Gr9677 Solutions

The Physics GRE Solution Guide

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GR9677 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

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Contents

i xv

Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xv

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Newton’s Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xvi

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Work & Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xvii

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Oscillatory Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xviii

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Rotational Motion about a Fixed Axis . . . . . . . . . . . . . . . . . . . . xxii

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Dynamics of Systems of Particles . . . . . . . . . . . . . . . . . . . . . . . xxiv

0.7

Central Forces and Celestial Mechanics . . . . . . . . . . . . . . . . . . . xxiv

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Three Dimensional Particle Dynamics . . . . . . . . . . . . . . . . . . . . xxvi

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Fluid Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxvi

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0.10 Non-inertial Reference Frames . . . . . . . . . . . . . . . . . . . . . . . . xxvii 0.11 Hamiltonian and Lagrangian Formalism . . . . . . . . . . . . . . . . . . . xxvii

Electromagnetism

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0.12 Electrostatics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxix 0.13 Currents and DC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxiv 0.14 Magnetic Fields in Free Space . . . . . . . . . . . . . . . . . . . . . . . . . xxxiv 0.15 Lorentz Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxiv 0.16 Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxiv 0.17 Maxwell’s Equations and their Applications . . . . . . . . . . . . . . . . . xxxiv 0.18 Electromagnetic Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxiv

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Contents 0.19 AC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxiv 0.20 Magnetic and Electric Fields in Matter . . . . . . . . . . . . . . . . . . . . xxxiv 0.21 Capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxv 0.22 Energy in a Capacitor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxv 0.23 Energy in an Electric Field . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxv 0.24 Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxv 0.25 Current Destiny . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxv 0.26 Current Density of Moving Charges . . . . . . . . . . . . . . . . . . . . . xxxv 0.27 Resistance and Ohm’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . xxxv

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0.28 Resistivity and Conductivity . . . . . . . . . . . . . . . . . . . . . . . . . . xxxvi 0.29 Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxvi 0.30 Kirchoff’s Loop Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxvi 0.31 Kirchoff’s Junction Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxvi 0.32 RC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxvi

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0.33 Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxvi 0.34 Speed of Propagation of a Light Wave . . . . . . . . . . . . . . . . . . . . xxxvii 0.35 Relationship between E and B Fields . . . . . . . . . . . . . . . . . . . . . xxxvii 0.36 Energy Density of an EM wave . . . . . . . . . . . . . . . . . . . . . . . . xxxviii 0.37 Poynting’s Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxviii Optics & Wave Phonomena

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0.38 Wave Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxix

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0.39 Superposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxix 0.40 Interference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxix 0.41 Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxix 0.42 Geometrical Optics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxix 0.43 Polarization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxxix 0.44 Doppler Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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0.45 Snell’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Thermodynamics & Statistical Mechanics

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0.46 Laws of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . xli 0.47 Thermodynamic Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . xli David S. Latchman

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Contents v 0.48 Equations of State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xli 0.49 Ideal Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xli 0.50 Kinetic Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xli 0.51 Ensembles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xli 0.52 Statistical Concepts and Calculation of Thermodynamic Properties . . . xlii 0.53 Thermal Expansion & Heat Transfer . . . . . . . . . . . . . . . . . . . . . xlii 0.54 Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xlii 0.55 Specific Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xlii 0.56 Heat and Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xlii

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0.57 First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . xlii 0.58 Work done by Ideal Gas at Constant Temperature . . . . . . . . . . . . . xliii 0.59 Heat Conduction Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . xliii 0.60 Ideal Gas Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xliv 0.61 Stefan-Boltzmann’s FormulaStefan-Boltzmann’s Equation . . . . . . . . xliv . . . . . . . . . . . . . . . . . . . . . . . . . . xliv

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0.62 RMS Speed of an Ideal Gas

0.63 Translational Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . xliv 0.64 Internal Energy of a Monatomic gas . . . . . . . . . . . . . . . . . . . . . xliv 0.65 Molar Specific Heat at Constant Volume . . . . . . . . . . . . . . . . . . . xlv 0.66 Molar Specific Heat at Constant Pressure . . . . . . . . . . . . . . . . . . xlv 0.67 Equipartition of Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . xlv 0.68 Adiabatic Expansion of an Ideal Gas . . . . . . . . . . . . . . . . . . . . . xlvii

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0.69 Second Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . xlvii Quantum Mechanics

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0.70 Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xlix 0.71 Schrodinger Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xlix ¨ 0.72 Spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . liv 0.73 Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . liv 0.74 Wave Funtion Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . liv 0.75 Elementary Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . liv

Atomic Physics

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0.76 Properties of Electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ©2009

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Contents 0.77 Bohr Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lv 0.78 Energy Quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lvi 0.79 Atomic Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lvi 0.80 Atomic Spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lvi 0.81 Selection Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lvii 0.82 Black Body Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lvii 0.83 X-Rays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lviii

Special Relativity

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0.84 Atoms in Electric and Magnetic Fields . . . . . . . . . . . . . . . . . . . . lix lxiii

0.85 Introductory Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxiii 0.86 Time Dilation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxiii 0.87 Length Contraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxiii 0.88 Simultaneity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxiii 0.89 Energy and Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxiv

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0.90 Four-Vectors and Lorentz Transformation . . . . . . . . . . . . . . . . . . lxv 0.91 Velocity Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxvi 0.92 Relativistic Doppler Formula . . . . . . . . . . . . . . . . . . . . . . . . . lxvi 0.93 Lorentz Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxvi 0.94 Space-Time Interval . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxvii Laboratory Methods

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0.95 Data and Error Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxix 0.96 Instrumentation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxi 0.97 Radiation Detection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxi 0.98 Counting Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxi 0.99 Interaction of Charged Particles with Matter . . . . . . . . . . . . . . . . lxxii 0.100Lasers and Optical Interferometers . . . . . . . . . . . . . . . . . . . . . . lxxii 0.101Dimensional Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxii 0.102Fundamental Applications of Probability and Statistics . . . . . . . . . . lxxii

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0.103Discharge of a Capacitor . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxiii David S. Latchman

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Contents vii 0.104Magnetic Fields & Induced EMFs . . . . . . . . . . . . . . . . . . . . . . . lxxiii 0.105A Charged Ring I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxiv 0.106A Charged Ring II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxiv 0.107Forces on a Car’s Tires . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxv 0.108Block sliding down a rough inclined plane . . . . . . . . . . . . . . . . . lxxv 0.109Collision of Suspended Blocks . . . . . . . . . . . . . . . . . . . . . . . . . lxxvi 0.110Damped Harmonic Motion . . . . . . . . . . . . . . . . . . . . . . . . . . lxxvii 0.111Spectrum of the Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . lxxvii 0.112Internal Conversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxviii

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0.113The Stern-Gerlach Experiment . . . . . . . . . . . . . . . . . . . . . . . . . lxxviii 0.114Positronium Ground State Energy . . . . . . . . . . . . . . . . . . . . . . lxxviii 0.115Specific Heat Capacity and Heat Lost . . . . . . . . . . . . . . . . . . . . . lxxix 0.116Conservation of Heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxix 0.117Thermal Cycles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxix

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0.118Mean Free Path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxx 0.119Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxi 0.120Barrier Tunneling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxii 0.121Distance of Closest Appraoch . . . . . . . . . . . . . . . . . . . . . . . . . lxxxii 0.122Collisions and the He atom . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxiii 0.123Oscillating Hoops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxiii 0.124Mars Surface Orbit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxiv

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0.125The Inverse Square Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxiv 0.126Charge Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxv 0.127Capacitors in Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxvi 0.128Resonant frequency of a RLC Circuit . . . . . . . . . . . . . . . . . . . . . lxxxvi 0.129Graphs and Data Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxvii 0.130Superposition of Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxviii 0.131The Plank Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . lxxxix 0.132The Open Ended U-tube . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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0.133Sphere falling through a viscous liquid . . . . . . . . . . . . . . . . . . . .

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0.134Moment of Inertia and Angular Velocity . . . . . . . . . . . . . . . . . . . xci 0.135Quantum Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . xcii ©2009

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Contents 0.136Invariance Violations and the Non-conservation of Parity . . . . . . . . . xcii 0.137Wave function of Identical Fermions . . . . . . . . . . . . . . . . . . . . . xciii 0.138Relativistic Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xciii 0.139Relativistic Addition of Velocities . . . . . . . . . . . . . . . . . . . . . . . xciii 0.140Relativistic Energy and Momentum . . . . . . . . . . . . . . . . . . . . . xciv 0.141Ionization Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xciv 0.142Photon Emission and a Singly Ionized He atom . . . . . . . . . . . . . . . xcv 0.143Selection Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xcvi 0.144Photoelectric Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xcvi

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0.145Stoke’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xcvii 0.1461-D Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xcvii 0.147High Pass Filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xcvii 0.148Generators and Faraday’s Law . . . . . . . . . . . . . . . . . . . . . . . . xcviii 0.149Faraday’s Law and a Wire wound about a Rotating Cylinder . . . . . . . xcviii

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0.150Speed of π+ mesons in a laboratory . . . . . . . . . . . . . . . . . . . . . . xcix 0.151Transformation of Electric Field . . . . . . . . . . . . . . . . . . . . . . . . xcix 0.152The Space-Time Interval . . . . . . . . . . . . . . . . . . . . . . . . . . . . xcix c

0.154Spherical Harmonics of the Wave Function . . . . . . . . . . . . . . . . .

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0.155Decay of the Positronium Atom . . . . . . . . . . . . . . . . . . . . . . . .

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0.156Polarized Electromagnetic Waves I . . . . . . . . . . . . . . . . . . . . . .

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0.157Polarized Electromagnetic Waves II . . . . . . . . . . . . . . . . . . . . . .

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0.153Wavefunction of the Particle in an Infinte Well . . . . . . . . . . . . . . .

0.158Total Internal Reflection . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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0.159Single Slit Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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0.160The Optical Telescope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cii 0.161Pulsed Lasers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cii 0.162Relativistic Doppler Shift . . . . . . . . . . . . . . . . . . . . . . . . . . . . ciii 0.163Gauss’ Law, the Electric Field and Uneven Charge Distribution . . . . . civ 0.164Capacitors in Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cv 0.165Standard Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cv 0.166Nuclear Binding Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . cv 0.167Work done by a man jumping off a boat . . . . . . . . . . . . . . . . . . . cvi

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Contents ix 0.168Orbits and Gravitational Potential . . . . . . . . . . . . . . . . . . . . . . cvi 0.169Schwartzchild Radius . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cvi 0.170Lagrangian of a Bead on a Rod . . . . . . . . . . . . . . . . . . . . . . . . cvii 0.171Ampere’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cvii 0.172Larmor Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cviii 0.173The Oscilloscope and Electron Deflection . . . . . . . . . . . . . . . . . . cix 0.174Negative Feedback . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cix 0.175Adiabatic Work of an Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . .

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0.176Change in Entrophy of Two Bodies . . . . . . . . . . . . . . . . . . . . . .

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0.177Double Pane Windows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxi 0.178Gaussian Wave Packets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxii 0.179Angular Momentum Spin Operators . . . . . . . . . . . . . . . . . . . . . cxii 0.180Semiconductors and Impurity Atoms . . . . . . . . . . . . . . . . . . . . cxii 0.181Specific Heat of an Ideal Diatomic Gas . . . . . . . . . . . . . . . . . . . . cxii

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0.182Transmission of a Wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxiii 0.183Piano Tuning & Beats . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxiii 0.184Thin Films . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxiv 0.185Mass moving on rippled surface . . . . . . . . . . . . . . . . . . . . . . . cxv 0.186Normal Modes and Couples Oscillators . . . . . . . . . . . . . . . . . . . cxv 0.187Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxv 0.188Charged Particles in E&M Fields . . . . . . . . . . . . . . . . . . . . . . . cxvi

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0.189Rotation of Charged Pith Balls in a Collapsing Magnetic Field . . . . . . cxvi 0.190Coaxial Cable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxvii 0.191Charged Particles in E&M Fields . . . . . . . . . . . . . . . . . . . . . . . cxviii 0.192THIS ITEM WAS NOT SCORED . . . . . . . . . . . . . . . . . . . . . . . cxix 0.193The Second Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . cxix 0.194Small Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxix 0.195Period of Mass in Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . cxx 0.196Internal Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxxi 0.197Specific Heat of a Super Conductor . . . . . . . . . . . . . . . . . . . . . . cxxi 0.198Pair Production . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxxi 0.199Probability Current Density . . . . . . . . . . . . . . . . . . . . . . . . . . cxxii ©2009

David S. Latchman

x

Contents 0.200Quantum Harmonic Oscillator Energy Levels . . . . . . . . . . . . . . . . cxxii 0.201Three Level LASER and Metastable States . . . . . . . . . . . . . . . . . . cxxiii 0.202Quantum Oscillator – Raising and Lowering Operators . . . . . . . . . . cxxiv

Constants & Important Equations

cxxv

Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxxv

.2

Vector Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxxv

.3

Commutators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxxvi

.4

Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxxvii

D

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FT

.1

David S. Latchman

©2009

FT

List of Tables

0.67.1Table of Molar Specific Heats . . . . . . . . . . . . . . . . . . . . . . . . . xlvi 0.119.1 Table of wavefunction amplitudes . . . . . . . . . . . . . . . . . . . . . . lxxxii 0.181.1 Table of degrees of freedom of a Diatomic atom . . . . . . . . . . . . . . . cxiii

D

RA

.1.1 Something . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxxv

List of Tables

D

RA

FT

xii

David S. Latchman

©2009

FT

List of Figures

D

RA

0.201.1 Three Level Laser . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . cxxiv

List of Figures

D

RA

FT

xiv

David S. Latchman

©2009

Classical Mechanics

0.1.1

Kinematics Linear Motion

Average Velocity

∆x x2 − x1 = ∆t t2 − t1

RA

v=

FT

0.1

(0.1.1)

Instantaneous Velocity

∆x dx = = v(t) ∆t→0 ∆t dt

v = lim

(0.1.2)

Kinematic Equations of Motion

D

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

0.1.2

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

(0.1.3) (0.1.4) (0.1.5) (0.1.6)

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 Acceleration

Classical Mechanics

xvi Centripetal Acceleration a=

v2 r

(0.1.7)

ω=

v r

(0.1.8)

Angular Velocity

We can write eq. (0.1.7) in terms of ω

FT

a = ω2 r Rotational Equations of Motion

(0.1.9)

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

0.2 0.2.1

RA

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

(0.1.10) (0.1.11) (0.1.12) (0.1.13)

Newton’s Laws

Newton’s Laws of Motion

D

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

(0.2.1)

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

0.2.2

(0.2.2)

Momentum p = mv

David S. Latchman

(0.2.3) ©2009

Work & Energy

0.2.3

xvii

Impulse w

∆p = J =

0.3 0.3.1

Fdt = Favg dt

(0.2.4)

Work & Energy Kinetic Energy 1 K ≡ mv2 2

FT

0.3.2

The Work-Energy Theorem

The net Work done is given by

Wnet = K f − Ki

(0.3.2)

Work done under a constant Force

RA

0.3.3

(0.3.1)

The work done by a force can be expressed as

W = F∆x

(0.3.3)

W = F · ∆r = F∆r cos θ

(0.3.4)

In three dimensions, this becomes

D

For a non-constant force, we have

0.3.4

W=

wx f

F(x)dx

(0.3.5)

xi

Potential Energy

The Potential Energy is dU(x) dx for conservative forces, the potential energy is F(x) = −

U(x) = U0 −

wx

F(x0 )dx0

(0.3.6)

(0.3.7)

x0

©2009

David S. Latchman

Classical Mechanics

xviii

0.3.5

Hooke’s Law F = −kx

(0.3.8)

where k is the spring constant.

