Chemistry: A Molecular Approach, 1st Ed.
Nivaldo Tro
Chapter 7
The QuantumMechanical
Model of the
Atom
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
2007, Prentice Hall
The Behavior of the Very Small
• electrons are incredibly small
a single speck of dust has more electrons than the
number of people who have ever lived on earth
• electron behavior determines much of the
behavior of atoms
• directly observing electrons in the atom is
impossible, the electron is so small that
observing it changes its behavior
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A Theory that Explains Electron Behavior
• the quantum-mechanical model explains the manner
•
electrons exist and behave in atoms
helps us understand and predict the properties of atoms
that are directly related to the behavior of the electrons
 why some elements are metals while others are nonmetals
 why some elements gain 1 electron when forming an anion,
while others gain 2
 why some elements are very reactive while others are
practically inert
 and other Periodic patterns we see in the properties of the
elements
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The Nature of Light
its Wave Nature
• light is a form of electromagnetic radiation
 composed of perpendicular oscillating waves, one for the
electric field and one for the magnetic field
 an electric field is a region where an electrically charged particle
experiences a force
 a magnetic field is a region where an magnetized particle experiences
a force
• all electromagnetic waves move through space at the
same, constant speed
 3.00 x 108 m/s in a vacuum = the speed of light, c
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Speed of Energy Transmission
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Electromagnetic Radiation
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Characterizing Waves
• the amplitude is the height of the wave
the distance from node to crest
or node to trough
the amplitude is a measure of how intense the light
is – the larger the amplitude, the brighter the light
• the wavelength, (l) is a measure of the distance
covered by the wave
the distance from one crest to the next
or the distance from one trough to the next, or the
distance between alternate nodes
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Wave Characteristics
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Characterizing Waves
• the frequency, (n) is the number of waves that
pass a point in a given period of time
the number of waves = number of cycles
units are hertz, (Hz) or cycles/s = s-1
1 Hz = 1 s-1
• the total energy is proportional to the amplitude
and frequency of the waves
the larger the wave amplitude, the more force it has
the more frequently the waves strike, the more total
force there is
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The Relationship Between
Wavelength and Frequency
• for waves traveling at the same speed, the shorter
the wavelength, the more frequently they pass
• this means that the wavelength and frequency of
electromagnetic waves are inversely proportional
since the speed of light is constant, if we know
wavelength we can find the frequency, and visa versa
c 
n s  
l m 
-1
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m
s
10
Example 7.1- Calculate the wavelength of red
light with a frequency of 4.62 x 1014 s-1
Given: n = 4.62 x 1014 s-1
Find: l, (nm)
Concept Plan: n s-1)
l (m)
c
l
n
1 nm
l (nm)
109 m
Relationships: l∙n = c, 1 nm = 10-9 m
Solve:
c 3.00 108 m  s -1
7
l 

