Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Chapter 27: Early Quantum
Physics and the Photon
•Blackbody Radiation
•The Photoelectric Effect
•X-ray Production
•Compton Scattering
•Early Models of the Atom
•The Bohr Model for the Hydrogen Atom
•Pair Production/Annihilation
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§27.1 Quantization
A quantity is quantized if its possible values are limited to a
discrete set.
An example from classical physics is the allowed
frequencies of standing waves on a stretched string. Only
integer multiples of the fundamental frequency produce
standing waves.
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§27.2 Blackbody Radiation
A blackbody emits a
continuous spectrum
of radiation. The
spectrum is
determined only by
the temperature of
the blackbody.
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To correctly explain the shape of the blackbody spectrum
Planck proposed that the energy absorbed or emitted by
oscillating charges came in discrete bundles called quanta.
The energy of the quanta are
E0  hf
where h=6.62610-34 J s is
called Planck’s constant.
The quantum of EM radiation is the photon.
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§27.3 The Photoelectric Effect
Under certain circumstances EM radiation incident on a
metal will eject electrons from the metal. This is the
photoelectric effect.
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Experiments show:
1. Brighter light causes more electrons to be ejected, but
not with more kinetic energy.
2. The maximum KE of ejected electrons depends on the
frequency of the incident light.
3. The frequency of the incident light must exceed a certain
threshold, otherwise no electrons are ejected.
4. Electrons are ejected with no observed time delay
regardless of the intensity of the incident light.
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The wave theory of light says EM waves carry energy. The
energy is absorbed by electrons in the metal target which
are then ejected when they accumulate enough energy to
escape. However the wave theory is unable to completely
explain the photoelectric effect. Einstein proposed a particle
theory of light.
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Observation 1
Wave theory predicts a more intense beam of light, having
more energy, should cause more electrons to be emitted
and they should have more kinetic energy.
Particle theory predicts a more intense beam of light to have
more photons so more electrons should be emitted, but
since the energy of a photon does not change with beam
intensity, the kinetic energy of the ejected electrons should
not change.
The particle theory is consistent with observation 1.
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Observation 2
Wave theory cannot explain the frequency dependence of
the maximum kinetic energy.
Particle theory predicts the maximum kinetic energy of the
ejected electrons to show a dependence on the frequency
of the incident light. Each electron in the metal absorbs a
whole photon: some of the energy is used to eject the
electron and the rest goes into the KE of the electron.
The particle theory is consistent with observation 2.
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The maximum KE of an ejected electron is
KEmax  hf  
where  is called the work function and is the energy
needed to break the bond between the electron and the
metal.
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Observation 3
Wave theory can offer no explanation.
Particle theory predicts a threshold frequency is needed.
Only the incident photons with f>fthreshold will have enough
energy to free the electron from the metal.
The particle theory is consistent with observation 3.
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The electron is ejected from the metal when the energy
supplied by the photon exactly equals the work function.
This defines the threshold frequency.
hf threshold    0
f threshold 

h
Here it is often convenient to use h = 4.13610-15 eV s.
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Observation 4
Wave theory predicts that if the intensity of the light is low,
then it will take some time before an electron absorbs
enough energy to be ejected from the metal.
Particle theory predicts a low intensity light beam will just
have a low number of photons, but as long as f>fthreshold an
electron that absorbs a whole photon will be ejected; no
time delay should be observed.
The particle theory is consistent with observation 4.
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The particle theory of light is needed to explain the
photoelectric effect (and Compton scattering and pair
production). A wave theory of light is needed to explain
interference patterns. Both are correct (wave-particle
duality).
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Example (text problem 27.1): A 200 W infrared laser emits
photons with a wavelength of 2.010-6 m while a 200 W
ultraviolet laser emits photons with a wavelength of
7.010-8 m. (a) Which has greater energy, a single infrared
photon or a single ultraviolet photon?
E  hf 
hc

The UV photon has the greater
energy; its wavelength is smallest.
(b) What is the energy of a single infrared photon and
the energy of a single ultraviolet photon?
EUV 
EIR 
hc
UV
hc
IR
 2.8 10 18 J  18 eV
 9.9 10  20 J  0.62 eV
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Example continued:
(c) How many photons of each kind are emitted per
second?
energy emitted/se c
number of photons emitted per second 
energy/pho ton
For both lasers the energy emitted per second is 200 J.
The UV laser emits 7.01019 photons/sec and the IR
laser emits 2.0 1021 photons per second.
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Example (text problem 27.4): The photoelectric threshold
frequency of silver is 1.041015 Hz. What is the minimum
energy required to remove an electron from silver?
KEmax  hf threshold    0
  hf threshold


