Gr9277 Solutions

The Physics GRE Solution Guide

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GR9277 Test

http://groups.yahoo.com/group/physicsgre_v2

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November 3, 2009

Author: David S. Latchman

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David S. Latchman

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Preface

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David Latchman

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This solution guide initially started out on the Yahoo Groups web site and was pretty successful at the time. Unfortunately, the group was lost and with it, much of the the hard work that was put into it. This is my attempt to recreate the solution guide and make it more widely avaialble to everyone. If you see any errors, think certain things could be expressed more clearly, or would like to make suggestions, please feel free to do so.

Document Changes 05-11-2009

1. Added diagrams to GR0177 test questions 1-25

2. Revised solutions to GR0177 questions 1-25

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04-15-2009 First Version

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Preface Classical Mechanics

3 13

1.1

Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.2

Newton’s Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.3

Work & Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.4

Oscillatory Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.5

Rotational Motion about a Fixed Axis . . . . . . . . . . . . . . . . . . . . 20

1.6

Dynamics of Systems of Particles . . . . . . . . . . . . . . . . . . . . . . . 22

1.7

Central Forces and Celestial Mechanics . . . . . . . . . . . . . . . . . . . 22

1.8

Three Dimensional Particle Dynamics . . . . . . . . . . . . . . . . . . . . 24

1.9

Fluid Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

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Contents

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1.10 Non-inertial Reference Frames . . . . . . . . . . . . . . . . . . . . . . . . 25 1.11 Hamiltonian and Lagrangian Formalism . . . . . . . . . . . . . . . . . . . 25

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Electromagnetism

27

2.1

Electrostatics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.2

Currents and DC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.3

Magnetic Fields in Free Space . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.4

Lorentz Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.5

Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.6

Maxwell’s Equations and their Applications . . . . . . . . . . . . . . . . . 32

2.7

Electromagnetic Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.8

Contents AC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.9

Magnetic and Electric Fields in Matter . . . . . . . . . . . . . . . . . . . . 32

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2.10 Capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 2.11 Energy in a Capacitor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 2.12 Energy in an Electric Field . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 2.13 Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 2.14 Current Destiny . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 2.15 Current Density of Moving Charges . . . . . . . . . . . . . . . . . . . . . 33 2.16 Resistance and Ohm’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . 33

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2.17 Resistivity and Conductivity . . . . . . . . . . . . . . . . . . . . . . . . . . 34 2.18 Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 2.19 Kirchoff’s Loop Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 2.20 Kirchoff’s Junction Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 2.21 RC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

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2.22 Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 2.23 Speed of Propagation of a Light Wave . . . . . . . . . . . . . . . . . . . . 35 2.24 Relationship between E and B Fields . . . . . . . . . . . . . . . . . . . . . 35 2.25 Energy Density of an EM wave . . . . . . . . . . . . . . . . . . . . . . . . 36 2.26 Poynting’s Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 Optics & Wave Phonomena

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3.1

Wave Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.2

Superposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.3

Interference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.4

Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.5

Geometrical Optics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.6

Polarization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.7

Doppler Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.8

Snell’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

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Thermodynamics & Statistical Mechanics

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4.1

Laws of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

4.2

Thermodynamic Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

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Contents 7 4.3 Equations of State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 4.4

Ideal Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

4.5

Kinetic Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

4.6

Ensembles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

4.7

Statistical Concepts and Calculation of Thermodynamic Properties . . . 40

4.8

Thermal Expansion & Heat Transfer . . . . . . . . . . . . . . . . . . . . . 40

4.9

Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

4.10 Specific Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 4.11 Heat and Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

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4.12 First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . 40 4.13 Work done by Ideal Gas at Constant Temperature . . . . . . . . . . . . . 41 4.14 Heat Conduction Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 4.15 Ideal Gas Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 4.16 Stefan-Boltzmann’s FormulaStefan-Boltzmann’s Equation . . . . . . . . 42 . . . . . . . . . . . . . . . . . . . . . . . . . . 42

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4.17 RMS Speed of an Ideal Gas

4.18 Translational Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . 42 4.19 Internal Energy of a Monatomic gas . . . . . . . . . . . . . . . . . . . . . 42 4.20 Molar Specific Heat at Constant Volume . . . . . . . . . . . . . . . . . . . 43 4.21 Molar Specific Heat at Constant Pressure . . . . . . . . . . . . . . . . . . 43 4.22 Equipartition of Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 4.23 Adiabatic Expansion of an Ideal Gas . . . . . . . . . . . . . . . . . . . . . 45

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4.24 Second Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . 45 5

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Quantum Mechanics

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5.1

Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

5.2

Schrodinger Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 ¨

5.3

Spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

5.4

Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

5.5

Wave Funtion Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

5.6

Elementary Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . 53

Atomic Physics

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6.1

Properties of Electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

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David S. Latchman

6.3

Energy Quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

6.4

Atomic Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

6.5

Atomic Spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

6.6

Selection Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

6.7

Black Body Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

6.8

X-Rays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

6.9

Atoms in Electric and Magnetic Fields . . . . . . . . . . . . . . . . . . . . 59

Special Relativity

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7.1

Introductory Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

7.2

Time Dilation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

7.3

Length Contraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

7.4

Simultaneity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

7.5

Energy and Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

7.6

Four-Vectors and Lorentz Transformation . . . . . . . . . . . . . . . . . . 65

7.7

Velocity Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

7.8

Relativistic Doppler Formula . . . . . . . . . . . . . . . . . . . . . . . . . 66

7.9

Lorentz Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

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Contents Bohr Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

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7.10 Space-Time Interval . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 Laboratory Methods

69

8.1

Data and Error Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

8.2

Instrumentation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

8.3

Radiation Detection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

8.4

Counting Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

8.5

Interaction of Charged Particles with Matter . . . . . . . . . . . . . . . . 72

8.6

Lasers and Optical Interferometers . . . . . . . . . . . . . . . . . . . . . . 72

8.7

Dimensional Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

8.8

Fundamental Applications of Probability and Statistics . . . . . . . . . . 72

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GR9277 Exam Solutions 9.1

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Momentum Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

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Contents 9 9.2 Bragg Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 9.3

Characteristic X-Rays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

9.4

Gravitation I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

9.5

Gravitation II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

9.6

Block on top of Two Wedges . . . . . . . . . . . . . . . . . . . . . . . . . . 75

9.7

Coupled Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

9.8

Torque on a Cone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

9.9

Magnetic Field outside a Coaxial Cable . . . . . . . . . . . . . . . . . . . 77

9.10 Image Charges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

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9.11 Energy in a Capacitor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 9.12 Potential Across a Wedge Capacitor . . . . . . . . . . . . . . . . . . . . . 79 9.13 Magnetic Monopoles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 9.14 Stefan-Boltzmann’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . 80 9.15 Specific Heat at Constant Volume . . . . . . . . . . . . . . . . . . . . . . . 80

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9.16 Carnot Engines and Efficiencies . . . . . . . . . . . . . . . . . . . . . . . . 81 9.17 Lissajous Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 9.18 Terminating Resistor for a Coaxial Cable . . . . . . . . . . . . . . . . . . . 82 9.19 Mass of the Earth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 9.20 Slit Width and Diffraction Effects . . . . . . . . . . . . . . . . . . . . . . . 83 9.21 Thin Film Interference of a Soap Film . . . . . . . . . . . . . . . . . . . . 83 9.22 The Telescope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

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9.23 Fermi Temperature of Cu . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 9.24 Bonding in Argon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 9.25 Cosmic rays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 9.26 Radioactive Half-Life . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 9.27 The Wave Function and the Uncertainty Principle . . . . . . . . . . . . . 86 9.28 Probability of a Wave function . . . . . . . . . . . . . . . . . . . . . . . . . 86 9.29 Particle in a Potential Well . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 9.30 Ground state energy of the positronium atom . . . . . . . . . . . . . . . . 87 9.31 Quantum Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . 88 9.32 Electrical Circuits I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 9.33 Electrical Circuits II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 ©2009

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Contents 9.34 Waveguides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 9.35 Interference and the Diffraction Grating . . . . . . . . . . . . . . . . . . . 89 9.36 EM Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 9.37 Decay of the π0 particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 9.38 Relativistic Time Dilation and Multiple Frames . . . . . . . . . . . . . . . 90 9.39 The Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 9.40 Rolling Cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 9.41 Rotating Cylinder I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 9.42 Rotating Cylinder II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

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9.43 Lagrangian and Generalized Momentum . . . . . . . . . . . . . . . . . . 93 9.44 Lagrangian of a particle moving on a parabolic curve . . . . . . . . . . . 94 9.45 A Bouncing Ball . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 9.46 Phase Diagrams I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 9.47 Phase Diagrams II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

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9.48 Error Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 9.49 Detection of Muons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 9.50 Quantum Mechanical States . . . . . . . . . . . . . . . . . . . . . . . . . . 95 9.51 Particle in an Infinite Well . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 9.52 Particle in an Infinite Well II . . . . . . . . . . . . . . . . . . . . . . . . . . 96 9.53 Particle in an Infinite Well III . . . . . . . . . . . . . . . . . . . . . . . . . 96 9.54 Current Induced in a Loop II . . . . . . . . . . . . . . . . . . . . . . . . . 97

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9.55 Current induced in a loop II . . . . . . . . . . . . . . . . . . . . . . . . . . 97 9.56 Ground State of the Quantum Harmonic Oscillator . . . . . . . . . . . . 98 9.57 Induced EMF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 9.58 Electronic Configuration of the Neutral Na Atom . . . . . . . . . . . . . . 99 9.59 Spin of Helium Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 9.60 Cyclotron Frequency of an electron in metal . . . . . . . . . . . . . . . . . 99 9.61 Small Oscillations of Swinging Rods . . . . . . . . . . . . . . . . . . . . . 100 9.62 Work done by the isothermal expansion of a gas . . . . . . . . . . . . . . 101 9.63 Maximal Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 9.64 Gauss’ Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 9.65 Oscillations of a small electric charge . . . . . . . . . . . . . . . . . . . . . 102 David S. Latchman

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Contents 11 9.66 Work done in raising a chain against gravity . . . . . . . . . . . . . . . . 103 9.67 Law of Malus and Unpolarized Light . . . . . . . . . . . . . . . . . . . . 103 9.68 Telescopes and the Rayleigh Criterion . . . . . . . . . . . . . . . . . . . . 104 9.69 The Refractive Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 9.70 High Relativistic Energies . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 9.71 Thermal Systems I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 9.72 Thermal Systems II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 9.73 Thermal Systems III . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 9.74 Oscillating Hoops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

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9.75 Decay of the Uranium Nucleus . . . . . . . . . . . . . . . . . . . . . . . . 106 9.76 Quantum Angular Momentum and Electronic Configuration . . . . . . . 107 9.77 Intrinsic Magnetic Moment . . . . . . . . . . . . . . . . . . . . . . . . . . 108 9.78 Skaters and a Massless Rod . . . . . . . . . . . . . . . . . . . . . . . . . . 108 9.79 Phase and Group Velocities . . . . . . . . . . . . . . . . . . . . . . . . . . 109

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9.80 Bremsstrahlung Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 9.81 Resonant Circuit of a RLC Circuit . . . . . . . . . . . . . . . . . . . . . . . 110 9.82 Angular Speed of a Tapped Thin Plate . . . . . . . . . . . . . . . . . . . . 111 9.83 Suspended Charged Pith Balls . . . . . . . . . . . . . . . . . . . . . . . . . 111 9.84 Larmor Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 9.85 Relativistic Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 9.86 Voltage Decay and the Oscilloscope . . . . . . . . . . . . . . . . . . . . . 113

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9.87 Total Energy and Central Forces . . . . . . . . . . . . . . . . . . . . . . . . 113 9.88 Capacitors and Dielectrics . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 9.89 harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 9.90 Rotational Energy Levels of the Hydrogen Atom . . . . . . . . . . . . . . 115 9.91 The Weak Ineteraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 9.92 The Electric Motor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 9.93 Falling Mass connected by a string . . . . . . . . . . . . . . . . . . . . . . 116 9.94 Lorentz Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 9.95 Nuclear Scatering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 9.96 Michelson Interferometer and the Optical Path Length . . . . . . . . . . 117 9.97 Effective Mass of an electron . . . . . . . . . . . . . . . . . . . . . . . . . . 118 ©2009

David S. Latchman

12

Contents 9.98 Eigenvalues of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 9.99 First Order Correction Perturbation Theory . . . . . . . . . . . . . . . . . 119 9.100Levers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

A Constants & Important Equations

121

A.1 Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 A.2 Vector Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 A.3 Commutators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

D

RA

FT

A.4 Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

David S. Latchman

©2009

Chapter

1

1.1 1.1.1

FT

Classical Mechanics Kinematics Linear Motion

RA

Average Velocity

∆x x2 − x1 = ∆t t2 − t1

(1.1.1)

∆x dx = = v(t) ∆t→0 ∆t dt

(1.1.2)

v=

Instantaneous Velocity

v = lim

D

Kinematic Equations of Motion

The basic kinematic equations of motion under constant acceleration, a, are

1.1.2

v = v0 + at v2 = v20 + 2a (x − x0 ) 1 x − x0 = v0 t + at2 2 1 x − x0 = (v + v0 ) t 2

(1.1.3) (1.1.4) (1.1.5) (1.1.6)

Circular Motion

In the case of Uniform Circular Motion, for a particle to move in a circular path, a radial acceleration must be applied. This acceleration is known as the Centripetal

Classical Mechanics

14 Acceleration Centripetal Acceleration a=

v2 r

(1.1.7)

ω=

v r

(1.1.8)

Angular Velocity

FT

We can write eq. (1.1.7) in terms of ω

a = ω2 r Rotational Equations of Motion

(1.1.9)

The equations of motion under a constant angular acceleration, α, are

(1.1.10) (1.1.11) (1.1.12) (1.1.13)

Newton’s Laws

D

1.2

RA

ω = ω0 + αt ω + ω0 t θ= 2 1 θ = ω0 t + αt2 2 ω2 = ω20 + 2αθ

1.2.1

Newton’s Laws of Motion

First Law A body continues in its state of rest or of uniform motion unless acted upon by an external unbalanced force. Second Law The net force on a body is proportional to its rate of change of momentum. F=

dp = ma dt

(1.2.1)

Third Law When a particle A exerts a force on another particle B, B simultaneously exerts a force on A with the same magnitude in the opposite direction. FAB = −FBA David S. Latchman

(1.2.2) ©2009

Work & Energy

1.2.2

15

Momentum p = mv

1.2.3

Impulse w

∆p = J =

1.3.1

Fdt = Favg dt

Work & Energy Kinetic Energy

1 K ≡ mv2 2

1.3.2

(1.3.1)

The Work-Energy Theorem

The net Work done is given by

Wnet = K f − Ki

RA

1.3.3

(1.2.4)

FT

1.3

(1.2.3)

(1.3.2)

Work done under a constant Force

The work done by a force can be expressed as

W = F∆x

(1.3.3)

W = F · ∆r = F∆r cos θ

(1.3.4)

In three dimensions, this becomes

D

For a non-constant force, we have

1.3.4

W=

wx f

F(x)dx

(1.3.5)

xi

Potential Energy

The Potential Energy is dU(x) dx for conservative forces, the potential energy is wx U(x) = U0 − F(x0 )dx0 F(x) = −

(1.3.6)

(1.3.7)

x0

©2009

David S. Latchman

Classical Mechanics

16

1.3.5

Hooke’s Law F = −kx

(1.3.8)

where k is the spring constant.

1.3.6

Potential Energy of a Spring 1 U(x) = kx2 2

1.4.1

Oscillatory Motion

FT

1.4

(1.3.9)

Equation for Simple Harmonic Motion x(t) = A sin (ωt + δ)

(1.4.1)

1.4.2

RA

where the Amplitude, A, measures the displacement from equilibrium, the phase, δ, is the angle by which the motion is shifted from equilibrium at t = 0.