0.3.6

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

0.4.1

Oscillatory Motion

FT

0.4

(0.3.9)

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

(0.4.1)

0.4.2

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=

(0.4.2)

Total Energy of an Oscillating System

D

0.4.3

2π ω

Given that

x = A sin (ωt + δ)

(0.4.3)

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

(0.4.4)

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

(0.4.5) ©2009

Oscillatory Motion The Potential Energy is

xix

1 U = kx2 2 1 = kA2 sin2 (ωt + δ) 2 Adding eq. (0.4.5) and eq. (0.4.6) gives

(0.4.6)

1 E = kA2 2

0.4.4

(0.4.7)

Damped Harmonic Motion

dx (0.4.8) dt where b is the damping coefficient. The equation of motion for a damped oscillating system becomes dx d2 x − kx − b = m 2 (0.4.9) dt dt Solving eq. (0.4.9) goves x = Ae−αt sin (ω0 t + δ) (0.4.10)

RA

We find that

FT

Fd = −bv = −b

α=

b 2m r

k b2 − m 4m2

ω0 =

r

D

=

0.4.5

(0.4.11)

ω20 −

b2 4m2

q = ω20 − α2

(0.4.12)

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

(0.4.13)

Small Oscillations

The Energy of a system is

We can solve for v(x), r

2 (E − V(x)) (0.4.14) 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 ) + ··· (0.4.15) dx x=xe 2 dx2 x=xe v(x) =

©2009

David S. Latchman

xx 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

(0.4.16)

# ≥0

(0.4.17)

x=xe

As V(xe ) is constant, it has no consequences to physical motion and can be dropped. We see that eq. (0.4.16) is that of simple harmonic motion.

Coupled Harmonic Oscillators

FT

0.4.6

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

(0.4.18)

RA

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

(0.4.19) (0.4.20)

y¨ + ω0 y = 0

(0.4.21)

eq. (0.4.18) becomes

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

(0.4.22)

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

(0.4.23)

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

(0.4.24)

Add figure with coupled pendulum-spring system

David S. Latchman

©2009

Oscillatory Motion The equations of motion are

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

(0.4.25) (0.4.26)

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

(0.4.27)

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

(0.4.28) (0.4.29)

(0.4.30)

(0.4.31)

ω2 =

RA

 g  k   =   m ` ω2 =  (0.4.32)  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

(0.4.33)

D

Substituting this into the equation of motion yields y1 = y2

(0.4.34)

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

(0.4.35)

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

(0.4.36)

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

xxii

0.4.7

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

(0.4.37)

  v v = f 0 λ0 v − vs

(0.4.38)

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 | (0.4.39) The modified frequency becomes

  v0 vr = f0 1 + λ v

RA

f0 =

(0.4.40)

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

λ0 =

(0.4.41) (0.4.42)

To give a modified frequency of

  v0 v − vr f = 0 = f0 λ v − vs

D

0

0.5

0.5.1

(0.4.43)

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

0.5.2

R2 dm

(0.5.1)

Rotational Kinetic Energy 1 K = Iω2 2

David S. Latchman

(0.5.2) ©2009

Rotational Motion about a Fixed Axis

0.5.3

0.5.4

xxiii

Parallel Axis Theorem I = Icm + Md2

(0.5.3)

τ=r×F τ = Iα

(0.5.4) (0.5.5)

Torque

0.5.5

FT

where α is the angular acceleration.

Angular Momentum

L = Iω

(0.5.6)

dL dt

(0.5.7)

RA

we can find the Torque

τ=

0.5.6

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 (0.5.8) K = Krot = Icontact ω2 2 Icontact can be found from the Parallel Axis Theorem. Icontact = Icm + MR2

(0.5.9)

Substitute eq. (0.5.8) and we have  1 Icm + MR2 ω2 2 1 1 = Icm ω2 + mv2 2 2

K=

(0.5.10)

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

xxiv

0.6

Dynamics of Systems of Particles

0.6.1

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

(0.6.1)

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

FT

V=

(0.6.2)

Acceleration Vector of a System of Particles

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

0.7 0.7.1

RA

A=

(0.6.3)

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

D



0.7.2

0.7.3

(0.7.1)

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

GMm r

(0.7.2)

Escape Speed and Orbits

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

(0.7.3) ©2009

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

xxv (0.7.4)

Solving for vesc we find r vesc =

0.7.4

2GM Re

(0.7.5)

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

(0.7.6)

RA

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

0.7.5

Types of Orbits

The Energy of an Orbiting Body is defined in eq. (0.7.3), 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

(0.7.7)

The Orbital Velocity is r v=

GM r

(0.7.8)

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 

 (0.7.9)

where a is the semi-major axis ©2009

David S. Latchman

xxvi 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

(0.7.10)

The Orbital Velocity is r v = vesc =

2GM r

(0.7.11)

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

0.7.6

GM a

FT

v∞ =

(0.7.12)

Derivation of Vis-viva Equation

The total energy of a satellite is

RA

1 GMm E = mv2 − 2 r

(0.7.13)

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

GMm 2a

(0.7.14)

Equating we get

2 1 v = GM − r a 



(0.7.15)

D

2

0.8

Three Dimensional Particle Dynamics

0.9

Fluid Dynamics

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

0.9.1

Equation of Continuity ρ1 v1 A1 = ρ2 v2 A2

David S. Latchman

(0.9.1) ©2009

Non-inertial Reference Frames

0.9.2

xxvii

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

(0.9.2)

0.10

Non-inertial Reference Frames

0.11

Hamiltonian and Lagrangian Formalism

0.11.1

Lagrange’s Function (L)

FT

L=T−V

(0.11.1)

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

0.11.2

Equations of Motion(Euler-Lagrange Equation)

0.11.3

!

RA

d ∂L ∂L = ∂q dt ∂q˙

Hamiltonian

H =T+V ˙ = pq˙ − L(q, q)

D

where

©2009

(0.11.2)

∂H = q˙ ∂p ∂H ∂L =− ∂q ∂x = −p˙

(0.11.3)

(0.11.4)

(0.11.5)

David S. Latchman

Classical Mechanics

D

RA

FT

xxviii

David S. Latchman

©2009

0.12

Electrostatics

0.12.1

Coulomb’s Law

FT

Electromagnetism

The force between two charged particles, q1 and q2 is defined by Coulomb’s Law. ! q1 q2 1 rˆ12 F12 = (0.12.1) 4π0 r212 where 0 is the permitivitty of free space, where

0.12.2

RA

0 = 8.85 × 10−12 C2 N.m2

(0.12.2)

Electric Field of a point charge

D

The electric field is defined by mesuring the magnitide and direction of an electric force, F, acting on a test charge, q0 . F (0.12.3) E≡ q0 The Electric Field of a point charge, q is E=

1 q rˆ 4π0 r2

(0.12.4)

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

(0.12.5)

Electric Fields and Continuous Charge Distributions If a source is distributed continuously along a region of space, eq. (0.12.5) becomes Z 1 1 E(r) = rˆdq (0.12.6) 4π0 r2

xxx Electromagnetism If the charge was distributed along a line with linear charge density, λ, λ=

dq dx

(0.12.7)

The Electric Field of a line charge becomes λ rˆdx r2

Z

1 E(r) = 4π0

(0.12.8)

line

FT

In the case where the charge is distributed along a surface, the surface charge density is, σ dq Q σ= = (0.12.9) A dA The electric field along the surface becomes

σ rˆdA r2

Z

1 E(r) = 4π0

(0.12.10)

Surface

RA

In the case where the charge is distributed throughout a volume, V, the volume charge density is dq Q = (0.12.11) ρ= V dV The Electric Field is

ρ rˆdV r2

Z

1 E(r) = 4π0

(0.12.12)

D

Volume

0.12.3

Gauss’ Law

The electric field through a surface is I Φ=

I dΦ =

surface S

E · dA

(0.12.13)

surface S

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

Q 0

(0.12.14)

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

©2009

Electrostatics

0.12.4

xxxi

Equivalence of Coulomb’s Law and Gauss’ Law

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

q 0

(0.12.15)

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

0.12.5

q 4π0 r2

(0.12.16)

Electric Field due to a line of charge

top surface

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 (0.12.17) 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 (0.12.18) At the side, the electric field is parallel to the area vector, thus E · dA = EdA

Thus the flux becomes,

Z

(0.12.19)

Z

Φ=

E · dA = E

dA

(0.12.20)

side sirface

D

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

(0.12.21)

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

0.12.6

E=

λ 2π0 r

(0.12.22)

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 (0.12.23) 3 V πR 3 ©2009

David S. Latchman

Electromagnetism

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

Q 4π0 r2

(0.12.24)

As the charge is evenly distributed throughout the sphere’s volume we can say that the charge density is dq = ρdV (0.12.25)

0.12.7

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 (0.12.26) = 4 3 π R 0 3

0.12.8

1 qq0 r 4π0

RA

U(r) =

(0.12.27)

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)

(0.12.28)

1 q 4π0 r

(0.12.29)

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`

(0.12.30)

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`

(0.12.31)

a

David S. Latchman

©2009

Electrostatics This can be further expressed

xxxiii

Z

b

V(b) − V(a) =

(∇V) · d`

(0.12.32)

a

And we can show that

0.12.9

E = −∇V

(0.12.33)

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 = (0.12.34) 4π0 r dq = λdx

(0.12.35)

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

(0.12.36)

(0.12.37)

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. (0.12.34), 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  (0.12.38)  4π0  d ©2009

David S. Latchman

Electromagnetism

xxxiv

0.13

Currents and DC Circuits

2

0.14

Magnetic Fields in Free Space

3

Lorentz Force

4

5

0.17 6

Maxwell’s Equations and their Applications

Electromagnetic Waves

D

0.18

Induction

RA

0.16

FT

0.15

7

0.19

AC Circuits

8

0.20

Magnetic and Electric Fields in Matter

9 David S. Latchman

©2009

Capacitance

0.21

xxxv

Capacitance Q = CV

0.22

(0.21.1)

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

Energy in an Electric Field

U 0 E2 = volume 2

(0.23.1)

dQ dt

(0.24.1)

J · dA

(0.25.1)

u≡

0.24

Current

I≡

Current Destiny

D

0.25

0.26

Z I= A

Current Density of Moving Charges J=

0.27

I = ne qvd A

(0.26.1)

Resistance and Ohm’s Law R≡

©2009

(0.22.1)

RA

0.23

FT

U=

V I

(0.27.1) David S. Latchman

Electromagnetism

xxxvi

0.28

Resistivity and Conductivity L A

(0.28.1)

E = ρJ

(0.28.2)

J = σE

(0.28.3)

R=ρ

Power

FT

0.29

P = VI

Kirchoff’s Loop Rules

Write Here

0.31

Kirchoff’s Junction Rule

Write Here

RC Circuits

D

0.32

RA

0.30

(0.29.1)

E − IR −

0.33

Maxwell’s Equations

0.33.1

Integral Form

Q =0 C

(0.32.1)

Gauss’ Law for Electric Fields w closed surface

David S. Latchman

E · dA =

Q 0

(0.33.1)

©2009

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

xxxvii w

B · dA = 0

(0.33.2)

closed surface

Amp`ere’s Law

z

d w B · ds = µ0 I + µ0 0 dt

E · dA

(0.33.3)

surface

Faraday’s Law

z

d w E · ds = − dt

B · dA

(0.33.4)

0.33.2

Differential Form

Gauss’ Law for Electric Fields

FT

surface

ρ 0

(0.33.5)

∇·B=0

(0.33.6)

∇·E=

Gauss’ Law for Magnetism

RA

Amp`ere’s Law

∇ × B = µ0 J + µ0 0

Faraday’s Law

∇·E=−

∂B ∂t

(0.33.7)

(0.33.8)

Speed of Propagation of a Light Wave

D

0.34

∂E ∂t

c= √

1 µ0 0

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

(0.34.1)

(0.34.2)

where n is the refractive index.

0.35

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

©2009

(0.35.1) (0.35.2) David S. Latchman

Electromagnetism

xxxviii

0.36

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

0.37

! (0.36.1)

Poynting’s Vector 1 E×B µ0

(0.37.1)

D

RA

FT

S=

David S. Latchman

©2009

Optics & Wave Phonomena Wave Properties

1

0.39

Superposition

0.40 3

Interference

Diffraction

D

0.41

RA

2

4

0.42

Geometrical Optics

5

0.43 6

FT

0.38

Polarization

Optics & Wave Phonomena

xl

0.44

Doppler Effect

7

0.45

Snell’s Law

0.45.1

Snell’s Law n1 sin θ1 = n2 sin θ2

Critical Angle and Snell’s Law

FT

0.45.2

(0.45.1)

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. (0.45.1), θ1 = 90 and θ2 = θc and n2 > n1 .

(0.45.2)

D

RA

n1 sin 90 = n2 sinθc n1 sin θc = n2

David S. Latchman

©2009

Thermodynamics & Statistical Mechanics 0.46

Laws of Thermodynamics

0.47

FT

1

Thermodynamic Processes

0.48 3

Equations of State

Ideal Gases

D

0.49

RA

2

4

0.50

Kinetic Theory

5

0.51 6

Ensembles

Thermodynamics & Statistical Mechanics

xlii

0.52

Statistical Concepts and Calculation of Thermodynamic Properties

7

0.53

Thermal Expansion & Heat Transfer

0.54

FT

8

Heat Capacity

  Q = C T f − Ti

(0.54.1)

0.55

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

(0.55.1)

D

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

0.56

0.57

Heat and Work Z W=

Vf

PdV

(0.56.1)

Vi

First Law of Thermodynamics dEint = dQ − dW

(0.57.1)

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

0.57.1

xliii

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

(0.57.2)

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 (0.57.3) 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

(0.57.4)

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 (0.57.5)

0.58

Work done by Ideal Gas at Constant Temperature

D

Starting with eq. (0.56.1), we substitute the Ideal gas Law, eq. (0.60.1), to get

0.59

Vf

Z W = nRT

Vi

= nRT ln

dV V

Vf Vi

(0.58.1)

Heat Conduction Equation

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

Q TH − TC = kA t L

(0.59.1)

where k is the thermal conductivity. ©2009

David S. Latchman

Thermodynamics & Statistical Mechanics

xliv

0.60

Ideal Gas Law PV = nRT

(0.60.1)

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

(0.60.2)

where k is the Boltzmann’s Constant, such that

0.61

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

0.62

0.63

r vrms =

3RT M

(0.62.1)

Translational Kinetic Energy 3 K¯ = kT 2

0.64

(0.61.1)

(0.63.1)

Internal Energy of a Monatomic gas 3 Eint = nRT 2

David S. Latchman

(0.64.1) ©2009

Molar Specific Heat at Constant Volume

0.65

xlv

Molar Specific Heat at Constant Volume

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

(0.65.1)

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

(0.65.2)

At constant volume, W = 0, and we get

Substituting eq. (0.64.1), we get

1 ∆Eint n ∆T

FT

CV =

0.66

RA

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

(0.65.3)

(0.65.4)

Molar Specific Heat at Constant Pressure

Starting with

D

and

0.67

Q = nCp ∆T

(0.66.1)

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

(0.66.2)

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

(0.67.1)

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 0.67.1: Table of Molar Specific Heats

D

Thermodynamics & Statistical Mechanics xlvi

Adiabatic Expansion of an Ideal Gas

0.68

Adiabatic Expansion of an Ideal Gas

where γ = CCVP . We can also write

0.69

xlvii

PV γ = a constant

(0.68.1)

TV γ−1 = a constant

(0.68.2)

Second Law of Thermodynamics

D

RA

FT

Something.

