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Lecture Presentation
Chapter 7
The QuantumMechanical
Model of the Atom
Catherine MacGowan
Armstrong Atlantic State University
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The Dual Nature of Matter and Light
Light and matter are both single
entities, and the apparent duality
arises in the limitations of our
language. (Heisenberg)
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The Dual Nature of Matter and Light
•
Matter has both particulate behaviors and characteristics (has mass
and occupies space) and energy-like (light) behaviors and
characteristics.
•
Most subatomic particles, such as an electrons, behave as
PARTICLES and obey the physics of light waves.
– This behavior is more prevalent due to “size” of the particle.
•
All matter possess this dual personality.
•
The concept that describes the dual nature of matter (i.e., particulate
and energy behaviors) is known as wave duality.
•
The theory that describes an electron’s behavior is quantum
mechanics.
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Electromagnetic Radiation:
Wavelength and Frequency Relationship
Wavelength–frequency
relationship:
– Long wavelength (l)
 low frequency (n)
– Short wavelength (l)
 high frequency (n)
• Wavelength (l) and
frequency (n) have an
INVERSE relationship.
n = 1/l
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Electromagnetic Radiation and
Nature of Light
• Light (electromagnetic radiation) waves have a
frequency, n.
– Use the Greek letter “nu,” n, for frequency.
n units are cycles per second (1/s or s-1).
• The distance between light waves is referred to as
wavelength.
– Wavelength is given by the Greek letter “lambda,”
l.
l units are given in meters (m).
– Formulas:
l•n=c
or
n = c/l
where c = velocity of light, which is
3.00 x 108 m/s
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Problem
Red light has wavelength (l) = 700 nm.
What would be its frequency (n)?
Solution:
1. Change wavelength to meters from nanometers.
700 nm x (1 m/1 x 109 nm) = 7.00 x 10-7 m
2. Use the relationship of n = c/l.
n = (3.00 x 108 m/s)/(7.00 x 10-7 m)
n = 4.29 x 1014/s or 4.29 x 1014 s-1
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Electromagnetic Spectrum
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The Electromagnetic Spectrum
• Chart that illustrates the range of electromagnetic radiation
• Classification of electromagnetic radiation is based on
wavelength.
– Short-wavelength (high-frequency) light has high energy.
• Radio wave light has the lowest energy.
• Gamma ray light has the highest energy.
– High-energy electromagnetic radiation can
potentially damage biological molecules.
» Ionizing radiation
• Visible light comprises only a small fraction of all the
wavelengths of the electromagnetic spectrum.
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What Is Color?
•
All electromagnetic radiation has “color.”
– However, not all electromagnetic radiation
“color” is VISIBLE to humans.
•
The color of light can be determined by either its
wavelength or frequency.
•
Visible spectrum
– White light is a mixture of all the colors of
visible light.
– 700 to 400 nm
• Red Orange Yellow Green Blue Violet
– Color happens when an object absorbs some
of the wavelengths of white light while
reflecting others.
– The observed color is predominantly the
colors reflected.
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Color and Intensity
Color (electromagnetic radiation)
is determined by the
wavelength’s distance.
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Intensity (brightness) is dictated by
the wavelength’s amplitude.
Properties of Light Waves
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The Characteristics of Light
Wavelength: distance from peak to peak
Visible light
Amplitude: height of wave
Node: point where wave crosses zero
wavelength
Ultraviolet radiation
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Wave Behavior: Interference
Constructive Interference
• In phase: align overlap
• Reinforces amplitude
Destructive Interference
•
•
Out of phase
Cancels out wave
An interference pattern is a characteristic of all light waves.
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Wave Behavior: Diffraction
• Diffraction is the ending of light wave.
– Occurs when a light wave encounters an obstacle or travels
through a slit similar in size to its wavelength.
• NOTE: Particles do not diffract.
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Two-Slit Interference
Light diffracted
through two
slits separated
by a distance
comparable to
the wavelength
results in an
interference
pattern of the
diffracted
waves.
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The Quantization of Energy
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Nature of Light and the Photoelectric Effect
The behavior of light
• It is analogous to an ocean wave.
