Transcript Slide 1

Physics at the end of XIX Century
and
Major Discoveries of XX Century
Thompson’s experiment (discovery of electron)
Emission and absorption of light
Spectra:
•Continues spectra
•Line spectra
Three problems:
•“Ultraviolet catastrophe”
•Photoelectric effect
•Michelson experiment
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Continues spectra and “Ultraviolet catastrophe”
Stefan-Boltzmann law
for blackbody radiation:
I(λ)
I  T 4
I   I  

Wien’s displacement law:
maxT  2.90103 m  K
E  hf
Plank’s constant:
h  6.63 1034 J  s
 4.14  1015 eV  s
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Example 1: What is the wavelength the frequency corresponding to the
most intense light emitted by a giant star of surface temperature 5000 K?
maxT  2.90103 m  K
max  2.90  103 m  K / 5000K  0.580 106 m  580nm
f max  c / max  3  108 m / s / 0.580 106 m  5.2  1014 Hz
Example 2: What is the wavelength the frequency of the most intense
radiation from an object with temperature 100°C?
max  2.90  103 m  K / 273 100K  7.77  106 m  7.77m
f max  c / max  3  108 m / s / 7.77  106 m  3.9  1013 Hz
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Photoelectric effect
Experiment:
If light strikes a metal, electrons are emitted.
•The effect does not occur if the frequency of the light is too low
•The kinetic energy of the electrons increases with frequency
light
Classical theory can not explain these results.
If light is a wave, classical theory predicts:
• Frequency would not matter
• Number of electrons and their energy should increase with intensity
A
Quantum theory:
Einstein suggested that, given the success of Planck’s theory, light must be
emitted and absorbed in small energy packets, “photons” with energy:
E  hf
If light is particles, theory predicts:
• Increasing intensity increases number of electrons but not kinetic energy
• Above a minimum energy required to break atomic bond, kinetic energy
will increase linearly with frequency
• There is a cutoff frequency below which no electrons will be emitted,
regardless of intensity
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Photoelectric effect (quantum theory)
light
Photons!
E  hf
hf  W0  Kmax
A
Plank’s constant:
h  6.63 1034 J  s
 4.14  1015 eV  s
2
Kmax  12 mvmax
hfmin  W0
Stopping potential (V0):
eV0  Kmax  hf-W0
I
-V0
V
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Example: The work function for a certain sample is 2.3 eV. What is the
stopping potential for electrons ejected from the sample by 7.0*1014 Hz
electromagnetic radiation?
W0  2.3eV
eV0  hf  W0
f  7.0 1014 Hz
V0  ?
eV0  4.141015 eV  s 7.0 1014 Hz  2.3eV  0.6eV



V0  0.6V
Example: The work function for sodium, cesium, copper, and iron are
2.3, 2.1, 4.7, and 4.5 eV respectively. Which of these metals will not
emit electrons when visible light shines on it?
Visible light:
400 nm    700 nm
2.3  1014 Hz  f  7.5 1014 Hz
f  7.5  1014 Hz hfmin  W0 
W0  ?
W0  4.14  1015 eV  s 7.5  1014 Hz  3.1eV


Copper, and iron will not emit electrons

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The Atom
1. The Thomson model (“plum-pudding” model)
It was known that atoms were electrically neutral,
but that they could become charged, implying that
there were positive and negative charges and that
some of them could be removed.
This model had the atom consisting of a bulk positive
charge, with negative electrons buried throughout.
Later, Rutherford did an experiment that showed that the positively charged
nucleus must be extremely small compared to the rest of the atom.
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2. Rutherford’s scanning experiment and planetary model
Rutherford scattered alpha particles – helium nuclei – from a metal foil and
observed the scattering angle. He found that some of the angles were far
larger than the plum-pudding model would allow.
Rutherford’s (planetary)
model:
The only way to account for the large angles was to assume that all
the positive charge was contained within a tiny volume – now we know
that the radius of the nucleus is about 1/100000 that of the atom.
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3. Atomic line spectra (Key to the structure of the atom)
A very thin gas heated in a discharge tube
emits light only at characteristic frequencies.
•An atomic spectrum is a line spectrum –
only certain frequencies appear.
•If white light passes through such a gas,
it absorbs at those same frequencies.
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4. Hydrogen atom
The wavelengths of electrons emitted from hydrogen have a regular pattern:
1 
 1
 R 2  2 

n 
m
1
Rydberg constant:
Lymanseries : m  1; n  2,3,...
Balmerseries : m  2; n  3,4,...
Paschenseries : m  3; n  4,5...
A portion of the complete spectrum of hydrogen:
These lines cannot be explained by the Rutherford theory
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5. The Bohr Atom
Bohr proposed that the possible energy states (stationary states) for atomic
electrons were quantized – only certain values were possible. Then the spectrum
could be explained as transitions from one level to another.
1
 1
 R 2  2 

m n 
1
hf 
1 
 1
 hcR 2  2   E f  Ei

n 
m
hc
hcR
En   2
n
hcR  13.60eV
hc  1243 eV  nm
Example:
For H 2 :
E min  ?
E2  ?
hcR
13.60eV


n2
n2
 E1  13.60eV
En  
Emin
 21  ? E2  
13.60eV
 3.40eV
2
2
E  E 2  E1  13.60eV (1  14 )  10.20eV
hc 1243eV  nm


E
10.20eV
  122nm
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The Bohr Atom
The lowest energy level is called the ground state;
the others are excited states.
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Example: Franck- Hertz experiment
Franck and Hertz studied the motion of electrons through mercury vapor under the
action of an electric field. When the electron kinetic energy was 4.9eV or grater, the
vapor emitted ultraviolet light. What was the wave length of this light?
E  4.9eV
 ?
hf 

hc

 E
hc 1243eV  nm

 250nm
E
4.9eV
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