Chapter 2: Particle Properties of Waves
Electromagnetic Waves
coupled Electric and Magnetic Oscillations
harmonic waves a.k.a. sine waves
E  yˆE max sin(kx   t)
Ey
B  zˆB max sin(kx   t)
k
2


f 
k
  2f
v
1

Bz
x
E max  vB max
cp351c2:1
electromagnetic spectrum
 = c/f
105
100
10-5
10-10
1025
Gamma
1020
x-ray
visible
ultraviolet
infrared
1015
millimeter
1010
microwave
105
Radio
100
f (Hz)
10-15
 (m)
4.3 x1014 - 7.5 x1014 (Hz)
Visible Light
ROYGBIV
700 nm - 400 nm
cp351c2:2
Principle of Superposition: add instantaneous amplitudes
Constructive interference
Destructive interference
cp351c2:3
Diffraction of light waves
r1
S1
d
S2
Dr
r2
Dr  d sin 
constructive: n  d sin 
n  0,1,2,

cp351c2:4
Ideal thermal radiator: Blackbody Radiator
observed contiunuous spectrum
vs. classical theory:
IC
I
Ultraviolet Catastrophe

Observed:
P
I   T4
A
 max  T
I

cp351c2:5
Theoretical black body: standing wave modes in a (3-d) cavity
8 2
Density of States:G( )  3 d
c
Statistical Mechanics: P( )  e  kT
 : microscopic energy
k : Boltzmann's constant 1.281x1023 J/K
  kT : average energy,classically via E.T.
1
Equipartion Thm : kT per D.O.F.
2
Planck: Quantization of energy
  nh , n  0,1,2
h
  h kT
8 h  3 d
 u( )d  3 h kT
e
1
c e
1
34
h  6.63x10 J  s
 4.14x1015 eV  s
cp351c2:6
Example 2.1 A certain 660 Hz tuning fork can be considered as a harmonic oscillator
with a vibrational energy of 0.04J. Compare its energy quantum of energy for the tuning
fork with its vibrational energy. Compare the fork’s quantum of energy with those of an
atomic oscillator which emits a frequency of 5.00x14 Hz.
cp351c2:7
Photoelectric effect
Classical problems:
A
no delay in emission of electrons
KE of electrons indepentdent of intensity
KE of electrons depends upon frequency of light
effect occurs only above threshold frequency 0
Einstein: quantize light (photons)
1 photon absorbed => 1 photoelectron released
Conservation of Energy
hc
KEmax  e Vs  h     

h 0  
  work function~ a few eV
hc  1240eV  nm
cp351c2:8
Example 2.2 Ultraviolet light of wavelength 350 nm and an intensity of 1.00 W/m2 is
directed at a potassium surface ( = 2.2eV). (a) Find the maximum KE of the
photoelectrons. (b) If 0.50 percent of the incident photons produce photoelectrons, how
many are emitted per second if the potassium surface area is 1.0 cm2?
Thermionic Emission: kT ~ f
=> random motion kicks electrons loose
Wave-particle “duality”
interference and diffraction: wave phenomena
photoelectric effect, etc.: particle phenomena
=> intensity ~ probability for individual photons
cp351c2:9
X-Ray production: bremsstrahlung (braking radiation)
“inverse” photoelectric effect
1 electron (KE) in => photon (E = hf) out
V
maximum energy photon get all of electron’s KE
electron KE from accelerating potential
KE  e V
(qDV  DPE)
hc
eV  h max 
 min
cp351c2:10
Example 2.3 Find the shortest wavelength present in the radiation from an X-ray
machine whose accelerating potential is 50 kV.
cp351c2:11
relative intensity
wavelength 
relative intensity
Typical continuous x-ray spectrum
wavelength 
Some target materials produce sharp maximum in the x-ray
spectrum
cp351c2:12
X-ray diffraction: how to measure the wavelength of x-rays
i
r
Constructive Interference when i = r (0th order)
i
d
Path difference = 2d sin i
Constructive Interference n  = 2d sin i
n = 1, 2, 3 ...
note: path deflected by 2
cp351c2:13
Compton effect: an elastic collision between a phton and a
charged particle initially at rest
’
e
e



e
particle nature of light (photons)
+ (Relativistic) conservation of energy and momentum
2 4
2
2 2
E  p c  m0 c
E  KE  m0c2
E tot  h  m0 c
p tot
h

c
2
Etot  h ' m0 c 2  KE
ptot x
ptot y
h '

cos  pcos
c
h '

sin   psin
c
cp351c2:14
h  h '  KE
h h '

cos  pcos 
c
c
h '
0
sin   psin  
c
pccos  h  h 'cos
pcsin   h sin 
p c  h   2h h 'cos  h '
2
2 2
2
2

p c  E  m0 c  KE  m0 c
2
2
2
4
2

2
2
 m0 c 4  KE 2  2KE m0 c 2
p c  h  h '  2h  h 'm0 c 2
2
2
2
h  h 'm c
0
2
 h h '1 cos 
h
  ' 
1 cos   C 1 cos 

m0 c
C = 2.426 pm for
electrons
cp351c2:15
Example 2.4 X-rays of wavelength 10.0 pm are scattered from a target. (a) Find the
wavelength of the x-rays scattered through an angle of 45 degrees. (b) Find the
maximum wavelength of the scattered x-rays. (c) Find the maximum KE of the recoil
electrons.
cp351c2:16
pair production:
conservation of energy, momentum + other conservation laws
E = mc2

p
e
e
e
N
creation of particle and antiparticle (antimatter)
antiparticle has same mass, opposite charge etc.
particle/antiparticle pair can anihilate to create a pair of
photons: e  + e  > g + g
cp351c2:17
Example 2.5: Show that pair production cannot occur in empty space. (Hint: look at
relativistic conservation of energy and momentum)
Example 2.6: An electron and a positron are moving side by side in the +x direction at
0.500c when they annihilate each other. Two photons are produced that move along the
x -axis. (a) Do both photons move in the +x direction? (b) What is the energy of each
photon?
cp351c2:18
Photon Absorption
Three chief mechanisms for x-ray and gamma ray photons to
interact with matter
photoelectric effect (photon absorbed)
Compton scattering (photon energy decreased)
pair production (photon converted to pair)
The dominant mechanism depends upon material and
photon energy
A slab of material will reduce intensity:
dI
  dx
I
I = I0 e
 x
  linear attenuation coeefficient
or x 
lnI 0 I 

cp351c2:19
Example 2.7: The linear attenuation coefficient for 2.0 MeV gamma rays in water is
4.9 m-1. (a) Find the relative intensity of the gamma rays after it has passed through 10
cm of water. (b) How far must the beam travel in water before being reduced to 1
percent of its original value?
Problems: 2,5,6,8,11,12,15,17,19,20,21,22,23,26,27,29,32,39,43,45,46,47
skip 2.9 or read at your liesure
cp351c2:20