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Chapter 7
Quantum Theory and Atomic Structure
7-1
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Quantum Theory and Atomic Structure
7.1 The Nature of Light
7.2 Atomic Spectra
7.3 The Wave-Particle Duality of Matter and Energy
7.4 The Quantum-Mechanical Model of the Atom
7-2
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The Wave Nature
of
Light
Figure 7.1
Frequency and
Wavelength
c=ln
7-3
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Figure 7.2
7-4
Amplitude (intensity) of a wave.
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Figure 7.3
7-5
Regions of the electromagnetic spectrum.
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Sample Problem 7.1
PROBLEM:
PLAN:
Interconverting Wavelength and Frequency
Use c = ln
wavelength in units given
o
1 A = 10-10 m
1 cm = 10-2 m
1 nm = 10-9 m
wavelength in m
n = c/l
frequency (s-1 or Hz)
7-6
o
A dental hygienist uses x-rays (l= 1.00A) to take a series of dental
radiographs while the patient listens to a radio station (l = 325 cm)
and looks out the window at the blue sky (l= 473 nm). What is the
frequency (in s-1) of the electromagnetic radiation from each source?
(Assume that the radiation travels at the speed of light, 3.00x108 m/s.)
SOLUTION:
o
-10
-10
1.00A 10 o m = 1.00x10 m
1A
3x108 m/s
= 3x1018 s-1
n=
1.00x10-10 m
-2
325 cm 10 m = 325x10-2 m
1 cm
3x108 m/s
n=
= 9.23x107 s-1
325x10-2 m
10-9 m
473nm
= 473x10-9 m
1 nm
3x108 m/s = 6.34x1014 s-1
n=
473x10-9 m
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Figure 7.4
Different
behaviors
of waves
and
particles.
7-7
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The diffraction pattern caused by light
passing through two adjacent slits.
Figure 7.5
7-8
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Figure 7.6
Demonstration of the
photoelectric effect.
7-9
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Sample Problem 7.2
Calculating the Energy of Radiation from Its
Wavelength
PROBLEM: A cook uses a microwave oven to heat a meal. The wavelength of
the radiation is 1.20cm. What is the energy of one photon of this
microwave radiation?
PLAN: After converting cm to m, we can use the energy equation, E = hn
combined with n = c/l to find the energy.
SOLUTION:
E=
E = hc/l
6.626X10-34J*s x 3x108m/s
1.20cm
10-2m
cm
7-10
= 1.66x10-23J
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Figure 7.7
The line
spectra of
several
elements.
7-11
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Rydberg equation
1
l
=
R
1
n12
1
n22
R is the Rydberg constant = 1.096776x107 m-1
Figure 7.8
Three series of spectral lines of atomic hydrogen.
for the visible series, n1 = 2 and n2 = 3, 4, 5, ...
7-12
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Figure 7.9
Quantum staircase.
7-13
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Figure 7.10
The Bohr explanation of the three series of spectral lines.
7-14
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Figure 7.13
Wave motion in
restricted systems.
7-15
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l = h /mu
Table 7.1
The de Broglie Wavelengths of Several Objects
Substance
Speed (m/s)
l (m)
slow electron
9x10-28
fast electron
9x10-28
5.9x106
1x10-10
alpha particle
6.6x10-24
1.5x107
7x10-15
one-gram mass
1.0
0.01
7x10-29
baseball
142
25.0
2x10-34
6.0x1027
3.0x104
4x10-63
Earth
7-16
Mass (g)
1.0
7x10-4
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Sample Problem 7.3 Calculating the de Broglie Wavelength of an Electron
PROBLEM: Find the deBroglie wavelength of an electron with a speed of
1.00x106m/s (electron mass = 9.11x10-31kg; h = 6.626x10-34
kg*m2/s).
PLAN:
Knowing the mass and the speed of the electron allows to use the
equation l = h/mu to find the wavelength.
SOLUTION:
l=
6.626x10-34kg*m2/s
9.11x10-31kg x 1.00x106m/s
7-17
= 7.27x10-10m
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Figure 7.14
Comparing the diffraction patterns of x-rays and electrons.
x-ray diffraction of aluminum foil
7-18
electron diffraction of aluminum foil
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Figure 7.15
CLASSICAL THEORY
Matter
particulate,
massive
Energy
continuous,
wavelike
Summary of the major observations
and theories leading from classical
theory to quantum theory.
