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Chapter 33
Early Quantum Theory
and Models of Atom
Revolution of classical physics
World was well explained except a few puzzles?
“two dark clouds in the sky of physics”
M-M experiment
Black body radiation
theory of relativity
quantum theory
Two foundations of modern physics
Revolution of Q-theory: (1900 – 1926) → now?
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Blackbody radiation
All objects emit radiation → thermal radiation
1) Total intensity of radiation ∝ T 4
2) Continuous spectrum of wavelength
Blackbody: absorbs all the radiation falling on it
Idealized model
Blackbody radiation → easiest
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Classical theories
Wien’s law:
PT  2.90  10 3 m  K
Experiment
Intensity
Rayleigh-Jeans
Wien
Planck
Wavelength
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Planck’s quantum hypothesis
Planck’ formula (1900):
2 hc 2  5
I ( , T )  hc /  kT
e
1
Completely fit the data!
Planck’s constant:
Max Planck
(Nobel 1918)
h  6.626 1034 J  s
The energy of any molecular vibration could be
only some whole number multiply of hf.
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Concept of quantum
The energy of any molecular vibration could be
only some whole number multiply of hf.
E  n  hf
h  6.626 1034 J  s
f : frequency of oscillation
Quantum → discrete amount / not continuous
hf : quantum of energy
(a)
(b)
continuous
discrete
n : quantum number
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Photon theory of light
Little attention to quantum idea
Until Einstein’s theory of light
Molecular vibration → radiation
E  hf 
hc

Albert Einstein
(Nobel 1921)
→ quantum of radiation
The light ought to be emitted, transported, and
absorbed as tiny particles, or photons.
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Energy of photon
Example1: Calculate the energy of a photon with
  450nm (blue light).
E
Solution:
hc

 4.4  10 19 J  2.7eV
Example2: Estimate the number of visible light
photons per sec in radiation of 50W light bulb.
Solution: Average wavelength:   550nm
nE
hc

 1.4 1020
invisible light photons?
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Photoelectric effect
Photoelectric effect: electron emitted under light
If voltage V changes
photocurrent I also changes
Saturated photocurrent
Stopping potential / voltage:
Ek max
1 2
 mv  eV0
2
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Experimental results
1) Ekmax is independent of the intensity of light
2) Ekmax changes over the frequency of light
3) If f < f0 (cutoff frequency), no photoelectrons
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Explanation by photon theory
The result can’t be explained by classical theory
An electron is ejected from the metal by a collision
(inelastic) with a single photon.
photon
energy
electron
(be absorbed)
Minimum energy to get out: work function W0
hf  Ek max  W0
Photoelectric equation
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Compare with experiment
hf  Ek max  W0
1) Intensity of light ↗
n ↗, f doesn’t change
2) E
k max  hf  W0
linear relationship
W0
3) f  f 0 
?
h
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Energy of photon
Example3: The threshold wavelength for a metal
surface is 350 nm. What is the Ekmax when the
wavelength changes to (a) 280 nm, (b) 380 nm?
Solution: hf  Ek max  W0 , W0  hf 0  h c 0
 Ek max 
hc


hc
0
(a)   280nm, Ek max  1.4  10
(b)   380nm  350nm
19
J  0.89eV
No ejected electrons!
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Compton effect
Compton’s x-ray scattering experiment (Nobel 1927)
Scattering: light
propagate in
different direction
EM waves: forced vibration → same f (=0)
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Experimental results
1) Besides 0, another peak  > 0 ( f < f0 )
2) Δ=-0 depends on the scattering angle 
Ordinary scattering & Compton scattering
Can not be explained by model of EM waves
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Explanation by photon theory
What happens in the view of photon theory?
A single photon strikes an electron and knocks it
out of the atom. (elastic collision)
Conservation of energy:
hc
0
 m0 c 
2
hc

 mc 2
Energy loss →  > 0
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Compton shift
hc
0
 m0 c 
2
hc

 mc 2
Conservation of momentum:
x:
y:
h
0

0
h

h

cos   mv cos 
m
sin   mv sin 
     0  2C sin
Compton shift
2

2
m0
1  v2 / c2
h
C 
 2.43  10 12 m
m0 c
Compton wavelength
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X-ray scattering
Example4: X-rays with 0= 0.2 nm are scattered
from a material. Calculate the wavelength of the
x-rays at scattering angle (a) 45°and (b) 90°.
Solution:     0  2C sin
2

2
 C (1  cos  )
h
  0    0 
(1  cos  )
m0 c
(a)   45 :   0.2007nm
(b)   90 :   0.2024nm
Maximum shift?
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Some questions
An collision between photon and free electron
1) Why there is still a peak of 0 ?
2) What is the difference from photoelectric effect?
3) Why not absorb the photon ?
4) Why not consider p in photoelectric effect?
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*Photon interaction
Four important types of interaction for photons:
1) Scattered from an electron but still exist
2) Knock an electron out of atom (absorbed)
3) Absorbed by an atom → excited state
4) Pair production: such as electron and positron
Inverse process → annihilation of a pair
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Wave-particle duality
Sometimes light behaves like a wave
sometimes it behaves like a stream of particles
Wave-particle duality as a fact of life
Bohr’s principle of complementarity:
To understand any given experiment of light, we
must use either the wave or the photon theory, but
not both.
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Wave nature of matter
L. de Broglie extended the wave-particle duality
Symmetry in nature:
Wave
particle
It’s called de Broglie wave or matter-wave
For a particle with momentum p,
wavelength:
h

p
L. de Broglie
( Nobel 1929)
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de Broglie wavelength
Example5: Calculate the de Broglie wavelength
of (a) a 70kg man moving with speed 5m/s; (b)
an electron accelerated through 100V voltage.
Solution: (a)
h

