–23 J = 6.43 MeV

Show that Y = f e – i w t is a wave function of a stationary state. For Y to be the wave function of a stationary state, the value of | Y |2 at each point in space must be constant, i.e., independent of time. Now, | Y |2 = Y * Y = (f e – i w t)* (f e – i w t) or, | Y |2 = (f*) (e i w t) (f e – i w t) or, | Y |2 = | f |2 e° = | f |2 where | f |2 is not a function of time, so, | y |2 is also independent of time. Hence, Y = f e – i w t is a wave function of stationary state.

Example 7.20 Sol.

Example 7.21 Find the probability that a particle trapped in a box of size L (units) can be found between 0.45 L and 0.55 L for the ground and the first excited states. Sol. The part between 0.45 L and 0.55 L of the box is one-tenth of the size of the box and is centered on the middle of the box. Classically, we could expect the particle to be in this region 10 per cent of the time. The quantum mechanics gives quite different predictions that depend on the quantum number of the particles state. The probability of finding the particle between x1 and x2 when it is in the nth state is given by, x2

2 P = Ú | Yn |2 dx = __ L x1 or,

[

x2

npx dx Ú sin2 ____ L

x1

2n p x x 1 P = __ – ____ sin _____ L 2np L

(

x2

)]

x1

For the ground state n = 1 \ we have P = 0.198 = 19.8% For the 1st excited state n = 2 \ P = 0.0065 = 0.65% The low figure is consistent with the probability density | Yn |2 = 0 at x = L.

Quantum Mechanics

7.45

Example 7.22

A system has two energy eigen states e0 and 3e0. Y1 and Y2 are the corresponding normal1__ ized wave functions. At an instant the system is in a superposed state Y = c1Y1 + c2Y2 and c1 = ___ . ÷2 (i) Find the value of c2 if Y is normalized. (ii) What is the probability that an energy measurement would yield a value of 3e0? (iii) Find out the expectation value of the energy. [WBUT 2007] +

Sol.

(i)

Ú

Y *(x) Y (x) dx = 1

–

+

or,

Ú

–

(c1Y1 + c2Y2)* (c1Y1 + c2Y2) dx = 1

+

or,

Ú

–

+

c1* c1

dx + Ú

Y 1* Y1

–

+

or,

Ú

–

or, or,

+

c2* c2

Y *2

Y2 dx + Ú (c*1 c2 Y 1* Y2 + c1 c2* Y *1 Y2) dx = 1 –

+

| c1 |2 Y1* Y1 dx + Ú | c2 |2 Y*2 Y2 dx + 0 = 1 –

c12

+

c22

=1

1 __ + c22 = 1 2 1 c22 = __, 2

1__ Hence c2 = ___ ÷2 (ii) The probability of finding the energy 3e0 is given by or,

1__ ___ 1 1 c2* c2 = ___ ◊ __ = __ ÷2 ÷2 2 +

(iii)

Ú

–

(c1Y1 + c2Y2)* (c1Y1 + c2Y2) dx +

=

Ú

(c1* Y1* + c2* Y2) (c1

Ú

(c1* Y 1* + c2* Y 2*) (e0 c1Y1 + 3e0 c2 Y2) dx

Ú

(e0 c1* c1 Y 1* Y1 + e0 c2* c1 Y 2* Y1 + 3e0 c1* c2 Y1* Y2 + 3e0 c2* c2 Y 2* Y2) dx

Ú

(e0 c12 Y 1* Y1) + 0 + 0 + 3e0 c22 Y 2* Y2) dx

– +

=

– +

=

– +

=

–

+

= e0 =

Ú

–

e0 c12

Y1 + c2 Y2) dx

+ 2

| c1 Y1 | dx + 3e0 +

3e0 c22

Ú

–

| c2Y2 |2 dx

[ Y1 and Y2 are normalized]

7.46

Advanced Engineering Physics

1 1 = e0 ◊ __ + 3e0 ◊ __ 2 2 = 2e0 Example 7.23

Sol.

(i) (ii) (iii) (iv) (i)

A particle of mass m is confined within the range between x = 0 and x = L. Write down Schrödinger’s equation for the particle. Solve the equation to find out the normalized eigen functions. Show that the eigen functions corresponding to two different eigen values are orthogonal. If px be the momentum, then find < px > as well as < px2 > in the ground state. [WBUT 2008] As the particle is confined in the range between x = 0 and x = L, it may be in a deep potential well or in any potential dependent on x. In the first case V = constant and in the second case V = V(x). So, Schrödinger’s time-independent equation for the particle is given by

d2 2m ___ Y(x) + ___ [E – V(x)] Y(x) = 0 dx2 h2 (ii) If the particle be in deep potential well, then V = constant and let V = 0. So, the wave equation becomes d2 2mE ___ Y(x) + ____ Y(x) = 0 2 dx h2 d2Y(x) 2mE ______ or, + k2Y(x) = 0 where k2 = ____ dx2 h2 The general solution of the equation is given by Y(x) = A sin (kx) + B cos (kx) Now, the boundary conditions are at x = 0, Y(x) = 0 and at x = L, Y(x) = 0 \ when x = 0, we get B = 0 So, the equation becomes Y(x) = A sin (kx) Again, at x = L, Y(x) = 0, so we get Y(L) = A sin (kL) or, A sin (kL) = 0 or, sin (kL) = sin np or, kL = np np \ k = ___ where n = 1, 2, 3, ... L Hence the permissible wave function is given by npx Y(x) = A sin ____ L Let Yn (x) be the normalized wave function 1 \ Yn(x) = __ Y(x) where N is the norm. N

( )

7.47

Quantum Mechanics +

Ú

Now

Y *n(x) Yn(x) dx = 1

– +

1 ___ Ú Y *(x) Y(x) dx = 1 N2 –

or,

L

npx dx = N2 Ú A2 sin2 ( ____ L )

or,

0

L

2

npx N dx = ___2 Ú sin2 ( ____ L )

or,

A

0

L

2

2n p x 1 or, __ Ú 1 – cos _____ L 20

(

) ) dx = ___NA 2n p x 2N [x] – Ú cos ( _____ ) dx = ____ 2 A 2n p x 2N L L – [ ____ sin ( _____ ) ] = ____ L 2n p A (

2

L

or,

2

L 0

2

0

2

L

or,

0

2

or,

2N2 L L – ____ sin (2n p) = ____ 2n p A2

or,

2N2 L – 0 = ____ A2

\

2N2 L = ____ A2 2

2

2N = A L __

\

__

fi

÷

L N = __ A 2

( ) ÷ npx 2 Y (x) = __ sin ( ____ ) ÷L L npx 2 1 Yn(x) = __ ◊ __ A sin ____ L A L __

fi

n

(iii) Let us consider two eigen functions Yn (x) and Ym (x) where m π n +

Let

I=

Ú

Y m* (x) Yn (x) dx

Ú

mpx npx 2 __ sin _____ sin ____ dx L L L

– +

or,

I=

–

1 = __ L

(

L

Ú 0

) ( )

[ cos __Lp (m – n)x – cos __Lp (m + n)x ] dx

p p 1 L L = __ ________ sin __ (m – n)x – ________ sin __ (m + n) x L (m – n)p L L (m + n)p

[

]

L 0

7.48

Advanced Engineering Physics

1 L = __ × __ [0 + 0 + 0 + 0] L p +

or,

I=

Ú

–

Ym* (x) Yn (x) dx = 0

Hence Ym (x) and Yn (x) are orthogonal. (iv) The ground-state wave function is Y1 (x). __

\

÷

px 2 Y1(x) = __ sin ___ L L

( )

+

< px > =

Ú

–

Y *1 (x)

x

Y1 (x) dx

L

or,

px dx ( ) ( – i h ___∂x∂ ) sin ( ___ L ) p px px 2 = ( __ ) (– i h) ( __ ) Ú sin ( ___ ) cos ( ___ ) dx L L L 2

px 2 < px > = __ Ú sin ___ L 0 L

L

0

L

ihp = – ____ 2L

2p x dx ( sin ( ____ L )) ihp 2p x L = – ____ [ – ___ cos ( ____ ) ] L 2L 2p

Ú 0

L 0

ih = __ (1 – 1) = 0 4 < px > = 0 L

< p2x > = Ú Y1(x)

2 x

Y1 (x) dx

0 L

( )(

px 2 = __ Ú sin ___ L 0 L 2h2 = ___ L

L

Ú ( __Lp )

2

0

)

∂2 px – h2 ___2 sin ___ dx L ∂x

( )

px sin2 ___ dx L

( )

L

2 h2 p2 px = ______ dx Ú sin2 ___ 3 L 0 L

( )

h2 p2 = _____ L3

L

Ú 0

2p x dx [ 1 – cos ( ____ L )] L

h2 p2 h2 p2 2p x _____ = _____ × L + dx Ú cos ____ 3 3 L L L 0

( )

7.49

Quantum Mechanics

h2 p2 ____ h2p2 L 2p x = _____ + 3 × ___ sin ____ 2 L 2p L L

[ ( )]

L 0

2 2

hp < px2 > = ____ L2 h2 < px2 > = ___2 4L

\

Show that [xn, n] = + i h nxn – 1, n being a positive integer. Let Y be the state function.

Example 7.24 Sol.

\

x

∂ xn Y(x) = (– i h) ___ (xn Y(x)) ∂x ∂Y = (– i h) nxn – 1 Y + (– i h) xn ____ ∂x ∂ ∂ = – i h ___ (xn) Y + xn – i h ___ Y ∂x ∂x

{

(

}

or, or,

= – x xn Y + xn x Y + 2 n x x Y = [xn, n]Y + 2 n x Y [xn, x]Y = – x xn Y

or,

[xn,

n

)

xn Y

n

or, \

x]Y

[xn, [xn,

(

)

∂ = – – i h ___ xn Y ∂x

= (i h nxn–1) Y n–1 x] = i h nx

n]

Example 7.25 Sol.

Show that the eigen value of a hermitian operator is real. An operator is hermitian if it satisfies the condition,

Ú Y 1* (

Y2) dV = Ú ( Y1)* Y2 dV

where Y 1* is the complex conjugate of Y1. Y1 and Y2 are two eigen functions of the operator . A hermitian operator is linear and has real eigen value. Y = lY

Now

*Y * = l*Y * \

Ú

and

ÚY

\

l Ú Y * Y dx = l* Ú Y Y * dx

Y*

Y dx = Ú Y *l Y dx = l Ú Y * Y dx * Y*dx = Ú Y *l* Y * dx = l* Ú Y Y * dx [

is hermitian]

or, l = l* \ l is real, i.e., if the eigen value of a hermitian operator is real.

7.50

Advanced Engineering Physics

Review Exercises

Part 1: Multiple Choice Questions 1. The probability of finding a particle in a distance dx around the point x is given by (a) Y * (b) Y * Y dx (c) Y Y * (d) Y 2. For a stationary state, the probability density is (a) function of time (b) dependent on wave function (c) independent of time (d) independent of space coordinates 3. Schrödinger’s wave equation for a moving particle contains (a) second-order time derivative (b) first-order time derivative (c) third-order time derivative (d) None of above 4. A free particle has (a) definite momentum but indefinite energy (b) definite energy and indefinite momentum (c) definite energy and definite momentum (d) energy and momentum both indefinite 5. The energy, which a particle moving in a one-dimensional box can have, is (a) directly proportional to the quantum number (b) inversely proportional to the quantum number (c) directly proportional to the square of the quantum number (d) inversely proportional to the square of the quantum number 6. For a particle trapped in a box of length l, the value of the expected average is l 1 2 (a) __ (b) __ (c) __ (d) None of these 2 l l 7. The expected average of the momentum of a particle trapped in a box of length l is h h (a) __ (b) __ (c) 1 (d) 0 l 2l 8. A particle is freely moving inside a box. Which of the following is incorrect for the energy of the particle? (a) The energy is directly proportional to the square of the quantum number n. (b) The energy is inversely proportional to the length of the box l. (c) The energy is inversely proportional to the mass of the particle. (d) The energy is directly proportional to the potential of the box. d2 9. Which of the following functions is an eigen function of the operator ___2 ? dx c (a) Y = c ln x (b) Y = cx2 (c) Y = __x (d) Y = ce– mx

7.51

Quantum Mechanics

10. The ground-state energy of a particle moving in a one-dimensional potential box is given in terms of length l of the box by 2h2 h2 h (a) ____2 (b) ____2 (c) ____2 (d) zero 8ml 8ml 8ml 11. Which one of the following is not an acceptable wave function of a quantum particle? (a) Y = ex (b) Y = e– x (c) Y = xn (d) Y = sin x 12. The wave function of motion of a particle in a one-dimensional box of length l is given by A sin n p x Yn = _________ where A is the norm for a wave function. The value of A is L __

(a)

__

÷

÷

1 __ L

2 (b) __ L

1 (c) __ L

2 (d) __ L

13. The spacing between the nth energy state and the next energy state in a one-dimensional potential box increases by (a) (2n – 1) (b) (2n + 1) (c) (n – 1) (d) (n + 1) 14. If E1 be the energy of the ground state of a one-dimensional potential box of length l and E2 be the energy of the ground state when the length of the box is halved, then (a) E2 = 2E1 (b) E2 = E1 (c) E2 = 4 E1 (d) E2 = 3E1 15. The energy of a particle which is confined in a cubic box of side l is given by h2 E = ____2 (n2x + n2y + nz2 ). If nx, ny, nz may have either of the three values –1, 2, and 3, then the degree 8ml of degeneracy of this energy level is given by (a) 3 (b) 6 (c) 2 (d) 4 16. The normalized wave function for a particle in a rectangular box of dimensions a, b, c is given by ____

____

ny p x ny p x nz p z nz p x nx p x nx p x abc 2 _____ _____ ____ _____ _____ (a) Yn = ____ sin _____ sin sin (b) Y = sin sin sin _____ n a c a c 2 b abc b

÷ n px n px n px 1 _____ sin _____ (c) Y = ____ sin _____ sin c ÷abc a b ____

y

x

n

z

÷ n px n px n px 8 _____ sin _____ (d) Y = ____ sin _____ sin c ÷abc a b ____

y

x

z

n

17. The waves representing a free particle in three dimensions are (a) standing waves (b) progressive waves (c) transverse waves (d) polarized waves 18. If the quantum numbers nx, ny and nz are each equal to zero for a particle trapped in a rectangular potential well, then the particle (a) may be absent from the well (b) may be present in the well (c) may or may not be present in the well (d) None of these 19. Schrödinger’s time-independent wave equation is (a) Y = E Y 2 (b) Y 2 = E Y (c) Y 3 = E Y 2 (d) Y = E Y d 20. For the function eb x, the eigen value of the operator ___ is given by dx (a) b

(b) b2

b2 (c) __ 2

b (d) __ 2

7.52

Advanced Engineering Physics

21. Which of the following is a linear operator? d (c) ___ (d) exp ( ) dx 22. The expected average of momentum of a particle in an infinite well is h (c) k h (d) 0 (a) h (b) __ 2 23. The wave functions Ym (x) and Yn (x) are orthogonal to each other. Which of the following relations must hold for them? (a) log ( )

(b) ÷

+

(a)

Ú

–

+

Y *m Yn dx = 1

(b)

Ú

–

+

Ym Yn dx = 0

(c)

Ú

–

+

Y *m Yn dx = 0

(d)

Ú

–

Ym Yn dx = 1

24. If Y (x, t) be a normalized wave function we must have +

(a)

Ú

–

+

Y *Y dx = 1

(b)

Ú

–

+

Y * Y dx = 0

–

(c)

Ú

–

1 Y *Y dx = __ 2

(d)

Ú

Y Y dx = 1

–

25. Which of the following functions is an eigen function of the momentum operator? ____

(a) A ln x (b) A ÷x/m (c) Ax2 (d) None of these [Ans. 1 (b), 2 (c), 3 (b), 4 (c), 5 (c), 6 (c), 7 (d), 8 (d), 9 (d), 10 (c), 11 (c), 12 (b), 13 (b), 14 (d), 15 (b), 16 (d), 17 (b), 18 (a), 19 (d), 20 (a), 21 (c), 22 (d), 23 (c), 24 (a), 25 (d)]

Short Questions with Answers 1. What is a wave function? Mention four points on its physical significance. Ans. Schrödinger’s equation represents the wave behavior of a particle while it moves. The solution of this differential equation is a function of space and time coordinates. As this function [usually denoted by _ Y (r, t)] is related to the wave behavior of a particle it is called a wave function. _ The following are the four points regarding the physical significance of the wave function Y(r, t): (a) It gives information regarding the space-time behavior of a particle. _ _ _ (b) | Y(r, t) |2 = Y *(r, t) Y (r, t) measures the probability of finding a particle in space and time. (c) The magnitude of the wave function is large where the probability of finding the particle is high. (d) The wave function must be continuous and single-valued at each point in space. 2. Define adjoint of an operator. Ans. Let us consider two operators and † which satisfy the following relation: Ú f* Y dV = Ú ( † f)* Y dV In this case † is said to be the adjoint of . In other words so far as the value of the integral is concerned, it does not make any difference whether acts on y or its adjoint † acts on the other function. 3. Define self-adjoint or hermitian operator. Ans. The physical quantities are real. The eigen values of operators representing a physical quantity must be real. This condition limits the type of functions which can serve as operators for the physical quantities. An operator which corresponds to a physical quantity must be in a position to satisfy the following condition:

7.53

Quantum Mechanics

< q > = < q >*

Ú f*

or,

Y dV = Ú ( f)* Y dV

where f and Y are two arbitrary wave functions. The operators whose eigen values are real are known as real or hermitian or self-adjoint operators. An operator is considered to be self-adjoint or hermitian if = †. 4. Show that the expected average of hermitian operator is real. Ans. Let be a hermitian operator and Y1 and Y2 be two eigen functions of corresponding to the eigen values q1 and q2 respectively. Then Y1 = q1Y1 and Y2 = q2Y2 The condition of self-adjointness (or hermitianity) is given by

Ú Y *1

Y2 dV = Ú ( Y1)* Y2 dV

Ú Y1* q2 Y2 dV = Ú q*1 Y *1 Y2 dV

or,

(q2 – q1*) Ú Y 1* Y2 dV = 0

or,

This is true for any two eigen function of . So, if q2 = q1, then q1* = q2 Hence q1 is real. 5. Find the eigen function and eigen values of the operator Ans. The eigen value equation of z

or,

z

z

d = – i h ___ df

is given by

Y(f) = q Y(f)

fi

dY – i h ___ = qY df

q dY ____ = – __ df Y ih

fi

= + __ Ú df + ln c Ú ____ Y h

dY

iq

( )

Y = ce

or,

iq __ f h

The function Y is a periodic function of variable f with a period of 2p, i.e., Y(f) = Y(f + 2p). This implies that the eigen value q is an integral multiple of h and Y as given by the above mentioned formula is the eigen function of z. 6. (a) Write down Schrödinger’s time-dependent equation. [WBUT 2006] (b) State quantum mechanical postulates. Ans. (a) Refer to Section 7.7.1. (b) Refer to Section 7.6. 7. Show that a normalized wave function must have unit norm. Ans. We know that Schrödinger’s wave equation is a linear and homogeneous equation in Y and its deriva-

(

)

∂Y ∂Y tives like ____ and ____ are also linear. For this reason, if any solution of this equation is multiplied ∂x ∂t by a constant quantity then the new wave function so resulted will also be a solution. So, in order to avoid this arbitrariness, it becomes necessary to impose a normalization condition.

7.54

Advanced Engineering Physics _

Let us assume that Y1(r, t) is a solution of Schrödinger’s equation and +

_

Ú

| Y1(r, t) |2 dV = N2

–

_

_

...(1)

_

where | Y1(r, t) |2 = Y1* (r, t) Y1(r, t) is a positive real number and the value of the integral is also so. _ N is, therefore, a real positive number and it is called the norm (or normalization constant) of Y1(r, t). _ _ Let us now consider another wave function Y(r, t) which is related to the present function Y1(r, t) through the following equation: _ _ 1 Y(r, t) = __ Y1(r, t) N

...(2)

_

where Y(r, t) is also a solution of Schrödinger’s equation, as it merely differs from the present solu_ tion Y1(r, t) by a factor of (1/N) which is a constant. _ Now, substituting this value of Y1(r, t) in Eq. (1), one gets +

Ú

_

N2 | Y(r, t) |2 dV = N2

–

+

Ú

or,

_

| Y(r, t) |2 dV = 1

...(3)

–

_

The wave function Y(r, t) satisfies Eq. (3) and it is called the normalized wave function. Hence, a normalized wave function must possess a unit norm, i.e., the normalization constant of a normalized wave function is unity. 8. What do you mean by an orthogonal wave function? Discuss briefly. Ans. Let us consider a set of wave functions where each member of the set satisfies Schrödinger’s wave _ _ _ _ _ equation, namely, Y1(r), Y2(r), Y3(r), etc. Such that any two members [e.g., Ym(r) and Yn(r)] of the set satisfy the condition given below: _

_

Ú Y m* (r) Yn(r) dV = 0

9. Ans. 10. Ans.

... (1)

where m π n, then the functions in the aforesaid set are said to be orthogonal in analogy with the condition of orthogonality of two vectors. Derive Schrödinger’s wave equation from the fundamental postulates of quantum mechanics. Refer to Section 7.7.1. What is the need for normalizing a wave function? Write down the normalization condition of two wave functions. What will be the condition for single wave function? As we know Schrödinger’s wave function predicts the position of the related particle in space and time. The wave function must have a finite value at any point in space and time. The square of the wave function (i.e., | Y |2) is proportional to the probability of finding the particle in space. But according to the definition of probability in statistics, its value can range from 0 to +1 only. For this reason, the wave function needs to be normalized so that it can satisfy the demand of probability theory.

Quantum Mechanics

7.55

The normalization condition of any two wave functions which are the solutions of the same Schrödinger’s equation is given by, +

1 Yn = __ Y N

and

Ú

–

Y n* Yn dV = 1

where Yn is the normalized wave function and Y is unnormalized one. The normalization condition for a single wave function is given by +

Ú

–

Y n* Yn dV = 1

Part 2: Descriptive Questions 1. (a) What is the physical interpretation of wave function? (b) How is wave function related to the probability of finding a particle at any point in space at a given time? (c) What is probability density? (d) Discuss the probability density of a particle represented by a wave which is generated by the superposition of two de Broglie waves. 2. (a) What do you mean by normalization of a wave function? What is the use of normalization? (b) How can you normalize a wave function by multiplying it with a suitable constant? (c) Can you normalize every wave function? (d) Consider two wave functions Y and Y exp (i q) where q is a real quantity. What is the probability density of a particle for each of the states? 3. (a) What do you mean by stationary states? Why are they called so? (b) What is the condition for which the system can be represented by stationary states? 4. (a) Write down Schrödinger’s time-dependent wave equation for three-dimensional and one-dimensional cases. (b) Obtain the time-independent form of Schrödinger’s equation form its time-dependent form. (c) Under what condition can we obtain the time-independent form of Schrödingers’s equation? 5. (a) What are the limitations of an any acceptable wave function? (b) Consider a particle moving along the x direction and represented by a normalized wave function Yn. What is the probability of finding the particle in a region which is at infinity from the origin? 6. Write down Schrödinger’s equation for one-dimensional motion of a free particle in a one-dimensional potential box. Find the eigen function and eigen energy. [WBUT 2002] 7. (a) Write down Schrödinger’s equation for a particle of mass m in a rectangular box with infinitely hard walls whose edges are a, b, and c. What are the boundary conditions? What conditions are responsible for the quantization of energy of the particle? (b) The wave functions of a particle are given by ny p y nz p z nx p x Y (x, y, z) = A sin _____ sin _____ sin _____ a c b

(

)

( ) ( )

7.56

Advanced Engineering Physics

where nx, ny and nz belong to the set {1, 2, 3, …} Find the value of the normalization constant A. [WBUT 2003] 8. Obtain the time-independent Schrödinger’s equation. Show that the energy eigen values of a particle enclosed in a potential box is quantized. [WBUT 2004] 9. (a) Write down the time-independent Schrödinger’s equation for a free particle of mass m and _ momentum p. (b) Why in case of moving electron quantum mechanics is used while for moving cars we use newtonian mechanics? Explain. [WBUT 2005] 10. (a) Give the physical interpretation of the wave function Y(x). (b) Obtain the expression for the stationary energy levels for a particle of mass m which is free to move in a region of zero potential between two rigid walls with x = 0 and x = L. Are the energy levels degenerate? (c) Find the corresponding wave function pertaining to part (b) of this question and normalize it. [WBUT 2005] 11. Write down Schrödinger’s equation for a particle confined in a one-dimensional box. [WBUT 2006] 12. If the wave function Y(x) of a quantum mechanical particle is given by px Y(x) = a sin ___, for 0 £ x £ L L =0 for 0 ≥ x ≥ L Then determine the value of a. Also determine the value of x where the probability of finding the particle is maximum. [WBUT 2007] 13. A particle of mass m is confined within the space between x = 0 and x = L. (i) Write down Schrödinger’s equation for such a system. (ii) Solve the equation to find out the normalized eigen functions. (iii) Show that the eigen functions corresponding to two different eigen values are orthogonal. (iv) If px denotes the momentum then find < px > as well as < px2 > in the ground state. [WBUT 2008] 14. What are the values of

[ ]

∂ , ___ , [ x, ] and [ x, y]? ∂x

[WBUT 2004]

df 15. Show that (i) [ x, n] = – i h n xn – 1 and (ii) [ x, f (x)] = – i h ___. dx 16. (a) State the basic postulates of quantum mechanics. (b) Define degeneracy and non-degeneracy. Prove that the lowest state of a free particle in a cubical box is not degenerate. (c) What do you mean by orthogonal wave function? Write down the condition of orthogonality.

7.57

Quantum Mechanics

Part 3: Numerical Problems 1. A particle exists inside a one-dimensional potential box of length L. Calculate the probability of find3L L ing the particle in a region between __ and ___ when the particle is in the lowest energy state. 4 4 [Ans. 0.818] 2. Examine whether the operator is linear in the following two cases: (i) f (x) = f (x) + sin x (ii)

f (x) = f (– x)

[Ans. (i) non-linear, (ii) linear] 2

∂ ∂ 3. Show that ___ and ___2 are commutative. ∂x ∂x 4. A particle is moving along the x axis has the wave function Y(x) = ax for 0£x£1 =0 for 0≥x≥1

[

Find the expected average < x > of the position of the particle.

a2 Ans. __ 4

]

5. Prove that the wave function Y(x, t) = A cos (kx – w t) does not satisfy the time-dependent Schrödinger’s equation for a free particle. 6. For an electron of mass 9.1 × 10–31 kg moving in the one-dimensional infinitely deep potential well of width 0.1 nm, find (i) the least possible energy, (ii) the first three eigen values in electron volt, (iii) the energy difference between the ground state and the first excited state, and (iv) the frequency of the emitted radiation due to the transition between the two states. [Ans. (i) 37.73 eV, (ii) E1 = 37.73 eV, E2 = 150.95 eV, E3 = 339.63 eV, (iii) 113.22 eV, (iv) v = 27.3 × 1015 Hz] 7. Calculate the normalization constant for a wave function (at t = 0) given by 2 2 __ __1 Y(x) = Ae– s x /2 ◊ eikx Ans. ( s /÷p ) 2 8. Determine the probability density for the wave function

[

– s2 x2 ______ 2 eikx

]

[ Ans. p = ___sp exp (– s x ) ]

Y(x) = Ae

__

2 2

÷

_

_

9. Consider two stationary state solutions Y1 (r) and Y2 (r) corresponding to energy states E1 and E2 respectively for time-independent Schrödinger’s equation. Prove that

_

Ú Y 1* (r) Y2 (r) dV = 0 v

10. Prove that for two linear operators and , [ , ] + [ , ] = 0. 11. If , and are three linear operators, then prove that [ , [ , ]] + [ , [ , ]] + [ , [ , ]] = 0 12. The minimum energy possible for a particle trapped in a one-dimensional infinite potential well is 10 eV. What are the next three energy levels? [Ans. 40 eV, 90 eV, 160 eV]

7.58

Advanced Engineering Physics

13. Find the commutator [ , ] where

= x3 and

d = x ___. dx

[Ans. –3x3] __

÷

px 2 14. Calculate the expected average of px2 for the wave function Y = __ sin ___ in the region from x = 0 L L to x = L. 15. Prove the following relation: [ , + ]=[ , ]+[ , ]

( )

[Ans. (h p)2/L]

Chapter

8 Statistical Mechanics 8.1

IntroduCtIon

Thermodynamics is the study of the relationship between macroscopic properties of systems, such as temperature, volume, pressure, magnetization, compressibility, etc. Statistical mechanics, also called statistical physics, is concerned with understanding how the various macroscopic properties arise as a consequence of the microscopic nature of the system. In essence, it makes macroscopic deduction from microscopic models. According to Lev Landau, statistical physics and thermodynamics together form a unit. All the concepts and quantities of thermodynamics follow most naturally, simply and rigorously from the concepts of statistical physics. Although the general statement of thermodynamics can be formulated non-statistically, their application to specific cases always requires the use of statistical physics. Statistical approach can be applied only to such systems which contain a large number of individuals identical in some sense, but are distinguishable by means of some property. The purpose of statistical mechanics is to provide special statistical or probability laws for the distribution of identical microscopic particles (atoms and molecules) of a macroscopic system in thermal equilibrium, without considering the detailed calculation of position and velocity of individual particles and to derive all equilibrium properties of the system. Suppose, for example, 1 cm3 of a gas under normal temperature and pressure contains 2.7 × 1019 molecules. Assuming that each molecule has three degrees of freedom, we will have to set up and solve 3 × 2.7 × 1019 equations of motion for 1 cm3 of the gas. Theoretically, some equations can be solved and the solutions would be in terms of mass, position, velocity and energy of each molecule, etc., and not in terms of thermodynamic variables such as pressure, entropy, etc. Thus there is a gap between the results of classical mechanics and thermodynamics. Statistical mechanics provides the link between mechanical properties and thermodynamical properties of a system. In statistical mechanics, the properties and laws of motion of individual atoms, molecules and elementary particles, are assumed to be known. If these laws are governed by classical mechanics, we call it classical statistics and if they are governed by quantum mechanics, we call it quantum statistics. Classical statistics was developed by Maxwell in 1859 and then refined by Boltzmann and hence correctly called Maxwell–Boltzmann (or MB) statistics. Classical statistics is applicable to a system of identical, weakly interacting, structureless and neutral particles. Such hypothetical particles are called boltzons. Quantum particles are divided into two classes. One with integral spin (i.e., nh) and are called Bose particles

Advanced Engineering Physics

8.2

Statistical Mechanics

Quantum Statistics

Classical Statistics [Maxwell–Boltzmann Statistics]

Bose–Einstein Statistics

Fermi–Dirac Statistics

[ ( )]

1 (or simply bosons) and other with half integral spin i.e., n + __ h and are called Fermi particles (or simply 2 fermions). Bose statistics was refined by Einstein and hence called Bose – Einstein (or BE) statistics and Fermi statistics was refined by Dirac and hence called Fermi–Dirac (or FD) statistics.

8.2

ConCept of phase spaCe

In classical mechanics, the position of a point particle in three-dimensional space is determined by three cartesian coordinates x, y, z and its state of motion is given by the velocity components vx, vy and vz. For many purposes it is more convenient to use the corresponding momenta px py pz (for m , m and m where m is the mass of the particle). To describe both the positions and the state of motion of the particle, it is required to set up a six-dimensional space called the phase space, in which the six coordinates x, y, z, px, py, pz are marked out along six mutually perpendicular axes. A point in this space describes both the position and the motion of the particle at some particular instant. Therefore, if a position of a phase point is known at any instant of time, its trajectory is also known.

8.2.1

phase space of N particles system

For N particles system, the mechanical state of the whole system can be determined completely in terms of 3 N position coordinates q1 ... qN and 3 N momenta coordinates p1 ... pN. The 6-N-dimensional space is called the phase space or G space of the system. A point in G space represents a state of the entire system. For monoatomic gas it is also called molecular phase space or m space.

8.2.2

smallest Cell in phase space

In the m space of particle, a volume element dt = dx dy dz dpx dpy dpz is the volume of a six-dimensional cell having sides dx, dy, dz, dpx, dpy, dpz. Such a cell of minimum volume is called a unit cell in the m space. So, (dt)min = (dx dpx)min (dy dpy)min (dz dpz)min But uncertainty principle tells us that the minimum value of each of the products is approximately equal to Planck’s constant h. Therefore, (dt)min = (h)(h)(h) = h3 In classical description, h can be chosen as an arbitrarily small constant having the dimensions of angular momentum. But quantum theory imposes a limitation on the accuracy with which a simultaneous specification

Statistical Mechanics

8.3

of coordinate x and its corresponding momentum px is possible and from uncertainty principles, phase space is actually a cell of minimum volume h3.

8.3

ConCept of energy LeveLs and energy states

In classical mechanics, the ‘state’ of a system is defined by the values of the coordinates and momenta of all its constituent particles. In quantum statistics, a quantum state is a mathematical object that fully describes a quantum system. One typically imagines some experimental apparatus and procedure which prepares this quantum state; the mathematical object then reflects the set-up of the apparatus. Quantum states can be statistically mixed, corresponding to an experiment involving a random change of the parameters. States obtained in this way are called mixed states, as opposed to pure states, which cannot be described as a mixture of others. When performing a certain measurement on a quantum state, the result is in general described by probability distribution and the form that this distribution takes is completely determined by the quantum state and the observable describing the measurement. However, unlike in classical mechanics, the result of a measurement on even a pure quantum state is only determined probabilistically. A quantum system or particle that is bound, confined spacially, can only take on certain discrete values of energy, as opposed to classical particles, which can have any energy. These are called energy levels. The term is most commonly used for energy levels of electrons in atoms or molecules, which are bound by the electric field of the nucleus. The energy spectrum of a system with energy levels is said to be quantized. Energy levels are said to degenerate, if the same energy level is obtained by more than one quantum mechanical state. They are then called degenerate energy levels. There may be several energy states corresponding to the same energy.

8.4

MaCrostate and MICrostate

Macrostate The macrostate of a system may be defined as the number of phase points in each cell of the phase space. For a macrostate, we specify only bulk quantities which can be determined by macroscopic measurements, such as pressure, temperature, etc. Thus an isolated system consisting of a fixed number (N) of non-interacting identical particles, having a fixed internal energy (E) and occupying a fixed space of volume (V) is said to be macroscopic state or macrostate of the system. The macrostate is also called the thermodynamic state. Microstate A state of the system in which we specify the states of all the constituent particles is called a microstate. From a dynamical point of view, each state of a system can be defined as precisely as possible by specifying all of the dynamical variables of the system. Such a state is called a microscopic state. In general, the number of different meaningful ways in which the total energy (E) of the system can be distributed among its constituent particles called microstate. One of the fundamental hypothesis of statistical mechanics is that microstates are equally probable, which means one microstate would occur as often as another over a long period of time. Example Suppose we have four distinguishable particles a, b, c, d and we wish to distribute them into two exactly similar compartments in an open box. Since both the compartments are exactly alike, the particles have the same a priori probability of going into either of them. The following table will give the various macrostates and microstates.

Advanced Engineering Physics

8.4 Macrostates

0, 4 1, 3

2, 2

3, 1

4, 0

Possible arrangement

No. of microstates

Compartment 1

Compartment 2

O a b c d ab ac ad bc bd cd bcd cda dab abc abcd

abcd bcd cda dab abc cd bd bc ad ac ab a b c d O

1 4

6

4

1

Each distinct arrangement is known as the microstate of the system.

8.5

therModynaMIC probabILIty and entropy

Thermodynamical probability is defined as the number of possible microstate corresponding to any given macrostate. Thermodynamic probability is, in general, very large, unlike mathematical probability, which when normalized, as always is less than one. In our previous example, the thermodynamic probability corresponding to the macrostate (2, 2) is equal to the number of microstates in (2, 2) which is 6. We know that when an isolated system undergoes an irreversible process, there is a net increase of the entropy of the system. The maximum value of the entropy is reached when the system arrives at a state of equilibrium. Boltzmann related the entropy S and the thermodynamic probability W by the equation W S = KB ln ___ Wo where KB is the Boltzmann constant and Wo the initial probability. From Eq. (8.1), we have S = KB ln W – KB ln Wo = KB ln W – So The quantity So = KB ln Wo can be taken as the zero from which the entropy is counted. Hence, S = KB ln W

8.6

....(8.1)

...(8.2)

equILIbrIuM MaCrostate

A macroscopic state which does not tend to change in time except for random fluctuations is known as equilibrium macrostate. The macrostate of a system in equilibrium is time-independent, except for ever-present

Statistical Mechanics

8.5

fluctuations. The macrostate of a system can be described by certain macroscopic parameters, i.e., parameters which characterize the properties of the system on a large scale. The equilibrium macrostate of a system is independent of its past history. For example, consider an isolated gas of N molecules in a box. These molecules may originally have been confined by a partition to one half of the box. But after the partition is removed and equilibrium has been attained, the macrostate of the gas is the same in both cases; it corresponds merely to the uniform distribution of all the molecules through the entire box.

8.7

Mb, be and fd statIstICs

In statistical mechanics, Maxwell–Boltzmann statistics describes the statistical distribution of material particles over various energy states in thermal equilibrium, when the temperature is high enough and density is low enough to render quantum effects negligible. Fermi–Dirac (FD) and Bose–Einstein (BE) statistics apply when quantum effects have to be taken into N account. Quantum effects appear if the concentration of particles __ ≥ nq (where nq is the quantum conV centration). The quantum concentration is when the interparticle distance is equal to the thermal de Broglie wavelength, i.e., when the wave functions of the particles are touching but not overlapping.

( )

8.8 8.8.1

MaxweLL – boLtzMann (Mb) statIstICs basic postulates

(i) Particles are identical and distinguishable. (ii) They do not possess any kind of spin. (iii) They do not obey the Pauli exclusion principle. So any state can accommodate any number of particles. For example, the molecules of a gas. The particles that obey MB statistics are called boltzons. Example: Ideal gas molecules.

8.8.2 Maxwell–boltzmann (Mb) distribution function Consider a system having energy levels e1, e2, e3 ..... and the number of particles in these levels be n1, n2, n3 ...... Moreover, let each level be degenerated, i.e., let each level be associated with several sublevels. Let g1, g2, g3, ......, be the number of sublevels corresponding to the energies e1, e2, e3, ......, respectively. So the available states of the system are the sublevels. Let us now consider the ith level. The probability that a particular ni 1 __ sublevel (or state) is occupied is __ gi . Hence the average number of particles in that state is gi , which is the

( )

( )

th

occupation number for the i state. In classical statistics, the number of ways in which the total number of particles, N can be grouped into levels having respectively n1, n2, n3 ...., number of particles. The number of ways in which we can do so is equal to the number of permutations of N things, out of which n1 are alike, n2 others are alike, n3 others are alike and so on. This is given by N! N! M = __________ = _____ n1! n2! n3! ... P ni! i

...(8.3)

Advanced Engineering Physics

8.6

where P denotes the product of ni for all values of i. Since classical particles are assumed to be distinguishi

able, then the total number of ways in which ni particles occupy gi sublevels in gni i. If Pclassical is the number of ways in which g1 sublevels are occupied by n1 particles, g2 by n2, g3 by n3 and so on, then Pclassical = g1n1 g2n2 g3n3 ... = P gni i i

The total probability W of the distribution is the product of M and Pclassical gni i N! W = _____ P gni i = N! P ___ i ni! P ni! i

...(8.4)

i

Taking natural logarithm on both sides, we have gni i ln W = ln N! + ln P ___ i ni! ln W = ln N! + S ni ln gi – S ln ni! i

i

Applying Stirling’s theorem, [ln N! = N ln N – N (where N is very large)], we have ln W = N ln N – N + S ni ln gi – i

[ S (n ln n – n ) ] i

i

i

i

= N ln N – N + S ni ln gi – S ni ln ni + S ni i

i

i

+ S (ni ln gi – ni ln ni) +

= N ln N –

i

gi = N ln N + S ni ln __ n i i

( )

...(8.5)

The most probable distribution can be obtained by maximizing ln W, i.e., by setting ∂ ln W d ln W = S ______ d ni = 0 ...(8.6) ∂ ni subject to the constraints that the total number of particles and total energy of the system are conserved: N = n1 + n2 + n3 + ... = S ni = const. i

E = n1 e1 + n2 e2 + ... = S ni ei = const. i

S d ni = 0

...(8.7)

S ei d ni = 0

...(8.8)

In differential form

i

and

i

Differentiating Eq. (8.5) w.r.t. ni , we get from Eq. (8.5) gi d ni = 0 S ln __ n i i

( )

...(8.9)

Statistical Mechanics

From Eqs (8.7), (8.8) and (8.9), we have gi – a – b ei d ni = 0 S ln __ n i i

(

)

8.7

...(8.10)

where a, b are Lagrange’s multipliers. Since coefficients d ni are arbitrary, so from Eq. (8.10) for all values of i, the term in the bracket must be zero. gi Thus, ln __ ni – a – b ei = 0 gi ln __ n = a + b ei

or,

i

ni 1 – (a + b ei) __ = ______ gi = e a + b ei e g i or, ni = ______ ...(8.11) ea + b ei This result is known as the Maxwell – Boltzmann distribution of distinguishable particles among the energy levels. Assuming Maxwell – Boltzmann distribution, the average speed, root mean square speed, most probable speed of the molecules of an ideal gas can be calculated. The total internal energy, specific heat at constant volume of an ideal gas can also be derived very easily.

or,

8.8.3 Limitations of Maxwell – boltzmann statistics If for a given density, the temperature of a perfect gas is sufficiently low, then Boltzmann statistics is no longer applicable and a new statistics must be set up. The MB distribution function suffers from the following major objections: (i) The particles are assumed to be distinguishable, although in actual practice many elementary particles like electrons are indistinguishable. In BE statistics there is no restriction on the number of particles that can occupy the same energy state. (ii) Any number of particles are assumed to occupy the same quantum state while many particles such as electrons obey the Pauli exclusion principle which does not allow a quantum state to accept more than one particle. These objections are removed in the distribution function known as Fermi–Dirac (FD) distribution. The particles that obey the FD statistics are called Fermions. Bose–Einstein statistics (BE) give the statistical behavior of indistinguishable particles. The particles that obey BE statistics are known as boson, e.g., phonons, photons, etc.

