Chemistry: Atoms First
Julia Burdge & Jason Overby
Chapter 3
Homework: 1,3, 5, 7, 15, 25, 27,29,35, 45
49, 51, 55, 57, 59, 121, 123, 125 and127
Quantum Theory and
the Electronic
Structure of Atoms
Kent L. McCorkle
Cosumnes River College
Sacramento, CA
Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
3
Quantum Theory and the Electronic Structure of
Atoms
3.1 Energy and Energy Changes
Forms of Energy
Units of Energy
3.2 The Nature of Light
Properties of Waves
The Electromagnetic Spectrum
The Double-Slit Experiment
3.3 Quantum Theory
Quantization of Energy
Photons and the Photoelectric Effect
3.4 Bohr’s Theory of the Hydrogen Atom
Atomic Line Spectra
The Line Spectrum of Hydrogen
3.5 Wave Properties of Matter
The de Broglie Hypothesis
Diffraction of Electrons
3.6 Quantum Mechanics
The Uncertainty Principle
The Schrödinger Equation
The Quantum Mechanical Description of the Hydrogen Atom
Characteristics of an Atom
1. They posses mass
2. Contains a positive nuclei
3. Contains electrons found
in an electron cloud
4. Occupy Volume
determined by electron cloud
5. Possess Energy
3.1
Energy and Energy Changes
Energy is the capacity to do work or transfer heat.
All forms of energy are either kinetic or potential.
Kinetic energy (Ek) is the energy of motion.
 m is the mass of the object
 u is its velocity
One form of kinetic energy of particular interest to chemists is
thermal energy, which is the energy associated with the random
motion of atoms and molecules.
Forms of Energy
Potential energy is the energy possessed by an object by virtue of its
position.
There are two forms of potential energy of great interest to chemists:
 Chemical energy is energy stored within the structural units of
chemical substances.
 Electrostatic energy is potential energy that results from the
interaction of charged particles.
Q1 and Q2 represent two charges separated by the distance, d.
Energy and Energy Changes
Kinetic and potential energy are interconvertible – one can be
converted to the other.
Although energy can assume many forms, the total energy of the
universe is constant.
 Energy can neither be created nor destroyed.
 When energy of one form disappears, the same amount of
energy reappears in another form or forms.
 This is known as the law of conservation of energy.
Units of Energy
The SI unit of energy is the joule (J), named for the English
physicist James Joule.
It is the amount of energy possessed by a 2-kg mass moving at a
speed of 1 m/s.
Ek = ½ mu2 = ½(2 kg)(1 m/s)2 = 1 kg∙m2/s2 = 1 J
The joule can also be defined as the amount of energy exerted
when a force of 1 newton (N) is applied over a distance of 1 meter.
1J=1N·m
Because the magnitude of a joule is so small, we often express
large amounts of energy using the unit kilojoule (kJ).
1 kJ = 1000 J
Worked Example 3.1
Calculate the kinetic energy of a helium atom moving at a speed of 125 m/s.
Strategy Use Ek = ½mu2 to calculate the kinetic energy of an atom. Note that
for units to cancel properly, giving Ek in joules, the mass of the helium atom
(4.003 amu) must be in kilograms. The factor for conversion of amu to g is
1.661×10-24 g/1 amu. Therefore, the mass of a helium atom in kilograms is:
1 kg
1.661×10-24 g
4.003 amu ×
×
= 6.649×10-27 kg
3
1 amu
1×10 g
Solution Ek = ½mu2
= ½(6.49×10-27 kg)(125 m/s)
= 5.19×10-23 kg∙m2/s2 = 5.19×10-23 J
Think About It We expect the energy of a single atom, even a fast-moving
one, to be extremely small.
Worked Example 3.2
How much greater is the attraction between charges of +2 and -2 than attraction
between charges of +1 and -1 if the opposite charges are separated by the same
distance.
QQ
Strategy Use Eel α 1d 2 to compare the magnitudes of the two Eel values.
Because the distance between charges in the same in both cases, we can solve for
the ratio of Eel values without actually knowing the distance. Both the distance
and the proportionality constant cancel in the solution.
Think About It Doubling both charges causes a four-fold
in the magnitude of the electrostatic energy
Setup increase
Eel(+2,-2) = c Q1Q2 , Q1 = +2 and Q2 = -2
d particles.
between charged
Eel(+1,-1) = c Q1Q2 , Q1 = +1 and Q2 = -1
d
Solution
2×(-2)
d
1×(-1)
c× d
c×
=4
The attraction between charges of +2 and -2 is four times as large as the
attraction between charges +1 and -1.
