1
2 e e e e e e +

•
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
A
Na + Cl -
•All ions have the same charge (or simple multiples of that charge)
•Avogadro’s number was not known at this time
–
+

•Charged particle, bends in presence of magnetic field
F q qu

 e -
B
•Relativity not discovered until 1905
 p
  mu e m


B mu u
R
•Velocity measured with help of electric field
F

0
•Ratio of charge to mass now known e

E m B
2
R u

E
B
–
+

•Electrons have only tiny fraction of an atom’s mass
•Atoms have no net charge m e m a


 m e e


 eN m N a a a


 m e e



1 Faraday

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

e
•Atomizer produced tiny drops of oil; gravity pulls them down
•Atomizer also induces small charges
•Electric field opposes gravity
E

218 kV/m
•If electric field is right, drop stops falling
F
E

F g q E
 gm
 g

3 r
3

 r


851 kg/m
3
+ q g

9.80 m/s
2
4
 
3

3E



2

3

6
4 9.8 m/s 851 kg/m 10 m
 
 

 5
3 2.18 10 V/m 1 J/C/V


1.60 10

19
C
3

e
•Millikan always found the charge was an integer multiple of e e
 
19
1.602 10 C
•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 . . .

The application of statistics to the properties of systems containing a large number of objects
Mid – late 1900’s, Statistical Mechanics successfully explains many of the properties of gases and other materials
•Kinetic theory of gases
•Thermodynamics Gas molecules
The techniques of statistical mechanics: in a tall box:
•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

B k
B

1.3806 10

23
J/K
  
5

Day
Today
Friday
Monday
Test Friday:
•
Pencil(s)
•
Paper
•
Calculator
Read
Sec. 3-2
ASSIGNMENTS
Quiz Homework
Study For Test
Quiz H
Sec. 3-3 & 3-4 Quiz I
Hwk. H none
Hwk. I
Equations for Test:
•
Force and Work Equations added
•
Lorentz boost demoted
F
 dp dt x
    x vt
 y
  y
W E F d
9/16 t
 z
  z
   t
 vx c
2


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

U

 u
 
0

Goal - Predict: u

Energy/volume/wavelength

Energy/volume
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
   
E u

8


4
E

What does E mean?
•It is an expectation value
E
 
 
E
 e

  
Ce

Sum of all probabilities must be 1
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
6
     
3
1
 
2
2
 
$ 1
6
1
 
E
Ce
 C


E e

1
E
 
E
CEe
 

E

E
Ee e


What do we do with these sums over energy?

E


E

E
 e E e

Waves of varying strengths with the same wavelength
E

•Energy can be anything
•Replace sums by integrals?


0

 e E dE e
 dE

 k T
B k T
B

2
0

 k T
B u
U

 

0
8

 k T
B
4
8

 k T
B
4 d
  
!
The ultraviolet catastrophe

Theory
What went wrong?
•Not truly in thermal equilibrium?
•Possible state counting done wrong?
•Sum 
Integral not really valid?
u

8

 k T
B
4
E u


8

E

4

E

E
 e E e

Experiment
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

1
  x
2  x
3  
Take d / dx of this expression . . .
1
1

0 1 2 x

3 x
2  
Multiply by x . . .
 
2 x
2 
3 x
3  
1

1
 x

2 x

1
 x

2
 x

 n



0 n



0 x n nx n

1 n



0 nx n
Some math notation: exp
   e x E


E

E exp exp






E k T
B
E k T
B




E

E
E
 nhf
From waves: f
 c
 n

0,1, 2,
E
 nhc




E

E exp exp






E k T
B
E k T
B

 exp
 


E
 hc
 k T
B n



0
 n n



0 hc
 exp exp







 hcn hcn k T
B k T
B

 hc

1 exp
  hc
 k T
B
1 exp
 
1 hc
 k T
B


2
 hc



 hc

1 exp
 hc
 k T
B
 
1 n



0 n



0 n 

 exp 





 exp 


 hc
 k T
B hc
 k T
B





 n





 n

u

8


4
E

8
 hc

5 exp
1
 hc
 k T
B
 
1
 u
Max Planck’s strategy (1900):
•Take limit h

0 at the end
•Except, it fit the curve with finite h !
h

6.626 10

34  h

4.136 10

15 
Planck Constant
E
 nhf
“When doing statistical mechanics, this is how you count states”

U

 u
 
0
 u

8


5 hc exp

1 hc
 k T
B
 
1


8
 hc


0
8
 hc


0


5 exp
 hc xk T
B d
 hc
 k T
B

B

 

5 exp
 
1 

1 
Let

= hc/xk
B
T

8
 hc

 k T
B hc


4


0 e
3 x dx x 
1
U

8

5
15
 k T
B

4
 
3

4
15

u

0

8
 hc
8
 hc




5
5

6 exp
1
 hc
 k T
B
 
1 exp
1
 hc
 k T
B
 
1

1 hc
 
2 k T
B
0 5 hc
 k T
B exp exp

 hc
 k T
B hc
 k T
B



1 hc
 k T
B

5 1 exp
  hc
 k T
B
 hc
 k T
B

4.96511
For what wavelength is
 this maximum?
0
 exp
 d hc
 k T
B d

 exp
 hc
 k T
B
 
1

2 u



T
 hc
4.96511
k
B
 
3
2.8978 10 m K

Often, when describing things oscillating, it is more useful to work in terms of angular frequency
 instead of frequency f
 
2
 f
E
 hf
 h
2


 


2 h
 



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


2
15
 k T
B

 
3
4
• h and h -bar have units of kg*m^2/s – same as angular momentum

•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 e e e e -
Einstein, 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
E kin
= hf
– 

•Will the electron pass through a charged plate that repels electrons?
•Must have enough energy
•Makes it if: e -
E kin
 eV eV max
 hf
  hf eV
+
– –
V
+
V max
Nobel Prize,
1921 f

eV max
 hf
 
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 V max
= 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
 eV max f
 c


 8
3.00 10 m/s

9


4.136 10

15 

 15

1
 15

1
1.32 10 s

 e

2.57 V


5.46 eV 2.57 eV

2.89 eV=
 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?
eV max
 hf
   
2.89 eV
 f
 c
•The lowest frequency comes from
V max f min

2.89 eV
4.136 10

15
•Now we get the wavelength:
 
 f c
 h

8
14

1
•Need frequency for last part: eV max
 hf
  

 f


= 0
 14
0
 hf

1
 
7
4.29 10 m

429 nm c



 8

9
 min
 
 14

1
9.61 10 s
4.136 10

15   14

1 
2.89 eV

1.08 eV

•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
  m

  d
•Scattering strong only if waves are in phase
•Must be integer multiple of wavelength

•By 1920’s X-rays were clearly light waves
•1922 Arthur Compton showed they carried momentum
Photon in e e -
Atom

Photons carry energy and momentum, just like any other particle e h mc

•Conservation of momentum and energy implies a change in wavelength
 
Meanwhile . . .