Download Chapter 28: Quantum Physics

Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Chapter 28: Quantum Physics
•Wave-Particle Duality
•Matter Waves
•The Electron Microscope
•The Heisenberg Uncertainty Principle
•Wave Functions for a Confined Particle
•The Hydrogen Atom
•The Pauli Exclusion Principle
•Electron Energy Levels in a Solid
•The Laser
•Quantum Mechanical Tunneling
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§28.1 The Wave-Particle Duality
Interference and diffraction experiments show that light
behaves like a wave. The photoelectric effect, the Compton
effect, and pair production demonstrate that light behaves
like a particle.
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Consider a double slit
experiment in which only one
photon at a time leaves the
light source. After a long time,
the screen will show a typical
interference pattern (c).
Even though there is only one
photon emitted at a time, we
cannot determine which slit it
will pass through nor where it
will land on the screen.
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The intensity pattern on the screen is representative of the
probability that a photon will land in a given location
(higher intensity = higher probability).
For an EM wave IE2, so E2 probability of a photon
striking the screen at a given location. For an EM, wave E
represents the wave function.
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§28.2 Matter Waves
If a wave (EM radiation) can behave like a particle, might a
particle act like a wave?
The answer is yes. If a beam of electrons with appropriate
momentum is incident on a sample of material, a
diffraction pattern will be evident.
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On the right is a diffraction pattern made by x-rays incident
on a sample. On the left is a diffraction pattern made by an
electron beam incident on the same sample.
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Like photons, the wavelength of a matter wave is given by
h
 .
p
This is known as the de Broglie wavelength.
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Example (text problem 28.4): What are the de Broglie
wavelengths of electrons with the following values of kinetic
energy? (a) 1.0 eV and (b) 1.0 keV.
(a) The momentum of the electron is
p  2mK
 29.1110 31 kg 1.0 eV 1.60 10 31 J/eV 
 5.4 10  25 kg m/s
and
h
6.626 1034 Js
9
 

1
.
23

10
m  1.23 nm.
 25
p 5.4 10 kg m/s
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Example continued:
(b) The momentum of the electron is
p  2mK
 29.1110 31 kg 1.0 103 eV 1.60 10 31 J/eV 
 1.7 10  23 kg m/s
and
h
6.626 1034 Js
11
 

3
.
88

10
m  38.8 pm.
 23
p 1.7 10 kg m/s
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Example (text problem 28.7): What is the de Broglie
wavelength of an electron moving with a speed of 0.6c?
This is a relativistic electron with
1

2
 1.25.
v
1 2
c
Its wavelength is
h
h
 
p mv
6.626 10 34 Js
12


3
.
23

10
m.
31
8
1.25 9.1110 kg 1.8 10 m/s



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A beam of electrons may be used in a double slit experiment
instead of a light beam. If this is done, a typical interference
pattern will be produced on the screen indicating electrons
act like waves.
If a detector is placed to try to determine which of the two
slits the electron goes through, the interference pattern
disappears indicating the electron now behaves like a
particle.
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§28.3 Electron Microscope
The resolution of a light microscope is limited by diffraction
effects. The smallest structure that can be resolved is
about half the wavelength of light used by the microscope.
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An electron beam can be produced with much smaller
wavelengths than visible light, allowing for resolution of
much smaller structures.
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Example (text problem 28.15): An image of a biological
sample is to have a resolution of 5 nm. (a) What is the
kinetic energy of a beam of electrons with a de Broglie
wavelength of 5.0 nm?
p2
h2
K

2m 2m2
 21
 9.64 10 J  0.060 eV
(b) Through what potential difference should the electrons
be accelerated to have this wavelength?
K  U
 qV  eV
K 0.060 eV
V 

 0.060 Volts
e
e
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Example continued:
(c) Why not just use a light microscope with a wavelength
of 5 nm to image the sample?
An EM wave with = 5 nm would be an x-ray.
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§28.4 The Uncertainty Principle
The uncertainty principle sets limits on how precise
measurements of a particle’s momentum and position can
be.
1
xp x  
2
where
h

2
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The more precise a measurement of position, the more
uncertain the measurement of momentum will be and the
more precise a measurement of momentum, the more
uncertain the measurement of the position will be.
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The energy-time uncertainty principle is
1
Et  .
2
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Example (text problem 28.18): An electron passes through
a slit of width 1.010-8 m. What is the uncertainty in the
electron’s momentum component in the direction
perpendicular to the slit but in the plane containing the slit?
The uncertainty in the electron’s position is half the slit
width x=0.5a (the electron must pass through the slit).


