Transcript LECTURE 10 - University of Maryland, College Park

PHYS 420-SPRING 2006
Dennis Papadopoulos
LECTURE 10
LIGHT II
http://physics.berea.edu/~king/Teaching/ModPhys/QM/Blackbody/BlackBody.html
http://physics.berea.edu/~king/Teaching/ModPhys/QM/Photoelectric/Photoelectric.html
http://www.student.nada.kth.se/~f93-jhu/phys_sim/compton/Compton.htm
• Hertz showed that light is an electromagnetic phenomenon, and that
electromagnetic waves behave much like any other wave-- they can be
reflected, refracted, diffracted and polarized.
• Heated solids emit a continuous spectrum of radiation whose
intensity and spectral form depends only on their temperature
(blackbody radiation)
• Planck realized that the emission is caused by electrons oscillating
on the surface of the body and emitting radiation att the frequency of
their oscillation (little antennas)
• However, unless energy is quantized, the radiation of a blackbody
will continue to increase with frequency—a dilemma dubbed the
ultraviolet catastrophe—forcing Planck to theorize that light comes in
lumps and “oscillators” atomic walls must have quantized energy.
Black Body Radiation
e  J ( , T ) A

etotal   ev d  T 4
0
just a measure of
the area under
the curve
ENERGY PER UNIT VOLUME
PER UNIT
FREQUENCY(J/m3Hz) u(,T)
or
POWER PER UNIT AREA PER
UNIT FREQUENCY J(, T)
J(,T)=(c/4)u(,T)
u(,T)= (average Energy per mode)(number of modes per unit volume)
u ( , T )  (kT )(8 2 / c3 )
UV catastrophe. To save it
average energy per mode <E>
must be a function of .
<E>= kT F(,T)
N ( ) 
c3
u ( , T )  N ( )  E 
Wien

 E 
Exp[  / kT ]
Planck
 E 
8 2

Exp[  / kT ]  1
  6.6  1034 J  sec  h
Particles have identical physical
properties…but can be
distinguished by following their
(well defined) classical paths.
See Ch.10,
section 10.1
ASSUMPTIONS
In equilibrium, the energy distribution
of the particles will converge
to the most probable allowed.
P(E)  AExp[E / kT ]
In principle, there is no
limit on the number of particles
occupying each state.
Imagine 6 particles with 9 units of energy that can be distributed among nine
states. The system has total energy 9E.
•If the particles are indistinguishable, we only care about how many particles
are in each state, and there are 26 unique ways to distribute the energy
among them—26 unique combinations.
To find the average number of particles in each state
average number of
particles in the
jth energy level
n j  n j1 p1  n j 2 p2  
count the
number of
particles in
each state for
this
distribution
multiply by the number
of permutations that
can produce this
distribution divided by
the total number of
permutation for all
distributions
Energy
level
Average
number
0
2.143
1
1.484
2
0.989
3
0.629
4
0.378
5
0.210
6
0.105
7
0.045
8
0.015
9
0.003
Average Energy-Continuous case
 E 



0
0
0
 EP(E)dE /  P(E)dE  (1/ kT ) EExp (E / kT )dE  kT

 Exp[E / kT ]dE  kT
Average Energy – Discrete case
0


0
replace
 
E  h  

 E 


n 0

nExp[n / kT ]
n 0
Exp[n / kT ]


Exp[ / kT ]  1
EExp [ E / kT ]dE  (kT ) 2
Fig. 3-11, p. 78
Fig. 3-9, p. 74
Fig. 3-10, p. 75
• Inconsistency in Planck’s thinking in that he
• Quantized the oscillators emitting radiation in the walls of
the cavity
• Insisted that the cavity radiation was composed of classical
waves
• Concluded that radiation must be composed of lumps ( quantas)
consstent with the quantas emitted by the cavity walls
The observation that electromagnetic waves could
eject electrons from the surface of a metal was
first made by Hertz.
A simple experiment can be designed to measure
the energy and intensity of the electrons ejected.
•Light shines on a metal plate emitting
electrons
•The voltage on a battery can be gradually
turned up until the electric field just stop
the electrons from reaching the collector
plate, thereby giving a measure of the
kinetic energy.
•The energy in the light wave is spread out uniformly and continuously over the wavefront.
light intensity
The maximum kinetic energy of an
ejected electron is therefore
time
Kmax  CIAt  
absorption
coefficient
work function
cross sectional
area of atom
which depends on the light intensity and the time over which it is exposed.
•The intensity of a light wave is proportional to the square of the amplitude of the electric field.
…and therefore does not depend on frequency.
•The energy in the light wave is spread out uniformly and continuously over the wavefront.
•The number of photoelectrons
ejected depended on the intensity
(as expected) but their maximum
kinetic energy did not!
•The maximum kinetic energy
depended only on the frequency, the
slope of the linear relationship
between the energy and the
frequency gives “Planck’s constant”,
h.
•The electrons were ejected
immediately after the light started
shining—the electron
instantaneously absorbed enough
energy to escape-provided there
was enough energy to overcome the
binding energy or “work function”.
•Even a high intensity source of low
frequency light cannot liberate
electrons.
We have to change our way of thinking about this picture:
Instead of continuous
waves we have to think
of the energy as being
localized in quanta.
In the photoelectric
effect, these discrete
localized quanta of
energy, hv, are
transferred entirely to
the electron
Kmax  hv  
Fig. 3-16, p. 84
Table 3-1, p. 84
Kmax varies
linearly with f
Fig. 3-17, p. 84
Part 2: Compton scattering: when you have a higher energy
photon
Photoelectric effect- all of the incident photon’s
Energy is transferred to an electron, ejecting it.
Compton scatteringelectron is ejected, but
photon retains some
energy.
Pair-production-the
photon’s energy is
consumed to produce
an electron and a positron.
Fig. 3-18, p. 87
Fig. 3-19, p. 87
Fig. 3-23a, p. 91
Fig. 3-22a, p. 90
Fig. 3-23b, p. 91
Fig. 3-23, p. 91
Fig. 3-22b, p. 90
The unshifted peak comes
from tightly bound
electrons
Contrast, classical scattering:
Electrons would shake with the
frequency of the incident wave
The incident and scattered
wavelengths would be the
same
Bragg spectroscopy
(a) Constructive interference occurs
when:
n  2d sin 
n  1, 2, 3,...
(b) At other angles the waves do not
interfere constructively
This is an important tool in crystallography as it is a sensitive measure of the
spacing of the crystalline planes.
Light:
•Interferes like a wave
•Scatters like a particle
•Diffracts like a wave
•Can be polarized like a
wave
…and the answer is…drumroll please…
If light, which was previously thought of as a wave, has characteristics of particles,
could it be true that particles must also be thought of as waves in some contexts in
order to fully describe their behavior?
If you are making choices from n objects, then on your first pick you have n
choices. On your second pick, you have n-1 choices, n-2 for your third choice
and so forth. As illustrated before for 5 objects, the number of ways to pick
from 5 objects is 5! .
Suppose you are going to pick a subset r out of the total number of objects n,
like drawing 5 cards from a deck of 52. For the first pick, you have n choices,
then n-1 and so on down to n-r+1 for the last pick. The number of ways you
can do it is:
n!
n(n  1)(n  2)...(n  r  1) 
 n Pr
(n  r )!