©2009

David S. Latchman

Thermodynamics & Statistical Mechanics

D

RA

FT

xlviii

David S. Latchman

©2009

Quantum Mechanics 0.70

Fundamental Concepts

0.71

FT

1

Schrodinger ¨ Equation

Let us define Ψ to be

Ψ = Ae−iω(t− v ) x

(0.71.1)

RA

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

i(Et−px) ~

(0.71.2)

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

(0.71.3)

To determine E and p,

D

p2 ∂2 Ψ = − Ψ ~2 ∂x2 ∂Ψ iE = Ψ ~ ∂t

and

(0.71.4) (0.71.5)

p2 H= +V 2m

(0.71.6)

This becomes

EΨ = HΨ

(0.71.7)

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

EΨ = −

∂Ψ ~ 2 ∂2 Ψ =− + V(x)Ψ 2m ∂x2 ∂t The Time Independent Schrodinger’s ¨ Equation is i~

EΨ = −

~ 2 ∂2 Ψ + V(x)Ψ 2m ∂x2

(0.71.8)

(0.71.9)

Quantum Mechanics

l

0.71.1

Infinite Square Wells

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, 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. (0.71.9), we substitute V(x) ¨ and in the region (−a/2, a/2), we get ~2 d2 ψ = Eψ 2m dx2

This differential is of the form

RA

d2 ψ + k2 ψ = 0 dx2 where r 2mE k= ~2 We recognize that possible solutions will be of the form cos kx

(0.71.10)

FT

−

(0.71.11)

(0.71.12)

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

D

We have known boundary conditions for our square well. ψ(0) = ψ(a) = 0

(0.71.13)

⇒ A cos 0 + B sin 0 = 0 ∴A=0

(0.71.14)

It shows that

We are now left with

B sin ka = 0 ka = 0; π; 2π; 3π; · · · (0.71.15) While mathematically, n can be zero, that would mean there would be no wave function, so we ignore this result and say kn = David S. Latchman

nπ a

for n = 1, 2, 3, · · · ©2009

Schr¨odinger Equation Substituting this result into eq. (0.71.12) gives √ nπ 2mEn kn = = a ~

li

(0.71.16)

Solving for En gives

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 =

r ψn (x) =

0.71.2

Harmonic Oscillators

(0.71.18)

FT

So we can write the wave function as

(0.71.17)

  2 nπx sin a a

(0.71.19)

RA

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

(0.71.20)

So the force experienced by this particle is F=−

dV = −kx dx

(0.71.21)

D

where k is the spring constant. The equation of motion can be summed us as m

d2 x = −kx dt2

(0.71.22)

And the solution of this equation is   x(t) = A cos ω0 t + φ

(0.71.23)

where the angular frequency, ω0 is r ω0 =

k m

(0.71.24)

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. (0.71.20) we substitute into eq. (0.71.9) to get !   d2 ψ 2m k 2 mk 2 2E = 2 x −E ψ= 2 x − ψ dx2 ~ 2 ~ k ©2009

David S. Latchman

Quantum Mechanics

lii With some manipulation, we get √ r  d2 ψ  mk 2 2E m   ψ =  x − √  2 dx ~ ~ k mk ~

This step allows us to to keep some of constants out of the way, thus giving us √ mk 2 ξ2 = x (0.71.25) ~r 2E m 2E (0.71.26) and λ = = ~ k ~ω0 This leads to the more compact

FT

 d2 ψ  2 = ξ − λ ψ dξ2

(0.71.27)

where the eigenfunction ψ will be a function of ξ. λ assumes an eigenvalue anaglaous to E.

RA

From eq. (0.71.25), we see that the maximum value can be determined to be √ mk 2 2 ξmax = A (0.71.28) ~ Using the classical connection between A and E, allows us to say √ mk 2E 2 ξmax = =λ ~ k

(0.71.29)

From eq. (0.71.27), we see that in a quantum mechanical oscillator, there are nonvanishing solutions in the forbidden regions, unlike in our classical case.

D

A solution to eq. (0.71.27) is

ψ(ξ) = e−ξ /2 2

(0.71.30)

where

and

dψ 2 = −ξe−ξ /2 dξ   2 dψ 2 −xi2 /2 −ξ2 /2 2 = ξ e − e = ξ − 1 e−ξ /2 dξ2

This gives is a special solution for λ where λ0 = 1

(0.71.31)

Thus eq. (0.71.26) gives the energy eigenvalue to be E0 = David S. Latchman

~ω0 ~ω0 λ0 = 2 2

(0.71.32) ©2009

Schr¨odinger Equation liii −ξ2 /2 The eigenfunction e corresponds to a normalized stationary-state wave function !1 mk 8 − √mk x2 /2~ −iE0 t/~ e (0.71.33) Ψ0 (x, t) = 2 2 e π~ This solution of eq. (0.71.27) 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.

0.71.3

Finite Square Well

FT

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= ~

RA

−

where

This gives us solutions that are

ψ(x) = A exp(−κx) + B exp(κx)

D

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

(0.71.34)

Region II: −a < x < a In this region, our potential is V(x) = V0 . Substitutin this into the Schrodinger’s Equation, eq. (0.71.9), gives ¨ ~2 d2 ψ − V0 ψ = Eψ − 2m dx2 d2 ψ or = −l2 ψ 2 dx p 2m (E + V0 ) where l ≡ (0.71.35) ~ 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 (0.71.36)

©2009

David S. Latchman

liv Quantum Mechanics 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

0.71.4

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

Hydrogenic Atoms

c

3

0.73 4

(0.71.38)

Angular Momentum

Wave Funtion Symmetry

D

0.74

Spin

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

RA

0.72

(0.71.37)

FT

This gives us

for x > a

5

0.75

Elementary Perturbation Theory

6

David S. Latchman

©2009

Atomic Physics 0.76

Properties of Electrons

0.77

FT

1

Bohr Model

RA

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 1 e2 me v2 = 4π0 r2 r

(0.77.1)

The Total Energy is the sum of the potential and kinetic energies, so E=K+U =

p2 − | f race2 4π0 r 2me

(0.77.2)

D

We can further reduce this equation by subsituting the value of momentum, which we find to be p2 1 e2 = me v2 = (0.77.3) 2me 2 8π0 r Substituting this into eq. (0.77.2), we get E=

e2 e2 e2 − =− 8π0 r 4π0 r 8π0 r

(0.77.4)

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 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.

lvi

Atomic Physics 2. The energy of these stationary states are determined by their angular momentum which must take on quantized values of ~. L = n~

(0.77.5)

We can find the angular momentum of a circular orbit. L = m3 vr

(0.77.6)

From eq. (0.77.1) we find v and by substitution, we find L. r m3 r L=e 4π0

(0.77.7)

Solving for r, gives (0.77.8)

n2 ~2 = n2 a0 rn = 2 me e /4π0

(0.77.9)

a0 = 0.53 × 10−10 m

(0.77.10)

RA

where a0 is the Bohr radius.

FT

L2 me e2 /4π0 We apply the condition from eq. (0.77.5) r=

Having discreet values for the allowed radii means that we will also have discreet values for energy. Replacing our value of rn into eq. (0.77.4), we get ! me e2 13.6 En = − 2 = − 2 eV (0.77.11) 2n 4π0 ~ n

0.78

D

3

Energy Quantization

0.79

Atomic Structure

4

0.80

Atomic Spectra

0.80.1

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

David S. Latchman

(0.80.1) ©2009

Selection Rules where RH is the Rydberg constant.

lvii

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.

0.81

Selection Rules

6

Black Body Radiation

0.82.1

Plank Formula

FT

0.82

f3 8π~ u( f, T) = 3 h f /kT c e −1

Stefan-Boltzmann Formula

RA

0.82.2

P(T) = σT4

0.82.3

(0.82.2)

Wein’s Displacement Law

λmax T = 2.9 × 10−3 m.K

(0.82.3)

Classical and Quantum Aspects of the Plank Equation

D

0.82.4

(0.82.1)

Rayleigh’s Equation

8π f 2 kT (0.82.4) c3 We can get this equation from Plank’s Equation, eq. (0.82.1). 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 u( f, T) =

ex ' 1 + x

(0.82.5)

Thus the demonimator in eq. (0.82.1) becomes eh f /kT − 1 ' 1 + ©2009

hf hf −1= kT kT

(0.82.6) David S. Latchman

Atomic Physics

lviii Thus eq. (0.82.1) takes the approximate form 8πh 3 kT 8π f 2 = 3 kT u( f, T) ' 3 f c hf c

(0.82.7)

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

FT

At large frequencies, where h f > kT, quantum effects become apparent. We can estimate that eh f /kT − 1 ' eh f /kT (0.82.8) Thus eq. (0.82.1) becomes

u( f, T) '

X-Rays

0.83.1

Bragg Condition

RA

0.83

8πh 3 −h f /kT f e c3

2d sin θ = mλ

(0.82.9)

(0.83.1)

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

0.83.2

The Compton Effect

D

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 =

hc λ

(0.83.2)

The photon has an associated momentum E

= pc E hυ h ⇒p = = = c c λ

(0.83.3) (0.83.4)

The Relativistic Energy for the electron is

where

David S. Latchman

E2 = p2 c2 + m2e c4

(0.83.5)

p − p0 = P

(0.83.6) ©2009

Atoms in Electric and Magnetic Fields Squaring eq. (0.83.6) gives p2 − 2p · p0 + p02 = P2

lix (0.83.7)

Recall that E = pc and E 0 = cp0 , we have c2 p2 − 2c2 p · p0 + c2 p02 = c2 P2 E 2 − 2E E 0 cos θ + E 02 = E2 − m2e c4

(0.83.8)

Conservation of Energy leads to E + me c2 = E 0 + E

(0.83.9)

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

(0.83.10) (0.83.11)

FT

Solving

Solving leads to

∆λ = λ0 − λ = h me c

(0.83.12)

is the Compton Wavelength.

h = 2.427 × 10−12 m me c

RA

where λc =

h (1 − cos θ) me c

λc =

0.84

Atoms in Electric and Magnetic Fields

0.84.1

The Cyclotron Frequency

(0.83.13)

D

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

(0.84.1)

or more simply FB = qvB. As this traces a circular path, from Newton’s Second Law, we see that mv2 FB = = qvB (0.84.2) R Solving for R, we get mv R= (0.84.3) qB

We also see that

qB (0.84.4) 2πm The frequency is depends on the charge, q, the magnetic field strength, B and the mass of the charged particle, m. f =

©2009

David S. Latchman

Atomic Physics

lx

0.84.2

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. 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.

FT

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 (0.84.5) υ0 = 2π me 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

RA

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. (0.84.5), to emiminate k, to get

eB υ − υ0 = 0 2πme eB υ2 + υ − υ0 = 0 2πme

D

υ2 −

(Clockwise) (Counterclockwise)

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

(0.84.6)

for counterclockwise motion

(0.84.7)

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

eB 4πme

(0.84.8)

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. David S. Latchman

©2009

Atoms in Electric and Magnetic Fields

0.84.3

lxi

Franck-Hertz Experiment

D

RA

FT

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.2 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.

2

Put drawing of Franck-Hertz Setup

©2009

David S. Latchman

Atomic Physics

D

RA

FT

lxii

David S. Latchman

©2009

Special Relativity Introductory Concepts

0.85.1

Postulates of Special Relativity

FT

0.85

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

0.86

1 γ= q 1−

RA

We can define

(0.85.1)

u2 c2

Time Dilation

∆t = γ∆t0

(0.86.1)

D

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

0.87

Length Contraction L=

L0 γ

(0.87.1)

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.