Period of Simple Harmonic Motion T=

(1.4.2)

Total Energy of an Oscillating System

D

1.4.3

2π ω

Given that

x = A sin (ωt + δ)

(1.4.3)

and that the Total Energy of a System is E = KE + PE

(1.4.4)

The Kinetic Energy is 1 KE = mv2 2 1 dx = m 2 dt 1 = mA2 ω2 cos2 (ωt + δ) 2 David S. Latchman

(1.4.5) ©2009

Oscillatory Motion The Potential Energy is

17

1 U = kx2 2 1 = kA2 sin2 (ωt + δ) 2 Adding eq. (1.4.5) and eq. (1.4.6) gives

(1.4.6)

1 E = kA2 2

1.4.4

(1.4.7)

Damped Harmonic Motion

dx (1.4.8) dt where b is the damping coefficient. The equation of motion for a damped oscillating system becomes dx d2 x − kx − b = m 2 (1.4.9) dt dt Solving eq. (1.4.9) goves x = Ae−αt sin (ω0 t + δ) (1.4.10)

RA

We find that

FT

Fd = −bv = −b

α=

b 2m r

k b2 − m 4m2

ω0 =

r

D

=

1.4.5

(1.4.11)

ω20 −

b2 4m2

q = ω20 − α2

(1.4.12)

1 E = K + V(x) = mv(x)2 + V(x) 2

(1.4.13)

Small Oscillations

The Energy of a system is

We can solve for v(x), r

2 (E − V(x)) (1.4.14) m where E ≥ V(x) Let the particle move in the potential valley, x1 ≤ x ≤ x2 , the potential can be approximated by the Taylor Expansion " # " 2 # dV(x) 1 2 d V(x) V(x) = V(xe ) + (x − xe ) + (x − xe ) + ··· (1.4.15) dx x=xe 2 dx2 x=xe v(x) =

©2009

David S. Latchman

18 Classical Mechanics 2 At the points of inflection, the derivative dV/dx is zero and d V/dx2 is positive. This means that the potential energy for small oscillations becomes 1 V(x) u V(xe ) + k(x − xe )2 2 where

"

d2 V(x) k≡ dx2

(1.4.16)

# ≥0

(1.4.17)

x=xe

As V(xe ) is constant, it has no consequences to physical motion and can be dropped. We see that eq. (1.4.16) is that of simple harmonic motion.

Coupled Harmonic Oscillators

FT

1.4.6

Consider the case of a simple pendulum of length, `, and the mass of the bob is m1 . For small displacements, the equation of motion is θ¨ + ω0 θ = 0

(1.4.18)

RA

We can express this in cartesian coordinates, x and y, where x = ` cos θ ≈ ` y = ` sin θ ≈ `θ

(1.4.19) (1.4.20)

y¨ + ω0 y = 0

(1.4.21)

eq. (1.4.18) becomes

This is the equivalent to the mass-spring system where the spring constant is mg `

(1.4.22)

D

k = mω20 =

This allows us to to create an equivalent three spring system to our coupled pendulum system. The equations of motion can be derived from the Lagrangian, where L=T−V  
2 1 2 1 2 1 1 2 1 2 = m y˙ 1 + m y˙ 2 − ky1 + κ y2 − y1 + ky2 2 2 2 2 2  1  
2  1  2 = m y˙1 + y˙2 2 − k y21 + y22 + κ y2 − y1 2 2

(1.4.23)

We can find the equations of motion of our system ! d ∂L ∂L = dt ∂ y˙ n ∂yn 1

(1.4.24)

Add figure with coupled pendulum-spring system

David S. Latchman

©2009

Oscillatory Motion The equations of motion are

19
m y¨ 1 = −ky1 + κ y2 − y1
m y¨ 2 = −ky2 + κ y2 − y1

(1.4.25) (1.4.26)

We assume solutions for the equations of motion to be of the form y1 = cos(ωt + δ1 ) y2 = B cos(ωt + δ2 ) y¨ 1 = −ωy1 y¨ 2 = −ωy2

(1.4.27)

Substituting the values for y¨ 1 and y¨ 2 into the equations of motion yields   k + κ − mω2 y1 − κy2 = 0   −κy1 + k + κ − mω2 y2 = 0

FT

We can get solutions from solving the determinant of the matrix
−κ k + κ − mω2
= 0 −κ k + κ − mω2 Solving the determinant gives  2   mω2 − 2mω2 (k + κ) + k2 + 2kκ = 0 This yields

(1.4.28) (1.4.29)

(1.4.30)

(1.4.31)

ω2 =

RA

 g  k   =   m ` ω2 =  (1.4.32)  g 2κ  k + 2κ  = +  m ` m We can now determine exactly how the masses move with each mode by substituting ω2 into the equations of motion. Where k We see that m

k + κ − mω2 = κ

(1.4.33)

D

Substituting this into the equation of motion yields y1 = y2

(1.4.34)

We see that the masses move in phase with each other. You will also notice the absense of the spring constant term, κ, for the connecting spring. As the masses are moving in step, the spring isn’t stretching or compressing and hence its absence in our result.

ω2 =

k+κ We see that m

k + κ − mω2 = −κ

(1.4.35)

Substituting this into the equation of motion yields y1 = −y2

(1.4.36)

Here the masses move out of phase with each other. In this case we see the presence of the spring constant, κ, which is expected as the spring playes a role. It is being stretched and compressed as our masses oscillate. ©2009

David S. Latchman

Classical Mechanics

20

1.4.7

Doppler Effect

The Doppler Effect is the shift in frequency and wavelength of waves that results from a source moving with respect to the medium, a receiver moving with respect to the medium or a moving medium. Moving Source If a source is moving towards an observer, then in one period, τ0 , it moves a distance of vs τ0 = vs / f0 . The wavelength is decreased by vs v − vs − f0 f0

(1.4.37)

  v v = f 0 λ0 v − vs

(1.4.38)

λ0 = λ − The frequency change is

FT

f0 =

Moving Observer As the observer moves, he will measure the same wavelength, λ, as if at rest but will see the wave crests pass by more quickly. The observer measures a modified wave speed. v0 = v + |vr | (1.4.39) The modified frequency becomes

  v0 vr = f0 1 + λ v

RA

f0 =

(1.4.40)

Moving Source and Moving Observer We can combine the above two equations v − vs f0 0 v = v − vr

λ0 =

(1.4.41) (1.4.42)

To give a modified frequency of

  v0 v − vr f = 0 = f0 λ v − vs

D

0

1.5

1.5.1

(1.4.43)

Rotational Motion about a Fixed Axis Moment of Inertia Z I=

1.5.2

R2 dm

(1.5.1)

Rotational Kinetic Energy 1 K = Iω2 2

David S. Latchman

(1.5.2) ©2009

Rotational Motion about a Fixed Axis

1.5.3

1.5.4

21

Parallel Axis Theorem I = Icm + Md2

(1.5.3)

τ=r×F τ = Iα

(1.5.4) (1.5.5)

Torque

1.5.5

FT

where α is the angular acceleration.

Angular Momentum

L = Iω

(1.5.6)

dL dt

(1.5.7)

RA

we can find the Torque

τ=

1.5.6

Kinetic Energy in Rolling

D

With respect to the point of contact, the motion of the wheel is a rotation about the point of contact. Thus 1 (1.5.8) K = Krot = Icontact ω2 2 Icontact can be found from the Parallel Axis Theorem. Icontact = Icm + MR2

(1.5.9)

Substitute eq. (1.5.8) and we have  1 Icm + MR2 ω2 2 1 1 = Icm ω2 + mv2 2 2

K=

(1.5.10)

The kinetic energy of an object rolling without slipping is the sum of hte kinetic energy of rotation about its center of mass and the kinetic energy of the linear motion of the object. ©2009

David S. Latchman

Classical Mechanics

22

1.6

Dynamics of Systems of Particles

1.6.1

Center of Mass of a System of Particles

Position Vector of a System of Particles R=

m1 r1 + m2 r2 + m3 r3 + · · · + mN rN M

(1.6.1)

Velocity Vector of a System of Particles dR dt m1 v1 + m2 v2 + m3 v3 + · · · + mN vN = M

FT

V=

(1.6.2)

Acceleration Vector of a System of Particles

dV dt m1 a1 + m2 a2 + m3 a3 + · · · + mN aN = M

1.7 1.7.1

RA

A=

(1.6.3)

Central Forces and Celestial Mechanics Newton’s Law of Universal Gravitation  GMm rˆ F=− r2

D



1.7.2

1.7.3

(1.7.1)

Potential Energy of a Gravitational Force U(r) = −

GMm r

(1.7.2)

Escape Speed and Orbits

The energy of an orbiting body is E=T+U GMm 1 = mv2 − 2 r David S. Latchman

(1.7.3) ©2009

Central Forces and Celestial Mechanics The escape speed becomes 1 GMm E = mv2esc − =0 2 RE

23 (1.7.4)

Solving for vesc we find r vesc =

1.7.4

2GM Re

(1.7.5)

Kepler’s Laws

First Law The orbit of every planet is an ellipse with the sun at a focus.

FT

Second Law A line joining a planet and the sun sweeps out equal areas during equal intervals of time. Third Law The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. T2 =C R3

(1.7.6)

RA

where C is a constant whose value is the same for all planets.

1.7.5

Types of Orbits

The Energy of an Orbiting Body is defined in eq. (1.7.3), we can classify orbits by their eccentricities.

D

Circular Orbit A circular orbit occurs when there is an eccentricity of 0 and the orbital energy is less than 0. Thus 1 2 GM v − =E<0 2 r

(1.7.7)

The Orbital Velocity is r v=

GM r

(1.7.8)

Elliptic Orbit An elliptic orbit occurs when the eccentricity is between 0 and 1 but the specific energy is negative, so the object remains bound. r v=

2 1 GM − r a 

 (1.7.9)

where a is the semi-major axis ©2009

David S. Latchman

24 Classical Mechanics Parabolic Orbit A Parabolic Orbit occurs when the eccentricity is equal to 1 and the orbital velocity is the escape velocity. This orbit is not bounded. Thus 1 2 GM v − =E=0 2 r

(1.7.10)

The Orbital Velocity is r v = vesc =

2GM r

(1.7.11)

Hyperbolic Orbit In the Hyperbolic Orbit, the eccentricity is greater than 1 with an orbital velocity in excess of the escape velocity. This orbit is also not bounded. r

1.7.6

GM a

FT

v∞ =

(1.7.12)

Derivation of Vis-viva Equation

The total energy of a satellite is

RA

1 GMm E = mv2 − 2 r

(1.7.13)

For an elliptical or circular orbit, the specific energy is E=−

GMm 2a

(1.7.14)

Equating we get

2 1 v = GM − r a 



(1.7.15)

D

2

1.8

Three Dimensional Particle Dynamics

1.9

Fluid Dynamics

When an object is fully or partially immersed, the buoyant force is equal to the weight of fluid displaced.

1.9.1

Equation of Continuity ρ1 v1 A1 = ρ2 v2 A2

David S. Latchman

(1.9.1) ©2009

Non-inertial Reference Frames

1.9.2

25

Bernoulli’s Equation 1 P + ρv2 + ρgh = a constant 2

(1.9.2)

1.10

Non-inertial Reference Frames

1.11

Hamiltonian and Lagrangian Formalism

1.11.1

Lagrange’s Function (L)

FT

L=T−V

(1.11.1)

where T is the Kinetic Energy and V is the Potential Energy in terms of Generalized Coordinates.

1.11.2

Equations of Motion(Euler-Lagrange Equation)

1.11.3

!

RA

d ∂L ∂L = ∂q dt ∂q˙

Hamiltonian

H =T+V ˙ = pq˙ − L(q, q)

D

where

©2009

(1.11.2)

∂H = q˙ ∂p ∂H ∂L =− ∂q ∂x = −p˙

(1.11.3)

(1.11.4)

(1.11.5)

David S. Latchman

Classical Mechanics

D

RA

FT

26

David S. Latchman

©2009

Chapter

2

2.1

Electrostatics

2.1.1

Coulomb’s Law

FT

Electromagnetism

RA

The force between two charged particles, q1 and q2 is defined by Coulomb’s Law. ! q1 q2 1 F12 = rˆ12 4π0 r212

(2.1.1)

where 0 is the permitivitty of free space, where

0 = 8.85 × 10−12 C2 N.m2

Electric Field of a point charge

D

2.1.2

(2.1.2)

The electric field is defined by mesuring the magnitide and direction of an electric force, F, acting on a test charge, q0 . F (2.1.3) E≡ q0 The Electric Field of a point charge, q is E=

1 q rˆ 4π0 r2

(2.1.4)

In the case of multiple point charges, qi , the electric field becomes n 1 X qi E(r) = rˆi 4π0 i=1 r2i

(2.1.5)

28 Electric Fields and Continuous Charge Distributions

Electromagnetism

If a source is distributed continuously along a region of space, eq. (2.1.5) becomes Z 1 1 E(r) = rˆdq (2.1.6) 4π0 r2 If the charge was distributed along a line with linear charge density, λ, λ=

dq dx

(2.1.7)

The Electric Field of a line charge becomes λ rˆdx r2

Z

(2.1.8)

FT

1 E(r) = 4π0

line

RA

In the case where the charge is distributed along a surface, the surface charge density is, σ dq Q σ= = (2.1.9) A dA The electric field along the surface becomes Z 1 σ E(r) = rˆdA (2.1.10) 4π0 r2 Surface

In the case where the charge is distributed throughout a volume, V, the volume charge density is dq Q = (2.1.11) ρ= V dV The Electric Field is Z ρ 1 E(r) = rˆdV (2.1.12) 4π0 r2

D

Volume

2.1.3

Gauss’ Law

The electric field through a surface is I Φ= surface S

I dΦ =

E · dA

(2.1.13)

surface S

The electric flux through a closed surface encloses a net charge. I Q E · dA = 0

(2.1.14)

where Q is the charge enclosed by our surface. David S. Latchman

©2009

Electrostatics

2.1.4

29

Equivalence of Coulomb’s Law and Gauss’ Law

The total flux through a sphere is I E · dA = E(4πr2 ) =

q 0

(2.1.15)

From the above, we see that the electric field is E=

2.1.5

q 4π0 r2

(2.1.16)

Electric Field due to a line of charge

FT

Consider an infinite rod of constant charge density, λ. The flux through a Gaussian cylinder enclosing the line of charge is Z Z Z Φ= E · dA + E · dA + E · dA (2.1.17) top surface

bottom surface

side surface

RA

At the top and bottom surfaces, the electric field is perpendicular to the area vector, so for the top and bottom surfaces, E · dA = 0 (2.1.18) At the side, the electric field is parallel to the area vector, thus E · dA = EdA

Thus the flux becomes,

Z

(2.1.19)

Z

Φ=

E · dA = E

dA

(2.1.20)

side sirface

D

The area in this case is the surface area of the side of the cylinder, 2πrh. Φ = 2πrhE

(2.1.21)

Applying Gauss’ Law, we see that Φ = q/0 . The electric field becomes

2.1.6

E=

λ 2π0 r

(2.1.22)

Electric Field in a Solid Non-Conducting Sphere

Within our non-conducting sphere or radius, R, we will assume that the total charge, Q is evenly distributed throughout the sphere’s volume. So the charge density of our sphere is Q Q ρ= = 4 (2.1.23) 3 V πR 3 ©2009

David S. Latchman

Electromagnetism

30 The Electric Field due to a charge Q is E=

Q 4π0 r2

(2.1.24)

As the charge is evenly distributed throughout the sphere’s volume we can say that the charge density is dq = ρdV (2.1.25)

2.1.7

Electric Potential Energy

FT

where dV = 4πr2 dr. We can use this to determine the field inside the sphere by summing the effect of infinitesimally thin spherical shells Z E Z r dq E= dE = 2 0 0 4πr Z r ρ = dr 0 0 Qr (2.1.26) = 4 3 π R 0 3

2.1.8

1 qq0 r 4π0

RA

U(r) =

(2.1.27)

Electric Potential of a Point Charge

The electrical potential is the potential energy per unit charge that is associated with a static electrical field. It can be expressed thus U(r) = qV(r)

(2.1.28)

1 q 4π0 r

(2.1.29)

D

And we can see that

V(r) =

A more proper definition that includes the electric field, E would be Z V(r) = − E · d`

(2.1.30)

C

where C is any path, starting at a chosen point of zero potential to our desired point. The difference between two potentials can be expressed such Z b Z a V(b) − V(a) = − E · d` + E · d` Z b =− E · d`

(2.1.31)

a

David S. Latchman

©2009

Electrostatics This can be further expressed

31

Z

b

V(b) − V(a) =

(∇V) · d`

(2.1.32)

a

And we can show that

2.1.9

E = −∇V

(2.1.33)

Electric Potential due to a line charge along axis

The charge density is

FT

Let us consider a rod of length, `, with linear charge density, λ. The Electrical Potential due to a continuous distribution is Z Z dq 1 V= dV = (2.1.34) 4π0 r dq = λdx

(2.1.35)

RA

Substituting this into the above equation, we get the electrical potential at some distance x along the rod’s axis, with the origin at the start of the rod. 1 dq 4π0 x 1 λdx = 4π0 x

dV =

This becomes

  λ x2 V= ln 4π0 x1

(2.1.36)

(2.1.37)

where x1 and x2 are the distances from O, the end of the rod.