– It has a wavelength (l), an amplitude, and an associated
frequency (n).
• It can be explained by classical electromagnetic wave theory.
• A light wave’s energy is directly proportional to its amplitude
and wavelength.
– The shorter the wavelength, the more intense the light
wave, so more electrons can be emitted.
– Classic electromagnetic wave theory attributed this
effect to light’s energy being transferred to the electron.
The photoelectric effect
• Shining light (radiation energy) on a metallic surface can cause
electron(s) to be emitted from the surface.
• The photoelectric effect led to the understanding of the particulate
nature of energy.
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Photoelectric Effect
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Photoelectric Effect and Ejected Electrons
• One photon at the threshold frequency gives the electron just
enough energy for it to escape the atom.
– This energy is called the binding energy, f.
Ebinding = f
• When the electron is irradiated with a photon having a shorter
wavelength, the electron absorbs more energy than is necessary
to escape.
Ephoton = hn
• The excess energy becomes the kinetic energy of the ejected
electron.
Kinetic energy (KE) = Ephoton – Ebinding
• Therefore, KE = hn − f
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The Photoelectric Effect and
the Quantization of Energy
• The discovery of the photoelectric effect led to the
understanding of the particulate nature of matter,
specifically, the electron.
– 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, but inversely proportional to its wavelength.
E = (hn), where
n = c/l; then
E = hc/l
– The proportionality constant is called Planck’s constant
(h) and has the value 6.626 x 10-34 J∙s.
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Problem Using E = hn
The laser light used to read a CD has a
wavelength of 785 nm.
Determine the energy and frequency at 785 nm.
Solution:
785 nm x (1 m/1 x 109 nm) = 7.85 x 10-7 m
n = c/l
n = 3.0 x 108 m/s/7.85 x 10-7 m = 3.82 x 1014/s
E = hn
E = 6.63 x 10-34 J s x 3.82 x 1014/s
E = 2.53 x 10-19 J
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Quantization of Energy Summary
• An object gains or loses radiant energy (light) by either
absorbing or emitting radiant energy (light).
• Radiant energy is absorbed or emitted in discrete “packets”
called QUANTA (hn).
• Radiant energy is proportional or DIRECTLY related to its
frequency (n).
E (energy) = (hn)
where h is Planck’s constant, having a value of 6.6262 x 10-34 J·s
• Radiant energy is INVERSELY related to its wavelength (l).
• So:
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Light with large l (small n) has a small energy and
light with a short l (large n) has a large energy.
Duality of Matter
Question:
Why can’t we observe the energy (light)
characteristics of everyday objects?
Answer:
It has to do with the mass of the object.
Explanation:
From de Broglie’s investigation of the relationship of mass to
wavelength.
He proposed that all moving objects have wave properties.
From Einstein: E = mc2
where m is for mass and c is the speed of light
From Planck:
Therefore,
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E = hn
E = hc/l
mc = h/l
(mass)(velocity) = h/l
Wave Behavior of Electrons
•
de Broglie’s experiments led to the idea that particles could have a wavelike
character.
l = h /(mass x velocity)
where the units are l (meters); h (kg m2/s2); m (kg); and v (m/s)
•
He predicted that the wavelength of a particle is inversely proportional to its
momentum.
momentum = mass x velocity
l = h/(momentum)
•
Because electrons are incredibly small pieces of matter, their wave
character/behavior is significant.
» Electron mass: 9.28 x 10-28 kg
•
Proof of the wave nature of the electron came a few years later with the
demonstration that a beam of electrons would produce an interference pattern as
if the electrons were waves.
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• If electrons behaved only like
particles, there should be only
two bright spots on the target.
• However, electrons actually
present an interference pattern,
therefore demonstrating that
they behave like waves.
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A Problem Illustrating Duality of Matter
Why can we not see the wave properties of a large (in mass) object?
