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

Transcription

1 General Pysics (PHY 2140) Lecture 15 Modern Pysics Cater Quantum Pysics Te Comton Effect Potons and EM Waves Wave Proerties of Particles Wave Functions Te Uncertainty Princile Reminder: Exam 3 Friday, July questions. Sow your work for credit. Closed book. You may bring a age of notes. Bring a calculator and a en or encil tt:// Last lecture: Ligtning Review 1. Quantum ysics Blackbody radiation Planck s s yotesis Potoelectric effect X-rays 2 λmaxt = m K En = nf, n = 1,2,3,... KE = f Φ c λ min = e ΔV ( ) Review Problem: A xenon arc lam is covered wit an interference filter tat only transmits ligt of 400-nm wavelengt. Wen te transmitted ligt strikes a metal surface, a stream of electrons emerges from te metal. If te intensity of te ligt striking te surface is doubled, Te Comton Effect Comton directed a beam of x-rays x toward a block of graite He found tat te scattered x-rays x ad a sligtly longer wavelengt tat te incident x-raysx Tis means tey also ad less energy Te amount of energy reduction deended on te angle at wic te x-rays were scattered Te cange in wavelengt is called te Comton sift 1. more electrons are emitted in a given time interval. 2. te electrons tat are emitted are more energetic. 3. bot of te above. 4. neiter of te above. 1

2 Comton Scattering Comton Scattering Comton assumed te otons acted like oter articles in collisions Energy and momentum were conserved Te quantity /m e c is called te Comton wavelengt Comton wavelengt = nm Very small comared to visible ligt Te Comton sift deends on te scattering angle and not on te wavelengt Exeriments confirm te results of Comton scattering and strongly suort te oton concet Te sift in wavelengt is Δλ = λ λ o = mec (1 cosθ) Comton wavelengt = nm Problem: Comton scattering QUICK QUIZ 1 A beam of 0.68-nm otons (E=1828 ev) ) undergoes Comton scattering from free electrons. Wat are te energy and momentum of te otons tat emerge at a 45 angle wit resect to te incident beam? Δ λ = λ λo = (1 cos θ) mc e Δλ = nm x ( ) = 7.11x10-4 nm E=c/λ =c/0.6807nm = 1826 ev = /λ = /0.6807nm = 1826 ev/c An x-ray oton is scattered by an electron. Te frequency of te scattered oton relative to tat of te incident oton (a) increases, (b) decreases, or (c) remains te same. (b). Some energy is transferred to te electron in te scattering rocess. Terefore, te scattered oton must ave less energy (and ence, lower frequency) tan te incident oton. 2

3 QUICK QUIZ Wave Proerties of Particles A oton of energy E 0 strikes a free electron, wit te scattered oton of energy E moving in te direction oosite tat of te incident oton. In tis Comton effect interaction, te resulting kinetic energy of te electron is (a) E 0, (b) E, (c) E 0 E, (d) E 0 + E, (e) none of te above. (c). Conservation of energy requires te kinetic energy given to te electron be equal to te difference between te energy of te incident oton and tat of te scattered oton. In 1924, Louis de Broglie ostulated tat because otons ave wave and article caracteristics, eras all forms of matter ave bot roerties For instance, for a oton: c E = f = λ tus E c = = = c cλ λ De Broglie suggested tat tis formula is true for any article! Tus, te frequency and wavelengt of matter waves can be determined. I.e. de Broglie wavelengt of a article is or λ = λ= mv Wave Proerties of Particles Te frequency of matter waves can also be determined De Broglie ostulated tat all articles satisfy Einstein s relation Or, in oter words, E = f ƒ = E Te Davisson-Germer Exeriment Tey scattered low-energy electrons from a nickel target Tey followed tis wit extensive diffraction measurements from various materials Te wavelengt of te electrons calculated from te diffraction data agreed wit te exected de Broglie wavelengt Tis confirmed te wave nature of electrons Oter exerimenters ave confirmed te wave nature of oter articles 3