Properties of Waves
All forms of electromagnetic
radiation travel in waves.
Waves are characterized by:
Wavelength (λ; lambda) – the
distance between identical
points on successive waves
Frequency (ν; nu) – the number
of waves that pass through a
particular point in 1 second.
Amplitude – the vertical distance
from the midline of a wave to the
top of the peak or the bottom of
the trough.
The Nature of Light
The speed of light (c) through a vacuum is a constant:
c = 2.99792458×108 m/s
Normally rounded to, c = 3.00×108 m/s.
Speed of light, frequency and wavelength are related:
 λ is expressed in meters
 v is expressed in reciprocal seconds (s−1)
 s-1 is also known as hertz (Hz)
The Electromagnetic Spectrum
An electromagnetic wave has both an electric field component and
a magnetic component. Maxwell first proposed this in 1873
The electric and magnetic components have the same frequency and
wavelength.
3.2 The Nature of Light
Visible light is only a small component of the continuum of radiant energy
known as the electromagnetic spectrum.
The Double-Slit Experiment
When light passes through two closely spaced slits, an interference
pattern is produced.
Constructive
interference is a result
of adding waves that
are in phase.
Destructive
interference is a result
of adding waves that
are out of phase.
This type of interference is typical of waves and demonstrates the
wave nature of light.
Worked Example 3.3
A laser commonly used in the treatment of vascular skin lesions has a wavelength
of 532 nm. What is the frequency of this radiation?
Strategy We must convert the wavelength to meters and solve for frequency
using c = λν.
c
Setup Rearranging the equation to solve for frequency gives ν = λ . The speed
of light, c, is 3.00×108 m/s.
Solution
1×10-9 m
λ = 532 nm×
1 nm
= 5.32×10-7 m
3.00×108 m/s
ν=
= 5.64×1014 s-1
-7
5.32×10 m
Think About It Make sure your units cancel properly. A common error in
this type of problem is neglecting to convert wavelength to nanometers.
A photon has a frequency of 6.0 x 104 Hz. Convert
this frequency into wavelength (nm). Does this frequency
fall in the visible region?
l
ln=c
l = c/n
n
l = 3.00 x 108 m/s / 6.0 x 104 Hz
l = 5.0 x 103 m
l = 5.0 x 1012 nm
Radio wave
Mystery #1, “Black Body Problem”
Solved by Planck in 1900
Energy (light) is emitted or
absorbed in discrete units
(quantum).
E=hxn
Planck’s constant (h)
h = 6.63 x 10-34 J•s
Quantum Theory
The energy E of a single quantum of energy is
 h is called Planck’s constant: 6.63×10−34 J∙s
The idea that energy is quantized rather than continuous is like
walking up a staircase or playing the piano
 You cannot step or play anywhere (continuous), you can only
step on a stair or play on a key (quantized).
When copper is bombarded with high-energy electrons,
X rays are emitted. Calculate the energy (in joules)
associated with the photons if the wavelength of the
X rays is 0.154 nm.
E=hxn
E=hxc/l
E = 6.63 x 10-34 (J•s) x 3.00 x 10 8 (m/s) / 0.154 x 10-9 (m)
E = 1.29 x 10 -15 J
3.3
Quantum Theory
When a solid is heated, it emits electromagnetic radiation, known as
blackbody radiation, over a wide range of wavelengths.
The amount of energy given off at a certain temperature depends on
the wavelength.
Classical physics failed to completely explain the phenomenon.
 Assumed that radiant energy was continuous; that is, could be
emitted or absorbed in any amount.
Max Planck suggested that radiant energy is only emitted or
absorbed in discrete quantities, like small packages or bundles.
A quantum of energy is the smallest quantity of energy that can be
emitted (or absorbed).
Photons and the Photoelectric Effect
Einstein proposed that the beam of
light is really a stream of particles.
These particles of light are now called
photons.
Each photon (of the incident light)
must posses the energy given by the
equation:
Photons and the Photoelectric Effect
Shining light onto a metal surface can
be thought of as shooting a beam of
particles – photons – at the metal
atoms.
If the ν of the photons equals the
energy the binds the electrons in the
metal, then the light will have enough
energy to knock the electrons loose.
If we use light of a higher ν, then not
only will the electrons be knocked
loose, but they will also acquire some
kinetic energy.