8.9

bose–eInsteIn (be) statIstICs

Bose–Einstein and Fermi–Dirac statistics apply when quantum effects have to be taken into account and the particles are considered indistinguishable. BE statistics was developed in 1924 by S N Bose for light quanta (photons) and generalized by A Einstein to find energy distribution among indistinguishable and identical particles, each having a spin angular momentum ms h given by any of the values ms h = 0, h, 2h, 3h, ... where ms is the spin quantum number.

Advanced Engineering Physics

8.8

8.9.1

basic postulates

(i) The particles of the system are identical and indistinguishable. (ii) The Pauli exclusion principle is not applicable, so any quantum state can accommodate any number of particles. (iii) BE statistics is applicable to particles with integral spin angular momentum in units of h. (iv) These particles have symmetric wave function. The particles that obey BE statistics are called bosons. Examples Photons, phonons, a - particles, p - mesons, k - mesons, higgs-boson, 1H2, 2He4, 6C12, 8O16 are examples of bosons.

8.9.2

bose–einstein (be) distribution function

For BE statistics, since particles are indistinguishable, the number of ways in which ni bosons can occupy gi sublevels (no restriction on the number of particles occupying a sublevel) be obtained by taking the permutations of {ni + (gi – 1)} things, out of which a group of ni things and another group of (gi – 1) things are alike. So the number is (ni + gi – 1)! Wi = ___________ ...(8.12) ni! (gi – 1)! The total number W of independent ways obtaining a distribution of particles among the quantum states in the various energy levels is the product of expressions given by Eq. (8.12) for i = 1, 2, ..., n. (ni + gi – 1)! So, W = P ___________ ...(8.13) i ni! (gi – 1)! Since ni and gi are very large numbers compared to unity so the Eq. (8.13) can be written as (ni + gi)! W = P ________ ni! gi! i

...(8.14)

Taking natural logarithm on both sides and applying Stirling’s approximation, we get ln W = S [(ni + gi) ln (ni + gi) – ni ln ni – gi ln gi]

...(8.15)

i

∂ ln W The most probable distribution can be obtained by maximizing ln W, i.e., by setting d ln W = S ______ d ni i ∂ ni = 0 subject to the constraints that the total number of particles and total energy of the system are conserved: S ni = const.

S ei ni = const. i

In differential form

S d ni = 0

...(8.16)

i

and S ei d ni = 0 So by applying d ln W = 0, Eq. (8.15) gives

S [ln (ni + gi) d ni – ln ni d ni] = 0 i

...(8.17) [d gi = 0]

...(8.18)

Statistical Mechanics

Using Eqs (8.16), (8.17) and (8.18), we have ni + gi S ln ______ ni – a – b ei d ni = 0

[

]

8.9

...(8.19)

where a, b are Lagrange’s multipliers. Since coefficients d ni are arbitrary, so from Eq. (8.19) for all values of i, the term in the braket must be zero. ni + gi Thus, ln ______ – a – b ei = 0 ni ni _________ 1 __ or, ...(8.20) gi = a + b ei e –1 This relation gives the most probable distribution of particles for a system obeying BE statistics and is known as Bose–Einstein distribution law. If m is the chemical potential and KB is the Boltzmann constant, then in thermal equilibrium for bosons at temperature T, we can write 1 b = ____ and a = – m b KBT

(

)

Substituting this value in Eq. (8.20), we obtain gi ni = ___________ ...(8.21) (ei – m)/KBT e –1 The energy distribution function f (ei ) is the number of particles per quantum state in the energy level ei. Therefore, BE distribution function becomes (if we drop the index i) n ___________ 1 f (e) = __ g = e(e – m)/KBT – 1

...(8.22)

If e >> KBT, the BE distribution reduces to the MB distribution.

8.10 ferMI–dIraC (fd) statIstICs This statistics was developed by E Fermi for electrons and its relation to quantum mechanics was established by PAM Dirac in 1926. The objective of this statistics is to find the energy distribution among indistinguishable identical particles, each having a spin angular momentum msh given by any of the values 1 3 5 ms h = __ h, __ h, __ h ... 2 2 2

8.10.1 basic postulates (i) The particles of the system are identical and indistinguishable. (ii) The Pauli exclusion principle is applicable, so each quantum state can accommodate either no particle or only one particle. 1 (iii) FD statistics is applicable to particles with __ integral spin angular momentum in units of h. 2 (iv) These particles have antisymmetric wave function.

Advanced Engineering Physics

8.10

The particles those obey FD statistics are called fermions. Examples

8.10.2

Protons, neutrons, electrons, m - mesons, 1H3, 2He3, 3Li7, 6C13 are examples of fermions.

fermi–dirac (fd) distribution function

For FD statistics, since the particles are indistinguishable, we should know the number of ways in which ni fermions can occupy gi sublevels. Since fermions are governed by the Pauli exclusion principle (no two particles can occupy the same sublevel), gi must be greater than ni. Thus, the number of ways in which ni fermions can occupy gi sublevels is equal to the number of ways in which ni things can be taken at a time gi! from gi different things. This is equal to gi Cni = __________ considering them all, we can write the probability ni!(gi – ni)! of the entire distribution of the particles is given by gi W = P __________ i ni! (gi – ni)!

...(8.23)

We shall now turn to the derivation of the distribution for fermions, the FD distribution. We shall proceed exactly on the same line as in the case of MB and BE statistics. Taking logarithm of Eq. (8.23) and applying Stirling’s formula, we get ln W = S [gi ln gi – (gi – ni) ln (gi – ni) – ni ln ni] ...(8.24) Under the conditions stipulated by Eqs (8.7) and (8.8), we find the maximum of the quantity ln W, using Lagrange’s undetermined multipliers. For this purpose, we equate to zero the derivatives of the quantity ln W – a S ni – b S ni ei [see Eq. (8.10)].

S [ln ni – ln (gi – ni) + a + b ei] d ni = 0

...(8.25)

Hence, we obtain ni _________ 1 __ ...(8.26) gi = a + b ei e +1 This relation gives the most probable distribution of particles for a system obeying FD statistics and is known as Fermi–Dirac distribution law. For fermions in statistical equilibrium at temperature T, ef 1 a = – ____ and b = ____ KBT KBT where ef = Fermi energy of the system and KB is the Boltzmann constant Now substituting the values of b and a in Eq. (8.26), we get (if we drop the index i) n ___________ 1 f (e) = __ g = e(e – eF)/KBT +1 This is known as Fermi–Dirac distribution function.

8.11 CLassICaL statIstICs as a speCIaL Case of quantuM statIstICs MB, BE and FD statistics can be represented by a single equation as

...(8.27)

Statistical Mechanics

8.11

ni _________ 1 __ ...(8.28) gi = a + b ei e +k k = 0 gives classical or MB statistics k = –1 gives BE statistics k = +1 gives FD statistics ni ______ 1 When ea + b ei >> 1 gives __ gi = a + b ei which is the general form of classical distribution. The inequality in e ni a + b ei __ >> 1 shows that g < 1 which means that classical distribution is true only in the limiting case equation e i of small number of particles per quantum state. Classical distribution is thus valid only for rarefied gases. ni For higher temperatures, since __ gi T1 T1 T2

1 = 1 if e < eF f (e) = ______ e–µ + 1 e = ef e 1 ______ = 0 if e > eF and = µ Fig. 8.1 Fermi distribution at zero e +1 and non-zero temperature. Thus, at absolute zero (T = 0 K) of temperature all possible quantum states of energy less then eF are occupied and all those of energy more than eF are empty. Accordingly, the Fermi energy eF is defined as the energy of the highest occupied level at absolute zero. Expanding Eq. (8.34) near e = eF we get at finite (T > 0 K) temperature as 1 1 e – eF f (e) = ____________ ...(8.35) = __ – _____ + ... (e – eF)/KBT e + 1 2 4KBT (i) If e £ eF – 2KBT f (e) = 1 (ii) If e ≥ eF + 2KBT f (e) = 0 (iii) If e = eF f (e) = 1/2 The spread region of Fermi distribution, i.e., the region of e when f (e) changes from unity at zero (from eF – 2KBT to eF + 2KBT) narrows as the temperature decreases and at absolute zero becomes a sharp discontinuity. 1 Thus, Fermi level is that energy level for which the probability of occupation at T > 0 is __, i.e., 50% of the 2 quantum states are occupied and 50% are empty. The distribution is said to degenerate at low temperature and non-degenerate at high temperature when the step-like distribution is lost.

Statistical Mechanics

8.13

8.13.1 fermi temperature It may be defined as the ratio of the Fermi energy (eF) at absolute zero to Boltzmann constant KB. eF \ Fermi temperature qF = ____ KB

...(8.36)

8.13.2 free electrons in a Metal The conduction electrons in a metal may be considered as nearly free-moving in a constant potential field, like particles of an ideal gas. Electrons have half-integral spin and hence the formula of Fermi–Dirac statistics are applicable to an ideal gas electron. Since there can be maximum of one particle per quantum state, the function f (e) is the ratio of the number of the quantum state of energy e occupied by electrons to the total number of quantum states available in the energy level e. From Eq. (8.27) f (e) is given by 1 f (e) = ___________ (e – eF)/(KBT) e +1 The number of particles having energies in the range between e and e + de is given by n (e) de = f (e) g (e) de where g (e) de is the number of quantum states of energy between e and e + de. Substituting the expression for f (e) in Eq. (8.38), we get g (e) de n (e) de = ____________ (e – eF)/KBT e +1 The energy of an electron (non-relativistic) having momentum p is p2 e = ___ 2m

...(8.37)

...(8.38)

...(8.39)

So from Eq. (8.33) we can write the number of quantum states or cells lying within the energy range e and e + de is given by 8 p V ____ g (e) de = ____ ...(8.40) ÷2me m de h3 where V is the volume of the Fermi-phase space of conductor. So from Eqs (8.39) and (8.40), we can write 8 p V ____ ____________ 1 n (e) de = ____ ÷2me × (e – e )/K T 3 F B h e +1 __

or,

8 ÷2 p V ____________ m3/2 e1/2 de n (e) de = _______ 3 (e – eF)/KBT h e +1 This is the Fermi–Dirac law of energy distribution for free electrons in the metal.

...(8.41)

Advanced Engineering Physics

8.14

8.13.3 total number of particles and total energy at absolute zero of temperature At absolute zero temperature, the total number of electrons (N) is equal to the total number of energy states occupied by the electrons from 0 to eF since each energy state can have only one electron. eF

Therefore,

eF

N = Ú n(e) de = Ú g(e) de 0

0

eF

( )

2m = 4 pV ___ h2

[since f (e) = 1]

3/2

Ú e1/2 de

[using Eq. (8.40)]

0

( )

8 p V 2meF 3/2 N = ____ _____ 3 h2 This is the expression of total number of electrons in a metal at absolute zero. The total energy Et of an electron at absolute zero temperature is given by Et =

eF

eF

0

0

Ú e N ( e ) d e = Ú e g( e ) d e

eF

or,

Et =

Ê ˆ Ú e ÁË h3 ˜¯ 2me ( mde ) 0

or,

Et =

8 2p Vm3/ 2 h3

8p V

eF

Úe

Et =

8 2p V m3/ 2 h3

or,

Et =

8p V Ê 2 me F ˆ 3 ÁË h 2 ˜¯

or, Again, N is given by

Et =

3 Ne 5 F

Ê 2 5/ 2 ˆ ÁË 5 e F ˜¯

( )

3/ 2

Ê3 ˆ ÁË 5 e F ˜¯

3/ 2 È 8p V Ê 2me F ˆ ˘ ÍQ N = ˙ 3 ÁË h 2 ˜¯ ˙ ÍÎ ˚

N = 8p V Ê 2 me F ˆ 3 ÁË h 2 ˜¯ h2 3N eF = ___ ____ 2m 8 p V

de

0

or,

or,

3/ 2

[by Eq. 8.40]

3/ 2

2/3

...(8.42)

N If n = __ = number of free electrons per unit volume, i.e., the free electron density, then V h2 3n 2/3 eF = ___ ___ 2m 8p

( )

...(8.43)

This is the expression of the Fermi energy at absolute zero temperature. Equation (8.43) shows that Fermi N energy depends only on electrons concentration __ and it is totally independent on the size and volume of V

( )

Statistical Mechanics

8.15

the conductor. The values of eF calculated from Eq. (8.43) for a number of metals are of the order of several electron volts. But according to classical statistics all electrons in a metal at absolute zero would have zero energy. Thus we see that the results of quantum statistics are appreciably different from those of classical statistics. eF

8.13.4 fermi energy at non-zero temperature The variation of eF with temperature T is given by Somerfield by the following equation:

[

p2 KB2 T 2 eF (T) = ef (0) 1 – ________ 12e2F (0)

]

eF (0)

...(8.44)

where eF (T) and eF (0) are Fermi energy at finite temperature T and at zero temperature. The variation of eF with temperature T is shown in Fig 8.2. It is seen that as T is increased eF decreases. But the rate of decrease with temperature is very small over a large range of temperature. Hence for all practical purposes, eF(T) may be considered constant and equal to eF(0).

T

Fig. 8.2 Variation of eF with temperature for the free electrons in Cu.

8.13.5 average energy of free electrons in a Metal at zero Kelvin The average integral energy of free electron in a metal at absolute zero is given by _

µ

1 e = __ Ú e n (e) de N0

...(8.45)

eF

µ

1 1 = __ Ú e f (e) g(e) de + __ Ú e f (e) g(e) de N0 N eF Now at T = 0 K the value of f (e) = 1 if e £ eF and at T = 0 K the value of f (e) = 0 if eF ≥ e So, Again So,

_

eF

1 e = __ Ú e g(e) de N0

( ) ( ) Ú ( ) [ ( ) ]

2m g(e) de = 4p V ___ h2 _

3/2 1/2

e

2m 1 e = __ × 4p V ___ N h2 3/2

4p V 2m = ____ ___ N h2

×

3/2

de [from Eq. (8.40)] eF

e3/2 de

0

eF5/2 ____ 5/2

4p V 2m 3/2 __ 2 3/2 = ____ ___ eF eF 2 N 5 h We know, the Fermi energy at absolute zero (T = 0K) h2 3N eF = ___ ____ 2m 8p V

( )

2/3

...(8.46)

Advanced Engineering Physics

8.16

( )

h2 eF3/2 = ___ 2m

or,

3/2

3N ____ 8p V

Substituting the value of eF3/2 in Eq. (8.46), we have _ __ 3 e = eF 5 3 Therefore, the average electron energy is equal to __th of the Fermi energy at absolute zero. 5

...(8.47)

8.13.6 fermi velocity and average velocity of a free electron at zero Kelvin We know that

h2 3n 1 __ mvF2 = eF = ___ ___ 2 2m 8p

( )

2/3

N where vF is the Fermi velocity and n is free electron density, i.e., n = __. V

( )

3n h ___ vF = __ m 8p

So

1/3

...(8.48)

_

Now if v be the average speed of an electron at 0 K, then vF

_

1 v = __ Ú vn (v) dv N0

...(8.49)

where n(v) dv is the number of particles within the velocity range v and v + dv. Now Again at 0 K So,

Therefore,

1 e = __ mv2 or, de = mv dv 2 f (e) = 1 if e £ eF n(e) de = g(e) de 8p V ____ = ____ ÷2me mde h3 2m 3/2 __ 1 n(v) dv = 4p V ___ mv2 2 h2

( ) (

m = 8p V __ h

)

1/2

(mv dv)

( ) v dv. 3

2

So from Eq. (8.49), we have _

vF

m 1 v = __ Ú 8p V __ N0 h

( ) v dv

( )(

8p V m = ____ __ N h

3

3

3

vF4 ___ 4

)

( ) ( __mh ) v [ where __NV = n free electron density ]

3 8p = __ ___ 4 3n

3

4 F

Statistical Mechanics

8.17

Again from Eq. (8.48) 3n ( ) ( ___ 8p ) 3 8p m 3n h ___ v = __ ( ___ ) ( __ ) ( __ m ) ( 8p ) v 4 3n h

h vF3 = __ m Hence

3

_

3

3

F

3 = __ vF 4

...(8.50)

3 So the average speed of the electron in a metal is equal to __ times the Fermi velocity at absolute zero. 4

8.14 derIvatIon of pLanCK’s Law of radIatIon froM be statIstICs Let T be the absolute temperature of a black-body chamber of volume V. The chamber is supposed to be filled with photons. Each photon has unit spin angular momentum. Hence photons are bosons and so we can use the BE statistics to derive Planck’s law of radiation. Photons are continuously emitted and absorbed by the walls of the chamber at constant temperature and constant volume. So, the number of photons is not constant. Therefore, S ni = const. or S d ni = 0 is not valid i

for photon gas although the total energy of the photons remains constant. Hence the multiplier a = 0. Thus, we have from BE distribution law [i.e., Eq. (8.20)] gi ni = _______ e b ei – 1 gi 1 ...(8.51) where b = ____ = ________ ei/KBT K BT e –1 Again energy of each photon e = hv. So from Eq. (8.51) [omitting index i] g n = _________ hv/KBT e –1 The number of photon in the frequency range v and v + dv is obtained by replacing g by g (v) dv and n by n (v) dv. Hence we get from the above equation g(v) dv n(v) dv = _________ ...(8.52) hv/KBT e –1 where g(v) dv is the number of quantum states in the frequency range v and v + dv. The number of quantum states corresponding to the momenta in the range between p and p + dp for particles is given by [See Eq. (8.30)] 4p Vp2 dp g(p) dp = 2 ________ h3 For photons of frequency v, we have hv Energy e = hv and momentum p = ___ c

...(8.53)

8.18

Advanced Engineering Physics

On substituting these values in Eq. (8.53), the number of quantum states with frequency range between v and v + dv is given by

( )

hv 2 __ h 8p V ___ c c dv _____________ g(v) dv = h3 8p Vv2 dv = ________ c3 Hence, substituting the value of g(v) dv in Eq. (8.52), we have

...(8.54)

8p Vv2 dv _________ 1 n(v) dv = ________ 3 hv/KBT c e –1

...(8.55)

The energy of each photon = hv. The energy per unit volume or energy density, within the frequency range v and v + dv is given by hv Uv dv = ___ ◊ n(v) dv V or,

...(8.56)

hv 8p V v2 dv _________ 1 Uv dv = ___ ◊ ________ 3 hv/KBT V c e –1 8p hv3 _________ dv = ______ ...(8.57) 3 hv/KBT c e –1 This is Planck’s law of radiation in terms of the frequency v. In terms of l, Eq. (8.57) can be written as 8p hc _________ dl Ul dl = _____ 5 hc/lKBT l e –1

...(8.58)

(

)

hc This is Planck’s law of radiation in terms of the wavelength l. On longer wavelength side ___ > KBT, Eq. (8.58) becomes l 8p hc –hc/lKBT Ul dl = _____ e dl l5 This is the Wien radiation formula.

...(8.60)

Statistical Mechanics

8.19

8.15 CoMparatIve study of three statIstICaL dIstrIbutIon funCtIons MB Nature of particles Identical, distinguishable, Category of particles classical (called boltzons) Properties of particles Any spin, particles adequately far apart so wave functions do not overlap Examples Examples: molecules of a gas Distribution Function No. of particles per energy state

1 fMB = ______ a + b ei e No upper limit

BE

FD

Identical, indistinguishable (called bosons) 0 or integral spin (1, 2, 3, 4, ... etc.), wave functions are symmetric under interchange of two bosons Examples: Photon, meson, muon

Identical, indistinguishable (called fermions) Half integral spin (1/2, 3/2/ 5/2, ....), wavefunctions are antisymmetric under interchange of two fermions Examples: Proton, electron, neutron

1 fBE = _________ a + b ei e –1 No upper limit

1 fFD = _________ a + b ei e +1 Fermions obey the Pauli exclusion principle, i.e., maximum of one particle per quantum state

Worked Out Problems Example 8.1 Distribute two particles in three different states according to (i) MB Statistics, (ii) BE statistics, and (iii) FD Statistics. Sol. MB Statistics Since the particles are distinguishable, so total number of microstates will be N! gni i 32 _____ W= = 2! __ = 9 microstates [gi = 3, ni = 2, N = 2] ni! 2! In table form: Here A, B are two distinguishable particles. 1

2

3

AB 0 0 A B 0 0 A B

0 AB 0 B A A B 0 0

0 0 AB 0 0 B A B A

Advanced Engineering Physics

8.20

So total number of microstates is 9. BE Statistics Since the particles are indistinguishable, so total number of microstates will be (ni + gi –1)! (2 + 3 – 1)! 4.3.2.1 4! W = __________ = __________ = _____ = ______ = 6 microstates ni! (gi – 1)! 2! (3 – 1)! 2! 2! 2.2 In table form: Here, the two particles are indistinguishable. Let us denote them both as A. 1

2

3

AA 0 0 A 0 A

0 AA 0 0 A A

0 0 AA A A 0

So total number of microstates is 6. FD Statistics Since the particles are indistinguishable and according to the Pauli exclusion principle each state can accommodate only one particle so total number of microstates will be gi! 3! W = __________ = _________ = 3 microstates [gi = 3, ni = 2] ni! (gi – ni)! 2! (3 – 2)! In table form: Here, the two particles are indistinguishable. Let us denote them both as A. 1

2

3

A 0 0

0 A 0

0 0 A

So total number of microstates is 3. Example 8.2 Three distinguishable particles each of which can be in one of the e, 2e, 3e, 4e energy states have total energy 6e. Find all possible number of distributions of all particles in the energy states. Find the number of microstates in each case. [WBUT 2007] Sol. Let the particles be A, B and C. Particles are distinguishable. The possible microstates will be Macrostate

e

2e

3e

4e

Total energy

Microstates

(2, 0, 0, 1)

AB AC BC

0 0 0

0 0 0

C B A

6e 6e 6e

3

(0, 3, 0, 0)

0

ABC

0

0

6e

1

Statistical Mechanics (1, 1, 1, 0)

A B C B A C

B C A A C B

8.21 6e 6e 6e 6e 6e 6e

0 0 0 0 0 0

C A B C B A

6

So, total number of microstates = 10 and total number of macrostates = 3. Example 8.3 A system has non-degenerate single-particle states with 0, 1, 2, 3 energy units. Three particles are to be distributed in these states so that the total energy of the system is 3 units. Find the number of microstates if the particles obey (i) MB Statistics, (ii) BE Statistics, and (iii) FD Statistics. [WBUT 2008, 2012] Sol. (i) MB Statistics Since the particles are distinguishable, let the particles be A, B, C. The possible microstates will be 0e

1e

(0, 3, 0, 0)

0

ABC

(2, 0, 0, 1)

AB AC BC

0 0 0

(1, 1, 1, 0)

A A B B C C

B C A C A B

C B C A B A

Macrostate

2e

3e

Total energy

Microstates

0

0

3e

1

0 0 0

C B A

3e 3e 3e

3

0 0 0 0 0 0

3e 3e 3e 3e 3e 3e

6

So, total number of microstates = 10 (ii) BE Statistics Since the particles are indistinguishable. Let us denote them as A. Macrostate

0e

1e

2e

3e

Total energy

Microstates

(0, 3, 0, 0) (2, 0, 0, 1) (1, 1, 1, 0)

0 AA A

AAA 0 A

0 0 A

0 A 0

3e 3e 3e

1 1 1

So, total number of microstates = 3 (iii) FD Statistics Since the particles are indistinguishable and according to the Pauli principle, each state can accommodate only one particle, so total number of microstates will be

Advanced Engineering Physics

8.22 Macrostate

0

e

2e

3e

Total energy

Microstates

(1, 1, 1, 0)

A

A

A

0

3e

1

So total number of microstates = 1 Example 8.4 Consider a two-particle system, each of which can exist in states e1, e2, and e3. What are the possible states if the particles are (i) bosons, and (ii) fermions ? [WBUT 2006] Sol. (i) BE Statistics Particles are indistinguishable and any number of particles can be accommodated in one quantum state. Let us denote them as A. e1

e2

e3

Microstates

AA 0 0 A 0 A

0 AA 0 A A 0

0 0 AA 0 A A

1 1 1 1 1 1

So, total number of microstates = 6 (ii) FD Statistics Particles are indistinguishable and one state can accommodate only one particle. e1

e2

e3

Microstates

A 0 A

0 A A

A A 0

1 1 1

So, total number of microstates = 3 Example 8.5 Six distinguishable particles are distributed over three non-degenerate levels of energies 0, e and 2e. Calculate the total number of microstates of the system. Find the total energy of the distribution for which the probability is maximum. Sol. There is only one state associated with non-degenerate energy levels. Let N1, N2 and N3 be the number of the particles in three energy states. The total number of particles N1 + N2 + N3 = 6 (given). As the particles are distinguishable, the number of microstates, i.e., the number of ways of choosing N1, N2 and N3 particles from 6 particles is 6 W = __________ N1! N2! N3! where W is the thermodynamical probability. The probability will be maximum when N1! N2! N3! is minimum. This is true when N1 = N2 = N3 = 2. The corresponding total energy distribution is O × N1 + e × N2 + 2e × N3 = 2e + 4e = 6e.

Statistical Mechanics

8.23

Example 8.6 Evaluate the temperature at which there is one per cent probability that a state, with an energy 0.5 eV above the Fermi energy will be occupied by an electron. Sol.

1 The FD distribution is f (e) = ____________ e(e – eF)/KBT + 1 eF is the Fermi energy KB = Boltzmann constant Given e = (eF + 0.5) eV 1 1 f (e) = ____ = _________ 100 1 + e0.5/KBT

Thus, or, or,

1 0.01 = _____x 1+e 0.01 + 0.01x = 1

0.5 where x = ____ KBT

or,

0.99 ex = ____ = 99 0.01

or,

0.5 x = 2.303 log10 99 = ____ KBT

So,

0.109 × 1.6 × 10–19 T = ________________ = 1264 K 1.38 × 10–23

Example 8.7 Calculate the Fermi energy at 0 K of metallic silver containing one free electron per atom. The density and atomic weight of silver is 10.5 g/cm3 and 108 respectively. Sol.

Fermi energy

h2 3N eF = ___ ____ 2m 8pV

Here

N __________ 6.02 × 1023 __ = × 10.5 V 108

( )

2/3

= 5.9 × 1028 m–3 So,

(6.6 × 10–34)2 3 eF = _____________ × _______ × 5.9 × 1028 –31 8 × 3.14 2 × 9.1 × 10 8.8 × 10–19 = 8.8 × 10–19 Joules = _________ eV 1.6 × 10–19 = 5.5 eV

(

)

2/3

Example 8.8 The Fermi energy of silver is 5.51 eV (i) What is the average energy of the free electrons in silver at 0 K? (ii) What is the speed of the electron corresponding to the above average energy? Sol. The average electron energy at 0 K is _ __ 3 e = eF 5 3 = __ × 5.51 = 3.306 eV. 5

Advanced Engineering Physics

8.24

If v is the velocity of the electron, then its kinetic energy 1 __ mv2 = 3.306 × 1.6 × 10–19 2

(

2 × 3.306 × 1.6 × 10–19 v = ___________________ 9.1 × 10–31

or,

)

1/2

m/s

= (1.16 × 1012)1/2 = 1.08 × 106 m/s Example 8.9 Consider a free electron gas at zero degree Kelvin and show that the de Broglie wavelength associated with an electron is given by p lF = 2 ___ 3n0

( )

1/3

where n0 is the number of electrons per c.c. of the gas. Sol. The momentum (p) of the electrons is given by _____

p = ÷2meF 2

~

( )

h 3N eF = ___ ____ 2m 8pV

But

2/3

Again, de Broglie wavelength lF is given by h lF = __ p h h _____ = _______________ ______________ = ______ 2 2me 3N 2/3 h ____ ÷ F 2m ___ 2m 8pV

÷

( ) 8pV pV = ( ____ ) = 2 ( ___ ) 3N ÷ 3N = 2 ___ ( 3np ) [ where n = __NV ] ________ 2/3

1/3

1/3

0

0

Example 8.10 Sol.

Estimate the temperature of the sun, if lm for the sun is 490 nm. From Wien’s displacement law of radiation, we know that lm T = 0.2896 0.2896 0.2896 ~ T = ______ = _________ 5910 K lm 490 × 10–9 So, the temperature of the sun is 5910 K. Here,

Example 8.11 The Fermi energy for sodium at T = 0 K is 3.1 eV. Find its value for aluminum, given that the free electron density in aluminum is approximately 7 times that in sodium. Sol.

n2 Electron density in Al __ ___________________ =n =7 Electron density in Na 1

Statistical Mechanics

( ) ( )

h2 3n1 eF1 = ___ ___ 2m 8p

2/3

For Na, Fermi energy

h2 3n2 eF2 = ___ ___ 2m 8p

2/3

For Al

8.25

...(1) ...(2)

Dividing Eq. (2) by Eq. (1) e F2 n2 ___ = __ n1 eF1

( )

so,

2/3

= (7)2/3

eF2 = eF1 × 72/3 = 3.1 × 3.66 = 11.35 eV.

Review Exercises

Part 1: Multiple Choice Questions 1. Statistical methods give greater accuracy when the number of observations is (a) very large (b) very small (c) average (d) None of these 2. The relation between entropy and thermodynamical probability is given by (a) S = ln W (b) S = eW (c) S = KB ln W (d) None of these 3. The dimension of phase space volume are (a) (length × momentum)3 (b) (time × momentum)3 (c) (energy × time)3 (d) Both (a) and (c) are true 4. The number of macrostates for N particles in MB distribution are N (a) N (b) __ (c) N + 1 (d) None of these 2 5. The number of macrostates for N particles in two compartments obeying MB statistics are N (a) __ (b) N + 1 (c) 2N (d) N – 1 2 6. The number of possible arrangements of two fermions in 3 cells is [WBUT 2008] (a) 9 (b) 6 (c) 3 (d) 1 7. If ni is the number of identical and indistinguishable particles in the ith energy state with degeneracy gi then classical statistics can be applied if [WBUT 2008] n n n i i i (a) __ (b) __ (c) __ (d) gi = 0 gi = 1 gi > 1 8. A coin and a six-faced dice are thrown. The probability that the coin shows tail and the dice shows five is [WBUT 2005] 7 (a) ___ 12

1 (b) __ 8

1 (c) ___ 12

1 (d) __ 6

Advanced Engineering Physics

8.26

9. Fermi–Dirac distribution approaches Maxwell–Boltzmann distribution at (a) low temperature and high pressure (b) low temperature and low particle mass (c) high temperature and high particle mass (d) high density and low energy range 10. MB statistics is applicable for (a) ideal gas (b) electron (c) proton (d) photon 11. Amongst the following statements, which one is wrong? [WBUT 2006] (a) Maxwell–Boltzmann statistics deal with distinguishable particles. (b) Fermi–Dirac Statistics deal with indistinguishable particles. (c) Bose–Einstein statistics deal with indistinguishable particles. (d) Bosons are subjected to the Pauli exclusion principle. 12. Which one of the following are bosons? (a) Photon (b) Phonon (c) Electron (d) Both (a) and (b) 13. The spin angular momentum of photon h (a) h (b) __ (c) 0 (d) 2h 8 14. The FD distribution function is expressed as

15.

16. 17. 18. 19.

20. 21.

1 1 (a) f (e) = ____________ (b) f (e) = ___________ e(e – eF)/KBT + 1 e(e – eF)/KBT – 1 1 (c) f (e) = ________ (d) None of these (e – eF/KBT e _ [WBUT 2007] Average energy e of an electron in a metal at T = 0 K is 3 5 1 (a) eF (b) __ eF (c) __ eF (d) __ eF 2 3 5 Which one of the following is a fermion? (a) Photon (b) Electron (c) Phonon (d) a particle The number of ways in which 4 fermions can be arranged in 5 compartments is (a) 3 (b) 7 (c) 5 (d) 9 Planck’s radiation law for black-body radiation can be derived from (a) MB statistics (b) FD statistics (c) BE statistics (d) None of these The rest mass of the photon is (a) same as that of electron (b) zero (c) infinity (d) None of these The maximum energy that can be occupied by an electron at T = 0 K is (a) Fermi energy (b) Bohr energy (c) chemical potential (d) None of these The mathematical formula for Fermi energy at 0 K is h2 3N (a) ____ ___ 8p V 2m

( )

2/3

( )

8p V ___ 3N (b) ____ h2 2m

2/3

h2 3N (c) ___ ____ 2m 8p V

( )

2/3

(d) None of these

Statistical Mechanics

8.27

22. A perfect black body is one that (a) reflects all radiation (b) absorbs all radiation (c) transmits all radiation (d) None of these 23. For T > 0 K, the probability of occupancy of an electron at Fermi energy level 1 (a) 1 (b) 0 (c) __ (d) 2 2 [Ans. 1 (a), 2 (c), 3 (d), 4 (c), 5 (c), 6 (c), 7 (b), 8 (c), 9 (c), 10 (a), 11 (d), 12 (d), 13 (a), 14 (a), 15 (c), 16 (b), 17 (c), 18 (c), 19 (b), 20 (a), 21 (c), 22 (b), 23 (c)]

Short Questions with Answers 1. Ans. 2. Ans. 3. Ans. 4. Ans.

5. Ans.

6. Ans.

7. Ans.

Define macrostate and microstate. Refer to Section 8.4 Define thermodynamical probability. Refer to Section 8.5 Define phase space. Refer to Section 8.2 What is an ensemble? The collection of very large number of identical interacting macroscopic systems is called an ensemble. There are three types of ensemble (i) micro canonical ensemble – systems cannot exchange energy or matter with each other. N, V and E remain constant, (ii) Canonical ensemble – systems can exchange energy but not matter with each other. N, V and T remain constant. (iii) Grand canonical ensemble – systems can exchange both energy and matter with each other. m, V and T remain constant, where m is the chemical potential. What is the most probable distribution? The most probable distribution of the particles among the energy states in equilibrium is that for which the probability of occurrence is maximum. For mathematical convenience, we consider the condition for maximum value of ln W. The condition for most probable distribution is d ln W = 0. What are the limitations of MB statistics? MB statistics are applicable only to an isolated gas of identical molecules in equilibrium. The expression for MB distribution does not give a correct expression for the entropy of an ideal gas. MB statistics cannot be applied to a system of indistinguishable particles. Why may the number of photons inside an enclosure not remain constant? The total energy of photons inside an enclosure remains constant but the total number of photons at particular temperature T may not remain constant. Photons may be absorbed and emitted by the walls of the container without any restriction. Suppose a photon of energy 2hv may be absorbed by the well and two photons each of energy hv may be emitted. So the photon may be destroyed and created, i.e., the number of photons inside an enclosure may not remain constant.

Part 2: Descriptive Questions 1. Define (i) Phase space, (ii) Macrostate, and (iii) Microstate.

Advanced Engineering Physics

8.28

2. Define thermodynamical probability. Obtain a relation between entropy and thermodynamical probability. 3. Three distinguishable particles, each of which can be in one of the e, 2e, 3e, 4e energy states having total energy 6e. Find all possible distributions of particles in energy states. Find the number of microstates in each case. [WBUT 2007] 4. What are fermions and bosons? Give two examples of each. [WBUT 2004] 5. (a) Sketch the Fermi distribution for T = 0 K and T > 0 K and explain. (b) What is the occupation probability at e = eF? (c) Calculate how the degeneracy function g(e) depends on e for a fermionic gas. (d) Express the Fermi level in a metal in terms of free electron density. [WBUT 2008] 6. (a) Compare MB, BE and FD statistics mentioning at least three characteristics. (b) Draw the Fermi distribution curve for (i) T = 0 K and (ii) T > 0 K. Explain their significance. (c) Indicate which statistics will be applicable for [WBUT 2005] (i) 1H1 (ii) e– (iii) pº (iv) photon [Ans. (i) FD (ii) FD (iii) BE (iv) BE] 7. (i) What do you mean by Fermi energy? Show that it is independent of temperature. 3 (ii) Prove that at 0 K the average energy is __ eF. 5 8. Calculate the total number of particles in a fermionic gas in terms of the Fermi level at absolute zero temperature. [WBUT 2007] 9. Give an expression of BE statistics and hence obtain Planck’s formula for black-body radiation. [WBUT 2002]

Part 3: Numerical Problems 1. Consider a two-particle system, each of which can exist in states e1, e2, e3. What are the possible states, if the particles are (i) bosons, and (ii) fermions? [WBUT 2006] [Ans. 6, 3] 2. Three identical particles can be in any of the five states. What are the number of possible ways of distributing them in various states according to MB, FD and BE statistics? [Ans. 125, 10, 35] 3. Assume in tungsten (at. wt = 183.8, density = 17.3 g/cc) there are two free electrons per atom. Calculate the Fermi energy and electron density. [Ans. 7.2 eV, 1.264 × 1023/cc] 4. Find the Fermi level at absolute zero for Cu. Given that molar mass of Cu, M = 63.55 × 10–3 kg/mol. Its density r = 8.93 × 103 kg/m3. Avogadro’s number N = 6.023 × 1023/mol. h = 6.63 × 10–34 Js. Mass of electron m = 7.1 × 10–31 kg. [Ans. 7.03 eV] 5. Find the Fermi temperature for the valance electrons in Cu. Given eF for copper = 7.03 eV. Boltzmann constant KB = 1.38 × 10–23 J/K. [Ans. 81.5 × 103 K]

( )

N 6. For silver, the electron concentration __ is 5.86 × 1028 m–3. Find its Fermi energy. [Ans. 5.511 eV] V 7. Determine the wavelength lm corresponding to the maximum emissivity of a black body at temperature T equal to 300 K. [Ans. 9660 nm]

APPENDIX

A The Surface Integrals

_›

Let a small outward surface of a body in a vector field F be divided into a finite number n of elementary_› area DSi where i = 1… n and take a point Q (xi, yi, zi) within Dsi. If i is the positive unit normal to Dsi and Fi (xi, xyi,_ zi) is the vector field [Fig. A.1] then normal flux › Z of Fi at Q is F n n

S i=1

i

_›

Fi.

i

DSi

Q (xi ,yi,zi )

…(1) S

Taking the limit of this sum as n Æ , DSi Æ O, then _›

n

Lt

S

Ds Æ 0 i = 1

Fi.

_›

i

DSi = Ú Ú F. dS

i

DSi

…(2)

s

The projection of DSi on the plane XY is DSi | i . | = | i. | DSi Here | i. | DSi = Dxi Dyi

Dxi Dyi or, DSi = ______ | i. |

Thus the sum (1) becomes n _› Dxi Dyi S Fi. i ______ i=1 | i. | In the limit n Æ , then Dxi and Dyi approach zero, and we have _› dx dy Dxi Dyi ______ = ÚÚ F. _____ R | i. | | i. | So, from expression (2) _› _› dxdy Ú Ú F. ds = Ú Ú F. ____ s R | . |

Lt

Lt

n

S

Dx Æ 0 Dy Æ 0 i = 1

_›

Fi.

i

Y O

R Dxi Dyi

X

Fig. A.1

Projection of DSi on the plane XY.

APPENDIX

B Angular Concept What is an Angle? The answer of this easy question is not that easy. Let us try to understand its meaning. The concept of an angle does not exist in a one-dimensional case. In a two-dimensional case, by angle we mean separation of two intersecting lines. To be more specific – an angle is one sense of separation between two intersecting lines which remains same if even the lines are extended up to infinity [Fig. B.1]. C

B

B

C H F

O

O

q

d1

d2

E G A

D (a)

Fig. B.1

A

D (b)

Separation of two intersecting lines (a) in terms of angle (q), (b) in terms of the distance between two equidistant points on two lines from the point of intersection.

If one wants to express the separation in terms of distance one can follow the method given below: Let us select two pairs of equidistant points which are equidistant from the point of intersection O. For one pair of points E and F the distance is d1 and for the other pair of points G and H it is d2, where d1 π d2 but the angular separation q between the two lines remains the same throughout. The unit of measurement of angle is radian (symbol is c or rad) in all systems of measurement like second which is the unit for time in all systems of measurement. If the angle between the two intersecting lines is 90º pc p or 100g (g stands for grade), then in radian one can express it as __ or __ rad. Radian is called a pseudounit as 2 2 it does not have any representation in dimensional analysis.

