Transcript e- e+ Metal e- eee- xp  12 Faraday’s experiment (1833) • • • Dissolve one mole of some substance in water Let an electric current run through it Measure how.

e-
e+
Metal
e-
eee-
xp  12
Faraday’s experiment (1833)
•
•
•
Dissolve one mole of some substance in water
Let an electric current run through it
Measure how much charge runs through before it
stops Q  96,500 C  1 Faraday  N e
Na+ Cl-
A
•All ions have the same charge (or simple multiples
of that charge)
•Avogadro’s number was not known at this time
–
+
J.J. Thomson discovers electron: 1897
•Charged particle, bends in presence of magnetic field eBR  p   mu
F  q  qu 

 mu
e
u

m BR
•Relativity not discovered
until 1905
e-
•Velocity measured with help of electric field
F 0
E
u
B
•Ratio of charge to mass now
known
e
E
 2
m BR
–
+
The Plum Pudding Model: 1904
•Electrons have only tiny fraction of an atom’s mass
•Atoms have no net charge
me  me  eNa
 me  1 Faraday
  
  
ma  e  ma N a  e 

1904: J.J. Thomson proposes the “Plum Pudding” Model
•Electrons “imbedded” in the rest of the atom’s charge
•Rest of charge is spread throughout the atom
Millikan measures charge e: 1909
•Atomizer produced tiny drops of oil; gravity pulls them down
•Atomizer also induces small charges
E  218 kV/m
•Electric field opposes gravity
•If electric field is right, drop stops falling
r  1 m
3
4

g


r
3 
qE  gm
FE  Fg
  851 kg/m3
g  9.80 m/s
2
4 g  r 3
q
3E
+

4  9.8 m/s
851 kg/m 10 m 
3  2.18 10 V/m  1 J/C/V 
 1.60 10
2
3
5
19
C
6
3
Millikan measures charge e: 1909
•Millikan always found the charge was an integer multiple of e
e  1.602 1019 C
The atom in 1909:
•Strong evidence for atoms had been found
•Avogadro’s number, and hence the mass of atoms, was now known
•Electron mass and charge were known
•Atoms contained negatively charged electrons
•The electrons had only a tiny fraction of the mass of the atom
•Distribution and nature of the positive charge was unknown
Meanwhile . . .
Statistical Mechanics
The application of statistics to the properties of systems
containing a large number of objects
gravity
Mid – late 1900’s, Statistical Mechanics successfully explains
many of the properties of gases and other materials
•Kinetic theory of gases
Gas molecules
•Thermodynamics
in a tall box:
The techniques of statistical mechanics:
•When there are many possibilities, energy will be distributed among
all of them
•The probability of a single “item” being in a given “state” depends on
temperature and energy
 E kBT
P E  e
kB  1.3806 1023 J/K  8.6173105 eV/K
Announcements
Day
Today
Friday
Monday
ASSIGNMENTS
Read
Quiz
Homework
Sec. 3-2
Quiz H
Hwk. H
Study For Test
none
Sec. 3-3 & 3-4 Quiz I
Hwk. I
Test Friday:
•Pencil(s)
•Paper
•Calculator
Equations for Test:
•Force and Work Equations added
•Lorentz boost demoted
dp
F
dt
W  E  F  d
9/16
x    x  vt 
y  y
z  z
t     t  vx c 2 
Black Body Radiation: Light in a box
Consider a nearly enclosed container at uniform temperature:
u() = energy/
volume /nm
•Light gets produced in hot
interior
•Bounces around randomly
inside before escaping
•Should be completely random
by the time it comes out
•Pringheim measures spectrum,
1899
Black Body Radiation
Goal - Predict:
u     Energy/volume/wavelength

U   u    d   Energy/volume
0
Can statistical mechanics predict the outcome?
•Find effects of all possible electromagnetic
waves that can exist in a volume
•Two factors must be calculated:
•n(): Number of “states” with wavelength 
•E: Average energy
Finding n()
How many waves can you fit in a
given volume?
•Leads to a factor of 1/4
•What are all the directions light
can go?
•Leads to a factor of 4
•How many polarizations?
•Leads to a factor of 2
u    n   E
u   
8

4
E
How to find E
What does E mean?
•It is an expectation value
E   PE E
E
PE  e
Example: Suppose you roll a fair die. If you roll 1
you win $3, if you roll 2 or 3 you win $1, but if you
roll 4, 5, or 6, you lose $2. What is the expectation
value of the amount of money you win?
1


$
$   3  1   2
6
 E k BT
1
6
P  E   Ce  E kBT
Sum of all probabilities must be 1
1
3
1   Ce E kBT
E
E   CEe E kBT
E
 Ee

e
E
E
 E k BT
 E k BT
1
2
1
C
 E k BT
e
E
What do we do with
these sums over
energy?
What do we do with the sums?
Waves of varying
strengths with
e E
the same
E
wavelength
e
 E k BT
E
 E k BT
E
•Energy can be anything
•Replace sums by integrals?

E
 E k BT
e
E dE

0

e
0
 E k BT
dE
k BT 


k BT
2
 kBT
u   
8 k BT
4

8 k BT
U 
d  !
4

0
The ultraviolet
catastrophe
Comparison Theory vs. Experiment:
Theory
What went wrong?
•Not truly in thermal equilibrium?
•Possible state counting done wrong?
•Sum  Integral not really valid?
u   
Experiment
8 k BT
4
Max Planck’s strategy (1900):
•Assume energy E must always be an integer multiple
of frequency f times a constant h
•E = nhf, where n = 0, 1, 2, …
•Perform all calculations with h finite
•Take limit h  0 at the end
u   
E
8

4
E
 E k BT
e
E

E
e
E
 E k BT
Math Interlude:

1
1  x  x 2  x3  
  xn
1  x n 0
Take d/dx of this expression . . .