p x 
  1.110  26 kg m/s
2x a
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Example (text problem 28.19): At a baseball game, a radar
gun measures the speed of a 144 gram baseball to be
137.320.10 km/hr. (a) What is the minimum uncertainty of
the position of the baseball?
px = mvx and vx = 0.10 km/hr = 0.028 m/s.
1
xp x  mxvx  
2

x 
 1.3 10 32 m
2mvx
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Example continued:
(b) If the speed of a proton is measured to the same
precision, what is the minimum uncertainty in its position?

6
x 
 1.110 m
2m p vx
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§28.5 Wave Functions for a
Confined Particle
A particle confined to a region of space will have quantized
energy levels.
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Consider a particle in a box of width L that has impenetrable
walls, that is, the particle can never leave the box.
Since the particle cannot be found outside of the box, its
wave function must be zero at the walls. This is analogous
to a standing wave on a string.
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This particle can have
2L
n 
n
With n=1,2,3,…
h
nh
pn 

.
n 2L
The kinetic energy of the particle is
p2
KE 
.
2m
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And its total energy is E  K  U
p2
n2h2

0 
.
2
2m
8mL
The energy of the particle is quantized. The ground state
(n=1) energy is
h2
E1 
8mL2
so that
En  n E1.
2
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Example (text problem 28.29): A marble of mass 10 g is
confined to a box 10 cm long and moves with a speed of 2
cm/s. (a) What is the marble’s quantum number n?
1 2
6
The total energy of the marble is En  mv  0  2.0 10 J.
2
In general
Solving for n:
En  n 2 E1
n
h2
E1 
.
2
8mL
En
8mEn L2
28


6

10
.
2
E1
h
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Example continued:
(b) Why do we not observe the quantization of the marble’s
energy?
The difference in energy between the energy levels n and
n+1 is
En 1  En  n  1 E1  n 2 E1
2
 2nE1  E1
 2n  1E1  6.6 1035 J.
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Example continued:
The change in kinetic energy of the marble would be
K 
1 2 1 2
mv f  mvi
2
2


1
 m v 2f  vi2
2
1
 mv f  vi v f  vi   mvi v f  vi .
2
Assume vfvi.
To make a transition to the level n+1, the ball’s speed must
change by
K
v f  vi    3.3 1031 m/s.
mvi
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If a container has walls of finite height, a particle in the box
will have quantized energy levels, but the number of bound
states (E < 0 ) will be finite.
In this situation the wave functions of the particle in the box
extend past the walls of the container. This means there is a
nonzero probability that the particle can “tunnel” its way
through the walls and escape the box.
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The probability of finding a particle is proportional to the
square of its wave function.
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§28.6 The Hydrogen Atom: Wave
Functions and Quantum Numbers
In the quantum picture of the atom the electron does not
orbit the nucleus. Quantum mechanics can be used to
determine the allowed energy levels and wave functions for
the electrons.
The wave function allows the determination of the
probability of finding the electron in a given region of space.
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The allowed energy levels in the hydrogen atom are
mk 2 e 4
2
En  