Mystery #2, “Photoelectric Effect”
Solved by Einstein in 1905 hn
Light
has
Shows
howboth:
the energy of light depends on
its frequency and intensity
1. wave nature
2. particle nature
Photon is a “particle” of light
hn = KE + BE (W)
KE = hn – BE(W)
Reveals how the behavior of
electrons is related
to the characteristics of light
KE e-
Photons and the Photoelectric Effect
This is summarized by the equation
 KE is the kinetic energy of the
ejected electron
 W is the binding energy of the
electron
Photons and the Photoelectric Effect
Albert Einstein used Planck’s theory
to explain the photoelectric effect.
Electrons are ejected from the surface
of a metal exposed to light of a certain
minimum frequency, called the
threshold frequency.
The number of electrons ejected is
proportional to the intensity.
Below the threshold frequency no
electrons were ejected, no matter how
bright (or intense) the light.
1. Below a characteristic threshold frequency (no) no electrons
are observed, regardless of the lights intensity
2. Above the threshold frequency, the maximum kinetic energy
of the ejected electrons increases linearly with the
frequency of light
3. Above the threshold frequency, the number of emitted electrons
increases with the lights intensity, but the Kinetic Energy per
electron does not depend on the lights intensity
4. All metals exhibit the
same pattern but each
metal has a different
threshold frequency
Variation in the maximum kinetic
energy of electrons ejected from
two different metal surfaces
(a and b) by light of various
frequencies.
Einstein provides an elegant explanation
Of the photoelectric effect
He first postulated that light comes in packets (bundles) called
photons and each photon has an energy that is
directly proportional to the frequency
Ephoton = hnphoton
= hC
l
Planck’s constant (h) = 6.626 x 10-34 J·s
Let’s work some problems…………..
Einstein then applied the law of conservation
of energy to the photoelectric effect
When a metal surface absorbs a photon the
energy of the photon is transferred to
an electron Eelectron = Ephoton
Part of this energy is used to overcome the
forces that bind the electron to the metal
and the rest is the kinetic energy of the
ejected electron
The threshold frequency is equal to the
minimum energy needed to overcome
the forces that bind the electron to the
metal
Ekinetic = hn - hno
Let’s work some problems………….
Worked Example 3.4
Calculate the energy (in joules) of (a) a photon with a wavelength of 5.00×104
nm (infrared region) and (b) a photon with a wavelength of 52 nm (ultraviolet
region). (c) Calculate the maximum kinetic energy of an electron ejected by the
photon in part (b) from a metal with a binding energy of 3.7 eV.
Strategy We must use c = λν and E = hν to determine the energy of each
photon. In part (c), we will use hν = KE + W to find the kinetic energy of an
ejected electron. The binding energy, given in eV, must be converted to J.
Solution
1×10-9 m
4
(a) 5.00×10 nm ×
= 5.00×10-5 m
1 nm
c
3.00×108 m/s
ν=
=
= 6.00×1012 s-1
-5
λ
5.00×10 m
E = hν = (6.63×10-34 J∙s)(6.00×10-12 s-1) = 3.98×10-21 J
Worked Example 3.4 (cont.)
Calculate the energy (in joules) of (a) a photon with a wavelength of 5.00×104
nm (infrared region) and (b) a photon with a wavelength of 52 nm (ultraviolet
region). (c) Calculate the maximum kinetic energy of an electron ejected by the
photon in part (b) from a metal with a binding energy of 3.7 eV.
Think About It Remember that frequency and wavelength
Solution
-9 m
are inversely
Thus,
as wavelength decreases,
1×10proportional.
-8
(b) 52 nm ×
= 5.2×10 m
frequency and
1 nmenergy increase. Note that in part (c),
subtracting
the 8binding
energy made
a relatively small
c
3.00×10
m/s
15
-1
ν=
=
= 5.8×10 s
-8 m
change
to the energy
of the incident photon. If the incident
λ
5.2×10
photon had been-34in the X-ray region
of the spectrum,
the
15
-1
-18
E = hν = (6.63×10 J∙s)(5.8×10 s ) = 3.8×10 J
difference between its energy and the kinetic energy of the
ejected electron would have been negligible.
1.602×10-19 J
= 5.9×10-19 J
(c) W = 3.7 eV ×
1 eV
KE = hv - W = 3.8×10-18 J - 5.9×10-19 J = 3.2×10-18 J
3.4
Bohr’s Theory of the Hydrogen Atom
Sunlight is composed of various color
components that can be recombined to
produce white light.
The emission spectrum of a substance can
be seen by energizing a sample of material
with some form of energy.
The “red hot” or “white hot” glow of an
iron bar removed from a fire is the visible
portion of its emission spectrum.
The emission spectrum of both sunlight and
a heated solid are continuous; all
wavelengths of visible light are present.
