Download Wave as particle 2

Electromagnetic radiation behaves as
particles
1.
2.
3.
Notes of the problem discussed Tuesday.
Quiz 9.11 and a few comments on quiz 9.09.
Topics in EM waves as particles:

Tuesday



today


Blackbody radiation and Planck’s constant, Planck’s Nobel
Prize in physics.
The Photoelectric Effect and Einstein's Nobel Prize in physics.
The X-rays and Roentgen’s Nobel Prize in physics.
The Compton Effect and Compton’s Nobel Prize in physics.
Pair production, energy to mass conversion and Anderson’s
Nobel Prize in physics.
The wave-particle duality and the door to yet another new
world.
One page review of Special
Relativity

S and S’ system:
z'
z
v
S’ moves with velocity v in S along the x-axis.
y'
y
S'
x'
v 
1
1  2
v
c
x
S
x'   v  x  vt 
y'  y
z'  z
L  Lproper  v
uv
u x '   u x  v  1  x 2 
c 

t   v tproper


uv
u y '  u y  v 1  x 2  
c 
 
1


uv
u z '  u z  v 1  x 2  
c 
 
1
 v

t'   v   2 x  t 
 c


, 
1
For a particle with velocity u in S:
p   u mu E   u mc 2 KE   u 1 mc2   1  u c   E 2  p 2c 2  m2c 4
The Doppler effect: f  f
When θ =0, the course is moving
1
2 2
u

source
obs
 v 1   cos  away from the observer.
problem 33, page 64

Step 1, choose reference systems
v
Ground: S

1
 0.9950  0.9954   0.9952c,  v  10.22
2
On the muon: S’
Step 2, interpreter proper length and time in their systems
in S: L0  1910m
in S’:  0  2.2μs
L  L0  v  186.89m
   v 0  22.48μs
Solve it in S:
Solve it in S’:
The time it takes the muons to cover
the distance L0:
The time it takes the muons to cover
the distance L:
t
L0
1910

s  6.40μs
v 0.9952  3 108
Now verify: N  N e
0
N 395

 0.75
N 0 527
So
N  N0e
e


t


t' 
L
186.89

s  0.63μs
v 0.9952  3 108
t

6.40
22.48
Since
e

 0.75
is verified.
N  N0e
0.63
2.2

t'
0
 0.75
and
is verified.
N 395

 0.75
N 0 527
The Compton Effect


When two waves meet, they superposition each other.
When two particles meet, they collide.
When X-rays meet electrons  Compton scatter,
behave exactly as two particles collide:

The energy of a photon with frequency f is: E  hf
So its momentum
E hf h
p

c

c


2
2 2
2 4
hint E  p c  m c
In the x-ray electron system, the initial momentum is
ph 


The final moment is
h  ' cos   u meucos
x-component
y-component
h  ' sin   u meusin
Based on moment conservation:
h   h  ' cos   u meucos
0  h  ' sin   u meusin

And energy conservation:
ch   me c 2  ch  '   u me c 2


Solve the above three equations (all algebraic    '    h 1  cos


manipulation) We have:
me c
y
x
Example 3.3
Compton scattering and incident photon’s wavelength is
known.
Asked: momenta of incident photon, scattered photon and
the electron.
Relevant formulas and concept:
ph 
   '   
h
1  cos 
me c
Work on the blackboard.
Momentum conservation
Example 3.4
Employ energy and momentum conservation laws.
h
c

 mHe c 2  2   0.6c mD c 2
p photon 
h

 2    0.6c mD 0.6c  cos
Work on the blackboard.
Pair production, energy to mass conversion




When photon with energy above the rest
mass of two electrons ( 2me c 2 ) interact with
the electric field of a nucleus, this photon
may be turned into a pair of electron and
positron. This process is called pair
production through which energy gets
turned into mass.
Positron is the anti-particle of electron: it
has the same mass as an electron but the
opposite charge.
When a particle and anti-particle meet, they
annihilate into a photon, the process that
mass converts into energy.
Example 3.5: energy conservation.
F  qE  qv  B
What is the B-field direction?
Pair production in a bubble chamber

What is the magnetic field direction?
electron
positron
The wave-particle duality

Wave or particle? Depends on the wavelength (  ) and
the dimensions ( D ) in your experiment.



D: particle
D: wave
The double-slit experiment
ph 
E  hf
Wave
Particle as Wave
Example 3.6

How do we solve this problem?
For (a), we have wavelength and the power intensity, and this formula:
E  nh
c

The number of photons is what in the question.
For (b), the minimum at the interference pattern has no light  no
photons
For (c), from
1 
I  I 0 cos 2   
2 
We calculate the power intensity at this place, and then repeat the
procedures in (a).
Review questions



List the arguments in this chapter that photon is
a particle.
List the arguments in this chapter that photon is
a wave.
When you treat photon as a wave? When as a
particle?
Preview for the next class

Text to be read:

In chapter 4:




Section 4.1
Section 4.2
Section 4.3
Questions:



What is the Bragg Law?
An electron is a particle. One cannot have half an electron.
How do you interpret the amplitude of an electron wave?
What is the form of the Shroedinger equation for free
particle?
Homework 5, due by 9/23
1.
2.
3.
4.
Problem 31 on page 94.
Problem 35 on page 94.
Problem 40 on page 94.
Problem 42 on page 94.