B.2

Advanced Engineering Physics

Solid angle As has been stated above the counterpart of a two-dimensional angle (planar angle or simply angle) is the solid angle in three dimension. It is usually denoted by the Greek letter W. It is the cone that an object subtends at a point in space. It is a measure of how big an object appears to an observer looking from the said point in space. For example, a small object, which is nearer to the observer could subtend the same solid angle as a large object which is far away from the observer. The solid angle is proportional to the surface area S, of a projection of that object onto a sphere centered at the point where the observer lies, divided by the square of the radius, R, of the said sphere, or it is exactly equal to the surface area, S, of the aforesaid projection of an object on to a sphere of unit radius centered at the point where the observer lies. Symbolically, one can represent it by the following equation: S W = K ___2 R where K is a constant of proportionality. A solid angle is related to the surface of a sphere in the same way as an ordinary angle or planar angle is related to the circumference of a circle. If the proportionality constant is chosen to be unity, then the unit of solid angle in the SI system is steradian (denoted by sr). Thus, the solid angle of a sphere measured from a point in its interior is 4p sr, and the solid angle subtended at the center of a cube by one of its six surfaces is one-sixth of that (i.e., 4p/6 sr) or 2p/3 sr. In general, one can define a solid angle as follows [Fig. B.2]: dA cos q dA¢ ________ d W = ___ = 2 r r2 _

__

dA¢

dA

P

dr dA¢ _____ r ◊ dA q d W = ___ = 3 2 r r r O¢ O R q where d W is the solid angle subtended by any surn face area dA at the point O in space. Here dA¢ is the projection of dA on the plane which is normal to OR and passes through the point O¢ r Q i.e., dA¢ = dA cos q. Fig. B.2 A solid angle d W at the point O For a spherical surface whose radius of curvature subtended by the surface dA¢ which is normal is OO¢ = r to the bisector OR of angle –POQ. dA¢ = 4pr2 and q = 0, so the solid angle subtended by the surface of the sphere having radius r is given by

or,

dA cos q W = Ú ________ = 4p sr. s r2

APPENDIX

C Useful Physical Constants Rest mass of electron Charge of electron Specific charge of electron Rest mass of proton Charge of proton Speed of light in vacuum Planck’s constant Boltzmann constant Universal gas constant Gravitational constant Avogadro’s number Refractive index of water Refractive index of glass Viscosity of water (20°C) Standard atmospheric pressure Bohr radius Compton wavelength of electron Permittivity of free space Permeability of free space Stefan’s radiation constant Rydberg constant Young’s modulus of iron Modulus of rigidity (iron)

me e e/me mp ep c h k R G NA mw mg hw PS r lc e0 m0 s Rd YS h

9.1 × 10–31 kg –1.6 × 10–19 C 1.758 × 10–11 C/kg 1.673 × 10–27 kg + 1.6 × 10–19 C 3 × 108 m/s 6.63 × 10–34 Js 1.38 × 10–23 J/K 8.314 J/(k mol) 6.6720 × 10–11 Nm2/kg2 6.023 × 1023 mol–1 1.33 1.50 1.002 × 10–3 Ns/m2 1.013 × 105 Nm–2 5.29 × 10–11 m = 0.53 Å 2.42 × 10–12 m 8.85 × 10–12 F/m 4p × 10–7 H/m 5.67 × 10–8 W/m2 K4 1.09737 × 107 m 19.5 × 1011 Nm–2 7.5 × 1010 Nm–2

APPENDIX

D Derivation of Gradient, Divergence, Curl and Laplacian Cartesian Coordinate System _›

Let y(x, y, z,) and A(x, y, z) be respectively a scalar and a vector function of x, y, and z. (i) The gradient of y is given by ∂y ∂y ∂y — y = ___ + ___ + ___ ∂x ∂y ∂z __

_›

(1)

(ii) The divergence of A is given by __ __

—◊A=

(

)

∂ ∂ ∂ ___ + ___ + __ . ( Ax + Ay + Az) ∂x ∂y ∂z

__ __

or, __

∂A ∂A ∂A — ◊ A = ____x + ____y + ____z ∂x ∂y ∂z

(2)

(iii) Curl of A is given by

| |

__

__ ∂ ∂ ∂ — × A = ___ ___ __ ∂x ∂y ∂z Ax Ay Az __

or,

__

—×A=

(

) (

∂A ∂Az ____ ____ – y – ∂y ∂z

) (

∂Az ____ ∂A ____ – x + ∂x ∂z

∂Ay ____ ∂A ____ – x ∂x ∂y

)

(3)

(iv) Laplacian of y is given by __ __

—2 y = — ◊ —y =

or,

(

)(

∂ ∂ ∂ ___ + ___ + __ ◊ ∂x ∂y ∂z

∂ 2 y ____ ∂ 2y ____ ∂ 2y —2 y = ____ + + ∂x2 ∂y2 ∂z2

∂y ∂y ∂y ___ + ___ + ___ ∂x ∂y ∂z

) (4)

Advanced Engineering Physics

D.2

y

Cylindrical Polar Coordinate System _›

Let y (r, f, z) and A(r, f, z) be respectively a scalar and a vector function of r, f and z.

j

Before derivation of the expressions of del (—), gradient, divergence, curl and Laplacian in cylindrical polar coordinate system, let us find the expression of the unit vectors , and . From Fig. D.1, we can write

and

= cos f + sin f

(5)

= – sin f + cos f

(6)

r

f

__

r

90° f 90° – f

O

x i

z k

Fig. D.1 and

Unit vectors

in the XY plane.

Now, differentiating Eqs (5) and (6), we obtain, ∂ ∂ ∂ ___ = 0, ___ = , ___ = 0 ∂r ∂j ∂z ∂f ∂ ∂ ___ = 0, ___ = – , ___ = 0 ∂r ∂j ∂z

(7)

∂ ∂ ∂ ___ = 0, ___ = 0, ___ = 0 ∂r ∂j ∂z (i) Gradient of y By using the method of partial differentiation on the scalar function y, we get ∂y ∂y ∂y dy = ___ dr + ___ df + ___ dz ∂r ∂j ∂z _

(8)

__

dy = (d r) ◊ (— y)

Again,

(9)

_

where d__r = dr + (rdf) + dz Since —y is a vector quantity, we can express it as follows: __

__

__

__

—y = (—y)r + (—y)f + (—y)z \

_

dy = (d r) ◊ (—y) =

[

][

dr + (rdf) + dz ◊ __

or,

(10)

__

__

__

__ ∂y (—y)r = ___ ∂r __ __ ∂y 1 ∂y (—y)f r = ___ fi (—y)j = __ ___ r ∂j ∂f

__

]

__

dy = (—y)r dr + (—y)f rdf + (—y)z dz Now, comparing Eqs (8) and (11), we can write

__

(—y)r + (—y)f + (—y)z

(11)

Appendix D

D.3

__ ∂y (—y)z = ___ ∂z

and

__

__

__

__

∂y ___ + r

Putting these values of (—y)r, (—y)j and (—y)z in Eq. (10), we get —y =

∂y ∂y 1 ___ __ + ___ r ∂f ∂z

(12)

Equation (12) in the expression of gradient of y in cylindrical coordinate system. __

\

∂ ∂ 1 ∂ ___ + __ ___ + __ r ∂j ∂r ∂z

—=

(13)

__

where Eq. (13) is the expression of del (—) in cylindrical coordinate system. __

(ii) Divergence of A __ __

—◊A=

(

)

∂ ∂ 1 ∂ ___ + __ ___ + __ ◊ ( Ar + Aj + Az) r ∂f ∂r ∂z

Now, using the set of Eq. (7), we can expand the above equation to obtain, __ __ 1 ∂ 1 ∂Af ∂Az — ◊ A = __ ___ (rAr) + __ ____ + ___ r ∂r r ∂f ∂z

[

◊

◊ = ◊

=

= 0]

(14)

__

This is the expression of divergence of A in cylindrical polar coordinate system. __

(iii) Curl of A The set of unit vectors { , , ) is an orthogonal set of unit vectors. \

×

= ,

× = and

×

=

(15)

__

curl of A is given by __

__

—×A=

(

∂ ___ + ∂r

)

∂ ∂ 1 ___ __ + __ × ( Ar + Aj + Az) r ∂j ∂z

Now, by using sets of Eqs (7) and (15) we can expand the above equation, and ultimately we will obtain, __ __ ∂ (Af) ∂A ∂Az ∂Ar 1 ∂Az ∂Aj — × A = __ ___ – ____ – ___ – ____ + __ _____ – ___ r ∂f r ∂r ∂z ∂r ∂z ∂f 1 = __ r

[ [{

] [ ] [ ] } { } { }]

∂(rAj) ∂Az ______ ___ – –r ∂j ∂z

∂Ar ∂Az ____ ___ – + ∂r ∂z

∂ (rAj) ∂Ar _______ – ____ ∂r ∂j

(16)

Advanced Engineering Physics

D.4

__

1 ∂ — × A = __ ___ r ∂r Ar

or,

r ∂ ___ ∂j rAf

| |

__

∂ __ ∂z Az

(17)

(iv) Laplacian To obtain the expression of the laplacian in cylindrical polar coordinate system, let us take divergence of the gradient of y (r, f, z). __

__

__ __

— ◊ (—y) = — ◊ —y = —2y __

__

or,

—2y = — ◊ (—y)

or,

—2y =

or,

∂2y 1 ∂2y ∂2y —2y = ____2 + __2 ____2 + ____ ∂r r ∂f ∂z2

)(

(

∂ ∂ 1 ∂ ___ + __ ___ + __ ◊ r ∂r ∂f ∂z

∂y ∂y 1 ∂y ___ + __ ___ + ___ r ∂r ∂f ∂z

)

[By using set of Eq. (7) and the fact that 2

2

◊

= ◊ = ◊

= 0]

2

∂ ∂ 1 ∂ —2 = ___2 + __2 ___2 + ___2 ∂r r ∂f ∂z

Hence,

(18)

(19)

which is the expression the laplacian in cylindrical polar coordinate system

Spherical Polar Coordinate System __

Let y (r, q, f) and A (r, q, f) be respectively a scalar and a vector function of r, q and f. Before derivation of the expressions of del (—), gradient divergence, curl and laplacian in spherical polar coordinate system, let us first find the expression of the unit vectors , and in terms of the cartesian unit vectors , and . r The positional vector r is given by r r = r sin q cos f + r sin q sin f + r cos q So, the unit vector is given by r = r /r = sin q cos f + sin q sin f + cos q _

∂r ___ = sin q cos f + sin q sin f + cos q ∂r So, in general, we can express the unit vectors as follows: Again,

_

_

r› ∂___ ∂r

[ ] /[| |] [ ] /[| |]

∂r = ___ ∂r _

∂r = ___ ∂q

= sin q cos f + sin q sin f + cos q

(20)

r› ∂___ = cos q cos f + cos q sin f – sin q ∂q

(21)

_

Appendix D _

D.5

_

[ ] /[| |]

∂r = ___ ∂f

and

∂r ___ = – sin q sin f + sin q cos f ∂f

(22)

As , and form a set of orthogonal set of unit vectors, we can write

and

×

= ,

×

=

(23)

× =

Also, we have ◊

= ◊ = ◊ =0

(24)

(i) The gradient of y Differentiating y (r, q, f) partially with respect to r, q and f, we get ∂y ∂y ∂y dy = ___ dr + ___ dq + ___ df ∂r ∂q ∂j We have already expressed dy in cartesian coordinate system as

(25)

_ __

dy = (d r ◊ —y) _

_

_

_

The direction of d r may be any direction including that of r. Let us denote d r by dl to emphasize on this _ nature of d r. _ __

\

_ __

dy = (d r ◊ —y) = (dl ◊ —y) _

dl = dr + r dq + r sin qdf

Again,

(26)

__

Since —y is a vector quantity, thus we can write it as __

__

__

__

—y = (—y)r + (—y)q + (—y)f __

__

__

(27)

__

where (—y)r , (—y)q and (—y)j are the components of —y along , and respectively. _

__

dy = (dl) ◊ (—y) = ( dr + r dq + r sin qdf) ◊ __

or,

__

[

__

__ ∂y (—y)r = ___ ∂r

__

(—y)f

]

__

dy = (—y)r dr + (—y)q r dq + (—y)j r sin qdf Comparing Eqs (25) and (28), we can write

__

(—y)r + (—y)q +

(28)

Advanced Engineering Physics

D.6

and

__ __ ∂y 1 ∂y r (—y)q = ___ fi (—y)q = __r ___ ∂q ∂q __ __ ∂y 1 ∂y r sin q (—y)j = ___ fi (—y)j = ______ ___ r sin q ∂j ∂j

Therefore, Eq. (27) can be written as ∂y 1 ∂y —y = ___ + __r ___ + ∂r ∂q

∂y 1 ___ ______ r sin q ∂j

__

∂ 1 ___ ______ y r sin q ∂j

__

—y =

or, __

(

(29)

)

∂ 1 ∂ __ + __r ___ + ∂r ∂q

So, the del operator (—) in spherical polar coordinate system can be written as __

∂ 1 ∂ — = __ + __r ___ + ∂r ∂q

∂ 1 ___ ______ r sin q ∂j

(30)

We have the expressions of unit vectors , and as follows: = sin q cos f + sin q sin f + cos q, = cos q cos f + cos q sin f – sin q = – sin f + cos f Let us now differentiate , relations after differentiation,

and

with respect to various coordinates; we are left with the following

∂ ∂ ∂ ___ = 0, ___ = , ___ = sin q ∂r ∂q ∂j ∂ ∂ ∂ ___ = 0, ___ = – , ___ = cos q ∂r ∂q ∂j

__

(31)

∂ ∂ ∂___ = 0, ___ = 0, ___ = – sin q – cos f ∂r ∂q ∂j

(ii) Divergence of A __ __

—◊A=

(

∂ __ ___ ∂ __ +r + ∂r ∂q

)(

∂ 1 ___ ______ ◊ r sin q ∂q

Ar + Aq + Aj

)

Now, using the Eqs (24) and (31) and expanding the dot product, we get __ __

∂ 1 ∂ — ◊ A = ◊ __ ( Ar + Aq + Aj) + ◊ __r ___ ( Ar + Aq + Aj) ∂r ∂q +

∂ 1 ◊ ______ ___ ( Ar + Aq + Aj) r sin q ∂j

Appendix D

D.7

∂ 1 ∂ 1 1 ∂Aj — ◊ A = __2 __ (r 2Ar) + ______ ___ (sin qAq) + ______ ____ r sin q ∂q r sin q ∂j r ∂r __ __

or, __

(32)

(iii) Curl of A __ In order to find the expression of (curl A) one may use the Eqs (23) and (31). __

__ ∂ 1 ∂ — × A = × __ ( Ar + Aq + Aj) + × __r ___ ( Ar + Aq + Aj) ∂r ∂q

+

[

∂ 1 × ______ ___ ( Ar + Aq + Aj) r sin q ∂j

] [

__ ∂Aq ∂ 1 1 ∂ 1 ∂Ar — × A = ______ ___ (sin q Aj) – ____ – __r __ (r Aj) – _____ ___ sin q ∂j ∂q ∂j ∂r r sin q __

or,

] [

∂Ar ∂ __ ___ + __ r ∂r (r Aq) – ∂q __

__ 1 — × A = _______ r 2 sin q

or,

[{

}

∂ ∂ ___ (r sin qAj) – ___ (rAq) – r sin q ∂q ∂j

{

∂ 1 ∂Ar __ (rAj) – _____ ___ sin q ∂j ∂r

{

+ r sin q __

__ 1 — × A = _______ 2 r sin q

or,

[{

} {

∂ ∂ ___ (r sin qAj) – ___ (r Aq) – r ∂q ∂j

{

+ r sin q

|

r sin q r ∂ ∂ ∂ 1 ___ __ ___ — × A = _______ 2 ∂f ∂r ∂q r sin q rAr rAq r sin qAj __

or,

__

|

__

or,

__ __

y = —2y = — ◊ —y —2y =

(

)(

∂ __ ___ ∂ ______ ___ ∂ __ + + ◊ ∂r r ∂q r sin q ∂j

)

∂ ______ ___ ∂ __ ___ ∂ __ + y + ∂r r ∂q r sin q ∂j

Now, by expanding using dot product rules, one can get ∂y ∂y ∂2y ∂ 1 ∂ 1 ___ ___ ____ —2y = __2 __ r2 ___ + _______ sin q + ∂q ∂r r ∂r r 2 sin q ∂q ∂f2

( )

}]

}

∂Ar ∂ __ (r Aq) – ___ ∂r ∂q

}]

(33)

(iv) The laplacian ( —2) and —2y Laplacian of

}

∂Ar ∂ __ (r Aq) – ___ ∂r ∂q

∂Ar ∂ __ (r sin qAj) – ___ ∂r ∂j

]

(

)

Advanced Engineering Physics

D.8

∂ ∂ ∂2 ∂ 1 ∂ 1 ___ —2 = __2 __ r 2 __ + _______ sin q ___ + ____2 2 ∂r ∂q r ∂r r sin q ∂q ∂j

( )

or, Problem 1

(

)

(34)

Show that in spherical polar coordinates

where , and

∂ ___ = – sin q – cos q ∂j are unit vectors

Ans. We know that

= sin q cos f + sin q sin f + cos q = cos q cos f + cos q sin f – sin q = – sin f + cos f ∂ ___ = – cos f – sin f ∂f

or,

∂ sin q ___ = – sin q cos f – sin q sin f ∂f

or,

∂ sin q ___ = – ( sin q cos f + sin q sin f + cos q) + cos q ∂f

or,

∂ sin q ___ = – + cos q ∂j

or,

∂ sin2 q ___ = – sin q + sin q cos q ∂j

(i)

(ii)

Multiplying Eq. (i) by cos q, we get ∂ cos q ___ = – cos q cos f – cos q sin f ∂j or,

∂ cos q ___ = – ( cos q cos f + cos q sin f – k sin q) – sin q ∂j

or,

∂ cos q ___ = – sin q ∂j

or,

∂ cos2 q ___ = – cos q – sin q cos q ∂j Adding Eqs (ii) and (iii), we get ∂ ___ = – sin q – cos q ∂j

(iii)

Appendix D __

D.9

__

Problem 2 Derive the expression of — × A where __

Ans.

__

__

__ _

__

(

__

A = A (r) = A (r, q, j)

__

curl A = — × A __

—×A=

or,

∂ 1 ∂ __ + __r ___ + ∂r ∂q

)

∂ 1 ___ ______ × ( Ar + Aq + Aj) r sin q ∂j

Now, using Eqs (23) and (31), we get __ __ ∂ 1 ∂ — × A = × __ ( Ar + Aq + Aj) + × __r ___ ( Ar + Aq + Aj) ∂r ∂q ∂ 1 + × ______ ___ ( Ar + Aq + Aj) r sin q ∂j ∂Aj ∂Aq ∂ ∂ ∂ 1 ∂Ar = Ar × __ + × ____ + Aq × ___ + × ____ + Aj × ___ + × __r ___ ∂r ∂r ∂r ∂r ∂r ∂q Aq ___ Ar ___ ∂ ∂ ___ + × __ r ∂q + × r ∂q + × Ar + ______ r sin q __

or,

__

—×A=

∂r × ___ + ∂j

∂Aj ∂Aq ____ – ____ – ∂r ∂r

∂Aj A ∂ 1 ____ __ __j ___ r ∂q + × r ∂q +

1 ∂Aq × ______ ____ + r sin q ∂j

∂Ar 1 ___ __ r ∂q +

1 ∂Ar × ______ ___ r sin q ∂j Aj ∂ × ______ ___ r sin q ∂j

Aq ∂ × ______ ___ + r sin q ∂j

∂Aj Aq ∂Aq ∂Ar 1 ____ 1 ____ ___ __ ______ ___ _____ r + r ∂q + r sin q ∂j – sin q ∂j Af cos q Aj + _______ – ___ r r sin q

__

or,

__

—×A=

(

) (

Aj cos f ∂Aq _______ ∂Aj ______ 1 ____ 1 ____ __ – + + r ∂q r sin q ∂j r sin q

Aj ∂Aj ∂Ar ___ 1 ___ ______ – r – ____ r sin q ∂j ∂r

(

+

(

) (

+

(

) (

∂ (sin qAj) ∂Aq __ 1 1 1 ∂Ar ∂ — × A = ______ _________ – ____ + __r _____ ___ – __ (rAj) r sin q sin q ∂j ∂r ∂q ∂j __

or,

∂Aq __ 1 ∂Ar Aq ____ – r ___ + ___ r ∂r ∂q

∂Aj ∂Aq ∂Aj __ 1 1 1 ∂Ar — × A = ______ sin q ____ – ____ + Aj cos q + __r _____ ___ – Aj – r ____ r sin q sin q ∂j ∂q ∂j ∂r __

or,

) )

)

(

∂Aq ___ ∂Ar 1 ____ __ r r ∂r – ∂q + Aq

)

) +

(

∂ (rAq) ___ ∂Ar 1 ______ __ – r ∂r ∂q

)

Advanced Engineering Physics

D.10

[{

__

or,

__ 1 — × A = _______ 2 r sin q

} {

∂ (r sin qAj) ∂ (r Aq) ___________ – ______ – r ∂q ∂j

∂ (r sin q Aj) ∂Ar ___________ – ___ ∂r ∂j + r sin q

or,

|

r ∂ ___ ∂ 1 _______ __ —×A = 2 r sin q ∂r ∂q Ar r Aq __

__

r sin q ∂ ___ ∂j r sin qAj

}

Ï ∂(rAq ) ∂Ar ¸˘ Ì ˝˙ ∂q ˛ ˚ Ó ∂r

|

Problem 3 Derive the expression for laplacian in spherical coordinates. Ans. Laplacian of y is given by __ __

—2y = — ◊ —y

(

)(

∂ __ ___ ∂ ______ ___ ∂ __ + + ◊ ∂r r ∂q r sin q ∂j

or,

—2y =

or,

∂ —2y = ◊ __ ∂r

(

∂ __ ___ ∂ __ + + ∂r r ∂q

)

∂ __ ___ ∂ ______ ___ ∂ __ + + y ∂r r ∂q r sin q ∂j

)

∂ 1 ___ 1 ∂ ______ y + ◊ __r ___ r sin q ∂j ∂q

∂ __ ___ ∂ __ + + ∂r r ∂q

∂ 1 ◊ ______ ___ r sin q ∂j

+

or,

(

(

)

∂ 1 ___ ______ y r sin q ∂j

)

∂ __ ___ ∂ ______ ___ ∂ __ + + y ∂r r ∂q r sin q ∂j

∂ 2y ____ ∂2y _ ___ ∂y ∂ ∂y ∂ 1 ◊ ____ ◊ 1 ∂y +r◊ – __r ◊ ___ – _______ —2y = ◊ ___ ___ + ◊ ____ + + __r ___ r 2 2 ∂r ∂r ∂r∂q ∂r ∂q ∂r ∂r r sin q 2 ∂y __ ∂y _______ ∂y ∂ 2y _______ cos q ___ 1 ∂ y __ 1 _____ 1 1 ___ ____ + __2 ____ + + + + 2 2 (– sin q – cos q) ___ r 2 2 sin q ∂q 2 2 ∂r ∂j r ∂q r r sin q ∂j r sin q

∂ 2y __ ∂y __ ∂ 2y _______ ∂y _______ ∂ 2y cos q ___ 1 ____ 1 2 ___ ____ = ____ + + + + ∂r 2 r ∂r r 2 ∂q 2 r 2 sin q ∂q r 2 sin q ∂j 2

(

)(

)(

∂ 2y __ ∂y ∂ 2y _______ ∂y ∂ 2y cos q ___ 2 ___ 1 ____ 1 __ _______ ____ = ____ + + + + ∂r 2 r ∂r r 2 ∂q 2 r 2 sin q ∂q r 2 sin q ∂j2 \

∂y ∂ 2y ∂ ∂ 1 ∂ 1 ___ ___ ____ y —2y = __2 __ r2 ___ + _______ sin q + ∂r ∂q r ∂r r 2 sin q ∂q ∂j 2

or,

∂ ∂ ∂2 ∂ 1 ∂ 1 ___ —2 = __2 __ r 2 __ + _______ sin q ___ + ____2 2 ∂r ∂q r ∂r r sin q ∂q ∂j

( ) ( )

( (

)

)

)

APPENDIX

E Set of Experiments (For Practical Classes) EXPERIMENT NO. 1 Aim: Determination of Planck’s constant (h) by measuring consisting filament radiation in a fixed spectral range. Apparatus: A Planck’s constant measurement kit which consists of a filament bulb with series resistance to control its supply current, an ammeter, a voltmeter, a small optical bench, a photovoltaic cell and a microammeter to measure the photocurrent. Filament current A Power supply

Filament voltage

V

Filament

Control (resistance) (a)

Filament

Photovoltaic cell

Fig. E.1 (a) Filament and power supply circuit with resistance, ammeter and voltmeter. (b) Filament and photovoltaic cell arrangement.

E.2

Advanced Engineering Physics

Theory: For a black body at absolute temperature T, the total radiation and the spectral distribution for this radiation are functions of the temperature alone, the spectral distribution is given by

[ ( ) ]

–1

8p hc hc Eldl = _____ exp ____ – 1 5 lkT l

...(1)

dl

While working with visible light and temperature up to 2500 K, we have hc ____ >> 1 lkT Hence Eq. (1) gets reduced to 8p hc – hc × exp ____ dl Eldl = _____ 5 lkT l

[ ( ) ]

...(2)

If the radiation is received through a filter on a photovoltaic cell, and the galvanometer response q is measured, we have Bl hc q = 8p hc A Ú ___5 exp – ____ dl ...(3) lkT l where, A is a factor depending on the geometry of the arrangement and the sensitivity of the galvanometer and the factor Bl is a function of l which includes (i) transmission characteristics of the filter and (ii) the frequency response characteristic of the photovoltaic cell. For a very narrow transmission band of the filter, Eq. (3) reduces to the form – hc q = 8p hc ACl0 exp ____ Dl0 l0kT

(

)

( )

(

)

hc = h exp – _____ l0kT where

h = 8p hc ACl0 Dl0

or,

– hc q = h exp _____ l0kT

or,

hc ln q = ln h – ______ l0kT

( )

...(4)

l0kD (ln q) ...(5) h = – __________ c D(T–1) where the slope of the curve (ln q versus T –1) is s = D(ln q)/D(T –1) l0ks h = – ____ ...(6) c Here l0 is the mean effective wavelength of the filtered light. For measurement of the temperature of the filament the Draper resistance of the given tungsten wire is measured. The Draper temperature and the temperature coefficients of resistance of tungsten being known a priori, the variation of the ratio (R/Rg) with temperature (in K) can be predicted and therefore supplied in the manual. (Rg is the Draper resistance) The temperature of the bulb can be measured by measuring its resistance and using the graph of T versus R/Rg. Hence,

Appendix E

E.3

Procedure: 1. Turn on the system. Keep the voltage control knob in minimum position. Initially vary the voltage very slowly and observe carefully, when the bulb exactly starts glowing. A dark room will help this part of the experiment because stray light will produce an erroneous result. Note down the current and voltage and hence the resistance of the bulb wire. This is the Draper resistance Rg. 2. Increase the voltage in small steps and record the current and hence calculate the resistance. These are the values of the resistance R. Note down also the photocurrent q. Take at least ten readings in the full range of the voltmeter. 3. From the supplied table of R/Rg and T draw a graph and refer it as graph 1. Use the above graph to determine the value of the temperature T of the filament from the experimentally observed value of 1 R/Rg obtained in steps 1 and 2. Hence, calculate __ . T 1 4. Draw a graph of lnq versus __ and determine the value of Planck’s constant h from the slope of this T curve from Eq. (5). Observations: Table 1 Supplied data for variation of R/Rg with T.

(

R Ratio of R and Rg i.e., ___ Rg

S. No.

)

Absolute Temperature T

1. 2. 3. 4. 5. Table 2

To find Draper resistance.

S. No.

Voltage, V (V)

Current, I (A)

Resistance Rg (W)

Average, Rg (W)

1. 2. 3. Table 3

To find the resistance and hence the temperature of the filament wire.

S. No. Voltage, V (V)

1. 2. 15.

Current I (A)

Resistance, R (W)

R ___ Rg

Temperature, T (K)

1 __ × 10–4 T –1 (K )

Photocurrent, q (A)

ln q

Advanced Engineering Physics

E.4

Calculations: Results: Percentage error: Discussion: Aids to Viva Voce: 1. What is Planck’s constant? Ans. It is a fundamental physical constant. It gives the order of energy exchange in case of quantum mechanical action. It is the ratio of the energy of a photon to its frequency. 2. What is the value of Planck’s constant? State its unit. Ans. The value of Planck’s constant is given by h = 6.6262 × 10–34 J-s. Its unit is joule-second. 3. Write down Planck’s radiation law. 8 p hc dl Ans. Planck’s radiation law is given by uldl = ◊ 5 l exp Ê hc ˆ - 1 Ë l kT ¯

4. Ans. 5. Ans. 6. Ans. 7. Ans.

where ul = the energy density at work-length l, h = Planck’s constant, c = velocity of light in the vacuum, k = Boltzmann’s constant, and T = Absolute temperature. What is quantum theory of light? It is one of the dual theories of light. It states that energy can be exchanged between two elementary particles in discrete quantum only. From which statistical distribution function Planck’s radiation law is derived? It is derived from Bose-Einstein statistical distribution function. How can you classify quantum particles? Quantum particles can be classified into two categories depending on their spin angular momentum. Namely, bosons and fermions. What are the values of spin angular momentum for bosons and fermions? The spin angular momentum is given by Ls = ms for bosons ms = n and for fermions ms = Ê n + 1 ˆ where n is an integer. Ë 2¯

8. What are bosons and fermions? Ans. Bosons are the particles which have integral values for their spin angular momentum while in case of fermions, spin angular momentum values are half-integral. 9. Cite some examples of bosons and fermion? Ans. Examples of bosons are phonon, photon, a-particles, etc., while examples of fermions are electron, proton, neutron, 2He3 nuclei, etc.

Appendix E

E.5

10. What are the values of spin in case of photons and electrons? Ans. For photons it is unity and for electrons it is ± 1 . 2 11. Why do we take help of B–E statistical distribution function to determine the value of Planck’s constant? Ans. In the present experiment, we use a filament which emits light photons. As photons are bosons, we take help of B–E statistics to determine Planck’s constant. 12. What is photovoltaic cell? Ans. It is a pn-junction which can convert light energy into electrical energy.

EXPERIMENT NO. 2 Aim: Determination of Stefan’s constant by using a vacuum tube diode. Apparatus: Stefan’s constant apparatus with a diode value. Cathode

Anode

Filament

A

K F (a)

(b)

Fig. E.2 (a) Diagram of a vacuum tube diode. (b) Circuit diagram of a vacuum tube diode.

Theory: In the experiment of Stefan’s constant one usually uses a commercially available vacuum tube diode which has a cylindrical cathode made of nickel (Fig. E.2). Inside the cathode, a tungsten filament is used as a heater. The cathode is heated by passing electric current through the tungsten filament. The temperature of the filament can be determined by using a known resistance (RT) and temperature (T) relationship for tungsten as given below: RT T ____ = ____ R2T3 273

( )

1.2

where RT is the resistance of the tungsten filament at temperature T K, and R273 is the resistance of the filament at temperature 273 K.

...(1)

Advanced Engineering Physics

E.6

The above Eq. (1) can be approximated as RT T ____ = ____ R300 300

( )

1.2

...(2)

where, R300 is the resistance of the filament at 300 K, a value very close to the room temperature. It is possible to estimate the value of the filament resistance at room temperature from a plot of the resistance of the wire for very small voltage range following an extrapolation to zero volt. From a current, voltage data for the tungsten filament resistance RT (and hence RT/R300) can be estimated. From a supplied data table, temperature T versus (RT /R300) graph can be drawn which can be used to estimate the temperature T from the measured value of (RT /R300). On the other hand, the power generated by the filament is given by VI, where V is the voltage across the filament and I is the current through it. Neglecting other modes of heat loss, this is equal to the power radiated by the filament. From Stefan’s law, one has, P = E s ST n Taking logarithm of both sides of the Eq. (3), one gets log10 P = log10 (E s S) + n log10 T where,

...(3) ...(4)

s = Stefan’s constant, P = Power radiated by the filament, E = Emissivity of the cathode,

and

S = 2p rl = surface area of the cathode, r = Radius of the cathode, l = Length of the cathode.

Procedure: 1. Switch on the vacuum tube diode. Apply a small voltage across the filament. Note down the current. 2. Increase the voltage in small steps up to the maximum possible value. Note down the voltage in each step. Take at least ten readings within the range of the voltage. 3. Determine the resistance in each case. 4. Plot a graph of resistance Rt versus voltage V only for small values of V. Extrapolate the curve to V = 0 volt. This will be a good approximation of R300. 5. Calculate the values of (RT /R300). 6. Plot the supplied data of the variation of (T) versus (RT /R300) in a graph. 7. Use the above graph to estimate the values of (T) corresponding to the values of calculated (RT /R300) in a graph. 8. Calculate the values of the power by multiplying the current and the voltage obtained in Step 2. 9. Plot a graph of logP10 versus logT10. The slope of the curve (in this case a straight line) gives the estimate of the power n of temperature T in Stefan’s law. 10. Choose a suitable point on the curve and use Eq. (4) to estimate the Stefan’s constant.

Appendix E

E.7

Observations: Table 1 Resistance-Temperature (K) data (supplied). S. No.

RT/R300

T(K)

– – –

– – –

– – –

Table 2 I-V (current-voltage) data of the filament wire. S. No.

Vf (V)

If (A)

P = Vf If (W)

RT (W)

RT/R300

TK

logT10

logP10

– – –

– – –

– – –

– – –

– – –

– – –

– – –

– – –

– – –

Calculations: Results: Percentage error: Discussion: *Aids to Viva Voce: 1. What is Stefan’s law of radiation? Ans. This law states that the total heat radiated per second per unit surface area of a perfectly blackbody is directly proportional to the fourth power of its absolute temperature. 2. What is Stefan’s constant? Ans. If the total energy radiated per second per unit surface area of a perfectly blackbody be E and its absolute temperature be T, then from Stefan’s law, we get, E μ T4 or, E = sT4, where s is the constant of proportionality. This constant s is called Stefan’s constant. 3. Write down Stefan’s law for any general radiating body. Ans. Stefan’s law can in general be written as P = EsSTn where P = Power radiated by the hot body, E = Emissivity of the radiator, s = Stefan’s constant, S = Surface area of the radiator, T = The absolute temperature of the radiator, and n = a constant.

Advanced Engineering Physics

E.8

4. What will be the value of n in the equation P = EsSTn for a perfectly blackbody? Ans. The value of n in the above equation will be exactly 4 (four) for a perfectly blackbody radiator and n < 4 for all real bodies. 5. Does Stefan’s constant depend on the wavelength of radiation? Ans. No. It is applicable to whole range of wavelength. 6. What is the relation between the temperature of a filament and its resistance? Ans. If RT be the resistance of a filament at absolute temperature T, then RT = C T1.2 where C is a constant of proportionality. 7. What wavelengths are present in blackbody radiation? Ans. All wavelengths from zero to infinity. 8. What is the standard value of s? Ans. It lies between 5.32 × 10–8 to 6.15 × 10–8 Wm–2K–4. 9. What is the application of the value of the Stefan’s constant s? Ans. It is used to determine the temperature of heavenly bodies such as the sun and other stars. 10. What has been used in the present experiment as black body radiator? Ans. In this experiment, a vacuum tube diode has been used as a blackbody radiator. 11. What is a diode? Ans. It is an electronic device which allows current to flow in one direction only in a circuit. 12. How many technologies are available for making a diode? Name them. Ans. There are two technologies to build a diode, namely, vacuum tube technology and semiconductor technology. 13. How can you build a vacuum tube diode? Ans. A vacuum tube diode is built by using two coaxial glass cylinders. A tungsten filament in kept along the common axis. The space between the two coaxial cylinders is evaculated. The outer surface of the inner cylinder is coated with a mixture of barium and strontium oxide which can emit electrons easily on being heated. The inner surface of the outer cylinder is coated with some metal to make it conductive. When the cylinders are connected externally though an ammeter and the circuit of the filament is closed, the filament heats up the inner cylinder (the cathode). It then starts emitting thermions. The device behaves as a diode. If the cathode is kept at a lower potential it conducts and if the anode (outer cylinder) is kept at lower potential does not conduct.

EXPERIMENT NO. 3 Aim: Determination of dielectric constant (i.e., relative permittivity) of a solid. Apparatus: An electrical signal generator, an oscilloscope and a parallel plate capacitor with circular plates.

Appendix E

E.9

C

M

Attenuation

N

P

Vo (P )

Q

DPM

Ri

R.O.

Peak reading (voltmeter)

Vs (PP )

Fig. E.3

C: Vs(PP) : V0(PP) : V0(P) : a:

aVs (PP)

Vo (PP)

Circuit diagram for measurement of relative permittivity.

Capacitor Peak-to-peak voltage of the signal Peak-to-peak voltage of the output signal Voltage of the output signal from reference (zero) level to peak Attenuation factor

Theory: The capacitance of a parallel-plate capacitor having air as dielectric medium between the plates is given by Œ0 A C0 = _____ (in farad) d where Œ0 = Permittivity of air = 109/(36p) A = Area of each of the plates of the parallel-plate capacitor = pr2 (for round disc) d = Distance between the parallel plates When a dielectric material is introduced between two plates of a parallel-plate capacitor, it is found that the capacitance increases by a factor ‘Œr’ which is the relative dielectric constant of the material concerned. It is also called relative permittivity which is the ratio of absolute permittivity of the medium to that of air. Now we get,

r2 1 r2 C0 = ____ × ___9 F = ____ nF 36 d 10 36 d

where, ‘r’ represents the radius of the gold-plated brass disc and ‘d’ represents the thickness of the sample dielectric material in meter. Cd Œr = ___ C0 where C0 represents the capacitance of the capacitor with the plates separated by air whose thickness is the same as the thickness of the sample dielectric material and Cd is the capacitance of the capacitor when a sample dielectric is introduced between its plates (e.g., glass, mica, P&T, plywood, etc.). Then,

Procedure: 1. Measure the diameter and hence radius ‘r’ of the discs (plates) of standard capacitor. 2. Measure the thickness ‘d’ of the dielectric slab (i.e., disc). 3. Put the disc of the dielectric material whose relative permittivity is to be measured in the dielectric cell.

Advanced Engineering Physics

E.10

4. Connect the standard capacitor, dielectric cell and the resistance (Ri) in series with the oscillator (i.e., signal generator). 5. Turn on the oscillator, set a frequency and an input voltage. Note down the voltage across the capacitor and the dielectric cell. 6. Change the input voltage to few other values, note down the voltage across the standard capacitor and the dielectric cell. 7. Change the input frequency, and repeat Step 6. Observations: Radius r of the disc is given. The thickness d of the disc is given. r2 C0 = ____ = … farad 36 d

\

Table 1 For recording data. S. No. Time period, T (ms) Frequency, f (kHz) Vs (PP) (V)

a ◊ Vs (PP) (V)

V0 (PP) (V)

V0 (P) (V)

1. 2. 3.

Attenuation factor a is given. The series resistance (connected) Ri is given. The peak reading of the voltmeter given V0(P) with V0(PP) = 2 V0(P). Vs(PP) × a × T Cd = _____________ 8 R × V0 (P) Results: Percentage error:

Cd Œr = ___ C0

Discussion: Aids to Viva Voce: 1. What is a dielectric? Ans. A dielectric is an insulating material in which all electrons are lightly bound to the nuclei of the atoms and there is no electron available for conduction of current. 2. What is dielectric constant? Ans. It is the ratio of the permittivity of a dielectric to that of the vacuum. It is also known as relative permittivity. It is given by k = Œr = Œ Œ0 where k is the dielectric constant, Œr is relative permittivity and the two quantities Œ and Œ0 which are permittivities of the dielectric and vacuum respectively.

Appendix E

3. Show that relative permittivity is given by Œr =

E.11

C C0

where C and C0 are the capacitances of a capacitor in presence and in absence of a dielectric material? Ans. Let us consider a parallel plate capacitor in an electrical circuit as shown below:

In absence of the dielectric between the plates the capacitance of the capacitor is given by C0 =

Œ0 A d

(1)

where A is the area of each plate of the capacitor and d is separation of the plates. And in the presence of a dielectric its capacitance is given by C=

ŒA d

(2)

So, from Eqs (1) and (2), we get C = Œ = Œr C0 Œ0

Hence,

Œr =

C . C0

4. What is attenuation? Ans. It is the reduction in the level of a physical quantity, such as the intensity of a wave, over an interval of a variable, such as the distance from a source. 5. What is attenuation constant? Ans. It is a rating for a line or medium through which a plane wave is being transmitted, equal to the relative rate of decrease of the amplitude of a field component, voltage or current in the direction of propagation. 6. What is an operational amplifier? Ans. An operational amplifier (OP AMP) is a high gain direct coupled amplifier which can be used to implement a wide variety of linear and non-linear operations by merely changing a few external circuit elements such as resistors, capacitors and diodes. 7. Write down the circuit symbol of an operational amplifier and state the characteristics of an ideal OP AMP.

