Download Chapter 7 Quantum Theory of the Atom

Chapter 7
Quantum Theory
of the Atom
Contents and Concepts
Light Waves, Photons, and the Bohr Theory
To understand the formation of chemical bonds,
you need to know something about the electronic
structure of atoms. Because light gives us
information about this structure, we begin by
discussing the nature of light. Then we look at the
Bohr theory of the simplest atom, hydrogen.
1. The Wave Nature of Light
2. Quantum Effects and Photons
3. The Bohr Theory of the Hydrogen Atom
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Quantum Mechanics and Quantum Numbers
The Bohr theory firmly establishes the concept of
energy levels but fails to account for the details of
atomic structure. Here we discuss some basic
notions of quantum mechanics, which is the theory
currently applied to extremely small particles, such
as electrons in atoms.
4. Quantum Mechanics
5. Quantum Numbers and Atomic Orbitals
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Learning Objectives
Light Waves, Photons, and the Bohr Theory
1. The Wave Nature of Light
a. Define the wavelength and frequency of a
wave.
b. Relate the wavelength, frequency, and
speed of light.
c. Describe the different regions of the
electromagnetic spectrum.
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2. Quantum Effects and Photons
a. State Planck’s quantization of vibrational
energy.
b. Define Planck’s constant and photon.
c. Describe the photoelectric effect.
d. Calculate the energy of a photon from its
frequency or wavelength.
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3. The Bohr Theory of the Hydrogen Atom
a. State the postulates of Bohr’s theory of the
hydrogen atom.
b. Relate the energy of a photon to the
associated energy levels of an atom.
c. Determine the wavelength or frequency of a
hydrogen atom transition.
d. Describe the difference between emission
and absorption of light by an atom.
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Quantum Mechanics and Quantum Numbers
4. Quantum Mechanics
a. State the de Broglie relation.
b. Calculate the wavelength of a moving
particle.
c. Define quantum mechanics.
d. State Heisenberg’s uncertainty principle.
e. Relate the wave function for an electron to
the probability of finding the electron at a
location in space.
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5. Quantum Numbers and Atomic Orbitals
a. Define atomic orbital.
b. Define each of the quantum numbers for an
atomic orbital.
c. State the rules for the allowed values for
each quantum number.
d. Apply the rules for quantum numbers.
e. Describe the shapes of s, p, and d orbitals.
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A wave is a continuously repeating change or
oscillation in matter or in a physical field.
Light is an electromagnetic wave, consisting of
oscillations in electric and magnetic fields traveling
through space.
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A wave can be characterized by its wavelength
and frequency.
Wavelength, symbolized by the Greek letter
lambda, l, is the distance between any two
identical points on adjacent waves.
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Frequency, symbolized by the Greek letter nu, n,
is the number of wavelengths that pass a fixed
point in one unit of time (usually a second). The
unit is 1/S or s-1, which is also called the Hertz (Hz).
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Wavelength and frequency are related by the wave
speed. The speed of light, c, is 3.00 x 108 m/s.
c = nl
The relationship between wavelength and
frequency due to the constant velocity of light is
illustrated on the next slide.
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When the
wavelength is
reduced by a
factor of two,
the frequency
increases by
a factor of
two.
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?
What is the wavelength of blue light
with a frequency of 6.4  1014/s?
n = 6.4  1014/s
c = 3.00  108 m/s
c = nl so
l = c/n
m
3.00 x 10
c
s
λ 
n
14 1
6.4 x 10
s
8
l = 4.7  10−7 m
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?
What is the frequency of light having a
wavelength of 681 nm?
c = nl so
n = c/l
l = 681 nm = 6.81  10−7 m
c = 3.00  108 m/s
m
3.00  10
c
s
n 
l 6.81  107 m
8
l = 4.41  1014 /s
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The range of frequencies and wavelengths of
electromagnetic radiation is called the
electromagnetic spectrum.
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Concept Check 7.1
Laser light of a specific frequency falls on a crystal
that converts a portion of this light into light with
double the original frequency. How is the wavelength
of this frequency- doubled light related to the
wavelength of the original laser light? Suppose the
original laser light was red. In which region of the
spectrum would the frequency-doubled light be? (If
this is in the visible region, what color is the light?)
When frequency is doubled, wavelength is halved.
The light would be in the blue-violet region.
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One property of waves is that they can be
diffracted—that is, they spread out when they
encounter an obstacle about the size of the
wavelength.
In 1801, Thomas Young, a British physicist,
showed that light could be diffracted. By the early
1900s, the wave theory of light was well
established.
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The wave theory could not explain the
photoelectric effect, however.
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The photoelectric
effect is the
ejection of an
electron from the
surface of a metal
or other material
when light shines
on it.
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Einstein proposed that light consists of quanta or
particles of electromagnetic energy, called
photons. The energy of each photon is
proportional to its frequency:
E = hn
h = 6.626  10−34 J  s (Planck’s constant)
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Einstein used this understanding of light to explain
the photoelectric effect in 1905.
Each electron is struck by a single photon. Only
when that photon has enough energy will the
electron be ejected from the atom; that photon is
said to be absorbed.
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Light, therefore, has properties of both waves and
matter. Neither understanding is sufficient alone.
This is called the particle–wave duality of light.
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?
The blue–green line of the hydrogen
atom spectrum has a wavelength of
486 nm. What is the energy of a photon
of this light?
l = 486 nm = 4.86  10−7 m
c = 3.00  108 m/s
h = 6.63  10−34 J  s
E = hn and
c = nl so
E = hc/l

