Modern Physics Unit 1: Classical Models and the Birth of Modern Physics Lecture 1.6: Compton Effect Ron Reifenberger Professor of Physics Purdue University 1 V. The Compton Effect (1923) Energy Loss Θ=30o Carbon block source of high energy photons Compton’s Apparatus (schematic) What was expected? - e f hf f hf No energy loss ! Source: Tipler and Llewellyn, Modern Physics, 4th edition2 Compton applies conservation of energy and momentum principles to explain puzzling data. Recall collision theory: Before After vi1 vi2 vf1 vf2 m m m m2 1 m22 1 Conservation of momentum: m1vf1 + m2vf2 = m1vi1 + m2vi2 Conservation of kinetic energy: 2 2 2 2 ½ m1vf1 + ½ m2vf2 = ½ m1vi1 + ½ m2vi2 3 What’s the momentum of a photon? λ Classical mechanics: p = mv p=???? E but….photon mass = 0 ⋅ from Planck E = hf ⋅ from Maxwell, the momentum of an EM Wave is: U Calculate change (see L1.03) in momentum of a p = surface when hit c by EM wave hf hf h p = = = cfλλ EM Wave h=Planck’s constant = 6.626x10-34 m2 kg/s 4 Compton’s analysis considered an X-ray photon scattering from a single electron inside a target Before After photon photon λ f electron at loses λ i rest energy me electron recoils How is λ f related to λ i ? 5 Geometry of Compton Scattering λ Initial Scattered f position of photon electron Incident photon Θ h _ λλfi−=(1 − cos Θ) mce λ i me _ Recoiling electron h 6.626×⋅ 10−34 Js ≡= Compton wavelength −31 8 mce (9.109×× 10kg )( 2.998 10m / s) =×=×=2.43 10−−3 nm 2.43 1012 m 2.43 pm Note: 1 pm= 1 × 10−12 m 6 Summary: Compton Scattering h λλfi−=(1 − cos Θ) mce energy h loss Inelastic mce λf Scattering λi λf Increases λi λi λf number of π photons h Θ= detected 2 2 mce Scattered λi = λf λi λf Beam Elastic m λ e Scattering Θ=π Θ=0 Example: EM radiation with differing wavelengths is scattered by an electron through an angle of 37o (w.r.t. incident beam). What is the wavelength of the scattered radiation? h Ef λλ−= − Θ fi (1 cos ) λ f mce E i o _ Θ=37 λi Fixed once θ is set! _ Recoiling electron If λ = h (1− cosθ ) then λ = Fractional Shift i mce f -6 420 nm 0.489 pm 420.000489 nm 1.16x10 S M -5 42 nm λ f0.489= 4.2 pm pm 42.000489 nm 1.16x10 A L -4 4.2 nm 0.489 pm 4.200489 nm 1.16x10 L -2 0.042 nm 0.489 pm 0.042489 nm 1.16x10 0.0042 nm 0.004689 nm -1 0.489 pm 1.16x10 (4.20 pm) (4.69 pm) 8 Why assume incident EM radiation is scattered only by an electron? Compton’s analysis is completely general. Evaluate the Compton wavelength for scattering by different particles. h Compton wavelength ≡ mc Possible Scattering Mass m (kg) Compton Wavelength Center Carbon atom 1.993 x 10-26 1.11x10-16 m really proton 1.672 x 10-27 1.32x10-15 m small! electron 9.109 x 10-31 2.43x10-12 m 9 Physical Significance of Compton Wavelength? h Compton wavelength ≡=λ C mc 11 rewriting: mc= h ⇒= mc2 hc = hf λλCC = m E=hf E=mc2 What is smallest wavelength possible for an EM wave? Photons with a wavelength smaller than λC tend to split into electron-positron pairs. Historically, Compton’s experiment was important because it finally convinced the last skeptics that light can behave as a particle. We are discussing different wavelengths . the wavelength of EM radiation . the de Broglie wavelength of a particle (next set of lectures) . the Compton wavelength of a particle Can you associate the correct wavelength with each experiment? Experiment Which wavelength? Maxwell’s EM radiation Young’ 2-slit experiment Blackbody radiation Photoelectric effect Electron diffraction Discrete line spectra X-ray diffraction Compton Scattering .