Atomic Line Spectra
Line spectra are the emission of light only at specific wavelengths.
Absorption Spectrum
Measures the frequencies of the photons that an atom absorbs
Emission Spectrum
Measures the energies of the photons emitted by atoms
Bohr’s Theory of the Hydrogen Atom
Every element has its own unique emission spectrum.
Bohr’s Theory of the Hydrogen Atom
The Rydberg equation can be used to calculate the wavelengths of
the four visible lines in the emission spectrum of hydrogen.
 R∞ is the Rydberg constant (1.09737317 x 107 m−1)
 λ the wavelength of a line in the spectrum
 n1 and n2 are positive integers where n2 > n1.
The Line Spectrum of Hydrogen
Neils Bohr attributed the emission of radiation by an energized
hydrogen atom to the electron dropping from a higher-energy orbit
to a lower one.
As the electron dropped, it gave up a quantum of energy in the form
of light.
Bohr showed that the energies of the electron in a hydrogen atom
are given by the equation:
 En is the energy
 n is a positive integer
The Line Spectrum of Hydrogen
As an electron gets closer to the nucleus, n decreases.
En becomes larger in absolute value (but more negative) as n gets
smaller.
En is most negative when n = 1.
Called the ground state, the lowest energy state of the atom
For hydrogen, this is the most stable state
The stability of the electron decreases as n increases.
Each energy state in which n > 1 is called an excited state.
The Line Spectrum of Hydrogen
Bohr’s theory explains the line spectrum of the hydrogen atom.
Radiant energy absorbed by the atom causes the electron to move
from the ground state (n = 1) to an excited state (n > 1).
Conversely, radiant energy is emitted when the electron moves
from a higher-energy state to a lower-energy excited state or the
ground state.
The quantized movement of the electron from one
energy state to another is analogous to a ball moving
and down steps.
nf is the final state
Bohr’s Theory of the Hydrogen Atom
Suppose an electron is initially in an excited state, ni.
During emission, the electron drops to a lower energy state, nf.
The energy difference between the initial and final states is
Bohr’s Theory of the Hydrogen Atom
To calculate wavelength, substitute c/λ for ν and rearrange:
Light acts both as a wave and a particle
Light had been explained as a wave for years before
Einstein came along but………
Einstein’s explanation of the photoelectric effect was
Revolutionary because it was the first time that light
Was described as a particle
Light and Atoms
When an atom absorbs a photon it
gains the photons energy
When an atom emits energy it
loses the photons energy
Energy level diagrams: represents atomic
energy transformations
Eatom = hnphoton
Ground State: the lowest energy state
of an atom (most stable state)
Excited State: higher energy state
Electrons are restricted to certain energy levels therefore
the energy of the electron is said to be quantized
Bohr’s Model of
the Atom (1913)
1. e- can only have specific
(quantized) energy
values
2. light is emitted as emoves from one energy
level to a lower energy
level
En = -RH (
1
n2
)
n (principal quantum number) = 1,2,3,…
RH (Rydberg constant) = 2.18 x 10-18J
Bohr Theory
E = -b
n2
b = 2.18 x 10-18 J
DE = Ef - Ei
DE = - b
(
1
n2f
-
1
n2i
)
ni = 3
ni = 3
ni = 2
nf = 2
nnf f==11
Ephoton = DE = Ef - Ei
1
Ef = -RH ( 2
nf
1
Ei = -RH ( 2
ni
1
DE = RH( 2
ni
)
)
1
n2f
)
Worked Example 3.5
Calculate the wavelength (in nm) of the photon emitted when an electron
transitions from the n = 4 state to the n = 2 state in a hydrogen atom.
Setup
Think About It Look again at the line spectrum of
h = 6.63×10-34 J∙s and c = 3.00×108 m/s
hydrogen and make sure your result matches one of them.
Note that for an emission, ni, is always greater than nf, and
-18 J
Solution
2.18×10
1
1
the 1equation
gives
a
positive
result.
=
λ
(6.63×10-34 J∙s )(3.00×108 m/s)
22
42
= 2.055×106 m-1
λ = 4.87×10-7 m ×
1 nm
1×10-9 m
= 487 nm
3.5
Wave Properties of Matter
Louis de Broglie reasoned that if light can behave like a stream of
particles (photons), then electrons could exhibit wavelike properties.
According to de
Broglie, electrons
behave like standing
waves.
Only certain
wavelengths are
allowed.
At a node, the
amplitude of the wave
is zero.
The deBroglie Equation
De Broglie (1924) reasoned
that e- is both particle and
wave.