Advanced Engineering Physics

E.12

Ans. The following is the circuit symbol of and OP AMP:

An ideal OP AMP has the following characteristics: (i) Input resistance Ri = (ii) Output resistance Ro = 0 (iii) Voltage gain |Av| = (iv) Band width Dw = (v) Perfect balance, i.e., v0 = 0 when v1 = v2 (vi) Characteristics do not drift with temperature. However, a practical OP AMP deviates from these ideal characteristics. 8. Why is the capacitor connected in parallel with OP AMP in the circuit? Ans. It is connected in parallel with an OP AMP because it will ensure a continuous current in the circuit and maintain a constant potential difference between the plates of the capacitor. 9. What is polarization? Ans. Polarization is the process of creating or inducing dipoles in a dielectric material by an external electric field. 10. What is dielectric strength? Ans. The dielectric strength of a dielectric is defined as the maximum value of the electric field that can be applied to the dielectric without its electric breakdown. 11. What is polarizability? Ans. It is the ability of an atom or a molecule to become polarized in the presence of an electric field.

EXPERIMENT NO. 4 Aim: Determination of the specific charge (e/m) of an electron by using Thomson’s (i.e., bar magnet) method. Apparatus: Two bar magnets, a cathode ray tube, a magnetic compass box, a wooden frame with scales to hold magnetic compass and cathode ray tube. S

Electron gun

N

x1 x2

Magnet on scale in wooden frame Fluorescent screen

y1 y2 S

Cathode ray tube

Fig. E.4

Deflecting plates

N

Experimental set-up for measurement of e/m of electron.

Appendix E

E.13

Theory: In the cathode ray tube, electrons ejected from the electron gun are allowed to pass through horizontally and vertically deflecting plates X1-X2 and Y1-Y2 respectively. The deflection of electrons due to electric field (applied by deflecting plates) is nullified by the magnetic field due to placement of two bar magnets on the scales of the wooden frame. And hence specific charge (e/m) is determined. The working formula is given by e/m = V × 107 × Y/(lLH2d) (emu of charge/gm) [The standard value of e/m = 1.7592 × 107 emu of charge/gm] where l = Length of the deflecting plate, L = Distance of the screen from the edges of the deflecting plates, V = Voltage applied between the plates, Y = Deflection of the spot (due to striking of electrons) on screen, and d = Separation between the deflecting plates. [The values of d, l and L are given and the horizontal magnetic field component of earth He = 0.345 oersted.] H = He tan q Procedure: 1. Put the compass box at the center of the two armed wooden frame. Rotate the wooden frame till the arms are parallel to the east-west direction. For your convenience you may mark the boundary of the apparatus on the table with a chalk so that if the east-west alignment is disturbed any time during performance of the experiment it can easily be restored. 2. Take out the compass box and insert the cathode ray tube. Due to the alignment as said in Step 1, the electron beam coming out of the electron gun will lie along north-south direction. 3. Turn on the voltage supply unit. Then adjust the brightness and sharpness of the spot. Move the spot horizontally until it reaches this middle of the scale. 4. Put the Forward/Reverse knob in the forward position. Note down the position of the spot on the vertical scale. This is the initial position. 5. Apply a voltage to cause a deflection of the spot of about 10 mm. Note down the position of the final position of the spot. 6. Place two bar magnets on the two scales of the wooden frame. North pole of one of the magnets will face the south pole of the other one. 7. Adjust the position of the magnets in such a way that they remain equidistant from the cathode ray tube and the deflection produced in Step 5 is nullified and the spot returns back to its initial position. Note the position of the magnets on the arms. 8. Repeat the steps 5 to 7 several times for different values of the deflecting voltages. Reverse the direction of deflection and repeat the steps 5 to 7 as done earlier. 9. Take out the cathode ray tube and place the compass box again. Adjust the dial to read zero in absence of the bar magnets. Bring the two bar magnets on the positions recorded in Step 7. Note down the deflection of both the ends of the indicator. Interchange the position of the magnets on the arms of the wooden frame and also the polarities of the magnets so that the resultant magnetic field is in opposite direction. Note down the deflections. The average of there four readings will give the value of the angle q (The deflection angle of the compass needle).

Advanced Engineering Physics

E.14

10. Draw a graph H2/Y versus V. The slope of the curve when used in the working formula will give you the estimated value of specific charge (e/m). Observations: Table 1 For calculation of specific charge (e/m). S. No. Applied voltage, V (Volt)

Spot deflection, Position of magnet Y (cm) to nullify the spotdeflection, x(cm)

Average compass deflection, q (degree)

Applied magnetic field, H = He tan q

H2/Y

1. 2. 3. 4. 5.

Results: Supplied values L= l= d= He = 0.345 oersted From the graph: H2/Y H2 1 VY Slope S = ____ = ___ fi __ = ___2 V VY S H \

107 1 e/m = ___ . __ emu of charge/gm lLd S

Percentage error: Discussion: Aids to Viva Voce: 1. What is specific charge? Ans. It is the charge to mass ratio of a charged fundamental particle like electron, proton, etc. 2. What type of electrons do you require to determine their specific charge? Ans. Free electrons are required to determine the specific charge of electrons, protons, etc. 3. How can you generate free electrons? Ans. One can generate free electrons in the process of thermionic emission, in the process of photoelectric effect and through operation of a cathode-ray tube. 4. What method of free electron generation is used in the present experiment? Ans. In the present experiment, a beam of freely moving electrons is generated by using a cathode-ray tube.

Appendix E

E.15

5. What is a cathode-ray tube? Why is it named so? How does it work? Ans. It is an evacuated hard glass tube fitted with two electrodes (namely, the cathode and the anode) which are externally connected in an electrical circuit which is capable of developing a high voltage between the electrodes. The cathode is kept at lower potential and the anode is kept at higher potential. When the potential difference crosses almost 4000 volt, a beam of free electrons starts moving from the cathode towards the anode. If the potential difference between the electrodes increases, the electrons in the beam are speeded up. The pressure in the tube usually remains at a fraction of one mm of mercury column. It is known as cathode-ray tube because the ray was discovered before formal discovery of electrons. As the ray emerges from the cathode and moves towards the anode it was maned cathode ray and the tube in which it is generated is known as cathode ray tube (CRT). After formal discovery of electrons, it was realized that the cathode ray is nothing but a beam of moving electrons with high speed. But the ray is still called cathode ray. 6. How is the cathode ray treated to measure the specific charge of electron? Ans. To perform the experiment the cathode ray is passed through two uniform (as well as variable) vertical and horizontal electric fields and a horizontal uniform (as well as variable) magnetic field. During the experiment the aforesaid fields are varied and the corresponding deflection of the ray is measured. 7. What is Lorentz force? Ans. It is the exerted force on a charged particle moving in electric and magnetic fields. It is given by Fl = Fe + Fm Fl = qE + q ( v ¥ B ) Fl = Lorentz force, Fe = electric force, Fm = magnetic force, E = electric field, B = magnetic field, q = charge of the particle, and v = velocity of the particle. 8. How is the effect of interference of the earth’s horizontal magnetic field overcome in this experiment? Ans. By using a magnetometer the CRT is held in such a way that the electrons move from south to north or from north to south direction. In the first case, the angle between the horizontal component of earth’s magnetic field and the velocity of the moving electrons is pc and in the second case it is 0c. So, the value of Fm becomes zero, since or, where

Fm = q ( v ¥ B ) \ Fm = qvB sin q, Fm = 0 for q = 0c or pc 9. Electrons are not visible particles, then how can one sense the presence of the electrons? Ans. In this experiment the cathode ray is deflected using one horizontal and one vertical electric field. The deflected ray is allowed to fall on a fluorescent screen (a screen coated with zinc sulfide). The kinetic energy of electrons of the cathode ray is transferred to the molecules of the fluorescent material. As a result of this, they get excited and immediately get deexcited (within 10–8 to 10–4 second) by emitting visible light making a bright coloured spot on the screen. The deflection of the spot gives the deflection of the cathode ray.

Advanced Engineering Physics

E.16

10. What is fluorescence? Ans. When a beam of light is incident on certain substances (e.g., zinc sulfide) they emit visible light or radiations. This phenomenon is known as fluorescence and the substances showing this phenomenon are known as fluorescent substances. The phenomenon of fluorescence is instantaneous and starts immediately after the absorption of light and stops as soon as the incident light is cut off. 11. What is phosphorescence? Ans. When light radiation is incident on certain substances, they emit light continuously even after the incident light is cut off. This type of delayed fluorescence is called phosphorescence and the substances responsible for this phenomenon are called phosphorescent substances. 12. Compare the lifetime of fluorescence to that of phosphorescence. Ans. Materials exhibiting fluorescence generally reemit excess radiation within 10–8 to 10–4 second of absorption. On the other hand, materials exhibiting phosphorescence reemit excess radiation within 10–4 to 20 seconds or longer. Thus the lifetime of phosphorescence is much longer than that of fluorescence. 13. What is the value of specific charge of an electron? Ans. The specific charge of an electron is given by e = 1.7592 × 107 emu of charge/g. m

EXPERIMENT NO. 5 Aim: Determination of Lande’s g factor (or spectroscopic splitting factor) by using electron spin resonance spectrometer. Apparatus: Lande’s g-factor set up which consists of a CRO, the main oscillator, detector unit, Helmholtz coil, sample tank containing the sample, and an external RF oscillator Theory: Spin, the intrinsic angular momentum, S, couples with the orbital angular-momentum, L to give a resultant angular momentum J. When it is kept in an external magnetic field, H0, 2J + 1 magnetic sublevels are generated with equal energy difference. DE = gm0H0

...(1)

where, m0 is the Bohr magneton and g is a factor known as Lande’s g factor, which is given by J(J + 1) + S(S + 1) + L(L + 1) g = 1 + ________________________ 2J (J + 1)

...(2)

(g is 2 for free electrons). Now, if the energy level is perturbed by an alternating magnetic field with frequency v1 such that the quantum of energy hv1 is exactly equal to DE as given in Eq. (1) and if the direction of the alternating magnetic field is normal to the static magnetic field then there will be transitions between neighboring sublevels according to selection rule Dm = ±1. Therefore, at resonance, DE = gm0H0 = hv1

...(3)

Now, if I be the current flowing through the Helmholtz coil with a number of turns n in each coil of radius ‘a’ then the peak-to-peak magnetic field so produced is given by __ 32p × n ____ I ...(4) H = 2÷2 ________ 10÷125 a

Appendix E

E.17

E S R Spectrometer

inprent Phase Sensitivity H-coil Y E X a

RF oscillator

(a) RF oscillator

AF amplifier

Detector

Oscilloscope

Sample

y

x Helmholtz coils 50 Hz Sweep unit

x y

50 Hz Phase shifter (b)

Fig. E.5

(a) Diagram of the set up. (b) Block diagram of the set up.

which produces P division of deflection in the x axis on the CRO screen and if the distance between the two absorption peaks is 2Q divisions on the said screen, then Q H0 = __ H P Using Eqs (3) and (4), we get hv1 hv1P g = _____ = ______ m0H0 m0kIQ where

...(5)

__ 32p n ____ k = 2÷2 ________ 10÷125 a

The product IQ may be obtained from the slope (s) of the curve Q versus 1/I, whereas the RF frequency v1 can be estimated by formation of beat by an external RF oscillator.

Advanced Engineering Physics

E.18

Procedure: 1. Connect the X and Y ports of the spectrometer unit to the corresponding ports of the cathode ray oscilloscope (CRO). Apply a current through the Helmholtz coil. Note down the current in the circuit. Maximise the horizontal sensitivity of the CRO and with this setting vary the frequency, detection level and vertical sensitivity of the spectrometer until the four peaks superpose into two distinct very well-separated peaks. 2. Now measure the maximum X-deflection (P) and the separation between the absorption peaks (dips) 2Q in terms of the CRO screen divisions. 3. Change the value of current and record it. Adjust the X-sensitivity in order to produce the same value of P as in case of step 2. Measure 2Q. 4. Repeat the last step (i.e., step 3) for few other values of the current I. 5. Now, connect the external radio frequency (RF) oscillator in the relevant socket of the spectrometer unit. Rotate the dial until a bit is observed on the screen of the CRO. Read out the dial reading. 6. Now a graph of Q versus 1/I. Find its slope and hence determine the product of QI. Use Eq. (5) to calculate the value of g. Observations: Given parameters: n= a= m0 = h= __ 32p n ____ k = 2÷2 ________ 10÷125 a

Table 1 Containing I-Q data. S. No.

I (A)

P (div.)

Calculations: From the graph, slope Lande’s g factor

DQ S = IQ = _____ = 1 D __ I hv1p ____ g= m0ks

()

2Q (div)

Q (div)

Appendix E

E.19

Results: Percentage error: Discussion: Aid to Viva Voce: 1. What is Lande’s g factor? Ans. It is the negative ratio of the magnetic moment of an electron or atom (in units of Bohr magneton) to its angular momentum (in units of Planck’s constant divided by 2p). It is also known as Lande’s splitting factor or spectroscopic splitting factor. 2. What is Bohr magneton? Ans. It is given by me = eh 2 me where e = charge of an electron, me = mass of an electron, and h = Planck’s constant divided by 2p (i.e., h = h/2p). In fact, it is the magnetic dipole moment of an electron. 3. What is resonance? Ans. It is a phenomenon exhibited by a physical system when acted upon by an external periodic driving force in which the resulting amplitude of oscillation of the system becomes large when the frequency of the driving force approaches the natural frequency of vibration of the system. 4. What is electron spin resonance? Ans. It is the magnetic resonance arising from the magnetic moment of unpaired electrons in a paramagnetic substance when the substance is kept in a variable external magnetic field. 5. What is a magnetic moment? Ans. It is a vector associated with a magnet (or current loop or a particle) whose cross-product with the magnetic induction (or field strength) of a magnetic field is equal to the torque exerted on the system by the field. i.e., t =m¥B where m is the magnetic moment of the system and B is the field strength of the magnetic field. If the system is a particle its magnetic moment is given by m = eh 2 mc where m is the mass of the particle, e is its charge and c is the speed of light in free space. 6. What is paramagnetism? Ans. It is a property exhibited by substances which, when placed in a magnetic field, are magnetized parallel to the field to an extent proportional to the field. 7. What is paramagnetic substance? Ans. It is a substance within which an applied magnetic field is increased by the alignment of electron orbits. 8. How does electron-spin comes under resonance in some materials in the presence of an external magnetic field? Explain.

Advanced Engineering Physics

E.20

Ans. ESR is observed in paramagnetic substances in the presence of an external magnetic field. These substances have unpaired electrons. The interaction of the unpaired electron spin with an external magnetic field depends on the magnetic moment associated with the said electron spin. And the nature of an unpaired electron spin is such that two or only two orientations are possible. The application of an external magnetic field then provides a magnetic potential energy which splits the spin states by an amount proportional to the magnetic field, and then a radio frequency radiation of the appropriate frequency can cause a transition from one spin-state to the other. When the frequency of oscillation of the radio oscillator becomes equal or an integral multiple of the natural frequency of transition of the electron-spin resonance takes place. The splitting electron spin in the presence of an external magnetic field is shown below:

9. Ans. 10. Ans.

Name a paramagnetic substance which is usually used as a sample in ESR spectrometer. 1, 1, diphenyl-2-picrylhydrazyl free radical (DPPH). What is its chemical structure? Its structure is given below:

DPPH is a chemically stable material having the spectroscopic splitting factor g = 2.0036. DPPH contains 1.53 × 1021 unpaired electrons per gram. 11. How can you use ESR spectroscopy to calculate the number of unpaired electrons in a sample? Ans. In order to calculate the number of unpaired electrons in an unknown sample, a comparison can be made with a standard sample (e.g., DPPH) having a known number of unpaired electrons and possessing the same line shape as the unknown (gaussian or lorentzian).

EXPERIMENT NO. 6 Aim: To observe discrete energy levels of an atom and hence to determine the first excitation potential of an atom by using Frank–Hertz experimental kit. Apparatus: Frank–Hertz experimental kit with a (vacuum tube) tetrode filled with some inert gas vapour. Theory: It was known from Bohr’s postulates that the internal energies of an atom are quantized. This can be proved directly by this experiment.

Appendix E

E.21

Nanoammeter

VG 2A

A G2 G1 K

VG 2K VG1K

F

Fig. E.6

Frank–Hertz experimental kit.

In the set-up for experiment (Fig. E.6) there is a tetrode filled with argon vapour. Electrons emitted by the heated filament F are accelerated by the potential VG2K (applied between the cathode and the grid G2). Initially some electrons reach the anode A provided that their kinetic energy is sufficient to overcome the returding potential VG1K (applied between the cathode K and the grid G1). So, in the beginning it is seen that the anode current increases with the increase of voltage VG2K. But as the voltage increases further, the energy of the electrons reaches the threshold value to excite the atoms to their first excited state and in the process many thermions lose their kinetic energy, for this reason the current in the circuit falls abruptly. And then the voltage VG2K is increased further. Again the current starts to increase and when the voltage VG2K reaches to a value twice that of its first excitation potential again, the current drops abruptly. In this way the current drops a few times in a certain voltage range. This observation proves that the energies of the atom are quantized. Procedure: 1. First of all confirm that all the control knobs are in their minimum position. 2. Then turn on the experimental kit. Now turn on the “manual/auto” switch to manual mode. 3. Turn on the voltage display selector to VG1K and adjust the VG1K control knob to 1.5 V. 4. Select the voltage display of VG2A and adjust it to 7.5 V. 5. Now change the value of VG2K in small steps (say, 0.5 V) and record the corresponding current reading. 6. Now draw a graph showing the variation of current as a function of the accelerating voltage. 7. Turn on the “manual/auto” switch to auto mode. 8. Connect the Y, G, K sockets of the instrument to the corresponding ports of the oscilloscope. Set the oscilloscope to X-Y mode and the trigger to external X. 9. Now adjust the ‘shift’ and the ‘gain’ switches to obtain a clear waveform. Apply the maximum scan range through the instrument. 10. Get the average horizontal distance measured between the peaks. This would give the value of gas atom’s first excitation potential in eV. 11. The excitation potential can also be obtained from the current versus voltage graph drawn in Step 6. And they may be compared. Observations: VG1K = 1.5 V, VG2A = 7.5 V

Advanced Engineering Physics

E.22

Table 1 For recording of voltage and current. S. No.

Gride voltage (V) × 2 V

Current (I) × 0.02 × 10–8 mA

– – – –

– – – –

– – – –

Calculations: Results: Discussions: Aids to Viva Voce: 1. What is the structure of an atom? Explain. Ans. An atom consists of three types of fundamental particles namely, proton, neutron and electrons. The atom has a solid central core consisting of protons and neutrons. It is called the nucleus of the atom. The electrons are distributed among different spherical shells (called energy levels) according to Pauli’s exclusion principle. The energy levels of an atom are discrete and well defined. According to Pauli, the maximum number of electron allowed in an energy level is given by Nn = 2n2 where n is the serial/order number of the shell and Nn is the number of maximum allowed electrons in the nth shell. The protons are positively charged particles, the electrons are negatively charged particles and the neutrons are neutral particles. So, the nucleus of an atom is positively charged. As the electrons are negatively charged, they experience a strong electrostatic attractive force towards the nucleus. Because their force the electrons keep on revolving around the nucleus to maintain a dynamic balance. This gives the idea of discrete energy levels. Other than the first energy level, all other energy levels are divided into a number of sublevels of energy. An electron belonging to an energy level can jump to a higher energy level if it is supplied with a definite amount of energy given by E = E2 – E1 = hn where E1 and E2 are the energies of the first and second energy levels and n is the frequency of the radiation and h is Planck’s constant. 2. What do you mean by excitation of an atom? Ans. When an assembly of atoms of a matter is exposed to a radiation, some of the atoms may absorb a quantum of energy (hn) and sends one of its electrons from a lower energy level to a higher energy level which satisfies the following condition: hn = E2 – E1 This phenomenon is known as excitation of an atom.

Appendix E

E.23

3. What is electron volt? Ans. Electron volt (eV) is a unit of energy. It is defined as follows: When an electron moves from a point of lower potential to a point of higher potential it gains energy. If the potential difference between the two points is one volt, then energy gained by the electron is called one electron volt (eV). 1 eV = 1.6 × 10–19 joule 4. What is a vacuum tube used in electronics? Ans. It is an electronic device like diode, triode, tetrode, etc., which are used in an electronic circuit. These devices can be built by using vacuum tube technology. 5. What is a tetrode? What is its counterpart in semiconductor technology? Ans. It is a vacuum tube which has four electrodes called cathode, anode, grid one and grid two. It does not have any counterpart in semiconductor technology. 6. Which vacuum tube is used in this experiment? Ans. In this experiment a tetrode is used. 7. Why are the energy levels of an atom discrete? Ans. The atoms have permanent atomic structure which does not collapse. An atom has positive charge at the central nucleus and negatively charged electrons revolve around the nucleus. In this process they are supposed to lose energy through radiation. But practically it does not happen. To support this situation Neils Bohr suggested one atomic model called the quantum model of atom. According to this model, the electrons of an atom are distributed in different discrete energy levels having well-defined value of energy. 8. What is excitation potential? Ans. It is the energy required to shift one electron from the ground state to higher energy state. 9. What do you mean by first excitation potential? Ans. It is the energy required to shift one electron from the valance level to the first unfilled level. 10. What is a grid? Ans. It is a mesh cylinder which is placed between the cathode and the anode in a vacuum tube (e.g., a triode). It allows the thermions to pass through it and controls the flow of thermions.

EXPERIMENT NO. 7 Aim: Determination of the energy of the band gap of a semiconductor by using the four probe method. Apparatus: A four probe device, an oven, a thermometer, a sample semiconductor crystal, a voltmeter, an ammeter and connecting leads. Theory: The highest filled energy band including electrons shared in covalent bonds or transferred in ionic bonds in a semiconductor is known as valence band. The energy level which corresponds to the top of the valence band in an intrinsic semiconductor is denoted by Ev.

E.24

Advanced Engineering Physics

The energy band which remains fully unfilled or includes some free electrons is called the conduction band. The energy level which corresponds to the bottom of the conduction band in an intrinsic semiconductor is denoted by Ec. The energy gap between the top of the valence band and the bottom of the conduction band is called band gap. So, the energy of the band gap (Eg) is given by Eg = Ec – Ev

...(1)

The band gap of a semiconductor can be calculated from the following formula: Eg ln r = ____ – ln A 2kT i.e.,

Eg loge r = ____ – loge A 2kT

...(2)

where r is the resistivity of the semiconductor, Eg is the band gap energy of a semiconductor, k is the Boltzmann constant (k = 1.38 × 10–23 J/K), and T is the absolute temperature and A is a constant. r0 The resistivity of the given semiconductor is given by r = _______ G7(W/s) where G7(W/S) is the correction factor. And it is expressed as 2S G7(W/s) = ___ loge2 W V Again r0 = __ (2p S) I where V is the applied voltage, I is the current, W is the thickness of the crystal, whose value is given (/ supplied) and S is the distance between the probes, the value of which is also supplied. Procedure: 1. Insert one sensitive thermometer through the hole of the oven. 2. Turn on the apparatus and set a current of the order of 5 mA. 3. Note down the temperature in the thermometer and the voltage in the voltmeter. 4. Now turn on the oven. Thus record the temperature and the corresponding voltage at an interval of 5ºC up to at least 120ºC. 5. Turn off the oven. Record the temperatures and the corresponding voltages at intervals of 5ºC while the temperature decreases. 6. Calculate average voltage at each temperature. Calculate r. Convert the temperatures into absolute scale. 1 7. Plot a graph of loge r versus __. Determine the slope from the linear part of the graph. T

Appendix E

E.25

8. Calculate the value of G7(W/S) from the given value of W and S.

Four probe set-up Model dep-02

ln Y ln A

x1

x 10

Oven

L

||

Voltage

SCIENTIFIC Current EQUIPMENT ROORKEE

Oven

ON

Spring loaded stand

Spring loaded probe (a) Galvanometer Potentiometer Millivoltmeter

Milliammeter Direct current source

V I

I Probes Semiconductor crystal

(b)

Fig. E.7 (a) Experimental set-up of band gap energy measurement. (b) Circuit diagram of four probes.

Observations: Current I = mA (constant), W = S= , G7(W/S) = Table 1 Records of voltage and temperature. S. No.

– –

Voltage (V)

– –

Temperature (ºC) Average while while temperature increasing decreasing T (K)

– –

– –

– –

T–1 × 10–3 (K–1)

– –

r0 (W-cm)

– –

r (W-cm)

– –

loge r

– –

E.26

Advanced Engineering Physics

Calculations: Results: Percentage error: Discussion: Aids to Viva Voce: 1. What is a semiconductor? Ans. It is a solid crystalline material whose electrical conductivity is intermediate between that of a conductor and an insulator, ranging from about 105 mho to 10–7 mho per meter. It is usually strongly temperature dependent. 2. What is an intrinsic semiconductor? Ans. If the electric conductivity of a semiconductor is entirely determined by the carriers which are generated by thermal excitations from the valence band to the conduction band, the semiconductor is referred to as a pure or intrinsic semiconductor. 3. How energy bands are generated in a semiconductor? Ans. A semiconductor (like silicon or germanium) remains in crystalline form. In such a crystal, the constituent atoms are orderly arranged, so the unfilled energy levels of the crystal atoms merge together to form an energy band called the conduction band and the filled and partially filled energy levels merge together to form another energy band called valence band. In a semiconductor there remains a gap between conduction band and the valence band. 4. What is a band gap? Ans. The difference between the lowest energy level of the conduction band and uppermost energy level of the valence band is called band gap of the semiconductor. 5. How much is the value of band gap in case of germanium and silicon? Ans. In case of germanium, the band is 0.785 eV and in case of silicon it is 1.21 eV at 0 K. 6. Comment on the band gap of insulator and metal. Ans. The band gap of insulator is larger than that of a semiconductor, while in case of metal (i.e., conductor) the band gap is zero because the conduction band and the valence band of a metal overlap each other. 7. What is a probe? Ans. It is a small device which can be brought into contact with or inserted into a system in order to make measurements on the system, ordinarily it is designed so that it does not significantly disturb the system. 8. Why is the present method of determination of band gap of semiconductor called a four-probe method? Ans. The present method of determination of band-gap energy is called four probe method because four probes are required to determine the band-gap energy in this method. 9. On the basis of what parameter can you compare one curve and one straight line? Explain. Ans. We can compare a curve with a straight line on the basis of the radius of curvature. A straight line is a special case of a curve whose radius of curvature is infinity.

Appendix E

E.27

10. What is the slope of a curve? What is it in case of a straight line? Ans. The slope of a curve is the tangent of the angle of inclination of the line, which touches the curve at the point of intersect, with the x-axis. In case of a straight line, it is tangent of the angle of inclination of the line itself with x-axis because a line and its tangent coincide with each other.

EXPERIMENT NO. 8 Aim: Determination of the thermoelectric power of a given thermocouple at a certain temperature. Apparatus: A copper-constantan thermocouple, a potentiometer, a heater, thermometers, an beaker, a resistance box and a galvanometer. Copper wire Constantan wire

R

Hot

Cold

Constantan wire

P

E

G

Fig. E.8

A copper-constantan thermocouple.

Theory: When two wires of two different materials are soldered together, having put them in contact, a thermocouple is prepared. A small voltage in the millivolt range is generated when the two junctions of the wires are kept at two different temperatures. The induced emf is dependent on temperature. A small current flows around the circuit made of the wire loop. In fact, a temperature difference produces a potential difference and the phenomenon is known as thermoelectric effect. When one junction of the thermocouple is kept at 0°C and the other junction at some higher temperature (t), a thermo emf (e) develops in the thermocouple. Measuring the thermo emf at different temperatures of the hot junction a graph of emf (e) versus temperature (t) can be drawn. The thermoelectric power is the slope of the aforesaid graph. It is defined by the following relation: de P = ___ (in mv/ºC) ...(1) dt The thermo emf (e) is measured with the help of a potentiometer, which has a dc source of emf e and a resistance R is the driving circuit. If l be the position of the null point from the start of the first wire, and L be the total length of the potentiometer wire (usually 10 meter) of resistance r, then one can have.

Advanced Engineering Physics

E.28

Erl e = ________ (R + r) L

...(2)

So, the potential drop per unit length of the wire is given by e Er r = __ = ________ ...(3) l (R + r) L Procedure: 1. Measure the resistance of the potentiometer wire (r) with the help of a multimeter calculate the value of the series resistance R so that you can find out the potential drop per unit length of the wire (r) by using the formula of Eq. (3). 2. Connect the circuit as shown in Fig. E.8. Apply a value of R calculated from the formula of Eqs (2) or (3). Fill the funnel of the cold junction with crushed ice. Care should be taken to avoid any air pocket surrounding the junction. 3. Dip the hot junction of the thermocouple in a beaker containing water at room temperature. Measure the position of the null point. Take care so that a high current does not pass through the galvanometer/ nanoammeter. It can be a good idea to keep a high resistance in series to delimit the current which is then removed when the position of the null point is roughly known. 4. Turn on the spirit lamp (or use a hot plate) and note the null point at an interval of 5ºC up to at least 80ºC. 5. Draw a graph having plotted e versus t. Then find out the slope of the graph. Observations: Resistance of the potentiometer r = (ohm) Value the series resistance (from Eq. (3)), R = (ohm) Actual value of the series resistance Ra = (ohm) Table 1 emf of the driving cell. E (volt) Before expt. After expt. Average Table 2 Null-point-temperature data. S. No.

Temperature, t (ºC)

Position of the null point Wire No.

Calculations: Percentage error: Results: Discussion: Aids to Viva Voce: 1. What is a thermocouple?

Length

Total, l (cm)

1000 Erl e = ________ (mV) (R + r) L

Appendix E

E.29

Ans. When two wires of different metals are joined at their ends and a temperature difference is maintained between the junctions, an electric current flows in the circuit. This device is called thermocouple. The emf which is generated between the two junctions of a thermocouple is known as thermo-emf. 2. What is Seeback effect? Ans. When two wires of different metals are joined at their ends and a temperature difference is maintained between the two junctions, a current flows in the circuit. This effect is called Seeback effect. 3. What is Peltier effect? Ans. When a battery is inserted in a thermocouple circuit whose two junctions are initially at the same temperature, one of the junctions becomes hot and the other becomes cold. This phenomenon is known as Peltier effect. 4. What is Thomson effect? Ans. When the two points of a conductor are kept at different temperatures, a potential difference develops between the said two points. This phenomenon is called Thomson effect. 5. Explain the existence of an emf at the junction of two metals? Ans. When two different metals are joined at their ends, the free electrons of one metal will flow to the other because the pressure of electrons of the metals is different. As a result of flow of the electrons, one metal becomes positive as compared to the other and a potential difference is created at the junction. The magnitude of this potential difference, referred to as the contact potential difference, increases with increase of temperature. 6. How is a thermocouple constructed? Name some pairs of metals which are generally used for the construction of thermocouple. Ans. To construct a thermocouple, say, copper-constantan thermocouple, one piece of constantan wire and one piece of copper wire are taken. After cleaning their ends with sand paper, one end of the copper wire is spot-welded with one end of the constantan wire. Similarly other end of each one is joined. Thus, the junctions of a thermocouple are formed, copper-constantan, copper-iron, platinum-rodium, etc., are the pairs of metals which are usually used for construction of thermocouple. 7. What is the neutral temperature of thermocouple? Ans. The neutral temperature is the temperature of the hot junction at which the generated thermo emf is a maximum, when the cold junction remains at 0°C. 8. Mention two applications of thermocouple? Ans. (i) It is used to measure the temperature of a point, and (ii) to measure the radiant heat. 9. Can one use an ordinary voltmeter to measure the thermo emf? Ans. No, one cannot use a voltmeter because the emf developed in thermocouple is in millivolt range. Also, a voltmeter actually gives the potential difference and not the emf. 10. What is the value of the thermoelectric power at the neutral temperature? Ans. The thermoelectric power at the neutral temperature is zero. 11. Explain the difference between joule effect and Peltier effect? Ans. In the case of joule effect the heat generated is proportional to the square of the current and hence is independent of the direction of current. On the other hand, the Peltier effect produces heating or

Advanced Engineering Physics

E.30

cooling at the junction that is proportional to the current. Thus, a junction which is heated by a current will be cooled when the direction of the current is reversed. 12. What is the general nature of the thermo emf versus temperature curve? What is the nature of the curve you have used? Ans. The general nature of the thermo emf versus temperature curve is parabolic. We get a straight line, because the temperature of the hot junction is much less than the neutral temperature of the couple. That is we get the straight portion of the parabola.

EXPERIMENT NO. 9 Aim: Determination of Rydberg constant by studying the spectrum of hydrogen gas. Apparatus: A spectrometer, a spirit level, a hydrogen discharge tube, a power supply for the discharge tube and a magnifying lens. Theory: Balmer series is the emission spectrum related to the transition of electrons from states with principal quantum number n > 2 to the state with n = 2. The wavelengths of the corresponding lines are expressed by a formula 1 1 1 __ = R __2 – __2 l nf ni

( )

where nf = 2 and ni > 2. In the spectrum of hydrogen gas there are a few bright lines, namely, Ha : ni = 3: color = red Hb : ni = 4: color = blue-green Hg : ni = 5: color = violet Hd : ni = 6: color = violet In this experiment wavelengths of these lines are determined by measuring the angle of diffraction by a transmission grating using the following formula: (a + b) sin q = n l where n is an integer. Here ‘a’ is the width of the opaque line in the grating and b is the with of the transparent region lying 1 between two opaque lines and consequently (a + b) = __, where N is the number of rulings per unit length of N the grating, n is the order of the spectrum and q is the angle of diffraction. Observations: 1 SDMS Vernier constant VC = __________________ No. of divisions in VS [SDMS is the smallest division in the main scale and VS is the vernier scale] No. of rulings

N = º per cm.

Appendix E

E.31

Table 1 Measurement of angle of diffraction. Order of the spectrum Color

Vernier (V)

Left spectrum

Right spectrum

Difference 2q

Avg. q

MSR VSR Total MSR VSR Total First

Red Blue-green Violet1 Violet2

V1 V2 V1 V2 V1 V2 V1 V2

Calculations: lred = lblue-green = lviolet 1 = lviolet 2 = Rydberg constant for red line R1 = Rydberg constant for blue-green line R2 = Rydberg constant for violet (1) line R3 = Rydberg constant for violet (2) line R4 = Average Rydberg constant R = Standard value of Rydberg constant Rs = Percentage error: Results: Discussion: Aids to Viva Voce: 1. What is a Rydberg constant (R)? Ans. It is a physical constant relating to atomic spectra in the science of spectroscopy. This constant R is named after the Swedish physicist Johannes Rydberg. The value of Rydberg constant is given by R = 1.097 × 107 m–1 i.e., R = 0.01097 nm–1 2. What is a wave number? Ans. It is inverse of the wavelength of a photon. Sometimes inverse of wavelength is multiplied by 2p to the wave number. 3. What does Rydberg constant represents? Ans. Rydberg constant represents the limiting value of the highest wave number of any photon that can be emitted from the hydrogen atom or alternatively the wave number of the photon (having lowest energy) capable of ionizing the hydrogen atom from its ground state.

Advanced Engineering Physics

E.32

4. What is Rydberg unit of energy? Ans. Rydberg unit of energy (symbol Ry) is closely related to the Rydberg constant. It corresponds to the energy of the photon whose wave-number is Rydberg constant, i.e., the ionization energy of the hydrogen atom. 5. Comment about the definition of the wave number. Is it unique? Ans. Definition of wave number is not unique. It is defined as k = 1 which means the number of waves l 2 present per unit length. But sometimes it is expressed as k = p . l 6. What is spectrum? Explain. Ans. A spectrum is a condition that is not limited to a specific set of values but can vary infinitely within a continuum. As for example we see the graphical representation of the spectrum of continuous X-ray given below.

7. Ans. 8. Ans.

What is a spectrometer? It is a device which is used to study the spectrum. It is also called a spectroscope. Write down Rydberg formula for any hydrogen-like atom and explain the symbols used? The Rydberg formula for hydrogen-like atom is given by 1 = Rz 2 Ê 1 - 1 ˆ ÁË n2 n2 ˜¯ lv f i

where lv is the wavelength of light emitted in the vacuum, R is the Rydberg constant for the concerned element, z is the atomic number of the element, and nf and ni are integers such that nf < ni 9. Express Rydberg constant in terms of other physical constants for an atom of hydrogen? Ans. The Rydberg constant for hydrogen when expressed in terms of other constants is given by me e 4

= 1.0973 × 107 m–1 8 e 02 h3 C 10. Express Rydberg unit of energy in terms of electron volt. Comment on it. Ans. 1 Ry = 13.605 eV [1 Ry = hcR] where R = 1.097 × 107 m–1 R=

one Rydbergy energy is the excitation potential of a hydrogen potential.

Appendix E

E.33

EXPERIMENT NO. 10 Aim: To study current-voltage characteristics, load response, areal characteristics and spectral response of a photovoltaic (or solar) cell. Apparatus: A solar panel, a voltmeter, a milliammeter, a decade resistance box, a 100 W lamp fitted with an intensity control, area choppers of different sizes and color filters. Theory: The photovoltaic cell is basically a p-n junction diode hn which converts solar energy into electrical energy. It is also known as solar cell. As its name is photovoltaic cell, the phenomn P enon is known as photovoltaic effect. When this is illuminated, – + the photons incident on the cell generate moving electrons and accordingly holes are also generated. At the junction, the barrier V K field separates the positive and negative charge carriers. Under the action of the electric field, the electrons (i.e., minority carriRL mA ers) from the p-region are swept into the n-region. Similarly the holes from the n-region are swept into the p-region. It leads to an Fig. E.9 Circuit diagram of a solar cell. increase of holes on the p-side and elections on the n-side of the junction. The accumulation of charges on the two sides of the junction produces an emf which is called photoemf. It is also known as open circuit voltage. It is proportional to the illuminated area. When an external circuit is connected across the solar cell terminals, the minority carriers return to their original side through the said circuit. The solar cell behaves as a battery with the n-side as negative terminal of the circuit and the p-side as the positive terminal. The photo-emf or voltage can be measured with a voltmeter. Procedure: 1. Illumination characteristics: Set the load resistance (RL) to zero. Then vary the voltage across the light bulb and record the photocurrent. Draw a graph by plotting photocurrent (Ip) versus lamp voltage (Vl). 2. Current-voltage characteristics: Keep the intensity of the lamp at a moderate level. Note the open circuit voltage of the photocell (Vopen). Short the output of the solar cell and note down the short circuit current (Is). ●

Change the load resistance in steps of 100 ohm and record the corresponding photocurrent and voltage. Draw a graph with voltage V along the y axis and current I along the x axis. 3. Power load characteristics: Calculate power from the current-voltage data obtained in Step 2 and plot a graph of power against load resistance. Find out the optimum value of load for which the power dissipation is the maximum. 4. Areal characteristics: Set the load at the optimum value obtained in Step 3. Set the chopper in the slot provided in front of the solar cell in different settings and measure the current and voltage. Calculate the power. Draw a graph of power versus area of the chopper settings. 5. Frequency characteristics: Keeping the load at its optimum value, put several fillers and take the current-voltage data to get the power. Plot a graph of power versus frequency of the used filter. ●

Advanced Engineering Physics

E.34

6. To obtain the load line on the current-voltage characteristic curve refer to the power versus load resistance curve and find the value of maximum power (Pmax) corresponding to the optimum load (Rop). Then calculate the corresponding value of voltage (V) from the following relation: V2 P = ___ R Now, find the point corresponding to V on the I-V characteristic curve. Draw a straight line passing through the origin and the point referred to above and extend it this is the required load line. Observations: Table 1 Illumination characteristics. No. of obs.

Filament voltage, Vf (V)

– – Table 2

Current through solar cell, I (A)

– –

– –

Current-voltage characteristics

No. of obs.

Filament voltage,

Load resistance,

Load voltage,

Current through

Vf (V)

R (W)

V (V)

solar cell, I (A)

– –

– –

– –

– –

No. of obs.

Chopper area, A (cm2)

Voltage, V (V)

– –

– –

Power, P = VI (W)

– –

– –

Optimum resistance, Rop =

Table 3 Areal characteristics.

– –

Current, I (A)

– –

Power, P (W)

– –

Appendix E

E.35

Table 4 Frequency characteristics. No. of obs.

