1

•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
2

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.
3

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.
4

Like photons, the wavelength of a matter wave is given by
  h p
.
This is known as the de Broglie wavelength .
5

Example (text problem 28.8): 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


2 mK
2

9 .
11

10

31 kg
 
1 .
0 eV
 
1 .
60

10

31 J/eV


5 .
4

10

25 kg m/s and
  h p

6 .
626

10

34
Js
5 .
4

10

25 kg m/s

1 .
23

10

9 m

1.23
nm.
6

Example continued:
(b) The momentum of the electron is p


2 mK
2

9 .
11

10

31 kg

1 .
0

10
3 eV

1 .
60

10

31
J/eV


1 .
7

10

23 kg m/s and
  h p

6 .
626

10

34
Js
1 .
7

10

23 kg m/s

3 .
88

10

11 m

38 .
8 pm.
7

Example (text problem 28.5): What is the de Broglie wavelength of an electron moving with a speed of 0.6c?
This is a relativistic electron with
 
1
1
 v
2 c
2

1 .
25 .
Its wavelength is
  h p
 h
 mv


1 .
25

6 .
626

9 .
11

10

31

10 kg

34
Js

1 .
8

10
8 m/s
 
3 .
23

10

12 m.
8

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.
An electron beam can be produced with much smaller wavelengths than visible light, allowing for resolution of much smaller structures.
9

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?
K

 p
2
2 m
 h
2
2 m

2
9 .
64

10

21
J

0.060
eV
(b) Through what potential difference should the electrons be accelerated to have this wavelength?

K

V
 
U
  q

V


K
 e
 e

V
0 .
060 eV

0 .
060 Volts e
10

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.
11

The uncertainty principle sets limits on how precise measurements of a particle’s momentum and position can be.
 x
 p x

1
2
 where   h
2

12

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.
13

The energy-time uncertainty principle is

E
 t

1
2
 .
14

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


2
 x

 a

1 .
1

10

26 kg m/s
15

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?
 p x
= m
 v x and
 v x
= 0.10 km/hr = 0.028 m/s.
 x
 p x
 m
 x
 v x

1
2

 x


2 m
 v x

1 .
3

10

32 m
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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?
 x


2 m p
 v x

1 .
1

10

6 m
17

A particle confined to a region of space will have quantized energy levels.
18

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
 n

2 L n p n
 h
 n
 nh
2 L
.
With n = 1, 2, 3,…
The kinetic energy of the particle is KE
 p
2
2 m
.
20

And its total energy is E

K

U
 p
2
2 m

0
 n
2 h
2
8 mL
2
.
The energy of the particle is quantized. The ground state
(n = 1) energy is h
2
E

1
8 mL
2 so that E n
 n
2
E
1
.
21

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?
The total energy of the marble is
E n

1
2 mv
2 
0

2 .
0

10

6
J.
In general E n
 n
2
E
1
E

1 h
2
8 mL 2
.
Solving for n: n

E n
E
1

8 mE n
L
2 h
2

6

10 28 .
22

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
E n

1

E n



 n

1

2
E
1
2 nE
1

E
1

2 n

1

E
1

 n
2
E
1
6 .
6

10

35
J.
23

Example continued:
The change in kinetic energy of the marble would be

K



1
2
1
1 mv
2 f
 mv i
2
2 m
 v
2 f
 v i
2

2
1
2 m
 v f
 v i
 v f
 v i

 mv i
 v f
 v i
.

Assume v f
 v i.
To make a transition to the level n+1, the ball’s speed must change by
 v f
 v i



K mv i

3 .
3

10

31 m/s.
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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.
25

The probability of finding a particle is proportional to the square of its wave function.
26

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.
27

The allowed energy levels in the hydrogen atom are
E n
  mk
2 e
4
2  2
 n
2
E
1 where E
1
=

13.6 eV.
n is the principle quantum number.
Even though the electron does not orbit the nucleus, it has angular momentum.
L

 l
 

 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.
L z
 m l
 m l
=
 l,
 l+1,…,  1, 0, +1,…l 
1, l m l is the orbital magnetic quantum number.
29

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
 m s
 m s
=
 ½ for an electron m s is the spin magnetic quantum number.
30

Electron cloud representations of the electron probability density for an H atom:
31

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 ,m l
, and m s
.
32

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.
33

Each subshell consists of one or more orbitals specified by the quantum numbers n, l, and m l
. 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 2n 2 .
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The subshells are filled by electrons in order of increasing energy.
1 s , 2 s , 2 p , 3 s , 3 p , 4 s , 3 d , 4 p , 5 s , 4 d , 5 p , 6 s , 4 f , 5 d , 6 p , 7 s
Beware! There are exceptions to this rule.
35

The electron configuration for helium is:
1 s
2 specifies the number of electrons in this orbital
Specifies n
Specifies l
36

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 m l
=

1, 0, 1 and since 2 electrons can be placed in each orbital, there can be 6 electron states.
3
3 n
3
3
3
3
1
1 l
1
1
1
1 m l

1

1
0
0
+1
+1 m s

½
+ ½

½
+ ½

½
+ ½
37

Example (text problem 28.46): (a) Find the magnitude of the angular momentum L for an electron with n = 2 and l = 1?
L
 l
 
 
1
 
 
2 
(b) What are the allowed values of L z
?
The allowed values of m l are +1,0,

1 so that L z can be

1 
0 

1  .
38

Laser is an acronym for L ight A mplification by S timulated
E mission of R adiation.
39

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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41

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.
42

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.
43

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 n

694 .
3 nm
1 .
75

396 .
7 nm number of wavelengt hs

L


1 .
51

10
5
.
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•Matter as a Wave
•The Uncertainty Principle
•What Is a Wave Function?
•The Hydrogen Atom
•The Pauli Exclusion Principle
•The Laser
•The Electron Microscope
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