6
.
49

10
m
14
1
n
4.62 10 s
1 nm
7
6.49 10 m  9  6.49 102 nm
10 m
Check: the unit is correct, the wavelength is appropriate
for red light
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Practice – Calculate the wavelength of a radio
signal with a frequency of 100.7 MHz
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Practice – Calculate the wavelength of a radio
signal with a frequency of 100.7 MHz
Given: n = 100.7 MHz
Find: l, (m)
Concept Plan: n MHz)
6 -1
10 s
1 MHz
n (s-1)
l
c
l (m)
n
Relationships: l∙n = c, 1 MHz = 106 s-1
Solve:
106 s -1
100.7 MHz 
 1.007 108 s -1
1 MHz
c 3.00 108 m  s -1
l 
 2.98 m
8
1
n
1.007 10 s
Check: the unit is correct, the wavelength is appropriate
for radiowaves
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Color
• the color of light is determined by its wavelength
 or frequency
• white light is a mixture of all the colors of visible light
 a spectrum
 RedOrangeYellowGreenBlueViolet
• when an object absorbs some of the wavelengths of
white light while reflecting others, it appears colored
 the observed color is predominantly the colors reflected
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Amplitude & Wavelength
15
Electromagnetic Spectrum
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Continuous Spectrum
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The Electromagnetic Spectrum
• visible light comprises only a small fraction of
all the wavelengths of light – called the
electromagnetic spectrum
• short wavelength (high frequency) light has high
energy
radiowave light has the lowest energy
gamma ray light has the highest energy
• high energy electromagnetic radiation can
potentially damage biological molecules
ionizing radiation
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Thermal Imaging using Infrared Light
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Using High Energy Radiation
to Kill Cancer Cells
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Interference
• the interaction between waves is called
interference
• when waves interact so that they add to make a
larger wave it is called constructive interference
waves are in-phase
• when waves interact so they cancel each other it is
called destructive interference
waves are out-of-phase
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Interference
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Diffraction
• when traveling waves encounter an obstacle or opening
in a barrier that is about the same size as the
wavelength, they bend around it – this is called
diffraction
 traveling particles do not diffract
• the diffraction of light through two slits separated by a
•
distance comparable to the wavelength results in an
interference pattern of the diffracted waves
an interference pattern is a characteristic of all light
waves
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Diffraction
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2-Slit Interference
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The Photoelectric Effect
• it was observed that many metals emit electrons when a
light shines on their surface
 this is called the Photoelectric Effect
• classic wave theory attributed this effect to the light
•
energy being transferred to the electron
according to this theory, if the wavelength of light is
made shorter, or the light waves intensity made
brighter, more electrons should be ejected
 remember: the energy of a wave is directly proportional to its
amplitude and its frequency
 if a dim light was used there would be a lag time before
electrons were emitted
 to give the electrons time to absorb enough energy
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The Photoelectric Effect
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The Photoelectric Effect
The Problem
• in experiments with the photoelectric effect, it
was observed that there was a maximum
wavelength for electrons to be emitted
called the threshold frequency
regardless of the intensity
• it was also observed that high frequency light
with a dim source caused electron emission
without any lag time
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Einstein’s Explanation
• Einstein proposed that the light energy was
delivered to the atoms in packets, called quanta
or photons
• the energy of a photon of light was directly
proportional to its frequency
inversely proportional to it wavelength
the proportionality constant is called Planck’s
Constant, (h) and has the value 6.626 x 10-34 J∙s
E  hn 
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hc
l
29
Example 7.2- Calculate the number of photons in a laser
pulse with wavelength 337 nm and total energy 3.83 mJ
Given: l = 337 nm, Epulse = 3.83 mJ
Find: number of photons
Concept Plan: lnm)
E
l (m)
9
hc photon
10 m
1 nm
E photon 
l
E pulse
number
photons
E photon
Relationships: E=hc/l, 1 nm = 10-9 m, 1 mJ = 10-3 J, Epulse/Ephoton = # photons
Solve:
9
3.37 10 2 nm 
E photon 
hc
l

10 m
 3.37 10 7 m
1 nm
6.626 1034 J  s 3.00 108 m  s -1


3.37 10
3
10 J
3.83 mJ 
 3.83 10 3 J
1 mJ
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7
m
  5.8985 10
19
J
number of photons 
3.83  10 3 J
5.8985 10 3 J
 6.49  1015 photons
30
Practice – What is the frequency of radiation
required to supply 1.0 x 102 J of energy from
8.5 x 1027 photons?
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What is the frequency of radiation required to supply
1.0 x 102 J of energy from 8.5 x 1027 photons?
Given: Etotal = 1.0 x 102 J, number of photons = 8.5 x 1027
Find: n
Concept Plan: number
E
n (s-1)
photon
photons E total
E photon
n
number of photons
h
Relationships: E=hn, Etotal = Ephoton∙# photons
Solve:
2
E photon 
n
E photon
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h

1.0 10 J
8.5 1027
 1.176 1026 J
1.176 10   1.8 10
26
6.626 10
34
J
7 -1
s
Js
32
Ejected Electrons
• 1 photon at the threshold frequency has just
enough energy for an electron to escape the atom
binding energy, f
• for higher frequencies, the electron absorbs more
energy than is necessary to escape
• this excess energy becomes kinetic energy of the
ejected electron
Kinetic Energy = Ephoton – Ebinding
KE = hn - f
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Spectra
• when atoms or molecules absorb energy, that energy is
often released as light energy
 fireworks, neon lights, etc.
• when that light is passed through a prism, a pattern is
seen that is unique to that type of atom or molecule –
the pattern is called an emission spectrum
 non-continuous
 can be used to identify the material
 flame tests
• Rydberg analyzed the spectrum of hydrogen and found
that it could be described with an equation that
involved an inverse square of integers
 1
1 
 1.097 10 m
 2
2


l
n
n
2 
 1
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1
7
-1 
34
Emission Spectra
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Exciting Gas Atoms to Emit Light
with Electrical Energy
Hg
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He
H
36
Examples of Spectra
Oxygen spectrum
Neon spectrum
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Identifying Elements with
Flame Tests
Na
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K
Li
Ba
38
Emission vs. Absorption Spectra
Spectra of Mercury
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Bohr’s Model
• Neils Bohr proposed that the electrons could only have
very specific amounts of energy
 fixed amounts = quantized
• the electrons traveled in orbits that were a fixed
distance from the nucleus
 stationary states
 therefore the energy of the electron was proportional the
distance the orbital was from the nucleus
• electrons emitted radiation when they “jumped” from
an orbit with higher energy down to an orbit with lower
energy
 the distance between the orbits determined the energy of the
photon of light produced
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Bohr Model of H Atoms
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Wave Behavior of Electrons
• de Broglie proposed that particles could have wave-like
•
•
character
because it is so small, the wave character of electrons is
significant
electron beams shot at slits show an interference
pattern
 the electron interferes with its own wave
• de Broglie predicted that the wavelength of a particle
was inversely proportional to its momentum
l m 
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h
 