0.88 4

Simultaneity

Special Relativity

lxiv

0.89

Energy and Momentum

0.89.1

Relativistic Momentum & Energy

In relativistic mechanics, to be conserved, momentum and energy are defined as Relativistic Momentum p¯ = γmv¯

(0.89.1)

E = γmc2

(0.89.2)

0.89.2

FT

Relativistic Energy

Lorentz Transformations (Momentum & Energy) E = γ px − β c 0 py = py p0x





= pz   E E =γ − βpx c c

0.89.4

(0.89.4) (0.89.5) (0.89.6)

Relativistic Kinetic Energy

D

0.89.3

RA

p0z 0

(0.89.3)

K = E − mc2    1 2 = mc  q  1−
= mc2 γ − 1

v2 c2

   − 1 

(0.89.7) (0.89.8) (0.89.9)

Relativistic Dynamics (Collisions)   ∆E ∆P0x = γ ∆Px − β c 0 ∆P y = ∆P y ∆P0z = ∆Pz   ∆E0 ∆E =γ − β∆Px c c

David S. Latchman

(0.89.10) (0.89.11) (0.89.12) (0.89.13) ©2009

Four-Vectors and Lorentz Transformation

0.90

lxv

Four-Vectors and Lorentz Transformation

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

(0.90.1)

FT

A different Lorents frame, S0 , corresponds to another set of space time axes so that  0   x   y0  (0.90.2) s0 =  0   z   0  ict

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

RA

The Lorentz Transformation is related by the matrix  0   0 0 iγβ  x   γ  y0   0 1 0 0     0  =  0 1 0  z   0 ict0 −iγβ 0 0 γ

(0.90.3)

We can express the equation in the form

s0 = L s

(0.90.4)

D

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   = x2 + y2 + z2 − c2 t2 sT s = x y z ict  (0.90.5)  z    ict

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

(0.90.6)

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  (0.90.7)  bz    ibt this can be transformed into a set of quantities of b0 in another frame S0 such that it satisfies the transformation b0 = L b (0.90.8) ©2009

David S. Latchman

lxvi Looking at the momentum-Energy four vector, we have    px   p    p =  y   pz    iE/c

Special Relativity

(0.90.9)

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

(0.90.10)

We can also get a Lorentz-invariation relation between momentum and energy such that p0T p0 = pT p (0.90.11)

02 02 p02 x + p y + pz −

0.91

FT

The resulting equality gives E02 E2 2 2 2 = p + p + p − x y z c2 c2

Velocity Addition

0.92

υ¯ = υ0

D

υ¯ receding = rυ0 υ0 υ¯ approaching = r

0.93

(0.91.1)

Relativistic Doppler Formula r

We have

v−u 1 − uv c2

RA

v0 =

(0.90.12)

c+u c−u

r

let r =

c−u c+u

(0.92.1)

red-shift (Source Receding)

(0.92.2)

blue-shift (Source Approaching)

(0.92.3)

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, x0 = γ (x − ut) y0 = y z0 = y   u 0 t = γ t − 2x c David S. Latchman

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

(0.93.1) (0.93.2) (0.93.3) (0.93.4) ©2009

Space-Time Interval

0.94

lxvii

Space-Time Interval
(∆S)2 = (∆x)2 + ∆y 2 + (∆z)2 − c2 (∆t)2

(0.94.1)

Space-Time Intervals may be categorized into three types depending on their separation. They are Time-like Interval c2 ∆t2 > ∆r2

(0.94.2)

∆S > 0

(0.94.3)

2

Light-like Interval

c2 ∆t2 = ∆r2

(0.94.4)

S2 = 0

(0.94.5)

RA

Space-like Intervals

FT

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

(0.94.6) (0.94.7)

D

c2 ∆t2 < ∆r2 ∆S < 0

©2009

David S. Latchman

Special Relativity

D

RA

FT

lxviii

David S. Latchman

©2009

Laboratory Methods Data and Error Analysis

0.95.1

Addition and Subtraction

FT

0.95

x=a+b−c

(0.95.1)

(δx)2 = (δa)2 + (δb)2 + (δc)2

(0.95.2)

0.95.2

RA

The Error in x is

Multiplication and Division x=

a×b c

The error in x is

D



0.95.3

2

δa = a 

2

δb + b

!2

δc + c 

2 (0.95.4)

Exponent - (No Error in b)

The Error in x is

0.95.4

δx x

(0.95.3)

x = ab

(0.95.5)

  δa δx =b x a

(0.95.6)

x = ln a

(0.95.7)

Logarithms

Base e

lxx We find the error in x by taking the derivative on both sides, so

Laboratory Methods

d ln a · δa da 1 = · δa a δa = a

δx =

(0.95.8)

Base 10 x = log10 a

δx =

FT

The Error in x can be derived as such

d(log a) δa da ln a ln 10

δa da 1 δa = ln 10 a δa = 0.434 a

0.95.5 Base e

RA

=

(0.95.9)

(0.95.10)

Antilogs

x = ea

(0.95.11)

ln x = a ln e = a

(0.95.12)

D

We take the natural log on both sides.

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

(0.95.13)

Base 10 x = 10a David S. Latchman

(0.95.14) ©2009

Instrumentation 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

Instrumentation

2

0.97

Radiation Detection

3

Counting Statistics

RA

0.98

(0.95.15)

FT

0.96

lxxi

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 (0.98.1)

D

Thus we can report our results as

Number of counts = N ±

We can find the count rate by dividing by T, so √ N N R= ± T T The fractional uncertainty of our count is rate. δR = R

©2009

δN . N

δN T N T

√ N

(0.98.2)

(0.98.3)

We can relate this in terms of the count

δN N √ N = N 1 = N =

(0.98.4) David S. Latchman

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

0.99

Interaction of Charged Particles with Matter

5

0.100

Lasers and Optical Interferometers

0.101

FT

6

Dimensional Analysis

0.102

Fundamental Applications of Probability and Statistics

D

8

RA

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.

David S. Latchman

©2009

GR9677 Exam Solutions 0.103

Discharge of a Capacitor

FT

The voltage of a capacitor follows an exponential decay   t V(t) = V0 exp − RC

(0.103.1)

Answer: (B)

Magnetic Fields & Induced EMFs

D

0.104

RA

When the switch is toggled in the a position, the capacitor is quickly charged and the potential across its plates is V. r is small and we assume that the potential difference across it is negligible. When the switch is toggled on the b position, the voltage across the capacitor begins to decay. We can find the current through the resistor, R, from Ohm’s Law   V(t) t = V0 exp − I(t) = (0.103.2) R RC At t = 0 V0 = V. Graph B, shows an exponential decay.

We have a circuit loop that is placed in a decaying magnetic field where the field direction acts into the page. We have two currents in the circuit. The first is due to the battery and the other is an induced current from the changing magnetic field. We can easily determine the current of the cell from Ohm’s Law. Ic =

V 5.0 = A R 10

(0.104.1)

The induced EMF from the magnetic field is found from Faraday’s Law of Induction E =−

dΦ dt

(0.104.2)

where Φ is the magnetic flux. dB dΦ =A dt dt

(0.104.3)

lxxiv GR9677 Exam Solutions 2 The area of our loop, A = 10 × 10cm . So the induced EMF is E = −100 × 10−4 × 150 = 1.5 V

(0.104.4)

The field acts into the page, we consider this a negative direction, it’s decaying, also negative. So negative × negative = positive (0.104.5) Faraday’s Law of Induction has a negative sign. So we expect our EMF to be negative. Using the Right Hand Grip Rule, and pointing our thumb into the page, our fingers curl in the clockwise direction. So we see that the current from our cell goes in the counter-clockwise direction and the induced current in the clockwise direction; they oppose each other. The total EMF is

The current through the resistor is

3.5 = 0.35 A 10

I=

RA

Answer: (B)

0.105

(0.104.6)

FT

V = 5.0 − 1.5 = 3.5 volt

(0.104.7)

A Charged Ring I

The Electric Potential is

V=

Q 4π0 r

(0.105.1)

The distance, r, of P from the charged ring is found from the pythagorean theorem

D

r2 = R2 + x2

(0.105.2)

Plugging this into the above equation yields V=

Q √ 4π0 R2 + x2

(0.105.3)

Answer: (B)

0.106

A Charged Ring II

The force a small charge, q experiences if placed in the center of the ring can be found from Coulomb’s Law qQ (0.106.1) F= 4π0 R2 David S. Latchman

©2009

Forces on a Car’s Tires If it undergoes small oscillations, R >> x, then

lxxv

F = mRω2

(0.106.2)

Equating the two equations, and solving for ω, we get r qQ ω= 4π0 mR3

(0.106.3)

Answer: (A)

Forces on a Car’s Tires

FT

0.107

The horizontal force on the car’s tires is the sum of two forces, the cetripetal force and the frictional force of the road. The cetripetal force acts towards the center, FA , while the frictional force acts in the forwards direction, FC . If it’s not immediately clear why it acts in the forward direction, the tires, as they rotate, exert a backward force on the road. The road exerts an equal and opposite force on the tires, which is in the forward direction3 So the force on the tires is the sum of these forces, FA and FC , which is FB

0.108

RA

Answer: (B)

Block sliding down a rough inclined plane

We are told several things in this question. The first is that the block attains a constant speed, so it gains no kinetic energy; all its potential energy is lost due to friction. Answer: (B)

Calculation

D

0.108.1

If you’d like a more rigorous proof, not something you might do in the exam. The work done by the frictional force, Fr is Z W = Fr dx (0.108.1) Fr acts along the direction of the incline and is equal to Fr = mg sin θ The distance the force acts is x= 3

(0.108.2)

h sin θ

(0.108.3)

Sometimes the frictional force can act in the direction of motion. This is one such case.

©2009

David S. Latchman

GR9677 Exam Solutions

lxxvi So the work done is W = Fr · x = mg sin θ × = mgh

h sin θ

Answer: (B)

0.109

Collision of Suspended Blocks

FT

We are told that the ball collides elastically with the block, so both momentum and energy are conserved. As the ball falls from a height, h, its potential energy is converted to kinetic energy 1 mgh = mv21 2 2 v1 = 2gh

(0.109.1)

mv1 = mv2 + 2mv3 v1 = v2 + 2v3

(0.109.2)

1 2 1 2 1 mv = mv + 2mv23 2 1 2 2 2 v21 = v22 + 2v23

(0.109.3)

RA

Momentum is conserved, so

Energy is also conserved, so

D

Squaring eq. (0.109.2) and equating with eq. (0.109.3) gives v21 = (v2 + 2v3 )2 ∴ 2v2 = −v3

(0.109.4)

Substituting this into eq. (0.109.2) gives v1 = v2 + 2v3 = 3v2 ⇒ v21 = 9v22

(0.109.5)

The 2m block’s kinetic energy is converted to potential energy as it rises to a height of h2 . Thus 1 mgh2 = mv22 2 ∴ v22 = 2gh2 David S. Latchman

(0.109.6) ©2009

Damped Harmonic Motion We see that

lxxvii 2gh = 2gh2 9 h ⇒ = h2 9

(0.109.7)

Answer: (A)

0.110

Damped Harmonic Motion

FT

From section 0.4.4, we see that the frequency of a damped oscillator is s !2 b ω0 = ω20 − 2m

(0.110.1)

This shows that the damped frequency will be lower than the natural frequency, ω0 , or its period, T0 , will be longer.

0.111

RA

Answer: (A)

Spectrum of the Hydrogen Atom

The hydrogen spectrum can be found by the emperical Rydberg equation    1  1 1 = RH  2 − 2  λ n f ni

(0.111.1)

where ni and n f are the intial and final states respectively. The longest wavelength, or the smallest energy transition, would represent the transition n1 = n f + 1.

D

For the Lyman series, n f = 1, which lies in the ultra-violet spectrum, we have   1 1 3 1 = RH 2 − 2 = RH (0.111.2) λL 1 2 4

For the Balmer series, n f = 2, which lies in the optical spectrum, we have   1 1 1 5 = RH 2 − 2 = RH λB 2 3 36

(0.111.3)

Dividing eq. (0.111.3) by eq. (0.111.2), we get λL = λB

5 R 36 H 3 R 4 H

=

5 27

(0.111.4)

Answer: (A)4 4

The other transition, the Paschen series, n f = 3, lies in the infra-red region of the spectrum.

©2009

David S. Latchman

GR9677 Exam Solutions

lxxviii

0.112

Internal Conversion

We are lucky that they actually tell us what the internal conversion process is. From this we gather that the inner most electron has left its orbit and the most likely outcome will be for the remaining electrons to ‘fall’ in and take its place. These transitions will result in the emission of X-ray photons5 . Answer: (B)

0.113

The Stern-Gerlach Experiment

FT

The description of the experiment in the question is the Stern-Gerlach Experiment. In this experiment, we expect the electrons to be deflected vertically into two beams representing spin-up and spin-down electrons. Answer: (D)

Positronium Ground State Energy

RA

0.114

The positronium atom consists of an electron and a positron in a bound state. Classically, this atom looks like two planets orbiting a central point or center of mass. We need to reduce this system to an equivalent one where an electron cirlces a central mass. We call this equivalent system the reduced mass of the two body system. This is µ=

me M M + me

(0.114.1)

D

The energy levels in terms of the reduced mass is defined as Z2 µ En = − 2 E0 n me

(0.114.2)

The reduced mass of positronium is µ me 1 = = me 2me 2

(0.114.3)

and the ground state of this atom, Z = 1 and n = 1. The ground state energy of Hydrogen is 13.6eV. eq. (0.114.2) becomes 1 E1 = − 13.6eV = −6.8 eV 2

(0.114.4)

Answer: (C) 5

This can also result in the emission of an Auger Electron

David S. Latchman

©2009

Specific Heat Capacity and Heat Lost

0.115

lxxix

Specific Heat Capacity and Heat Lost

In this question, you are being asked to put several things together. Here, we are told, a heater is placed into the water but the water does not boil or change temperture. We can assume that all of the supplied heat by the heater is lost and we infer from the power of the heater that 100 Joules is lost per second. The energy to change water by one degree is derived from its specific heat capacity. E = SHC × Mass × Temp. Diff. = 4200 × 1 × 1 = 4200 J

(0.115.1)

So the time to loose 4200 Joules of heat is

Answer: (B)

(0.115.2)

Conservation of Heat

RA

0.116

E 4200 = = 42 s P 100

FT

t=

Assuming that little to no heat is lost to the environment, the two blocks will exchange heat until they are both in thermal equilibrium with each other. As they have the same masses we expect the final temperature to be 50°C. We can, of course, show this more rigorously where both blocks reach a final temperature, T f . The intial temperatures of blocks I and II are T1 = 100°Cand T2 = 0°C, respectively.

D

Heat Lost by Block I = Heat gained by Block II     0.1 × 103 × 1 × 100 − T f = 0.1 × 103 × 1 × T f − 0   2 0.1 × 103 T f = 10 × 103 ∴ T f = 50 °C

Thus the heat exchanged is 0.1 × 103 × 1 × (100 − 50) = 5 kcal

(0.116.1)

Answer: (D)

0.117

Thermal Cycles

We are told the cycle is reversible and moves from ABCA. We can examine each path and add them to get the total work done. ©2009

David S. Latchman

GR9677 Exam Solutions

lxxx Path A → B is an isothermal process Z WA→B =

V2

where P =

P · dV V1

nRT V

V2

Z

dV V1 V   V2 = nRTh ln V1

= nRTh

(0.117.1)

Path B → C is an isobaric process V1

Z WB→C =

= P2 (V1 − V2 ) = nR (Tc − Th ) and Path C → A Z WC→A =

P1 dV

and P2 V1 = nRTc

(0.117.2)

where

dV = 0

(0.117.3)

RA

=0

where P2 V2 = nRTh

FT

P2 dV V2

Adding the above, we get

W = WA→B + WB→C + WC→A   V2 + nR (Tc − Th ) = nRTh ln V1

where n = 1 mole.



(0.117.5)

D

Answer: (E)

 V2 W = RTh ln − R (Th − Tc ) V1

(0.117.4)

0.118

Mean Free Path

The mean free path of a particle, be it an atom, molecule or photon, is the average distance travelled between collisions. We are given the equation as `=

1 ησ

(0.118.1)

where η is the number desnity and σ is the collision cross section. The number density works out to be N η= (0.118.2) V David S. Latchman

©2009

Probability lxxxi where N is the number of molecules and V is the volume. We can determine this from the ideal gas law, PV = NkT P N = ∴η= V kT

(0.118.3)

The collision cross section is the area through which a particle can not pass without colliding. This works out to be σ = πd2 (0.118.4) Now we can write eq. (0.118.1) in terms of variables we know kT πPd2

(0.118.5)

FT

`=

As air is composed mostly of Nitrogen, we would have used the diameter of Nitrogen in our calculations. This is approximately d = 3.1 Å. Plugging in the constants given we have (1.38 × 1023 )(300) π × 1.0 × 105 (3.1 × 1010 )2 = 1.37 × 10−7 m

RA

`=

As we don’t have a calculator in the exam, we can estimate by adding the indices in our equation, − 23 + 2 − 5 + 20 = −6 (0.118.6) So we expect our result to be in the order of 1 × 10−6 m. We choose (B).

