Transcript Document
Chapter 2: Particle Properties of Waves
Electromagnetic Waves coupled Electric and Magnetic Oscillations harmonic waves a.k.a. sine waves
E
ˆ max
B
ˆ max
k
f
2
k
sin(
kx
sin(
kx v
1 2
f t
)
t
)
B
z
E
y
E
max
vB
max
x
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electromagnetic spectrum
= c/f
f
(Hz) 10 0 10 5 10 10 10 15 10 20 10 25 10 5 10 0 10 -5 (m) 10 -10 4.3 x10 14 - 7.5 x10 14 (Hz) Visible Light R O Y G B I V 700 nm - 400 nm 10 -15 cp351c2:2
Principle of Superposition: add instantaneous amplitudes Constructive interference Destructive interference cp351c2:3
Diffraction of light waves
r
1
S
1
d r
2
S
2 D
r
D
r
d
sin
constructive
: 0, 1, 2,
n
d
sin cp351c2:4
Ideal thermal radiator: Blackbody Radiator observed contiunuous spectrum vs. classical theory: Ultraviolet Catastrophe
I
Observed:
I
max
P A
T T
4
I
I C cp351c2:5
Theoretical black body: standing wave modes in a (3-d) cavity Density of States:
G
( ) 8
c
3 2
d
Statistical Mechanics:
P
( )
e
kT
: microscopic energy
k
: Boltzmann's constant 1.281
x
10 23 J/K
kT
: average energy, classically via E.T. Equipartion Thm : 1 2
kT
per D.O.F.
Planck: Quantization of energy
nh
,
n
0,1,2
h
h e h
kT
1 6.63
x
10 34
J
s
u
( )
d
8
h c
3 3
d
e h
kT
1 4.14
x
10 15
eV
s
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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 5.00x
14 Hz.
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Photoelectric effect Classical problems: 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) A 1 photon absorbed => 1 photoelectron released Conservation of Energy
KE
max
e
V s
h
hc
h
0 work function~ a few
eV
hc
1240
eV
nm
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Example 2.2 Ultraviolet light of wavelength 350 nm and an intensity of 1.00 W/m 2 directed at a potassium surface ( = 2.2eV). (a) Find the maximum KE of the is 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 cm 2 ?
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
(
q
D
V
D
PE
)
eV
h
max
hc
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.
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wavelength 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 (0 th order) i d Path difference =
2d
sin i Constructive Interference
n
= 2
d
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
p
2
c
2
m
0 2
c
4
E
KE
m
0
c
2
E tot p tot
h
h
c
m
0
c
2
E tot p tot x p tot y
h
'
m
0
c
2
h
'
c h
' cos sin
c
KE p
cos
p
sin cp351c2:14
h
h
c
0
p
2
c
2
h
'
KE h
' cos
h c
'
c h
sin 2
p
cos
p
sin 2 cos
pc
cos
pc
sin 2
h
h
'cos
h
sin
p
2
c
2
p
2
c
2
E
2
m
0 2
c
4
h
h
' 2
KE
2
h
m
0
c
2 2
h
'
m
0
c
2
m
0 2
c
4
KE
2 2
KE m
0
c
2
h
h
'
m
0
c
2 1 cos '
h m
0
c
1 cos
C
1 cos
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.
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pair production: conservation of energy, momentum + other conservation laws
E
=
mc
2
p
e
N e
e
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?
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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 = I 0
e
x
or linear attenuation coeefficient
x
ln 0
I
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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