### Ch 7

```Chapter 7
Quantum Theory and
Atomic Structure
7-1
Dr. Wolf’s CHM 101
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
Dr. Wolf’s CHM 101
Electromagnetic Radiation (light) - Wave like
Wavelength, l, the
distance from one
crest to the next in
the wave. Measured
in units of distance.
Frequency, n, the
number of complete
cycles per sec., cps, Hz
Speed of Light, C, same
3.00 x 1010 cm/sec in
vacuum
c=ln
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Dr. Wolf’s CHM 101
Amplitude (Intensity) of a Wave
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Dr. Wolf’s CHM 101
c=ln
Regions of the Electromagnetic Spectrum
n =c/l
3m
100 MHz
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300 m
1000 kHz
Sample Problem 7.1
Interconverting Wavelength and Frequency
PROBLEM: 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 = 325cm)
and looks out the window at the blue sky (l= 473nm). What is the
frequency (in s-1) of the electromagnetic radiation from each source?
(Assume that the radiation travels at the speed of light,
3.00x108m/s.)
PLAN:
Use c = ln
wavelength in units given
1A = 10-10m
1cm = 10-2m
1nm = 10-9m
wavelength in m
n = c/l
frequency (s-1 or Hz)
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Dr. Wolf’s CHM 101
SOLUTION:
1.00A 10-10m = 1.00x10-10m
1A
3x108m/s
= 3x1018s-1
n=
1.00x10-10m
-2
325cm 10 m = 325x10-2m
1cm
3x108m/s
n=
= 9.23x107s-1
325x10-2m
10-9m
473nm
= 473x10-9m
1nm
3x108m/s = 6.34x1014s-1
n=
473x10-9m
Different behaviors of waves and particles.
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Dr. Wolf’s CHM 101
The diffraction pattern caused by light passing through two adjacent slits.
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Dr. Wolf’s CHM 101
The view that EM was wavelike could not
explain certain phenomena like :
1) Blackbody radiation - when objects are
heated, they give off shorter and more intense
radiation as the temperature increases, e.g. dull
red hot, to hotter orange, to white hot.......
But wavelike properties would predict hotter
temps would continue to give more and more
of shorter and shorter wavelengths. But
instead a bell shape curve of intensities is
obtained with the peak at the IR- Vis part of the
spectrum.
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Dr. Wolf’s CHM 101
E = hn
E = hc/l
Planck’s constant
h = 6.626 x 10-34 J-s
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Dr. Wolf’s CHM 101
The view that EM was wavelike could not
explain certain phenomena like :
2) Photoelectron Effect - light shinning on
certain metal plates caused a flow of electrons.
However the the light had a minimum
frequency to cause the effect, i.e. not any color
would work.
And although bright light caused more electron
flow than weak light, electron flow started
immediately with both strong or weak light.
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Dr. Wolf’s CHM 101
Demonstration of the
photoelectric effect
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Dr. Wolf’s CHM 101
The better explanation for these experiments was
that EM consisted of packets of energy called
photons (particle-like) that had wave-like properties
as well. And that atoms could have only certain
quantities of energy, E = nhn , where n is a positive
integer, 1, 2, 3, etc. This means energy is quantized.
Ephoton = hn = Eatom
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Dr. Wolf’s CHM 101
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
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
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Dr. Wolf’s CHM 101
= 1.66x10-23J
The view that EM was wavelike could not explain
certain phenomena like :
3) Atomic Spectra - Electrical discharges in tube of
gaseous elements produces light (EM).
But not all wavelengths of light were produced but
just a few certain wavelengths (or frequencies).
And different elements had different wavelengths
associated with them. Not just in the Visible but
also IR and UV regions.
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Dr. Wolf’s CHM 101
The line spectra
of several
elements
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Dr. Wolf’s CHM 101
Three Series of Spectral Lines of Atomic Hydrogen
Looking for an equation that would predict the wavelength seen in H spectrum
Rydberg equation
1
l
=
R is the Rydberg constant = 1.096776 x
R
107
1
2
n1
m-1
But WHY does this equation work?
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Dr. Wolf’s CHM 101
1
n22
for the visible series, n1 = 2
and n2 = 3, 4, 5, ...
Bohr Model of the Hydrogen Atom
Assumed the H atom has only certain allowable energy levels for the electron
orbits. (quantized because it made the equations work))
When the electron moves from one orbit to another, it has to absorb or emit a
photon whose energy equals the difference in energy between the two orbits.
By assuming the electron traveled in circular orbits, the energy level for each
orbits was
E = -2.18 x 10-18 J / n2 where n = 1, 2, 3,....
Note: because of negative sign, lowest energy (most stable) when n = 1 and
highest energy is E = 0 when n = .