D

Answer: (B)

0.119

Probability

The probability of finding a particle in a finite interval between two points, x1 and x2 , is Z 4 P(2 ≤ x ≤ 4) = (0.119.1) |Ψ(x)|2 dx 2

with the normalization condition, Z

+∞

|Ψ(x)|2 dx = 1

(0.119.2)

−∞

We can tally the values given to us on the graph ©2009

David S. Latchman

GR9677 Exam Solutions

lxxxii x

Ψ

Ψ2

1 2 3 4 5 6

1 1 2 3 1 0

1 1 4 9 1 0

Total

16

FT

Table 0.119.1: Table of wavefunction amplitudes

The probability of finding the particle between (2 ≤ x ≤ 4) is

22 + 32 12 + 12 + 22 + 32 + 12 + 02 4+9 = 1+1+4+9+1+0 13 = 16

Answer: (E)

0.120

(0.119.3)

RA

P(2 ≤ x ≤ 4) =

Barrier Tunneling

D

Classically, if a particle didn’t have enough kinetic energy, it would just bounce off the wall but in the realm of Quantum Mechanics, there is a finite probability that the particle will tunnel through the barrier and emerge on the other side. We expect to see a few things. The wave function’s amplitide will be decreased from x > b and to decay exponentially from a < x < b. We see that choice (C) has these characteristics. Answer: (C)

0.121

Distance of Closest Appraoch

This question throws a lot of words at you. The α-particle with kietic energy 5 MeV is shot towards an atom. If it goes towards the atom it will slow down, loosing kinetic energy and gaining electrical potential energy. The α-particle will then be repelled by the Ag atom. Thus 1 Q1 Q2 = KE (0.121.1) U= 4π0 D David S. Latchman

©2009

Collisions and the He atom lxxxiii Where D is the distance of closest approach, q1 = ze and q2 = Ze. We are given z = 2 for the alpha particle and Z = 50 for the metal atom. Plugging in all of this gives us 1 (2e)(50e) 4π0 5 × 106 e 1 100e = 4π0 5 × 106 1 100 × 1.6 × 10−19 = 5 × 106 4π × 8.85 × 10−12 ≈ 0.3 × 10−13 m

D=

(0.121.2)

0.122

FT

Answer: (B)

Collisions and the He atom

As the collision is elastic, we know that both momentum and kinetic energy is conserved. So conservation of momentum shows

RA

4uv = (−0.6)(4u)v + MV ⇒ 6.4 uv = MV

(0.122.1)

Conservation of Energy shows that

1 1 1 (4 u)v2 = (4 u)(0.6v)2 + MV 2 2h 2 i 2 2 2 4 u 0.64v = MV

(0.122.2)

Solving for M

6.42 u = 16 u 4(0.64) We see that this corresponds to an Oxygen atom, mass 16 u.

(0.122.3)

D

M=

Answer: (D)

0.123

Oscillating Hoops

We are given the period of our physical pendulum, where s I T = 2π mgd

(0.123.1)

where I is the moment of inertia and d is the distance of the pivot from the center of mass. The moment of inertia of our hoop is Icm = Mr2 ©2009

(0.123.2) David S. Latchman

lxxxiv GR9677 Exam Solutions The moment of inertia of the hoop hanging by a nail is found from the Parallex Axis Theorem I = Icm + Md2 = Mr2 + Mr2 = 2Mr2 (0.123.3)

FT

Plugging this into the first equation gives s 2Mr2 T = 2π Mgr s 2r = 2π g r 2 × 20 × 10−2 ≈ 2π 10 −1 = 4π × 10 ≈ 1.2 s Answer: (C)

0.124

Mars Surface Orbit

RA

If a body travels forward quickly enough that it follows the planet’s curvature it is in orbit. We are told that in the case of Mars, there is a 2.0 meter drop for every 3600 meter horizontal distance. We are also told that the acceleration due to gravity on Mars is gM = 0.4g. So the time to drop a distance of 2.0 meters is 1 gM t2 2 ⇒ t = 1s s=

(0.124.1)

So the horizontal speed is

vx =

(0.124.2)

D

Answer: (C)

3600 m/s 1

0.125

The Inverse Square Law

Choice A Energy will be conserved. This isn’t dependent on an inverse square law. Choice B Momentum is conserved. This also isn’t dependent on the inverse square law. Choice C This follows from Kepler’s Law  2 Gm1 m2 2π mr = T r2+ ⇒ T ∝ r(3+)/2 David S. Latchman

(0.125.1) ©2009

Charge Distribution lxxxv 6 Choice D This is FALSE. This follows from Bertrand’s Theorem , which states that only two types of potentials produce stable closed orbits 1. An inverse square central force such as the gravitational or electrostatic potential. −k (0.125.2) V(r) = r 2. The radial Harmonic Oscillator Potential 1 V(r) = kr2 2

(0.125.3)

FT

Choice E A stationary circular orbit occurs under special conditions when the central force is equal to the centripetal force. This is not dependent on an inverse square law but its speed. Answer: (D)

0.126

Charge Distribution

RA

An inportant thing to keep in mind is that charge will distribute itself evenly throughout the conducting spheres and that charge is conserved. Step I: Uncharged Sphere C touches Sphere A A Q 2

B Q

C Q 2

Step II: Sphere C is touched to Sphere B

B 3Q 4

D

A Q 2

C 3Q 4

The initial force between A and B is F=

kQ2 r2

(0.126.1)

The final force between A and B is Ff =

k Q2 3Q 4 r2

3 = F 8

(0.126.2)

Answer: (D) 6

Add reference here

©2009

David S. Latchman

GR9677 Exam Solutions

lxxxvi

0.127

Capacitors in Parallel

We have one capacitor, C1 connected to a battery. This capacitor gets charged and stores a charge, Q0 and energy, U0 . Q0 = C1 V (0.127.1) 1 U0 = C1 V 2 (0.127.2) 2 When the switch is toggled in the on position, the battery charges the second capacitor, C2 . As the capacitors are in parallel, the potential across them is the same. As C1 = C2 , we see that the charges and the energy stored across each capacitor is the same. Thus Q2 = C2 V2 Q2 = Q1 1 U2 = C2 V22 2 U2 = U1

FT

Q1 = C1 V1 1 U1 = C1 V12 2 We see that

U1 + U2 = C1 V12 = 2U0

(0.127.3)

RA

We can also analyze this another way. The two capacitors are in parallel, so their net capacitance is CT = C1 + C2 = 2C1 (0.127.4) So the total charge and energy stored by this parallel arrangemt is Q = CT V1 = 2C1 V1 1 UT = 2C1 V12 = 2U0 2 Of all the choices, only (E) is incorrect.

D

Answer: (E)

0.128

Resonant frequency of a RLC Circuit

The circuit will be best ‘tuned’ when it is at its resonant frequence. This occurs when the impedances for the capacitor and inductor are equal. Thus XC =

1 ωC

and

XL = ωL

(0.128.1)

When they are equal XC = XL 1 = XL = ωL ωC 1 ∴ω= √ LC David S. Latchman

(0.128.2) ©2009

Graphs and Data Analysis Solving for C,

lxxxvii

C=

ω2 L

1 = π2 × (103.7 × 106 )2 × 2.0 × 10−6 4 ≈ 0.125 × 10−11 F

(0.128.3)

Answer: (C)

0.129

Graphs and Data Analysis

FT

It is best to analyse data is they are plotted on straight line graphs of the form, y = mx+c. This way we can best tell how well our data fits, etc.7 A We want a plot of activity, dN vs. time, t. If we were to plot this as is, we would dt get an exponential curve. To get the straight line graph best suited for further analysis, we take the logs on both sides.

RA

dN ∝ e−2t dt " # dN log = log e−2t dt " # dN log = −2t (0.129.1) dt h i We have a Semilog graph with a plot of log dN on the y-axis, t on the x-axis with dt a gradient of 2. B This is already a linear equation we can plot with the data we already have. No need to manipulate it in any way.

D

C We take logs on bot sides of the equation to get

s ∝ t2 log s = 2 log t

(0.129.2)

We can plot log s vs. log t. This gives a linear equation with log s on the y-axis and log t on the x-axis and a gradient of 2.

D Again, we take logs on both sides of the equation Vout 1 ∝ Vin ω " # V log out = − log ω Vin 7

This is of course with nothing but a sheet of graph paper and calculator and without the help of computers and data analysis software.

©2009

David S. Latchman

GR9677 Exam Solutions

lxxxviii V



out on the y-axis and log ω on the x-axis with a in gradient of -1. We see that this choice is INCORRECT.

We have a log-log plot of log

V

E As with the other choices, we take logs on both sides and get P ∝ T4 log P = 4 log T This can be plotted on a log-log graph with log P on the y-axis and log T on the x-axis and a gradient of 4. Answer: (D)

Superposition of Waves

FT

0.130

RA

As the question states, we can see the superposition of the two waves. For the higher frequency wave, we see that the period on the oscilloscope is about 1cm. This works out to be a period of 1 cm = 2.0 ms (0.130.1) T= 0.5 cm ms−1 The frequency is 1 2.0 × 10−3 = 500 Hz

f =

(0.130.2)

We can measure the amplitude of this oscillation by measuring the distance from crest to trough. This is approximately (2 − 1)/2, thus8 A = 1 cm × 2.0 V cm−1 ≈ 2.0 V

(0.130.3)

D

For the longer period wave, we notice that approximately a half-wavelength is displayed, is 2(4.5 − 1.5) = 6 cm. The period becomes 6.0 cm 0.5 cm/ms = 12.0 ms

T=

(0.130.4)

Thus the frequency is

f =

1 T

1 = 83 Hz 12.0 × 10−3 We see that (D) matches our calculations. =

(0.130.5)

Answer: (D) 8

If you happened to have worked this one first you’ll notice that only choice (D) is valid. You can stop and go on to the next question.

David S. Latchman

©2009

The Plank Length

0.131

lxxxix

The Plank Length

This question is best analysed through dimensional analysis; unless of course you’re fortunate to know the formula for the Plank Length. We are told that G = 6.67 × 10−11 m3 kg−1 s−2 6.63 × 10−34 −1 Js 2π c = 3.0 × 108 m s−1

~=

FT

We can substitute the symbols for Length, L, Mass, M and Time, T. So the dimensions of our constants become G = L3 M−1 T−2 ~ = ML2 T−1 c = LT−1 `p = L Our Plank Length is in the form

RA

`p = Gx ~ y cz Dimensional Analysis Gives

z y  x   L = L3 M−1 T−2 ML2 T−1 LT−1

We get L

D

3x + 2y + z = 1

M

−x+y=0

T

z = −3x

Solving, we get x=

1 2

y=

1 2

z=−

3 2

Thus r `p =

G~ c3

Answer: (E) ©2009

David S. Latchman

GR9677 Exam Solutions

xc

0.132

The Open Ended U-tube

We recall that the pressure throughout a fluid is equal throughout the fluid. As the system is in equlibrium, the pressure on the left arm is equal to the pressure on the right arm. We can set up an equation such that ρ2 g5 + ρ1 g (h1 − 5) = ρ1 gh2

(0.132.1)

where whater, ρ1 = 1.0 g/cm3 , some immiscible liquid, ρ2 = 4.0 g/cm3 . Solving, gives us h2 − h1 = 15 cm

(0.132.2)

FT

Let’ call the height of the water column on the left side of the tube, x1 . We get h2 − (x1 + 5) = 15 ∴ h2 − x1 = 20

Answer: (C)

0.133

RA

We expect the water column to go down on the left side of the tube as it goes up on the right side of the tube; conservation of mass. So we infer the change in height on both sides is 10 cm. We conclude that since the intial height is 20cm, then h2 = 30 cm and x1 = 10 cm. So h2 30 = =2 (0.132.3) h1 15

Sphere falling through a viscous liquid

D

Our sphere falls through a viscous liquid under gravity and experiences a drag force, bv. The equation of motion can be expressed ma = mg − bv

(0.133.1)

We are also told that the buoyant force is negligible. Armed with this information, we can analyze out choices and emiminate. A This statement will be incorrect. We have been told to ignore the buoyant force, which if was present, would act as a constant retarding force and slow our sphere down and reduce its kinetic energy. INCORRECT

B This is also incorrect. In fact if you were to solve the above equation of motion, the speed, and hence kinetic energy, would monotonically increase and approach some terminal speed. It won’t go to zero. INCORRECT C It may do this if it was shot out of a gun, but we were told that it is released from rest. So it will not go past its terminal speed. David S. Latchman

©2009

Moment of Inertia and Angular Velocity xci D The terminal speed is the point when the force due to gravity is balanced by the retarding force of the fluid. Setting ma = 0 in the above equation, we get 0 = mg − bv

(0.133.2)

Solving for v yields,

mg (0.133.3) b We see that our terminal velocity is dependent on both b and m. This choice is INCORRECT v=

E From the above analysis, we choose this answer. CORRECT

0.134

FT

Answer: (E)

Moment of Inertia and Angular Velocity

The moment of inertia of an object is I=

N X

mi r2i

RA

i=1

where ri is the distance from the point mass to the axis of rotation. The moment of inertia about point A is found by finding the distances of each of the three masses from that point. The distance between the mass, m and A is ` r= √ 3

D

Thus the moment of inertia is

` IA = 3m √ 3

!2

= m`2

The Moment of Inertia about B can be found by the Parallel Axis Theorem but it may be simpler to use the formula above. As the axis of rotation is about B, we can ignore this mass and find the distances of the other two masses from this point, which happens to be `. Thus IB = 2m`2

The rotational kinetic energy is 1 K = Iω2 2 So the ratio of the kinetic energies at fixed, ω becomes KB IB 2m`2 = = =2 KA IA m`2 Answer: (B) ©2009

David S. Latchman

GR9677 Exam Solutions

xcii

0.135

Quantum Angular Momentum

The probability is P=

32 + 22 13 = 38 38

(0.135.1)

NOT FINISHED Answer: (C)

Invariance Violations and the Non-conservation of Parity

FT

0.136

RA

Electromagnetic and strong interactions are invariant under parity transformations. The only exception to this rule occurs in weak interactions, the β-decay bring one such example. It had always been assumed that invariance was a “built-in” property of the Universe but in the 1950s there seemed to be some puzzling experiments concerning certain unstable particles called tau and theta mesons. The “tau-theta puzzle” was solved in 1956 by T.D. Lee9 and C.N. Yang10 when they proposed the nonconservation of parity by the weak interaction. This hypothesis was confirmed experimentally through the beta decay of Cobalt-60 in 1957 by C.S. Wu11 . 60