kg m 2
s2
mass (kg)  velocity (m  s -1 )
42
Electron Diffraction
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however, electrons actually
if electrons behave like
present an interference
particles, there should
pattern, demonstrating the
only be two bright spots
behave like waves
on the target
43
Example 7.3- Calculate the wavelength of an electron
traveling at 2.65 x 106 m/s
Given: v = 2.65 x 106 m/s, m = 9.11 x 10-31 kg (back leaf)
Find: l, m
Concept Plan:
m, v
l (m)
l
h
mv
Relationships: l=h/mv
Solve:
2
k
g

m


34
6
.
626

10


2
s
h


l

mv

-31
6 m
9.1110 kg  2.65 10 
s 



 2.74 10 10 m
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Practice - Determine the wavelength of a neutron
traveling at 1.00 x 102 m/s
(Massneutron = 1.675 x 10-24 g)
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Practice - Determine the wavelength of a neutron
traveling at 1.00 x 102 m/s
Given: v = 1.00 x 102 m/s, m = 1.675 x 10-24 g
Find: l, m
Concept Plan:
m(g) 1 kg m (kg), v
h l (m)
l
103 g
mv
Relationships: l=h/mv, 1 kg = 103 g
Solve:
2
k
g

m


34
6
.
626

10


2
1 kg
s
h


1.675 10 24 g  3
l

10 g
mv

-27
1.675 10 kg 1.00 10 2

 1.675 10 27 kg
 3.96 10 9 m

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
m
s


46
Complimentary Properties
• when you try to observe the wave nature of the
electron, you cannot observe its particle nature –
and visa versa
wave nature = interference pattern
particle nature = position, which slit it is passing
through
• the wave and particle nature of nature of the
electron are complimentary properties
as you know more about one you know less about
the other
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Uncertainty Principle
h
Dx  Dv 
4
1
 
m
• Heisenberg stated that the product of the uncertainties
in both the position and speed of a particle was
inversely proportional to its mass
 x = position, Dx = uncertainty in position
 v = velocity, Dv = uncertainty in velocity
 m = mass
• the means that the more accurately you know the
position of a small particle, like an electron, the less
you know about its speed
 and visa-versa
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Uncertainty Principle Demonstration
any experiment
designed to observe the
electron results in
detection of a single
electron particle and no
interference pattern
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Determinacy vs. Indeterminacy
• according to classical physics, particles move in a path
determined by the particle’s velocity, position, and
forces acting on it
 determinacy = definite, predictable future
• because we cannot know both the position and velocity
of an electron, we cannot predict the path it will follow
 indeterminacy = indefinite future, can only predict
probability
• the best we can do is to describe the probability an
electron will be found in a particular region using
statistical functions
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Trajectory vs. Probability
51
Electron Energy
Ηψ  Eψ
• electron energy and position are complimentary
 because KE = ½mv2
• for an electron with a given energy, the best we can
do is describe a region in the atom of high probability
of finding it – called an orbital
 a probability distribution map of a region where the
electron is likely to be found
 distance vs. y2
• many of the properties of atoms are related to the
energies of the electrons
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Wave Function, y
• calculations show that the size, shape and
orientation in space of an orbital are determined
be three integer terms in the wave function
added to quantize the energy of the electron
• these integers are called quantum numbers
principal quantum number, n
angular momentum quantum number, l
magnetic quantum number, ml
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Principal Quantum Number, n
• characterizes the energy of the electron in a particular
orbital
 corresponds to Bohr’s energy level
• n can be any integer  1
• the larger the value of n, the more energy the orbital has
• energies are defined as being negative
 an electron would have E = 0 when it just escapes the atom
• the larger the value of n, the larger the orbital
• as n gets larger, the amount of energy between orbitals
gets smaller
E n  -2.1810
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-18
 1 
J 2  for an electron in H
n 
54
Principal Energy Levels in Hydrogen
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Electron Transitions
• in order to transition to a higher energy state, the
•
electron must gain the correct amount of energy
corresponding to the difference in energy between the
final and initial states
electrons in high energy states are unstable and tend to
lose energy and transition to lower energy states
 energy released as a photon of light
• each line in the emission spectrum corresponds to the
difference in energy between two energy states
56
Predicting the Spectrum of Hydrogen
• the wavelengths of lines in the emission spectrum of
•
•
hydrogen can be predicted by calculating the
difference in energy between any two states
for an electron in energy state n, there are (n – 1)
energy states it can transition to, therefore (n – 1) lines
it can generate
both the Bohr and Quantum Mechanical Models can
predict these lines very accurately

