PH253 Lecture 10: Photons, but more so Compton scattering P. LeClair Department of Physics & Astronomy The University of Alabama Spring 2020 LeClair, Patrick (UA) PH253 Lecture 9 February 3, 2020 1 / 33
Exam 1 Raw average: 71.4% 1 Average excluding results under 60% (21/73): 87.2% 2 Very bimodal distribution . . . 3 Exams returned, solution out when makeups done. 4 PH253 Exam 1 30 25 20 15 10 5 0 <50 50-59 60-69 70-79 80-89 90-100 LeClair, Patrick (UA) PH253 Lecture 9 February 3, 2020 2 / 33
Outline Compton Scattering 1 Problems 2 LeClair, Patrick (UA) PH253 Lecture 9 February 3, 2020 3 / 33
Last time: Light looks like a particle on large scales ( ≫ λ ) 1 But it looks like a wave on small scales ( � λ ) 2 Photoelectric effect details supports this idea 3 Photons have momentum by virtue of energy, but no mass 4 Better evidence light = photons? Scattering. 5 Energy & Momentum for photons: E = h f = hc p = E c = h f c = h λ λ LeClair, Patrick (UA) PH253 Lecture 9 February 3, 2020 4 / 33
Outline Compton Scattering 1 Problems 2 LeClair, Patrick (UA) PH253 Lecture 9 February 3, 2020 5 / 33
Compton scattering if light = particles = photons . . . 1 . . . scatter the photons off of another particle, e.g., e − 2 if photon=particle, specific angular dispersion, energy loss 3 Energy loss of photon = red shift = observable 4 classical (EM waves) - incident / scattered photons ∼ same f 5 LeClair, Patrick (UA) PH253 Lecture 9 February 3, 2020 6 / 33
Compton scattering An incident photon of frequency f i , energy E i = h f i , and momentum p i = h / λ i strikes an electron (mass m ) at rest. The photon is scattered through an angle θ , and the scattered photon has frequency f f , energy E f = h f f , and momentum p f = h / λ f . Electron recoils at angle ϕ relative to the incident photon, acquires kinetic energy KE e and momentum p e . recoiling electron e − incident photon ϕ e − θ target electron at rest scattered photon LeClair, Patrick (UA) PH253 Lecture 9 February 3, 2020 7 / 33
Compton scattering Conserve energy and momentum: � m 2 c 4 + p 2 e c 2 − mc 2 h f i = h f f + KE e = h f f + Noted KE e − is total energy minus its rest energy mc 2 . Conservation of p in both directions: p i = p e cos ϕ + p f cos θ p e sin ϕ = p f sin θ recoiling electron e − incident photon ϕ e − θ target electron at rest scattered photon LeClair, Patrick (UA) PH253 Lecture 9 February 3, 2020 8 / 33
Compton scattering Solution made simpler by defining dimensionless energy parameters. Recognizes e − rest energy mc 2 is the key energy scale α i = incident photon energy = h f i mc 2 electron rest energy = h f f α f = scattered photon energy mc 2 electron rest energy ǫ = electron kinetic energy = E e mc 2 electron rest energy LeClair, Patrick (UA) PH253 Lecture 9 February 3, 2020 9 / 33
Compton scattering Substitutions change our energy and momentum equations to: � p 2 e α i = α f + m 2 c 2 + 1 − 1 � p e � α i = α f cos θ + cos ϕ mc � p e � α f sin θ = sin ϕ mc LeClair, Patrick (UA) PH253 Lecture 9 February 3, 2020 10 / 33
Compton scattering Experiment gets incident and scattered photons’ energy and angle So eliminate the electron’s momentum p e and scattering angle ϕ Rearrange energy equation, square it, solve for p e : � p 2 e α i − α f + 1 = m 2 c 2 + 1 p 2 � 2 − 1 e � m 2 c 2 = α i − α f + 1 e = m 2 c 2 � � p 2 α 2 i − 2 α i α f + α 2 f + 2 α i − 2 α f LeClair, Patrick (UA) PH253 Lecture 9 February 3, 2020 11 / 33
Compton scattering �� �� � 2 + 2 p 2 e = m 2 c 2 α i − α f � α i − α f Now square and add the two momentum equations to eliminate ϕ � p e � e cos 2 ϕ = m 2 c 2 � � 2 p 2 cos ϕ = α i − α f cos θ = ⇒ α i − α f cos θ mc � p e � e sin 2 ϕ = m 2 c 2 α 2 f sin 2 θ p 2 sin ϕ = α f sin θ = ⇒ mc � � f sin 2 θ + � 2 p 2 e = m 2 c 2 α 2 � α i − α f cos θ e = m 2 c 2 � f sin 2 θ + α 2 f cos 2 θ � p 2 α 2 i − 2 α i α f cos θ + α 2 LeClair, Patrick (UA) PH253 Lecture 9 February 3, 2020 12 / 33
Compton scattering e = m 2 c 2 � � p 2 α 2 f + α 2 i − 2 α i α f cos θ Comparing this with our previous equation for p 2 e : e = m 2 c 2 � � p 2 α 2 i − 2 α i α f + α 2 f + 2 α i − 2 α f α 2 i − 2 α i α f + α 2 f + 2 α i − 2 α f = α 2 f + α 2 i − 2 α i α f cos θ − α i α f + α i − α f = − α i α f cos θ 1 − 1 α i − α f = α i α f ( 1 − cos θ ) or = 1 − cos θ α f α i LeClair, Patrick (UA) PH253 Lecture 9 February 3, 2020 13 / 33