 1.9  10 36 m
mv
Much too small to be measured
(b)
p  2mE  2m  eV
h
 
p
h
2meV
 1.2  1010 m
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Experiments of de Broglie wave
1) Davisson-Germer experiment
2) G. P. Thomson’s experiment (Nobel 1937)
3) Other experiments & other particles
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What is an electron?
An electron is neither a wave nor a particle
It is the set of its properties that we can measure
“A logical construction” —— B. Russell
de Broglie wave → a wave of probability
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Early models of atom
1) J. J. Thomson’s plum-pudding model
α particle scattering experiment
2) Rutherford’s planetary model (nuclear model)
Stability of atom & discrete spectrum
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Atomic spectra
Light spectrum of atom: line spectrum (discrete)
Emission
spectrum
& Absorption
spectrum
Characteristic of the material → “fingerprint”
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Spectrum of Hydrogen
Hydrogen: simplest atom → simplest spectrum
UV range
I R range
Visible light
(UltraViolet ray)
(Infrared Ray)
Balmer’s formula for visible lines:
1
1
1
 R ( 2  2 ), n  3, 4,...

2
n
Rydberg constant:
Balmer series
R  1.097 10 m
7
1
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General formula
There are other series in the UV and IR regions
1
1
1
 R ( 2  2 ), n  k  1, k  2,...

k
n
k = 1 → Lyman series ( ultraviolet )
k = 2 → Balmer series ( visible )
k = 3 → Paschen series ( infrared ) …
Lyman
Balmer
Paschen
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Bohr’s three postulates
Rutherford’s model + quantum idea
1) Stationary states:
stable & discrete energy level
2) Quantum transition: (“jump”)
emit or absorb a photon:
Neils Bohr
(Nobel 1922)
hf  En  Ek
3) Quantum condition: (for angular momentum)
h
( 
)
L  mvr  n , n  1, 2, ...
2
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Bohr model (1)
Rutherford’s model + quantum idea
e2
v2
m ,
2
r
4 ε0 r
nh
L  mvr 
2
r
2

h
 rn  n 2  0 2  n 2 r1 , n  1, 2, ...
 me
Bohr radius:
 0 h2
10
r1 

0.529

10
m
2
 me
The orbital radius of electron is quantized
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Bohr model (2)
2

h
e2
v2
2
0
r

n

m
,
n
2
 me 2
r
4 ε0 r
r
1 2
e2
Kinetic energy: Ek  mv 
2
8 0 rn
Potential energy: U  eV  
Total energy:
e2
4 0 rn
e 2
1 me 4
E  Ek  U 
 2 ( 2 2)
8 0 rn
n 8 0 h
13.6eV
En 
n2
Energy is also quantized
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Energy levels
13.6eV
En 
, n  1, 2, ...
2
n
1) Quantization of energy (energy levels)
n = 1: ground state, E1= -13.6eV;
n = 2: 1st exited state, E2= -3.4eV;
n = 3: 2nd exited state, E3= -1.51eV; …
Negative energy → bound state
2) Binding / ionization energy → E = 13.6eV
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Transition & radiation
Jumping from upper state n to lower state k :
me 4 1
1
1
1
hf  En  Ek  2 2 ( 2  2 )  hc  R ( 2  2 )
8 0 h k
n
k
n
Theoretical value of R :
me 4
7
1
R

1.097

10
m
8 0 2 h 3 c
1 me 4
En   2 ( 2 2 )
n 8 0 h
1
1
1
 R( 2  2 )

k
n
In accord with the experimental value!
34
Energy level diagram
13.6eV
En 
,
2
n
En  Ek
f 
,
h
1
1
 R( 2  2 )

k
n
…
E=0
-0.85eV
-1.51eV
Paschen
-3.4eV
-13.6eV
1
Balmer
Lyman
n=4(3rd exited)
n=3(2nd exited)
n=2(1st exited)
n=1(ground)
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Transition of atom
Example6: Hydrogen atom in 3rd excited state,
(a) how many types of photon can it emit? (b)
What is the maximum wavelength?
Solution: (a) n = 4
6 types of photon
1
1 1
(b)
 R( 2  2 )

3 4
   1.9μm
-0.85eV
4
-1.51eV
3
-3.4eV
2
-13.6eV
1
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Single-electron ions
Example7: Calculate (a) the ionization energy of
He+; (b) radiation energy when jumping from n=6
to n=2. (c) Can that photon be absorbed by H?
Solution: (a) For single-electron ions:
Ze 2
v2
m
2
r
4 ε0 r
( e 2  Ze 2 , Z: number of proton )
1 me 4
En   2 ( 2 2 )  En  Z 2 En
n 8 0 h
  4  13.6  54.4eV
 Eion
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(b) radiation energy if jumping from n=6 to n=2
En  Z En ,
2
13.6eV
En 
n2
1 
 1
 E  4  13.6   2  2   12.09eV
6 
2
(c) Can that photon be absorbed by H?
1 
1
 1

E  4  13.6   2  2   13.6  1  2 
2 6 
 3 
So it can be absorbed by Hydrogen atom
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Value of Bohr’s theory
1) Precisely explained the discrete spectrum
2) Lyman series & Pickering series
3) Ensures the stability of atoms
Semi-classical: other atoms, line intensity, …
New theory → quantum mechanics
Niels Bohr Institute & Copenhagen School
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*de Broglie’s hypothesis applied to atoms
Stable orbit for electron → “standing wave”
de Broglie wave:
h

mv
r
Circular standing wave:
2 r  n
Combine two equations: L  mvr  nh
2
It is just the quantum condition by Bohr!
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