D

Now consider that we are some distance, y, from the axis of the rod of length, `. We again look at eq. (2.1.34), where r is the distance of the point P from the rod’s axis. Z dq 1 V= 4π0 r Z ` 1 λdx = 4π0 0 x2 + y2
12    12 ` λ = ln x + x2 + y2 4π0 0    12  λ = ln ` + `2 + y2 − ln y 4π0 
1   ` + `2 + y2 2  λ  = ln  (2.1.38)  4π0  d ©2009

David S. Latchman

Electromagnetism

32

2.2

Currents and DC Circuits

2

2.3

Magnetic Fields in Free Space

3

Lorentz Force

4

5

2.6 6

Maxwell’s Equations and their Applications

Electromagnetic Waves

D

2.7

Induction

RA

2.5

FT

2.4

7

2.8

AC Circuits

8

2.9

Magnetic and Electric Fields in Matter

9 David S. Latchman

©2009

Capacitance

2.10

33

Capacitance Q = CV

2.11

(2.10.1)

Energy in a Capacitor Q2 2C CV 2 = 2 QV = 2

Energy in an Electric Field

U 0 E2 = volume 2

(2.12.1)

dQ dt

(2.13.1)

J · dA

(2.14.1)

u≡

2.13

Current

I≡

Current Destiny

D

2.14

2.15

Z I= A

Current Density of Moving Charges J=

2.16

I = ne qvd A

(2.15.1)

Resistance and Ohm’s Law R≡

©2009

(2.11.1)

RA

2.12

FT

U=

V I

(2.16.1) David S. Latchman

Electromagnetism

34

2.17

Resistivity and Conductivity L A

(2.17.1)

E = ρJ

(2.17.2)

J = σE

(2.17.3)

R=ρ

Power

FT

2.18

P = VI

Kirchoff’s Loop Rules

Write Here

2.20

Kirchoff’s Junction Rule

Write Here

RC Circuits

D

2.21

RA

2.19

(2.18.1)

E − IR −

2.22

Maxwell’s Equations

2.22.1

Integral Form

Q =0 C

(2.21.1)

Gauss’ Law for Electric Fields w closed surface

David S. Latchman

E · dA =

Q 0

(2.22.1)

©2009

Speed of Propagation of a Light Wave Gauss’ Law for Magnetic Fields

35 w

B · dA = 0

(2.22.2)

closed surface

Amp`ere’s Law

z

d w B · ds = µ0 I + µ0 0 dt

E · dA

(2.22.3)

surface

Faraday’s Law

z

d w E · ds = − dt

B · dA

(2.22.4)

2.22.2

Differential Form

Gauss’ Law for Electric Fields

FT

surface

ρ 0

(2.22.5)

∇·B=0

(2.22.6)

∇·E=

Gauss’ Law for Magnetism

RA

Amp`ere’s Law

∇ × B = µ0 J + µ0 0

Faraday’s Law

∇·E=−

∂B ∂t

(2.22.7)

(2.22.8)

Speed of Propagation of a Light Wave

D

2.23

∂E ∂t

c= √

1 µ0 0

In a material with dielectric constant, κ, √ c c κ = n

(2.23.1)

(2.23.2)

where n is the refractive index.

2.24

Relationship between E and B Fields E = cB E·B=0

©2009

(2.24.1) (2.24.2) David S. Latchman

Electromagnetism

36

2.25

Energy Density of an EM wave 1 B2 u= + 0 E2 2 µ0

2.26

! (2.25.1)

Poynting’s Vector 1 E×B µ0

(2.26.1)

D

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S=

David S. Latchman

©2009

Chapter

3

3.1

Wave Properties

1

2

3.3

Interference

D

3

Superposition

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3.2

3.4

Diffraction

4

3.5

Geometrical Optics

5

3.6 6

FT

Optics & Wave Phonomena

Polarization

Optics & Wave Phonomena

38

3.7

Doppler Effect

7

3.8

Snell’s Law

3.8.1

Snell’s Law n1 sin θ1 = n2 sin θ2

Critical Angle and Snell’s Law

FT

3.8.2

(3.8.1)

The critical angle, θc , for the boundary seperating two optical media is the smallest angle of incidence, in the medium of greater index, for which light is totally refelected. From eq. (3.8.1), θ1 = 90 and θ2 = θc and n2 > n1 .

(3.8.2)

D

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n1 sin 90 = n2 sinθc n1 sin θc = n2

David S. Latchman

©2009

Chapter

4

4.1

FT

Thermodynamics & Statistical Mechanics Laws of Thermodynamics

1

2

4.3

Equations of State

D

3

Thermodynamic Processes

RA

4.2

4.4

Ideal Gases

4

4.5

Kinetic Theory

5

4.6 6

Ensembles

Thermodynamics & Statistical Mechanics

40

4.7

Statistical Concepts and Calculation of Thermodynamic Properties

7

4.8

Thermal Expansion & Heat Transfer

4.9

FT

8

Heat Capacity

  Q = C T f − Ti

(4.9.1)

4.10

RA

where C is the Heat Capacity and T f and Ti are the final and initial temperatures respectively.

Specific Heat Capacity

  Q = cm T f − ti

(4.10.1)

D

where c is the specific heat capacity and m is the mass.

4.11

4.12

Heat and Work Z W=

Vf

PdV

(4.11.1)

Vi

First Law of Thermodynamics dEint = dQ − dW

(4.12.1)

where dEint is the internal energy of the system, dQ is the Energy added to the system and dW is the work done by the system. David S. Latchman

©2009

Work done by Ideal Gas at Constant Temperature

4.12.1

41

Special Cases to the First Law of Thermodynamics

Adiabatic Process During an adiabatic process, the system is insulated such that there is no heat transfer between the system and its environment. Thus dQ = 0, so ∆Eint = −W

(4.12.2)

If work is done on the system, negative W, then there is an increase in its internal energy. Conversely, if work is done by the system, positive W, there is a decrease in the internal energy of the system.

FT

Constant Volume (Isochoric) Process If the volume is held constant, then the system can do no work, δW = 0, thus ∆Eint = Q (4.12.3) If heat is added to the system, the temperature increases. Conversely, if heat is removed from the system the temperature decreases. Closed Cycle In this situation, after certain interchanges of heat and work, the system comes back to its initial state. So ∆Eint remains the same, thus ∆Q = ∆W

(4.12.4)

RA

The work done by the system is equal to the heat or energy put into it.

Free Expansion In this process, no work is done on or by the system. Thus ∆Q = ∆W = 0, ∆Eint = 0 (4.12.5)

4.13

Work done by Ideal Gas at Constant Temperature

D

Starting with eq. (4.11.1), we substitute the Ideal gas Law, eq. (4.15.1), to get

4.14

Vf

Z W = nRT

Vi

= nRT ln

dV V

Vf Vi

(4.13.1)

Heat Conduction Equation

The rate of heat transferred, H, is given by H=

Q TH − TC = kA t L

(4.14.1)

where k is the thermal conductivity. ©2009

David S. Latchman

Thermodynamics & Statistical Mechanics

42

4.15

Ideal Gas Law PV = nRT

(4.15.1)

where n = Number of moles P = Pressure V = Volume T = Temperature and R is the Universal Gas Constant, such that

We can rewrite the Ideal gas Law to say

FT

R ≈ 8.314 J/mol. K

PV = NkT

(4.15.2)

where k is the Boltzmann’s Constant, such that

4.16

R ≈ 1.381 × 10−23 J/K NA

RA

k=

Stefan-Boltzmann’s FormulaStefan-Boltzmann’s Equation P(T) = σT4

RMS Speed of an Ideal Gas

D

4.17

4.18

r vrms =

3RT M

(4.17.1)

Translational Kinetic Energy 3 K¯ = kT 2

4.19

(4.16.1)

(4.18.1)

Internal Energy of a Monatomic gas 3 Eint = nRT 2

David S. Latchman

(4.19.1) ©2009

Molar Specific Heat at Constant Volume

4.20

43

Molar Specific Heat at Constant Volume

Let us define, CV such that Q = nCV ∆T

(4.20.1)

Substituting into the First Law of Thermodynamics, we have ∆Eint + W = nCV ∆T

(4.20.2)

At constant volume, W = 0, and we get

Substituting eq. (4.19.1), we get

1 ∆Eint n ∆T

FT

CV =

4.21

RA

3 CV = R = 12.5 J/mol.K 2

(4.20.3)

(4.20.4)

Molar Specific Heat at Constant Pressure

Starting with

D

and

4.22

Q = nCp ∆T

(4.21.1)

∆Eint = Q − W ⇒ nCV ∆T = nCp ∆T + nR∆T ∴ CV = Cp − R

(4.21.2)

Equipartition of Energy ! f CV = R = 4.16 f J/mol.K 2

(4.22.1)

where f is the number of degrees of freedom. ©2009

David S. Latchman

0 2 3n − 5 3n − 6

3 R 2 5 R 2 3R 3R

CV

5 R 2 7 R 2 4R 4R

CP = CV + R

Predicted Molar Specific Heats 3 5 6 6

FT

RA

Degrees of Freedom 0 2 3 3

Translational Rotational Vibrational Total ( f ) 3 3 3 3

©2009

David S. Latchman

Molecule Monatomic Diatomic Polyatomic (Linear) Polyatomic (Non-Linear)

Table 4.22.1: Table of Molar Specific Heats

D

Thermodynamics & Statistical Mechanics 44

Adiabatic Expansion of an Ideal Gas

4.23

Adiabatic Expansion of an Ideal Gas

where γ = CCVP . We can also write

4.24

45

PV γ = a constant

(4.23.1)

TV γ−1 = a constant

(4.23.2)

Second Law of Thermodynamics

D

RA

FT

Something.

©2009

David S. Latchman

Thermodynamics & Statistical Mechanics

D

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FT

46

David S. Latchman

©2009

Chapter

5

5.1

Fundamental Concepts

1

Schrodinger ¨ Equation

RA

5.2

FT

Quantum Mechanics

Let us define Ψ to be

Ψ = Ae−iω(t− v ) x

(5.2.1)

Simplifying in terms of Energy, E, and momentum, p, we get Ψ = Ae−

i(Et−px) ~

(5.2.2)

D

We obtain Schrodinger’s Equation from the Hamiltonian ¨ H =T+V

(5.2.3)

To determine E and p,

p2 ∂2 Ψ = − Ψ ~2 ∂x2 ∂Ψ iE = Ψ ~ ∂t

and H=

p2 +V 2m

(5.2.4) (5.2.5)

(5.2.6)

This becomes EΨ = HΨ

(5.2.7)

48

~ ∂Ψ ∂Ψ p2 Ψ = −~2 2 i ∂t ∂x The Time Dependent Schrodinger’s ¨ Equation is EΨ = −

i~

Quantum Mechanics

2

∂Ψ ~ 2 ∂2 Ψ + V(x)Ψ =− 2m ∂x2 ∂t

(5.2.8)

The Time Independent Schrodinger’s ¨ Equation is EΨ = −

(5.2.9)

Infinite Square Wells

FT

5.2.1

~ 2 ∂2 Ψ + V(x)Ψ 2m ∂x2

Let us consider a particle trapped in an infinite potential well of size a, such that ( 0 for 0 < x < a V(x) = ∞ for |x| > a,

RA

so that a nonvanishing force acts only at ±a/2. An energy, E, is assigned to the system such that the kinetic energy of the particle is E. Classically, any motion is forbidden outside of the well because the infinite value of V exceeds any possible choice of E. Recalling the Schrodinger Time Independent Equation, eq. (5.2.9), we substitute V(x) ¨ and in the region (−a/2, a/2), we get

This differential is of the form

D

where

~2 d2 ψ = Eψ − 2m dx2

(5.2.10)

d2 ψ + k2 ψ = 0 2 dx

(5.2.11)

r k=

2mE ~2

(5.2.12)

We recognize that possible solutions will be of the form cos kx

and sin kx

As the particle is confined in the region 0 < x < a, we say ( A cos kx + B sin kx for 0 < x < a ψ(x) = 0 for |x| > a We have known boundary conditions for our square well. ψ(0) = ψ(a) = 0 David S. Latchman

(5.2.13) ©2009

Schr¨odinger Equation It shows that

49 ⇒ A cos 0 + B sin 0 = 0 ∴A=0

(5.2.14)

We are now left with B sin ka = 0 ka = 0; π; 2π; 3π; · · · (5.2.15)

kn =

FT

While mathematically, n can be zero, that would mean there would be no wave function, so we ignore this result and say nπ a

for n = 1, 2, 3, · · ·

Substituting this result into eq. (5.2.12) gives nπ kn = = a

RA

Solving for En gives

√ 2mEn ~

n2 π2 ~2 2ma2 We cna now solve for B by normalizing the function Z a a |B|2 sin2 kxdx = |A|2 = 1 2 0 2 So |A|2 = a En =

(5.2.16)

(5.2.17)

(5.2.18)

D

So we can write the wave function as

5.2.2

r ψn (x) =

  2 nπx sin a a

(5.2.19)

Harmonic Oscillators

Classically, the harmonic oscillator has a potential energy of 1 V(x) = kx2 2

(5.2.20)

So the force experienced by this particle is F=− ©2009

dV = −kx dx

(5.2.21) David S. Latchman

50 Quantum Mechanics where k is the spring constant. The equation of motion can be summed us as d2 x m 2 = −kx dt

(5.2.22)

  x(t) = A cos ω0 t + φ

(5.2.23)

And the solution of this equation is

where the angular frequency, ω0 is r ω0 =

k m

(5.2.24)

With some manipulation, we get

FT

The Quantum Mechanical description on the harmonic oscillator is based on the eigenfunction solutions of the time-independent Schrodinger’s equation. By taking V(x) ¨ from eq. (5.2.20) we substitute into eq. (5.2.9) to get !   d2 ψ 2m k 2 mk 2 2E x − E x − ψ = ψ = dx2 ~2 2 ~2 k

RA

√ r  d2 ψ  mk 2 2E m   ψ =  x − √ 2 ~ ~ k  mk dx ~

This step allows us to to keep some of constants out of the way, thus giving us √ mk 2 x (5.2.25) ξ2 = ~r 2E m 2E and λ = = (5.2.26) ~ k ~ω0

D

This leads to the more compact

 d2 ψ  2 = ξ − λ ψ dξ2

(5.2.27)

where the eigenfunction ψ will be a function of ξ. λ assumes an eigenvalue anaglaous to E. From eq. (5.2.25), we see that the maximum value can be determined to be √ mk 2 2 ξmax = A ~ Using the classical connection between A and E, allows us to say √ mk 2E 2 ξmax = =λ ~ k

David S. Latchman

(5.2.28)

(5.2.29) ©2009

Schr¨odinger Equation 51 From eq. (5.2.27), we see that in a quantum mechanical oscillator, there are nonvanishing solutions in the forbidden regions, unlike in our classical case. A solution to eq. (5.2.27) is ψ(ξ) = e−ξ /2 2

(5.2.30)

where

and

dψ 2 = −ξe−ξ /2 dξ   2 dψ 2 −xi2 /2 −ξ2 /2 2 = ξ e − e = ξ − 1 e−ξ /2 2 dξ

This gives is a special solution for λ where

FT

λ0 = 1

(5.2.31)

Thus eq. (5.2.26) gives the energy eigenvalue to be E0 =

~ω0 ~ω0 λ0 = 2 2

(5.2.32)

The eigenfunction e−ξ /2 corresponds to a normalized stationary-state wave function 2

! 18

√

e−

mk x2 /2~ −iE0 t/~

RA

mk Ψ0 (x, t) = 2 2 π~

e

(5.2.33)

This solution of eq. (5.2.27) produces the smallest possibel result of λ and E. Hence, Ψ0 and E0 represents the ground state of the oscillator. and the quantity ~ω0 /2 is the zero-point energy of the system.