n
E1
2
2
where E1=-13.6 eV.
n is the principle quantum number.
Even though the electron does not orbit the nucleus, it
has angular momentum.
L
 ll 1
Where l=0, 1, 2,…n-1
l is known as the orbital angular
momentum quantum number.
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For a given n and l, the angular momentum about the z-axis
(an arbitrary choice) is also quantized.
Lz  ml 
ml=-l, -l+1,…, -1, 0, +1,…l-1, l
ml is the orbital magnetic quantum number.
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The spectrum of hydrogen can only be fully explained if the
electron has an intrinsic spin. It is useful to compare this to
the Earth spinning on its axis. This cannot be truly what is
happening since the surface of the electron would be
traveling faster than the speed of light.
S z  ms 
ms=½ for an electron
ms is the spin magnetic quantum number.
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Electron cloud representations of the electron probability
density for an H atom:
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§28.7 The Pauli Exclusion Principle
The Pauli Exclusion Principle says no two electrons in an
atom can have the same set of quantum numbers. An
electron’s state is fully described by four quantum numbers
n, l ,ml, and ms.
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In an atom:
A shell is the set of electron states with the same
quantum number n.
A subshell is a unique combination of n and l. A
subshell is labeled by its value of n and quantum
number l by using spectroscopic notation.
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Each subshell consists of one or more orbitals specified
by the quantum numbers n, l, and ml. There are 2l+1
orbitals in each subshell.
The number of electron states in a subshell is 2(2l+1),
and the number of states in a shell is 2n2.
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The subshells are filled by electrons in order of increasing
energy.
1s,2s,2 p,3s,3 p,4s,3d ,4 p,5s,4d ,5 p,6s,4 f ,5d ,6 p,7 s
Beware! There are exceptions to this rule.
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The electron configuration for helium
is:
1s
2
specifies the number of
electrons in this orbital
Specifies n
Specifies l
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Example (text problem 28.36): How many electron states of
the H atom have the quantum numbers n=3 and l=1? Identify
each state by listing its quantum numbers.
Here ml=-1,0,1 and since 2 electrons can be placed in
each orbital, there can be 6 electron states.
n
l
ml
ms
3
1
-1
-½
3
1
-1
+½
3
1
0
-½
3
1
0
+½
3
1
+1
-½
3
1
+1
+½
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Example (text problem 28.38): (a) Find the magnitude of the
angular momentum L for an electron with n=2 and l=1?
L  l l  1  11  1  2
(b) What are the allowed values of Lz?
The allowed values of ml are +1,0,-1 so that Lz can be
 1
0
 1.
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Example continued:
(c) What are the angles between
the positive z-axis and L so that
the quantized components, Lz,
have allowed values?
Lz
1
1
When l=1, ml=-1,0,+1
0
2
3
ml 
cos  

L
l l  1
Lz
 1
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Example continued:
 1
1
cos 1 

 1  45
2
2
0
cos  2 
 0   2  90
2
 1
1
cos  3 

  3  45  135
2
2
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§28.8 Electron Energy Levels in a
Solid
An atom in isolation will only be able to emit photons of
energy E that correspond to the difference in energies
between the energy levels in the atom (a line spectrum).
When atoms are not in isolation, the wave functions overlap
which causes the energy levels to split. As a result, a solid
(a large collection of atoms close together) will emit a
continuous spectrum.
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In a solid, because of the
large number of atoms (N)
present, each energy level
becomes a band of N
closely spaced energy
levels. Solids also show
band gaps where there
are no allowed electron
energy levels.
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A material is a conductor if
the highest energy electron
state filled at T= 0 is in the
middle of the band (the
band is only partially filled).
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If electrons fill their allowed states right to the top of the
band, the material is either a semiconductor or an
insulator.
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§28.9 Lasers
Laser is an acronym for Light Amplification by Stimulated
Emission of Radiation.
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An electron can go to a higher energy level by the
absorption of a photon.
When an electron is in an excited state, it can go into a
lower energy level by the spontaneous emission a photon.
An electron in an excited state can also go into a lower
energy level by the stimulated emission of a photon.
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A photon of energy E can stimulate the emission of a
photon (by interacting with the excited electron). The
emitted photon will have the same energy, phase, and
momentum of the stimulating photon.
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Typically the excited states of electrons have lifetimes of
about 10-8 seconds. To make a laser, the material must have
metastable states with lifetimes of about 10-3 seconds. This
allows for a population inversion in which more electrons
are in a higher energy state rather than in a lower energy
state.
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Example (text problem 28.52): In a ruby laser, laser light of
wavelength 694.3 nm is emitted. The ruby crystal is 6.00
cm long, and the index of refraction of the ruby is 1.75.
Think of the light in the ruby crystal as a standing wave
along the length of the crystal. How many wavelengths fit
in the crystal?
The wavelength of light in the crystal is
0
694.3 nm
 
 396.7 nm
n
1.75
number of wavelengt hs 
L

 1.51105.
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§28.10 Tunneling
For a wide barrier, the probability per unit time of a particle
tunneling through the barrier is
Pe
2a
where a is the width of barrier and  is a measure of the
barrier height.
2m
U 0  E 

2

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Summary
•Matter as a Wave
•The Uncertainty Principle
•What Is a Wave Function?
•The Hydrogen Atom
•The Pauli Exclusion Principle
•Quantum Mechanical Tunneling
•Electron Energy Levels in a Solid
•The Laser
•The Electron Microscope
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