Compton Effect

• 1923 Compton performed an experiment which supported this idea • directed a beam of x-rays of wavelength  onto a carbon target • x-rays are scattered in different directions    = 71.1 pm (10 -12 m)  `  ` has 2 peaks

Compton Scattering

• Wavelength  ` • these occur at  of scattered x-rays has two peaks and  +   •   >0 is the Compton shift • classical physics predicts   =0 • Quantum picture: • a

single

photon interacts with electrons in the target • light behaves like a ‘particle” of energy E=hf=hc/  and momentum p=h/  => a collision

Compton Scattering

E`=hf `=hc/  `  E=hf=hc/  K=m e c 2 (  -1) • Conservation of energy E = E` + K • => E` < E => f ` < f • X-ray momentum p=h/  =>  ` >  p`= h/  ` • electron momentum p e =  m e v

Compton Scattering

E`=hf `=hc/  `  E=hf=hc/  K=m e c 2 (  -1) • Conservation of energy E = E` + K • => E` < E => f ` < f • X-ray momentum p=h/  =>  ` >  p`= h/  ` • electron momentum p e =  m e v

X-ray scattering

• Energy and momentum are conserved • Momentum is a vector!

F

=d

p

/dt=0 =>

p

= constant

h

 

h

 

cos

 

m

v cos



0



h

 

sin

m

v sin



hc

 

hc

  

mc

2

(

 

1 )

b

p

x



constant

g d

p

y



constant

i 

constant

g

h

 

h

 

cos X-ray Scattering

m

e

v cos

 b

p

x



constant

g

0



h

 

sin

m

e

v sin

 d

p

y



constant

i

hc

 

hc

  

m c

e

2

(

 

1 )

• 3 equations in 5 variables:  ,  `,v,  ,  

constant

• • • eliminate the electron variables  `  v,  => find  (v) =(h/m e c) (1 - cos  ) =  c (1 - cos  )  c is Compton wavelength of the electron g

Compton Scattering

• • • •  `  =  c (1 - cos  =0 ==>  ` =   )   = =   /2 ==> ==>   ` = ` =   +  c + 2  c • why are there two peaks?

Compton Scattering

• “loosely” bound electrons in Carbon are ejected and the x-rays are scattered •  `  =(h/m e c) (1 - cos  ) • “tightly” bound electrons are not ejected => photon interacts with entire carbon atom • mass ~ 22,000 m e =>  reduced by this factor

Problem

• An x-ray beam of wavelength 0.01 nm strikes a target containing free electrons. Consider the xrays scattered back at 180 0 • Determine (a) change in wavelength of the xrays (b) change in photon energy between incident and scattered beams (c) the kinetic energy transferred to the electron (d) electron’s direction of motion the

• • •

Solution

• X-ray beam has  =.01 nm = 10 pm  =180 0  `  =(h/m e c) (1 - cos  ) =  c (1 - cos  )  c is Compton wavelength of the electron • (a)  =(h/cm e )(1-cos(180))= 2h/cm e =2(6.63x10

-34 )/[(3x10 8 )(9.11x10

-31 )] =2(2.43 pm)= 4.86 pm • (b)  E={ hc/  ` -hc/  } =(6.63x10

-34 )(3x10 8 ){1/14.86 -1/10}/(10 -12 ) =-.65x10

-14 J = -.41x10

5 eV = -41 keV

Solution

• (c) K (electron) = 41 keV • (d) direction of electron?

• Momentum conserved => electron moves forward  p

Photons Revealed • Can we devise an experiment where both the wave and photon characteristics of light are involved?

• Photon has 50% chance of being transmitted or reflected at B • reduce light beam energy to that of a single photon • if photon picture is correct we get anticoincidences • experiments were not convincing

Designated Photons • 1986 Grangier, Roger, Aspect replaced source S by beam of calcium ions • Calcium excited by a laser and emits two photons • trigger photon turns detectors on and off (emitted and absorbed!) • designated photon demonstrated anticoincidences!

• Supported the photon picture! - no wave interpretation possible • other modifications are possible to demonstrate both wave and photon properties of light - see section 45-6

Slowing Atoms by Photon Bombardment • A gas of atoms at room temperature is in constant motion • for argon gas at T=300 0 K, v rms = (3kT/m) 1/2 = 430 m/s !

• How can we use photons to slow these atoms down?

• Consider that the atom absorbs a photon with p=h/ 

mv i



h

/  

mv f h

/  

i



v f

) Typically a change of few cm/s due to a single photon • Atoms moving in the same direction as the laser are speeded up => no net slowing down

Laser Cooling • When a photon is absorbed, the atom moves from the ground state to an excited state • excited state does not have a well defined energy since it only exists for a short time => uncertainty principle Probability of absorption by atom at rest Atom illuminated by two beams  laser greater than peak value Photons absorbed from L slow it down and those absorbed from R speed it up => do these effects cancel?

Laser Cooling • Doppler effect: • atom detects L as a higher f or lower  L <  laser • atom detects R as a lower f or higher  R >  laser • probabilities are not the same!

• net reduction in speed results • experiments use 6 laser beams trap