Color of the filter

– –

Calculations: From the graphs: Voltage

– –

Frequency of the filter

– –

V=,

Current I = ,

Voltage, V (V)

– –

Current, I (A)

– –

Power, P (W)

– –

Power P = VI =

Results: Percentage error: Discussion: Aids to Viva Voce: 1. What is a solar cell? Ans. A solar cell (also called photovoltaic cell or photoelectric cell) is a solid state electrical device that converts light energy directly into electricity by photovoltaic effect. 2. What is photovoltaics? Ans. It is the field of technology and research related to the practical application of photovoltaic cells in producing electricity from light, though it is often used specifically to refer to the generation of electricity from sunlight. 3. What is optimum load resistance? Ans. It is the load resistance used in the circuit of a solar cell for which the power output of the circuit is maximum. 4. How can one determine the optimum load resistance? Ans. By plotting a graph of output power versus the load resistance in the circuit, one can determine the optimum resistance. The resistance in the aforesaid graph for which the output power is maximum in the optimum load resistance. 5. How can you build a solar cell? Ans. One can prepare a photovoltaic cell by doping a semiconductor chip. One half of the chip is doped positively and the other half is doped negatively: In fact, a photovoltaic cell is a p-n junction. 6. What is the difference between excitation and ionization? Which one is responsible for the generation of photovoltaic effect? Ans. In case of excitation an electron which absorbs the light photon does not leave the atom. It jumps from a lower energy level to a higher energy level of the atom, while in case of ionization, the electron, which absorbs the light photon, becomes completely free by separating itself from the atom. Ionization is responsible for generation of photovoltaic effect.

Advanced Engineering Physics

E.36

7. What are open circuit voltage and short circuit current? Ans. Open circuit voltage is the voltage across the source when the circuit remains open. And short-circuit current is the current in circuit when there is no external resistance present in the circuit. 8. What is the difference between a p-n junction diode and a solar cell? Ans. So far as construction is concerned a p-n junction diode and a solar cell are same. But when it is used as diode it is operated with the help of electrical energy and when it is used as solar cell it is operated with the help of light energy. 9. What is a hole? Ans. The absence of an electron in the valence band of p-type region is referred to as a hole. 10. Can one establish the presence of a hole practically? Ans. Yes, by performing one experiment, one can practically establish the presence of holes.

EXPERIMENT NO. 11 Aim: To determine Hall coefficient of a semiconductor. Apparatus: Electromagnet, constant power supply, semiconductor, digital millivoltmeter, gaussimeter, etc. Theory: A static magnetic field does not have any effect on charged particles if they are at rest. But the moving charged particles are affected by a static magnetic field. When a charged particle moves in a magnetic field which is perpendicular to the direction of the moving charge, a force acts on the charge which is perpendicular to both, the direction of the magnetic field and the direction of the velocity of the charged particle. When a semiconductor chip is placed in a constant magnetic field and kept its surface perpendicular to the magnetic field and a current is allowed to flow through its surface in a direction which is perpendicular to the said field, then the electrons and the holes of the semiconductor are separated by opposite forces. They will r in turn produce an electric field ( Eh ) which depends on the cross product of the magnetic field intensity ( H ) and the current density ( J ) . Mathematically it can be expressed as follows: Eh = R ( J ¥ H )

(1)

where R is known as Hall coefficient or Hall constant. Let us consider a bar of semiconductor having dimensions x, y and z. Let J be directed along x-axis and H along the z-axis, then Eh will be directed along the y-axis. In general, the Hall voltage is not a linear function of the magnetic field applied, i.e., the Hall coefficient is not generally constant, but it is a function of the applied magnetic field. The working formula is given by R=

Vh t Ih H

where, Ih = the current flowing through the semiconductor due to Hall effect. t = thickness of the semiconductor. Vh = Hall voltage

(2)

Appendix E

E.37

B Z Y P

Q M

Jx X

N

(a)

–Y Conductor Ea

Q

P

d M

N

Jx

(b) Q+ + + + + + + + + + + + + + P Eh

Eh

Eh

M– – – – – – – – – – – – – – – N (c)

r Fig. E.10 rHall effect is shown here. (a) B : magnetic field, J x current density, (b) E a : applied electric field related to J x and Ø: direction of electron shift, (c) ≠: direction of Hall field Eh, ¨: direction of electron flow.

To measure the magnetic field intensity H, a standard Hall prove is used and calibrated against the magnetizing current passed through the coil. Procedure: 1. Place the Hall probe in the central region of the gap between the poles of the electromagnet. Turn on the current source and the gaussimeter. Turn on and adjust the knob so that the current from this source is zero. The Hall probe should read zero magnetic field. 2. Increase the current in small steps and note down the magnetic field in each case. The current should be increased up to a level so as to produce 3000 gauss magnetic field. 3. Remove the Hall probe and insert the semiconductor chip for which you wish to measure the Hall constant. 4. Apply a magnetizing current, which can produce a magnetic field of 1000 gauss. 5. Increase the current through the semiconductor (Hall current) in small steps and note down the voltage (Hall voltage) in each case. 6. Take three sets of reading for Hall current (Ih) and Hall voltage (Vh) by fixing the value of the magnetizing current (Im) at three values respectively (say 1 ampere, 2 amperes and 3 amperes). 7. Draw a graph of magnetic field (H) against the magnetizing current (Im) (Table 1). 8. Draw three more curves of Hall voltage (Vh) against Hall current (Ih) and take slopes of them DVh ˆ Ê ÁË S = DI ˜¯ . h

Advanced Engineering Physics

E.38

9. Use these three slopes in equation (2) to calculate three values of Hall coefficients for three values of magnetizing current (Im) (1 ampere, 2 amperes and 3 amperes). Then find the average value of R. Observations: Table 1 Current Im (amp)

S. No.

Magnetic field

H (gauss)

1. 2. 3. 4. 5. 6. 7. Table 2 S. No.

—

Im = x amp

Im = y amp

Im = z amp

1. 2. 3. 4. 5.

Ih (mA) ... ... ... ... ...

Vh (mV) ... ... ... ... ...

Vh (mV) ... ... ... ... ...

Vh (mV) ... ... ... ... ...

Results: Discussion: Aids to Viva Voce: 1. What is Hall effect? Ans. When a sheet of material (metal or semiconductor) carrying an electric current is placed in a magnetic field, an electric field is developed inside the material sheet in a direction which is normal to both the current and the magnetic field. This effect is known as Hall effect. This effect was discovered by Edwin H. Hall in 1879. 2. Define Hall coefficient? Ans. Hall coefficient is the inverse of the density of charge (i.e., charge per unit volume) of the charge carriers when Hall effect takes place in a material sample. It is given by Vt 1 Rh = = h ne I h H where n is the number charge carriers per unit volume e is the amount of charge possessed by a single charge carrier Vh is the Hall voltage. Ih is the Hall current.

Appendix E

3. Ans. 3. Ans. 4. Ans. 5. Ans. 6. Ans. 7. Ans.

E.39

H is the acting magnetic field. t is the thickness of material sample. (a) What is Hall voltage? It is the voltage developed across a material sample when Hall effect takes place in it. (b) What is Hall current? It is the rate of flow of charge with respect to time to develop Hall voltage in a material sample. What is a Hall probe? It is a probe which is placed in the central region within the gap of the two pole pieces of a magnet to measure the magnetic field strength with the help of a gaussimeter. What is a gaussimeter? It is an instrument which is used to measure the magnetic field strength at a point in a magnetic field. What is gauss? It is a unit of magnetic field strength. 1 gauss = 10–4 tesla = 10–4 weber/m2 What is Lorentz force? It is the force experienced by a moving charge while it passes through an electric and a magnetic field. The mathematical expression of the Lorentz force is given by r r r r r Fl = Fe + Fm fi Fl = qE + q(v ¥ B) where e is the charge of the charge carrier particle, n is its velocity and B is magnetic field. According to some authors the Lorentz force is given by

r r r Fl = q(v + B) where E is the electric field. 8. Explain effect of a static magnetic field on static and moving charges. Ans. A static magnetic field does not affect a static electric charge while a moving electric charge is deflected by a static magnetic field. The deflecting force on the moving charge is normal to both the velocity (n ) of the charge and the static magnetic field ( B ) . 9. What is a an electromagnet? How is it constructed? Ans. When a soft-iron rod is wrapped with a coil made of insulated metalic wire and current is passed through the wire, the iron rod behaves as a temporary magnet. Such an iron rod showing magnetic field is called an electromagnet. On withdrawal of the current the soft-iron rod loses its magnetic property. 10. What is the magnetizing current? Ans. The current which is passed through a coil to magnetize a soft-iron rod is called magnetizing current. 11. What is the unit of the Hall co-efficient? Ans. The unit of Hall coefficient in SI system is m3/coulomb or in cgs system it is cm3/statcoul. [1 statcoul = 1 esu of charge] A practical unit of Hall coefficient is cm3/abcoul. [1 abcoul = 1 emu of charge]

Model Question Paper 1

Group A Multiple Choice Questions __ _

1. (i) The value of — ◊ r is given by (a) 1 __

__

(b) 0

__

(c) 2

(d) 3

__

(ii) If A and B are two rotational vectors, then A × B is (a) solenoidal

(b) irrotational

(c) rotational

(iii) In free space, Poisson’s equation is given by r r (a) —2V = – ___ (b) —2V = ___ (c) —2V = 0 Œ0 Œ0 __ __

(iv) If — ◊ B = 0, then __

__ __

__

__

__

__

__

__

__

__

__

(a) B = — ◊ A (b) B = — × A (c) B = —2 A (v) In free space, modified Ampere’s circuital law is given by (a) — × E = 0 __ __

(d) None of these

(d) —2V = rŒ0

(d) None of these

__

∂E (b) — × B = m0Œ0 ___ ∂t __ __

(d) — ◊ B = 0 (c) — ◊ E = 0 (vi) Electromagnetic wave is propagated through a region of vacuum which does not contain any charge __

or current. If the electric vector is given by E = E0 exp i(kx – wt). Then magnetic vector is (a) in the x direction (b) in the y direction (c) in the z direction (d) rotating uniformly in the xy plane (vii) If y (x, t) be a normalized wave function, it must have +

(a)

Ú

–

+

y y dx = 1

+

(b) Ú y y dx = 1 (c) *

–

Ú

–

+ *

y y dx = 0

(d)

Ú

–

1 y *y dx = __ 2

(viii) For T > OK, The probability of occupancy of an electron at Fermi energy level is 1 (a) 1 (b) 0 (c) __ (d) 2 2

Advanced Engineering Physics

M1.2

(ix) The one-dimensional time-independent Schrödinger wave equation for a free particle is given by d 2y (x) ___ 2m (a) ______ + 2 Ey (x) = 0 2 dx h

d2y ___ 2m (b) ____ + 2 [E – V] y (x) = 0 2 dx h

2 ∂y h2 ∂ y (c) ih ___ = – ___ ____ 2m ∂x2 ∂t

d 2y ____ 2mh (d) ____ + y (x) = 0 2 E dx

__

__

(x) If the vectors A and B are conservative, then __ __ __ __ (a) A × B is solenoidal (b) A × B is conservative __ __ __ __ (c) A + B is solenoidal (d) A – B is solenoidal _ (xi) The value of dl along a circle of radius 2 unites is c (a) zero (b) 2p (c) 4p (d) p (xii) Which one of the following operators is associated with energy? ∂ h2 h h __ (a) – ___ —2 + V (b) – ___ —2 (c) __ — (d) ih __ i 2m 2m ∂t

Group B Short Answer Questions Answer any three of the following questions: 2. (a) Sketch the Fermi distribution function for T = 0 and T > OK. (b) Assume that in tungsten (at. wt. = 183.8, density = 19.3 g/cc) There are two free electrons per atom. Calculate the Fermi energy and electron density. [9.2 eV, 1.264 × 1023 g/cc] d2 3. Which of the following functions are eigen functions of the operator ___2 ? What are the eigen dx values? (a) A sin kx (b) A ln x (c) Beikx __ 4. What do you mean by a scalar field? If f (x, y, z) = 3x2y – y3z2, then find —f at the point (1, – 2, 1). 5. (a) What is an electric line of force? What is its importance? (b) Sketch the electric lines of force due to point charger (i) q > 0 and (ii) q < 0. 6. (a) What is electric field intensity? (b) Prove that newtons per coulomb is dimensionally same as volt per meter. 7. State Ampere’s law in magnetostatics in integral form and from that deduce its differential form.

Model Question Paper 1

M1.3

Group C Long Answer Questions Answer any three questions from the following: 8. (a) State Gauss’ law in electrostatics and hence obtain its differential form. (b) What is scalar field? Give example. (c) Show that the vector field

( )

x+ y ______ is a sink field. V= – _______ ÷x2 + y2 __

(d) Three charges: 1 mc, 1 mc and – 4 mc are placed at the three corners of an equilateral triangle with side 10 cm. Calculate electric potential energy of the system. 9. (a) State Biot–Savart law. Calculate the magnetic field at a point on the axis of a circular coil carrying current by using Biot–Savart law. (b) What is motional emf? Write down its expression. (c) Write down Maxwell’s equation for vacuum and derive wave equation for electric field in vacuum. 10. (a) Explain the terms surface integral and volume integral. (b) Name the theorem which relates surface and volume integrals of a system. Give its mathematical expression in terms of unit normal vector. __ __ __ (c) If H = — × A, then prove that __

_

Ú Ú H ◊ d s = 0 for any closed surface S. s

11. (a) Define (i) phase space and (ii) thermodynamical probability. (b) Give an expression of BE statistics and hence obtain Planck’s formula for black-body radiation. (c) Determine the wavelength lm corresponding to the maximum emissivity of a black body at temperature T equal to 300 K.

Model Question Paper 2

Group A Multiple Choice Questions __ _

1. (i) The value of — ◊ r is (a) 1 __

__

(b) 0

__

(c) 2

__

(d) 3

(ii) If A and B are irrotational, then A × B is (a) solenoidal (b) irrotational (c) rotational (d) None of these (iii) In free space, Poisson’s equation is r r (a) —2V = – ___ (b) —2V = ___ (c) —2V = 0 (d) —2V = rŒ0 Œ0 Œ0 _ _ (iv) The magnetic field at the origin due to a current element i dl placed at a position r is _ _ _ _ _ m0 idl × r m0 idl × r m0 idl × (a) ___ ______ (b) ___ ______ (c) ___ ______ (d) zero 2 3 4p r 4p r 4p r3 (v) Displacement current arises due to (a) positive charge only (b) negative charge only (c) time-varying electric field (d)__ None of these __ (vi) If — ◊ B = 0, then __

__ __

__

__

__

__

__

__

__

(a) B = — ◊ A (b) B = — × A (c) B = —2 A (vii) In free space, modified Ampere’s circuital law

(d) None of these __

__ __ ∂E (b) — × B = m0Œ0 ___ ∂t __ __ __ __ (c) — ◊ E = 0 (d) — ◊ B = 0 __ (viii) The angle between —f and the normal to the surface f = constant is p (a) __ (b) 0 (c) p 2 __ __ _ (ix) Condition of coplanarity of three vectors a, b and g is

(a) — × E = 0

__

_

__

__

_

__

__

_

(a) a ◊ (b + g ) = 0

(b) a ◊ (b + g ) = 0

(c) a × (b + g)

(d) a ◊ (b – g) = 0

__

__

_

p (d) __ 4

Advanced Engineering Physics

M2.2

__

(x) The magnitude of electric field E in the annular region of a charged cylindrical capacitor 1 1 (a) is same anywhere (b) varies as __r (c) varies as __2 (d) None of these r (xi) For a particle trapped in a box of length l, the value of the expected average is l 1 2 (a) __ (b) __ (c) __ (d) None of these 2 l l (xii) Which one of the following is not an acceptable wave function of a quantum particle? (a) y = e x (b) y = e– x (c) y = x n (d) y = sin x (xiii) A non-polar molecule is the one in which the center of gravity of positive and negative charges (a) coincides (b) gets separated by 10–8 m (c) gets separated by 1 Å (d) None of these

GROUP B Short Answer Questions Answer any three of the following questions: __ 2. Show that the wave equation in free space of electric field E is given by __

∂2E — E = m0Œ0 ____ . ∂ 2t 3. Distribute two particles in three different states according to (i) MB statistics, (ii) BE statistics, and (iii) FD statistics. 4. The wave function of a particle in a one-dimensional box of length L is given by 2

__

__

( )

npx 2 f = __ sin ____ L L

÷

Find the expectation value of x and x2.

[

L __ L2 L2 __ , – _____ 2 3 2p 2n2

]

5. State Stokes theorem. Find the vector perpendicular to x2 + y2 – z2 = 100 at the point (1, 2, 3). 6. Derive the differential form of the Gauss’ law from its integral form.

GROUP C Long Answer Questions Answer any three questions from the following: 7. (a) Define electric flux. (b) A spherical gaussian surface encloses a charge 8.55 × 10–8 C. Calculate the electric flux passing through the surface.

Model Question Paper 2

M2.3

(c) State Ampere’s circuital law. (d) In case of a long solenoid, prove that the intensity of the magnetic field at a point inside the solenoid is double of that at the end. (e) What is Lorentz force? 8. (a) Two straight conductors, each 10 cm long, are parallel to each other and separated by a distance of 2 cm. They carry current of 30 A and 40 A, respectively. Calculate the force on conductor. (b) What are scalar and vector potentials? __

(c) If the vector potential A = (x2 + y2+ z2) at the position (x, y, z), then find the magnetic field at (1, 1, 1). (d) Write down__Poisson’s and Laplace’s equations. __ (e) Show that — ◊ B = 0 9. (a) What is a dielectric? What is dielectric constant? (b) Discuss different polarisation mechanisms in dielectrics. __ __ __ (c) Define and explain the three electric vectors __ P, E__ and __ D. Why electric field inside a dielectric decreases due to polarization? Show that D = Œ0 E + P. Also give their units. 10. (a) State a comparative study among MB, BE and FD statistics. (b) Define macrostate and microstate with suitable example. (c) Draw the Fermi distribution function for T = OK and T > OK. (d) Calculate the total member of particles in a Fermionic gas in terms of the Fermi level at absolute zero. (e) Indicate which statistics will be applicable for (i) e– (ii) photon (iii) 2He4 11. (a) Write down Schrödinger’s equation for one-dimensional motion of a free particle in one-dimensional potential box. Find its eigen function and eigen energy. (b) Show that y = Ax + B (where A and B are constants) is a solution to the Schrödinger equation for E = 0 energy of a particle in a box. (c) Show that y (x, t) = y (x) e– iwt is a wave function of a stationary state. (d) Show that [x, px] = ih.

Model Question Paper 3

GROUP A Multiple Choice Questions 1. (i) A vector field is said to be sodenoidal only when (a) its divergence is zero (b) its curl is zero (c) both of its curl and divergence are zero (d) None of these (ii) The angle between and 2 + is __ ÷2 1 1 2 –1 __ –1 ___ –1 __ (b) cos (c) cos (d) cos–1 __ (a) cos 2 _ 3 5 5 (iii) For incompressible fluid of velocity v satisfies __ _ __ __ _ __ _ __ __ _ (a) — × v = 0 (b) — ◊ (— × v) = 0 (c) — ◊ v = 0 (d) — × (— × v) = 0 (iv) In a gaussian surface, the electric field intensity at energy point on the surface is (a) constant (b) not constant (c) zero (d) None of these (v) Flux of the electric field for a point charge (+q) at the origin through a spherical surface centered at the same origin is q q 2q 4q (a) ___ (b) _____ (c) ___ (d) ___ Œ0 Œ Œ 4pŒ0 0 0 (vi) The direction of propagation of electromagnetic wave is given by __ __

__

__

_

__

__

_

__ __

_

(a) — ◊ B = 0 (b) — × B = m0 J (c) — × B = J (d) — ◊ B = m0 J (vii) The direction of propagation of electromagnetic wave is given by __ __ (a) along the direction of E (b) along B __ __ (c) along E × B (d) None of these __ (viii) The relation between scalar potential f and vector potential A is __ __ __ __ ∂f ∂f (b) — × A = – m0Œ0 ___ (a) — × A = m0Œ0 ___ ∂t ∂t __ __ ∂f (c) — ◊ A = m0Œ0 ___ ∂t

__ __ ∂f (d) — ◊ A = – m0Œ0 ___ ∂t

Advanced Engineering Physics

M3.2

(ix) The FD distribution function is given by 1 1 (a) f (Œ) = ____________ (b) f (Œ) = ____________ (Œ – ŒF)/(KBT) (Œ – ŒF)/(KBT) e +1 e –1 1 __________ (c) f (Œ) = (Œ – Œ )/(K T) (d) None of these F B e (x) Choose the function which can be a physically acceptable wave function: __

2 2

(b) e– a x

(a) ax

2

(d) ea x

(c) ÷x

(xi) Which one of the following is a Fermion? (a) Photon (b) Electron (c) Phonon (d) Alpha-particle (xii) The normalized wave function for a particle moving in a one-dimensional potential box of length L is __

np 2 (a) y = __ sin ___ L xL

÷

__

÷

npx L (b) y = __ sin ____ L 2

__

__

npx px 2 2 (c) y = __ sin ____ (d) y = __ sin ___ L L L nL

÷

÷

(xiii) The net charge inside a dielectric before and after polarization remains (a) negative (b) positive (c) same (d) None of these

GROUP B Short Answer Questions Answer any three of the following: 2. Find the displacement current within a parallel-plate capacitor in series with a resistor which carries current I. Area of the capacitor plates is A and dielectric is vacuum. __

3. Show that F = (2xy + z3) + x2 + 3zx2 is a conservative force field. Hence obtain its scalar potential. 4. A particle is located in a two-dimensional square potential well with absolutely impenetrable walls L (0 < x < L, 0 < y < L). Find the probability of finding the particle within a region 0 < x < __, 0 < y 3 L < __. 3 5. What is a dielectric substance? Discuss the importance of dielectrics. 6. Distribute two particles in three different states according to (a) MB statistics (b) BE statistics, and (iii) FD statistics.

GROUP C Long Answer Questions Answer any three questions. 7. (a) Evaluate the expectation value of x for a one-dimensional potential box of length L in the ground state.

Model Question Paper 3

M3.3

(b) State the basic postulates of quantum mechanics. (c) Write down the Schrödinger wave equation for a particle of mass m in a rectangular box with infinitely rigid walls with edges a, b and c. What are the boundary conditions? What conditions are responsible for the quantization for energy of the particle? (d) Prove that the lowest state of a free particle in a cubical box is not degenerate. (e) Compute the lowest energy of a neutron confined in the nucleus where the nucleus in considered as a box with a size of 10–14 m (h = 6.2 × 10–34 Js, mn = 1.6 × 10–24 kg). 8. (a) State Gauss’ law in [6.43 MeV] electrostatics and extend the same to Poisson’s equations. (b) A spherical symmetric charge distribution is given by

( )

r2 r(r) = r0 1 – __2 a

for

0£r £a

=0 for u > a (c) Calculate (i) the total charge and (ii) the electric field intensity both inside and outside the charge distribution. __ (d) What is vector potential? Verify that the vector potential A is given by __ 1 _ __ A = __ (r × B) 2 9. (a) State Ampere’s circuital law. Show that the magnetic field inside a toroid having n member of turns per unit length and carrying a current 1 is m0nI. (b) What is Lorentz force? Show that Lorentz force does not work on a charged particle. __ _ ∂r (c) Show that — ◊ J + ___ = 0, where the symbols have their usual meanings. ∂t 10. (a) Write down Maxwell’s field equations, explaining the terms used. (b) Prove that electromagnetic waves are transverse in nature. (c) What is displacement current? __ __ __ ∂A ___ (d) Show that E = – —f – . ∂t 11. (a) What are the three electric vectors in dielectrics? Name and find the relation between them. (b) Explain why the introduction of a dielectric slab between the plates of a capacitor changes its capacitance? (c) State and prove Gauss’ Law in case of dielectrics. (d) Deduce an expression for energy stored in dielectric in electrostatic field.

Model Question Paper 4

GROUP A Multiple Choice Questions

__

1. (i) If be the unit vector in the direction A, then __ __ __ __ |A | A ___ ___ (a) = _› (c) = __ (d) None of these (b) = A |A| A | A | __ _ › (ii) — ◊ r is equal to (a) 2 (b) 0 (c) 1 (d) 3 (iii) Flux of the electric field for a point charge (q) at origin through a spherical surface centered at the origin is 2q q q (a) ___ (b) ___ (c) _____ (d) 0 Œ0 Œ0 4pŒ0 (iv) In free space, Poisson’s equation is Œ0 (c) —2V = (a) —2V = 0 (b) —2V = ___ (d) None of these r (v) A moving charge produces (a) electric field only (b) magnetic field only (c) both of these (d) None of these (vi) A current-carrying straight wire cannot move, but a current-carrying square loop adjacent to it can move under the influence of a magnetic force. The square loop will remain (a) remain stationary (b) move towards the wire (c) move away from the wire (d) None of these (vii) The magnetic flux linked with a coil at any instant t is given by f(t) = 5t3 – 100t + 200, the emf induced in the coil at t = 2 seconds is (a) 200 V (b) 40 V (c) 20 V (d) –20 V (viii) The differential form of Faraday’s law of electromagnetic induction is __

__

__

__

__ __ __ __ ∂B __ __ ∂E __ __ ∂B ∂E (b) — × E = ___ (c) — × B = ___ (d) — × B = – ___ (a) — × E = – ___ ∂t ∂t ∂t ∂t (ix) A non-polar molecule is the one in which the center of gravity of positive and negative charges

Advanced Engineering Physics

M4.2

(a) coincide (b) get separated by 10–8 m (c) get separated by 1Å (d) None of these (x) The ground state energy of a particle moving in a one-dimensional potential box is given in terms of length l of the box by 2h2 h h2 (a) ____2 (b) ____2 (c) ____2 (d) 0 8ml 8ml 8ml (xi) The waves representing a free particle in three dimensions are (a) standing waves (b) progressive waves (c) transverse waves (d) polarized waves (xii) The spin angular momentum of photon is h (a) h (b) __ (c) 0 (d) 2h 8

Group B Short Answer Questions Answer any three questions: 2. State Stokes theorem. Find the vector perpendicular to x2 + y2 – z2 = 100 at the point (1, 2, 3). 3. Derive Poisson’s and Laplace’s law from Gauss’ law of electrostatics. __ __ __ ∂2 E 2 ____ 4. Show that the wave equation in free space of electric field E is given by — E = m0Œ0 2 . ∂t __ _ _ ∂r 5. Show that the equation of continuity is given by — ◊ J + ___ = 0, where J and r have their usual ∂t meaning. __ npx 2 6. The wave function of a particle in one-dimensional box of length L is given by fn = __ sin ____. Find L L the expectation value of x and x2.

÷

GROUP C Long Answer Questions Answer any three questions out of the following: 7. (a) Find if the work done in moving an object in the field F = (2xy + z3) + x2 + 3xz2 from the point (1, – 2, 1) to (3, 1, 4) is independent of the path chosen. __

(b) If the vector potential A = (x2 + y2 – z2) at position (x, y, z), find the magnetic field at (1, 1, 1). (c) Find a unit vector perpendicular to x2 + y2 – z2 = 100 at point (1, 2, 3).

Model Question Paper 4

M4.3

__

(d) Find the magnetic induction B, at a point on the axis of an infinitely long solenoid carrying current 1, number of turns per unit length being n. 8. (a) Write down Laplace’s equation in spherical coordinate system and hence find the solution. (b) Show that the potential V = V0 (x2 – 2y2 + z2) satisfies Laplace’s function where V0 in a constant. (c) Charge is distributed along the x axis from x = 0 to x = L = 500 cm in such a way that its linear charge density is given by l = ax2, where a = 18.0 m cm–3. Calculate the total charge in the region 0 £ x £ 1. 9. (a) Write down the Schrödinger’s equation for a particle confined in a one-dimensional box. (b) If the wave function y (x) of a quantum mechanical particle is given by px y (x) = a sin ___, for 0 £ x £ L L = 0, for 0 ≥ x ≥ L Then determine the value of a. Also determine the value of x where the probability of finding the particle is maximum. ∂f (c) Show that [ x, n] = – ih nxn – 1 and [ x, f (x)] – ih ___. ∂x ∂ ∂2 (d) Show that ___ and ___2 are commutative. ∂x ∂x 10. (a) What are three electric vectors in dielectrics? Name them and find a relation among them. (b) Explain the phenomenon of polarization of dielectric medium and show that K = 1 + ce. Hence the symbols have their usual meaning. (c) Explain why the introduction of a dielectric slab between the plates of a capacitor changes its capacitance? (d) Determine the electrical susceptibility at 0°C for a gas whose dielectric constant at 0°C is 1.000041. 11. (a) Sketch the Fermi distribution for T = OK and T > OK and explain. (b) What is the occupation probability at Œ = ŒF? (c) Calculate how the degeneracy function g(Œ) depends on Œ for a Fermionic gas. (d) Express the Fermi level in a metal in terms of free electron density.

Model Question Paper 5

GROUP A Multiple Choice Questions __

__

1. (i) Two vectors A and B are parallel where __

__ __

__

(a) A × B = 0

__

(b) A ◊ B = 0

__ __

(c) A ◊ B = 1

(d) None of these

(ii) The angle between —f and surface of f = constant is p p (a) __ (b) __ (c) p (d) 0 2 4 (iii) The electric flux through each of the four faces of a cube of 1 m side if a charge q coulomb is placed at its center is Œ0 q q (a) ____ (b) 4Œ0 q (c) ____ (d) ___ 4Œ0 6Œ0 6q (iv) Laplace’s equation for an electrostatic field is r r (a) —2V = ___ (b) —2V = 0 (c) —2V = – ___ Œ0 Œ0

__ r (d) —V = ___ Œ 0

(v) The equation of continuity in a steady charged distribution is __ _

__

(a) — ◊ J = 0

__ _

_

(b) — × J = 0

__ _ r (d) — ◊ J = ___ Œ

(c) — ◊ J = r

__

0

(vi) The work done by the Lorentz force F on a charge particle is __ q _ (a) F ◊ d r (b) zero (c) ___ (d) qF Œ0 (vii) Displacement current arises due to (a) positive charge only (b) negative charge only (c) time-varying electric field (d) None of these __

∂2 B (viii) The solution of a plane electromagnetic wave — B = m0Œ0 ____ is ∂t2 2

__

__

_ _

__

__

_ _

(a) B = B0 ei (wt – k ◊ r) (b) B = B0ei (wt + k ◊ r)

__

__

__

_ _

(c) B = B0ei ( k . r – wt) (d) None of these

Advanced Engineering Physics

M5.2

__ __

__

(ix) The relation between three electric vectors E, D and P is __

__

__

(a) D = Œ0 E + P

__

__

__

(b) D = Œ0 (E + P)

__

__

__

(c) D = E + Œ0 P

__ 1 __ __ (d) D = ___ ΠE+P 0

(x) The spacing between the nth energy state and the next energy state in a one-dimensional potential box increases by (a) (2n – 1) (b) (2n + 1) (c) (n – 1) (d) (n + 1) (xi) The wave function ym (x) and yn (x) are orthogonal to each other. Which of the following relations must hold for them? +

(a)

Ú

–

+

ym*yn dx = 1

+

(b) Ú ymyn dx = 0 –

(c) Ú ym*yn dx = 0 (d) –

(xii) Average energy Œ of an electron in a metal at T = OK is 3 1 (a) EF (b) __ ŒF (c) __ ŒF 2 5

Ú

–

ymyn dx = 1

5 (d) __ ŒF 3

GROUP B Short Answer Questions Answer any three of the following questions: __

2. Prove that Ú —y dV = Ú y ds. V

3. 4. 5. 6.

s

What is an electrostatic field? Why is it called a conservative field? What are polar and non-polar molecules? Discuss different polarization methods in dielectrics. What is a wave function? Mention four points on its physical significance. Sketch the Fermi distribution function for T = OK and T > OK and explain. What is the occupation probability at Œ = ŒF?

GROUP C Long Answer Questions Answer any three questions from the following: 7. (a) The gradient of a scalar quantity is a vector quantity. Explain. d2f (r) __ 2 df (r) (b) Prove that —2f (r) = _____ + r _____. 2 dr dr _

(c) Find the work done to move an object along a vector r = 3 + 2 – 5 if the applied force is __

F=2 – – .

__

(d) For what value of ‘a’ is the vector A = (x + 3y) + (y – 2z) + (x + 9z) solenoidal?

Model Question Paper 5

M5.3

8. (a) The gradient of a scalar quantity is a vector quantity explain. 1 (b) Show that —2(logie r ) = __2 . r __ (c) If a rigid body rotates about an axis passing through the origin with angular velocity w and with _› __ _› __ 1 __ _› linear velocity v = w × r , then prove that w = __ (— × v ). 2 __ __ (d) Find the torque about the point 0(3, – 1, 3) of a force F(4, 2, 1) passing through the point A (5, 2, 4). 9. (a) Write down Schrödinger’s equation for one-dimensional motion of a free particle in one dimensional potential box. Find the eigen function and eign energy.

[ ]

∂ , ___ , [ x, ] and [ x, y]? ∂x (c) Prove that the wave function y (x, t) = A cos (kx – wt) does not satisfy equation for a free particle.

(b) What are the values of

2 2

(d) Calculate the normalization constant for a wave function (at t = 0) given by y (x) = Ae– 6 x /2. eikx. 10. (a) What is a dielectric? What are dielectric losses? (b) What are polar and non-polar molecules? Discuss the effect of electric field on polar dielectrics. What is meant by polarization of a dielectric? (c) State and prove Gauss’ law in case of dielectrics. (d) Dielectric constant of a gas at NTP is 1.00074. Calculate dipole moment of each atom of the gas when it is held in an external field of 3 × 104 V/m. 11. (a) What are fermions and bosom? Given two examples of each. (b) Three distinguishable particle’s, each of which can be in one of the Œ, 2Œ, 3Œenergy states, have total energy 6Œ. Find all possible distributions of particles is energy states. Find the number of microstate in each case. (c) Indicate which statistics will be applicable for (i) 1H1 (ii) e – (iii) p ° (iv) photon (d) The Fermi energy for sodium at T = OK is 3.1 eV. Find its value for aluminum given that the free electron density in aluminum is approximately 7 times that of sodium.

Solved WBUT Question Paper (2010)

Group A Multiple-Choice Questions 1. (i) The average energy of a free electron in a metal at absolute zero is 1 (a) zero (b) more than __ Ef 2 1 (c) less than or equal to __ Ef (d) equal to Ef when Ef is the Fermi level 2 Ans. (b) (ii) If f (x) denotes the wave function of a particle in a one-dimensional box then the dimension of f (x) is of 1 ______ (a) length (b) _______ ÷length 1 (c) ______ (d) None of these (f (x) is dimensionless) length Ans. (b) (iii) If x component of the momentum operator, and

2 x

( i.e., [

∂2 (a) – h 2 ___2 ∂x Ans. (b)

,

2 x]

) is represented by (b) ih

x,

∂ is represented by – ih ___, then the commutator of ∂x ∂ (c) 2h2 ___ ∂x

(d) 0

(c) proton

(d) positron

(iv) Example of a Boson is (a) neutron Ans. (b)

(b) deuterium

S1.2

Advanced Engineering Physics

Group B Short-Answer Questions 1. If a system has two eigen states y1 and y2 with eigen values E1 and E2, under what condition, will linear combination (y = ay1 + by2) be also an eigen state. Ans. y1 and y2 are two eigen states with eigen values E1 and E2 respectively. The linear combination of then y = ay1 + by2 will also be an eigen state of the system +•

if

Ú

–•

(ay1 + by2)* (ay1 + by2) dV = 1 •

+•

or, if

Ú

–•

2

|a|

y*1 y1 dV

+•

or, if or, if

Ú

–•

+•

+•

+ Ú a*b y*1 y2 dV + Ú ab*y1y*2 dV + Ú |b|2 y*2 y2 dV = 1 –•

–•

–•

+•

|a|2 y*1 y1 dV + Ú |b|2 y*2 y2 dV = 1 –•

a2 + b2 = 1

So, if a2 + b2 = 1, then y = ay1 + by2 will be an eigen state. This is the required condition. 2. If the wave function y (x) of quantum mechanical particle is given by px y (x) = a sin ___ for 0 £ x £ L L

( )

=0

otherwise

Then determine the value of a. Also determine the value of x where probability of finding the particle is maximum. •

Ú y *(x) y (x) dx = 1

Ans.

–• L

or,

Ú y *(x) y (x) dx = 1 0 L

or,

px px a sin ( ___ ) dx = 1 Ú a* sin ( ___ L) L 0

L

or,

px |a|2 Ú sin2 ___ dx = 1 L 0

or,

px a2 Ú 2 sin2 ___ dx = 2 L 0

or,

2px a2 Ú 1 – cos ____ dx = 2 L 0

L

L

(

)

[because of the given condition]

S1.3

Solved WBUT Question Paper (2010)

[

a2

or,

L

L

0

0

2px dx Ú dx – Ú cos ____ L

[

(

2px L a2 L – ___ sin ____ L 2p

or,

]

=2

)| ] = 2 L 0

2

or,

La = 2 __ 2 2 or, a2 = __ fi a = __ L L Now, the wave function is given by

÷

__

÷

px 2 y (x) = __ sin ___ L L The probability density P of the particle is given by the following equation:

( )

P = |y (x)|2 = y *(x) y (x) or, \ Since

px 2 P = __ sin2 ___ L L px 2 Pmax = __ and sin2 ___ = 1 L L px px p 2 ___ sin = 1, ___ = (2n + 1) __ L L 2

( ) ( )

( )

where

n = 0, 1, 2, 3, …

Hence,

L x = (2n + 1) __ 2

L \ the probability will be maximum at x = __ . 2 3. A system has non-degenerate single particle states with energy levels 0, E, 2E and 3E. Three particles are to be distributed in these states such that the total energy is 6E. Write down in each case all possible microstates assuming that the particles obey MB statistics or FD statistics. Ans. (i) MB statistics Since the particles are distinguishable, let the particles be A, B and C. The possible microstates will be Microstate

0E

1E

2E

(0, 0, 3, 0) (0, 1, 1, 1)

(1, 0, 0, 2)

0

0

0 0 0 0 0 0

A A B B C C

A B C

0 0 0

So, total number of microstates = 10

3E

Total energy

Microstates

ABC

0

6E

1

B C A C A B

C B C A B A

6E 6E 6E 6E 6E 6E

6

0 0 0

BC AC AB

6E 6E 6E

3

S1.4

Advanced Engineering Physics

(ii) FD statistics Since the particles are indistinguishable and according to the Pauli exclusion principle, each state can accommodate only one particle, so the total number of states (microstates) will be as follows: Macrostate 0E

1E

2E

3E

Totalenergy

Microstates

(0, 1, 1, 1)

A

A

A

6E

1

0

So total member of microstates = 1

Group C Long-Answer Questions 1. (a) The ground state and the excited state normalized wave functions of an atom are y 0 and y1 respectively, and the corresponding energy is E0 and E1. If the probability of finding the atom in the ground state is 90% and that for the excited state is 10%, then find the average energy of the atom. Also determine the normalized wave function of the atom. Ans. In this case the state of the particle is generated by superpositioning of two states y0 and y1, so the wave function should be a linear combination of the two wave functions y0 and y1. \ the wave function could be written as y = C0y0 + C1y1 The wave functions y0, y1 and y are normalized and the functions y0 and y1 are orthogonal to each other So, one can write

Ú

y *y dV = 1

[where V represents entire space]

V

or,

Ú

(C0y0 + C1y1)* (C0y0 + C1y1) dV = 1

V

or,

C*0C0 Ú y*0 y0 dV + C1*C1 Ú y1*y1 dV + C*0C1 Ú y*0y1 dV + C*1C0 Ú y*1y0 dV = 1 V

V

V

V

Since y0 and y1 are orthogonal wave functions, the third and fourth terms on the left side of the equation are zero. \

C0*C0 Ú y0* y0 dV + C1*C1 Ú y1*y1 dV = 1 V

V

Because of the normalization property of the wave functions y0 and y1, the integrals are unity. \

C0*C0 + C1*C1 = 1

or,

C21 + C20 = 1

[

C0 and C1 are real constants.]

The sum of probabilities of all possible states must be unity. So, C20 and C21 may be regarded as probabilities of finding the atom in the ground and first excited states respectively.