5.2.3

Finite Square Well

D

For the Finite Square Well, we have a potential region where ( −V0 for −a ≤ x ≤ a V(x) = 0 for |x| > a We have three regions

Region I: x < −a In this region, The potential, V = 0, so Schrodinger’s Equation be¨ comes ~2 d2 ψ = Eψ 2m dx2 d2 ψ ⇒ 2 = κ2 ψ √ dx −2mE κ= ~ −

where ©2009

David S. Latchman

Quantum Mechanics

52 This gives us solutions that are ψ(x) = A exp(−κx) + B exp(κx)

As x → ∞, the exp(−κx) term goes to ∞; it blows up and is not a physically realizable function. So we can drop it to get ψ(x) = Beκx

for x < −a

(5.2.34)

Region II: −a < x < a In this region, our potential is V(x) = V0 . Substitutin this into the Schrodinger’s Equation, eq. (5.2.9), gives ¨

FT

~2 d2 ψ − V0 ψ = Eψ − 2m dx2 d2 ψ or = −l2 ψ 2 dx p 2m (E + V0 ) where l ≡ ~

(5.2.35)

RA

We notice that E > −V0 , making l real and positive. Thus our general solution becomes ψ(x) = C sin(lx) + D cos(lx) for −a < x < a (5.2.36) Region III: x > a Again this Region is similar to Region III, where the potential, V = 0. This leaves us with the general solution ψ(x) = F exp(−κx) + G exp(κx)

As x → ∞, the second term goes to infinity and we get ψ(x) = Fe−κx

for x > a

(5.2.37)

for x < a for 0 < x < a for x > a

(5.2.38)

D

This gives us

5.2.4

 κx  Be    D cos(lx) ψ(x) =     Fe−κx

Hydrogenic Atoms

c

5.3

Spin

3 David S. Latchman

©2009

Angular Momentum

5.4

53

Angular Momentum

4

5.5

Wave Funtion Symmetry

5

Elementary Perturbation Theory

FT

5.6

D

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6

©2009

David S. Latchman

Quantum Mechanics

D

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FT

54

David S. Latchman

©2009

Chapter

6

6.1

FT

Atomic Physics Properties of Electrons

1

Bohr Model

RA

6.2

To understand the Bohr Model of the Hydrogen atom, we will take advantage of our knowlegde of the wavelike properties of matter. As we are building on a classical model of the atom with a modern concept of matter, our derivation is considered to be ‘semi-classical’. In this model we have an electron of mass, me , and charge, −e, orbiting a proton. The cetripetal force is equal to the Coulomb Force. Thus

D

1 e2 me v2 = 4π0 r2 r

(6.2.1)

The Total Energy is the sum of the potential and kinetic energies, so p2 E=K+U = − | f race2 4π0 r 2me

(6.2.2)

We can further reduce this equation by subsituting the value of momentum, which we find to be p2 1 e2 = me v2 = (6.2.3) 2me 2 8π0 r Substituting this into eq. (6.2.2), we get E=

e2 e2 e2 − =− 8π0 r 4π0 r 8π0 r

(6.2.4)

At this point our classical description must end. An accelerated charged particle, like one moving in circular motion, radiates energy. So our atome here will radiate energy

56 Atomic Physics and our electron will spiral into the nucleus and disappear. To solve this conundrum, Bohr made two assumptions. 1. The classical circular orbits are replaced by stationary states. These stationary states take discreet values. 2. The energy of these stationary states are determined by their angular momentum which must take on quantized values of ~. L = n~

(6.2.5)

We can find the angular momentum of a circular orbit.

FT

L = m3 vr

(6.2.6)

From eq. (6.2.1) we find v and by substitution, we find L. r

L=e Solving for r, gives

m3 r 4π0

(6.2.7)

L2 me e2 /4π0

(6.2.8)

n2 ~2 = n2 a0 me e2 /4π0

(6.2.9)

RA

r=

We apply the condition from eq. (6.2.5) rn =

where a0 is the Bohr radius.

a0 = 0.53 × 10−10 m

(6.2.10)

D

Having discreet values for the allowed radii means that we will also have discreet values for energy. Replacing our value of rn into eq. (6.2.4), we get ! me e2 13.6 = − 2 eV (6.2.11) En = − 2 2n 4π0 ~ n

6.3

Energy Quantization

3

6.4

Atomic Structure

4 David S. Latchman

©2009

Atomic Spectra

6.5 6.5.1

57

Atomic Spectra Rydberg’s Equation   1 1 1 = RH 02 − 2 λ n n

(6.5.1)

where RH is the Rydberg constant.

6.6

Selection Rules

6

6.7.1

Black Body Radiation

RA

6.7

FT

For the Balmer Series, n0 = 2, which determines the optical wavelengths. For n0 = 3, we get the infrared or Paschen series. The fundamental n0 = 1 series falls in the ultraviolet region and is known as the Lyman series.

Plank Formula

f3 8π~ u( f, T) = 3 h f /kT c e −1

Stefan-Boltzmann Formula

D

6.7.2

6.7.3

6.7.4

(6.7.1)

P(T) = σT4

(6.7.2)

Wein’s Displacement Law λmax T = 2.9 × 10−3 m.K

(6.7.3)

Classical and Quantum Aspects of the Plank Equation

Rayleigh’s Equation 8π f 2 u( f, T) = 3 kT c ©2009

(6.7.4) David S. Latchman

58 Atomic Physics We can get this equation from Plank’s Equation, eq. (6.7.1). This equation is a classical one and does not contain Plank’s constant in it. For this case we will look at the situation where h f < kT. In this case, we make the approximation ex ' 1 + x

(6.7.5)

Thus the demonimator in eq. (6.7.1) becomes eh f /kT − 1 ' 1 +

hf hf −1= kT kT

(6.7.6)

Thus eq. (6.7.1) takes the approximate form 8πh 3 kT 8π f 2 f = 3 kT c3 hf c

FT

u( f, T) '

(6.7.7)

As we can see this equation is devoid of Plank’s constant and thus independent of quantum effects. Quantum

RA

At large frequencies, where h f > kT, quantum effects become apparent. We can estimate that eh f /kT − 1 ' eh f /kT (6.7.8) Thus eq. (6.7.1) becomes

u( f, T) '

6.8.1

(6.7.9)

X-Rays

D

6.8

8πh 3 −h f /kT f e c3

Bragg Condition 2d sin θ = mλ

(6.8.1)

for constructive interference off parallel planes of a crystal with lattics spacing, d.

6.8.2

The Compton Effect

The Compton Effect deals with the scattering of monochromatic X-Rays by atomic targets and the observation that the wavelength of the scattered X-ray is greater than the incident radiation. The photon energy is given by E = hυ = David S. Latchman

hc λ

(6.8.2) ©2009

Atoms in Electric and Magnetic Fields The photon has an associated momentum

59

E

= pc E hυ h ⇒p = = = c c λ

(6.8.3) (6.8.4)

The Relativistic Energy for the electron is (6.8.5)

p − p0 = P

(6.8.6)

p2 − 2p · p0 + p02 = P2

(6.8.7)

where Squaring eq. (6.8.6) gives

Recall that E = pc and E 0 = cp0 , we have

FT

E2 = p2 c2 + m2e c4

c2 p2 − 2c2 p · p0 + c2 p02 = c2 P2 E 2 − 2E E 0 cos θ + E 02 = E2 − m2e c4 Conservation of Energy leads to

E + me c2 = E 0 + E

(6.8.9)

E − E 0 = E − me c2 E 2 − 2E E 0 + E 0 = E2 − 2Eme c2 + m2e c4 2E E 0 − 2E E 0 cos θ = 2Eme c2 − 2m2e c4

(6.8.10) (6.8.11)

RA

Solving

(6.8.8)

Solving leads to

D

∆λ = λ0 − λ =

where λc =

6.9 6.9.1

h me c

h (1 − cos θ) me c

(6.8.12)

is the Compton Wavelength. λc =

h = 2.427 × 10−12 m me c

(6.8.13)

Atoms in Electric and Magnetic Fields The Cyclotron Frequency

A test charge, q, with velocity v enters a uniform magnetic field, B. The force acting on the charge will be perpendicular to v such that FB = qv × B ©2009

(6.9.1) David S. Latchman

6.9.2

FT

60 Atomic Physics or more simply FB = qvB. As this traces a circular path, from Newton’s Second Law, we see that mv2 = qvB (6.9.2) FB = R Solving for R, we get mv R= (6.9.3) qB We also see that qB (6.9.4) f = 2πm The frequency is depends on the charge, q, the magnetic field strength, B and the mass of the charged particle, m.

Zeeman Effect

The Zeeman effect was the splitting of spectral lines in a static magnetic field. This is similar to the Stark Effect which was the splitting in the presence in a magnetic field.

RA

In the Zeeman experiment, a sodium flame was placed in a magnetic field and its spectrum observed. In the presence of the field, a spectral line of frequency, υ0 was split into three components, υ0 − δυ, υ0 and υ0 + δυ. A classical analysis of this effect allows for the identification of the basic parameters of the interacting system. The application of a constant magnetic field, B, allows for a direction in space in which the electron motion can be referred. The motion of an electron can be attributed to a simple harmonic motion under a binding force −kr, where the frequency is r k 1 (6.9.5) υ0 = 2π me

D

The magnetic field subjects the electron to an additional Lorentz Force, −ev × B. This produces two different values for the angular velocity. v = 2πrυ

The cetripetal force becomes me v2 = 4π2 υ2 rme r

Thus the certipetal force is 4π2 υ2 rme = 2πυreB + kr

for clockwise motion

4π2 υ2 rme = −2πυreB + kr

for counterclockwise motion

We use eq. (6.9.5), to emiminate k, to get eB υ − υ0 = 0 2πme eB υ2 + υ − υ0 = 0 2πme

υ2 −

David S. Latchman

(Clockwise) (Counterclockwise) ©2009

Atoms in Electric and Magnetic Fields 61 As we have assumed a small Lorentz force, we can say that the linear terms in υ are small comapred to υ0 . Solving the above quadratic equations leads to eB 4πme eB υ = υ0 − 4πme υ = υ0 +

for clockwise motion

(6.9.6)

for counterclockwise motion

(6.9.7)

We note that the frequency shift is of the form δυ =

eB 4πme

(6.9.8)

6.9.3

Franck-Hertz Experiment

FT

If we view the source along the direction of B, we will observe the light to have two polarizations, a closckwise circular polarization of υ0 + δυ and a counterclosckwise circular polarization of υ0 − δυ.

D

RA

The Franck-Hertz experiment, performed in 1914 by J. Franck and G. L. Hertz, measured the colisional excitation of atoms. Their experiement studied the current of electrons in a tub of mercury vapour which revealed an abrupt change in the current at certain critical values of the applied voltage.1 They interpreted this observation as evidence of a threshold for inelastic scattering in the colissions of electrons in mercury atoms.The bahavior of the current was an indication that electrons could lose a discreet amount of energy and excite mercury atoms in their passage through the mercury vapour. These observations constituted a direct and decisive confirmation of the existence os quantized energy levels in atoms.

1

Put drawing of Franck-Hertz Setup

©2009

David S. Latchman

Atomic Physics

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FT

62

David S. Latchman

©2009

Chapter

7

7.1 7.1.1

FT

Special Relativity Introductory Concepts

Postulates of Special Relativity

RA

1. The laws of Physics are the same in all inertial frames. 2. The speed of light is the same in all inertial frames. We can define

(7.1.1)

u2 c2

Time Dilation

D

7.2

1 γ= q 1−

∆t = γ∆t0

(7.2.1)

where ∆t0 is the time measured at rest relative to the observer, ∆t is the time measured in motion relative to the observer.

7.3

Length Contraction L=

L0 γ

(7.3.1)

where L0 is the length of an object observed at rest relative to the observer and L is the length of the object moving at a speed u relative to the observer.

Special Relativity

64

7.4

Simultaneity

4

7.5 7.5.1

Energy and Momentum Relativistic Momentum & Energy

FT

In relativistic mechanics, to be conserved, momentum and energy are defined as Relativistic Momentum

7.5.2

(7.5.1)

E = γmc2

(7.5.2)

RA

Relativistic Energy

p¯ = γmv¯

Lorentz Transformations (Momentum & Energy)

E = γ px − β c 0 py = py

D

p0x

7.5.3





= pz   E E =γ − βpx c c p0z 0

(7.5.3) (7.5.4) (7.5.5) (7.5.6)

Relativistic Kinetic Energy K = E − mc2    1 = mc2  q  1−
= mc2 γ − 1

David S. Latchman

(7.5.7)

v2 c2

   − 1 

(7.5.8) (7.5.9) ©2009

Four-Vectors and Lorentz Transformation

7.5.4

65

Relativistic Dynamics (Collisions) ∆E = γ ∆Px − β c 0 ∆P y = ∆P y 

∆P0x

 (7.5.10) (7.5.11)

∆P0z 0

= ∆Pz   ∆E ∆E =γ − β∆Px c c

(7.5.13)

Four-Vectors and Lorentz Transformation

FT

7.6

(7.5.12)

We can represent an event in S with the column matrix, s,    x   y   s =    z  ict

(7.6.1)

RA

A different Lorents frame, S0 , corresponds to another set of space time axes so that  0   x   y0    s0 =  0  (7.6.2)  z   0  ict

D

The Lorentz Transformation is related by the matrix  0   0 0 iγβ  x   γ  y0   0 1 0 0     0  =   z   0 0 1 0  0   ict −iγβ 0 0 γ

    x    y          z    ict

(7.6.3)

We can express the equation in the form s0 = L s

(7.6.4)

The matrix L contains all the information needed to relate position four–vectors for any given event as observed in the two Lorentz frames S and S0 . If we evaluate    x  h i  y  T 2 2 2 2 2   x y z ict s s= (7.6.5)  z  = x + y + z − c t   ict

Similarly we can show that s0T s0 = x02 + y02 + z02 − c2 t02 ©2009

(7.6.6) David S. Latchman

66 Special Relativity We can take any collection of four physical quantities to be four vector provided that they transform to another Lorentz frame. Thus we have    bx   b    b =  y   bz    ibt

(7.6.7)

this can be transformed into a set of quantities of b0 in another frame S0 such that it satisfies the transformation b0 = L b (7.6.8) Looking at the momentum-Energy four vector, we have       

FT

  px  p  p =  y  pz  iE/c

(7.6.9)

Applying the same transformation rule, we have p0 = L p

(7.6.10)

RA

We can also get a Lorentz-invariation relation between momentum and energy such that p0T p0 = pT p (7.6.11) The resulting equality gives

02 02 p02 x + p y + pz −

7.8

(7.6.12)

Velocity Addition

D

7.7

E02 E2 2 2 2 = p + p + p − x y z c2 c2

v0 =

v−u 1 − uv c2

(7.7.1)

Relativistic Doppler Formula r υ¯ = υ0

c+u c−u

r let r =

c−u c+u

(7.8.1)

We have υ¯ receding = rυ0 υ0 υ¯ approaching = r David S. Latchman

red-shift (Source Receding)

(7.8.2)

blue-shift (Source Approaching)

(7.8.3) ©2009

Lorentz Transformations

7.9

67

Lorentz Transformations

Given two reference frames S(x, y, z, t) and S0 (x0 , y0 , z0 , t0 ), where the S0 -frame is moving in the x-direction, we have,

7.10

Space-Time Interval

x = (x0 − ut0 ) y = y0 y0 = y   u 0 0 t = γ t + 2x c

FT

x0 = γ (x − ut) y0 = y z0 = y   u 0 t = γ t − 2x c


(∆S)2 = (∆x)2 + ∆y 2 + (∆z)2 − c2 (∆t)2

(7.9.1) (7.9.2) (7.9.3) (7.9.4)

(7.10.1)

Space-Time Intervals may be categorized into three types depending on their separation. They are

c2 ∆t2 > ∆r2

(7.10.2)

∆S2 > 0

(7.10.3)

RA

Time-like Interval

When two events are separated by a time-like interval, there is a cause-effect relationship between the two events. Light-like Interval

c2 ∆t2 = ∆r2

(7.10.4)

S =0

(7.10.5)

c2 ∆t2 < ∆r2 ∆S < 0

(7.10.6) (7.10.7)

D

2

Space-like Intervals

©2009

David S. Latchman

Special Relativity

D

RA

FT

68

David S. Latchman

©2009

Chapter

8

8.1.1

Data and Error Analysis Addition and Subtraction

x=a+b−c

(8.1.1)

(δx)2 = (δa)2 + (δb)2 + (δc)2

(8.1.2)

RA

8.1

FT

Laboratory Methods

The Error in x is

Multiplication and Division

D

8.1.2

x=

a×b c

(8.1.3)

The error in x is

8.1.3



δx x

2

δa = a 

2

δb + b

!2

δc + c 

2 (8.1.4)

Exponent - (No Error in b)

The Error in x is

x = ab

(8.1.5)

  δx δa =b x a

(8.1.6)

Laboratory Methods

70

8.1.4

Logarithms

Base e x = ln a

(8.1.7)

We find the error in x by taking the derivative on both sides, so d ln a · δa da 1 = · δa a δa = a

Base 10

FT

δx =

x = log10 a The Error in x can be derived as such

=

ln a ln 10

δa da 1 δa = ln 10 a δa = 0.434 a

(8.1.10)

Antilogs

D

8.1.5

(8.1.9)

d(log a) δa da

RA

δx =

(8.1.8)

Base e

x = ea

(8.1.11)

ln x = a ln e = a

(8.1.12)

We take the natural log on both sides.

