Quantum Physics Lecture 2

Particle-like properties of wavesCompton effect - photon/particle scattering

Pair production & annihilation

Wave-like properties of particles – De Broglie wavelength

Electron diffraction – Davisson & Germer experiment

TCD SF PYP20 2016 J.B.Pethica

Compton Effect Photon collides with stationary electron(must do full relativistic treatment)

Result: Electron scattered through angle θ and gains kinetic energy KEPhoton is scattered through angle φ and increases wavelength λ to λs i.e. decreases frequency ω to ωs where

Evidence that Photon has momentum relate λ change to angle φ?

KE = !ω − !ω s

Compton Effect analysis Photon momentum

Electron acquires momentum pe

Conservation of momentum:

Horizontal momentum:

so

Vertical momentum:

Square expressions and add, using sin2 + cos2 =1

!ωc+ 0 = !ω s

ccosφ + pe cosθ

=!ωc= !k = h

λ

peccosθ = !ω − !ω s cosφ

0 = !ω s

csinφ − pe sinθ

pecsinθ = !ω s sinφ

pe2c2 = !ω( )2 − 2!ω !ω s cosφ + !ω s( )2

Compton Effect analysis Can write energy (total energy) of a particle in two ways:

Recall and substitute….

subtract to give..

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E = mo2c 4 + p2c 2 and E = KE + moc

2

mo2c4 + p2c2 = KE +moc2( )2

p2c2 = KE2 + 2moc2KE

KE = !ω − !ω s

pe2c2 = !ω( )2 − 2!ω !ω s cosφ + !ω s( )2

pe2c2 = !ω( )2 − 2!ω !ω s + !ω s( )2 + 2m0c

2 !ω − !ω s( )

Compton Effect analysis

More usual to convert to wavelength (exercise…)

for electron λc ~ 2.4 x 10-12 m

“named the photon”

ʹ λ − λ =hmoc

1− cosφ( )

Compton wavelength λc =hmoc

ʹ λ − λ = λc 1− cosφ( )

2m0c2 !ω − !ω s( ) = 2!ω !ω s 1− cosφ( )

Pair Production Photon energy exchange:

Total (Photoelectric effect) or Partial (Compton effect)Also “materialise” into electron plus positron (+ve electron) Conserve charge, energy and momentum (needs 3rd party!)

Charge: 0 = (+e) + (-e)

Energy:

2 × rest mass energy KE carried off by e-/e+ pair (γ-ray)

Momentum: need 3rd object - nearby nucleus - try without it?

!ω = 2m0c2 + KE

!ωmin = 2m0c2 = 1.02MeV

Pair Production (momentum) Photon momentume-/e+ momenta = p

Vertical: 0 = psinθ - psinθ (OK)

Horizontal:

using p = mv . Since v/c < 1 and cosθ ≤ 1Then

But, conservation of energy requires hence need 3rd party to take up the unused portion of photon momentum!

= !ωc

!ωc= 2pcosθ

!ω = 2pccosθ = 2mc2 v c( )cosθ

!ω < 2mc2

!ω = 2mc2

Pair Annihilation Inverse of pair production

e- + e+ ⇒ γ + γ Differences? 2 photons result

3rd party not needed to conserve momentum

more complex, can haveTotal energy of 2moc2 + sum of electron KEcom

converts into sum of photon energies. In symmetric case

each photon energy = 0.51 MeV + (KEcom)/2

com: centre of mass

!ω1 ≠ !ω2

!ω1 = !ω2

Electromagnetic radiation interacts with matter in 3 ways: Photoelectric, Compton, Pair production

Relative strength depends on energy, in order as shown

If light (waves) “particle-like” are particles “wave-like”?

Wave Properties of Particles Recall, photon has momentum:

De Broglie (1924) suggested that this relation is general:particles which have momentum have an associated (or de Broglie) wavelength:

where m is relativistic mass. m~mo at non-relativistic speed.Suggested in De B. PhD thesis & implied by Bohr model of atom

Direct Experimental evidence??

N.B. Significance is for SMALL objects…

or λ =hp

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λ =hp

=hmv

p = !ω

c= !k = h

λ

h = 2π! = 6.625 ×10−34 Js

Davisson & Germer Experiment “Diffraction of Electrons by a Crystal of Nickel”

The Physical Review 1927

“The investigation reported in this paper was begun as the result of an accident which occurred in this laboratory in April 1925…..”

D & G had been working on electron scattering (1921) frompolycrystalline nickel: during the accident the target became oxidised. After removing the oxide by heating, the scatteringwas dramatically altered :

The target was now more crystalline: electrons were diffracted by the crystal, just like x-rays!

Analysis of D & G Experiment just like x-rays…

(planes drawn in blue) 2dsinθ = nλ electron energy KE = 54 eV

normal incidenceθ : angle with planes enhanced reflectivity at 50˚d : plane separation to find θ bisect 50˚ (⇒25˚)

(Bragg) θ = 90˚ - 25˚ = 65˚

crystal planes

θ θ d

Analysis of D & G Expt. cont.

2dsinθ = nλ λ = h/p θ = 65˚ find p from KE KE = 54 eV (<< 0.51 MeV (= moc2) so non-relativistic)

KE = mov2/2 = (mov)2/2mo so p = mov = √(2moKE)

(n=1) λ= 2dsin 65˚ = (2)(0.091 nm)(0.906) = 0.165 nm

crystal planes

θ θ

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λ =hmov

=h

2moKE

λ =6.63x10−34 J s

2( ) 9.1x10−31 kg( ) 54eV( ) 1.6x10−19C( )= 0.166nm

d

Conclude:

Particles can act as if they are also waves

Diffraction becomes significant if λ (= h/p) is similar in size to aperture/spacing

What if the particle (wave) is confined?

e.g. In a ‘box’