S1.5

Solved WBUT Question Paper (2010) ___

___

\ we have C0 = ÷0.9 and C1 = ÷0.1 . Then the wave function of the atom is given by ___

___

y = C0 y0 + C1y1 = ÷0.9 y0 + ÷0.1 y1 If Pn be the probability of finding the atom at the energy level En, then the average energy is given by EaV = S PnEn = P0E0 + P1E1 n

or,

[ P0 = C20 and P1 = C21]

EaV = 0.9 E0 + 0.1 E1

(b) A particle, of mass m, moving in three dimensions is confined within a box 0 < x < a, 0 < y < b, 0 < z < c. (The potential is zero inside and infinite outside.) Write down Schrödinger’s equation for the particle. By considering a stationary state wave function of the form u (x, y, z) = f1 (x) f2 (y)

(

)

2 2 2 p 2h2 n1 n2 n3 f3 (z), show that the allowed energies are E = ____ __2 + __2 + __2 . What is the degeneracy of the 2m a b c first excited energy level when a = b = c? Explain. What is ground state energy? Ans. As the potential inside the box is zero, it is independent of time. So, the required Schrödinger equation is given by

– h2 2 ____ — u (x, y, z) = Eu (x, y, z) 2m where

(1)

y (x, y, z, t) = u (x, y, z) f (t) u (x, y, z) = f1(x) f2(y) f3(z)

Equation (1) can be written as

or,

(

2mE —2 + ____ u (x, y, z) = 0 h2

)

(

∂2 ___ ∂2 ___ ∂2 ____ 2mE ___ + + + 2 f1 (x) f2 (y) f3 (z) = 0 2 2 2 ∂x ∂y ∂z h

)

∂ 2f3 ____ ∂2f1 ∂2f2 2mE f2 f3 ____2 + f1 f3 ____2 + f1 f2 ____ + 2 f1 f2 f3 = 0 2 ∂x ∂y ∂z h Dividing this equation by f1 f2 f3 and rearranging the terms, we get or,

∂ 2f3 ______ ∂ 2f1 __ ∂f2 __ – 2mE 1 ____ 1 ___ 1 ____ __ + + = f1 ∂x2 f2 ∂y2 f3 ∂z2 h2 Since f1 is a function of x only, we have ∂ 2f1 ____ d2f1 ____ = . ∂x2 dx2 For similar reasons, we can write ∂ 2f2 ____ d2f2 ____ = ∂y2 dy2

∂ 2f3 ____ d2f3 and ____ = . ∂z2 dz2

(2)

S1.6

Advanced Engineering Physics

Then Eq. (2) can be written as d2f3 ______ d2f1 __ d2f2 __ – 2mE 1 ____ 1 ____ 1 ____ __ + + = f1 dx2 f2 dy2 f3 dz2 h2

(3)

In Eq. (3), the three terms on the left-hand side are completely independent of each other. There are respectively functions of x, y and z and their sum is independent of x, y and z. This is possible only when each term is separately equal to a constant. 2mE Let K be a constant so that K2 = ____ . h2 So, Eq. (3) takes the following form: d2f3 d2f1 __ d2f 2 __ 1 ____ 1 ____ 1 ____ __ + + = – K2 f1 dx2 f2 dy2 f3 dz2 Let K2 = K21 + K22 + K23, such that d2f1 1 ____ __ = – K21 f1 dx2

(4)

d2f2 1 ____ __ = – K22 f2 dy2

(5)

d2f3 1 ____ __ = – K23 f3 dz3

(6)

Now, we have 2mE K2 = K21 + K22 + K23 = ____ h2 Equation (4) can be written as d2f1 ____ + K21 f1 = 0 dx2 The general solution of this equation is f1 = A1 sin K1x + B1 cos K1x

(7)

Similarly, the general solutions of Eqs (5) and (6) are: and

and

f2 = A2 sin K2 y + B2 cos K2 y

(8)

f3 = A3 sin K3 z + B3 cos K3 z

(9)

f1 = 0 at

x=0

and x = a

(10)

f2 = 0 at

y=0

and y = b

(11)

f3 = 0 at

z=0

and

(12)

z=c

Applying the condition (10) upon Eq. (7), we get n1p B1 = 0 and K1 = ___ a

(13)

S1.7

Solved WBUT Question Paper (2010)

where

n1 = 1, 2, 3, º

Similarly, we can write n2p B2 = 0 and K2 = ___ b where

n2 = 1, 2, 3, º

and

n3p B3 = 0 and K3 = ___ c

where

n3 = 1, 2, 3, º

(14)

(15)

The expressions for f1, f2 and f3 can be written as n1px f1 = A1 sin ____ a

(

)

(16)

( )

(17)

n2py f2 = A2 sin ____ b

n3pz f3 = A3 sin ____ c Now, substituting these expressions in equation,

( )

(18)

u (x, y, z) = f1 (x) f2 (y) f3 (z) we get n3pz n1px n2py ____ ____ u (x, y, z) = A1 A2 A3 sin ____ a sin b × sin c Due to the dependence of u on the quantum numbers n1, n2, n3, the expression for u (x, y, z) is written as n3pz n1px n2py ____ ____ un1n2n3 = A1 A2 A3 sin ____ a sin b sin c

(

or, where

un1n2n3

) (

( )

)

( ) ( ) ( ) n pz n px n py ____ ____ = A sin ( ____ a ) sin ( b ) sin ( c ) 1

2

3

(19)

A = A1 A2 A3 n1 = 1, 2, 3, ... n2 = 1, 2, 3, ... n3 = 1, 2, 3, ...

2mE Using the equation K 2 = K21 + K22 + K23 = ____ h2 we get (K21 + K22 + K23) 2 K2h2 ____________ ____ E= = h 2m 2m

(20)

S1.8

Advanced Engineering Physics

Substituting in it the expressions for K1, K2, and K3 from Eqs (13), (14) and (15) we get

(

)

(K21 + K22 + K23) p 2h2 n21 n22 n23 ____________ E= = _____ __2 + __2 + __2 2m 2m a b c Due to its dependence on the quantum numbers n1, n2, n3, the energy expression is written as,

(

2 2 2 p2h2 n1 n2 n3 En1n2n3 = ____ __2 + __2 + __2 2m a b c

)

(21)

Degeneracy of Energy States: In a cubical box, the total energy is proportional to the sum of squares of the quantum numbers (n1, n2, n3). The particle can have the same value of energy for more than one set of values of the quantum numbers (n1, n2, n3). It means that the energy values corresponding to more than one wave functions may be same. The states, represented by these wave functions, are called degenerate states. When a = b = c = L (say), the energy value is given by p 2h2 En1n2n3 = _____2 (n21 + n22 + n23) 2mL In the first excited state the values of n1, n2 and n3 can have either of the following three sets: (1, 1, 2), (1, 2, 1), (2, 1, 1). So, the first excited state has three-fold degeneracy with energy as follows: 6h2 E112 = E121 = E211 = _____2 8mL The ground state energy is given by 3h2 E111 = _____2 8mL (c) Define the follow: (i) Microstate

with no degeneracy.

(ii) Macrostate

(iii) Thermodynamic probability

Ans. Refer to Sections 8.4 and 8.5. 2. (a) Write down the time-dependent Schrödinger equation explaining the symbols used. Derive time-independent Schrödinger equation assuming that the potential is time independent. Ans. The time-dependent Schrödinger equation is given by – h2 2 ____ — y (x, y, z, t) + V (x, y, z, t) y (x, y, z, t) 2m ∂ = ih __ y (x, y, z, t) ∂t where —2 is the laplacian operator and ∂2 ∂2 ∂2 —2 = ___2 + ___2 + ___2 ∂x ∂y ∂z y (x, y, z, t) is the wave function of the particle which represented by it.

(1)

Solved WBUT Question Paper (2010)

S1.9

V(x, y, z, t) is the potential function. If the potential is independent of time, then V = V(x, y, z). Now, Eq. (1) can be written as ∂y – h2 2 ____ — y + Vy = ih ___ 2m ∂t

(2)

As the potential energy V is a function of position only, the total energy is a constant. So, the equation is separable into the time-dependent and time-independent parts. So, we can write y (x, y, z, t) = f (x, y, z) f (t) Now, putting this value of y in Eq. (2), one can get _ _ – h2 2 _ ____ — f (r) f(t) + V(r) f (r) f (t) 2m _ ∂ = ih __ { f (r ) f (t) } ∂t _

Now, dividing both sides by f(r) f (t), one gets _ _ – h2 ____ 1 1 ∂f (t) 2 ____ _ — f (r) + V(r) = ih ___ ____ 2m f (r ) f (t) ∂t _

The left-hand side of Eq. (3) is a function of r only while the right-hand side is a function of t only. this is possible only when both of the sides are equal to a constant separately. This constant is the total energy E of the v particle. Thus, one can write two equations, when one equation the left side to E, the following equation is obtained: _ _ – h2 2 _ ____ — f (r) + V(r ) f (r ) = E f (r) 2m _ _ _ 2m —2 f (r ) + ___ (4) [ E – V(r ) ] f (r) = 0 h2 Equation (4) is the required time-independent Schrödinger equation. What are Fermions? Give two examples. Fermions are the particles which obey Fermi–Dirac statistics. Fermions are spin-half-integral particles. Examples of Fermions are electrons and protons. Write the expression of BE distribution law and derive Planck’s black-body radiation law from BE statistics. Bose–Einstein distribution law is given by

or,

(b) Ans.

(c) Ans.

1 f (Œ) = ___________ (Œ – m)|(KBT) e –1

S1.10

Advanced Engineering Physics

where m is the chemical potential and KB is the Boltzmann constant. T is the absolute temperature and e is the energy level. For the second part, refer to Section 8.14. (d) Sketch the Fermi distribution for T = 0 and T > 0 and explain. Ans. Refer to Section 8.13.

Solved WBUT Question Paper (2011) Group A Multiple-Choice Questions 1. Choose the correct alternatives for any ten of the following: 10 ¥ 1 = 10 (i) Which of the following functions is acceptable as a wave function of a one-dimensional quantum mechanical system? 2 A (a) y = (b) y = Ae - x (c) y = Ax2 (d) y A sec x x Ans. (b) (ii) A system with time independent potential is in an energy state E. The wave function of this state is (a) independent of time (b) an exponentially decaying function of time (c) a periodic function of time with time period proportional to E (d) a periodic function of time with time period inversely proportional to E Ans. (b) (iii) For a free particle in cubical box the degeneracy of the first excited state is (a) two-fold (b) three-fold (c) six-fold (d) nine-fold Ans. (b) (iv) Wave function of a certain particle is given by y = A cos x for |x| £ =0

p 2

otherwise

Then the value of A is (a)

2 p

Ans. (b)

(b)

2 p

(c) 1

(d) zero

S2.2

Advanced Engineering Physics

(v) The physical interpretation of — ◊ B = 0 is ( B is the magnetic field) (a) magnetic monopole cannot exist (b) magnetic field is irrotational (c) magnetic field is conservative (d) magnetic lines of force are open curves Ans. (a) (vi) If a vector field, F is given by F =

sin h(r ) r then — ¥ F is given by 1 + 2r 3

(a) 3r (b) zero Ê cos h(r ) ˆ (c) Á - 3r ˜ qˆ, qˆ is the unit vector along q-line Ë 1 + 2r 3 ¯ cos h(r ) ˆ (d) ÊÁ - 3r ˜ F 3 Ë 1 + 2r ¯ Ans. (b) (vii) A proton traveling vertically downwards experiences a southward force due to a magnetic field directed at right angles to its path. An electron traveling northward in the same magnetic field will experience a magnetic force directed. (a) upwards (b) downwards (c) towards east (d) towards west Ans. (a) (viii) Ampere’s circuital law is applicable when the current density is (a) constant over space (b) time independent (c) solenoidal (d) irrotational Ans. (b) (ix) Dimension of m0 e0 (symbols with their usual significance) is (a) L–2T–2 (b) LT–1 (c) L–2T–1

(d) L–2T2

Ans. (d) (x) Electrostatic field may be obtained as the gradient of a scalar potential (a) because it is a conservative field (b) because it is always solenoidal (c) when the field is linear in x, y, z (d) where there is no magnetic field in the same location Ans. (a) (xi) If P represents the polarization vector then — ◊ P = –r, where r is the density of (a) free charge (b) bound charge (c) free charge at the boundary of the dielectric (d) sum of free and bound charge Ans. (b)

S2.3

Solved WBUT Question Paper (2011)

(xii) Two conducting spheres of radii r and 2r contain charges q and 2q. Electric field just outside the surface of these two spheres are E1 and E2 respectively. Then (a) E1 = E2 (b) E1 = 2E2 (c) 2E1 = E2 (d) E1 = 4E2 Ans. (b) (xiii) Which of the following particles is not a Fermion? (a) Proton (b) Neutron (c) Alpha-particle

(d) Electron

Ans. (c) (xiv) For T > 0 the probability of occupancy of a state of a Fermion with energy equal to Fermi energy is (a) 1 (b) decreases linearly with T 1 (c) (d) decreases exponentially with T 2 Ans. (c) (xv) Number of ways a total of N distinguishable particles may be placed in different energy levels (Ni particles in energy level Ei with degeneracy gi) is (a)

 ( N i !)

N

N! Â N i gi

(b) N ! Â i

i

gi i Ni !

(c)

i

Â

N gi i

N! (d)

i

 ( N i + gi - 1)! i

Ans. (b)

Group B Short-Answer Questions 3 ¥ 5 = 15

Answer any three of the following. 1 1. (a) Show that — 2 ÊÁ ˆ˜ = 0 when r π 0. Ë r¯

3

∂2 ∂2 ˆ Ê 1 Ê 1 ˆ Ê ∂2 ˆ Ans. — 2 Á ˜ = Á 2 + 2 + 2 ˜ Ë r ¯ Ë ∂x ∂y ∂z ¯ ÁË x 2 + y 2 + z 2 ˜¯ 1 ∂2 2 ∂ È 1 ˘ ( x + y 2 + z 2 )- 2 = Í- ( x 2 + y 2 + z 2 )-3/2 (2 x )˙ ∂x Î 2 ˚ ∂x 2

∂ [x(x2 + y2 + z2)–3/2] ∂x = –(x2 + y2 + z2)–3/2 + 3x2 (x2 + y2 + z2)–5/2 =

\ Similarly,

1 ∂2 2 2 x 2 - y2 - z2 ( x + y 2 + z 2 )- 2 = 2 2 ∂x ( x + y 2 + z 2 )5/2 1 ∂2 2 2 y2 - z2 - x 2 ( x + y 2 + z 2 )- 2 = 2 2 ∂y ( x + y 2 + z 2 )5/2

S2.4

Advanced Engineering Physics

and

∂2 2 2 z2 - x 2 - y2 2 2 - 21 ( + + ) = x y z ∂z 2 ( x 2 + y 2 + z 2 )5/2

\

1 ∂2 ∂2 ˆ Ê 1 ˆ Ê ∂2 — 2 Á ˜ = Á 2 + 2 + 2 ˜ ( x 2 + y 2 + z 2 )- 2 Ë r ¯ Ë ∂x ∂y ∂z ¯

= Hence,

2 x 2 - y2 - z2 + 2 y2 - z2 - x 2 + 2 z2 - x 2 - y2 ( x 2 + y 2 + z 2 )5/2

Ê 1ˆ —2 Á ˜ = 0 Ë r¯

(Proved)

(b) Find the angle between the normal to the x – y plane and the normal to surface x2 + y2 = z2. 2 Ans. The unit vector normal to the X – Y plane is kˆ . And the normal to the surface x2 + y2 = z2 is given by 2 2 2 n = — (x + y – z ) ∂ ˆ ∂ˆ 2 Ê ∂ = Á iˆ + ˆj + k ˜ (x + y2 +– z2) Ë ∂x ∂y ∂z ¯ or,

n = 2 xiˆ + 2 yjˆ - 2 zkˆ

\ The unit vector normal to the surface is given by nˆ =

2 xiˆ + 2 yjˆ - 2 zkˆ 4 x 2 + 4 y2 + 4 z2

Let q be the required angle \

ˆ ˆ + ˆjy - kz (ix xˆ ◊ kˆ = (1) (1) cos q = kˆ ◊ x 2 + y2 + z2

\

cos q =

-z x + y2 + z2 2

Now, the set of values of (x, y, z) like (1, \

cos q =

or,

cos q = -

-2 1+ 3+ 4

3 , 2) will be on the plane given,

fi cos q =

-2 2 2

1 fi cos q = 135° 2

Hence, q = 135° 2. Starting from Maxwell’s equation in a charge free conducting media arrive at the concept of skin depth and express it in terms of signal frequency and conducting for a very good conductor. Ans. Refer to Sections 5.13 and 5.14.

S2.5

Solved WBUT Question Paper (2011)

3. A capacitor is made with two infinitely long cylindrical conductors of radii a and b (a > b), with vacuum in the intervening space. If the internal cylinder is kept at ground and the outer cylinder has a charge density s then solve Laplace equation to find the electrostatic potential in the space between the cylinders. z Ans. The diagram given in Fig. 1 explains the phenomenon. s A B The radii of the inner and outer cylinders are given by a ri = b and ro = a. VA = 0 b And the relative permittivity in the space between the cylinders is eo. Since the potential exists only along the radial direction, we get, Y

1 ∂ Ê ∂V ˆ Ár ˜ =0 r ∂r Ë ∂r ¯

(1)

È∵ V is independent of q and z here ˘ Í ˙ 2 2 Í we have ∂ V = ∂ V = 0 ˙ ÍÎ ˙˚ ∂q 2 ∂z 2

X

Fig. 1

Now, integrating Eq. (1) with respect to r, we get, ∂V =c where c is a constant of integration r ∂r ∂V c = ∂r r Integrating again with respect to r, we get V = c ln r + D (2) where D is another constant of integration C and D both are arbitrary. Now, applying appropriate boundary conditions (namely when r = a, V = VA and when r = b, V = VB = 0) in Eq. (2), we have VA = c ln a + D VB = c ln b + D \ VA – VB = c ln a – c ln b or,

or,

VA = c(ln a – ln b)

or,

VA = c ln

Again or, or,

VB = c ln b + D D + c ln b = 0 D = –c ln b

a fic= b

[∵ VB = 0] VA Ê aˆ ln Á ˜ Ë b¯ [∵ VB = 0]

S2.6

Advanced Engineering Physics

or,

D= -

VA ◊ ln b Ê aˆ ln Á ˜ Ë b¯

From Eq. (2), we get VA VA ◊ ln r – ¥ ln b Ê aˆ Ê aˆ ln Á ˜ ln Á ˜ Ë b¯ Ë b¯

V=

VA (ln r – ln b) Ê aˆ ln Á ˜ Ë b¯

or,

V=

or,

Ê ln r /b ˆ V = VA Á Ë ln a /b ˜¯

(3)

Equation (3) is the solution of Laplace’s equation in cylindrical coordinate, which gives the potential in the space between the cylinders of the cylindrical capacitor. 4. Write down the lagrangian of a simple harmonic oscillator moving in one dimension. Calculate the generalized momentum. Write down Lagrange’s equation of motion. Find out the hamiltonian of this system. Write down Hamilton’s equations for this system. 1+1+1+1+1 Ans. The lagrangian of a simple harmonic oscillator is given by 1 1 m L = T – V = mx 2 – kx2 2 2 k where

T=

x 1 1 2 mx and V = - Ú Fdx = kx 2 2 2 0

x

The generalized momentum is given by px =

∂T ∂ Ê1 ˆ = Á mx 2 ˜ ¯ ∂x ∂x Ë 2

fi px = mx The lagrange’s equation of motion is given by d Ê ∂L ˆ ∂L =0 Á ˜dt Ë ∂x ¯ ∂x or,

d Ê ∂ Ê 1 2 1 2ˆˆ ∂ Ê 1 2 1 2ˆ Á Á mx - kx ˜¯ ˜¯ - ÁË mx - kx ˜¯ dt Ë ∂x Ë 2 2 ∂x 2 2

or,

d (mx ) + kx = 0 dt

or,

mx + kx = 0

Fig. 2

S2.7

Solved WBUT Question Paper (2011)

The hamiltonian is given by

or,

H = px x -

1 2 1 2 mx + kx 2 2

H = mx 2 -

1 2 1 2 mx + kx 2 2

È∵ H = Â pi qi - L ˘ i Î ˚

1 2 1 2 px2 1 2 mx + kx = + kx 2 2 2m 2 The Hamilton’s equations of motion given by or,

H=

x=

∂H ∂p x

or

x=

px m

and

px = -

fi

∂H ∂x

x=

fi

∂ Ê px2 1 2 ˆ + kx ˜ Á ¯ ∂p x Ë 2 m 2

px = -

∂ Ê px2 1 2 ˆ + kx ˜ ∂x ÁË 2 m 2 ¯

fi

px = - kx

5. In how many ways 2 indistinguishable particles can be distributed in three distinct non-degenerate states, if the particles obey (i) F-D statistics, (ii) B-E statistics? Write down the expression of occupation probability of photon in the frequency interval n to n + dn. (2 + 2) + 1 Ans. Refer to Example 8.1. The answer of the last part of the question is as follows: The probability of a boson to occupy an energy state (e) at absolute temperature T is given by 1 f(e) = e(e - m )/( kT ) - 1 As in the present case the bosons are photons, the energy level corresponding to frequency n is given by e = hn and the energy level corresponding to frequency n + dn is given by h(n + dn) = e + de \ the required occupation probability is given by p(e) = f(e) de dΠor, p(e) = (e - m )/( kT ) e -1 or,

p(e) =

hdn e

( hn - m )/( kT )

-1

S2.8

Advanced Engineering Physics

Group C Long-Answer Questions 1. (a) Prove that for any closed curve

Ú B ◊ ds = 0 where B is the magnetic field.

3

s

Ans. The differential form of Gauss’ law in magnetostatics is given by —◊B =0 where B is the magnetic field. The magnetic flux fB through a closed surface is given by fB =

(1)

Ú B ◊ ds s

Now, by applying Gauss’ divergence theorem we get,

Ú B ◊ d s = Ú (— ◊ B)dv s

or,

v

Ú B ◊ ds = 0 s

[by Eq. (1)]

(b) Given n = w ¥ r , where r is the position vector and w is a constant angular velocity vector, then find out — ¥ n . 4 Ans. Refer to Section 1.11. (c) Use the expression of gradient in the spherical coordinate system to find the normal to the surface r sin q = 1. 3 Ans. The equation of the surface is given by r sin q = 1 Let y represent the surface. \ y = r sin q – 1 = 0 In spherical polar coordinate system ∂ 1 ∂ 1 ∂ — = rˆ + qˆ + fˆ ∂r r ∂q r sin q ∂j \ the normal to the surface is given by 1 ∂ 1 ∂ˆ Ê ∂ —y = Á rˆ + qˆ + fˆ y Ë ∂r r ∂q r sin q ∂f ˜¯ or,

1 ∂ 1 ∂ˆ Ê ∂ —y = Á rˆ + qˆ + fˆ (r sin q - 1) Ë ∂r r ∂q r sin q ∂f ˜¯

or,

—y = rˆ sin q + qˆ cos q

\ the normal to the surface is nˆ = rˆ sin q + qˆ cos q

[∵ | rˆ sin q + qˆ cos q | = 1]

Solved WBUT Question Paper (2011)

(d) If A is a constant vector, show that —(r ◊ A) = A , where r is the position vector.

S2.9

3

Ans. As A is a constant vector, we can write ˆ ˆ + ˆjA + kA A = iA 1 2 3 where A1, A2 and A3 are three constants. \

ˆ ) ◊ (iA ˆ ) ˆ + ˆjy + kz ˆ + ˆjA + kA r ◊ A = (ix 1 2 3

or,

r ◊ A = A1x + A2y + A3z

So,

—(r ◊ A) = — (A1x + A2y + A3z)

or,

∂ ∂ ˆ ∂ˆ —(r ◊ A) = ÊÁ iˆ + ˆj + k ˜ (A1x + A2y + A3z) Ë ∂x ∂y ∂z ¯

or,

ˆ ˆ + ˆjA + kA —(r ◊ A) = iA 1 2 3

or,

—( r ◊ A ) = A

(e) If the electric field is given by E =

(proved) 1 ˆ ( xi + yjˆ - 2 zkˆ ) , then find the charge density. Œ0

2

Ans. The differential form of Gauss’ law is given by r —◊E = e0 or,

r ʈ ∂ ˆ ∂ ˆ ∂ ˆ 1 ˆ ˆ ˆ ÁË i ∂x + j ∂y + k ∂z ˜¯ ◊ e ( xi + yj - 2 zk ) = e 0 0

or,

1 1 (1 + 1 – 2) = r e0 e0

or, r=0 2. (a) Define polarization vector and displacement vector. Rewrite differential form of Coulomb’s law in terms of displacement vector. 1+1+2 Ans. Refer to Sections 3.3.4 and 3.4. The answer of the last part of the question is as follows: Coulomb’s law is given by F=

1 q0 q rˆ 4pe 0 r 2

where F is the electric force between the two changes q and q0 and r is the distance between them. 1 q0 q F = q0 E = rˆ 4pe 0 r 2 Considering q as field generating charge and q0 as test charge. \

E=

1 q rˆ 4pe 0 r 2

S2.10

Advanced Engineering Physics

q Ê rˆ ˆ —◊Á 2˜ Ër ¯ 4pe 0

or,

—◊E =

or,

— ◊ (e 0 E ) =

or,

—◊D=

q Ê rˆ ˆ —◊Á 2˜ Ër ¯ 4p

q Êr ˆ —◊Á 3˜ Ër ¯ 4p

(1) [∵ D = e 0 E ]

So, Eq. (1) is the differential form of Coulomb’s law in terms of the displacement vector D . (b) Define electronic polarizibility. Discuss how monatomic gases can be polarized. 1+1 Ans. Refer to Sections 3.5 and 3.7. (c) State Gauss’ law in electrostatics and derive Poisson’s equation from that. 2+2 Ans. Refer to Sections 2.12 and 2.13. (d) Starting from the principle of charge conservation establish the equation of continuity. 5 Ans. Refer to Section 4.4. 3. (a) Find magnetic field at a point (1, 1, 1) if vector potential at that position is A = (10x2 + y2 = z2) ˆj . 2 Ans. Let B and A represent respectively the magnetic field and magnetic vector potential. [Given A = (10x2 + y2 – z2) ˆj ]

\

B=—¥ A

or,

∂ ˆ ∂ˆ Ê ∂ 2 2 2 B = Á i + ˆj + k ˜ ¥ (10x + y – z ) ˆj Ë ∂x ∂y ∂z ¯

or,

B = kˆ(20 x ) + 0 + (– iˆ )(-2 z )

or,

B = 2 ziˆ + 20 xkˆ

The value of B at the point (1, 1, 1) is given by B(1, 1, 1) = 2iˆ + 20 kˆ (b) A very long cylinder of radius a carries current I, which is uniformly distributed over the crosssection of the cylinder. If the current flows along the axis of the cylinder then find out the electric field in the conductor, the magnetic field just outside the conductor and the Poynting vector at the surface of the cylinder. You may assume that the conductivity of the material is s. 2 + 2+ 2 Ans. Refer to Sections 2.12. 3(c), 4.13.1, and 5.15. (c) Calculate the magnetic field at the center of a circular lamp carrying current I. 5 Ans. Refer to Section 4.8(B). (d) If E = E0 ei(kx – wt) denotes the electric vector of an electromagnetic field in vacuum then find out the magnetic vector. 2 Ans. From the set of Maxwell’s equations we know that —¥E=-

∂B ∂t

S2.11

Solved WBUT Question Paper (2011)

or,

B = - Ú— ¥ E dt

or,

B = - Ú — ¥ E0 ei ( kx - w t ) dt [∵ E = E0 ei ( kx - w t ) ] i(kx – wt)

or,

B = - — ¥ E0 Úe

or,

B = - — ¥ E0 ei kx Úe

or,

Ê 1 ˆ ikx - iw t B = - — ¥ E0 Á (e )(e ) Ë -iw ˜¯

or,

B=-

or,

dt

–i wt

dt

i — ¥ E0 ei ( kx - w t ) w i B=- —¥E w

[∵ E = Eo ei(kx – wt)] 4. (a) What do you mean by cyclic coordinate? Explain with an example. 2 Ans. If the lagrangian (L) of a system does not contain a generalized coordinate explicitly, then that coordinate is called a cyclic coordinate. In case of a cyclic coordinate (qi), we get ∂L =0 ∂q j Example: In case of a free particle (i.e. when the potential energy V = 0, the coordinate x, y and z are cyclic. (b) Show that if generalized force for a conservative system is zero then the generalized momentum will be conserved. 3 Ans. The generalized force is given by Qj =

∂V ∂q j

As the given system is conservative and the generalized force (Qj) is zero, we get ∂V =0 ∂q j \ the potential energy V is not a function of qj, i.e., V π V(qj) Hence the generalized coordinate qj is cyclic [∵ T π T(qj) and V π V(qj) fi L π L(qj)] For such a coordinate the Lagrange’s equation

d Ê ∂L ˆ ∂L = 0 reduces to dt ÁË ∂q j ˜¯ ∂q j

S2.12

Advanced Engineering Physics

d Ê ∂L ˆ =0 dt ÁË ∂q j ˜¯ or,

∂L = constant ∂q j

(1)

Now, we can write, ∂L ∂ ∂T = (T - V ) = ∂q j ∂q j ∂q j or,

[∵ V π V(qj)]

∂L = pj = constant [by eqn. (1)] ∂q j

Hence, the generalized momentum pj is conserved. (c) The wave function in a one-dimensional potential box with rigid walls is given by 2 np x , |x| £ l, sin l l =0 otherwise ˆ p ˆ x Find the expectation value of and . Ans. The expected value xˆ is given by yn(x) =

3+3

l

< x > = Ú y n* ( x ) xˆ y n ( x )dx 0 l

or,

< x> = Ú

0

2 np x 2 np x sin x sin dx l l l l

or,

< x> =

2l np x x sin 2 dx Ú l0 l

or,

< x> =

1l Ê 2 np x ˆ x Á 1 - cos ˜ dx Ú Ë l0 l ¯

or, or, or,

l 1l 2 np x x dx - Ú x cos dx Ú l0 l 0 l = – 0 2 l = 2

< x> =

l

Now

= Ú y n* ( x ) pˆ xy n ( x )dx 0

[∵ the problem is one-dimensional pˆ = pˆ x ]

Solved WBUT Question Paper (2011)

S2.13

∂ˆ 2 2 np x Ê np x Ú l sin l ÁË -i ∂x ˜¯ l sin l dt 0 l

or,

=

or,

< px > = -

2i l

or,

< px > = -

2i np l np x np x ¥ Ú sin l cos l dx l l 0

or,

< px > = -

i p x l Ê 2 np x ˆ ˜ dx Ú sin Á l2 0 Ë l ¯

or,

i np È l Ê 2 np x ˆ ˘ < px > = - 2 ¥ Í cos Á Ë l ˜¯ ˙˚ 0 2 n p l Î

or,

< px > =

l

np x d Ê

np x ˆ

Ú sin l dx ÁË sin l ˜¯ dx 0

l

l

i È 2 np x ˘ cos 2l ÍÎ l ˙˚ 0

i (1 - 1) 2l or, = 0 (d) Consider j = c1j1 + c2j2, where j1 and j2 are orthonormal energy eigenstates of a system cor1 responding to energy E1 and E2 at t = 0. If j is normalized and c1 = then what is the value 3 of c2? or,

< px > =

Find the expectation value of E2. Write the wave function at a subsequent time. Ans. f = c1f1 + c2f2 As f is normalized, we can write

+

Ú f * f dV = 1

+

or,

* * * * Ú (c1 f1 + c2f2 ) (c1f1 + c2f2)dV = 1

-

or,

+

+

-

-

* * * * Ú c1 c1f1 f1dV + Ú c1 c2f1 f2 dV

+

+

+

-

-

* * * * Ú c1c2f1f2 dV + Ú c2 c2f2f2 dV = 1

or,

c12 + 0 + 0 + c22 = 1

or,

c12 + c22 = 1

or,

1 + c2 = 1 3 2

1+2+1

S2.14

Advanced Engineering Physics

or,

c22 =

\

c2 =

2 3 2 3

ˆ = Ef We know that Hf Hˆ ◊ Hˆ f = Hˆ ( Ef ) fi Hˆ ◊ Hˆ f = E2f \

< E 2> =

+

* * * * 2 Ú (c1 f1 + c2f2 ) Hˆ (c1f1 + c2f2 )dV

-

or,

< E 2> =

+

* * * * 2 2 Ú (c1 f1 + c2f2 )( E1 c1f1 + E2 c2f2 )dV

-

or,

< E2> =

+

2 * * 2 * * 2 * * 2 * * Ú ( E1 c1 c1f1 f1 + E2 c1 c2f1 f2 + E1 c1c2f1f2 + E2 c2 c2f2f2 ) dV

-

or,

< E 2 > = E12

or,

=

or,

< E 2> =

or,

2

+

+

* * 2 * * Ú c1 c1f1 f1dV + 0 + 0 + E2 Ú c2 c2f2f2 dV

-

c12 E12

-

+

c22 E22

E12 2 E22 + 3 3 1 < E 2 > = ( E12 + 2 E22 ) 3

The wavefunction at a subsequent line t is given by f (r , t ) = f (r )e -

iEt

5. (a) Calculate the total number of particles in a fermionic gas in terms of Fermi level at absolute zero. 6 Ans. Refer to Section 8.13.3. (b) Calculate the total energy of particles in a fermionic gas at absolute zero. Ans. The total energy Et of an electron gas at absolute zero temperature is given by eF

Et =

Ú

eF

Ú eg(e)de

e N(e)de =

0

0 eF

or,

Et =

Ê ˆ Ú e ÁË h3 ˜¯ 2me mde 0 8p V

8p V È ˘ ÍÎ∵ g(e )de = h 3 2 me mde ˙˚ or,

Et =

8 2p Vm3/2 h3

eF

3/2 Ú e de 0

Solved WBUT Question Paper (2011)

or,

Et =

8 2p Vm3/2 Ê 2 3/2 ˆ ÁË e F ˜¯ 5 h3

or,

Et =

16 2p Vm3/2 3 5/2 ¥ eF 5 3h 3

or,

Et =

8p V Ê 2 m e F ˆ Á ˜ 3 Ë h2 ¯

or,

3 Et = N ¥ eF 5

or,

Et =

3/2

S2.15

3 ¥ eF 5 3/2 È 8p V Ê 2 m e F ˆ ˘ Í N= Á ˜ ˙ 3 Ë h2 ¯ ˚ Î

3 N eF 5

(c) Show that both F-D statistics and B-E statistics approach MB statistics at a certain limit. When does that happen? 3 Ans. Bose-Einstein’s distribution function is given by n 1 f(e) = = (e - m )/( kT ) g e -1 Hence where

ni = b=

e

gi (e i - m )/( kT )

-1

1 m and a = – kT kT

If ei >> kT, then e(ei - m )/( kT ) >> 1 \

ni =

or,

ni =

e

gi (e i - m )/( kT )

e

gi ei b + a

=

e

gi (e i /( kT ) - m /( kT )

which is same as MB distribution function. \ if ei >> kT, BE statistics reduces to MB statistics. The Fermi–Dirac distribution function is given by n 1 f(e) = = g e(e - e j )/( kT ) + 1 Hence where

ni =

gi (e i - e f )/( kT )

+1 eF 1 a=– and b = kT kT e

If ei >> kT, then e(ei – ef)/(kT) >> 1

S2.16

Advanced Engineering Physics

\

ni =

or,

e

gi e i /( kT ) - e f /( kT ) gi b ei + a

ni =

e which is same as MB distribution function. \ if ei >> kT, FD statistics reduces to MB statistics. (d) Two distinguishable particles are distributed in 2 non-degenerate states of energy 0 and Œ. List all the microstates and find the total energy corresponding to the states with maximum number of microstates. 3 Ans. As the particles are distinguishable, let them be denoted by a and b. The possible states will be as follows: Macrostate

e=0

e=e

Total energy

Microstates

(2, 0)

ab

0

0

1

(1, 1)

a

b

e

2

b

a

e

0

ab

2e

(0, 2)

The 1

The

The macrostate (1, 1) has maximum number of microstates with energy e. 6. (a) Use separation of variables technique to calculate the eigenfunctions and eigenvalues of a rigid box of side L. Calculate the wavelength of the photon that must be absorbed for a transition from the lowest to the second excited state. 4+1 Ans. Refer to Sections 7.8.1 (ii) (3D box). If the wavelength of the photon is given by l, then its energy will be given by E = hn. This energy will be stansferred by the photon to the particle in the box to go to the second excited state from the ground state. \ E = E221 – E111 6h 2 3h 2 2 8 mL 8 mL2

or,

hn =

or,

hc 3h 2 = l 8 mL2

\

l=

hc ¥ 8 mL2 2 3h

or,

l=

8 mcL2 units 3h

where L is the side of the box and m is the mass of the particle (b) Starting from Maxwell’s equation show that the electric field can be written in terms of a scalar potential and the magnetic vector potential as E = - —f -

∂A ∂t

5

Solved WBUT Question Paper (2011)

S2.17

Ans. Maxwell’s third equation is given by —¥E=-

∂B ∂t

(1)

But magnetic vector potential is given by B=—¥ A

(2)

where A is the magnetic vector potential From Eq. (1) and Eq. (2), we get ∂ — ¥ E = - (— ¥ A ) ∂t or,

Ê ∂A ˆ — ¥ E = — ¥ Á- ˜ Ë ∂t ¯

or,

Ê ∂A ˆ — ¥ E = - — ¥ —f + — ¥ Á - ˜ Ë ∂t ¯

[∵ curl of gradient of any scalar point function is zero. i.e., — ¥ —f = 0] Ê ∂A ˆ So, — ¥ E = — ¥ Á -—f ˜ Ë ∂t ¯ ∂A (proved) ∂t (c) Using Gauss’s law in integral form obtain the electric field due to the following charge distribution:

Hence,

E = - —f -

Ê r2 ˆ r = r0 Á 1 - ˜ , 0 < r £ a Ë a2 ¯ =0 a T1 > T0

T0

0.5

T1

Fermi distribution at zero and non-zero temperature

Fig. 1

T2

S3.6

Advanced Engineering Physics

5. Show that if the lagrangian does not depend on time, then the hamiltonian is a constant of motion. Write down the hamiltonian and obtain the equation of motion for a simple harmonic oscillator. 3+1+1 Ans. The Hamilton’s equations of motion are given by ∂H (1) qj = ∂p j ∂H ∂q j

(2)

∂L ∂H = ∂t ∂t

(3)

pj = -

where j = 1, 2, 3,…, 2f Now in this case lagrangian L is independent of t. So from Eqn. (3), we find that the hamiltonian H is also independent of time t. So,

dH =0 dt

Hence, H is constant of motion. A simple harmonic oscillator has one-dimensional motion. So, its kinetic and potential energies are given by T = 1 mx 2 2 1 2 kx 2

and

V=

\

the lagrangian L is given by 1 2 1 L = T – V = mx 2 – kx 2 2

\

px =

∂L = mx ∂x

px m The hamiltonian is given by or,

x=

H = Â pi qi - L = px x i

1 2 1 2 mx + kx 2 2

1 2 1 2 1 2 1 2 mx + kx = mx + kx 2 2 2 2

or,

H = mx 2 -

or,

H=

\

∂H p x ∂H = and = kx ∂p x m ∂x

1 px2 1 2 + kx 2 m 2

Solved WBUT Question Paper (2012)

S3.7

Now, the Hamilton’s equations are given by, x= and

∂H ∂p x

px = -

∂H ∂x

fi fi

x=

px m

px = - kx

fi mx + kx = 0 md 2 x or + kx = 0 dt 2

6. (a) Find the value of [ Lˆ x , zˆ ] . Ans. Refer to Example 7.11. (b) Show that the eigenvalues of a hermitian operation are real. Give an example of a hermitian operator in quantum mechanics. 2+2+1 Ans. Refer to Example 7.25. Example of hermitian operators are hamiltonian operator 2 Ê ˆ ˆ — 2 + V (rˆ )˜ and position operator (rˆ ). ÁË H = ¯ 2m

Group C Long-Answer Questions Answer any three of the following. 3 ¥ 15 = 45 1. (a) Distinguish between holonomic and non-holonomic constraints. Ans. If the relation of the constraints can be expressed as an equation and if they are independent of velocity then such constraints are called holonomic constraints. On the other hand, if the constraints, cannot be expressed in equational form and they are dependent on velocity then such constraints are called non-holonomic constraints. (b) Write down the equation of constraint, specify the nature of constraint and calculate the degrees of freedom in each case: (i) A particle constrained to move on the surface of a sphere (ii) A simple pendulum with a fixed support. 3+3 Ans. (i) It is a holonomic as well as scleronomic constraint. The equation is x2 + y2 + z2 = a2 Degrees of freedom f = 3 ¥ 1 – 1 = 2 (ii) The constraint is holonomic and scleronomic. The equation is given by x2 + y2 = l2 Degrees of freedom f = 2 ¥ 1 – 1 = 1 (c) Show that if a generalized coordinate is cyclic in lagrangian, then the corresponding generalized momentum will be conserved. 3 Ans. Refer to Section 6.2.6. (d) Find the equation of motion using Hamilton’s canonical equation for a system comprising masses m1 and m2 connected by a massless string of length L through a frictionless pulley such that m1 < m2. 4

S3.8

Advanced Engineering Physics

Ans. Figure 2 shows the device consisting of two masses m1 and m2 suspended over a frictionless pulley P of radius ‘a’ and connected by an inextensible string of length l. Let Q1A = x and Q2B = x1 From the above diagram we can write l = x + x1 + pa [∵ a is the radius] or, x = l – x1 – pa \ x = - x1

Q P Q1

Q2

a Q

x

x1

B

Now the kinetic energy of the masses m1 and m2 are given by 1 1 m1 x 2 and T2 = m2 x12 2 2 \ the total kinetic energy is given by 1 1 T = T1 + T2 = m1 x 2 + m2 x12 2 2 T1 =