4 Problem: te wavelengt of a roton Calculate te de Broglie wavelengt for a roton (m =1.67x10-27 kg ) moving wit a seed of 1.00 x 10 7 m/s. Calculate te de Broglie wavelengt for a roton (m =1.67x10-27 kg ) moving wit a seed of 1.00 x 10 7 m/s. Given: v = 1.0 x 10 7 m/s Find: λ =? Given te velocity and a mass of te roton we can comute its wavelengt Or numerically, s λ = mv 34 ( J s) 31 7 ( kg )( m s) λ = = m QUICK QUIZ 2 A non-relativistic electron and a non-relativistic roton are moving and ave te same de Broglie wavelengt. Wic of te following are also te same for te two articles: (a) seed, (b) kinetic energy, (c) momentum, (d) frequency? λ = mv (c). Two articles wit te same de Broglie wavelengt will ave te same momentum = mv. If te electron and roton ave te same momentum, tey cannot ave te same seed because of te difference in teir masses. For te same reason, remembering tat KE = 2 /2m, tey cannot ave te same kinetic energy. Because te kinetic energy is te only tye of energy an isolated article can ave, and we ave argued tat te articles ave different energies, Equation ( f = E/ ) tells us tat te articles do not ave te same frequency. Te Electron Microscoe Te electron microscoe deends on te wave caracteristics of electrons Microscoes can only resolve details tat are sligtly smaller tan te wavelengt of te radiation used to illuminate te object Te electrons can be accelerated to ig energies and ave small wavelengts λ e m (5 m) for 50 kv acceleration otential. 4

5 27.8 Te Wave Function In 1926 Scrödinger roosed a wave equation tat describes te manner in wic matter waves cange in sace and time Scrödinger dinger s s wave equation is a key element in quantum mecanics ΔΨ i = H Ψ Δt Te Wave Function Te wave function deends on te article s s osition and te time Te value of Ψ 2 at some location at a given time is roortional to te robability of finding te article at tat location at tat time Scrödinger dinger s s wave equation is generally solved for te wave function, Ψ Orbitals of Atomic Hydrogen Comuter generated figures of atomic orbitals (electron wave functions) for te Hydrogen atom Te Uncertainty Princile Wen measurements are made, te exerimenter is always faced wit exerimental uncertainties in te measurements Classical mecanics offers no fundamental barrier to ultimate refinements in measurements Classical mecanics would allow for measurements wit arbitrarily small uncertainties 5

6 Te Uncertainty Princile Quantum mecanics redicts tat a barrier to measurements wit ultimately small uncertainties does exist In 1927 Heisenberg introduced te uncertainty rincile If a measurement of osition of a article is made wit recision Δx and a simultaneous measurement of linear momentum is made wit recision Δ, ten te roduct of te two uncertainties can never be smaller tan /4π Te Uncertainty Princile Matematically, ΔxΔ x 4π It is ysically imossible to measure simultaneously te exact osition and te exact linear momentum of a article Anoter form of te rincile deals wit energy and time: ΔEΔt 4π Tougt Exeriment te Uncertainty Princile Problem: macroscoic uncertainty A 50.0-g g ball moves at 30.0 m/s.. If its seed is measured to an accuracy of 0.10%, wat is te minimum uncertainty in its osition? A tougt exeriment for viewing an electron wit a owerful microscoe In order to see te electron, at least one oton must bounce off f it During tis interaction, momentum is transferred from te oton to te electron Terefore, te ligt tat allows you to accurately locate te electron ectron canges te momentum of te electron 6

7 A 50.0-g g ball moves at 30.0 m/s. If its seed is measured to an accuracy y of 0.10%, wat is te minimum uncertainty in its osition? Problem: Macroscoic measurement Given: v = 30 m/s Δv/v = 0.10% m = 50.0 g Find: δx =? Notice tat te ball is non-relativistic. Tus, = mv, and uncertainty in measuring momentum is δv Δ = m( Δ v) = m v v = kg m s = kg m s ( )( ) Tus, uncertainty relation imlies J s Δx = = π ( Δ ) π ( kg m s) 32 m A 0.50-kg block rests on te icy surface of a frozen ond, wic we can assume to be frictionless. If te location of te block is measured to a recision of 0.50 cm, wat seed must te block acquire because of te measurement rocess? Recall: ΔxΔ and x = mv 4π Scanning Tunneling Microscoe (STM) Scanning Tunneling Microscoe, cont Allows igly detailed images wit resolution comarable to te size of a single atom A conducting robe wit a sar ti is brougt near te surface Te electrons can tunnel across te barrier of emty sace By alying a voltage between te surface and te ti, te electrons can be made to tunnel referentially from surface to ti Te ti samles te distribution of electrons just above te surface Te STM is very sensitive to te distance between te surface and te ti Allows measurements of te eigt of surface features witin nm 7

8 Limitation of te STM STM Images Tere is a serious limitation to te STM since it deends on te conductivity of te surface and te ti Most materials are not conductive at teir surface An atomic force microscoe as been develoed tat overcomes tis limitation It measures te force between te ti and te samle surface Has comarable sensitivity More STM Images 8