Applaying the same general method, we see d ln x δx = δa dx δx ⇒ = δa x David S. Latchman

(8.1.13) ©2009

Instrumentation Base 10

71

x = 10a

(8.1.14)

We follow the same general procedure as above to get log x = a log 10 log x δx = δa dx 1 d ln a δx = δa ln 10 dx δx = ln 10δa x

FT

8.2

Instrumentation

3

8.4

RA

2

8.3

(8.1.15)

Radiation Detection

Counting Statistics

D

Let’s assume that for a particular experiment, we are making countung measurements for a radioactive source. In this experiment, we recored N counts in time T. The ¯ counting rate for this trial is R = N/T. This rate should be close to the average √ rate, R. The standard deviation or the uncertainty of our count is a simply called the N rule. So √ σ= N (8.4.1) Thus we can report our results as Number of counts = N ±

√ N

(8.4.2)

We can find the count rate by dividing by T, so √ N N R= ± T T ©2009

(8.4.3) David S. Latchman

72 The fractional uncertainty of our count is rate.

δN . N

δN T N T

δR = R

Laboratory Methods We can relate this in terms of the count

δN N √ N = N 1 = N =

(8.4.4)

We see that our uncertainty decreases as we take more counts, as to be expected.

Interaction of Charged Particles with Matter

FT

8.5 5

6

8.7

Lasers and Optical Interferometers

RA

8.6

Dimensional Analysis

D

Dimensional Analysis is used to understand physical situations involving a mis of different types of physical quantities. The dimensions of a physical quantity are associated with combinations of mass, length, time, electric charge, and temperature, represented by symbols M, L, T, Q, and θ, respectively, each raised to rational powers.

8.8

Fundamental Applications of Probability and Statistics

8

David S. Latchman

©2009

Chapter

9

9.1

Momentum Operator

We recall the momentum operator is

FT

GR9277 Exam Solutions

RA

~ p= ∇ i

(9.1.1)

So the momentum of the particle is

(9.1.2)

2d sin θ = mλ

(9.2.1)

D

~ pψ = ∇ψ i ~ ∂ = ψ i ∂x ~ = · ikei(kx−ωt) i = ~kψ

Amswer: (C)

9.2

Bragg Diffraction

We know Bragg’s Law to be

where d is the distance between the atomic layers and λ is the wavelength of the incident X-ray beam. We know that sin θ falls between the 0 and 1 and our longest wavelength will occur at our first order, m = 1. So 2d = λ Answer: (D)

(9.2.2)

GR9277 Exam Solutions

74

9.3

Characteristic X-Rays

Mosley1 showed that when the square root of an element’s characteristic X-rays are plotted agianst its atomic number we get a straight line. X-ray spectra is associated with atoms containg many electrons but in the X-ray regime, excitation removes tightly bound electrons from the inner orbit near the atom’s nucleus. As these emitted X-rays are the result of transitions of a single electron, Bohr’s Hydrogen model proves useful. Thus Mosley’s adaption of Bohr’s model becomes

FT

   1  1 1 = Z2 RH  2 − 2  eff λ n f ni

(9.3.1)

where RH is the Rydberg number and Zeff is the effective charge parameter that replaces the nuclear charge index, Z. The effective charge comes into play because the transition electron sees a nuclear charge that is smaller than Ze as the other electrons shield the nucleus from view. Thus our effective “shielding” constant, z f is Zeff = Z − z f

RA

For the Kα series, z f = 1, we get

2

EKα = hυKα = (Z − 1)

So



 3 1 1 − 2 13.6 eV = (Z − 1)2 13.6eV 2 1 2 4

(9.3.2)

(9.3.3)

D

(ZC − 1)2 EC = EMg Z − 12 Mg

1 4

=

(6 − 1)2

(12 − 1)2 25 1 = ≈ 121 5

(9.3.4)

is our closest answer.

Answer: (A) 1

Henry G. J. Moseley (1887-1915) was described by Rutherford as his most talented student. In his early 20’s, he measured and plotted the X-ray frequencies for 40 elements of the periodic table and showed that the K-alpha X-rays followed a straight line when the atomic number Z versus the square root of frequency was plotted. This allowed for the sorting of the elements in the periodic table by atomic number and not mass as was popular at the time. Moseley volunteered for combat duty with the Corps of Royal Engineers during World War I and was killed in action by a sniper at age 27 during the attack in the Battle of Gallipoli. It is widely speculated that because of his death, British and other world governments bagan a policy of no longer allowing scientists to enlist for combat.

David S. Latchman

©2009

Gravitation I

9.4

75

Gravitation I

The Force due to gravity, Newton’s Law of Gravitation, follows an inverse square law F= or rather F(R) ∝

1 R2

GMm r2

F(2R) ∝

(9.4.1)

1 2R

(9.4.2)

Dividing, we get

FT

F(R) (2R)2 = F(2R) R2 =4 Answer: (C)

Gravitation II

RA

9.5

(9.4.3)

Newton’s Law of Gravitation becomes a linear law inside a body. So F∝r

Thus

F(R) ∝ R

F(2R) ∝ 2R

(9.5.1)

(9.5.2)

D

Dividing the two equations gives us

F(R) R = R F(2R) 2 =2

(9.5.3)

Answer: (C)

9.6

Block on top of Two Wedges

The principles of equilibrium of forces tells us that 1. All forces are balanced 2. Total torque is zero ©2009

David S. Latchman

GR9277 Exam Solutions

76 We can determine the reaction on the two wedges to be 2R = 2mg + Mg Mg ⇒ R = mg + 2

(9.6.1)

As the block rests on the wedges, its weight causes it to push out on the two blocks. There are horizontal and vertical components to this force; the horizontal component being of interest to us FB cos θ = Mg

(9.6.2)

FT

where θ = 45◦ . Since the wedges aren’t moving, this is also equal to the frictional force. This force on the wedge has a horizontal component where Fx = F sin θ =

Mg 2

FR = µR We have, eq. (9.6.4) equal to eq. (9.6.3)

Answer: (D)

(9.6.4)

(9.6.5)

Coupled Pendulum

D

9.7

RA

Fx = FR " # Mg Mg =µ m+ 2 2 2µm ∴M= 1−µ

(9.6.3)

From what we know about coupled pendulums, there are two modes in which this system can oscillate. The first is when the two pendulum masses oscillate out of phase with each other. As they oscillate, there is a torsional effect on the tube and we expect to see the effects of its mass, M somewhere in the equation. The second occurs when the two masses swing in phase with each other. As they are in phase, there is no torsional effect on the connecting tube and the mode would be that of a single pendulum. So our modes of oscillation are r ω=0

;

ω=

g (M + 2m) `M

r and

ω=

g `

(9.7.1)

Answer: (A) David S. Latchman

©2009

Torque on a Cone

9.8

77

Torque on a Cone

The Torque is defined τ=r×F

(9.8.1)

We are looking for a negative kˆ vector so we are looking for answers with ˆi and ˆj cross products. With this in mind we can easily eliminate answers (A), (B) and (E). From the above, the torque is defined iˆ jˆ kˆ τ = rx r y rz (9.8.2) Fx F y Fz

FT

Choice C iˆ jˆ kˆ τ = −b 0 c 0 a 0 = −aciˆ − abkˆ

Choice D

RA

This is the answer

iˆ jˆ kˆ τ = b 0 c 0 a 0 = −acˆi + abkˆ

(9.8.3)

(9.8.4)

This is NOT the answer

D

Answer: (C)

9.9

Magnetic Field outside a Coaxial Cable

If we were to draw an Amperian loop around the outside of the cable, the enclosed current is zero. We recall Ampere’s Law, I B · ds = µ0 Ienclosed

(9.9.1)

As Ienclosed = 0, then the magnetic induction at P(r > c), is also zero2 . Answer: (A) 2

Think of the two resulting fields cancelling each other out.

©2009

David S. Latchman

GR9277 Exam Solutions

78

9.10

Image Charges

The grounded conducting plate will act as a ‘charge mirror’ to our two positive charges, producing two negative charges, −q at −0.5a and −2q at −1.5a. We can solve this using the vector form of Coulomb’s Law but it is simpler to realize that the forces on the q charge all act in the negative-x direction thusl allowing us to sum their magnitudes. Force between the −2q and q charges

Force between the −q and q charges

FT

F(−2q)(q)

−2q2 = 4π0 (2.0a)2

F(−q)(q) = Force between the +2q and q charges

−q2 4π0 (a)2

(9.10.2)

2q2 4π0 (a)2

(9.10.3)

RA

F(2q)(q) =

(9.10.1)

Adding eq. (9.10.1), eq. (9.10.2) and eq. (9.10.3) gives 1 7q2 F= 4π0 2a2

Answer: (E)

Energy in a Capacitor

D

9.11

(9.10.4)

The Energy stored in a Capacitor is 1 U = CV 2 2

(9.11.1)

The time for the potential difference across a capacitor to decrease is given by   t V = V0 exp − RC

(9.11.2)

The energy stored in the capacitor is half of its initial energy, this becomes 1 U = U0 2 David S. Latchman

(9.11.3) ©2009

Potential Across a Wedge Capacitor where U0 is the initial stored energy. We can find eq. (9.11.3) in terms of C and V, 1 1 CV 2 = CV02 2 2 V0 2 ⇒V = 2

79

(9.11.4)

Substituting eq. (9.11.2) into the above equation gives   V2 2t V02 exp − = 0 RC 2 Solving for t gives

Answer: (E)

9.12

RC ln 2 2

(9.11.5)

FT

t=

Potential Across a Wedge Capacitor

RA

We are told that the plates of this capacitor is large which allows us to assume that the field is uniform between the plates except maybe at the edges. As the capacitor is sufficiently large enough we can ignore the edge effects. At α the potential is V0 . As we have a linear relationship, the potential is proportional to the angle, ϕ. Thus V0 ϕ (9.12.1) V= α Answer: (B)

Magnetic Monopoles

D

9.13

The Maxwell Equations deal with electric and magnetic fields to the motion of electric charges and disallow for magnetic charges. If we were to allow for a ‘magnetic charge’ or magnetic monopole, we would also have to allow for a ‘magnetic current’. As we do have electrical charges and currents and equations describing them, we can observe how they differ of our magnetic equations to come up with an answer.

Equation I.: Amp`ere’s Law This relates the magnetic field to an electrical current and a changing electric field. Equation II: Faraday’s Law of Induction This equation is similar to Amp`ere’s Law except there is no ‘magnetic current’ component. As we have stated above, the presence of a magnetic charge will lead us to assume a magnetic current, this is one of the equations that would be INCORRECT. ©2009

David S. Latchman

80 GR9277 Exam Solutions Equation III.: Gauss’ Law Here we see that the distribution of an electric charge gives us an electric field. Equation IV.:Gauss’ Law for Magnetism This is similar to Gauss’ Law above except that we have no ‘magnetic charge’. If we did assume for monopoles, this equation would not be zero, so this equation would also be INCORRECT Answer: (D)

9.14

Stefan-Boltzmann’s Equation

FT

The Stefan-Boltzmann’s Equation says that the total energy emitted by a black body is proportional to the fourth power of its temperature, T. E = σT4

(9.14.1)

If we were to double the temperature of this blackbody, the energy emitted would be E1 = σ(2T)4 = 8σT4 = 8E

(9.14.2)

RA

Let C be the heat capacitance of a mass of water, the energy to change its temperature by a half degree is E = C∆T (9.14.3) where δT = 0.5 K. At T1 = 2T, the energy used is 8E, so E1 = 8E = 8C∆T

(9.14.4)

So the temperature change is eight degrees. Answer: (C)

Specific Heat at Constant Volume

D

9.15

To determine the specific heat at constant volume we identify the degrees of freedom or the ways the molecule can move; translational and rotational. The question also adds that we are looking at high temperatures, so we have to add another degree of freedom; vibrational. Translational = 3 Rotational = 2 Vibrational = 2 We recall the formula, eq. (4.22.1) we used to determine the specific heat per mole at constant volume, CV ! f R = 4.16 f J mol−1 K−1 (9.15.1) CV = 2 David S. Latchman

©2009

Carnot Engines and Efficiencies where f = 7 Thus, CV is

81 CV =

  7 R 2

(9.15.2)

Answer: (C)

9.16

Carnot Engines and Efficiencies

The efficiency of an engine is the ratio of the work we get out of it to the energy we put in. So Wout e= (9.16.1) Ein

FT

A Carnot Engine is theoretically the most efficient engine. Its efficiency is e=1−

Tc Th

(9.16.2)

where Tc and Th are the temperatures of our cold and hot reservoirs respectively. We must remember that this represents absolute temperature, so

RA

Tc = 527 + 273 = 800 K Th = 727 + 273 = 1000 K

Plugging this into eq. (9.16.2), we get

e=1−

800 = 0.2 1000

(9.16.3)

So the work our engine performs is

Wout = e × Ein = 0.2 × 2000 = 400 J

D

Answer: (A)

(9.16.4)

9.17

Lissajous Figures

This question deals with Lissajous figures. These can be drawn graphically with the use of your favorite programming language or with the use of an oscilloscope. Before the days where digital frequency meters were prevalent, this was a common method to determine the frequency of a signal. To generate these figures, one signal was applied across the horizontal deflection plates of an oscilloscope and the other applied across the vertical deflection plates. The resulting pattern traces a design that is the ratio of the two frequencies.3 3

If you’ve never seen this in the lab, one of the best examples where this can be seen is during the opening sequence of the 1950’s TV series, “The Outer Limits”. “We will control the horizontal. We will control the vertical. We can roll the image, make it flutter....”

©2009

David S. Latchman

82 GR9277 Exam Solutions With this basic understanding we can determine signals in the X and Y inputs to our oscilloscope. Each figure represents a trace over a full period. Answer: (A)

9.18

Terminating Resistor for a Coaxial Cable

Terminating a coaxial cable is important as it reduces refelctions and maximizes power transfer across a large bandwidth. The best way to think of this is to think of a wave propagating along a string, any imperfections will cause the wave’s energy to either be attenuated or reflected across this boundary.4

FT

Choice A The terminating resistor doesn’t prevent leakage, the outer core of the cable was designed to confine the signal. Your coaxial cable is a waveguide. Choice B The cable doesn’t transmit enough power to cause over heating along it’s length.

RA

Choice C This is correct. The resistor essentially attenuates the remaining power across itself, making it seem that the wave gets propagated across an infinite length i.e. no reflections. Choice D This won’t prevent attenuation across the cable. The cable has a natural impedance which will attenuate the signal to a degree. The terminating resistor absorbs the remaining power so no signal gets refelcted back5 . Choice E Improbable since the outer sheath’s purpose is to cancel out these currents. Answer: (C)

Mass of the Earth

D

9.19

There are many ways to tackle this question which depends on what you know. You may already know the mass of the earth for one, which may make things convenient and a time saver. The density of the earth is approximately that of Iron, if you knew that and the volume of the earth, you would get an answer where M = ρV = 7870 × 109 × 1021 = 8 × 1024 kg

(9.19.1)

4

You may be familiar or have seen the use of terminating resistors if you’ve dabbled in computers or electronics for the past couple of years. SCSI cables made use of terminating resistors as you daisy chained your drives across the cable. In the time when 10BASE2 ethernet networks were prevalent, the use of a 50 Ω BNC Terminator was of utmost importance or your computers would have lost connectivity. 5 Think Maximum Power Transfer

David S. Latchman

©2009

Slit Width and Diffraction Effects 83 Of course, if you want to apply some real Physics, we start with Newton’s Gravitation Equation GM (9.19.2) g= 2 RE Solving for M, we get M=

gR2E

G (9.8)(6.4 × 106 )2 = kg 6.67 × 10−11 1 + 12 + 11 = 24

(9.19.4)

FT

Adding the indices, gives

(9.19.3)

So our answer is of the magnitude 1024 kg. Answer: (A)

9.20

Slit Width and Diffraction Effects

RA

We are familar with the Double slit experiment and interference and can recall the equation where d sin θ = mλ (9.20.1) where d is the distance between the slit centers and m is the order maxima. As the width of the slit gets smaller, the wave gets more and more spread out due to diffraction effects. If it gets wide enough, it will spread out the widths of the interference pattern and eventually ‘erase’ it. We recall the equation for the single slit interference w sin θ = nλ (9.20.2)

D

For this ‘swamping out’ effect to occur, the mth and nth orders must align. The ratios between them will be mλ d sin θ = w sin θ nλ m d ∴ = >1 w n

(9.20.3)

As m and n are integers and m > n, answer (D) fits this criteria. Answer: (D)

9.21

Thin Film Interference of a Soap Film

Based on our knowledge of waves and interference we can eliminate choices ©2009

David S. Latchman

84 GR9277 Exam Solutions Choice I This is an impossibility. Light isn’t ‘absorbed’. Destructive interference can take place and energy is ‘redistributed’ to areas of constructive interferece. The soap film can’t absorb energy without it going somewhere. Choice II This is CORRECT. At the front of the soap film, there is a phase change of 180°as the soap film as a refractive index greater than air. The part that gets transmitted through this film gets reflected by the back part of the film with no phase change as air has a lower refractive index. This means that the two waves are out of phase with each other and interfere destructively. Choice III Yes, this is true. Light comes from a less dense medium, air, and bounces off a more dense medium, soap, there is a phase change. There is no phase change for the transmitted wave through the soap film.

