electronic structure
TRANSCRIPT
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7-1
Chapter 4
Quantum Theory and Atomic Structure
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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
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Figure 7.1
Frequency and Wavelength
c =
The Wave Nature The Wave Nature ofof
LightLight
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Figure 7.2 Amplitude (intensity) of a wave.
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Figure 7.3 Regions of the electromagnetic spectrum.
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Sample Problem 7.1
SOLUTION:PLAN:
Interconverting Wavelength and Frequency
wavelength in units given
wavelength in m
frequency (s-1 or Hz)
= c/
Use c =
10-2 m1 cm
10-9 m1 nm
= 1.00x10-10 m
= 325x10-2 m
= 473x10-9 m
=3x108 m/s
1.00x10-10 m= 3x1018 s-1
=
=
3x108 m/s
325x10-2 m= 9.23x107 s-1
3x108 m/s
473x10-9 m= 6.34x1014 s-1
PROBLEM: A dental hygienist uses x-rays (= 1.00A) to take a series of dental radiographs while the patient listens to a radio station ( = 325 cm) and looks out the window at the blue sky (= 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.)
o
1 A = 10-10 m1 cm = 10-2 m1 nm = 10-9 m
o
325 cm
473nm
1.00Ao 10-10 m
1Ao
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7-7
Figure 7.4
Different behaviors of waves and particles.
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Figure 7.5
The diffraction pattern caused by light passing through two adjacent slits.
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Figure 7.6 Blackbody radiationE = n h
E = n h
E = h when n = 1
smoldering coal
electric heating element light bulb filament
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7-10
Figure 7.7
Demonstration of the photoelectric effect.
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Sample Problem 7.2
SOLUTION:
PLAN:
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?
After converting cm to m, we can use the energy equation, E = h combined with = c/ to find the energy.
E = hc/
E =6.626X10-34J*s 3x108m/s
1.20cm 10-2m
cm
x= 1.66x10-23J
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7-12
Figure 7.8
The line spectra of
several elements.
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Figure 7.10
Quantum staircase.
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Figure 7.11
The Bohr explanation of the three series of spectral lines.
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Figure 7.14
Comparing the diffraction patterns of x-rays and electrons.
x-ray diffraction of aluminum foil electron diffraction of aluminum foil
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7-16
Figure B7.2
Figure B7.1
Emission and absorption spectra of sodium atoms.
Flame tests.
strontium 38Sr copper 29Cu
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Figure 7.13
Wave motion in restricted systems.
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The Dual Property of Matter
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7-20
Table 7.1 The de Broglie Wavelengths of Several Objects
Substance Mass (g) Speed (m/s) (m)
slow electron
fast electron
alpha particle
one-gram mass
baseball
Earth
9x10-28
9x10-28
6.6x10-24
1.0
142
6.0x1027
1.0
5.9x106
1.5x107
0.01
25.0
3.0x104
7x10-4
1x10-10
7x10-15
7x10-29
2x10-34
4x10-63
h /mu
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7-21
Sample Problem 7.3
SOLUTION:
PLAN:
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).
Knowing the mass and the speed of the electron allows to use the equation = h/mu to find the wavelength.
= 6.626x10-34kg*m2/s
9.11x10-31kg x 1.00x106m/s= 7.27x10-10m
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7-22
Bohr’s Model of The Hydrogen Atom
• Ground state• Excited state• When an electron falls
back to the lower level, it emits a definite amount of energy in the form of a quantum of light.
• Ex. X-rays
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Limitations of the Bohr Model → Quantum Mechanical Model
• Unfortunately, the Bohr Model failed for all other elements that had more than one proton and one electron. (The multiple electron-nuclear attractions, electron–electron repulsions, and nuclear repulsions make other atoms much more complicated than hydrogen.)
• In 1920s, a new discipline, quantum mechanics, was developed to describe the motion of submicroscopic particles confined to tiny regions of space.
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CLASSICAL THEORYCLASSICAL THEORY
Matter particulate,
massive
Energy continuous,
wavelike
Since matter is discontinuous and particulate perhaps energy is discontinuous and particulate.
Observation Theory
Planck: Energy is quantized; only certain values allowed
blackbody radiation
Einstein: Light has particulate behavior (photons)photoelectric effect
Bohr: Energy of atoms is quantized; photon emitted when electron changes orbit.
atomic line spectra
Figure 7.15
Summary of the major observations and theories leading from classical theory to quantum theory.
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Since energy is wavelike perhaps matter is wavelike
Observation Theory
deBroglie: All matter travels in waves; energy of atom is quantized due to wave motion of electrons
Davisson/Germer: electron diffraction by metal crystal
Since matter has mass perhaps energy has mass
Observation Theory
Einstein/deBroglie: Mass and energy are equivalent; particles have wavelength and photons have momentum.
Compton: photon wavelength increases (momentum decreases) after colliding with electron
Figure 7.15 continued
QUANTUM THEORY
Energy same as Matterparticulate, massive, wavelike
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The Heisenberg Uncertainty Principle
x * m u ≥
h
4
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7-27
The Schrödinger Equation
H = E
d2dy2
d2dx2
d2dz2
+ +82m
h2(E-V(x,y,z)(x,y,z) = 0+
how changes in space
mass of electron
total quantized energy of the atomic system
potential energy at x,y,zwave function
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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
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Quantum Numbers
Energies
n 1 < 2 < 3 < 4…
l s < p < d < f
ml Orbitals in the same subshell have equal energies
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Azimuthat quantum number, l
Subshell designation Maximum number of electrons per subshell
0 s 2
1 p 6
2 d 10
3 f 14
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Table 7.2 The Hierarchy of Quantum Numbers for Atomic Orbitals
Name, Symbol(Property) Allowed Values Quantum Numbers
Principal, n(size, energy)
Angular momentum, l
(shape)
Magnetic, ml
(orientation)
Positive integer(1, 2, 3, ...)
0 to n-1
-l,…,0,…,+l
1
0
0
2
0 1
0
3
0 1 2
0
0-1 +1 -1 0 +1
0 +1 +2-1-2
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7-32
Sample Problem 7.5
SOLUTION:
PLAN:
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?
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.
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.
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Sample Problem 7.6
SOLUTION:
PLAN:
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
Combine the n value and l designation to name the sublevel. Knowing l, we can find ml and the number of orbitals.
n l sublevel name possible ml values # of orbitals
(a)
(b)
(c)
(d)
3
2
5
4
2
0
1
3
3d
2s
5p
4f
-2, -1, 0, 1, 2
0
-1, 0, 1
-3, -2, -1, 0, 1, 2, 3
5
1
3
7
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Figure 7.18
The 2p orbitals.
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Figure 7.19 The 3d orbitals.
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Figure 7.19 continued
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Figure 7.20 One of the seven possible 4f orbitals.