A

m2 m1

Fig. 2

1 (m1 + m2) x 2 [∵ x1 = - x ] 2 With reference to the horizontal line passing through Q, the potential energy of m1 is given by V1 = –m1gx and that of m2 is given by V2 = –m2g(l – pa – x) \ V = V1 + V2 = –m1gx – m2g(l – pa – x) \ the lagrangian L is given by L=T–V 1 or, L= (m1 + m2) x 2 + m1gx + m2g (l – pa – x) 2 1 or, L= (m1 + m2) x 2 + (m1 – m2)gx + m2(l – pa)g 2 ∂L \ px = fi px = (m1 + m2) x ∂x The hamiltonian is given by H = Â pjq˙ j – L or,

T=

j

or,

H = px x – L

or,

H = (m1 + m2) x 2 –

or,

H=

1 (m1 + m2) x 2 – (m1 – m2)gx – m2g (l – pa) 2

or,

H=

px2 – (m1 – m2)gx – m2g(l – pa) 2(m1 + m2 )

1 (m1 + m2) x 2 – (m1 – m2)gx – m2(l – pa)g 2

Solved WBUT Question Paper (2012)

S3.9

The Hamilton’s equations are given by qj = From q j =

∂H ∂p j

and

pj = -

∂H ∂q j

we get x=

And from p j = -

∂H ∂p j

∂H , ∂q j

px (m1 + m2 )

(1)

we get

or,

(m1 + m2) x = –[–(m1 – m2)g]

or,

x=

m1 - m2 g m1 + m2

(2)

2. (a) What do you mean by m and G-phase space? Find the area in the phase space of a one-dimensional harmonic oscillator of mass m whose total energy is E. 2+2 Ans. For N particle system, the mechanical state of the whole system can be determined completely in terms of 3N position coordinates (q1, q2, ..., qN) and 3N momentum coordinates p1, p2, ... pN. The six-N dimensional space is called the phase space or G-space of the system. A point in the G-space represents a state of the entire system. For monoatomic gas it is also called molecular phase space or m-space. For a particle which is executing simple harmonic motion, the total energy E is given by E = Ep + Ek or,

E=

1 2 1 2 kx + mx 2 2

where x is the displacement from the equilibrium position, k is force constant, m is the mass and the velocity v = x . \

2E = k(x2) + m( x 2 )

or,

kx2 + mx 2 = 2E

or,

kx 2 mx 2 + =1 2E 2E

or,

x2 x2 + =1 Ê 2E ˆ Ê 2E ˆ ÁË ˜¯ Á ˜ Ë m¯ k

or,

x2 x2 + =1 (2 E / k ) (2 E / m)

S3.10

Advanced Engineering Physics

or,

px2 x2 + =1 (2 E / k ) (2 mE )

(1)

[∵ px = mx ] Equation (1) is an equation of an ellipse. So, the phase-space is two-dimensional and x–px plane forms a curve known as phase trajectory. The phase trajectory of a simple harmonic oscillator is an ellipse with semi-major axis a = 2 E / k and the semiminor axis b = 2 mE . So, the area in phase space (A) is given by A = pab 2E ¥ 2 mE k

or,

A=p

or,

A = 2pE

m k

(b) Derive Planck’s radiation law from BE statistics. State clearly the assumptions made in the theory. 3+2 Ans. Refer to the Section 8.14. (c) What is Fermi energy? Calculate the degeneracy function g(E) as a function of energy E for an ideal Fermi gas. 1+3 Ans. The Fermi energy eF is defined as the energy of the highest occupied level at absolute zero. For calculation of the degeneracy function g(E) refer to Section 8.12. (d) Evaluate the temperature at which there is one per cent probability that a state with energy of 0.6 eV above the Fermi energy will be occupied by an electron. 2 Ans. Fermi–Dirac distribution function is given by 1

f (e) = e

(e - e f ) / ( kT )

+1

This function is the required probability function through which we can calculate the temperature. Here De = e – ef = 0.6 eV. or, De = 0.6 ¥ 1.6 ¥ 10–19 J or, De = 0.96 ¥ 10–19 J 1 The probability is given by p = 100 1 e De /( kT ) + 1

\

p = f(e) =

or,

1 1 = 100 e De /( kT ) + 1

or, or,

eDe/(kT) + 1 = 100 eDe/(kT) = 99

S3.11

Solved WBUT Question Paper (2012)

or,

De = ln 99 kT

or,

T=

De k ln 99

fi T=

0.96 ¥ 10 -19 1.38 ¥ 10 -23 ¥ 4.595

Hence, T = 1513.93 K 3. (a) Give the physical interpretation of the wave function y(x). Ans. Refer to Section 7.4. 3. (b) Show that for a stationary state given by the wave function y(x, t) = y(x)e e value of energy is equal to the energy eigenvalue. Ans. The given wave function is y(x, t) = y(x) e - i The energy operator is given by ∂ Eˆ n = i ∂t

2 iE t n h

, the expectation 3

En t

\

E t ∂ Eˆ ny (x, t) = i y ( x )e ∂t

or,

∂ Eˆ ny ( x, t ) = i y ( x ) e ∂t

or,

Ê iE ˆ Eˆ ny ( x, t ) = i y ( x ) Á - n ˜ e Ë ¯

or, or,

Eˆ n y(x, t) = En y(x) e Eˆ n y(x, t) = En y(x, t)

[

n

\

e j iEn t

iEnt

i En t

\ En is the eigenvalue of Eˆ n . The expectation value En is given by +•

=

* Ú y (x, t) Eˆ n y (x, t)dx

-•

+•

or,

= Ú y * ( x )e -•

i En t

i

∂ ∂t

[y ( x)e \dx -

i En t

or,

+• Ê iE ˆ = Ú y *( x )(i ) Á - n ˜ y ( x )dx Ë ¯ -•

or,

= En

+•

or,

Ú y*(x) y(x)dx

-•

= En(1)

È +• * ˘ Í∵ Ú y y dx = 1˙ ÍÎ -• ˙˚

S3.12

Advanced Engineering Physics

or, = En \ the expectation value is equal to the eigenvalue En. (c) A particle is in a cubic box with infinitely hard walls whose edges are L units long. The wave function is given by np x ˆ Ê np y ˆ Ê np z ˆ y(x, y, z) = A sin ÊÁ sin sin Ë L ˜¯ ÁË L ˜¯ ÁË L ˜¯ Find the value of A. Find the ground state and first excited energy eigenvalues. Are they nondegenerate? Explain. 2+2+2 Ans. The normalized wave function is given by Ê n p xˆ Ê n p xˆ Ê n p xˆ y(x, y, z) = A sin Á 1 ˜ sin Á 2 ˜ ¥ sin ÁË 3 ˜¯ Ë L ¯ Ë L ¯ L As the wave function is normalized we have xyz

Ú Ú Ú y*y dxdydz = 1

000

xyz

Ê n p xˆ

Ê n p yˆ

Ê n pzˆ

2 2 2 3 1 2 Ú Ú Ú sin ÁË L ˜¯ sin ÁË L ˜¯ sin ÁË L ˜¯ ¥ dxdydz = 1 000

or,

A2

or,

Ê Lˆ Ê Lˆ Ê Lˆ A2 Á ˜ Á ˜ Á ˜ = 1 Ë 2¯ Ë 2¯ Ë 2¯

or,

A=

2 fi A = ÊÁ ˆ˜ Ë L¯

2 2

3/2

L3 \ the wave function can be written as y(x, y, z) =

2 2 L3

Now,

Ê n pzˆ n px n py sin ÊÁ 1 ˆ˜ sin ÊÁ 2 ˆ˜ ¥ sin Á 3 ˜ Ë L ¯ Ë L ¯ Ë L ¯

n12p 2 n22p 2 n32p 2 2 mE + 2 + 2 = 2 L2 L L

[For details, refer to article 7.8.1(b)] p È n12 n22 n32 ˘ + ˙ Í + 2 m Î L2 L2 L2 ˚ 2 2

or,

E=

As E is dependent on n1, n2 and n3, we can write En1n2 n3 =

2 2 2 h 2 ÈÊ n1 ˆ Ê n2 ˆ Ê n3 ˆ ˘ ÍÁ ˜ + Á ˜ + Á ˜ ˙ 8m ÎË L ¯ Ë L ¯ Ë L ¯ ˚

For ground state energy, n1 = n2 = n3 = 1

S3.13

Solved WBUT Question Paper (2012)

3h 2 8mL2 For first excited (n1, n2, n3) will be either (1, 1, 2) or (1, 2, 1) or (2, 1, 1) \

E111 =

\

E112 = E121 = E211 =

6h 2 8mL2

The ground state eigenvalue E111 is non-degenerate. But first excited state is three fold degenerated. (d) Show that the function y(x) = Cx e corresponding eigenvalue.

-

x2 2

Ê 2 d2 ˆ is an eigenfucntion of the operator ÁË x - 2 ˜¯ . Find the dx 3+1

Ans. The given wave function is given by y(x) = cx e - x

2

/2

and the operator given is

Ê 2 d2 ˆ ÁË x - 2 ˜¯ dx Now, let us operate on the wave function by the operator. \

Ê 2 d2 ˆ ÁË x - 2 ˜¯ y(x) dx Ê 2 d2 ˆ = Á x 2 - 2 ˜ (cx e - x /2 ) Ë ¯ dx = cx 3 e - x = cx 3 e - x

/2

-

2 d2 (cx e - x /2 ) 2 dx

/2

-

x 2 d ce - x /2 - cx 2 e - 2 dx

2

2

= ( cx 2 e - x

2

= 3c x e - x i.e.,

2

e

2

/2

+ cx e - x

/2

= 3y(x)

2

/2

+ 2cx e - x

2

j /2

- cx 3e - x

2

/2

)

Ê 2 d2 ˆ ÁË x - 2 ˜¯ y(x) = 3y(x) dx

2 ˆ Ê Hence, y is an eigen function of Á x 2 - d ˜ and the eigenvalue is 3. Ë dx 2 ¯

q 4. (a) If aˆ and bˆ are unit vectors and q is the angle between them, show that 2 sin = | aˆ - bˆ | . 2 2 ˆ ˆ Ans. | aˆ - b | = (aˆ - b) or,

| aˆ - bˆ | = (aˆ - bˆ ) ◊ (aˆ - bˆ )

or,

| aˆ - bˆ | = aˆ ◊ aˆ - aˆ ◊ bˆ - bˆ ◊ aˆ + bˆ ◊ bˆ

or,

| aˆ - bˆ | = 1 - cos q - cos q + 1

2

S3.14

Advanced Engineering Physics

or,

| aˆ - bˆ | = 2 - 2 cos q

or,

2 1 - cos q = | aˆ - bˆ |

or,

2 ¥ 2 sin 2

or,

2 sin

q = | aˆ - bˆ | 2

q = | aˆ - bˆ | 2

(b) Show that the electric field is always perpendicular to the equipotential surface. 3 Ans. Let us consider a sphere. And let charge q be placed at the center of it (vide Fig. 3): As the surface of the sphere is equidistant from the centre C, the surface of the sphere is an equipotential surface. Now, if a charge q1 be shifted from the point P to the point Q on the surface, the work done is given by

Q q

C

P c

Fig. 3

Q

W=

Ú q1E ◊ dl

P Q

=

Ú q1E dl cos q

P

Q

or,

W=

Ú q1E cos q dl

(1)

P

The potential energy possessed by the charge q1 at point P is Ep = Vq1 and the same at Q is E¢p = Vq1. \ DEp = Ep – E¢p = Vq1 – Vq1 = 0 \ the work done on the charge q1 while moving it from P to Q is zero. i.e., W = DEp = 0 (2) Q

or,

Ú q1 E cos q dl = 0

[from Eq. (1) and Eq. (2)]

(3)

P

The equation (3) can be true only if q = 90°, i.e., if E and dl are normal to each other. As dl is tangent to the surface, E is normal to the surface. In this case dl is arbitrary, so the electric field is always perpendicular to the equipotential surface. (c) Show that the vector A = (6xy + z3) iˆ + (3x2 – z) ˆj + (3xz2 – y) kˆ is irrotational. Find f such that A = —f . 2+3 Ans. A vector can be irrotational only if the curl of it is zero. Now,

— ¥ A = — ¥ {(6xy + z3) iˆ + (3x2 – z) ˆj + (3xz2 – y) kˆ }

Solved WBUT Question Paper (2012)

or,

iˆ ∂ ∂x

—¥ A=

6 xy + z 3 or,

ˆj ∂ ∂y

S3.15

kˆ ∂ ∂z

3 x 2 - z 3 xz 2 - y

∂ Ï∂ ¸ — ¥ A = iˆ Ì (3 xz 2 - y) - (3 x 2 - z ) ˝ ∂z Ó ∂y ˛ ∂ Ï∂ ¸ - ˆj Ì (3 xz 2 - y) - (6 xy + z 3 ) ˝ ∂z Ó ∂x ˛

or,

∂ Ï∂ ¸ + kˆ Ì (3 x 2 - z ) - (6 xy + z 2 ) ˝ ∂ x ∂ y Ó ˛ 2 2 ˆ — ¥ A = iˆ (–1 + 1) – ˆj (3z – 3z ) + k(6x – 6x)

or,

—¥ A=0

\

A is an irrotational vector.

Now,

A = —f

or,

ˆ = iˆ ∂f + ˆj ∂f + kˆ ∂f ˆ + ˆjA + kA iA x y z ∂x ∂y ∂z

\

∂f ∂f ∂f = Ax , = Ay , = Az ∂x ∂y ∂z

f = ÚAx dx f = ÚAy dy f = ÚAz dz By using Eq. (1), we get f = Ú(6xy + z3)dx or, f = 3x2y + z3x By using Eq. (2), we get f = Ú(3x2 – z)dy or, f = 3x2y – yz By using Eq. (3), we get f = Ú(3xz2 – y)dz or, f = xz3 – yz Now, observing Eqs (4), (5) and (6), we can express f as follows: f = 3x2y – yz + z3 x + c where c is an arbitrary constant. Hence,

(1) (2) (3)

(4)

(5)

(6)

(d) Calculate the work done in moving a particle in a force field given by F = 3 xyiˆ - 4 zjˆ + 8 ykˆ along the curve x = t2 + 1, y = t2, z = t3 from t = 0 to t = 1. 3

S3.16

Advanced Engineering Physics

Ans. F = 3xy iˆ – 4z ˆj + 8y kˆ \ Fx = 3xy, Fy = –4z, Fz = 8y The work done is given by

or,

W = Ú F ◊ dr = Ú(Fxdx + Fydy + Fzdz) W = Ú(3xydx – 4zdy + 8ydz) 1

=

2 2 2 2 2 Ú 3(t + 1) ◊ t ◊ 2t dt – Ú4t ◊ 2t dt + Ú8t ◊ 3t dt

0 1

or,

W=

5 3 4 4 Ú {6(t + t ) – 4t + 24t }dt

0

1

or,

W = Ú (6t5 + 6t3 + 16t4)dt 0

1

or,

W = Ú (6t5 + 16t4 + 6t3)dt 0

1

or,

16 6 W = ÈÍt 6 + t 5 + t 4 ˘˙ 5 4 ˚0 Î

or,

È 16 3 ˘ W = Í1 + + 5 2 ˙˚ Î

or,

W = 57 = 5.7 10

4p (e) Show Ú (ax iˆ + by ˆj + cz kˆ ) ◊ dS = (a + b + c), where S is the surface of the sphere 3 S x2 + y2 + z2 = 1. Ans. The equation of the sphere is given by x2 + y2 + z2 = 1 \ its radius r = 1 I = Ú (axiˆ + byjˆ + czkˆ ) ◊ d s

2

S

= Ú {— ◊ (axiˆ + byjˆ + czkˆ )}dV V

[by using Gauss’ divergence theorem] ∂ ∂ˆ Ê ∂ = Ú Á iˆ + ˆj + kˆ ˜ ◊ (axiˆ + byjˆ + czkˆ ) dV ∂y ∂z ¯ V Ë ∂x or,

I=

Ú (a + b + c)dV

V

or,

I = (a + b + c)

Ú dV

V

Solved WBUT Question Paper (2012)

or,

4p r 3 3

I = (a + b + c)

S3.17

[∵ the volume is spherical] r =1

4p (a + b + c) 3

or,

I=

\

Ú (axiˆ + byjˆ + czkˆ ) ◊ d S = 3 (a + b + c) S

4p

Ê r2 ˆ 5. (a) A spherically symmetric charge distribution is given by r(r) = r0 Á 1 - 2 ˜ for 0 £ r £ a, where Ë a ¯ r0 is a constant and r(r) = 0, for r > a. Calculate the (i) total charge (ii) the electric field intensity E and potential V both inside (r < a) and outside (r > a) regimes. 1+2+2 Ê r2 ˆ Ans. (a) (i) r(r) = r0 Á 1 - 2 ˜ for 0 £ r £ a Ë a ¯ =0 for r > a The total charge Q is given by Q = Úr (r)dV or,

a Ê r2 ˆ Q = Ú r0 Á 1 - ˜ dV Ë a2 ¯ 0

or,

Q = r0

or,

Q = 2pr0

or,

aÊ r2 ˆ Q = 2pr0 = Ú Á 1 - ˜ r2dr [ - cos q ]p0 Ë a2 ¯ 0

or,

aÊ r4 ˆ Q = 4pr0 Ú Á r 2 - ˜ dr Ë a2 ¯ 0

or,

5 ˘ È 3 Q = 4pr 0 Í r - r ˙ Î 3 5a 2 ˚ 0

or,

3 3 Q = 4pr0 ÊÁ a - a ˆ˜ Ë 3 5¯

or,

Ê 3ˆ Q = 4pr0 Á 2 a ˜ Ë 15 ¯

\

Q=

a

p

2p

Ê

r2 ˆ

Ú Ú Ú ÁË 1 - a 2 ˜¯ r2 sin q dqdfdr r = 0 q =f f = 0 Ê r2 ˆ 2 1 Á Ú Ú Ë a 2 ˜¯ r sin q dqdr r =0q =0 a

p

a

8p r0 a3 15

S3.18

Advanced Engineering Physics

(ii) Case I 8p When a < r < • charge Q = r0 a3 15 From Gauss’ law, we have Q Ú E ◊ ds = e 0 s

a

Fig. 4

1 8p ¥ r a3 e 0 15 0

or,

E(4pr2) =

or,

E=

1 8p r0 a3 ¥ 2 15 4pe 0 r

or,

E=

2 r0 a 3 15e 0 r 2

Again,

E=–

or,

dv dr Údv = –ÚE dr

or,

V = -Ú

\

V=

2 r0 a 3 dr 15e 0 r 2

2 r0 a 3 15e 0 r

Case II When r = a \

E=

2 r0 a 3 15e 0 a 2

And

V=

2 r0 a 3 15e 0 a

fi E= fi

V=

2 r0 a 15e 0

2 r0 a 2 15e 0

Case III 2ˆ Ê r(r) = r0 Á 1 - r ˜ Ë a2 ¯ The charge within the gaussian surface of radius r is given by Q¢ = Ú rdv v

r

p

2p

r

p

Ê

r2 ˆ

Ú Ú Ú ÁË 1 - a 2 ˜¯ r2 sin q dqdfdr r =0q =0f =0

or,

Q¢ = r0

or,

Q¢ = 2pr0

2 2 2 Ú Ú (1 – r /a )r (sin qdq)dr

r =0q =0

r a

Fig. 5

Solved WBUT Question Paper (2012)

or,

rÊ r2 ˆ Q¢ = 2pr0 Ú Á 1 - ˜ [ - cos q ]p0 r 2 dr Ë a2 ¯ 0

or,

rÊ r4 ˆ Q¢ = 4pr0 Ú Á r 2 - ˜ dr Ë a2 ¯ 0

S3.19

r

Èr3 Ê r3 r5 ˘ r5 ˆ Q¢ = 4pr0 Í fi Q¢ = 4pr 0 Á ˙ Ë 3 5a 2 ˜¯ Î 3 5a 2 ˚ 0 Again from Gauss’ law, we can write Q¢

Ú E ◊ ds = e 0 s or,

E(4pr2) =

\

E=

or,

V = –ÚE dr

or,

V = -Ú

\

V= -

4pr0 Ê r 3 r5 ˆ ÁË - 2 ˜¯ e0 3 5a

r0 Ê r r3 ˆ ÁË - 2 ˜¯ e 0 3 5a r0 Ê r r3 ˆ ÁË - 2 ˜¯ dr e 0 3 5a

r0 Ê r 2 r4 ˆ ÁË ˜ e 0 6 20 a 3 ¯

(b) If f is a scalar potential associated with the electric field E and A is the vector potential associated with the magnetic induction B , show that they must satisfy the equation —2f + ∂ r 5 (— ◊ A) = - . ∂t e0 Ans. Since the potential f is related to the electric field E , we can write E = - —f

(1)

Since A is the vector potential related to the magnetic field B , we get, B=—¥ A

(2)

∂ (— ◊ A) = - r / e 0 ∂t

Given,

—2f +

or,

Ê ∂A ˆ — ◊ ( —f ) + — ◊ Á ˜ = - r / e 0 Ë ∂t ¯

or,

r Ï ∂A ¸ — ◊ Ì —f + ˝=- 0 ∂t ˛ e0 Ó

or,

Ê ∂A ˆ r — ◊ Á -E + ˜ =Ë ∂t ¯ e0

[by Eq. (1)]

S3.20

Advanced Engineering Physics

or,

Ê ∂A ˆ — ◊ Á -E + ˜ = -— ◊ E Ë ∂t ¯

or,

Ê ˆ ∂A — ◊ Á -E + + E˜ = 0 Ë ¯ ∂t

or,

—◊

∂A =0 ∂t

or,

∂ (— ◊ A ) = 0 ∂t 0=0

\

L.S. = R.S.

or,

[∵ — ◊ A = 0] ∂ r Hence, —2f + (— ◊ A) = ∂t e0

(c) The intensity of sunlight reaching the earth’s surface is about 1300 W/m2. Calculate the strength of the electric and magnetic fields of the incoming sunlight. 3 Ans. The intensity of the sunlight reaching the earth surface is given by I = 1300 W/m2 \ the pointing vector P is given P = I = 1300 W/m2 We know that, P=E¥H or, or, or, \

|P| = |E ¥ H | P = E H sin 90° P = EH EH = 1300 W/m2

Again,

E = H

or,

m0 4p ¥ 10 -7 = e0 8.85 ¥ 10 -12

E = 376.82 H E EH ¥ = 1300 ¥ 376.82 H fi E = 699.90 Volt/m

or,

E2 = 489866

Again,

Ê Eˆ EH Á ˜ = 1300/376.82 Ë H¯

or,

EH ¥

or,

H2 = 345 fi H = 1.86 amp/m

H 1300 = E 376.82

[∵ E ^ H ]

S3.21

Solved WBUT Question Paper (2012)

(d) N charged spherical water drops, each having a radius r and charge q, coalesce into a single big drop. What is the potential of the big spherical drop? 2 Ans. Let the volume of a single water drop be represented by 4 Vs = pr2 [r = radius] 3 So, the volume of the coalesced drop of water is given 4 Vb = N Vs = pR3 [R = radius] 3 or,

Ê4 ˆ 4 N Á p r 3 ˜ = p R3 Ë3 ¯ 3

or,

R3 = Nr3

\ R= 3Nr The charge of the coalesced drop Q is given by Q = Nq Now from Gauss’ law, we get Q Ú E ◊ ds = e 0 S or,

(E)(4pr¢2) =

or,

E=

But

E= -

or,

(1)

dV ¢ dr ¢

(2)

dV ¢ 1 Nq = ◊ 2 dr ¢ 4pe 0 r ¢ dr ¢

1

Ú dV ¢ = 4pe Ú ( Nq) r ¢ 2 V¢ =

\ the potential, V ¢ =

Fig. 6

1 Nq ◊ 2 4pe 0 r ¢

0

or,

r¢ > R

Nq e0

From Eqs (1) and (2), we get -

r

1 Nq 4pe 0 r ¢

Nq 4pe 0 r ¢

The potential on the surface of the coalesced drop is given by Nq V ¢(R) = 4pe 0 R

[V ¢ = potential]

S3.22

Advanced Engineering Physics

or,

V ¢(R) =

Nq 4pe 0 N 1/3r

or,

V ¢(R) =

N 2/3q 4pe 0r

Solved WBUT Question Paper (Dec. 2012) Group A Multiple-Choice Questions 1. Choose the correct alternatives for any ten of the following: 10 ¥ 1 = 10 (i) The angle between the vectors iˆ + ˆj and iˆ - ˆj is (a) 90° (b) 60° (c) 30° Ans. (a) (ii) The value of

(d) 0°

Ú r ◊ dl on any arbitrary closed curve C is C

(a) 3 (b) 1 (c) –1 Ans. (d) (iii) In free space Poisson’s equation reduces to r r (a) —2u = 0 (b) —2u = (c) —2u = e0 e0

(d) 0

(d) —2u = •

Ans. (a) (iv) The continuity equation for steady current is (a) — ◊ j +

∂r =0 ∂t

(b) — ◊ j = 0

(c)

∂r =0 ∂t

(d) — ¥ j = 0

Ans. (b) (v) The electrostatic potential energy of a system of two charges q1 and q2 separated by distance r is (a)

1 q1q2 4pe 0 r 2

(b)

1 q1q2 4pe 0 r

(c)

1 q12 q2 4pe 0 r 2

(d)

e 0 q1q2 4p r

Ans. (b) (vi) If B = — ¥ A, B and A are any vectors then (a) — ◊ B = 0 Ans. (a)

(b) — ◊ B = 1

(c) — ◊ B = - 1

(d) — ◊ B = | A |

Advanced Engineering Physics

S4.2

(vii) The energy associated with a magnetic field H is (a)

1 2 H 2

(b) m0H2

(c)

1 m0H2 2

(d)

1 H2 2 m0

Ans. (c) (viii) Skin depth for a conductor in reference to electromagnetic wave varies (a) inversely as frequency (b) directly as frequency (c) inversely as square root of frequency (d) directly as square root of frequency Ans. (c) (ix) Schroendinger time independent wave equation is 1 1 (a) Hˆ y = Ey2 (b) Hˆ y2 = Ey2 (c) Hˆ = E (d) Hˆ y = Ey y y Ans. (d) (x) The ground state energy of a particle moving in a one-dimensional potential box is given in terms of length L of the box by 2 h2 2 2 (b) (c) (d) 0 8mL2 8mL2 8mL2 Ans. (c) (xi) The communication bracket [ pˆ y , yˆ ] is equal to (a) i (b) –i (c) i 2 (d) i/ Ans. (b) (xii) The electric dipole moment of a particle (atom or molecule) per unit polarizing electric field is termed as (a) polarization (b) polarizability (c) net dipole moment (d) susceptibility Ans. (b) (xiii) A system is called strongly degenerate if Ni Ni Ni (a) =1 (b) >> 1 (c) m2. (3 + 3 + 3) + 6 Ans. From Fig. 2, Let Q1A = x and Q2B = x1 From the diagram we can write L = x + x1 + pa [a is the radius] x1 = L – pa – x or,

x1 = - x

and

x1 = - x

a

Q1

Q2

Q x

x1

B m2

A m1

Fig. 2

Solved WBUT Question Paper (Dec. 2012)

S4.9

The kinetic energy of m2 is T2 =

1 1 m2 x12 = m2 x 2 2 2

Total kinetic energy T = T1 + T2 =

1 1 m x 2 + m2 x 2 2 1 2

1 (m + m2 ) x 2 2 1 With reference to the horizontal line passing through Q, the potential energy of m1 is given by V1 = –m1gx and that of m2 is V2 = –m2g(L – pa – x) \ total potential energy V = V1 + V2 = –m1gx + m2g (L – pa – x) The lagrangian of the system is given by 1 L = T – V = (m1 + m2) x 2 + m1gx + m2g(L – pa – x) 2 The generalized momentum px can be written as =

∂L = (m1 + m2) x ∂x The hamiltonian of the system is given by px =

or,

H = x px - L = (m1 + m2) x 2 – Again and x=

x =

x=

px (m1 + m2 )

1 (m1 + m2) x 2 – m1gx – m2g(L – pa – x) 2

px ∂H = ∂px (m1 + m2 )

px = -

∂H = m1g – m2g = (m1 – m2)g ∂x

px Ê m - m2 ˆ =Á 1 g , which is the hamiltonian equations of motion. m1 + m2 Ë m1 + m2 ¯˜

5. (a) State Gauss’ law of electrostatics. Ans. Refer to Section 2.12. (b) Derive an expression for the electric field between two infinite extent parallel plate capacitors carrying charge density s and mutual separation d. Draw the necessary diagram. Ans. Same as Q. 1(a) of Short Answer Type Questions. (c) Stating from the definition of current density derive the equation of continuity in current electricity. What is the condition of steady current? Ans. Refer to Section 4.4. (d) Two parallel wires carry equal current of 10A along with the same direction and are separated by a distance of 2.0 cm. Find the magnetic field at a point which is two cm away from any of these wires. 2 + 4 + (4 + 1) + 4

Advanced Engineering Physics

S4.10

Ans. The magnetic field at a point which is 2 cm away from any of these wires is B=

mo Ê 2i ˆ mo Ê 2i ˆ Á ˜+ Á ˜ 4p Ë d ¯ 4p Ë 2d ¯

Ê 2i ˆ Ê = 10 -7 ¥ Á ˜ Á 1 + Ë d¯Ë

Ans.

Ans.

Ans.

1ˆ ˜ 2¯

10 Tesla = 1.5 ¥ 10–4 Tesla 0.02 (a) Deduce density of states of free electrons having energy between E and E + dE in the phase space. Refer to Section 8.12. (b) Write down the postulates of B-E statistics and write down the B-E distribution function explaining the symbols. At what condition BE-statistics will yield classical statistics? Refer to Sections 8.9.1 & 8.9.2. (c) A system has non-degenerate single particle states with 0, 1, 2, 3 energy units. Three particles are to be distributed in these states such that the total energy of the system is 3 units. Find the number of microstates if the particles obey (i) MB statistics (ii) BE statistics (iii) FD statistics. 4 + (3 + 1 + 1) + (2 + 2 + 2) Refer to Worked out examples, Example 8.3 of Chapter 8. = 3 ¥ 10–7 ¥

6.

m0 È -7 ˘ ÍÎHere d = 2 cm = 0.02 m, I = 10A and 4p = 10 ˙˚

Solved WBUT Question Paper (June 2013)

GROUP A Multiple-Choice Questions 1.

Choose the correct alternatives for any ten of the following: 10 × 1 = 10 (i) If the constraint relations can be made independent of velocity then the constraints are called (a) sclerenomic (b) bilateral (c) holonomic (d) conservative

Ans. (c) (ii) If a system had f degree of freedom, then the number of Lagrange’s equations for the system is (a) 3 (b) f (c) 2f (d) f/2 Ans. (b) �� �� (iii) The physical interpretation of V ◊ B = 0 is (a) magnetic monopole does not exist (c) magnetic field is conservative

(b) magnetic field is irrotational (d) magnetic lines of force are open curves

Ans. (a) (iv) The velocity of electromagnetic waves in free space is (a) equal to velocity of light (b) greater than the velocity of light (c) less than the velocity of light (d) zero Ans. (a) (v) Dielectrics are substances which are (a) semiconductors (b) conductors (c) insulators (d) none of these Ans. (c) (vi) Ampere’s circuital law is applicable when the current density is (a) constant over space (b) time independent (c) solenoidal (d) irrotational Ans. (b) (vii) The waves representing a free particle in three dimensions are (a) standing waves (b) progressive waves (c) transverse waves (d) polarized waves Ans. (c)

S5.2

Advanced Engineering Physics

(viii) In an electromagnetic wave in free space the electric and magnetic fields are (a) parallel to each other (b) perpendicular to each other (c) inclined at an angle (d) inclined at an obtuse angle Ans. (b) (ix) A moving charge producers (a) electric field only (b) magnetic field only (c) both of them (d) static electric field only Ans. (c) (x) When the Hamiltonian operator operates on a wave function y(r) then the corresponding eigenvalue is (a) potential energy of the system (b) kinetic energy of the system (c) total energy of the system (d) none of these Ans. (c) (xi) The value of probability of an event cannot be (a) 1 (b) negative (c) zero (d) positive Ans. (b) Ê x2 ˆ (xii) If a wave packet is described by y(x) = A exp Á then the normalization constant is Ë 2s 2 ˜¯ ps (a) 2s (b) s (c) (d) none of these 2 Ans. (d) (xiii) He3 and muon are (a) Fermions (b) Bosons (c) Fermions and Bosons respectively (d) classical particles Ans. (a) (xiv) The spin angular momentum of a photon is (a) h (b) h/8 (c) 0 (d) 2 h Ans. (c) (xv) The maximum energy that can be occupied by an electron at T = 0 K is known as (a) band gap energy (b) Fermi energy (c) radiation energy (d) potential energy Ans. (b)

GROUP B Short-Answer Questions Answer any three of the following. 2. (a) If the vectors A and B be irrotational then show that the vector A × B is solenoidal. Ans. Refer to Example 1.35. (b) Prove that i ¥ ( j ¥ k ) = j ¥ (k ¥ i ) = k ¥ (i ¥ j ) = 0

3 × 5 = 15

2½ + 2½

Solved WBUT Question Paper (June 2013)

S5.3

Ans. The vector products of three unit vectors iˆ , ˆj , and kˆ which act along X, Y and Z-axis respectively are iˆ × ˆj = – ˆj × iˆ = kˆ ; ˆj × kˆ = – kˆ × ˆj = iˆ and kˆ × iˆ = – iˆ × kˆ = ˆj also iˆ × iˆ = ˆj × ˆj = kˆ × kˆ = 0. 3. (a) Write down Laplace’s equation. Show that the potential function x2 – y2 + z satisfies Laplace’s equation. 1+2 Ans. The Laplace’s equation is —2V = 0 Here, V = x2 – y2 + z �� ∂ ˆ ∂ˆ 2 Ê ∂ or —V = Á iˆ + ˆj + k ˜ ( x - y 2 + z ) = (2 xiˆ - 2 yjˆ + kˆ ) ∂y ∂z ¯ Ë ∂x Again

(b) Ans. 4. (a) Ans. (b) Ans. 5. (a)

�� �� ∂ ˆ ∂ˆ Ê ∂ —2V = — ◊ —V = Á iˆ + ˆj + k ˜ ◊ (2 xiˆ - 2 yjˆ + kˆ ) ∂y ∂z ¯ Ë ∂x

=2–2+0=0 or —2V = 0 So, the potential function V satisfied Laplaces’ equation. Show that when a dielectric is placed in an electric field, the field within the dielectric becomes weaker than the original field. 2 Refer to Section 3.4. Calculate the magnetic field along the axis of the current-carrying circular coil. 4 Refer to Section 4.8(ii). What is the value of the magnetic field at the centre of the coil? 1 Refer to Section 4.8(ii). What do you mean by a commutator? Prove that [x, Px] = i�. 1+2

Ans. If aˆ and bˆ be two operators, the commutator of this pair of operators is denoted by [a, b] and it is defined as [ aˆ , bˆ ] = aˆ bˆ – bˆ aˆ Let y be the state function \

[ xˆ , pˆ x ]y = È x, -i� ∂ ˘ y ÍÎ ∂x ˙˚ ∂y ∂ + i� ( xy ) ∂x ∂x ∂y ∂y = -i�x + i�x + i �y ∂x ∂x = i�y = -i�x

[ xˆ , pˆ x ] = i� \ (b) Write the basic postulates of wave mechanics. 2 Ans. Refer to Section 7.6. 6. (a) Show that the average energy of an electron in a metal at 0 K is given by 3/5 EF, where EF is the Fermi energy. 3

S5.4

Advanced Engineering Physics

Ans. Refer to Section 8.13.5. (b) Show that both FD and BE statistics approach MB statics at a certain limit. When does that happen? 2 Ans. Refer to Section 8.11.

GROUP C Long-Answer Questions Answer any three of the following. 3 × 15 = 45 7. (a) If in a region of space, the electric field is always in the x-direction then prove that (i) the potential is independent of y and z coordinates, (ii) if the field is constant, there is no free charge in that region 2+1 �� �� Ans. (a) (i)We know that E = -—V �� Ê ∂V ˆ ∂V ˆ ∂V ˆ i.e., E = - Á iˆ +j +k ∂y ∂z ˜¯ Ë ∂x �� Here, E is along x-direction, i.e., E = iˆ Ex ∂V So Ex = ∂x ∂V ∂V = =0 ∂y ∂z So, V is independent of y and z coordinates. �� �� (ii) Again we know — ◊ E = r Œ0 �� �� �� If E is constant V ◊ E = 0 i.e., r=0 So, if the field is constant, there is no free charge in the region. (b) Write down ��the differential form of Gauss’s law. Suppose that electric field in some region is found to be E = a r 3rˆ in spherical coordinates (a is a constant). Find the electric charge density. 1+3 Ans. The differential form of Gauss’s Law is �� �� r —◊ E = Œ0 But

In spherical polar coordinate system, �� ∂ 1 ∂ 1 ∂ — = rˆ + qˆ + jˆ ∂r r ∂q r sin q ∂j �� and E = a r 3rˆ

S5.5

Solved WBUT Question Paper (June 2013)

(c)

Ans. (d) Ans. 8. (a) Ans. (b) Ans. (c) Ans. 9. (a) Ans. (b)

Now

�� �� ∂ r V ◊E = (a r 3 ) = ∂r Œ0

or

3 a r2 =

r Œ0

So, the electric charge density r = 3 Œ0 a r 2 . A very long cylindrical object carries charge distribution proportional to the distance from the axis (r). If the cylinder is of radius a, then find the electric field both at r > a and r < a, by the application of Gauss’ law in electrostatics. 4 Refer to Section 2.12.3(iii). �� �� �� What is electric displacement vector? Establish the relation D = e 0 E + P where symbols have their usual meanings. 1+3 Refer to Section 3.4. State Biot–Savart’s law and obtain the magnetic field induction due to a wire carrying current I at a point P situated at a distance R from it. 2+3 Refer to Sections 4.7 and 4.8(i). Find the magnetic field at a point (1, 1, 1) if the vector potential at that position is �� 3 A = (10 x 2 + y 2 - z 2 ) ˆj Refer to Example 4.16. �� Obtain the magnetic field induction B at a point on the axis of a current circular conductor (loop) with n turns. 7 Refer to Section 4.8(ii). State the Poynting theorem. 2 Refer to Section 5.15. �� �� �� ∂B Prove that V ¥ E = . 3 ∂t

Ans. Refer to Section 5.5. (c) A conducting wire in the shape of an equilateral triangle of each side a carries a current I. Calculate the magnetic field at its centroid. 3 Ans. Here, ABC is an equilateral triangle of each side a carrying a current I [Fig. 1] in the anti-clockwise direction. a The distance OD = and –AOD = –BOD = 60°. 2 3 The magnetic field at the centroid O due to side AB is m0 I 3m 0 I [sin 60∞ + sin 60∞] = 2p a Ê a ˆ 4p Á Ë 2 3 ˜¯

Fig. 1

Since the current is in the anti-clockwise direction, the magnetic field at O due to AB, BC and CA is

S5.6

Advanced Engineering Physics

3¥

9 m0 I 3m 0 I = 2p a 2p a

and its direction is perpendicular (outward) to the paper. �� �� (d) If f is a scalar, potential associated with�� the electric field E and A is is the vector potential associated with the magnetic induction B , show that they must satisfy the equation ∂ �� �� r —2f + (V ◊ A) = 5 ∂t e0 �� �� �� Ans. We know that magnetic field is solenoidal, i.e., , so B , in terms of vector potential — ◊ B = 0 �� �� �� B = —¥ A. Again from Faraday’s laws of electromagnetic induction, �� �� �� ∂B ∂ �� �� = - (— ¥ A) —¥E = ∂t ∂t �� �� Ê �� ∂ A ˆ Therefore, — ¥ Á E + ˜ =0 Ë ∂t ¯ Since the curl of the gradient of a scalar function is zero, so �� �� ∂ A �� E+ = -—j [j is the scalar potential] ∂t �� �� �� ∂A or E = -—j ∂t Now taking divergence on both sides, we have �� �� È 2 ∂ �� �� ˘ — ◊ E = - Í—j + (— ◊ A)˙ ∂t Î ˚ �� �� r Again, from the differential form of Gauss’s Law, — ◊ E = Œ0 So, —j 2 +

r ∂ �� �� . (— ◊ A) = Œ0 ∂t

(e) A long solenoid of 40 cm length has 300 turns. If the solenoid carries a current of 3.5 A, find the magnetic field at one end of the solenoid. 2 Ans. Refer to Example 4.10. 10. (a) Calculate the total number of particles in a Fermionic gas in terms of the Fermi level at absolute zero temperature. 4 Ans. Refer to Section 8.13.3. (b) Apply B-E statistics to a photon and deduce Planck’s law of spectral energy density of blackbody radiation. 3 Ans. Refer to Section 8.14. (c) Define microstates and macrostates with suitable examples. 3 Ans. Refer to Section 8.4.