So energy released when the electron moves from one n level to another is
E= hn = hc/l = -2.18 x 10-18 J ( 1 / n2final - 1 / n2initial )
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Quantum staircase
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The Bohr explanation of the three series of spectral lines.
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Dr. Wolf’s CHM 101
But why must the electron’s energy
be quantized?
If EM can have particle-like properties in
addition to being wave-like, what if the
electron particles have wave-like
properties?
Quantization is a natural consequence of
having wave-like properties.
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Dr. Wolf’s CHM 101
Wave motion in
restricted systems
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Dr. Wolf’s CHM 101
de Broglie Wavelength E = hc/l
giving particles wave-like properties
so l = hc/ E
but E = muc
l = h /mu
The de Broglie Wavelengths of Several Objects
Substance
Speed, u, (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
7-23
Mass (g)
Dr. Wolf’s CHM 101
1.0
7x10-4
Sample Problem 7.3
PROBLEM:
PLAN:
Calculating the de Broglie Wavelength of an
Electron
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 l = h/mu to find the wavelength.
SOLUTION:
l=
6.626x10-34kg-m2/s
9.11x10-31kg x 1.00x106m/s
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Dr. Wolf’s CHM 101
= 7.27x10-10m
CLASSICAL THEORY
Matter
particulate,
massive
Energy
continuous,
wavelike
Summary of the major observations
theory to quantum theory.
Since matter is discontinuous and particulate
perhaps energy is discontinuous and particulate.
Observation
7-25
Theory
photoelectric effect
Energy is quantized; only certain values
allowed
Einstein: Light has particulate behavior (photons)
atomic line spectra
Bohr:
Dr. Wolf’s CHM 101
Planck:
Energy of atoms is quantized; photon
emitted when electron changes orbit.
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
Theory
Compton: photon
Einstein/deBroglie: Mass and energy are
wavelength increases
equivalent; particles have wavelength and
(momentum decreases)
photons have momentum.
after colliding with electron
QUANTUM THEORY
Energy same as Matter
particulate, massive, wavelike
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Dr. Wolf’s CHM 101
The Heisenberg Uncertainty Principle
Heisenberg Uncertainty Principle expresses a
limitation on accuracy of simultaneous measurement
of observables such as the position and the
momentum of a particle.
x
7-27
Dr. Wolf’s CHM 101
.mu 
h
4p
Sample Problem 7.4
PROBLEM:
PLAN:
Applying the Uncertainty Principle
An electron moving near an atomic nucleus has a speed 6 x 106 ± 1%
m/s. What is the uncertainty in its position ( x)?
The uncertainty ( u) is given as ±1% (0.01) of 6 x 106 m/s. Once
we calculate this, plug it into the uncertainty equation.
SOLUTION:
 u = (0.01) (6 x 106 m/s) = 6 x 104 m/s
 x
. m u 
h
4p
6.626 x 10-34 kg-m2/s
x
4p (9.11 x 10-31 kg) (6 x104 m/s)
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Dr. Wolf’s CHM 101
 10-9m
The Schrödinger Equation Quantum Numbers and Atomic Orbitals
A complicated equation with multiple solutions
which describes the probability of locating an
electron at the various allowed energy levels.
Solutions involve three interdependent variables
to describe an electron orbital.
i.e., 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
ms the spin quantum number, + 1/2 or - 1/2
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Electron probability in the
ground-state H atom
n=1 l=0 m=0
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“Orbital” showing 90%
of electron probability
Table 7.2 The Hierarchy of Quantum Numbers for Atomic Orbitals
Name, Symbol
(Property)
Allowed Values
Principal, n
Positive integer
(size, energy)
(1, 2, 3, ...)
Angular
momentum, l
0 to n-1
(shape)
sublevel names
l = 0, called “s” l = 1, “p”
l = 2, “d” l = 3, “f”
Quantum Numbers
1
2
0
0
0
0
3
1
0
-1 0 +1
-l,…,0,…,+l
-2
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Dr. Wolf’s CHM 101
2
0
-1 0 +1
Magnetic, ml
(orientation)
1
-1
0
+1 +2
Sample Problem 7.5
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.
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Dr. Wolf’s CHM 101
Sample Problem 7.6
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
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sublevel name possible ml values # of orbitals
s orbitals
1s
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2s
3s
p orbitals - three of them
Combination
The 2p orbitals n = 2, l = 1
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d orbitals - five of them
The 3d orbitals n = 3 l = 2
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d orbitals - five of them
Combination
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f orbitals - seven of them
One of the seven
possible 4f orbitals
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End of Chapter 7
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Dr. Wolf’s CHM 101
```