8 m
6.63  10 Js  3.00  10

hc
s


E

λ
4.86  107 m

34



l = 4.09  10−19 J
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In the early 1900s, the atom was understood to
consist of a positive nucleus around which
electrons move (Rutherford’s model).
This explanation left a theoretical dilemma:
According to the physics of the time, an electrically
charged particle circling a center would continually
lose energy as electromagnetic radiation. But this
is not the case—atoms are stable.
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In addition, this understanding could not explain
the observation of line spectra of atoms.
A continuous spectrum contains all wavelengths
of light.
A line spectrum shows only certain colors or
specific wavelengths of light. When atoms are
heated, they emit light. This process produces a
line spectrum that is specific to that atom. The
emission spectra of six elements are shown on the
next slide.
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In 1913, Neils Bohr, a Danish scientist, set down
postulates to account for
1. The stability of the hydrogen atom
2. The line spectrum of the atom
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Energy-Level Postulate
An electron can have only certain energy values,
called energy levels. Energy levels are quantized.
For an electron in a hydrogen atom, the energy is
given by the following equation:
E
RH
n2
RH = 2.179  10−18 J
n = principal quantum number
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Transitions Between Energy Levels
An electron can change energy levels by
absorbing energy to move to a higher energy
level or by emitting energy to move to a lower
energy level.
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For a hydrogen electron, the energy change is
given by
ΔE  E f  E i
 1

1
ΔE  RH  2  2 
n

n
i 
 f
RH = 2.179  10−8 J, Rydberg constant
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The energy of the emitted or absorbed photon is
related to DE:
E photon  ΔE electron  hn
h  Planck' s constant
We can now combine these two equations:
 1

1
hn   R H  2  2 
n

n
i 
 f
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Light is absorbed by an atom when the electron
transition is from lower n to higher n (nf > ni). In this
case, DE will be positive.
Light is emitted from an atom when the electron
transition is from higher n to lower n (nf < ni). In this
case, DE will be negative.
An electron is ejected when nf = ∞.
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Energy-level diagram
for the hydrogen atom.
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Electron transitions for
an electron in the
hydrogen atom.
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?
What is the wavelength of the light emitted
when the electron in a hydrogen atom
undergoes a transition from n = 6 to n = 3?
 1
ΔE  RH  2 
n
 f
hc
ΔE 
so λ 
λ
ni = 6
nf = 3
RH = 2.179  10−18 J