Solved WBUT Question Paper (June 2013)

S5.7

(d) A box contains 5 red balls and 3 white balls. The balls, except their colours, are identical. What is the probability that, on two independent draws, 1 ball is red and 1 ball is white? 3 5

Ans. The probability of 1 ball is red =

8

C1

=

5 8

C1

=

3 (with replacement) 8

=

3 (without replacement) 7

C1 3

The probability of 1 ball is white =

=

8

C1

3

C1

7

C1

So, on two independent draws, the probability of 1 ball is red and 1 ball is white will be

(e) Ans. 11. (a) Ans. (b)

5 3 15 5 3 15 ¥ = (for with replacement) or ¥ = (for without replacement). 8 8 64 8 7 56 What do you mean by macro-canonical and micro-canonical ensembles? 2 Refer to Chapter 8, Part 1; Short Questions with Answers, Question no. 4. If a system has two eigenstates y1 and y2 with eigenvalues E1 and E2, under what condition will linear combination (y = a y1 + v y2) be also an eigenstate? 2 Refer to Solved Question Paper (2010), Short Answer Questions (Group B), Question 1. If the wave function y(x) of quantum mechanical particle is given by Ê pxˆ y(x) = a sin Á ˜ for 0 £ x £ L Ë L ¯

Ans. (c) Ans. (d) Ans.

= 0, otherwise, then determine the value of a. Also determine the value of x where probability of finding the particle is maximum. 5 Refer to Solved Question Paper (2010), Short-Answer Questions (Group B), Question 2. Write down the Schrödinger equation for one-dimensional motion of a free particle in a onedimensional potential box. Find its eigenfunction and eigenenergy. Refer to Section 7.8.1(i). Prove that the first excited energy state of a free particle in a cubical box has three fold degeneracy. 3 Refer to Section 7.8.1(ii).

Solved WBUT Question Paper (Dec. 2013)

GROUP A Multiple-Choice Questions 1.

Choose the correct alternatives for any ten of the following questions: (i) Two vector A and B are parallel when (a) A × B = 0

(b) A × B = 1

(c) A · B = 0

10 × 1 = 10

(d) A · B = 1

Ans. (a) (ii) The volume element in spherical polar coordinates is (a) r sin q dr d q df (b) r2 sin q dr d q df (c) sin q dr d q df (d) r2 sin2 q dr d q df Ans. (b) (iii) The Hamiltonian of a system is a function of (a) qj, pj, t (b) pj, p˙j, t (c) qj, q˙j, t

(d) qj, p˙j, t

Ans. (a) (iv) When a particle is placed on the surface of a sphere and moves on the surface, the constraint is (a) holonomic (b) non-holonomic (c) scleronomic (d) rheonomic Ans. (a) ∂D represents ∂t (b) Maxwell’s modification of Ampere’s Law (d) Coulomb’s law

(v) The equation — ¥ H = J c +

(a) Gauss’s law (c) Faraday’s law Ans. (b) (vi) Work done to displace a charge q from a point P1 having potential V1 to a point P2 having potential V2 is (a) q(V2 – V1) (b) q(V1V2) (c) q(V2 + V1) (d) q(V2/V1) Ans. (a) (vii) The 3-dimensional momentum operator is i (a) Pˆ = — (b) Pˆ = i — Ans. (a)

i (c) Pˆ = —

(d) P = (i —)-1

S6.2

Advanced Engineering Physics

(viii) In a plane, e.m. wave is (a) E ¥ B = 0

(b) E || B

(c) E ^ B

(d) k ¥ B = 0

Ans. (c) (ix) If a proton of charge +q is accelerated through the potential V then the kinetic energy is 1 2 qV (b) qV2 2 Ans. (d) (x) B – E statistics is applicable for (a) photons (b) electrons (a)

Ans. (a) (xi) The energy of a free electron is given by (a) E = 2k2/2m (b) E = k/m

1 qV 2

(c) qV

(d)

(c) protons

(d) positrons

(c) E = 2 2k2/m

(d) 0

(c) –2

(d) 1

Ans. (a) È ∂y ˘ (xii) The value of Í x, ˙ is Î ∂x ˚ (a) –1

(b) 0

Ans. (a) (xiii) Which of the following is a fermion? (a) Photon (b) Electron (c) Phonon (d) Alpha particle Ans. (b) (xiv) For T > 0 K, the probability of occupancy of an electron at Fermi energy level is (a) 1

(b) 0

(c)

1 2

(d) 2

Ans. (c) (xv) In a homogenous, isotropic conducting medium of permittivity e, permeability m and conductivity s the skin depth d is (a)

1 wsm

(b)

2 wsm

(c)

m ws

(d)

2m ws

Ans. (b)

GROUP B Short-Answer Questions Answer any three of the following. 2. (a) Define canonical and micro-canonical ensemble. Ans. Refer to Chapter 8, Review Exercises (Short Questions with Answers, Question 4).

3 × 5 = 15 2

S6.3

Solved WBUT Question Paper (Dec. 2013)

(b) There are three distinguishable particles, each of which can be in one of the energy states e, 2e, 3e, 4e, having total energy 6e. Find the possible number of distributions of all particles in the energy states. Find the number of microstates in each case. 3 Ans. Refer to Example 8.2. 3. (a) What are the constraints of motion? Give an example. 3 Ans. Refer to Section 6.1.2. (b) Give the expression for generalized momentum. 2 Ans. Refer to Section 6.1.8. 4. (a) Calculate the magnetic field intensity just outside and inside of a hollow cylinder of 4 cm radius carrying 50 A current. 2 Ans. The magnetic field intensity just outside the hollow cylinder is given by m I B= 0 2p R Here, m0 = 4p × 10–7 H/m, R = 4 cm = 4 × 10–2 m, I = 50 A 4p ¥ 10 -7 ¥ 50

= 25 ¥ 10 -5 Tesla 2p ¥ 4 ¥ 10 -2 The magnetic field intensity inside the hollow cylinder is zero. (b) If a charged particle of 0.4 C charge is moving with a velocity 4i – j – 2k ms–1 through an electric field E = 10i + 10k and magnetic field of induction B = 2i – 6j – k, then find the Lorentz force experienced by the particle. 3 Ans. Refer to Example 4.6. 5. Obtain the expression for stationary energy levels for a particle of mas m which is free to move in a region of zero potential between two rigid walls at x = 0 and x = L. Are the energy levels degenerate? Explain. Ans. Refer to Section 7.8.1(i). 6. What is meant by ignorable or cyclic coordinates? Explain with an example. What are the limitations of Newtonian mechanics? 1+2+2 \

B=

Ans. Refer to Solved Question Paper 2011 (Question 4a and for second part, Section 6.1.1.

GROUP C Long-Answer Questions Answer any three of the following. 7. (a) Ans. (b) Ans. (c) Ans.

3 × 15 = 45

State ‘D’ Alembert’s principle. Refer to Section 6.2.3. Derive the Lagrangian equation from D’Alembert’s principle for a conservative system. Refer to Section 6.2.4. What is a cyclic coordinate? Give an example. Same as Question no. (6).

S6.4

Advanced Engineering Physics

(d) Find out Hamilton equation of motion for a system comprising masses m1 and m2 connected by a massless string of length L through a frictionless pulley such that m1 > m2. 3+6+2+4 Ans. Refer to Solved Question Paper (Dec. 2012), Question 4(b). 8. (a) Write down the Schrödinger equation for one-dimensional motion of a free particle in a onedimensional potential box. Find its eigenfunction and eigenenergy. Ans. Refer to Section 7.8.1 (i). (b) Prove that the first excited state of a free particle in a cubical box has threefold degeneracy. Ans. Refer to Section 7.8.1 (ii). (c) If the wave function y(x) of a quantum mechanical particle is given by ’x y(x) = a sin for 0 £ x £ L L = 0 otherwise

Ans. 9. (a) Ans. (b) Ans. (c) Ans. (d) Ans. 10. (a) Ans. (b)

then find the value of x where the probability of finding the particle is maximum. (1+5)+4+5 Refer to Solved Question Paper (2011), Question 4(c). If a system has two eigenstates y1 and y2 with eigenvalues E1 and E2, under what condition, will a linear combination (y = ay1 + by2) be also an eigenstate? Refer to Solved Question Paper (2010), Short-Answer Question (1). What are bosons and ferinons? Give examples. Refer to Sections 8.9.1 and 8.10.1. Derive the expression of partition function for an ideal gas and find out the Helmholtz free energy and the entropy. Out of syllabus. There are two particles in three quantum state gi = 1, 2, 3. Distribute the particle according to MB, BE and FD statistics. 2+(1+2)+(3+2+2)+3 Refer to Example 8.1. Using Gauss’s divergence theorem and Stoke’s theorem, prove that div(curl A) = 0. Refer to Example 1.33. Verify Stokes’ theorem for F = x2i + xyj, and S, the area bounded by x = 0, y = 0, x = a, y = a, in the XY plane.

iˆ ∂ Ans. Here, — ¥ F = ∂x

ˆj ∂ ∂y

kˆ ∂ = y kˆ ∂z

x2

xy

0

Now

ˆ = Ú (— ¥ F ) ◊ nds s

aa

Ú Ú y dx dy = 00

a3 2

Again, from Fig. 1,

Ú F ◊ dr

Fig. 1

=

Ú F ◊ dr + Ú

AB

BC

F ◊ dr +

Ú

CD

F ◊ dr +

Ú F ◊ dr

DA

Solved WBUT Question Paper (Dec. 2013)

Ú

For the first part,

S6.5

a

F ◊ dr =

AB

3 ˆ = x 2 dx = a F ◊ idx Ú Ú 3 0 a

For the second part,

a3 Ú F ◊ dr = Ú F ◊ ˆjdy = Ú aydy = 2 BC 0

For the third part,

Ú

0

F ◊ dr =

CD

Ú F ◊ dr

For the fourth part,

total

3

0

=

DA

So,

0

ˆ = x 2 dx = - a Ú F ◊ idx Ú 3 a a

Ú 0 dy = 0 a

Ú F ◊ dr

=

a3 a3 a3 a3 + +0 = 3 2 3 2

So, Stokes’ theorem is verified. (c) Determine the constant a so that the vector V = ( x + 3 y)i + ( y - 2 z ) j + ( x + az ) k is solenoidal. Ans. Same as Example 1.26. (d) In a region of space, the electric field is given by E = 8i + 4 j + 3k . Calculate the electric flux through a surface area of 100 units in XY plane. Ans. Same as Example 2.14. (e) Show that j = 1/r is a solution of Laplace’s equation.

3+4+3+2+3

Ans. Refer to Example 1.40. 11. (a) Apply Biot–Savart’s law to calculate the field at any axial point due to a circular current-carrying conductor carrying 2A current, and calculate B at the centre of the circle with radius a = 10 cm. Ans. Refer to Section 4.8(ii). (b) Two concentric spheres of radii a and b are kept at potentials Va and Vb. If the intervening space is vacuum, then write the appropriate differential that the electrostatic potential satisfies. Solve this equation to find the potential at any point between the spheres and also calculate the total charge outside the surface. Ans. The electric potential satisfies the Laplace’s equation, —2V = 0 Since the variation of potential exists only along the radial direction then from Laplace’s equation, È ˘ ∂V ∂ 2 V 1 ∂ Ê 2 ∂V ˆ = 0 Here, = 2 = 0˙ r Í ˜ 2 ∂r Á Ë ∂r ¯ ∂q ∂f ÍÎ ˙˚ r By integration twice with respect to r, V= -

È where C and D are ˘ C +DÍ ˙ r Î integrating constants˚

S6.6

Advanced Engineering Physics

Now applying boundary conditions, i.e., r = a, V = Va r = b, V = Vb So, C Va = - + D a C Vb = - + D b Ê V - Va ˆ On solving, we get C = Á b ab Ë b - a ˜¯ and

D=

Vb b - Va a b-a

So,

V=

˘ 1 È (Va - Vb )ab + (Vb b - Va a )˙ b - a ÍÎ r ˚

which is the solution of Laplace’s equation and gives the potential at any point between the spheres. The capacitance of the spherical capacitor formed by two concentric spheres of radius a and b (b > a) is given by ab b-a Now the total charge outside the surface is C = 4p Œ0

Ê ab ˆ Q = CVb = 4p Œ0 Á V Ë b - a ˜¯ b (c) Find the potential at the centre of a square of 1 m side with charges q, 4q, –3q and –2q at its corner. Given q = 1 nC. Ans. Here, AB = BC = CD = DA = 1 m 1 Now AD = m 2 From fig. 2, So, the potential at O is 1 È q 4q 3q 2q ˘ + 4p Œ0 Í Ê 1 ˆ Ê 1 ˆ Ê 1 ˆ Ê 1 ˆ ˙ ÍÁ ˜ Á ˜ Á ˜ Á ˜˙ ÍÎ Ë 2 ¯ Ë 2 ¯ Ë 2 ¯ Ë 2 ¯ ˙˚ =0

V=

Fig. 2

So, the potential at centre O is zero. (d) Derive Coulomb’s law from Gauss’s law. 4+4+4+3 Ans. Refer to Section 2.12.2. 12. (a) Given that for a particle in a rigid box (spanning from x = 0 to x = a) the eigenfunctions are given by

S6.7

Solved WBUT Question Paper (Dec. 2013)

j(x) = A sin (2px/L)e–iEt/h for –L/2 £ x £ L/2 =0 elsewhere Write down the wave equation of motion of the particle. Also obtain the energy eigenvalue and the normalized wave function. Ans. Refer to Section 7.8.1 (i). (b) Find the expectation value of x for the waver function given by, j(x) = Ae–bx. Ans. The expectation value of x is given by +μ

=

Ú

+μ

j * ( x ) xj ( x ) dx =

-μ

Ú

A2 x e -2 bx dx

-μ +μ

= A2

Ú

xe -2bx dx

-μ

=0 (c) Considering an electron is confined between two rigid walls 0.30 nm apart, find the energy levels for the states n = 1, 2 and 3. Ans. Energy associated with an electron confined between two rigid walls of 0.30 nm is 2 2 E= n h 8mL2

For n = 1,

E1 =

h2 8mL2

=

(6.63 ¥ 10 -34 )2 8 ¥ 9.1 ¥ 10 -31 ¥ (0.3 ¥ 10 -9 )2

= 0.67 ¥ 10 -18 J =

0.67 ¥ 10 -18

For n = 2,

1.6 ¥ 10 -19 2 E2 = n E1 = 4 × 4.2 = 16.8 eV

For n = 3,

E3 = n2E1 = 9 × 4.2 = 37.8 eV

= 4.18 eV

4.2 eV

(d) Show that the eigenfunctions corresponding to two different eigenvalues are orthogonal. (2+3+2)+2+3+3 Ans. Let us consider two eigenvalues equation of the operator A such that Ayn(x) = anyn(x) Aym(x) = amym(x)

and

* Multiplying first equation from left by ym and integrating in the limit –μ to +μ, we get +μ

Ú y m Ay n dx *

-μ

= an Ú y m* y n dx

If operator A is hermitian, then +μ

Ú

-μ

+μ

A*y *m y n dx = an

Ú y my n dx *

-μ

S6.8

Advanced Engineering Physics +μ

or,

am

Ú

+μ

y m* y n dx = an

-μ

Ú y my n dx *

-μ +μ

or, (am - an ) Ú y m* y n dx = 0 -μ

Since the eigenvalues are different, i.e., am – an π 0, So +μ

Ú y my n dx = 0 *

-μ

Hence ym and yn are the orthogonal.

Solved WBUT Question Paper (2014)

GROUP A Multiple-Choice Questions 1.

Choose the correct alternatives for any ten of the following: (i) The angle between i and (2 iˆ + ˆj ) is? (a) cos–1 2/5

(b) cos -1 2 / 5

(c) cos -1 2 / 5

Ans. (b) (ii) A rigid body whose equation of constraint is given by r1 - rj (a) (b) (c) (d)

10 × 1 = 10

2

(d) cos -1 (2 / 5) = constant is governed by

holonomic, rheonomic, dissipative and bilateral constraints non-holonomic, conservative and schleronomic constraints holonomic, schleronomic and conservative constraints non-holonomic, rheonomic and unilateral constraints

Ans. (c) (iii) The electric flux through each face of the cube of 1 m side if a charge q coulomb is placed at its centre is (a) q/4e0 (b) 0 (c) e0/6q (d) q/6e0 Ans. (d) (iv) Displacement current arises due to (a) positive charges only (b) negative charges only (c) time varying electric field (d) time varying magnetic field Ans. (c) (v) The dimension of 1/m0e0 is? (a) L–2T–2 (b) L–2T2 (c) LT–1 (d) L2T–2 Ans. (d) (vi) The relation between electronic polarisability and atomic radius for monoatomic gases is (a) ae μ a3 (b) ae μ 1/a3 (c) ae μ a2 (d) ae μ a3/2 Ans. (a) (vii) The magnetic flux linked with a coil at any instant ‘t’ is given by ft = 5 t3 – 100t + 200, the emf induced in the coil at t = 2 seconds is

Advanced Engineering Physics

S7.2

(a) 20 V (b) 40 V (c) 100 V (d) 200 V Ans. (b) (viii) Depth of penetration of a wave in a lossy dielectric increases with increase in (a) conductivity (b) wavelength (c) permeability (d) permittivity Ans. (b) (ix) Which one of the following is a Fermion? (a) a particle (b) m meson (c) photon (d) g particle Ans. (b) (x) The constraint associated with the motion of a pendulum is (a) dissipative (b) holonomic (c) rheonomic (d) non-holonomic Ans. (b) (xi) If E1 be the energy of the ground state of a one-dimensional potential box length of I and E2 be the energy of the ground state when the length of the box is halved then (a) E2 = 2E1 (b) E2 = 2E1 (c) E2 = 4E1 (d) E2 = 3E1 Ans. (c) (xii) The pair of observable that cannot be measured simultaneously is (a) pˆ x , pˆ y (b) xˆ , pˆ y (c) yˆ , pˆ y (d) xˆ , yˆ Ans. (c) (xiii) Which of the following functions is acceptable as a wave function of a one-dimensional quantum mechanical system. A - x2 (a) y = (b) y = Ae (c) y = Ax 2 (d) y = A sec x x Ans. (b) È ˆ ˘ (xiv) The commutator bracket Í xˆ , ∂ ˙ is equal to Î ∂x ˚ (a) 1 (b) –1 (c) 2 (d) 0 Ans. (b) (xv) Average energy ·E Ò of electron in a metal at T = 0 K is (a) EF

(b) EF/2

(c)

Ans. (c)

3 E 5 F

(d)

3 EF 2

GROUP B Short-Answer Questions Answer any three of the following.

3 × 5 = 15 ˆ ˆ ˆ 2. (a) Show that the fluid motion given by vector V = ( y + z )i + ( z + x ) j + ( x + y)k , is solenoidal and irrotational.

Solved WBUT Question Paper (2014)

Ans.

S7.3

— ◊ V = — ◊ [( y + z ) iˆ + ( z + x ) ˆj + ( x + y) kˆ ] ∂ ˆ ∂ˆ Ê ∂ = Á iˆ + ˆj + k ˜ ◊ [( y + z ) iˆ + ( z + x ) ˆj + ( x + y) kˆ ] ∂y ∂z ¯ Ë ∂x =

∂ ∂ ∂ ( y + z ) + ( z + x ) + ( x + y) ∂x ∂y ∂z

=0 i.e., — ◊ V = 0 Hence, the fluid motion is solenoidal. ∂ ˆ ∂ˆ Ê ∂ Again — ¥ V = Á iˆ + ˆj + k ˜ ¥ [( y + z ) iˆ + ( z + x ) ˆj + ( x + y)kˆ ] ∂y ∂z ¯ Ë ∂x iˆ =

kˆ

ˆj

∂ ∂ ∂x ∂y y+z z+ x

∂ ∂z x+y

È∂ ˘ ∂ = iˆ Í ( x + y)˙ - ( z + x )] Î ∂y ˚ ∂z

ˆj È ∂ ( x + y) - ∂ ( y + z )˘ + kˆ È ∂ ( z + x ) - ∂ ( y + z )˘ Í ˙ Í ∂x ˙ ∂z ∂y Î ∂x ˚ Î ˚

= iˆ(1) - iˆ(1) - ˆj (1) + ˆj (1) + kˆ(1) - kˆ (1) =0 i.e., — ¥ V = 0 Hence, the fluid motion is also irrotational. (b) Using Gauss’ divergence theorem, prove surface S.

Ú r ◊ ds = 3V , where V is the volume bounded by the

3+2

s

Ans. The given equation is

Ú r.ds ÈÍ= ÚÚ r. ds˘˙ = 3V s

Î

˚

s

LHS =

ÚÚ r ◊ ds s

È using Gauss' divergence theorem ˘ ˙ = ÚÚÚ — ◊ r dV Í i.e. A ◊ ds = ÚÚÚ — ◊ A dV Í ˙ ÚÚ V ÍÎ ˙˚ s V =

Ê ∂

∂

V

=

∂ˆ

ÚÚÚ ÁË iˆ ∂x + ˆj ∂y + kˆ ∂z ˜¯ ◊ (ixˆ + ˆjy + kzˆ ) dV ÚÚÚ (1 + 1 + 1) dV V

Advanced Engineering Physics

S7.4

= 3ÚÚÚ dV = 3V = R.H.S. V

Hence,

ÚÚ r ◊ ds = 3V (Proved) S

3. (a) Derive Poisson’s equation and Laplace’s equation from fundamentals. Ans. The differential form of a Gauss’ law is given by r —◊ E = Œ0

(1)

where r is volume charge density. Again, the electric field ( E ) at any point is equal to the negative gradient of the potential V E = -—V

i.e.,

(2)

Replacing E in Eq. (1) from Eq. (2), we get r — ◊ ( -—V ) = Œ0 or,

2

— V = –r/Œ0

(3)

Equation (3) is Poisson’s equation. Now in a charge-free region r = 0. Hence, Poisson’s equation becomes —2V = 0

(4)

Equation (4) is known as Laplace’s equation and is valid only in the charge free region. (b) Show that the function V = 2x2 + 7y – 2z2 represents the potential function in a charge-free region. 3+2 Ans. The given function V is given by V = 2x2 + 7y – 2z2 \

Ê ∂2 ∂2 ∂2 ˆ —2V = Á 2 + 2 + 2 ˜ (2 x 2 + 7 y - 2 z 2 ) ∂y ∂z ¯ Ë ∂x

or,

—2V =

or,

—2V = 4 + 0 – 4

or,

—2V = 0

∂2 ∂x 2

(2 x 2 + 7 y - 2 z 2 ) +

∂2 ∂y 2

(2 x 2 + 7 y - 2 z 2 ) +

∂2 ∂z 2

(2 x 2 + 7 y - 2 z 2 )

(1)

Equation (1) is nothing but Laplace’s equation. Hence, the function V represents the electric potential function in free space, where charge density r = 0.

Solved WBUT Question Paper (2014)

S7.5

4. (a) Write Lorentz force equation and explain the symbols used. Ans. The Lorentz force is given by the following equation: Fe = Fe + Fm F e = qE + q(v ¥ B)

or,

where F e = qE is the electric force experienced by a moving charge q and F m = q(v ¥ B) is the magnetic force experienced by the same moving charge q. (b) Deduce an expression of the force experienced by a current element placed in a magnetic field. 2+3 Ans. Refer to Section 4.6. 5. (a) A particle is falling vertically under the influence of gravity. Obtain the Hamiltonian and the equation of motion corresponding to the Hamiltonian. Ans. Refer to Example 6.16. (b) What is meant by cyclic coordinate? Derive its significance.

3+2

Ans. A generalized co-ordinate (qk) which does not appear in the expression of the Lagrangian is called an ignorable or cyclic co-ordinate. For the answer of the rest part of the question, refer to the Section 6.2.6. 6. A particle of mass m is confined by infinite potentials within x = 0 to x = L. (a) Write the Schrödinger’s equation to describe the motion of the particle. Ans. The Schrödinger’s equation to describe the motion of a particle which is confined in a potential well where the potential V = 0 within the well and V = μ outside the walls of the well is given by d 2y dx

2

+

2 mE 2

y =0

where y is the wavefunction describing the particle and the walls of the well lie at x = 0 and x = L. (b) Solve the equation to find out the normalised eigen functions.

1+4

Ans. For the answer to this question, refer to Section 7.8.1.

GROUP C Long-Answer Questions Answer any three of the following.

3 × 15 = 45

7. (a) Prove that A ¥ ( B ¥ C ) = B( A ◊ C ) - C ( A ◊ B) Ans. Let the three vectors be

__

A = Ax + Ay + Az ,

__

B = Bx + By + Bz

Advanced Engineering Physics

S7.6 __

and

C = Cx + Cy + Cz __

Now,

__

__

__

|

A × (B × C) = A × Bx Cx

or,

__

__

__

__

__

__

__

__

__

By Cy

Bz Cz

|

[

A × (B × C) = (Ax + Ay + Az ) × (ByCz – BzCy) + (BzCx – BxCz) + (BxCy – ByCx)

or,

A × (B × C) =

or,

|

Ay Ax (ByCz – BzCy) (BzCx – BxCz)

Az (BxCy – ByCx)

]

|

A × (B × C) = (AyBxCy – AyByCx – AzBzCx + AzBxCz) + (AzByCz – AzBzCy – AxBxCy __ __ __

__ __ __

__ __ __

__ __ __

+ AxByCx) + (AxBzCx – AxByCz – AyByCz + AyBzCy)

Again, B(A ◊ C) – C (A ◊ B) = (Bx + By + Bz ) (AxCx + AyCy + AzCz) – (Cx + Cy + Cz ) (AxBx + AyBy + AzBz)

(1)

B (A ◊ C) – C (A ◊ B) = (Ay BxCy – Ay ByCx – Az BzCx + Az BxCz) + (Az ByCz – Az BzCy

or,

– Ax Bx Cy + Ax ByCx) + (Ax Bz Cx – Ax Bx Cz – Ay By Cz + Ay Bz Cy) Now, having compared Eqs (1) and (2), we can write __

__

__

__ __ __

__ __ __

A × (B × C) = B (A ◊ C) – C (A ◊ B) (Proved). (b) If A is a constant vector, show that —(r. A) = A , where r is the position vector. Ê ∂ ˆ ∂ ˆ ∂ˆ ˆ ˆ ˆ ) ◊ (iA ˆ ) ˆ + ˆjA + kA —(r ◊ A) = Á iˆ +j + k ˜ (ix + jy + kz 1 2 3 ∂y ∂z ¯ Ë ∂n

Ans.

∂ ˆ ∂ˆ Ê ∂ + k ˜ ( xA1 + yA2 + zA3 ) = Á iˆ + ˆj ∂y ∂z ¯ Ë ∂x ˆ [∵ A is independent of x, y and z ] ˆ + ˆjA + kA = iA 1 2 3 = A Hence, —(r ◊ A) = A (Pr oved) (c) Find — . F and — × F where F = — (x3 + y3 + z3 + 3xyz). Ans. Let f = (x3 + y3 + z3 + 3xyz) \

F = —f

(2)

Solved WBUT Question Paper (2014)

S7.7

∂f = 3x2 + 3yz, ∂f = 3 y 2 + 3 xz and ∂f = 3z 2 + 3 xy ∂x ∂y ∂z Now,

— ◊ F = — ◊ —f Ê ∂f ˆ ∂f ˆ ∂f ˆ = — ◊ Á iˆ +j +k ˜ ∂y ∂z ¯ Ë ∂x = — ◊ (iˆ )(2 x 2 + 3 yz ) + — ◊ ( ˆj )(2 y 2 + 3zx ) + — ◊ (kˆ )(2 z 2 + 3 xy) =

∂ ∂ ∂ (3 x 2 + 3 yz ) + (3 y 2 + 3zx ) + (3z 2 + 3 xy) ∂x ∂y ∂z

= 6x + 6y + 6z \ Again,

— ◊ F = 6(x + y + z) — ¥ F = — ¥ —f È ∂f ˆ ∂f ˆ ∂f ˘ = — ¥ Íiˆ +j +k ˙ ∂y ∂z ˚ Î ∂x = — ¥ [iˆ(2 x 2 + 3 yz ) + ˆj (2 y 2 + 3zx ) + kˆ(2 z 2 + 3 xy)] ∂ ˆ ∂ˆ ˆ 2 Ê ∂ = Á iˆ + ˆj + k ˜ [i (2 x + 3 yz ) + ˆj (2 y 2 + 3zx ) + kˆ(2 z 2 + 3 xy)] ∂y ∂z ¯ Ë ∂x iˆ ∂ ∂x

=

ˆj ∂ ∂y

kˆ ∂ ∂z

2 x 2 + 3 yz 2 y 2 + 3zx 2 z 2 + 3 xy ∂ È∂ ˘ ∂ È∂ ˘ 2 2 = iˆ Í (2 z + 3 xy) - (2 y + 3zx )˙ - ˆj Í (2 z 2 + 3 xy) - (2 x 2 + 3 yz )˙ ∂ y ∂ z ∂z Î ∂x ˚ Î ˚

= iˆ(3 x - 3 x ) - ˆj (3 y - 3 y) + kˆ(3z - 3z ) = iˆ(0) + ˆj (0) + kˆ(0) \

∂ È∂ ˘ + kˆ Í (2 y 2 + 3zx ) - (2 x 2 + 3 yz )˙ ∂y Î ∂x ˚

—¥F = 0

[Alternatively — ¥ F = — ¥ —f = 0 ∵ curl of gradien of a scalar is zero] 1 (d) The electrostatic potential at a point P( r ) is given by V (r ) = . Show that it satisfies Laplace’s 2 r equation — V( r ) = 0. 3+3+4+5 Ans. Laplace’s equation is given by —2V = 0 where V is the electric potential at a point in charge-free space.

(1)

Advanced Engineering Physics

S7.8

Here

V( r ) =

1 (Given) r

2 Ê 1ˆ —2V( r ) = — Á ˜ Ë r¯ 2 Ê 1ˆ But — Á ˜ = 0 [See Example 1.40] Ë r¯

\

\

—2V( r ) = 0

(2)

1 Equation (1) and (2) are identical. So, the given electrostatic potential V( r ) = satisfies r Laplace’s equation. 8 (a) Deduce the D’lembert’s principle from the principle of virtual work. Ans. Refer to Section 6.2.3. 1 1 (b) The Lagrangian of a system is given by L = a q 2 - b q 2 , where a and b are constants. Obtain 2 2 Lagrangian equation of motion and also find the Hamiltonian of the system. 1 2 1 2 aq - bq 2 2 The Lagrange’s equations are given by d ∂L ∂L = 0 for j = 1, 2, 3, ..., f dt ∂qi ∂q j

Ans. The Lagrangian is given by L =

(1)

(2)

In this case there is only one generalised variable (q), so there will be only one equation of motion. ∂L ∂L = a q and = -b q ∂q ∂q \ The Lagrange’s equation of motion is given by

or

d (a q ) + b q = 0 [by Eq. (2)] dt a q¨ + bq = 0

1 1 The hamiltonian H is given by the lagrangian is L = a q 2 - b q 2 2 2 ∂L \ pq = = aq ∂q \

q˙ =

pq

a \ the hamiltonian is given by H=

 pjqj - L j

or,

1 Ê1 ˆ H = pq q - Á a q 2 - b q 2 ˜ Ë2 ¯ 2

Solved WBUT Question Paper (2014)

S7.9

1 1 = a q2 - a q2 + b q2 2 2 1 1 H = a q2 + b q2 2 2

\

(c) Prove that for a conservative system, the Hamiltonian represents the total energy of the system. Ans. The hamiltonian H is defined as H =

 qj pj - L

(1)

j

For a conservative system, the potential energy (V) is a function of the generalised coordinates (qj) only. It is independent of generalized velocities (q·j). Hence, we can write ∂L ∂T ∂V ∂T ∂T ˘ È p= = = Í∵ p j = ∂q ˙ ∂q j ∂q j ∂q j ∂q j j ˚ Î Hence,

∂T

 p j =  ∂q j

j

or,

j

∂T

 p j q j =  ∂q j

j

qj

j

The hamiltonian can be expressed as ∂T H= Â q j - L [by Eq. (1)] j ∂q j or,

H=

j

or,

H=

∂ Ê 1 2ˆ ÁË mq j ˜¯ - L j 2

 q j ∂q

1 È 2˘ ÍÎ∵ T = 2 mq j ˙˚

 q j (mq j ) - L j

or,

1 H = 2Â ( mq 2j ) - L j 2

or, H = 2T – (T – V) or, H = T + V = E, the total energy Hence, for a conservative system, the hamiltonian represents the total energy of the system j, i.e., the hamiltonian is equal to the total energy. (d) Find out Hamilton’s equation of motion for a system comprising masses m1 and m2 (m1 > m2) connected by a mass less string of length ‘L’ through a frictionless pulley. 3+4+3+5 Ans. Refer to Solved WBUT Question Paper (2012); the answer of Question no. 1 under Group C. 9. (a) What are bosons and fermions? Give examples. Ans. Bosons: The particles which obey Bose-Einstein statistics are called bosons. Fermions: The particle’s which obey Fermi-Dirac statistics are called fermions. Example of bosons: Photons, phonons, a-particles (2He4), p-mesons, k-mesons, deuterons (1H2), 12 6C , etc. Example of fermions: Protons, neutrons, electrons, m-mesons, trisium (1H3), 2He3 etc.

Advanced Engineering Physics

S7.10

(b) Apply B-E statistics to a photon and deduce Planck’s law of spectral energy density of a blackbody radiation. Ans. Refer to Section 8.14. (c) Draw the Femi distribution function for (i) T = 0 K, (ii) T > 0 K. Ans. Refer to Section 8.13. (d) Find out the numbers of possible arrangement of three particles in two cells for (i) M-B statistics, (i) B-E statistics, and (iii) F-D statistics. Ans. In MB statistics the particles are distinguishable and any member of particles can be put in one cell. Let the particles be a, b and c. So, the distribution can be shown in the following table. cell 1

abc

ab

bc

ca

a

b

c

0

cell 2 0 c a b bc ca ab abc So, there may be eight possible arrangements (i.e., MB distributions). In BE statistics the particles are indistinguishable and any number of particles can be placed in a cell. Let the particles be a, a and a. So the distribution can be shown as follows: cell 1

aaa

aa

0

a

cell 2 0 a aa aaa Hence, there are 4 possible arrangements in BE distribution. In FD statistics also, the particles are indistinguishable and only one particle can be placed in one cell. Let the particles be a, a and a. Since, we have two cells available, so we cannot place all the three particles in cells simultaneously. Hence, using all the three particles no distribution is possible in FD statistics. (e) What is thermodynamic probability? 2+3+3+(2+2+2)+1 Ans. The thermodynamic probability is defined as the number of possible microstates corresponding to a given macrostate. It is usually a large quantity unlike the statistical probability which when normalized (as always done) is less than one. 10. (a) What are the basic postulates of quantum mechanics? Ans. Refer to Section 7.6 (b) Find expectation value of position for the wave function given by y(x) = ax for 0 £ x £ 1. Ans. The required expectation value is given by 1

= Ú y * ( x )( x )y ( x ) dx 0

where y(x) = ax 1

\

=

Úa

*

x( x ) ax dx

0

1

= (a* a )Ú x 3 dx 0

(1)

Solved WBUT Question Paper (2014)

S7.11

1

È x4 ˘ = |a| Í ˙ Î 4 ˚0 2

=

| a |2 4 (1 - 0 4 ) 4

| a |2 4 (c) Show that for a particle in a rigid box (spanning from x = 0 to x = a) the eigen function is given 2 np x by y ( x ) = . Also find the energy eigen values. sin a a Ans. Refer to Section 7.8.1. (one-dimensional case). Hence,

=

(d) Prove that the first excited state of a free particle in a cubical box has threefold degeneracy. Ans. The normalized wave function of a particle in a three dimensional cubic box of length L is given by y n( nx ny nz ) ( x, y, z ) =

Ê ny p y ˆ Ê n pzˆ Ê n pxˆ sin Á x ˜ sin Á sin Á z ˜ ˜ Ë L ¯ Ë L ¯ Ë L ¯ L 8

3

and the energy eigenvalue is given by Enx, ny, nz =

h2 2

(nx2 + ny2 + nz2 )

8mL As one can see from the expression of the wave function, the value of the wave function (yn) is dependent on the values of nx, ny and nz . And when nx = ny = nz = 1, the value of energy 3h 2 Enx, ny, nz = which is the minimum possible value of Enx ny nz. So, it is the ground state 8mL2 value. And for this set of value [i.e., (1, 1, 1)], the wave function has single value. Hence the ground state is non-degenerate. The next possible value of the energy Enx ny nz is obtained when (nx, ny, nz) has either of the three values given by (1, 1, 2) or (1, 2, 1) or (2, 1, 1). 6h 2 . For 8mL this value of energy, there are three values of the wave function yn(nx, ny, nz)(x, y, z). Hence, the first excited state is three fold degenerated.

For such a set of values, the value of the 1st excitation energy is E112 = E121 = E211 =

(e) Find the energy of the ground state of an electron moving in a cubical box having each side equal to 1 Å (Mass of electron = 9.1 × 10–31 kg, h = 6.63 × 10–34 Js). Ans. The energy of a particle moving in a cubical box is given by Enx ny nz =

h2 8mL2

(nx2 + ny2 + nz2 )

Given h = 6.63 × 10–34 Js, m = 9.1 × 10–31 kg and L = 1 Å = 10–10 m

Advanced Engineering Physics

S7.12

The ground-state energy is given by Eg = E111 =

(6.63 ¥ 10 -34 )2 8 ¥ 9.1 ¥ 10 -31 ¥ (10 -10 )2

(6.63)2 ¥ 10 -68

or

Eg =

or

Eg = 6.038 × 10–18 J

72.8 ¥ 10 -31 ¥ 10 -20

¥ (12 + 12 + 12 )

J

11. (a) State Ampere’s circuital law. Write down its differential form. Ans. Ampere’s circuital law states that the line integral of the magnetic field ( B ) around any closed path is given by m0 times the net current enclosed by the path. Mathematically,

Ú B ◊ dl = m0 I c

The symbols carry their usual meaning the differential form of Ampere’s law is given by — ¥ B = m0 J where J is current density. (b) Express Ampere’s circuital law in terms of magnetic vector potential. Ans. The ampere’s circuital law is given by

Ú B ◊ dl

= m0I

(1)

c

when Eq. (1) is expressed in differential form, we get — ¥ B = m0 J

(2)

The magnetic vector potential A is given by B = — × A

(3)

Now, by using Eq. (3), Ampere’s circuital law (i.e., Eq. (2)) can be written as — × ( — × A ) = m0 J or

— × — × A = m0 J

(4)

But for any vector A ; we have — × — × A = — (— · A ) – —2 A Hence, Eq. (4) can be written as — ( — · A ) – — 2 A = m0 J – — 2 A = m0 J

[∵ for steady current — · A = 0]

Hence, — 2 A = m0 J

(5)

or,

Equation (5) is the expression of Ampere’s circuital law in terms of the magnetic vector potential. ∂B (c) State Poynting theorem. Prove that — ¥ E = from Maxwell’s equation. ∂t

Solved WBUT Question Paper (2014)

S7.13

Ans. The Poynting theorem states that, the rate at which electromagnetic energy in a finite volume decreases with respect to time is equal to the rate of dissipation of energy in the form of Joule heat plus the rate at which energy flows out of the volume. Derivation of — ¥ E = -

∂B ∂t

Faraday’s law can be expressed as df e= dt

(1)

where f is is the flux linked with the closed coil at time t. If E be the field induced in space then the induced emf e around the closed path c is given by e=

Ú E ◊ dl

(2)

c

\ from equations (1) and (2), we get df Ú E ◊ dl = - dt

(3)

c

The total flux through the circuit is equal to the integral of normal component of flux density B over the surface bounded by the circuit, f=

ÚÚ B ◊ ds

(4)

s

From equation (3) and (4), we get

Ú E ◊ dl c

= - ÚÚ

∂B ◊ ds dt

(5)

Equation (5) is the integral form of Faraday’s law of electromagnetic induction. Here, we use dB ∂B instead of , because B is a function of both position and time. dt ∂t Using Stokes’s theorem (given below)

Ú E ◊ dl

=

ÚÚ (— ¥ E ) ◊ ds S

Equation (5), can be written as

ÚÚ (— ¥ E ) ◊ ds S

or

Ê

ÚÚ ÁË — ¥ E + S

= - ÚÚ S

∂B ˆ ˜ ◊ ds = 0 ∂t ¯

∂B ◊ ds ∂t (6)

Advanced Engineering Physics

S7.14

Equation (6) must hold for any arbitrary surface S \

—¥E +

∂B =0 ∂t

∂B ∂t Equation (7) in the required equation (i.e., Faraday’s equation of em induction). or

—¥E = -

(d) Give the physical significance of the Maxwell’s equation —. B = 0 .

(7)

(1+2)+4+(2+4)+2

Ans. The physical significance of — · B = 0. This equation is Gauss’ law in magnetostatics and states that there are no magnetic flux sources and magnetic flux lines always close upon themselves. Magnetic field is solenoidal means, it has no sources or sinks. The total magnetic flux through a closed surface is equal to zero, i.e., the magnetic flux entering into the volume is equal to the magnetic flux leaving the volume. The magnetic monopole does not exist, magnetic poles exist in pairs.