FT

Choice IV Inside the soap film, the wave meets an interface from an optically more dense medium to a less dense one. There is no phase change. From the above, we see that choices II, III and IV are all true. Answer: (E)

The Telescope

RA

9.22

The Magnification of the Telescope is

M=

fo fe

(9.22.1)

D

where fo and fe are the focal lengths of the objective and eyepiece lens respectively. From the information given to us, we know that fe =

fo = 0.1 m 10

(9.22.2)

The optical path length is simply d = f0 + fe = 1.0 + 0.1 = 1.1 m

(9.22.3)

Answer: (D)

9.23

Fermi Temperature of Cu

The Fermi Temperature is related to the Fermi Energy by EF = kTF David S. Latchman

(9.23.1) ©2009

Bonding in Argon This is also the kinetic energy of the system,

85

1 EF = me v2 2

(9.23.2)

Solving for v r

2kTF me

r

2(1.38 × 10−23 )(8 × 104 ) 9.11 × 10−31

v= =

(9.23.3)

Adding the indices, we get

Answer: (E)

Bonding in Argon

(9.23.4)

RA

9.24

FT

−23 + 4 + 31 =6 2 We are looking for speeds to be in the order of 106 m s−1 .

We recall that Argon is a noble gas. This means that its outermost electron shell is filled and thus unable to bond by conventional means. It won’t bond ionically as this method involves either giving up or receiving electrons. A full electron shell means that this will be difficult without some effort. In covalent bonds, electrons are shared between atoms. Again, because the shell is filled, there is no place to share any electrons. A metallic bond involves a positive charge, the nucleus, ‘swimming in a sea of free electrons. Again, with Argon, all of its electrons are tightly bound and hence have no free electrons.

D

This leaves us with van der Waal bonds. van der Waal forces occur between atoms or molecules of the same type and occur due to variances in charge distribution in the atoms. This is our only obvious choice. Answer: (E)

9.25

Cosmic rays

For our cosmic rays to reach deep underground, we are looking for particles that are essentiall massless and pass through matter easily. A Alpha particles and neutrons have high kinetic energy but very short penetrating depth. This is primarily due to their masses. So we can eliminate this. B protons and electrons don’t penetrate much as they interact easily with matter. ©2009

David S. Latchman

86 GR9277 Exam Solutions C Iron and Carbon nuclei are very heavy and interact very easily with matter. We would not expect them to penetrate far, much less deep underground. D muons and neutrinos. A muon is esseentially a ‘heavy’ electron they don’t emit much brenstraalung radiation and hence are highly penetrating. Neutrinos have almost no mass and travel close to the speed of light. In the beginning it was dispited whether they had any mass at all. These two particles fit our choices. E positrons and electrons. Thes two are highly interacting with matter. Answer: (D)

Radioactive Half-Life

FT

9.26

The radioactive decay is

N = N0 exp[−λt]

(9.26.1)

Answer: (B)

9.27

RA

Where λ is the decay constant. At t = 0 on the graph, the count rate is N = 6 × 103 counts per minute. We are looking for the time when N = 3 × 103 counts per minute, which falls at around t = 7 minutes.6

The Wave Function and the Uncertainty Principle

NOT FINISHED

Probability of a Wave function

D

9.28

NOT FINISHED

You may wonder what the presence of the log10 = 0.03 and log10 e = 0.43 is there for. It simply means that you will go down three divisions on the y-axis to get the half life count. We see that from eq. (9.26.1), dN = −λNdt (9.26.2) 6

Integrating gives us N 2

Z N

dN = −λ N

Z

t1

dt t0

ln 2 = t2 − t1 log10 2 ∴ = t2 − t1 = ∆t log10 e

(9.26.3)

So at any point on the graph, a change in three divisions on the y-axis will give us the half life.

David S. Latchman

©2009

Particle in a Potential Well

9.29

87

Particle in a Potential Well

This tests our knowledge of the properties of a wave function as well as what the wavefunction of a particle in a potential well looks like. We expect the wave function and its derivative to be continuous. This eliminates choices (C) & (D). The first is not continuous and the second the derivative isn’t continuous. We also expect the wave to be fairly localized in the potential well. If we were to plot the function |ψ|2 , we see that in the cases of choices (A) & (E), the particle can exist far outside of the well. We expect the probability to decrease as we move away from the well. Choice (B) meets this. In fact, we recognize this as the n = 2 state of our wave function.

9.30

FT

Answer: (B)

Ground state energy of the positronium atom

RA

The positronium atom consists of a positron and an electron bound together. As their masses are the same, we can’t look at this as a standard atom consisting of a proton and an electron. In our standard atom, it’s center of mass is somewhere close to the center of mass of the proton. In the case of our positronium atom, it’s center of mass is somewhere in between the electron and positron. To calculate the energy levels of the positronium atom, we need to “reduce” this two mass system to an effective one mass system. We can do this by calculating its effective or reduced mass. 1 1 1 = + (9.30.1) µ mp me

D

where µ is the reduced mass and mp and me are the masses of the positron and the electron respectively. As they are the same, µ is µ=

me 2

(9.30.2)

We can now turn to a modified form of Bohr’s Theory of the Hydrogen Atom to calculate our energy levels En =

Z2 µ · 13.6eV n2 me

(9.30.3)

where Z is the atomic mass and n is our energy level. We have calculated that the reduced mass of our system is half that of Hydrogen. So Z = 1. For the n = 2 state we have 1 13.6 E0 E2 = · 2 eV = eV (9.30.4) 2 2 8 Answer: (E) ©2009

David S. Latchman

GR9277 Exam Solutions

88

9.31

Quantum Angular Momentum

Not FINISHED

9.32

Electrical Circuits I

The power dissipated by a resistor, R, is P = I2 R

(9.32.1)

Answer: (A)

Electrical Circuits II

RA

9.33

FT

where I is the current through the resisitor. In this case the current through the R1 resistor is the current from the battery, I. After the current passes through R1 , the current divides as it goes through to the other resistors, so the current passing through R1 is the maximum current. As the other resistors are close to R1 but have smaller currents passing through them, the power dissipated by the R1 resistor is the largest.

We can find the voltage across the R4 resisitor by determining how the voltage divids across each resisitors. R3 and R4 are in parallel and so the potential difference across both resistors are the same. The net resistance is R3 R4 R3 + R4 60 · 30 = = 20Ω 90

R3k4 =

(9.33.1)

D

This is in parallel with the R5 resistor, so

R5+(3k4) = 20 + 30 = 50Ω

(9.33.2)

This net resistance is the same as the R2 resistor. This reduces our circuit to one with two resistors in series where RT = 25Ω. The voltage across RT is found by using the voltage divider equation 25 VT = 3.0 V = 1.0 V (9.33.3) 75 This means that the potential across the R3 , R4 and R5 combination is 1.0 V. The voltage across the R4 resistor is the voltage across the R3 k R4 resistor V4 =

20 1.0 V = 0.4 V 50

(9.33.4)

Answer: (A) David S. Latchman

©2009

Waveguides

9.34

89

Waveguides

NOT FINISHED Answer: (D)

9.35

Interference and the Diffraction Grating

Interference maxima for the diffraction grating is determined by the equation d sin θm = mλ

(9.35.1)

FT

where d is the width of the diffraction grating. We are told that our grating has 2000 lines per cm. This works out to 1 × 10−2 = 0.5 × 10−5 m d= 2000

(9.35.2)

As θ is very small, we can approximate sin θ ≈ θ. The above equation can be reduced to dθ = λ (9.35.3)

RA

Plugging in what we know

λ d 5200 × 10−10 = 0.5 × 10−5 = 0.1 radians

θ=

(9.35.4)

As the answer choices is in degrees,

D

x=

18 180 · 0.1 = ≈ 6◦ π π

(9.35.5)

Answer: (B)

9.36

EM Boundary Conditions

NOT FINSHED

9.37

Decay of the π0 particle

Relativity demands that th photon will travel at the speed of light. Answer: (A) ©2009

David S. Latchman

GR9277 Exam Solutions

90

9.38

Relativistic Time Dilation and Multiple Frames

For this question we recall the time dilation equation where ∆T ∆T0 = q 2 1 − uc2

(9.38.1)

where T is the time measure in the fram at rest and T0 is the time measured in the frame moving at speed u relative to the rest frame. With the information given in the question we can see that

FT

∆t1 ∆t2 = q v2 1 − c122 ∆t1 ∆t3 = q v2 1 − c132

(9.38.2)

(9.38.3)

Answer: (B)

9.39

RA

You may think that Answer: (C) is a possible answer but it would be incorrect the leptons are not in the S2 frame, they are in the S1 frame, so this possibility has no physical consequence.

The Fourier Series

D

In most cases, we rarely see pure sine waves in nature, it is often the case our waves are made up of several sine functions added together. As daunting as this question may seem, we just have to remember some things about square waves, 1. We see that our square wave is an odd function so we would expect it to be made up of sine functions. This allows us to eliminate all but two of our choices; answers (A) and (B)7 .

2. Square waves are made up of odd harmonics. In choice (A), we see that both even and odd harmonics are included. In the case of choice (B), only odd harmonics will make up the function. As choice (B) meets the criteria we recall, we choose this one. Answer: (B) 7

If you got stuck at this point, now would be a good time to guess. The odds are in your favor.

David S. Latchman

©2009

The Fourier Series

9.39.1

91

Calculation

As we have the time, we see that the function of out square wave is    1 V(t) =   −1

0 < t < ωπ π < t < 2π ω ω

(9.39.1)

Our square wave will take the form,

We can solve for an and bn , where 1 an = π

Z

1 bn = π

Z

(9.39.2)

2π/ω

V(t) cos nωtdt

(9.39.3)

V(t) sin nωtdt

(9.39.4)

2π/ω

RA

0

FT

1 V(t) = a0 + a1 cos ωt + a2 cos 2ωt + · · · + an cos nωt+ 2 + b1 sin ωt + b2 sin 2ωt + · · · + bn sin nωt

0

Solving for an , gives us

D

"Z π/ω # Z 2π/ω 1 an = (1) cos(nωt)dt + (−1) cos(nωt)dt π 0 π/ω  R π/ω R 2π/ω   1  dt − π/ω dt for n = 0   π " 0 # π π =   ω   ω  1 1 1  for n , 0  π nω sin(nωt) − nω sin(nωt) 0 0  h  i 2π π 1 π   − − for n = 0  π ω   ω ω π  2π =  1  for n , 0  nπω sin(nωt)|0ω − sin(nωt)| ωπω    for n = 0 0 =  1  nπω [sin(nπ) − (sin(2nπ) − sin(nπ))] = 0 for n , 0

We can see that8 a0 = 0

and

an = 0

8

We expected this as we can see that the functions is an odd function. Odd functions are made up of sine functions.

©2009

David S. Latchman

GR9277 Exam Solutions

92 Now we solve for bn , 1 bn = π

π/ω

"Z

2π/ω

Z

#

1 · sin(nωt)dt −

sin(nωt)dt   ωπ   2πω  −1 1  −1  cos(nωt) − cos(nωt)  =  π π nω nω 0

0

0

ω

We can write this as V(t) = or we can say

4 sin(nωt) nπω n = 2m + 1

This leads to9

∞

for all odd values of n

for all values of m

4 X 1 sin((2m + 1)ωt) πω m=0 2m + 1

(9.39.5)

(9.39.6) (9.39.7) (9.39.8)

RA

V(t) =

9.40

FT

1 [cos(2nπ) − 2 cos(nπ) + 1] = nπω 2 − 2(−1)n =  nπω  0 for even n  =  4  nπω for odd n

Rolling Cylinders

At the pont of contact, the cylinder is not moving. We do see the center of the cylinder moving at speed, v and the top of the cylinder moving at speed, 2v. Thus the acceleration acting at the point of contact is the cetripetal acceleration, which acts in an upwards direction.

D

Answer: (C)

9.41

Rotating Cylinder I

We recall that the change in kinetic energy of a rotating body is  1  ∆K = I ω2f − ω2i 2

(9.41.1)

This becomes  1  ∆K = 4 802 − 402 2 = 9600 J 9

(9.41.2)

Definitely not something you have lots of time to do in the exam.

David S. Latchman

©2009

Rotating Cylinder II Answer: (D)

9.42

93

Rotating Cylinder II

Again we recall our equations of motion of a rotating body under constant angular acceleration. Fortunatley, they are similar to the equations for linear motion ω = ω0 + αt

(9.42.1)

Plugging in the values we were given, we can find the angular acceleration, α, α = −4 rad.s-2

Which works out to be Answer: (D)

τ = Iα

(9.42.3)

τ = 16 Nm

(9.42.4)

Lagrangian and Generalized Momentum

RA

9.43

FT

The torque is

(9.42.2)

We recall that the Lagrangian of a system is

L=T−V

(9.43.1)

and in generalized coordinates, this looks like

1 L = mq˙ 2n − U(qn ) 2

(9.43.2)

D

We can find the equations of motion from this

We are told that

or rather

! d ∂L ∂L = dt ∂q˙ n ∂qn

(9.43.3)

∂L =0 ∂qn

(9.43.4)

! d ∂L =0 dt ∂q˙ n

(9.43.5)

We can see that the generalized momentum is ∂L = mq˙ n = pn ∂q˙ n ©2009

(9.43.6) David S. Latchman

GR9277 Exam Solutions

94 and that

d mq˙ n = mq¨n = 0 dt So we expect the generalized momentum, pn is constant.

(9.43.7)

Answer: (B)

9.44

Lagrangian of a particle moving on a parabolic curve

As the particle is free to move in the x and y planes, Lagrangian of the system is

FT

L=T−V 1 1 = m y˙ 2 + mx˙ 2 − mgy 2 2 We are given a realtinship between y and x, where y = ax2

(9.44.1)

(9.44.2)

Differentiating this with respect to time gives

y˙ = 2axx˙

Answer: (A)

9.45

RA

Subsitutung this into eq. (9.44.1), gives us " # y˙ 2 1 2 L = m y˙ + − mgy 2 4ay

(9.44.3)

(9.44.4)

A Bouncing Ball

D

As the ball falls from a height of h, its potential energy s converted into kineic energy. 1 E = mgh = mv2i (9.45.1) 2 Upon hitting the floor, the ball bounces but some of the energy is lost and its speed is 80% of what it was before v f = 0.8vi (9.45.2)

As it rises, its kinetic energy is converted into potential energy 1 2 mv = mgh2 2 f 1 = m(0.8vi )2 2 = mgh2 ⇒ h2 = 0.64h

(9.45.3) (9.45.4)

Answer: (D) David S. Latchman

©2009

Phase Diagrams I

9.46

95

Phase Diagrams I

NOT FINISHED Answer: (B)

9.47

Phase Diagrams II

NOT FINISHED

9.48

FT

Answer: (B)

Error Analysis

The error for Newton’s equation, F = ma, would be  σ 2 f

F

σm = m 

9.49

σa + a 

2

RA

Answer: (C)

2

(9.48.1)

Detection of Muons

The muons travel a distance of 3.0 meters. As muons move at relativistic speeds, near the speed of light, the time taken for a photon to traverse this distance is the time needed to distinguish between up travelling muons and down travelling muons. 3.0 x = seconds c 3.0 × 108

(9.49.1)

D

t=

Answer: (B)

9.50

Quantum Mechanical States

NOT FINISHED

9.51

Particle in an Infinite Well

We recall that the momentum operator is p= ©2009

~ d i dx

(9.51.1) David S. Latchman

96 and the expectation value of the momentum is

GR9277 Exam Solutions

hpi = hψ∗n |p|ψn i ! Z ∗ ~ d ψn dx = ψn i dx Given that

r ψn =

  nπx 2 sin a a

(9.51.2)

(9.51.3)

eq. (9.51.2) becomes r

2 a

a

Z 0

Answer: (A)

9.52

   nπx nπx cos dx sin a a 

FT

~ nπ hpi = i a =0

(9.51.4)

Particle in an Infinite Well II

Answer: (B)

9.53

RA

This is the very defintion of orthonormality.