ΔE  2.179  10
18
1 
ni 2 
hc
ΔE
1
 1
J  2  2 = −1.816  10−19 J
6 
3


8 m
6.626  10 J  s  3.00  10

s

λ

1.094  10−6 m
19
1.816  10 J

34

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

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Concept Check 7.2
An atom has a line spectrum consisting of a red line
and a blue line. Assume that each line corresponds
to a transition between two adjacent energy levels.
Sketch an energy-level diagram with three energy
levels that might explain this line spectrum, indicating
the transitions on this diagram. Consider the
transition from the highest energy level on this
diagram to the lowest energy level. How would you
describe the color or region of the spectrum
corresponding to this transition?
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A minimum of three energy levels are required.
n=3
n=2
n=1
The red line corresponds to the smaller energy
difference in going from n = 3 to n = 2.
The blue line corresponds to the larger energy
difference in going from n = 2 to n = 1.
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Planck
Vibrating atoms have only certain energies:
E = hn or 2hn or 3hn
Einstein
Energy is quantized in particles called photons:
E = hn
Bohr
Electrons in atoms can have only certain values of
energy. For hydrogen:
RH
E 2
n
RH  2.179  1018 J, n  principal quantum number
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Light has properties of both waves and
particles (matter).
What about matter?
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In 1923, Louis de Broglie, a French physicist,
reasoned that particles (matter) might also have
wave properties.
The wavelength of a particle of mass, m (kg), and
velocity, v (m/s), is given by the de Broglie relation:
h
λ
mv
34
h  6.626  10 J  s
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?
Compare the wavelengths of
(a) an electron traveling at a speed that
is one-hundredth the speed of light and
(b) a baseball of mass 0.145 kg having
a speed of 26.8 m/s (60 mph).
Electron
me = 9.11  10−31 kg
v = 3.00  106 m/s
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Baseball
m = 0.145 kg
v = 26.8 m/s
h
λ
mv
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Electron
me = 9.11  10−31 kg
v = 3.00  106 m/s
6.63  1034 J  s
λ
 2.43  10−10 m

31
6 m
9.11  10 kg  3.00  10

s




Baseball
m = 0.145 kg
v = 26.8 m/s
34
6.63  10 J  s
λ
 1.71  10−34 m
m

 0.145 kg  26.8 
s

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Concept Check 7.3
A proton is approximately 2000 times heavier than an
electron. How would the speeds of these particles
compare if their corresponding wavelengths were
about equal?
λ electron  λ proton
λ electron
h

melectronv electron
h
λ proton 
mproton v proton
h
h

mproton v proton
melectronv electron
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melectronv electron  mproton v proton
mproton
v electron