Example:
Baseball
m = 115 g with v = 100 mph
l = 1.3 x 10-25 nm A VERY SHORT WAVELENGTH
Example:
Electron
m = 9.28 x 10-31 kg with v = 1.9 x 108 cm/s
l = 0.388 nm A VERY LONG WAVELENGTH
NOTE: The experimental work of Einstein, Planck, and de Broglie led
scientists to propose that
Matter and energy are one and the same entity –
dual nature of matter
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Problem
Calculate the energy associated with 1.00 mol of photons of green light (555 nm).
Solution:
1. Determine frequency (n).
n = c/l
l = 555 nm x (1 m/1 x 109 nm) = 5.55 x 10-7 m
n = 3.00 x 108 m/s/5.55 x 10-7 m
n = 5.40 x 1014 /s
2. Calculate energy.
E = h·n
E = (6.63 x 10-34 J·s)(5.40 x 1014 s-1)
E = 3.58 x 10-19 J per photon
3. Determine the energy per mol.
(3.58 x 10-19 J/photon)(6.02 x 1023 photons/mol)
= 2.16 x 105 J/mol or 216 kJ/mol
This is in the range of energies that can break bonds.
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Atomic Spectroscopy
and the Modern Atom Model
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Spectra and Radiant Energy
•
Atoms or molecules can absorb energy.
– The energy can be released as light energy.
•
If this emitted energy is passed through a prism, a pattern of
particular wavelengths of light is observed.
– This pattern of line spectra is unique to the material absorbing
the radiant light.
•
This pattern observed is called an emission spectrum.
– The pattern is not a continuous spectrum as in a rainbow but is
line specific.
•
Line spectra can be used to identify the chemical substances.
– Each chemical spectrum has a unique line spectrum.
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Line Spectra
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Line Spectra of Hydrogen
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Quantum Leaps
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The Evolution of the
Atom Model
From Rutherford
to the Quantum Atom
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Experimental Data vs. Rutherford’s Model
Rutherford model of the atom:
The electrons of an atom orbit around a dense, positively charged center of
mass called the nucleus.
electron
orbiting
atom nucleus
Problems with Rutherford model:
•
•
Problem #1: Classical physics doesn’t work.
– Electrons are small, negatively charged particles spinning around a
densely packed, positive charged center (nucleus)
• Positive and negative charges attract.
– Attraction should slow down electron, resulting in the electron(s)
collapsing into nucleus and imploding atom.
– Electrons as moving charged particles should give off energy, causing the
atom to glow.
No experimental evidence of this happening
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Experimental Data vs. Rutherford’s Model
•
Problem #2: Line Spectra vs. Continuous Spectrum
– If an electron “excited” according to Rutherford’s
model, a continuous spectrum (rainbow effect)
would be emitted.
– Experimental evidence showed a line spectrum.
• Line spectra: light is emitted in discrete
“packets” and only at specific wavelengths
• Every element has its OWN line emission
spectrum.
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The Bohr Atom: Concentric Circles
•
Bohr proposed that the classical physics (Newtonian) view of matter cannot
adequately explain the behavior of the electron in an atom.
•
Needed a new theoretical approach that would:
– explain the microscopic behaviors and characteristics of small pieces of
matter and
– the relationship between an atom’s electron and its nucleus (proton) as that
was observed experimentally
•
Bohr’s model of the atom assumed that:
– electrons can only exist in certain discrete orbits called stationary
states/energy levels
–
– Electrons are restricted to QUANTIZED energy states or energy levels
around the nucleus.
• These energy levels are designated by the symbol n, where n (quantum
no.) = whole integers 1, 2, 3, 4, ...
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Bohr’s Model:
The Good and the Not So Good
GOOD
– Explained line spectra of elements and compounds
– Acknowledged energy absorption/emission is discrete packets of
energy
E = hn
NOT SO GOOD
– Theory only successful for the element hydrogen
– Bohr introduced quantum idea artificially, but never fully explained
theoretically
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Schrodinger’s Model: The Quantum Atom
Using mathematics and probabilities, Schrodinger applied the idea of
electrons behaving as a wave to the problem of electrons in atoms.
•
He developed the WAVE EQUATION
Hyi (r) = ep y i(r)
•
Its solution gives set of math expressions called WAVE FUNCTIONS, 
 describes an area that the electron can be found in in relationship to
the nucleus.
 does not denote the EXACT location of the electron in an orbital.
