Introduction. Learning Objectives. On completion of this chapter you will be able to:

Similar documents
Final exam: Tuesday, May 11, 7:30-9:30am, Coates 143

DUAL NATURE OF RADIATION AND MATTER

UNIT-1 MODERN PHYSICS

Notes on wavefunctions II: momentum wavefunctions

Problem Set 4 Solutions

Problem Set 3: Solutions

Reminder: Exam 3 Friday, July 6. The Compton Effect. General Physics (PHY 2140) Lecture questions. Show your work for credit.

nucleus orbital electron wave 2/27/2008 Quantum ( F.Robilliard) 1

Test on Nuclear Physics

Assignment Solutions- Dual Nature. September 19

CHAPTER 4 QUANTUM PHYSICS

Preview from Notesale.co.uk Page 2 of 42

ATOMIC PHYSICS PREVIOUS EAMCET QUESTIONS ENGINEERING

7. QUANTUM THEORY OF THE ATOM

Physics Teach Yourself Series Topic 15: Wavelike nature of matter (Unit 4)

The Doppler Factor and Quantum Electrodynamics Basics in Laser-Driven Light Sailing

QUESTIONS ) Of the following the graph which represents the variation of Energy (E) of the photon with the wavelength (λ) is E E 1) 2) 3) 4)

Graviton Induced Nuclear Fission through Electromagnetic Wave Flux Phil Russell, * Jerry Montgomery

The derivative function

A Reconsideration of Matter Waves

2016 PRELIM 2 PAPER 2 MARK SCHEME

Dual Nature of matter and radiation: m v 1 c

Quantum Mechanics and Atomic Theory

Analysis: The speed of the proton is much less than light speed, so we can use the

2.2 WAVE AND PARTICLE DUALITY OF RADIATION

Quantum Numbers and Rules

38. Photons and Matter Waves

Extracting Atomic and Molecular Parameters From the de Broglie Bohr Model of the Atom

2.3 Product and Quotient Rules

Numerical evidence of ultrarefractive optics in photonic crystals

Solution for the Homework 4

SIMG Solution Set #5

M12/4/PHYSI/HPM/ENG/TZ1/XX. Physics Higher level Paper 1. Thursday 10 May 2012 (afternoon) 1 hour INSTRUCTIONS TO CANDIDATES

Mathematics 5 Worksheet 11 Geometry, Tangency, and the Derivative

Lecture XVII. Abstract We introduce the concept of directional derivative of a scalar function and discuss its relation with the gradient operator.

Tutorial 2 (Solution) 1. An electron is confined to a one-dimensional, infinitely deep potential energy well of width L = 100 pm.

6.4: THE WAVE BEHAVIOR OF MATTER

APPENDIXES. Let the following constants be established for those using the active Mathcad

Chemistry. Slide 1 / 63 Slide 2 / 63. Slide 4 / 63. Slide 3 / 63. Slide 6 / 63. Slide 5 / 63. Optional Review Light and Matter.

How to Find the Derivative of a Function: Calculus 1

Microstrip Antennas- Rectangular Patch

Lecture: Experimental Solid State Physics Today s Outline

Order of Accuracy. ũ h u Ch p, (1)

Problem Set 4: Whither, thou turbid wave SOLUTIONS

Practice Problem Solutions: Exam 1

Derivatives of Exponentials

Everything comes unglued

Consider a function f we ll specify which assumptions we need to make about it in a minute. Let us reformulate the integral. 1 f(x) dx.

Quantum Mechanics Chapter 1.5: An illustration using measurements of particle spin.

Solution. Solution. f (x) = (cos x)2 cos(2x) 2 sin(2x) 2 cos x ( sin x) (cos x) 4. f (π/4) = ( 2/2) ( 2/2) ( 2/2) ( 2/2) 4.

Logarithmic functions

5.74 Introductory Quantum Mechanics II

Why gravity is not an entropic force

Exam 1 Review Solutions

Phy 231 Sp 02 Homework #6 Page 1 of 4

MATH CALCULUS I 2.1: Derivatives and Rates of Change

The distance between City C and City A is just the magnitude of the vector, namely,

NUMERICAL DIFFERENTIATION. James T. Smith San Francisco State University. In calculus classes, you compute derivatives algebraically: for example,

λ φ φ = hc λ ev stop φ = λ φ and now ev stop λ ' = Physics 220 Homework #2 Spring 2016 Due Monday 4/11/16

Preface. Here are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed.