 6.626 10 34 Js 1.04 1015 Hz
 6.89 10
19

J  4.30 eV
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Example (text problem 27.11): Two different monochromatic
light sources, one yellow (580 nm) and one violet (425 nm),
are used in a photoelectric effect experiment. The metal
surface has a photoelectric threshold frequency of
6.201014 Hz. (a) Are both sources able to eject
photoelectrons from the metal? Explain.
The frequency of each source is
f yellow 
f violet 
c
yellow
c
violet
 5.17  1014 Hz
 7.06 1014 Hz.
Only the violet light is above the threshold frequency.
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Example continued:
(b) How much energy is required to eject an electron
from the metal?
KEmax  hf threshold    0
  hf threshold


 6.626 10 34 Js 6.20 1014 Hz
 4.1110
19

J  2.56 eV
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§27.4 X-ray Production
When high energy electrons impact a target x-ray photons
can be emitted as the electrons are slowed. This process is
called bremsstrahlung (German for breaking radiation).
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There is a continuous spectrum of radiation emitted up to a
cutoff frequency.
The spikes in the
spectrum are called
characteristic x-rays.
These peaks depend
on the target material.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§27.5 Compton Scattering
Before Collision
After Collision
Photon
(E1, p1)
y


x
Photon
(E0, p0)
Free
electron
at rest
Free
electron
(K, p)
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Conserve momentum and energy during the collision:
pi  p f
x : p0  p1 cos   p cos 
y : 0  p1 sin   p sin 
Ei  E f
E0  me c 2  E1  K  me c 2
 K  E0  E1  c( p0  p1 )
E=pc for a photon
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Manipulating the previous expressions gives
 is the
Compton
shift.
h
1  cos  
1  0 
me c
  c 1  cos  .
h
c 
 2.426 pm
me c
The Compton
wavelength
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 27.21): A photon is incident on an
electron at rest. The scattered photon has a wavelength of
2.81 pm and moves at an angle of 29.5 with respect to the
direction of the incident photon. (a) What is the wavelength
of the incident photon?
The Compton shift is
  c 1  cos  
 2.43 pm 1  cos 29.5
 0.315 pm.
The incident wavelength is
0  1  
 2.81 pm  0.314 pm  2.50 pm.
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Example continued:
(b) What is the final kinetic energy of the electron?
The final kinetic energy of the electron is equal
to the change in the photon’s energy.
K  c( p0  p1 )
 h h
 1 1
 c    hc  
 0 1 
 0 1 
15
 8.77  10 J  55 keV
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§27.6 Spectroscopy and Early
Models of the Atom
A hot, solid object will emit a continuous spectrum. A hot gas
will show an emission or line spectrum (dark background
with bright lines). Each element has its own unique set of
spectral lines.
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Examples of emission spectra:
An absorption spectrum (bright background with dark lines)
is seen if a hot source is viewed though a gas.
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Before the structure of the atom was known, an empirical
result was derived for the wavelengths of the spectral lines
of hydrogen in the visible portion of the spectrum (the
Balmer series).
1 1 
 R  2 

4 n 
1
Where R = 1.097107, m-1 is the Rydberg constant and n
3.
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The Thomson model of the atom
had a volume of positive charge with
the negatively charged electrons
embedded within the volume.
Scattering experiments by Rutherford led to the conclusion
that an atom had a very small nucleus of positive charge
(10-5 times the size of the atom containing nearly all of the
mass) that was surrounded by the electrons.
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It was thought that the electrons in their orbits should radiate
(they are accelerated) causing the electron’s orbit to decay,
implying that atoms are not stable. This is obviously false.
Any model of the atom must also explain the line spectra of
the elements.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§27.7 The Bohr Model of the
Hydrogen Atom
The Bohr model assumes:
The electron is allowed to be in only one of a discrete
set of states called stationary states. The electron
orbits have quantized radii, energy, and angular
momentum.
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Newtonian physics applies to an electron in a stationary
state.
The electron can transition between one stationary state
and another provided it can absorb/emit a photon of
energy equal to the energy difference between the
states. E=hf.
The stationary states have quantized angular momentum
in the amount
h
Ln  n
 n;
2
n is an integer.
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The allowed radii are
n 2 2
2
rn 