Co −−→ 60Ni + e – + υ¯ e

D

The cobalt source was chilled to a temperature of 0.01 K and placed in a magnetic field. This polarized the nuclear spins in the direction of the magnetic field while the low temperatures inhibited the thermal disordering of the aligned spins. When the directions of the emitted electrons were measured, it was expected that there would be equal numbers emitted parallel and anti-parallel to the magnetic field, but instead more electrons were emitted in the direction opposite to the magnetic field. This observation was interpreted as a violation of reflection symmetry. Answer: (D) 9

Tsung-Dao Lee is a Chinese-born American physicist, well known for his work on parity violation, the Lee Model, particle physics, relativistic heavy ion (RHIC) physics, nontopological solitons and soliton stars. He and Chen-Ning Yang received the 1957 Nobel prize in physics for their work on parity nonconservation of weak interactions. 10 Chen-Ning Franklin Yang is a Chinese-American physicist who worked on statistical mechanics and particle physics. He and Tsung-dao Lee received the 1957 Nobel prize in physics for their work on parity nonconservation of weak interactions. 11 Chien-Shiung Wu was a Chinese-American physicist. She worked on the Manhattan Project to enrich uranium fuel and performed the experiments that disproved the conservation of parity. She has been known as the“First Lady of Physics”, “Chinese Marie Curie” and “Madam Wu”. She died in February 16, 1997

David S. Latchman

©2009

Wave function of Identical Fermions

0.137

xciii

Wave function of Identical Fermions

The behavior of fermions are described by the Pauli Exclusion Principle, which states that no two fermions may have the same quantum state. This is a results in the anti-symmetry in the wave funtion. Answer: (A)

0.138

Relativistic Collisions

FT

We are told that no energy is radiated away, so it is conserved; all of it goes into the composite mass. The relativistic energy is E = γmc2 Given that v = 3/5c

1 γ= q 1−

=

v2 c2

5 4

RA

So the energy of the lump of clay is

(0.138.1)

5 E = γmc2 = mc2 4

(0.138.2)

(0.138.3)

The composite mass can be found by adding the energies of the two lumps of clay ET = 2E 10 Mc2 = mc2 4 ∴ M = 2.5m = 2.5 × 4 = 10 kg

D

Answer: (D)

(0.138.4)

0.139

Relativistic Addition of Velocities

We recall that the relativistic addition formula u+v 1 + uv c2

(0.139.1)

0.9c 0.9 9 = c ≈ c = 0.75c 1 + 0.18 1.18 12

(0.139.2)

v0 = where u = 0.3c and v = 0.6c. This becomes v0 = Answer: (D) ©2009

David S. Latchman

GR9677 Exam Solutions

xciv

0.140

Relativistic Energy and Momentum

The Relativistic Momentum and Energy equations are p = γmv

E = γmc2

(0.140.1)

We can determine the speed by dividing the relativistic momentum by the relativistic energy equation to get

Answer: (D)

Ionization Potential

(0.140.2)

RA

0.141

FT

γmv p = E γmc2 v = 2 c 5MeV/c v ∴ = 2 10MeV c v 5 = 2 10c c 1 ⇒v= c 2

The Ionization Potential, or Ionization Energy, EI , is the energy required to remove one mole of electrons from one mole of gaaseous atoms or ions. It is an indicator of the reactivity of an element. 2 He 4

The Helium atom is a noble gas and has filled outermost electron shells as well as its electrons being close to the nucleus. It would be very difficult to ionize.

D

He = 1s2

7 N 14

8 O 16

Nitrogen has two outermost electrons. N = 1s2 , 2s2 , 2p6 , 2s2 , 2p2

Oxygen has four outermost electrons. O = 1s2 , 2s2 , 2p6 , 2s2 , 2p4

18 Ar 40

Another noble gas, this has filled outermost electrons and is not reactive.

55 Cs 133

We can see that Cs has a high atomic number and hence a lot of electrons. We expect the outmost electrons to be far from the nucleus and hence the attraction to be low. This will have a low ionization potential.

Answer: (E) David S. Latchman

©2009

Photon Emission and a Singly Ionized He atom

0.142

xcv

Photon Emission and a Singly Ionized He atom

The energy levels can be predicted by Bohr’s model of the Hydrogen atom. As a Helium atom is more massive than Hydrogen, some corrections must be made to our model and equation. The changes can be written En = −

Z2 µ E0 n2 me

(0.142.1)

where Z is the atomic number, n is the energy level, E0 is the ground state energy level of the Hydrogen atom and µ/me is the reduced mas correction factor.

FT

The emitted photon can also be found through a similar correction µ hc = Z2 ∆E = λe me

   1   − 1  13.6  2  n f n2i 

(0.142.2)

RA

As Helium’s mass is concentrated in the center, it’s reduced mass is close to unity12 . µ Z = ≈1 me Z + me

(0.142.3)

Plugging in the values we know into eq. (0.142.2), we get    1  6.63 × 10−34 × 3 × 108 1 2  −  = 2 13.6   n2 42  470 × 10−9 × 1.60 × 10−19 f

(0.142.4)

D

After some fudging and estimation we get

19 6.63 × 3 × 102 ≈ × 102 470 × 1.6 470 × 1.6 20 ≈ × 102 750 = 0.026 × 102 eV

(0.142.5)

and 22

1 2.6 ≈ · 13.6 20

12

It is helpful to know that in the case of atoms, the reduced mass will be close to unity and can be ignored from calculation. In the case of smaller bodies, e.g. positronium, this correction factor can not be ignored.

©2009

David S. Latchman

GR9677 Exam Solutions

xcvi Solving for n f , gives13 1 1 1 = 2 − 2 20 n f 4 1 1 9 1 + = = 2 20 16 80 nf ≈ ∴ nf = 3

1 9 (0.142.6)

Now we can calculate the energy level at n = 3 from eq. (0.142.1), which gives, 22 · 13.6 32 = −6.0 eV

FT

E3 = −

(0.142.7)

We get E f = −6.0 eV and n f = 3. This corresponds to (A). Answer: (A)

Selection Rules

RA

0.143

NOT FINISHED Answer: A

0.144

Photoelectric Effect

D

This question deals with the photoelectric effect which is essentially an energy conservation equation. Energy of a photon strikes a metal plate and raises the electrons to where they can leave the surface. Any extra energy is then put into the kinetic energy of the electron. The photoelectric equation is h f = eVs + K

(0.144.1)

As our choices are in electron-volts, our equation becomes hc = eVs + K eλ

(0.144.2)

where K is the kinetic energy of our photoelectrons. Plugging in the values we were given and solving for K, we get K = 0.2 eV (0.144.3) Answer: (B) 13

As n f = 3 is only in choice (A), we can forego any further calculation and choose this one.

David S. Latchman

©2009

Stoke’s Theorem

0.145

xcvii

Stoke’s Theorem

NOT FINISHED Answer: (C)

0.146

1-D Motion

A particle moves with the velocity v(x) = βx−n

(0.146.1)

FT

To find the acceleration, a(x), we use the chain rule

dv dx dv = · dx dt dx dv =v· dx

a(x) =

(0.146.2)

Differentiating v(x) with respect to x gives

RA

dv = −nβx−n−1 dx

Thus, our acceleration, a(x), becomes

a(x) = βx−n · −nβx−n−1 = −nβ2 x−2n−1

Answer: (A)

High Pass Filter

D

0.147

(0.146.3)

Capacitors and Inductors are active components; their impedances vary with the frequency of voltage unlike an ohmic resistor whose resitance is pretty much the same no matter what. The impedances for capacitors and inductors are XC =

1 ωC

XL = ωL

We see that in the case of capacitors, there is an inverse relationship with frequency and a linear one for inductors. Simply put, at high frequencies capacitors have low impedances and inductors have high inductances. NOT FINSIHED Answer: (E) ©2009

David S. Latchman

GR9677 Exam Solutions

xcviii

0.148

Generators and Faraday’s Law

The induced EMF in the loop follows Faraday’s Law E =−

dΦ dt

In this case, the magnetic field, B, is constant and the Cross Sectional Area, A, through which the magnetic field acts changes. Thus the above equation becomes E0 sin ωt = −B

dA dt

(0.148.1)

Let’s say that at t = 0 the loop is face on with the magnetic field,

Substituting this into eq. (0.148.1) gives

(0.148.2)

FT

A = πR2 cos ωt

dA dt  = −B · −ωπR2 sin ωt

E0 sin ωt = −B ·

RA

= ωBπR2 sin ωt Solving for ω gives

ω=

Answer: (C)

(0.148.4)

Faraday’s Law and a Wire wound about a Rotating Cylinder

D

0.149

E0 BπR2

(0.148.3)

The induced EMF of our system can be found from Faraday’s Law, where E =−

dΦ dt

(0.149.1)

Here the flux changes because the number of loops enclosing the field increases, so Φ = NBA

(0.149.2)

Substituting this into Faraday’s Equation we get dN dt = BπR2 N

E = BA

(0.149.3)

Answer: (C) David S. Latchman

©2009

Speed of π+ mesons in a laboratory +

0.150

xcix

Speed of π mesons in a laboratory

As the π+ meson travels through our laboratory and past the detectors, its half life is time dilated in our laboratory’s rest frame. We can also look at things in the π+ meson’s rest frame. In this case, the distance it travels will be length contracted in it rest frame. The speed of our π+ mesons is the length divided by the time dilation in the laboratory’s rest frame or the length contraction in the π+ meson’s rest frame divided by its half life. In either case, we get r L v2 v= 1− 2 T1/2 c Factorizing we get

FT

  2   L L2   v2 1 + 2 2  = 2 c T1/2 T1/2 We see that

L = 6 × 108 T1/2

(0.150.2)

Transformation of Electric Field

D

0.151

RA

Plugging this into eq. (0.150.1), the speed in terms of c   v2 36 36 1+ = 2 c 9 9 v2 (5) = 4 c2 2 ⇒v= √ c 5 Answer: (C)

(0.150.1)

NOT FINSHED Answer: (C)

0.152

The Space-Time Interval

We have two events, in the S-frame, S1 (x1 , t)

and S2 (x2 , t)

In the S0 -frame, the co-ordintes are S01 (x01 , t01 ) ©2009

and S02 (x02 , t02 ) David S. Latchman

GR9677 Exam Solutions

c The Space-Time Interval in the S-frame ∆S = ∆x2 = 3c minutes

(0.152.1)

In the S0 -frame, the Space-Time Interval is ∆S0 = ∆x02 − c2 ∆t02 = 5c minutes

(0.152.2)

The Space-Time Interval is invariant across frames, so eq. (0.152.1) is equal to eq. (0.152.2) (3c)2 = (5c)2 − c2 ∆t2 ⇒ ∆t = 4 minutes

0.153

FT

Answer: (C)

(0.152.3)

Wavefunction of the Particle in an Infinte Well

Answer: (B)

0.154

RA

The wave function has zero probability density in the middle for even wave functions, n = 2, 4, 6, · · · .

Spherical Harmonics of the Wave Function

NOT FINSIHED

D

Answer: (C)

0.155

Decay of the Positronium Atom

NOT FINSHED Answer: (C)

0.156

Polarized Electromagnetic Waves I

We are given an electromagnetic wave that is the superposition of two independent orthogonal plane waves where E = xˆ E1 exp [i (kz − ωt)] + yE ˆ 2 exp [i (kz − ωt + π)] David S. Latchman

(0.156.1) ©2009

Polarized Electromagnetic Waves II As we are looking at the real components and E1 = E2 , we have

ci

E = <(E1 eikz · e−iωt )ˆx + <(E1 eikz · e−iωt · e−iπ )yˆ = <(E1 eikz · e−iωt )ˆx − <(E1 eikz · e−iωt )yˆ We see that the xˆ and yˆ vectors have the same magnitude but opposite sign; they are both out of phase with each other. This would describe a trajectory that is 135°to the x-axis. Answer: (B)

Polarized Electromagnetic Waves II

FT

0.157

NOT FINISHED Answer: (A)

Total Internal Reflection

RA

0.158

Total internal reflectance will occur when the incident beam, reaches a critical angle, θi , such that the refracted angle just skims along the water’s surface, θr = 90◦ . We can find this critical angle using Snell’s Law n2 sin θi = n1 sin 90

where n2 = 1.33 and n1 = 1, we have

1 3 = 1.33 4 √ We know that sin 30° = 1/2 and sin 60° = 3 /2, so 30° < θ < 60°. sin θ2 =

D

(0.158.1)

Answer: (C)

0.159

Single Slit Diffraction

For a single slit, diffraction maxima can be found from the formula a sin θ = mλ

(0.159.1)

where a is the slit width, θ is the angle between the minimum and the central maximum, and m is the diffraction order. As θ is small and solving doe d, we can approximate the ©2009

David S. Latchman

GR9677 Exam Solutions

cii above equation to λ θ 400 × 10−9 = 4 × 10−3 = 0.1 × 10−3 m

d=

(0.159.2)

Answer: (C)

0.160

The Optical Telescope

FT

The magnification of the optical telescope can be found from the focal length of the eyepiece, fe and the objective, fo . Thus M= The focal length of the objective is

fo = 10 fe

RA

fe = 10 × 1.5 = 15 cm

(0.160.1)

(0.160.2)

To achieve this magnification, the lens must be placed in a position where the focal length of the eyepiece meets the focal length of the objective. Thus D = fe + fo = 15.0 + 1.5 = 16.5 cm

Answer: (E)

Pulsed Lasers

D

0.161

(0.160.3)

Lasers operating in pulsed mode delivers more energy in a short space of time as opposed to delivering the same energy over a longer period of time in a continuous mode. While there are several methods to achieve a pulsed mode, beyond what is needed to answer this question, we can determine the number of photons delivered by such a device. The energy of a photon is E = hf =

hc λ

(0.161.1)

The power is the energy delivered in one second. So for a 10kW laser, the total energy in 10−15 seconds is EL = Pt = 10 × 103 × 10−15 = 10 × 10−12 J David S. Latchman

(0.161.2) ©2009

Relativistic Doppler Shift So the total number of photons is

ciii

EL E 10 × 10−12 × λ = hc 10 × 10−12 × 600 × 10−9 = 6.63 × 10−34 × 3 × 108 10 × 600 = × 106 ≈ 3 × 108 6.63 × 3

n=

(0.161.3)

0.162

FT

Answer: (B)

Relativistic Doppler Shift

The relativistic doppler shift is

s

1+β 1−β

RA

fs λo = = λs fo

(0.162.1)

The redshift is calculated to be

z=

fs − fo λo − λs = λs fo

(0.162.2)

we can rewrite eq. (0.162.2) as

s

z=

1+β −1 1−β

(0.162.3)

D

In the non-relativistic limit, v << c, we can approximate eq. (0.162.3) to z≈β=

v c

(0.162.4)

This, equating eq. (0.162.2) and eq. (0.162.4), we see that v=

∆f c f

(0.162.5)

Substituting the values given into eq. (0.162.5), we have 0.9 × 10−12 × 3 × 108 122 × 10−9 ≈ 2.2 m s−1

v=

(0.162.6)

Answer: (B) ©2009

David S. Latchman

GR9677 Exam Solutions

civ

0.163

Gauss’ Law, the Electric Field and Uneven Charge Distribution

We can find the electric field in a non-conducting sphere by using Gauss’ Law I E · dA =

Qenclosed 0

(0.163.1)

The enclosed charge can be found from the charge density, which is Enclosed Charge Enclosed Volume q = 4 3 πr 3

FT

ρ=

(0.163.2)

We can find the enclosed charge by integrating within 0 to R/2. The charge density is dq dV dq = 4πr2 dr ∴ dq = ρ4πr2 dr

RA

ρ=

= 4πAr4 dr

(0.163.3) (0.163.4)

Gauss’ Law becomes

I

qenclosed 0 R Z   2 dq 2 E 4πr = 0 0 Z R2 4πA = r4 dr 0 0 R 4πA r5 2 = 0 5 0 R A r3 2 ∴E= 0 5 0

D

E · dA =

(0.163.5) (0.163.6)

Solving gives E=

AR3 400

(0.163.7)

Answer: (B) David S. Latchman

©2009

Capacitors in Parallel

0.164

cv

Capacitors in Parallel

We initially charge both of our capacitors in parallel across a 5.0V battery. The charge stored on each capacitor is Q1 = C1 V

Q2 = C2 V

where V = 5.0 V, C1 = 1µF and C2 = 2µF.