Since matter is discontinuous and particulate
perhaps energy is discontinuous and particulate.
Observation
blackbody radiation
7-19
Theory
Planck:
photoelectric effect
Energy is quantized; only certain values
allowed
Einstein: Light has particulate behavior (photons)
atomic line spectra
Bohr:
Energy of atoms is quantized; photon
emitted when electron changes orbit.
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Figure 7.15 continued
Since energy is wavelike perhaps matter is wavelike
Observation
Davisson/Germer:
electron diffraction
by metal crystal
Theory
deBroglie: All matter travels in waves; energy of
atom is quantized due to wave motion of
electrons
Since matter has mass perhaps energy has mass
Observation
Compton: photon
wavelength increases
(momentum decreases)
after colliding with
electron
Theory
Einstein/deBroglie: Mass and energy are
equivalent; particles have wavelength and
photons have momentum.
QUANTUM THEORY
Energy same as Matter
particulate, massive, wavelike
7-20
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The Heisenberg Uncertainty Principle
Dx * mDu ≥
7-21
h
4p
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The Schrödinger Equation
HY = EY
wave function
d2Y
dx2
+
d2Y
dy2
mass of electron
+
d2Y
dz2
how y changes in space
7-22
potential energy at x,y,z
8p2mQ
+
(E-V(x,y,z)Y(x,y,z) = 0
2
h
total quantized energy of
the atomic system
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Figure 7.16
Electron probability in the
ground-state H atom.
7-23
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Quantum Numbers and Atomic Orbitals
An atomic orbital is specified by three quantum numbers.
n the principal quantum number - a positive integer
l the angular momentum quantum number - an integer from 0 to n-1
ml the magnetic moment quantum number - an integer from -l to +l
7-24
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Table 7.2 The Hierarchy of Quantum Numbers for Atomic Orbitals
Name, Symbol
(Property)
Allowed Values
Quantum Numbers
Principal, n
Positive integer
(size, energy)
(1, 2, 3, ...)
1
Angular
momentum, l
0 to n-1
(shape)
0
0
0
0
Magnetic, ml
-l,…,0,…,+l
(orientation)
2
3
1
0
2
0
-1 0 +1
-1 0 +1
-2
7-25
1
-1
0
+1 +2
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Sample Problem 7.4
Determining Quantum Numbers for an Energy Level
PROBLEM: What values of the angular momentum (l) and magnetic (ml)
quantum numbers are allowed for a principal quantum number (n) of
3? How many orbitals are allowed for n = 3?
PLAN: Follow the rules for allowable quantum numbers found in the text.
l values can be integers from 0 to n-1; ml can be integers from -l
through 0 to + l.
SOLUTION: For n = 3, l = 0, 1, 2
For l = 0 ml = 0
For l = 1 ml = -1, 0, or +1
For l = 2 ml = -2, -1, 0, +1, or +2
There are 9 ml values and therefore 9 orbitals with n = 3.
7-26
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Sample Problem 7.5
Determining Sublevel Names and Orbital Quantum
Numbers
PROBLEM: Give the name, magnetic quantum numbers, and number of orbitals
for each sublevel with the following quantum numbers:
(a) n = 3, l = 2
(b) n = 2, l = 0
(c) n = 5, l = 1 (d) n = 4, l = 3
PLAN: Combine the n value and l designation to name the sublevel.
Knowing l, we can find ml and the number of orbitals.
SOLUTION:
n
l
(a)
3
2
3d
-2, -1, 0, 1, 2
5
(b)
2
0
2s
0
1
(c)
5
1
5p
-1, 0, 1
3
(d)
4
3
4f
-3, -2, -1, 0, 1, 2, 3
7
7-27
sublevel name possible ml values # of orbitals
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Figure 7.18
7-28
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Figure 7.19
7-29
The 2p orbitals.
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Figure 7.20
7-30
The 3d orbitals.
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Figure 7.20 continued
7-31
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Figure 7.21
7-32
One of the seven
possible 4f orbitals.