If things that behave like waves like waves (light) have
Particle characteristics then things that behave like
Particles hould also have wave characteristics
electrons have wave properties and particle properties
l = h/mu
u = velocity of em = mass of e- (in Kg)
What is the de Broglie wavelength (in nm)
associated with a 2.5 g Ping-Pong ball
traveling at 15.6 m/s?
l = h/mu
h in J•s m in kg u in (m/s)
l = 6.63 x 10-34 / (2.5 x 10-3 x 15.6)
l = 1.7 x 10-32 m = 1.7 x 10-23 nm
Wave Properties of Matter
De Broglie deduced that the particle and wave properties are
related by the following expression:
 λ is the wavelength associated with the particle
 m is the mass (in kg)
 u is the velocity (in m/s)
The wavelength calculated from this equation is known as the de
Broglie wavelength.
Worked Example 3.6
Calculate the de Broglie wavelength of the “particle” in the following two cases:
(a) a 25-g bullet traveling at 612 m/s and (b) an electron (m = 9.109×10-31 kg)
moving at 63.0 m/s.
Setup Think About It While you are new at solving these
problems, always write out the units of Planck’s constant
(J∙s) as kg∙m2/s. This will enable you to check your unit
-34 J∙s, or 6.63×10
-34 kg∙m
2/s; Remember
h = 6.63×10
m must
be expressed in
cancellations
and detect
common
errors such
as expressing
kg.
mass in grams rather than kilograms. Note that the
calculated wavelength of a macroscopic object, even one
Solution
as small
as a bullet, is extremely small. An object must be
1 kg
= 0.025 kg
(a) 25 g ×
at least
asgsmall as a subatomic particle in order for its
1000
wavelength to be large
enough for us to observe.
6.63×10-34 kg∙m2/s
h
=
λ =
= 4.3×10-35 m
(0.025 kg)(612 m/s)
mu
h
(b) λ =
mu
6.63×10-34 kg∙m2/s
=
= 1.16×10-5 m
-31
(9.109×10 kg)(63.0 m/s)
Diffraction of Electrons
Experiments have shown that electrons do indeed possess wavelike
properties:
X-ray diffraction pattern of
aluminum foil
Electron diffraction pattern of
aluminum foil.
Heisenberg’s Uncertainty Principle
The position and Energy of an electron
cannot be precisely defined.
The more accurately we know the position the
more uncertain we are about energy
3.6
Quantum Mechanics
The Heisenberg uncertainty principle states that it is impossible to
know simultaneously both the momentum p and the position x of a
particle with certainty.
 Δx is the uncertainty in position in meters
 Δp is the uncertainty in momentum
 Δu is the uncertainty in velocity in m/s
 m is the mass in kg
Worked Example 3.7
An electron in a hydrogen atom is known to have a velocity of 5×106 m/s + 1
percent. Using the uncertainty principle, calculate the minimum uncertainty in the
position of the electron and, given that the diameter of the hydrogen atom is less
than 1 angstrom (Å), comment on the magnitude of this uncertainty compared to
the size of the atom.
Strategy The uncertainty in the velocity, 1 percent of 5×106 m/s, is Δu.
Calculate Δx and compare it with the diameter of they hydrogen atom.
Setup The mass of an electron is 9.11×10-31 kg. Planck’s constant, h, is
6.63×10-34 kg∙m2/s.
Worked Example 3.7
An electron in a hydrogen atom is known to have a velocity of 5×106 m/s + 1
percent. Using the uncertainty principle, calculate the minimum uncertainty in the
position of the electron and, given that the diameter of the hydrogen atom is less
than 1 Think
angstrom
(Å), comment
on theerror
magnitude
of this uncertainty
to
About
It A common
is expressing
the masscompared
of
the sizethe
of particle
the atom.in grams instead of kilograms, but you should
discover this inconsistency if you check your unit cancellation
Solution
carefully.
Remember
if =
one
uncertainty
is small, the other
6 m/s
4 m/s
Δu = 0.01
× 5×10that
5×10
must be large. The uncertainty principle applies in a practical
h
way only
Δx = to submicroscopic particles. In the case of a
4πobject,
∙ mΔu where the mass is much larger than that
macroscopic
of an electron, small uncertainties, relative to the size of the
-34 kg∙m2/s
6.63×10
object,
are
possible
for
both
position and velocity.
Δx =
> 1×10-9 m
-31
4
4π(9.11×10 kg)(5×10 m/s)
The minimum uncertainty in the position x is 1×10-9 m = 10Å. The uncertainty
is 10 times larger than the atom!