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

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 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 ? λ Classical : p = mv p=???? E but….photon = 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 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

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 −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 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 4.2 nm 0.489 pm 4.200489 nm 1.16x10-4 L 0.042 nm 0.489 pm 0.042489 nm 1.16x10-2 0.0042 nm 0.004689 nm 0.489 pm 1.16x10-1 (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 . h Compton wavelength ≡ mc

Possible Scattering Mass m (kg) Compton Wavelength Center Carbon atom 1.993 x 10-26 1.11x10-16 m really 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 . 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 Discrete line spectra X-ray diffraction Compton Scattering