n
a0
2
me ke
where a0 = 52.9 pm is
the Bohr radius.
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The energy levels are given by
me k 2e 4 E1
En  
 2
2 2
2n 
n
where E1=-13.6 eV is the energy of the ground state, the
lowest possible energy of the electron. When n>1 the
electron is in an excited state. The quantity n is an integer
and is the principal quantum number.
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Energy level
diagram for
hydrogen
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The energy of a photon emitted (absorbed) by an electron
during a transition is
 1

1
E
 Ei  E f   E1  2  2 
n


n
f
i


hc
E1  1
1 

 2
2


hc  n f ni 
1
where 
E1
 1.097 107 m 1  R is the Rydberg constant.
hc
When nf=2, the above result reduces to the Balmer formula.
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The Bohr model correctly predicted the wavelengths of the
spectral lines of hydrogen in the visible. There are several
problems with the Bohr model.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Bohr’s model, while successful at predicting the spectrum of
hydrogen, fails at predicting the spectra of most other
elements. Only hydrogenic atoms (atoms that only have one
electron; Li2+ for example) can have their spectra computed
using the Bohr model.
The allowed radii are
n 2 a0
n22
rn 

.
2
me kZe
Z
me k 2 Z 2e 4 Z 2 E1H
The energy levels are En  

2 2
2n 
n2
where Z is the atomic number of
the atom and E1H = -13.6 eV .
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Example (text problem 27.34): Find the Bohr radius of
doubly ionized lithium (Li2+).
What is r1?
n 2 a0
rn 
Z
12 a0 1
r1 
 a0  17.6 pm
3
3
The inner most energy level is closer to the
nucleus than in an H atom.
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Example (text problem 27.35): Find the energy in eV
required to remove the remaining electron from a doubly
ionized lithium (Li2+) atom.
Z 2 E1H
En 
n2
The electron is in the ground state (n=1), so
32 E1H
E1  2  9 E1H  122 eV.
1
To remove the electron will require the electron be
given 122 eV of energy.
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Example (text problem 27.51): A hydrogen atom has an
electron in the n=5 level. (a) If the electron returns to the
ground state, what is the minimum number of photons that
can be emitted?
One photon; the electron may transition from
the n = 5 level to the n = 1 level.
(b) What is the maximum number that might be emitted?
Four photons; the electron may cascade
from n = 5 to 4 to 3 to 2 to n = 1.
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§27.8 Pair Annihilation and Pair
Production
A photon can interact with an atomic nucleus and change
itself into an electron-positron pair (or some particleantiparticle pair.) A positron is an antielectron. The nucleus
is needed to ensure that momentum is conserved.
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The energy of the photon must be at least 2mec2. If E >
2mec2 , then the additional energy goes into the kinetic
energy of the electron-positron pair. This is pair production.
The inverse process is


e  e  2 photons.
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The process of pair production created protons, neutrons,
and electrons in the earliest moments after the Big Bang.
To have enough energy, the photons must be “hot” enough.
Electrons need T~1010 K and for protons/neutrons T~1013 K.
The early Universe must have been much hotter than it is
today.
Pair production creates equal amounts of matter and
antimatter. Where in the Universe is all of the antimatter?
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Example (text problem 27.60): A muon and an antimuon,
each with mass 207 times greater than an electron, were at
rest when they annihilated and produced two photons of
equal energy. What is the wavelength of each of the
photons?
For an electron-positron pair
For a muon-antimuon pair
2me c  1.022 MeV.
2 m c 2
2


 207 2me c 2  212 MeV.
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Example continued:
The created photons each have 106 MeV of energy.
Their wavelengths are
hc 1240 eV nm
 
6
E 106 10 eV
 1.17 10 5 nm  1.17 10 14 m.
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Summary
•Quantization
•The Photoelectric Effect
•Compton Scattering
•Pair Production/Annihilation
•Spectroscopy
}
These processes
are explained by
light behaving like
a particle, not as a
wave.
•Bohr Model for Hydrogen
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