FT

The battery is then disconnected and the plates of opposite charges are connected to each other. This results in the excess charges cancelling each other out and then redistributing themselves until the potential across the new configuration is the same. The charge left after this configuration is QA = Q2 − Q1 The charges on each capacitor becomes

Q1A + Q2A = QA where

Q2A = C2 V f

RA

Q1A = C1 V f

and

(0.164.1)

(0.164.2)

(0.164.3)

Solving for V f gives us

Vf =

Answer: (C)

(0.164.4)

Standard Model

D

0.165

(C2 − C1 )V 5 = = 1.67 V C1 + C2 3

NOT FINSIHED Answer: (A)

0.166

Nuclear Binding Energy

Typically a heavy nucleus contains ∼ 200 nucleons. The energy liberated would be the difference in the binding energies 1 MeV ×200. Answer: (C) ©2009

David S. Latchman

GR9677 Exam Solutions

cvi

0.167

Work done by a man jumping off a boat

The work the man does is the sum of the kinetic energies of both the boat and himself. We can find the speeds of the man and the boat because momentum is conserved. mu = Mv Mv u= m

(0.167.1)

The total energy of the system, and hence the work the man does in jumping off the boat is

Answer: (D)

0.168

(0.167.2)

FT

1 1 W = mu2 + Mv2 2  2  1 2 M = Mv +1 2 m

Orbits and Gravitational Potential

RA

For an attractive potential, such as we would expect for the Gravitational Potential, we have Orbit

Total Energy

Ellipse Parabola Hyperbola

E<0 E=0 E>0

D

where E is the total energy of the system; potential and kinetic energy. As the potential energy of the system remains unchanged, the only difference is the is the kinetic energy, the orbit will be hyperbolic. When the spacecraft has the same speed as Jupiter, the orbit will be locked and will be elliptical. If the gravitational potential energy was equal to the kinetic energy, the orbit will no longer be locked and will be parabolic. We assume that the huge difference will cause the orbit to be hyperbolic. Answer: (E)

0.169

Schwartzchild Radius

Any mass can become a black hole if it is compressed beyond its Schwarzschild Radius. Beyond this size, light will be unable to escape from it’s surface or if a light beam were David S. Latchman

©2009

Lagrangian of a Bead on a Rod cvii to be trapped within this radius, it will be unable to escape. The Schwarzschild Radius can be derive by putting the Gravitational Potential Energy equal to a mass of kinetic energy travelling at light speed. GMm 1 2 = mc R 2

(0.169.1)

Solving for R yields R=

2GM c2

(0.169.2)

Plugging in the values given, we get 2 × 6.67 × 10−11 × 5.98 × 102 4 (3 × 108 )2

FT

R=

Our indices indicate we will get an answer in the order ≈ 10−3 meters. Answer: (C)

Lagrangian of a Bead on a Rod

RA

0.170

The Lagrangian of a system is defined

L=T−V

(0.170.1)

The rod can move about the length of the rod, s and in circular motion along a radius of s sin θ. The Lagrangian of this system becomes (0.170.2)

D

1 1 L = ms˙2 + m(s sin θ)2 ω2 − mgs cos θ 2 2

Answer: (E)

0.171

Ampere’s Law

We can use Ampere’s Law to tell us the magnetic field at point A. I B · ds = µ0 Ienclosed

(0.171.1)

The point A is midway between the center of the two cylinders and as the currents are in opposite directions, thier magnetic fields at A point in the +y-direction. We can use the right hand grip rule to determine this. This leaves us with choices (A) or (B). ©2009

David S. Latchman

GR9677 Exam Solutions

cviii The current density, J is I Area I = 2 πr

J=

(0.171.2)

We draw an Amperian loop of radius, r = d/2, thus the magnetic field becomes, I B · ds = µ0 Ienclosed   B · (2πr) = µ0 J πr2 µ0 πJr 2π µ0 πJ d = 2π 2

FT

B=

We expect B⊗ to be the same, thus

(0.171.3)

B = B + B⊗ µ  0 = πdJ 2π

0.172

RA

Answer: (A)

(0.171.4)

Larmor Formula

D

The Larmor Formula is used to calculate total power radiated by an accelerating nonrelativistic point charge. P=

e2 a2 6π0 c3

(0.172.1)

where a is the acceleration. For particles A & B, we are given A Charge qa = q Mass ma = m Velocity va = v Acceleration aa = a

B qb = 2q mb = m/2 vb = 3v ab = 4a

From the above, we see that PA ∝ q2 a2 David S. Latchman

PB = (2q)2 (4a2 ) ©2009

The Oscilloscope and Electron Deflection Thus

cix

(2q)2 (4a2 ) PB = PA q2 a2 = 64

(0.172.2)

Answer: (D)

0.173

The Oscilloscope and Electron Deflection

FT

As the electron passes through the deflection plates, the electric field, E exerts a force on the charge, pulling it up. Ignoring gravitational effects, the electric force is Fe = qE =

qV = ma y d

(0.173.1)

The time it takes to traverse this distance is

t=

(0.173.2)

vy vx

RA

The deflection angle, θ is determined by

L v

tan θ =

Now

D

vy = ayt qV = t me d qV L = me d v

(0.173.3)

(0.173.4)

The the horizontal speed is v The angle of deflection becomes qV L tan θ = md v v qVL = mdv2

(0.173.5)

Answer: (A)

0.174

Negative Feedback

All amplifiers exhibit non-linear behavior of some sort. Negative feedback seeks to correct some of these effects by sending some of the output back and subtracting it from ©2009

David S. Latchman

cx GR9677 Exam Solutions the input. This results in a decrease in gain. This tradeoff og gain improves linearity and hence the stability of the amplifier. This also allows for increased bandwidth response and decreased distortion. Answer: (A)

0.175

Adiabatic Work of an Ideal Gas

The adiabatic condition states that PV γ = C

(0.175.1)

The work done by an ideal gas is

FT

Z

W=

PdV

(0.175.2)

RA

Substituting the adiabatic condition into the work equation yields Z Vf dV W=C γ Vi V V V −γ+1 f =C 1 + γ Vi i C h −γ+1 −γ+1 = Vf − Vi 1−γ

(0.175.3)

The adiabatic condition is

γ

γ

C = PV γ = Pi Vi = P f V f

(0.175.4)

Substituting this into eq. (0.175.3), we get −γ+1

D

W=

=

CV f

−γ+1

− CVi

1−γ

γ −γ+1 Pf Vf Vf

γ

−γ+1

− Pf Vf Vf

1−γ P f V f − Pi Vi = 1−γ

(0.175.5)

Answer: (C)

0.176

Change in Entrophy of Two Bodies

The change in entrophy of a system is dS = David S. Latchman

dQ T

(0.176.1) ©2009

Double Pane Windows The change in heat of a system is

cxi dQ = nCdT

(0.176.2)

Substituting this into the above equation, we get Z Tf dT dS = mC T Ti " # Tf = mC ln Ti

(0.176.3)

We are told that the two bodies are brought together and they are in thermal isolation. This means that heat is not absorbed from or lost to the environment, only transferred between the two bodies. Thus

FT

Heat Lost by Body A = Heat Gained by Body B     mC 500 − T f = mc T f − 100 ∴ T f = 300 K

The Total Change in Entrophy is the sum of the entrophy changes of bodies A and B. Thus

Answer: (B)

0.177

RA

dS = dSA + dSB     300 300 + mC ln = mC ln 500 100   9 = mC ln 5

(0.176.4)

Double Pane Windows

D

The rate of heat transer is proportional to the temperature difference across the body, is stated in Newton’s Law of Cooling.

dQ = kxdT (0.177.1) dt where k is the thermal conductivity, x is the thickness of the material and dT is the temperature difference across the material. For Window A

dQA = 0.8 × 4 × 10−3 dT dt

(0.177.2)

dQB = 0.025 × 2 × 10−3 dT dt

(0.177.3)

0.8 × 4 × 10−3 dT Pa = = 16 PB 0.025 × 2 × 10−3 dT

(0.177.4)

PA = For Window B PB = The ratio of heat flow is

Answer: (D) ©2009

David S. Latchman

GR9677 Exam Solutions

cxii

0.178

Gaussian Wave Packets

We have a Gaussian wave packet travelling through free space. We can best think of this as an infinite sum of a bunch of waves initially travelling together. Based on what we know, we can eliminate the choices given. I The average momentum of the wave packet can not be zero as p = ~k. As we have a whole bunch of wave numbers present, the average can not be zero. INCORRECT

FT

II Our wave packet contains a bunch of waves travelling together each with a differnt wave vector, k. The speed of propagation of these individual wave vectors is defined by the group velocity, v g = dω/dk. So some waves will travel, some slower than others. As a result of these different travelling rates our wave packet becomes spread out or ‘dispersed’. This is the basis of our dispersion relation, ω(k), relative to the center of the wave packet. CORRECT III As we expect the wave packet to spread out, as shown above, the amplitude will decrease over time. The energy that was concentrated in this packet gets spread out or dispersed. INCORRECT

RA

IV This is true. This statement is the Uncertainty Principle and comes from Fourier Analysis. CORRECT We see that choices II and IV are CORRECT. Answer: (B)

0.179

Angular Momentum Spin Operators

NOT FINISHED

D

Answer: (D)

0.180

Semiconductors and Impurity Atoms

NOT FINISHED Answer: (B)

0.181

Specific Heat of an Ideal Diatomic Gas

The formula for finding the molar heat capacity at constant volume can be found by ! f R (0.181.1) cv = 2 David S. Latchman

©2009

Transmission of a Wave cxiii where f is the number of degrees of freedom. At very low temperatures, there are only three translational degrees of freedom; there are no rotational degrees of freedom in this case. At very high temperatures, we have three translational degrees of freedom, two rotational and two vibrational, giving a total of seven in all. For Low Temperatures

For High Temperatures

Translational Rotational Vibrational

3 0 0

Translational Rotational Vibrational

3 2 2

Total( f )

3

Total( f )

7

FT

Table 0.181.1: Table of degrees of freedom of a Diatomic atom We see that in the case of very low temperatures, 3 cvl = R 2 and at very high temperatures,

(0.181.2)

RA

7 cvh = R (0.181.3) 2 Thus, the ratio of molar heat capacity at constant volume at very high temperatures to that at very low temperatures is 7 R cvh 2 = 3 cvl R 2 7 = 3

D

Answer: (D)

(0.181.4)

0.182

Transmission of a Wave

NOT FINISHED Answer: (C)

0.183

Piano Tuning & Beats

The D2 note has a frequency of 73.416 Hz and the A4 note has a frequency of 440.000 Hz. Beats are produced when the two frequencies are close to each other; if they were the ©2009

David S. Latchman

cxiv GR9677 Exam Solutions same, there would be no beat frequency. So we can determine the D2 note where this happens by setting the beat frequency to zero. 440.000 − n(73.416) = 0

(0.183.1)

This works out to be 440.000 73.416 440 ≈ =6 72

n=

(0.183.2)

Thus the closest harmonic will be the 6th one. As we expect this to be very close to the A4 frequency, the number of beats will be small or close to zero. Answer (B) fits this.14

0.184

FT

Answer: (B)

Thin Films

RA

As light moves from the glass to air interface, it is partially reflected and partially transmitted. There is no phase change when the light is refected. In the case of the transmitted wave, when it reaches the air-glass interface, there is a change in phase of the reflected beam. Thus the condition for destructive interference is   1 (0.184.1) 2L = n + λ 2 where L is the thickness of the air film and n = 0, 1, 2 is the interference mode. Thus   2n + 1 L= λ (0.184.2) 4

D

We get

λ = 122 nm 4 3λ L1 = = 366 nm 4 5λ L2 = = 610 nm 4 L0 =

(0.184.3) (0.184.4) (0.184.5)

Answer: (E) 14

Incidentally, you can multiply the D2 frequency by six to determine the harmonic. This turns out to

be

73.416 × 6 = 440.496 Hz

(0.183.3)

Subtracting this from the A2 frequency gives 440.496 − 440.000 = 0.496 Hz

David S. Latchman

(0.183.4)

©2009

Mass moving on rippled surface

0.185

cxv

Mass moving on rippled surface

For the mass to stay on the rippled surface, the particle’s horizontal velocity should not be so great that it flies off the track. So as it falls, it must hug the track. The time for the particle to fall from the top of the track to the bottom is 2d = where

1 2 gt 2

(0.185.1)

s t=

4d g

(0.185.2)

We could also have said,

but this is the same as eq. (0.185.2).