E photonreleased  DE hydrogenelectron   E final  E initial


 1  
 1  
18 

18
    2.18 10 J
 
hn 
   2.18 10 J
2
2
n

 n final  
l

 initial  

 
hc
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Hydrogen Energy Transitions
58
Example 7.7- Calculate the wavelength of light emitted when
the hydrogen electron transitions from n = 6 to n = 5
Given: ni = 6, nf = 5
Find: l, m
Concept Plan: n , n
DEatom
i
f
 1 
E  R H  2 
n 
Ephoton
DEatom = -Ephoton
hc
l
E
l
Relationships: E=hc/l, En = -2.18 x 10-18 J (1/n2)
1 
Solve: DE
18  1
 20


2
.
18

10
J



2
.
6
6
44

10
J
 2
atom
2 
5 6 
Ephoton = -(-2.6644 x 10-20 J)
= 2.6644 x 10-20 J


3.00 10 
hc 6.626 10
l 
 7.46 10
E
2.6644 10 
34
8 m
s
Js
-20
6
m
J
Check: the unit is correct, the wavelength is in the infrared, which
is appropriate because less energy than 4→2 (in the visible)
Practice – Calculate the wavelength of light emitted when
the hydrogen electron transitions from n = 2 to n = 1
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60
Calculate the wavelength of light emitted when the hydrogen
electron transitions from n = 2 to n = 1
Given: ni = 2, nf = 1
Find: l, m
Concept Plan: n , n
DEatom
i
f
 1 
E  R H  2 
n 
Ephoton
DEatom = -Ephoton
hc
l
E
l
Relationships: E=hc/l, En = -2.18 x 10-18 J (1/n2)
1 
Solve:
18  1
DE atom  2.18 10 J 2  2   1.64 10 18 J
1 2 
Ephoton = -(-1.64 x 10-18 J) =
1.64 x 10-18 J


3.00 10 
hc 6.626 10
l 
 1.2110
E
1.64 10 
34
8 m
s
Js
-18
7
m
J
Check: the unit is correct, the wavelength is in the UV, which is
appropriate because more energy than 3→2 (in the visible)
Probability & Radial Distribution
Functions
• y2 is the probability density
 the probability of finding an electron at a particular point in
space
 for s orbital maximum at the nucleus?
 decreases as you move away from the nucleus
• the Radial Distribution function represents the total
probability at a certain distance from the nucleus
 maximum at most probable radius
• nodes in the functions are where the probability drops to 0
62
Probability Density Function
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Radial Distribution Function
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64
The Shapes of Atomic Orbitals
• the l quantum number primarily determines the
shape of the orbital
• l can have integer values from 0 to (n – 1)
• each value of l is called by a particular letter that
designates the shape of the orbital
s orbitals are spherical
p orbitals are like two balloons tied at the knots
d orbitals are mainly like 4 balloons tied at the knot
f orbitals are mainly like 8 balloons tied at the knot
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l = 0, the s orbital
• each principal energy state
has 1 s orbital
• lowest energy orbital in a
principal energy state
• spherical
• number of nodes = (n – 1)
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2s and 3s
2s
n = 2,
l=0
3s
n = 3,
l=0
67
l = 1, p orbitals
• each principal energy state above n = 1 has 3 p orbitals
 ml = -1, 0, +1
• each of the 3 orbitals point along a different axis
 px, py, pz
• 2nd lowest energy orbitals in a principal energy state
• two-lobed
• node at the nucleus, total of n nodes
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p orbitals
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l = 2, d orbitals
• each principal energy state above n = 2 has 5 d orbitals
 ml = -2, -1, 0, +1, +2
• 4 of the 5 orbitals are aligned in a different plane
 the fifth is aligned with the z axis, dz squared
 dxy, dyz, dxz, dx squared – y squared
• 3rd lowest energy orbitals in a principal energy state
• mainly 4-lobed
 one is two-lobed with a toroid
• planar nodes
 higher principal levels also have spherical nodes
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d orbitals
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l = 3, f orbitals
• each principal energy state above n = 3 has 7 d orbitals
 ml = -3, -2, -1, 0, +1, +2, +3
• 4th lowest energy orbitals in a principal energy state
• mainly 8-lobed
 some 2-lobed with a toroid
• planar nodes
 higher principal levels also have spherical nodes
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f orbitals
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Why are Atoms Spherical?
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Energy Shells and Subshells
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