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Lecture physics a2 wave particle duality huynh quang linh

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Lecture 5 Wave Particle Duality f (x1014 Hz) V st op (v ) 0 0 5 1 1 5 2 2 5 3 3 5 0 5 10 15 f0 S1 S2 + V Collector Metal Surface electrons vacuum A Content  Photoelectric Effect  light as particles[.]

Lecture 5:Wave-Particle Duality Collector A electrons S1 Metal Surface + V vacuum 3.5 S2 Vstop (v) 2.5 f0 1.5 0.5 0 10 f (x1014 Hz) 15 Content  Photoelectric Effect  light as particles (“Quantization of EM waves”)  Photon momentum/Compton scattering  Wave-particle Duality  Weird Quantumness Wave-particle Duality for Light and Matter  In previous lectures we viewed “light” as a wave (i.e it causes interference and diffraction)    Surprise: In the early 1900’s, it was discovered that light has “particle”-like properties in some situations! Furthermore, “matter” (i.e., electrons, protons, etc.) was found to exhibit “wave-like” properties under certain circumstances These two discoveries revolutionized science and technology What’s the evidence of the wave-particle duality? Photoelectric Effect (1)  Electrons in a metal are “bound” by the energy F, the “work function” If you shine light on a clean metal surface, electrons can emerge  the light gives the electrons enough energy (> F) to escape    Binding potential perform the experiment in vacuum measure the flow of emitted electrons with an ammeter How will the current depend on I and f? We might expect:    Increasing intensity I should increase the current (By increasing the electric field E, the force on electrons, F = eE, is increased, causing more electrons to be kicked out of the metal.) Increasing frequency f shouldn’t matter much Perhaps a decrease in current due to rapid oscillations With a weak light, there should be a time delay before current starts to flow (to build up enough energy) Photoelectric Effect (2) Incident Light (variable frequency f) Collector  Experiment 1: Measure the maximum energy of ejected electrons  A  electrons Metal Surface + vacuum V  Bias the “collector” with a negative charge to repel ejected electrons Increase negative bias voltage until flow of ejected electrons decreases to zero (Current = at V = Vstop) Measurement of Vstop tells the max kinetic energy, KEmax = eVstop The Result: The “stopping voltage” is independent of light intensity! Therefore, increasing the intensity I does not increase KE ! Photoelectric Effect (3) Incident Light (variable frequency f) Experiment 2: Measure the maximum energy vs f Vstop (v)  Collector A f0 electrons 0 10 15 f (x1014 Hz) Metal Surface + V vacuum The Results:   Stopping voltage Vstop (and the maximum kinetic energy of electrons) decreases with decreasing f (linear dependence) Below a certain frequency fo, no electrons are emitted, even for intense light! Makes no sense classically: Increasing E should have an effect Photoelectric Effect (4) Vstop (v) slope h/e f0 Collector A 0 10 electrons 15 f (x1014 Hz) Metal Surface + V Summary of Results: vacuum  Energy of electrons emitted depends on frequency, not intensity  Electrons have a probability to be emitted immediately  Electrons are not ejected for frequencies below f0 KEmax  e Vstop  hf  F Conclusion: h is Planck’s constant (measured here) F is the “work function” Light comes in “packets” of energy with Ephoton = hf Photons ! h = 6.626 x 10-34 J • s Increasing I simply increases # photons, not the photon energy Convenient Units for Ephoton Recall: For EM waves, frequency and wavelength are related by f = c/l New result:    Photons Light comes in “packets” of energy Ephoton = hf = hc/l  c = 2.9979 x 108 m/s h = 6.626 x 10-34 J • s hc = 1.986 x 10-25 J • m For light waves it is useful to define wavelength in nanometers (nm) For electrons it is useful to define energy in electron volts (eV) eV = energy an electron gains moving through a potential energy of one volt = (1.6022 x 10-19 Coulomb)(1 volt) = 1.6022 x 10-19 Joules Therefore, h = 4.14 x 10-15 eV-s, and hc = 1240 eV-nm E photon  1240 eV  nm l Ephoton in electron volts l in nanometers Example: A red photon with wavelength of 620 nm has an energy of eV Photoelectric Effect: Example When light of wavelength l = 400 nm shines on a piece of lithium, the stopping voltage of the electrons is Vstop = 0.21 V What is the work function of lithium? What is the maximum wavelength that can cause the photoelectric effect in lithium? Hint: What is Vstop at the maximum wavelength (minimum frequency)? Photoelectric Effect: Example When light of wavelength l = 400 nm shines on a piece of lithium, the stopping voltage of the electrons is Vstop = 0.21 V What is the work function of lithium? Solution: F = hf - eVstop = 3.1eV - 0.21eV = 2.89 eV E photon  hc 1240 eV  nm  l l  1240 eV  3.1 eV 400 For Vstop = 0.21 V, eVstop = 0.21 eV What is the maximum wavelength that can cause the photoelectric effect in lithium? Hint: What is Vstop at the maximum wavelength (minimum frequency)? Answer: lmax = 429 nm ... light as particles (“Quantization of EM waves”)  Photon momentum/Compton scattering  Wave- particle Duality  Weird Quantumness Wave- particle Duality for Light and Matter  In previous lectures... was found to exhibit ? ?wave- like” properties under certain circumstances These two discoveries revolutionized science and technology What’s the evidence of the wave- particle duality? Photoelectric... maximum wavelength that can cause the photoelectric effect in lithium? Hint: What is Vstop at the maximum wavelength (minimum frequency)? Photoelectric Effect: Example When light of wavelength

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