FT

  1 2π = gt2 d − d cos k · 2k 2 1 ∴ 2d = gt2 2

RA

To stay on the track, the particle must cover a horizontal distance of x = π/k in the same time. Thus the horizontal speed, v, is   x π g 1/2 v= = t k 4d  g 1/2 π = · 2 (0.185.3) 2 kd Any speed greater that 0.185.3 would result in the particle flying off the track. So r g v≤ (0.185.4) k2 d

D

Answer: (D)

0.186

Normal Modes and Couples Oscillators

NOT FINISHED Answer: (D)

0.187

Waves

NOT FINSIHED Answer: (B) ©2009

David S. Latchman

GR9677 Exam Solutions

cxvi

0.188

Charged Particles in E&M Fields

NOT FINISHED Answer: (B)

0.189

Rotation of Charged Pith Balls in a Collapsing Magnetic Field

I ∂S

FT

As the magnetic field, B, collapses, it indices a clockwise electric field, E, that causes the charged pith balls to rotate. The electric field can be found from Faraday’s Law E · d` = −

The magnetic flux covers an area, A, of

∂Φ ∂t

A = πR2

(0.189.1)

(0.189.2)

RA

and the electric field makes a loop of circumference d` = πd

(0.189.3)

Substituting eqs. (0.189.2) and (0.189.3) into eq. (0.189.1) gives us E · πd = πR2

dB dt

(0.189.4)

D

Solving for E, gives us

E=

R2 dB d dt

(0.189.5)

The question answers ask for the angular momentum, we recall that the torque is τ=r×F

(0.189.6)

where F = qE. Substituting eq. (0.189.5) into eq. (0.189.6) dL dB = qR2 dt dt

(0.189.7)

L = qR2 B

(0.189.8)

So it follows that

Answer: (A) David S. Latchman

©2009

Coaxial Cable

0.190

cxvii

Coaxial Cable

We expect there to be no magnetic field outside, r > c, the coaxial cable. Choices (D) and (E) make no sense. So choice (B) is our best one left. We can also use Ampere’s Law to show what the magnetic induction will look like as we ove away from the center. We recall Ampere’s Law I B · d` = µ0 Ienclosed (0.190.1) 0 < r < a The current enclosed by our Amperian Loop is πr2 πa2

(0.190.2)

FT

Ienclosed = I Ampere’s Law shows us

πr2 πa2 µ0 I r B(0
(0.190.3)

RA

a < r < b Within the shaded region, the enclosed current is, Ienclosed = I. Ampere’s Law becomes B (2πr) = µ0 I µ0 I B(a
(0.190.4)

D

b < r < c The area of the outer sheath is

  A = π c2 − b 2

(0.190.5)

The enclosed current will be πr2 − πb2 =I−I πc2 − πb2 " 2 # c − r2 =I 2 c − b2 "

Ienclosed

#

(0.190.6)

Ampere’s Law becomes "

# c2 − r2 B (2πr) = µ0 I 2 c − b2 " 2 # µ0 I c − r2 B(b
(0.190.7)

This will fall to an inversely with respect to r. ©2009

David S. Latchman

cxviii r rel="nofollow"> c We see that Ienclosed = 0, so from Ampere’s Law B(r>c) = 0

GR9677 Exam Solutions

(0.190.8)

The graph shown in (B) fits this15 . Answer: (B)

0.191

Charged Particles in E&M Fields

FT

A seemingly difficult question but it really is not16 . All we need to turn to is Pythagoras Theorem and relate the centripetal force to the Lorentz Force Law. From the Lorentz Force Law, we see that

mv2 = Bqv r

RA

We can simplify this to read,

p = Bqr

(0.191.1)

(0.191.2)

We know B and q. We can determine r in terms of s and ` by using Pythagoras Theorem. r2 = `2 + (r − s)2

(0.191.3)

! 1 `2 +s r= 2 s

(0.191.4)

D

Solving for r, gives us

As s << `, eq. (0.191.4) becomes r=

1 `2 2 s

(0.191.5)

Substituting eq. (0.191.5) into eq. (0.191.2), gives us Bq`2 2s

(0.191.6)

Answer: (D) 15

This is one of the reasons we use coaxial cables to transmit signals. No external magnetic field from our signals means that we can, theoretically, eliminate electromagnetic interference. 16 Draw Diagrams

David S. Latchman

©2009

THIS ITEM WAS NOT SCORED

cxix

0.192

THIS ITEM WAS NOT SCORED

0.193

The Second Law of Thermodynamics

This question deals with the Second Lar of Thermodynamics, which states that the entrophy of an isloated system, which is not in equilibrium, will increase over time, reaching a maximum at equilibrium. An alternative but equivalent statement is the Clausius statement, “Heat can not flow from cold to hot without work input”. Thusit would be impossible to attain the 900 K temperature the experimenter needs with a simple lens.

0.194

Small Oscillations

FT

Answer: (E)

We are given a one dimensional potential function

V(x) = −ax2 + bx4

(0.194.1)

D

RA

We can find the points of stability by differentiating the above equation to get and setting it to zero. dV = −2ax + 4bx3 = 0 (0.194.2) dx we see that x = 0; a/2b. Taking the second differential gives us the mass’s spring constant, k d2 V k= = −2a + 12bx2 (0.194.3) dx2 We can also use this to find the minimum and maximum points of inflection in our potential graph. We see that k=

d2 V(0) = −2a dx2

k=

d2 V(a/2b) = 4a dx2

We see that when x = a/2b, we are at a minima and hence at a point of stable equilibrium. We can now find the angular frequency r k ω= m r 4a = m r a =2 (0.194.4) m Answer: (D) ©2009

David S. Latchman

GR9677 Exam Solutions

cxx

0.195

Period of Mass in Potential

The total period of our mass will be the time it takes to return to the same point, say the origin, as it moves through the two potentials. for x < 0 The potential is 1 V = k2 2

(0.195.1)

The spring constant, k, is d2 V =k dx2 Thus the period of oscillation would be

(0.195.2)

FT

k=

r

T = 2π

m k

(0.195.3)

But this represents the period of the mass could also swing from x > 0. As a result, our period would just be half this. So r

m k

RA

T(x<0) = π

(0.195.4)

for x > 0 The potential is

V = mgx

(0.195.5)

D

We may recognize this as the gravitational potential energy. In this case, the ”period” would be the time for the mass to return the origin. This would simply be 1 s = v0 t − gt2 (0.195.6) 2 where s = 0. Solving for t, we get t = 0 and t = 2v0 /g We can solve this in terms of the total energy, E of the mass. The energy is E = 1/2mv20 . Our period works out to be s T(x>0) = 2

2E mg2

(0.195.7)

Thus our total period is T = T(x<0) + T(x>0) s r m 2E =π +2 k mg2

(0.195.8)

Answer: (D) David S. Latchman

©2009

Internal Energy

0.196

cxxi

Internal Energy

NOT FINISHED Answer: (D)

0.197

Specific Heat of a Super Conductor

FT

Superconductors experience an increase in their specific heat, Cv , at the transition temperature, Tc . At high temperatures, the specific heat falls linearly but at low temperatures, the specific heat falls below the linear dependence of a degenrate Fermi gas and is proportional to exp(−k/T). If the superconductor is placed in a magnetic field that exceeds the critical field strength, B > Bc , the specific heat reverts to the normal linear behavior. We see both the linear dependence and exponential decay below a certain value in Choice:(E). Answer: (E)

Pair Production

RA

0.198

We are given the equation of pair production

γ → e− + e+

(0.198.1)

D

Pair production occurs when a γ ray of high energy is absorbed in the vincinity of an atomic nucleus and particles are created from the absorbed photon’s energy. This takes place in the Coulomb field of the nucleus; the nucleus acts as a massive body to ensure the conservation of momentum and energy. The nucleus is an essential part of this process; if the photon could spontaneously decay into an electron-positron pair in empty space, a Lorentz frame could be found where the electron and positron have equal and opposite momenta and the photon will be at rest. This is a clear violation of the principles of Special Relativity. We choose (A). Based on what we know, we can determine the maximum wavelength as a matter or interest. As the nucleus is massive, we will ignore its recoil and consider that all of the photon’s energy goes into electron-positron creation and the particles’ kinetic energy. hυ = E− + E+     = K− + me c2 + K+ + me c2 = K− + K+ + 2me c2

(0.198.2)

where K− and K+ are the kinetic energies of the electron and positron respectively. Thus the minimum energy needed to initiate this process is hυmin = 2me c2 ©2009

(0.198.3) David S. Latchman

GR9677 Exam Solutions

cxxii The wavelength of this photon is λmax =

h 2me c2

(0.198.4)

which happens to be half the Compton wavelength. Answer: (A)

0.199

Probability Current Density

FT

The probability current density is defined as17 ~ ∂Ψ ∂Ψ∗ j(x, t) − Ψ∗ − Ψ 2im ∂x ∂x Thus given the wavefunction

!

  Ψ(x, t) = eiωt α cos kx + β sin kx

RA

Its complex conjugate is

  Ψ∗ (x, t) = e−iωt α∗ cos kx + β∗ sin kx

(0.199.1)

(0.199.2)

(0.199.3)

D

Differentiating the above two equations yields

  dΨ = eiωt −kα sin kx + kβ cos kx dx

(0.199.4)

  dΨ∗ = e−iωt −kα∗ sin kx + kβ∗ cos kx dx

(0.199.5)

Substituting eqs. (0.199.2) to (0.199.5) into eq. (0.199.1) yields j(x, t) =


k~ α∗ β − β∗ α 2im

(0.199.6)

Answer: (E)

0.200

Quantum Harmonic Oscillator Energy Levels

Given the potential 1 V(x) = mω2 x2 2 17

(0.200.1)

Add derivation in section.

David S. Latchman

©2009

Three Level LASER and Metastable States Solving the Schrodinger’s Equation for this potential ¨

−

cxxiii

~2 d2 ψ 1 + mω2 x2 ψ = Eψ 2m dx2 2

(0.200.2)

leaves us with mω ψn = π~ 

 14

√

1 2n n!

Hn e−ξ /2 2

(0.200.3)

FT

where H( n) are Hermite polynomials and with energy levels of18   1 En = n + ~ω 2

(0.200.4)

But we have placed an infinitely large barrier, V = ∞ at x ≤ 0. This serves to constrain ψn (0) = 0 at x = 0. This occurs at n = 1, 3, 5, . . . or rather only odd values of n are allowed. Thus

Answer: (D)

(0.200.5)

Three Level LASER and Metastable States

D

0.201

RA

5 7 11 3 En = ~ω, ~ω, ~ω, ~ω, . . . 2 2 2 2

Metastability describes a state of delicate equilibrium. Such a system is in a state of equilibrium but is susceptible to fall into a lower-energy state with a slight interaction.

The laser operation depends on an active medium of atoms whose energy states can be populated selectively by radiative means.19 In the three-level laser, a discharge of some sort raises atoms from the ground state, E1 , to a higher state, E3 . A rapid spontaneous decay then occurs, bringing the excited atoms down to the E2 state. This state is metastabe as it inhibits spontaneous decay back down to the ground state. It needs an incident photon of energy hυ = E2 −E1 to stimulate the desired laser transition back down to the ground state, E1 . 18 19

Add wavefunctions here Add laser explanation in one of the sections.

©2009

David S. Latchman

GR9677 Exam Solutions

cxxiv Pumping Transition

E3

E2

Metastable Level

E1

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Figure 0.201.1: Three Level Laser The Ruby Laser is an example of a three-level laser. Green light from a flash lamp pumps the chromium ions to an excited level and non-radiative de-excitation promptly brings the ions to a long lived metastable state. Stimulated emission then follows, generating a coherent beam of red light of 694 nm.

0.202

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Answer: (B)

Quantum Oscillator – Raising and Lowering Operators

We are given the lowering operator

r

aˆ =

pˆ mω0 xˆ + i 2~ mω0

! (0.202.1)

D

The raising operator is given by

r aˆ = †

pˆ mω0 xˆ − i 2~ mω0

! (0.202.2)

So we immediately see that aˆ† , aˆ

(0.202.3)

which we see from Choice: III. But what about the other two choices? In reality we only have to prove/disprove Choice: I but we will go through both. We know that aˆ is not Hermitian from eq. (0.202.3) and hence not observable. Hence we can eliminate Choice: II. This affects Choice: I as aˆ is not observable and hence won’t commute with the Hamiltonian, H. We eliminate Choice: I. Answer: (C) David S. Latchman

©2009

Constants & Important Equations 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

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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

.1

D

Table .1.1: Something

.2

.2.1

Vector Identities Triple Products

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

(.2.1) (.2.2)

Constants & Important Equations

cxxvi

.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

.2.3

∇ · (∇ × A) = 0
∇ × ∇f = 0

∇ × (∇ × A) = ∇ (∇ · A) − ∇2 A

.3.1

Commutators

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.3

D .3.3

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

Lie-algebra Relations

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

.3.2

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

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

Canonical Commutator [x, p] = i~

(.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

(.3.5) ©2009

Linear Algebra

.4

cxxvii

Linear Algebra

.4.1

Vectors

Vector Addition The sum of two vectors is another vector

Associative Zero Vector

|αi + |βi = |βi + |αi

(.4.2)



|αi + |βi + |γi = |αi + |βi + |γi

(.4.3)

|αi + |0i = |αi

(.4.4)

|αi + | − αi = |0i

(.4.5)

D

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Inverse Vector

(.4.1)

FT

Commutative

|αi + |βi = |γi

©2009

David S. Latchman

Constants & Important Equations

D

RA

FT

cxxviii

David S. Latchman

©2009

Bibliography

FT

[1] Wikipedia. Mean free path — wikipedia, the free encyclopedia, 2009. [Online; accessed 23-March-2009]. [2] John J. Brehm and William J. Mullin. Introduction to the Structure of Matter, chapter 12-3, page 594. Wiley, first edition, 1989. [3] John J. Brehm and William J. Mullin. Introduction to the Structure of Matter, chapter 15-5, page 772. Wiley, first edition, 1989.

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[4] John J. Brehm and William J. Mullin. Introduction to the Structure of Matter, chapter 5-10, pages 283–287. Wiley, first edition, 1989. [5] David J. Griffiths. Introduction to Quantum Mechanics, chapter 5.1.1, pages 203–205. Prentice Hall, second edition, 2005. [6] John J. Brehm and William J. Mullin. Introduction to the Structure of Matter, chapter 11-1, pages 539–540. Wiley, first edition, 1989.

D

[7] John J. Brehm and William J. Mullin. Introduction to the Structure of Matter, chapter 2-8, pages 114–116. Wiley, first edition, 1989.

Index

Bohr Model Hydrogen Model, lv

Parallel Axis Theorem, see Rotational Motion

RA

Celestial Mechanics, xxiv Circular Orbits, xxv Escape Speed, xxiv Kepler’s Laws, xxv Newton’s Law of Gravitation, xxiv Orbits, xxv Potential Energy, xxiv Circular Orbits, see Celestial Mechanics Commutators, cxxvi Canonical Commutators, cxxvi Kronecker Delta Function, cxxvi Lie-algebra Relations, cxxvi Compton Effect, lviii Counting Statistics, lxxi

Coupled Harmonic Oscillators, xx Damped Motion, xix Kinetic Energy, xviii Potential Energy, xix Simple Harmonic Motion Equation, xviii Small Oscillations, xix GR9677 Q92, cxix Total Energy, xviii

FT

Angular Momentum, see Rotational Motion

D

Doppler Effect, xxii

Franck-Hertz Experiment, lxi

Rolling Kinetic Energy, see Rotational Motion Rotational Kinetic Energy, see Rotational Motion Rotational Motion, xxii Angular Momentum, xxiii Moment of Inertia, xxii Parallel Axis Theorem, xxiii Rolling Kinetic Energy, xxiii Rotational Kinetic Energy, xxii Torque, xxiii

Gravitation, see Celestial Mechanics

Subject, xliv System of Particles, xxiv

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

Torque, see Rotational Motion

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

Vector Identities, cxxv Product Rules, cxxvi Second Derivatives, cxxvi Triple Products, cxxv