1. ATOMIC STRUCTURE. Specific Charge (e/m) c/g

LIMITATIONS OF EULER S METHOD FOR NUMERICAL INTEGRATION

Chapter 1 Functions and Graphs. Section 1.5 = = = 4. Check Point Exercises The slope of the line y = 3x+ 1 is 3.

A NEW INTERPRETATION OF PHOTON. Kunwar Jagdish Narain

Krazy Katt, the mechanical cat

Pre-lab Quiz/PHYS 224 Earth s Magnetic Field. Your name Lab section

1.4 The Compton Effect

10.1 VIBRATIONAL RELAXATION *

Section 3.1: Derivatives of Polynomials and Exponential Functions

Pre-Calculus Review Preemptive Strike

Chapter 4 The debroglie hypothesis

Cubic Functions: Local Analysis

2.8 The Derivative as a Function

Math 34A Practice Final Solutions Fall 2007

Physics 111 Homework Solutions Week #9 - Friday

1. State whether the function is an exponential growth or exponential decay, and describe its end behaviour using limits.

Robotic manipulation project

HOW TO DEAL WITH FFT SAMPLING INFLUENCES ON ADEV CALCULATIONS

SECTION 1.10: DIFFERENCE QUOTIENTS LEARNING OBJECTIVES

158 Calculus and Structures

Last lecture (#4): J vortex. J tr

DELFT UNIVERSITY OF TECHNOLOGY Faculty of Electrical Engineering, Mathematics and Computer Science

Possibility of High Performance Quantum Computation by using Evanescent Photons in Living Systems

3. Using your answers to the two previous questions, evaluate the Mratio

The Electron in a Potential

3.1 Extreme Values of a Function

Polynomial Interpolation

2.11 That s So Derivative

INTRODUCTION AND MATHEMATICAL CONCEPTS

Phase space in classical physics

Lecture 21. Numerical differentiation. f ( x+h) f ( x) h h

1240 ev nm nm. f < f 0 (5)

2.1 THE DEFINITION OF DERIVATIVE

Homework 1. L φ = mωr 2 = mυr, (1)

DO NOT OPEN THIS EXAM UNTIL TOLD TO DO SO.

Reflection of electromagnetic waves from magnetic having the ferromagnetic spiral

Physically Based Modeling: Principles and Practice Implicit Methods for Differential Equations

Derivation Of The Schwarzschild Radius Without General Relativity

Chapter 2 Problem Solutions

Chapters 19 & 20 Heat and the First Law of Thermodynamics

Transcription:

Introduction Learning Objectives On completion of tis capter you will be able to: 1. Define Compton Effect. 2. Derive te sift in incident ligt wavelengt and Compton wavelengt. 3. Explain ow te Compton wavelengt gets varied wit respect to various incident angles. 4. Discuss te experimental verification of Compton scattering. Material prepared by: < Pysics Faculty > Topic No: < 2 > Page 1 of 7

COMPTON EFFECT:- Compton (1923) measured intensity of scattered X-rays from solid target, as function of wavelengt for different angles. He got Nobel prize for tat in 1927. Wen a beam of monocromatic radiation suc as X-rays, γ-ray, etc., of ig frequency is allowed to fall on a fine scatterer, te beam is scattered into two components namely one component aving te same frequency or wavelengt as tat of te incident radiation, so called unmodified radiation and te oter component aving lower frequency or iger wavelengt compared to incident radiation, so called modified radiation. te energy Let us consider te collision wic occurs between te poton (aving initially c ν = ) and electron. Te electron is free and is at rest before collision wit te incident poton. During collision, a part of energy is given to te electron, wic in turn increases te K.E. of te electron and ence it recoils at an angle of Φ as sown in figure. Te scattered poton moves wit an energy te original direction. c ν = ( < ν ) at an angle θ wit respect to Material prepared by: < Pysics Faculty > Topic No: < 2 > Page 2 of 7

Obviously, one can use law of conservation of energy and momentum for suc a system. For tat, we ave to know te momentum and energy values before and after collision. Energy before collision Energy after collision c + mc c 2 2 e ' + were m is te mass of te electron at rest, m e is te mass of te scattering electron moving wit velocity v and c is te velocity of ligt. According to law of conservation of energy, Along x-axis, Momentum before collision c c + = ' + 2 2 e (1) Momentum after collision cos θ + m e v cos φ ' According to law of conservation of momentum, cos m e v cos = ' θ + φ (2) Similarly, on can write for y-axis, = sinθ mev sinφ (3) ' By solving te above tree equations*, one can get ' = (1 cos θ ) wic is te cange (or sift) in wavelengt. It is clear tat te wavelengt sift is independent of te wavelengt of te incident radiation as well as te nature of te scatterer and it is found to depend on te angle of scattering θ. * Te detailed derivation will be given at te class room session. Material prepared by: < Pysics Faculty > Topic No: < 2 > Page 3 of 7