Particle in an Infinite Well III

D

We recall that Schrodinger’s Equation is ¨

~2 d2 ψ − + V(x) = Eψ 2m dx2

(9.53.1)

within the region 0 < x < a, V(x) = 0, thus ~2 d2 ψ − = Eψ 2m dx2 d2 ψ 2mE ∴ 2 = −kn2 ψ = − 2 ψ dx ~

(9.53.2)

We see that E=

kn2 ~2 n2 π2 ~ = 2 2m a 2m

(9.53.3)

π2 ~2 2ma2

(9.53.4)

where n = 1, 2, 3 · · · . So E≥ Answer: (B) David S. Latchman

©2009

Current Induced in a Loop II

9.54

97

Current Induced in a Loop II

Recalling Faraday’s Law dΦ (9.54.1) dt This gives the relationship between the induced EMF and the change in magnetic flux. The minus sign in the equation comes from Lenz’s Law; the change in flux and the EMF have opposite signs. E =−

Induced Current Clockwise

FT

If we use the Right Hand Grip Rule we see that the magnetic flux goes into the page around the loop. Pulling the loop to the right, moving away from the wire, reduced the magnetic flux inside the loop. As a result, the loop induces a current to counteract this reduction in flux and induces a current that will produce a magnetic field acting in the same direction. Again, using the Right Hand Grip Rule, this induced magnetic field acting into the page induces a clockwise current in the loop. Force on Left Side To the left

Force on Right Side To the right

RA

Table 9.54.1: Table showing something

Now we know the current directions, we can determine the forces on the wires. As we have a current moving through the wire, a motor, we use Fleming’s Left Hand Rule. From this rule, we see the force on the wire is actin to the left. Using the same principle, on the right side, the force is acting on the right. Answer: (E)

Current induced in a loop II

D

9.55

The magnetic force on a length of wire, ` is FB = I` × B

(9.55.1)

We need to calculate the magnetic induction on the left and right sides of the loop. For this we turn to Ampere’s Law I B · ds = µ0 Ienclosed (9.55.2) On the left side of the loop, BL (2πr) = µ0 I µ0 I BL = 2πr ©2009

(9.55.3) David S. Latchman

GR9277 Exam Solutions

98 On the right side of the loop, we have BR [2π (r + a)] = µ0 I BR =

µ0 I 2π (r + a)

(9.55.4)

The Magnetic force on the left side becomes µ0 I FL = ib 2πr

! (9.55.5)

and the magnetic force on the right side is !

FT

µ0 I FR = ib 2π (r + a)

(9.55.6)

We know that from the above question, these forces act in oppsoite directions, so

Answer: (D)

9.56

RA

F = FL − FR  µ0 iIb  1 1 = − 2π " r r + #1 µ0 iIb a = 2π r (r + a)

(9.55.7)

Ground State of the Quantum Harmonic Oscillator

D

We recall that the ground state of the quantum harmonic oscillator to be 12 hν. Answer: (C)

9.57

Induced EMF

The induced EMF follows Lenz Law E =−

dΦ dt

(9.57.1)

where Φ = BA, where B is the magnetic field and A is the area of the coil.As the coil cuts into the field, an EMF is induced. As it starts leaving, an EMF the EMFwill change polarity. Answer: (A) David S. Latchman

©2009

Electronic Configuration of the Neutral Na Atom

9.58

99

Electronic Configuration of the Neutral Na Atom

The atomic mass, Z, of the neutral Na atom is 11. We want our superscripts to ad to 11. Thus 1s2 , 2s2 , 2p6 , 3s1 (9.58.1) Answer: (C)

9.59

Spin of Helium Atom

FT

The electronic configuration of the He atom is 1s2

The spin of this is

(9.59.1)

1 1 − =0 (9.59.2) 2 2 representing one electron in the spin up and the other in the spin down directions. Thus the total spin is zero making it a spin singlet. Answer: (A)

9.60

RA

S=

Cyclotron Frequency of an electron in metal

As a charged particle enters a transverse magnetic field, it experiences a centripetal force. Recalling the Lorents Force equation

D

mv2 = Bev r Be ⇒ ωc = m

where ω =

v r (9.60.1)

Plugging in what we know, we get ωc =

1 × 1.6 × 10−19 0.1 × 9.11 × 10−31

Adding the indices of the equation gives an order of magnitude approximation − 19 + 31 = 12 Our closest match is 1.8 × 1012 rad/s. Answer: (D) ©2009

David S. Latchman

GR9277 Exam Solutions

100

9.61

Small Oscillations of Swinging Rods

The period of an oscillator given its moment of inertia is10 r mgd ω= I

(9.61.1)

where m is the mass, r is the distance from the center of rotation and I is the moment of inertia. The moment of inertia is essentially dependent on how the mass is distributed and is defined by I = mr2 (9.61.2) In the first case, we have two masses at a distance, r. So its moment of inertia becomes

FT

I1 = 2mr2

(9.61.3)

In the second case, we have one mass at the distance 2r and another mass at r. We can get the total moment of inertia by adding the moment of inertias of both these masses  2 5mr2 r 2 = (9.61.4) I2 = mr + m 2 4

RA

Now the tricky part. The distance d in eq. (9.61.1) is the distance from the pivot to the center of mass. In the first case it’s the distance from the pivot to the two masses. In the second case the center of mass is between the two masses. The center of mass can be found by P mi ri (9.61.5) rcm = P mi Thus the center of mass for the second pendulum is r2 =

mr + m 2r 2m

3 = r 4

(9.61.6)

D

Now we can determine the angular frequencies of our pendulums. For the first pendulum r 2mgr ω1 = 2mr2 r g = (9.61.7) r

For the second pendulum v t ω2 =

10

  3r 4

5 mr2 4

r =

(2m)g

6g 5r

(9.61.8)

Show derivation of this equation.

David S. Latchman

©2009

Work done by the isothermal expansion of a gas Dividing eq. (9.61.8) by eq. (9.61.7) gives

101

 6g  21   12 ω2  5r  6 =  g  = ω1 5 r

(9.61.9)

Answer: (A)

9.62

Work done by the isothermal expansion of a gas

FT

The work done by the expansion of an ideal gas is Z

W=

V1

PdV

(9.62.1)

V0

As the temperature is constant, we substitute the ideal gas equation P=

RA

to give

nRT V

Z

V1

dV V V0   V1 = nRT ln V0

W=

(9.62.2)

(9.62.3)

For one mole of gas, n = 1, gives us

V1 W = RT0 ln V0

D





(9.62.4)

This gives11

Answer: (E)

9.63

Maximal Probability

NOT FINISHED12 Answer: (D) 11

It seems the inclusion of the specific heat ratio was not needed and was there to throw you off. As Prof. Moody would say, “CONSTANT VIGILANCE!!!” 12 Add something

©2009

David S. Latchman

GR9277 Exam Solutions

102

9.64

Gauss’ Law

We recall the principle of Gauss’ Law for electric fields. I E · dA = S

Q 0

(9.64.1)

The presence of a field means the presence of a charge somewhere. Answer: (B)

Oscillations of a small electric charge

FT

9.65

In this scenario, we have a charge, −q, placed between two charges, +Q. The net force on the small charge but if we were to slightly displace this charge it would be pulled back to its central axis. The force by which it is pulled back is F=

Qq 2Qq Qq + = 4π0 R2 4π0 R2 4π0 R2

RA

For small oscillations

(9.65.1)

k=

d2 V dx2

(9.65.2)

F=

dV dx

(9.65.3)

where k is the spring constant. We also recall that

D

Thus we can find k by differentiating 9.65.1.

! 2Qq Qq d k= = 2 dR 4π0 R 2π0 R3

(9.65.4)

The angular frequency, ω, of our point charge system is ω2 =

k m

(9.65.5)

where m is the mass of the small particle. Thus the angular frequency, ω, is "

Qq ω= 2π0 mR3

# 12 (9.65.6)

Answer: (E) David S. Latchman

©2009

Work done in raising a chain against gravity

9.66

103

Work done in raising a chain against gravity

We recall our familar Work equation L

Z W=

L

Z F · dx =

0

mg · dx

(9.66.1)

0

As the chain is pulled up, the mass changes with the length that is hanging. We are given the linear density so we know the relationship of the chain’s mass to its length13 ρ=

M m = L x

(9.66.2)

L

Z W=

FT

Substituting this into the above equation, we have ρx · dx 0

Answer: (C)

9.67

RA

L x2 = ρg 2 0 ρgL2 2(10)(100) = = = 1000 J 2 2

(9.66.3)

Law of Malus and Unpolarized Light

D

We suspect that the equation might have polarized and unpolarized components. Having no real idea and all the time in the world, we can derive the equations on what this might look like. We recall the Law of Malus Ip = Ipo cos2 θ =

Ipo [cos 2θ − 1] 2

(9.67.1)

Unpolarized light would have the same intensity through all θ so Iu = Iuo

(9.67.2)

13

There is another way to think of this problem. We are told that the steel chain is uniform, so its center of mass is in the middle of its length. The work done is the work done in raising the mass of the chain by this distance. Thus W = Mg

  L L 10 = ρLg = 2 · 10 · 10 · = 1000 J 2 2 2

As you can see, we get the same result. You may find this solution quicker.

©2009

David S. Latchman

104 The total intensity is the sum of the two intensities

GR9277 Exam Solutions

I = Iu + Ip Ipo [cos 2θ − 1] = Iuo + Ipo cos2 θ = 2 " # Ipo Ipo = Iuo − + cos 2θ 2 2 where A = Iuo − B

B=

(9.67.3)

Ipo 2

Given that A > B > 0, the above hypothesis holds.

9.68

FT

Answer: (C)

Telescopes and the Rayleigh Criterion

The resolution limit of a telescope is determined by the Rayleign Criterion λ d

RA

sin θ = 1.22

(9.68.1)

where θ is the angular resolution, λ is the wavelength of light and d is the len’s aperature diameter. As θ is small we can assume that sin θ ≈ θ. Thus λ θ 1.22 × 5500 × 10−10 = 8 × 10−6

d = 1.22

(9.68.2)

D

After some fudging with indices we get something in the order of 10−2 m.

Answer: (C)

9.69

The Refractive Index

The refractive index is the ratio of the speed of light in vacuum to the speed of light in the medium. c n= (9.69.1) v Given n = 1.5, c 2 v= = c (9.69.2) n 3 Answer: (D) David S. Latchman

©2009

High Relativistic Energies

9.70

105

High Relativistic Energies

Recalling the equation for relativistic energy E2 = pc2 + m2 c4

(9.70.1)

We are told that the energy of the particle is 100 times its rest energy, E = 100mc2 . Substituting this into the above equation gives  2  2 100mc2 = pc2 + mc2 (9.70.2)

FT

As we are looking at ultra-high energies, we can ignore the rest energy term on the right hand side, so14  2 100mc2 = pc2 ∴ p ≈ 100mc Answer: (D)

Thermal Systems I

RA

9.71

(9.70.4)

NOT FINISHED Answer: (B)

9.72

Thermal Systems II

NOT FINISHED

D

Answer: (A)

9.73

Thermal Systems III

NOT FINISHED Answer: (C) 14

We can find a solution by relating the relativistic energy and momentum equations. Given E = γmc2

p = γmc

This gives us E = pc ⇒ p = 100mc

©2009

(9.70.3)

David S. Latchman

GR9277 Exam Solutions

106

9.74

Oscillating Hoops

The angular frequency of our hoop can be found by the equation r mgd ω= (9.74.1) I where m is the mass of the hoop, d is the distance from the center of mass and I is the moment of inertia. The Moment of Inertia of our hoop is Icm = Mr2

(9.74.2)

where M is the mass of the hoop and r is its radius. As the hoop is hanging from a nail on the wall, we use the Parallel Axis Theorem to determine its new Moment of Inertia I = Icm + Mr2 = 2Mr2

FT

we are given that

(9.74.3)

MX = 4MY

dX = 4dY

We can now find the moment of inertias of our two hoops IX = 2MX R2X s ωX =

MX gRX

r

= =

Answer: (B)

ωY =

g 2RX

2π T

2MY R2Y

r =

MY gRY g 2RY

= 2ωX

Decay of the Uranium Nucleus

D

9.75

s

RA

2MX R2X

IY = 2MY R2Y

When the Uranium nucleus decays from rest into two fissile nuclei, we expect both nuclei to fly off in opposite directions. We also expect momentum to be conserved thus

we see that vHe Helium nuclei

MTh VTh = mHe vHe MTh ∴ vHe = VTh mHe ≈ 60VTh . We can calculate the kinetic energies of the Thorium and

1 2 KTh = MTh VTh 2

1 kHe = mHe v2He 2  1 MTh (60VTh )2 = 2 60 ≈ 60KTh

Now we can go through our choices and find the correct one. David S. Latchman

©2009

Quantum Angular Momentum and Electronic Configuration A This is clearly not the case. From the above, we see that kHe = 60KTh .

107

B Again, this is clearly not the case. We see that vHe ≈ 60VTh . C No this is not the case as momentum is conserved. For this to take place the two nuclei must fly off in opposite directions. D Again, this is incorrect. Momentum is conserved so the momentum of both nuclei are equal. E This is correct. We see from the above calculations that kHe ≈ 60KTh .

9.76

FT

Answer: (E)

Quantum Angular Momentum and Electronic Configuration

The total angular momentum is

J=L+S

(9.76.1)

RA

As none of the electron sub-shells are filled, we will have to add the individual angular momentum quantum numbers. For the 1s case, the spin, s, is

1 2 As this is in the s sub-shell, then the orbital quantum number is s1 =

(9.76.2)

`1 = 0

(9.76.3)

The total angular momentum for this electron is

D

J1 =

1 2

(9.76.4)

For the other two electron shells we get 1 2 `2 = 1 s2 =

j2 = `2 + s2 =

3 2

(9.76.5)

and similarly for the third electron shell s3 =

1 2

`3 = 1j3 ©2009

= `3 + s3 =

3 2 David S. Latchman

GR9277 Exam Solutions

108 Thus the total angulat momentum is j = j1 + j2 + j3 1 3 3 7 = + + = 2 2 2 2

(9.76.6)

Answer: (A)

9.77

Intrinsic Magnetic Moment

NOT FINISHED

9.78

FT

Answer: (E)

Skaters and a Massless Rod

RA

As the two skaters move towards the rod, the rod will begin to turn about its center of mass. The skaters linear momentum is converted to a combination of linear and rotational momentum of the rod. The rotational momentum can be calculated

L = m (r × V)

(9.78.1)

D

The total rotational momentum can be calculated,

L = Ltop + Lbottom ! ! b b =m (2v) + m v 2 2 3 = mbv 2

(9.78.2)

The rod will rotate with angular velocity, ω, L = Iω

(9.78.3)

where I is the Moment of Inertia, where b I = 2m 2

!2 (9.78.4)

Thus the angular frequency of the rod becomes ω= David S. Latchman

3v b

(9.78.5) ©2009

Phase and Group Velocities 109 As the two masses move towards the rod, the speed of the center of mass remains the same even after collision. This allows us to define the linear motion of the rod which we know to be at the center of the rod. n P

vcenter =

mi vi

i=1 n P

(9.78.6) mi

i=1

Thus the center of mass velocity is vcenter =

m(2v) + m(−v) v = 2m 2

(9.78.7)

FT

The position of the mass at b/2 is a combination of the translational and rotational motions. The translational motion is vtranslational = 0.5vt

(9.78.8)

vrotational = 0.5b sin(ωt)

(9.78.9)

And the rotational motion is

RA

where ω = 3v/b. This becomes

x = 0.5vt + 0.5b sin

Answer: (C)

9.79



3vt b

 (9.78.10)

Phase and Group Velocities

D

We recall that the group velocity is

vg =

dω dk

(9.79.1)

and the phase velocity is

ω (9.79.2) k From the graph, we see that in the region k1 < k < k2 , the region of the graph is a straight line with a negative gradient. So we can assume that dω < 0 and that ωk > 0. dk Thus the two velocities are in opposite directions. vp =

A This is correct as shown above. B They can’t be in the same directions. The phase velocity is moving in the opposite direction to the group velocity. ©2009

David S. Latchman

110 GR9277 Exam Solutions C Again incorrect, they are not travelling in the same direction. D Also incorrect, the phase velocity is finite as k , 0. E Again, they are not in the same direction. Answer: (A)

9.80

Bremsstrahlung Radiation

FT

As an electron is accelerated towards a target, it gets rapidly decelerated and emits emectromagnetic radiation in the X-ray spectrum. The frequency of this emitted radiation is determined by the magnitude of its decelertion. If it has been completly decelerated, then all of its kinetic energy is converted to EM radiation. We are told the kinetic energy of our accelerated electrons are K = 25 keV. The energy of a photon is E = hf = Solving for λ, we have

hc =K λ

hc K 6.63 × 10−34 × 3 × 108 = 25 × 103 × 1.60 × 10−19 6.63 × 3 = × 10−10 25 × 1.6 12 ≈ × 10−10 = 0.5 Å 25

Answer: (B)

(9.80.2)

Resonant Circuit of a RLC Circuit

D

9.81

RA

λ=

(9.80.1)

The maximum steady state amplitude will occur at its resonant frequency. This can be see if one were to draw a graph of E vs. ω. The resonant frequency occurs when the capacitive impedance, XC and the inductive impedance,XL are equal. Thus XL = ωL

XC =

1 ωC

Equating XL and XC together gives 1 ωC 1 ω= √ LC

ωL =

(9.81.1)

Answer: (C) David S. Latchman

©2009

Angular Speed of a Tapped Thin Plate

9.82

111

Angular Speed of a Tapped Thin Plate

Fortunately we have been given the angular impulse, H, Z H= τdt

(9.82.1)

where tau is the torque on the plate. We recall that τ = Iα where I is the moment of inertia and α is the angular acceleration. Thus the above equation becomes Z H ω= αdt = (9.82.2) I

FT

The moment of inertia of the plate is I = 31 Md215 . Thus ω= Answer: (D)

(9.82.3)

Suspended Charged Pith Balls

RA

9.83

3H md2

We can resolve the horizontal and vertical force components acting on the pith balls. We see that Horizontal Force: Vertical Force:

kq2 2d2 T cos θ = mg

T sin θ =

(9.83.1)

D

As we are looking at small values of θ we can make the following approximations

We see that sin θ =

d . 2L

sin θ ≈ tan θ ≈ θ cos θ ≈ 1

Subsituting the above, we get T = mg ! kq2 d ⇒ T sin θ = mg = 2 2L 2d ! 2 2kq L ∴ d3 = mg

(9.83.2)

Answer: (A) 15

Add derivation of moment of inertia for thin plate.