v proton
melectron
27
velectron 1.67262  10 kg

vproton 9.10938  1031 kg
v electron
3
 1.8361 5  10
v proton
The velocity of the electron is more
than 1800 times that of the proton.
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Building on de Broglie’s work, in 1926, Erwin
Schrödinger devised a theory that could be used to
explain the wave properties of electrons in atoms
and molecules.
The branch of physics that mathematically
describes the wave properties of submicroscopic
particles is called quantum mechanics or wave
mechanics.
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Quantum mechanics alters how we think about the
motion of particles.
In 1927, Werner Heisenberg showed how it is
impossible to know with absolute precision both
the position, x, and the momentum, p, of a particle
such as electron.
h
(Δx )(Δp) 
4π
Because p = mv this uncertainty becomes more
significant as the mass of the particle becomes
smaller.
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Quantum mechanics allows us to make statistical
statements about the regions in which we are most
likely to find the electron.
Solving Schrödinger’s equation gives us a wave
function, represented by the Greek letter psi, y,
which gives information about a particle in a given
energy level.
Psi-squared, y 2, gives us the probability of finding
the particle within a region of space.
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The wave function for
the lowest level of the
hydrogen atom is
shown to the left.
Note that its value is
greatest nearest the
nucleus, but rapidly
decreases thereafter.
Note also that it never
goes to zero, only to a
very small value.
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Two additional views are shown on the next slide.
Figure A illustrates the probability density for an
electron in hydrogen. The concentric circles
represent successive shells.
Figure B shows the probability of finding the
electron at various distances from the nucleus.
The highest probability (most likely) distance is at
50 pm.
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According to quantum mechanics, each electron is
described by four quantum numbers:
1.
2.
3.
4.
Principal quantum number (n)
Angular momentum quantum number (l)
Magnetic quantum number (ml)
Spin quantum number (ms)
The first three define the wave function for a
particular electron. The fourth quantum number
refers to the magnetic property of electrons.
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A wave function for an electron in an atom is called
an atomic orbital (described by three quantum
numbers—n, l, ml).
It describes a region of space with a definite shape
where there is a high probability of finding the
electron.
We will study the quantum numbers first, and then
look at atomic orbitals.
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Principal Quantum Number, n
This quantum number is the one on which the
energy of an electron in an atom primarily
depends. The smaller the value of n, the lower the
energy and the smaller the orbital.
The principal quantum number can have any
positive value: 1, 2, 3, . . .
Orbitals with the same value for n are said to be in
the same shell.
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Shells are sometimes designated by uppercase
letters:
Letter
n
K
1
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L
2
M
3
N
4
...
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Angular Momentum Quantum Number, l
This quantum number distinguishes orbitals of a
given n (shell) having different shapes.
It can have values from 0, 1, 2, 3, . . . to a
maximum of (n – 1).
For a given n, there will be n different values of l,
or n types of subshells.
Orbitals with the same values for n and l are said
to be in the same shell and subshell.
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Subshells are sometimes designated by lowercase
letters:
l
Letter
0
s
1
p
2
d
3
f
n≥
1
2
3
4
...
Not every subshell type exists in every shell. The
minimum value of n for each type of subshell is
shown above.
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Magnetic Quantum Number, ml
This quantum number distinguishes orbitals of a
given n and l—that is, of a given energy and shape
but having different orientations.
The magnetic quantum number depends on the
value of l and can have any integer value from –l
to 0 to +l. Each different value represents a
different orbital. For a given subshell, there will be
(2l + 1) values and therefore (2l + 1) orbitals.
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Let’s summarize:
When n = 1, l has only one value, 0.
When l = 0, ml has only one value, 0.
So the first shell (n = 1) has one subshell, an ssubshell, 1s. That subshell, in turn, has one orbital.
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When n = 2, l has two values, 0 and 1.
When l = 0, ml has only one value, 0. So there is a
2s subshell with one orbital.
When l = 1, ml has only three values, -1, 0, 1. So
there is a 2p subshell with three orbitals.
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When n = 3, l has three values, 0, 1, and 2.
When l = 0, ml has only one value, 0. So there is a
3s subshell with one orbital.
When l = 1, ml has only three values, -1, 0, 1. So
there is a 3p subshell with three orbitals.
When l = 2, ml has only five values, -2, -1, 0, 1, 2.
So there is a 3d subshell with five orbitals.
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We could continue with n = 4 and 5. Each would
gain an additional subshell (f and g, respectively).
In an f subshell, there are seven orbitals; in a
g subshell, there are nine orbitals.
Table 7.1 gives the complete list of permitted
values for n, l, and ml up to the fourth shell. It is on
the next slide.
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The figure
shows
relative
energies for
the hydrogen
atom shells
and
subshells;
each orbital
is indicated
by a dashedline.
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Spin Quantum Number, ms
This quantum number refers to the two possible
orientations of the spin axis of an electron.
It may have a value of either +1/2 or -1/2.
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?
Which of the following are permissible
sets of quantum numbers?
n = 4, l = 4, ml = 0, ms = ½
n = 3, l = 2, ml = 1, ms = -½
n = 2, l = 0, ml = 0, ms = ³/²
n = 5, l = 3, ml = -3, ms = ½
(a) Not permitted. When n = 4, the maximum
value of l is 3.
(b) Permitted.
(c) Not permitted; ms can only be +½ or –½.
(b) Permitted.
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Atomic Orbital Shapes
An s orbital is spherical.
A p orbital has two lobes along a straight line
through the nucleus, with one lobe on either side.
A d orbital has a more complicated shape.
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The cross-sectional view
of a 1s orbital and a
2s orbital highlights the
difference in the two
orbitals’ sizes.
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The cutaway diagrams
of the 1s and 2s orbitals
give a better sense of
them in three
dimensions.
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The next slide illustrates p orbitals.
Figure A shows the general shape of a p orbital.
Figure B shows the orientations of each of the
three p orbitals.
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The complexity of the d orbitals can be seen below:
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