Case (i): Case (ii): Case (iii): Wen θ = o, ' = no scattering along te direction of incidence Wen θ = 9, ' = 6.6 *1 =.2445 A 9 *1 * 3 *1 34 =. 31 8 Tis difference in wavelengt is known as Compton wavelengt wen θ = 9. Wen θ = 18, 2 ' = =.4849 A. Hence, as θ varies from to 18, te wavelengt of te scattered radiation increases from to 2 +. Experimental verification: Consider te experimental set up sown in te following figure. Te monocromatic X-rays from a molybdenum target of an X-rays tube are made to fall on a carbon scatterer (grapite) and te corresponding values are calculated using Bragg s law. Te scattered X-ray is observed at te ionization camber or X-ray detector to to measure its intensity. Compton found tat te spectrum recorded after scattering ad K α line (corresponding to ) wic is on te longer wavelengt side of primary K α line (corresponding to ) as sown in figure. He also observed tat te cange in wavelengt increased rapidly as te scattering angle θ increased; te cange in wavelengt was independent of. Material prepared by: < Pysics Faculty > Topic No: < 2 > Page 4 of 7

X-ray detector X-ray tube target (ligt atoms, e.g. grapite) e - A final word. We ve seen tat ligt can be described as a stream of particles called potons, but we don t ave to take tis picture too literally. Instead can tink of te situation as te electromagnetic wave excanging energy and momentum wit a carged particle in quantised amounts Material prepared by: < Pysics Faculty > Topic No: < 2 > Page 5 of 7

Ceck Your Understanding (CYU) * 1. Define Compton effect. 2. A poton of energy E strikes a free electron, wit te scattered poton of energy E moving in te direction opposite tat of te incident poton. In tis Compton effect interaction, te resulting kinetic energy of te electron is (a) E, (b) E, (c) E E, (d) E + E, (e) none of te above. 3. An X-ray poton wit =6pm is scattered over 15 by a target electron. (i) Find te cange of its wavelengt. (ii) Find te angle between te directions of motion of te recoil electron and te incident poton. (iii) Find te energy of te recoil electron. Summary On completion of tis capter you ave learned te following: (1) Te classical wave teory of ligt cannot explain te scattering of x-rays from electrons. (2) Compton effect: Te decrease in energy (increase in wavelengt) of an x-ray or gamma ray poton wen interacting wit matter. (3) Te sifted in wavelengt is caused by te scattering of x-rays from free electrons and is given by ' = (1 cos θ ). (4) For a large energy transfer (and tus a measurable wavelengt sift), te poton energy sould be muc iger tan tat in te visible range X-rays. * (Ceck te correct answers on next page) Material prepared by: < Pysics Faculty > Topic No: < 2 > Page 6 of 7

Activity On te basis of Compton Effect, it is apparent tat even toug ligt is a wave, it beaves like a particle. Can you guess? Is it possible to argue tis statement? If a wave beaves like a particle, in a reverse manner, is it possible for a particle (object) to exibit wave nature??? For example, as te bullet moving wit ig speed exibited wave nature or not? Just tink about it Suggested Reading 1. P.K. Palanisamy, Engineering Pysics, Scitec Publications Pvt Ltd, Cennai. 2. C. Santi et al., Engineering Pysics, Sonaversity, Salem. 3. M. Arumugam, Engineering Pysics, Anurada Agencies, Kumbakonam. (And some oter open resources from internet) Answers to CYU 1. Wen a beam of monocromatic radiation suc as X-rays, γ-ray, etc., of ig frequency is allowed to fall on a fine scatterer, te beam is scattered into two components namely one component aving te same frequency or wavelengt as tat of te incident radiation, so called unmodified radiation and te oter component aving lower frequency or iger wavelengt compared to incident radiation, so called modified radiation. 2. (c). Conservation of energy requires te kinetic energy given to te electron be equal to te difference between te energy of te incident poton and tat of te scattered poton. 3. (i) ' = 64.5 1 12 m (ii) ϕ = 14.4 (iii) 1.44keV Material prepared by: < Pysics Faculty > Topic No: < 2 > Page 7 of 7