©2009

David S. Latchman

GR9277 Exam Solutions

112

9.84

Larmor Formula

We recall the Larmor Formula which describes the total power of radiated EM radiation by a non-relativistic accelerating point charge. q2 a2 P= 6π0 c3

(9.84.1)

where q is the charge and a is the acceleration. We can use this to eliminate choices. A This says

FT

P ∝ a2

(9.84.2)

We see that this is TRUE from the above equation. B This says

P ∝ e2

C Also true. D False E True

D

Answer: (D)

RA

This is also TRUE.

(9.84.3)

9.85

Relativistic Momentum

We recall the Relativistic Energy Formula E2 = p2 c2 + m2 c4

(9.85.1)

We are given E = 1.5 MeV. Plugging into the above equation yields 1.52 = p2 c2 + 0.52

(9.85.2)

We find p2 = 2 MeV/c. Answer: (C) David S. Latchman

©2009

Voltage Decay and the Oscilloscope

9.86

113

Voltage Decay and the Oscilloscope

We recall that the voltage decay across a capacitor follows an exponential decay, such that   t V = V0 exp − (9.86.1) RC Solving for C, we see that C=−

  V0 t ln R V

(9.86.2)

We need to find t, which we can determine by how fast the trace sweeps, s. We need to find R, which we will be given. The ration VV0 can be read off the vertical parts of the scope.

9.87

FT

Answer: (B)

Total Energy and Central Forces

The total energy of a system is

RA

E=T+V

(9.87.1)

where T is the kinetic energy and V is the potential energy. We are told the particle moves under a circular orbit where F=

K mv2 = r3 r

D

The potential energy can be found Z Z dr 1K V= Fdr = K =− 2 3 r 2r

(9.87.2)

(9.87.3)

The kinetic Energy is

1 1K T = mv2 = 2 2 2r

Thus

E=

1K 1K − =0 2 r2 2 r2

(9.87.4)

(9.87.5)

Answer: (C)

9.88

Capacitors and Dielectrics

This question can be answered through the process of elimination and witout knowing exactly what the displacement vector is or what it does. ©2009

David S. Latchman

114 GR9277 Exam Solutions As the capacitor is connected to a battery with potential difference, V0 , an electric field, E0 forms between the plates and a charge, Q0 accumulates on the plates. We can relate this to the capacitance, C0 , of the capacitor Q0 = C0 V0

(9.88.1)

While still connected to the battery, a dielectric is inserted between the plates. This serves to change the electric field between the plates and as a result the capacitance. Qf = Cf Vf

(9.88.2)

As we have not disconnected the battery,

FT

V f = V0

(9.88.3)

The dieiectric has a dielectric constant, κ0 , such that

It follows that

C f = κC0

(9.88.4)

Q f = κQ0

(9.88.5)

RA

When a dielectric is placed inside an electric field, there is an induced electric field, that points in the opposite direction to the field from the battery. E f = E0 + Einduced

(9.88.6)

This results in

Ef =

E0 κ

(9.88.7)

From the above we can infer 1. V f = V0

D

2. Q f > Q0 3. C f > C0 4. E f < E0

Based on this we can eliminate all but choice (E). In the case of the last choice, the effect of the electric field places charges on the plates of the capacitor. Gauss’ Law tells us 0 ∇ · E = ρ

(9.88.8)

If we were to place the dielectric between the plates, the atoms in the dielectric would become polarized in the presence of the electric field. This would result in the accumulations of bound charges, ρb within the dielectric. The total charge becomes ρ = ρb + ρ f David S. Latchman

(9.88.9) ©2009

harmonic Oscillator This becomes This becomes

115 0 ∇ · E = −∇ · P + ρ f

(9.88.10)

∇ · (0 E + P) = ρ f

(9.88.11)

Where the term 0 E + P is the displacemet vector. In the beginning, there is no polarization vector, P so D0 = 0 E (9.88.12) But with the presence of the dielectric it becomes, (9.88.13)

D f > D0

(9.88.14)

Thus Answer: (E)

9.89

harmonic Oscillator

NOT FINISHED

9.90

RA

ANSWER: (E)

FT

D f = 0 E + P

Rotational Energy Levels of the Hydrogen Atom

NOT FINSIHED Answer: (B)

The Weak Ineteraction

D

9.91

NOT FINISHED Answer: (D)

9.92

The Electric Motor

For each rotation the wire makes, it crosses the the three pairs of magnets which results in three periods. Thus we expect f = 3 f0 = 30Hz

(9.92.1)

Answer: (D) ©2009

David S. Latchman

GR9277 Exam Solutions

116

9.93

Falling Mass connected by a string

When the mass is at the top of its swing, θ = 0°, the only force acting on it will be the one due to gravity; the tangential force acting downward. As it is let go, we expect the total acceleration to increase and reach a maximum at θ = 90°. We can use this to eliminate the choices given. g sin θ At θ = 0, the total acceleration, a, is zero. At θ = 90, the total acceleration is a maximum, a = g. We would expect this to be greater than g16 We can eliminate this choice.

FT

2g cos θ At θ = 0, the total acceleration is 2g.A bit of an impossibility considering the mass is stationary. At θ = 90 the total acceleration becomes zero. Another impossibility, we expect the total acceleration to increase. We can eliminate this choice.

RA

2g sin θ At θ = 0, the total acceleration is zero. Another impossibility. At θ = 90, the acceleration is a maximum at 2g. We can eliminate this choice. √ g 3 cos2 θ + 1 At θ = 0, our total acceleration is a maximum, a = 2g and a minumum at the bottom, a = g. We don’t expect this physically. We can eliminate this. √ g 3 sin2 θ + 1 At θ = 0 the total accleration is a = g. This we expect. At θ = 90, it is a maximum, a = 2g. Again this is what we expect. We choose this one. ANSWER: (E)

9.93.1

Calculation

D

As the mass falls, its gravitaional potential energy is converted to kinetic energy. We can express this as a function of θ. 1 mg` sin θ = mv2 2

(9.93.1)

where ` is the length of the rod. The radial or centripetal force, ar is mar = Solving for ar gives us

mv2 `

ar = 2g sin θ

(9.93.2)

(9.93.3)

16

Think an amusement park ride, something along ‘the Enterprise’ ride manufactured by the HUSS Maschinenfabrik company. The g-forces are at the greatest at the bottom about 2gs and lowest at the top. There are no restraints while you’re inside; you’re kept in place through centripetal forces. Your faith in the force should dispel any fears.

David S. Latchman

©2009

Lorentz Transformation The tangential accelertion is a component of the gravitational acceleration, thus at = g cos θ

117

(9.93.4)

We can find the total acceleration, a, by adding our tangential and radial accelerations. As these accelerations are vectors and they are orthogonal a=

q a2t + a2r

(9.93.5)

Plugging in and solving gives us

9.94

(9.93.6)

Lorentz Transformation

Answer: (C)

RA

NOT FINISHED

9.95

FT

q a = 4g2 sin2 θ + g2 cos2 θ p = g 3 sin2 θ + 1

Nuclear Scatering

NOT FINISHED

D

ANSWER: (C)

9.96

Michelson Interferometer and the Optical Path Length

The optical path length through a medium of refractive index, n, and distance, d, is the length of the path it would take through a vacuum, D. Thus D = nd

(9.96.1)

The Optical Path Difference is the difference in these two lengths. As the gas is evacuated, we observe 40 fringes move past our field of view. So our optical path difference is ∆ = nλ (9.96.2) NOT FINISHED ANswer: (C) ©2009

David S. Latchman

GR9277 Exam Solutions

118

9.97

Effective Mass of an electron

NOT FINISHED Answer: (D)

9.98

Eigenvalues of a Matrix

We are given a matrix

FT

  0 1 0   (9.98.1) A = 0 0 1   1 0 0 By finding the determinant of the characteristic matrix, we can find its eigenvalues 0 −λ 1 0 −λ 1 = 0 (9.98.2) 1 0 −λ Solving for λ yields Thus

RA

− λ[(−λ)(−λ) − (1)(0)] − 1[(0)(−λ) − (1)(1)] + 0[(0)(0) − (−λ)(1)] = 0 λ3 = 1

(9.98.3) (9.98.4)

This leaves us solutions where

D

λ3 = exp [2πni]   i2πn ⇒ λ = exp =1 3 eq. (9.98.5) has solutions of the form       i2πn 2πn 2πn = cos + i sin λn = exp 3 3 3 where n = 1, 2, 3.

(9.98.5)

(9.98.6)

This gives us solutions where    2π 2π λ1 = cos + i sin 3 3 √ 1 3 =− +i 2  2   4π 4π λ2 = cos + i sin 3 3 √ 1 3 =− −i 2 2 λ3 = cos (2π) + i sin (2π) = +1 

(9.98.7)

(9.98.8) (9.98.9)

Now we have our solutions we can eliminate choices, David S. Latchman

©2009

First Order Correction Perturbation Theory A We see that

119

λ1 + λ2 + λ3 = 0 √ ! √ ! 1 3 3 1 − +i + − −i +1=0 2 2 2 2 This is TRUE B We see that λ1 and λ2 are not real. This is NOT TRUE. C We see that, in our case,

So this is also true. D This is also true

Answer: (E)

9.99

(9.98.10)

RA

E Also TRUE

FT

√ ! √ ! 1 3 3 1 − −i λ1 λ2 = − + i 2 2 2 2 1 3 = + 4 4 =1

First Order Correction Perturbation Theory

NOT FINSHED Answer: (A)

Levers

D

9.100

As our system is in equlibrium, we know that the sum of the moments is equal to zero. We are also told that the rod is uniform, so the center of mass is in the middle of the rod. Thus, taking the clockwise and anticlockwise moments about the pivot 20x + 20(x + 5) = 40y

(9.100.1)

where x is the distance of the center of mass from the pivot and y is the distance of the 40kg mass from the pivot. We know

Solving for x results in

10 = 5 + x + y

(9.100.2)

x = 1.25 m

(9.100.3)

Answer: (C) ©2009

David S. Latchman

GR9277 Exam Solutions

D

RA

FT

120

David S. Latchman

©2009

Appendix

A

A.1

Constants

Symbol c G me NA R k e 0 µ0 1 atm a0

Value 2.99 × 108 m/s 6.67 × 10−11 m3 /kg.s2 9.11 × 10−31 kg 6.02 × 1023 mol-1 8.31 J/mol.K 1.38 × 10−23 J/K 1.60 × 10−9 C 8.85 × 10−12 C2 /N.m2 4π × 10−7 T.m/A 1.0 × 105 M/m2 0.529 × 10−10 m

D

RA

Constant Speed of light in a vacuum Gravitational Constant Rest Mass of the electron Avogadro’s Number Universal Gas Constant Boltzmann’s Constant Electron charge Permitivitty of Free Space Permeability of Free Space Athmospheric Pressure Bohr Radius

FT

Constants & Important Equations

Table A.1.1: Something

A.2

Vector Identities

A.2.1

Triple Products A · (B × C) = B · (C × A) = C · (A × B) A × (B × C) = B (A · C) − C (A · B)

(A.2.1) (A.2.2)

Constants & Important Equations

122

A.2.2

Product Rules


∇ f g = f ∇g + g ∇ f ∇ (A · B) = A × (∇ × B) + B × (∇ × A) + (A · ∇) B + (B · ∇) A

∇ · f A = f (∇ · A) + A · ∇ f ∇ · (A × B) = B · (∇ × A) − A · (∇ × B)

∇ × f A = f (∇ × A) − A × ∇ f ∇ × (A × B) = (B · ∇) A − (A · ∇) B + A (∇ · B) − B (∇ · A)

Second Derivatives

FT

A.2.3

∇ · (∇ × A) = 0
∇ × ∇f = 0

∇ × (∇ × A) = ∇ (∇ · A) − ∇2 A

Commutators

A.3.1

Lie-algebra Relations

RA

A.3

D

[A, A] = 0 [A, B] = −[B, A] [A, [B, C]] + [B, [C, A]] + [C, [A, B]] = 0

A.3.2

A.3.3

(A.2.3) (A.2.4) (A.2.5) (A.2.6) (A.2.7) (A.2.8)

(A.2.9) (A.2.10) (A.2.11)

(A.3.1) (A.3.2) (A.3.3)

Canonical Commutator [x, p] = i~

(A.3.4)

Kronecker Delta Function ( δmn =

For a wave function

0 1

if m , n; if m = n;

Z ψm (x)∗ ψn (x)dx = δmn

David S. Latchman

(A.3.5) ©2009

Linear Algebra

123

A.4

Linear Algebra

A.4.1

Vectors

Vector Addition The sum of two vectors is another vector

Associative Zero Vector

|αi + |βi = |βi + |αi

(A.4.2)



|αi + |βi + |γi = |αi + |βi + |γi

(A.4.3)

|αi + |0i = |αi

(A.4.4)

|αi + | − αi = |0i

(A.4.5)

D

RA

Inverse Vector

(A.4.1)

FT

Commutative

|αi + |βi = |γi

©2009

David S. Latchman

Constants & Important Equations

D

RA

FT

124

David S. Latchman

©2009

Bibliography

D

RA

FT

[1] David J. Griffiths. Introduction to Quantum Mechanics, chapter 4.4.3, pages 184–185. Prentice Hall, second edition, 2005.

Index

Parallel Axis Theorem, see Rotational Motion

RA

Kepler’s Laws, 23 Newton’s Law of Gravitation, 22 Orbits, 23 Potential Energy, 22 Circular Orbits, see Celestial Mechanics Commutators, 122 Canonical Commutators, 122 Kronecker Delta Function, 122 Lie-algebra Relations, 122 Compton Effect, 58 Counting Statistics, 71

FT

Angular Momentum, see Rotational Mo- Oscillatory Motion, 16 tion Coupled Harmonic Oscillators, 18 Damped Motion, 17 Bohr Model Kinetic Energy, 16 Hydrogen Model, 55 Potential Energy, 17 Simple Harmonic Motion Equation, 16 Celestial Mechanics, 22 Small Oscillations, 17 Circular Orbits, 23 Total Energy, 16 Escape Speed, 22

D

Doppler Effect, 20

Fourier Series GR9277 Q39, 90 Franck-Hertz Experiment, 61

Rolling Kinetic Energy, see Rotational Motion Rotational Kinetic Energy, see Rotational Motion Rotational Motion, 20 Angular Momentum, 21 Moment of Inertia, 20 Parallel Axis Theorem, 21 Rolling Kinetic Energy, 21 Rotational Kinetic Energy, 20 Torque, 21 Subject, 42 System of Particles, 22

Gravitation, see Celestial Mechanics

Torque, see Rotational Motion

Kepler’s Laws, see Celestial Mechanics Kronecker Delta Function, 122

Vector Identities, 121 Product Rules, 122 Second Derivatives, 122 Triple Products, 121

Linear Algebra, 123 Vectors, 123 Moment of Inertia, see Rotational Motion Newton’s Law of Gravitation, see Celestial Mechanics