1
n 1
0  1  2 x  3x 2  

nx

2
1  x  n0
Multiply by x . . .

x
2
3
n
0  1x  2 x  3x  

nx
2

1

x
  n 0
Some math notation:
exp  x   ex
 E 
E exp   k T  E

B 
E
 E 
E exp   k T 
 B 
Planck’s computation:
E  nhf
From waves:
f c 
E  nhc 
n  0,1, 2,

 hcn
 E 
hc
n exp  
E exp   k T  E 

 k BT
n 0


B 

E

 hcn 
 E 
exp  

E exp   k T 

n 0
  k BT 
 B 
exp   hc  k BT 
hc 1  exp   hc  k BT  

1

1  exp   hc  k BT 
2

hc

 hc  

n exp  



  k BT  
  hc n 0 
n



 hc  
exp  


n 0 
  k BT  

1
 exp  hc  kBT   1
n
Planck’s Black Body Law
u   
8

4
E 
8 hc

5
exp  hc  kBT   1
Max Planck’s strategy (1900):
•Take limit h  0 at the end
•Except, it fit the curve with finite h!
34
h  6.626 10
1
J s
h  4.136 1015 eV  s
Planck Constant
E  nhf
“When doing statistical mechanics,
this is how you count states”
 u  
Total Energy Density
u   
8 hc


5
1
exp  hc  kBT   1

d
U   u    d   8 hc  5
0
0  
exp  hc  k BT   1
d  hc xk B T 

 8 hc 
0
 hc
xk B T  exp  x   1
5
 k BT 
 8 hc 

hc


U
 k BT 
3
15  hc 
8
5
4 
4
x 3 dx
0 e x  1
4
15
Let  = hc/xkBT
u   
8 hc
Wien’s Law
1
For what wavelength is
this maximum? d
0
u  
 exp  hc  kBT   1
d
 5
exp  hc  kBT  
1
1 hc

0  8 hc  6
 5 2
2
  exp  hc  kBT   1   kBT  exp  hc  kBT   1 


hc exp  hc  kBT 
0  5 
 kBT exp  hc kBT   1
5
hc
 5 1  exp   hc  kBT  
 kBT
hc
 4.96511
 kBT
hc
T 
 2.8978 103 m  K
4.96511kB
Planck constant
Often, when describing things oscillating, it is more useful to work in terms of
angular frequency instead of frequency f
  2 f
  h 
E  hf  h



2  2 
This ratio comes up so often, it is given its own name and symbol.
It is called the reduced Planck constant, and is read as h-bar
h

2
E 
U
 kBT 
3
15  c 

2
Units of Planck constant
•h and h-bar have units of kg*m^2/s – same as angular momentum
4
Photoelectric Effect: Hertz, 1887
e-
Metal
•Metal is hit by light
•Electrons pop off
•Must exceed minimum frequency
•Depends on the metal
•Brighter light, more electrons
•They start coming off immediately
•Even in low intensity
ee-
eEinstein, 1905
•It takes a minimum amount of energy to free an electron
•Light really comes in chunks of energy hf
•If hf < , the light cannot release any electrons from the metal
•If hf > , the light can liberate electrons
•The energy of each electron released will be Ekin = hf – 
•Will the electron pass through a charged
plate that repels electrons?
•Must have enough energy
•Makes it if:
Ekin  eV
hf    eV
Metal
Photoelectric Effect
e-
eVmax  hf  
+
–
–
V
+
Vmax
Nobel Prize,
1921
f
eVmax  hf  
Sample Problem
When ultraviolet light of wavelength 227 nm strikes calcium metal,
electrons are observed to come off which can penetrate a barrier of
potential up to Vmax = 2.57 V.
1. What is the work function for calcium?
2. What is the longest wavelength that can free electrons from calcium?
3. If light of wavelength 312 nm were used instead, what would be the
energy of the emitted electrons?
We need the frequency:
f c
  hf  eVmax
3.00 10 m/s  1.32 1015 s1
f  
9
 227 10 m
c
8
  4.136 1015 eV  s 1.32 1015 s 1   e  2.57 V 
 5.46 eV  2.57 eV 2.89 eV=
Continued . . .
Sample Problem continued
2. What is the longest wavelength that can free electrons from calcium?
3. If light of wavelength 312 nm were used instead, what would be the
energy of the emitted electrons?
eVmax  hf     2.89 eV  f  c
•The lowest frequency comes from Vmax = 0
f min

0  hf min  
2.89 eV
14 1

6.99

10
s


15
h
4.136 10 eV  s
•Now we get the wavelength:
c 3.00 108 m/s
7
 

4.29

10
m  429 nm
14 1
f
6.99 10 s
c 3.00 108 m/s
14 1
f  

9.61

10
s
•Need frequency for last part:

9

312 10 m
eVmax  hf     4.136 1015 eV  s  9.611014 s 1   2.89 eV  1.08 eV
X-rays
•Mysterious rays were discovered by Röntgen in 1895
•Suspected to be short-wavelength EM waves
•Order 1-0.1 nm wavelength
•Scattered very weakly off of atoms
•Bragg, 1912, measured wavelength accurately
2d cos   m
 
d
•Scattering strong only if
waves are in phase
•Must be integer multiple of
wavelength
The Compton Effect
•By 1920’s X-rays were clearly light waves
•1922 Arthur Compton showed they carried momentum
Photon in
ee-

Atom
e-
Photons carry energy
and momentum, just
like any other particle
h
   
1  cos  
mc
•Conservation of momentum and energy implies a change in wavelength
Meanwhile . . .