2 is proportional to the probability of knowing the location of the
electron at a given place.
•
Solving  results in four values known as eigen values.
– Quantum numbers: n, l, ml, and ms
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Quantum Mechanics and
the Uncertainty Principle
•
What quantum mechanics tells about an atom’s electron(s):
•
It is the wave nature, not the particulate (matter), characteristics of the
electron that explain the chemical and physical properties of matter.
•
The electron no longer is viewed as a small piece of matter orbiting around
the nucleus but as a cloud of probability that is spread out over the orbit
area.
• Heisenberg’s uncertainty principle explains:
•
Why the wave-particle duality nature of the electron makes it difficult to
know its exact position and velocity it is traveling/orbiting around the
nucleus
• If you know the position with exactness, then only the probability of
the electron’s velocity is known, or if the velocity of the electron is
known with certainty, its position is probable.
Dx (position) Dmv (momentum) > h/4p
•
An electron’s energy can be defined with exactness, but its exact position
relative to the nucleus is known with limited certainty.
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Probability Density Function
The probability density function represents the total probability of
finding an electron at a particular point in space.
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Quantum Numbers
What do quantum numbers tell us?
•
Location of electron(s) in relationship to the atom nucleus.
• They are like a home address.
• They specify the area an electron occupies in relationship to
the nucleus.
• Each quantum number describes an allowed energy state of the
electron.
Pauli Exclusion Principle
•
States that no two electrons can have the same set of quantum
numbers
• Every electron surrounding an atom’s nucleus has its own set of
quantum numbers.
• Each orbital can only have TWO electrons with opposite spins.
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What do the four quantum numbers represent?
n:
Principal quantum number; indicates the energy
level the electron occupies.
– Values of n are whole integers; n = 1,2,3,4 …
– Also known as shell
l:
Angle quantum number; indicates the orbital shape
» Values of l are determined by (n - 1)
» Orbitals grouped in s, p, d (and f)
• Also known as subshell
ml:
Orientation quantum number; indicates the area the electron occupies
» Values of ml are determined by +/- value of l
» Explains why there is one type of s, three types of p, five types of d,
and seven types of f orbitals
ms:
Magnetic quantum number; indicates the spin of the electron in the
orbital
» Electron can spin with magnetic field ( ) or against a magnetic field ( ).
» Only has values of + ½ or – ½
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Quantum Numbers
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Electron Spin (ms) and Magnetism
•Diamagnetic
• NOT attracted to a magnetic field
• Substances have paired spins.
•Paramagnetic
• Substance is attracted to a magnetic field.
• Substances with unpaired electrons are
paramagnetic.
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Predicting Quantum Numbers
Which of the following sets of quantum numbers is not
allowed?
(4, 2, -1, 1/2)
(8, 4, -2, -1/2)
(2, 1, 2, 1/2)
(3, 0, 0, - 1/2)
Answer: (2, 1, 2, 1/2)
if n = 2, you can have l as 1 or 0
l=n-1
if l = 1, then ml can only be -1, 0, or 1
ml = +/- l
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Predicting Quantum Numbers
Determine a set of quantum numbers for an
electron in a 3p orbital.
n = 3 for an electron in the third level
l = 1 for a p orbital
ml = (+/-) l so can have -1, 0, 1
ms = can be either + 1/2 or - 1/2
Possible answers:
(3, 1, -1, 1/2)
(3, 1, 0, 1/2)
(3, 1, 1, 1/2)
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(3, 1, -1, -1/2)
(3, 1, 0, -1/2)
(3, 1, 1, -1/2)
The Angular Momentum Quantum Number (l)
Value of l (subshell) Letter Designation
l=1
s
p
l=2
d
l=3
f
l=0
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When l = 0, s orbital
• Every energy level (n) has an “s”
orbital.
• s orbitals have the lowest energy
within a principal energy level (n).
• s orbitals have a spherical
probability plot.
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Illustrations of 2s and 3s
2s
n = 2,
l=0
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3s
n = 3,
l=0
The p Orbitals
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The d Orbitals
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