Compton scattering End result: substitute back for α = h f / mc 2 = h / λ mc . 1 − 1 = 1 − cos θ α f α i λ f mc − λ i mc = 1 − cos θ h h λ f − λ i = ∆ λ = h mc ( 1 − cos θ ) This is the Compton equation. h / mc has units of length - the Compton wavelength λ c = h / mc ≈ 2.42 × 10 − 12 m. Scale at which quantum effects dominate LeClair, Patrick (UA) PH253 Lecture 9 February 3, 2020 14 / 33
Compton scattering Figure: Original data. Physical Review 21 , 483-502 (1923) LeClair, Patrick (UA) PH253 Lecture 9 February 3, 2020 15 / 33
Compton scattering Figure: It works! Physical Review 21 , 483-502 (1923) LeClair, Patrick (UA) PH253 Lecture 9 February 3, 2020 16 / 33
Electron kinetic energy e − kinetic energy = difference between incident & scattered photons: KE e = h f i − h f f = α i mc 2 − α f mc 2 = mc 2 � � α i − α f Solving the Compton equation for α f , we have α i α f = 1 + α i ( 1 − cos θ ) Combining these two equations � � α i mc 2 = mc 2 KE e = � α i − α f � α i − 1 + α i ( 1 − cos θ ) α 2 i ( 1 − cos θ ) ǫ = KE e mc 2 = 1 + α i ( 1 − cos θ ) LeClair, Patrick (UA) PH253 Lecture 9 February 3, 2020 17 / 33
Compton scattering � α i � mc 2 = mc 2 � � KE e = α i − α f α i − 1 + α i ( 1 − cos θ ) α 2 i ( 1 − cos θ ) ǫ = KE e mc 2 = 1 + α i ( 1 − cos θ ) e − kinetic energy can only be a fraction of the incident photon’s energy There will always be some energy left over for a scattered photon. Means that a stationary, free electron cannot absorb a photon! Absorption can only occur if the electron is bound to, e.g., a nucleus which can take away a bit of the net momentum and energy. LeClair, Patrick (UA) PH253 Lecture 9 February 3, 2020 18 / 33
Compton scattering Shift in wavelength ∆ λ independent of photon energy 1 Shift in photon energy is not 2 Change in photon energy is equal to e − KE 3 Strongly dependent on the incident photon energy! 4 Relevant energy scale set by the ratio of the incident photon 5 energy to e − rest energy If ratio is large, the fractional shift in energy is large 6 When the incident photon energy ∼ e − rest energy, Compton 7 scattering significant mc 2 ≈ 511 keV, hard x-rays or gamma rays 8 LeClair, Patrick (UA) PH253 Lecture 9 February 3, 2020 19 / 33
Compton scattering Fraction of the photon energy retained by the e − vs. incident photon energy for various photon scattering angles. 1.00 Fraction of incident photon energy retained by electron 0.75 0.50 Photon scattering angle 30 30 120 45 45 135 60 60 150 90 90 180 0.25 0 0.1 1 10 100 1000 photon energy / electron rest mass LeClair, Patrick (UA) PH253 Lecture 9 February 3, 2020 20 / 33
Compton scattering Maximum electron energy or photon energy shift? Set d ǫ / d θ = 0: α 2 i ( 1 − cos θ ) ǫ = KE e mc 2 = 1 + α i ( 1 − cos θ ) � � d ǫ − α i sin θ ( 1 − cos θ ) sin θ d θ = α 2 ( 1 + α i ( 1 − cos θ )) 2 + = 0 i 1 + α i ( 1 − cos θ ) 0 = sin θ [ − α i + α i cos θ + 1 + α i ( 1 − cos θ )] 0 = sin θ = ⇒ θ = { 0, π } θ = 0 can be discarded, this corresponds to the photon going right through the electron At θ = π , backward scattering of photon, electron has max energy LeClair, Patrick (UA) PH253 Lecture 9 February 3, 2020 21 / 33
Compton scattering Maximum energy of the electron at θ = π (photon backscattered) � 2 α i � KE max = h f i 1 + 2 α i 2 α 2 � � 2 α i i ǫ = α i = 1 + 2 α i 1 + 2 α i Max e − KE at most a fraction of the incident photon energy - absorption cannot occur for free electrons LeClair, Patrick (UA) PH253 Lecture 9 February 3, 2020 22 / 33
Compton scattering Compton wavelength sets fundamental limitation on measuring the position of a particle. Depends on the mass m of the particle. Can measure the position of a particle by bouncing light off it But! Need short wavelength for accuracy. That means higher p and E for the photon! (Which disturbs position . . . uncertainty) If photon energy > mc 2 , when it hits particle being measured there is enough energy to create a new particle of the same type! Meaning you still don’t know where the original one is. Or if photon energy > 2 mc 2 , photon can decay into particle-antiparticle pair LeClair, Patrick (UA) PH253 Lecture 9 February 3, 2020 23 / 33
Compton scattering e − ( β − ) E γ > 1.022 MeV e + ( β + ) 511 keV 511 keV Figure: Pair production A photon can “decay” into an electron and a positron (electron antiparticle). Try to measure electron with high energy photon? Now you have 3 particles. LeClair, Patrick (UA) PH253 Lecture 9 February 3, 2020 24 / 33
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