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278
c h a p t er 6 The Structure of Atoms
Now apply Planck’s equation to calculate the wavelength (Ephoton = hν = hc/λ, and so λ =
hc/Ephoton). (Recognize that, while the change in energy has a sign indicating the “direction” of
energy transfer, the energy of the photon emitted, Ephoton, does not have a sign.)
λϭ
hc
Ephoton
Jи s
Ϫ34
8
Ϫ1
6 .626 ϫ 10 photon (2 .998 ϫ 10 m и s )
ϭ
4 .086 ϫ 10Ϫ19 J/photon
= 4.862 × 10−7 m
= (4.862 × 10−7 m)(1 × 109 nm/m)
= 486.2 nm
Think about Your Answer You might recall that visible light has wavelengths of 400 to 700 nm.
The calculated value is in this region, and your answer has a value appropriate for the green line.
The experimentally determined value of 486.1 nm is in excellent agreement with this answer.
Check Your Understanding
The Lyman series of spectral lines for the H atom, in the ultraviolet region, arises from transitions
from higher levels to n = 1. Calculate the frequency and wavelength of the least energetic line in
this series.
rEvIEW & cHEcK FOr SEctIOn 6.3
1.
Based on Bohr's theory, which of the following transitions for the hydrogen atom will evolve
the most energy?
(a)
from n = 3 to n = 2
(c)
(b) from n = 4 to n = 2
2.
from n = 5 to n = 2
(d) from n = 6 to n = 2
Based on Bohr's theory, which of the following transitions for the hydrogen atom will occur
with emission of visible light?
(a)
from n = 3 to n = 1
(c)
(b) from n = 4 to n = 2
from n = 5 to n = 3
(d) from n = 6 to n = 4
© R. K. Bohn, Department of Chemistry,
University of Connecticut
6.4 Particle–WaveDuality:PreludetoQuantum
Mechanics
The photoelectric effect demonstrated that light, usually considered to be a wave,
can also have the properties of particles, albeit without mass. But what if matter,
which is usually considered to be made of particles, could have wave properties?
This was pondered by Louis Victor de Broglie (1892–1987), who, in 1925, proposed
that a free electron of mass m moving with a velocity v should have an associated
wavelength λ, calculated by the equation
λϭ
h
mv
(6.6)
The wave nature of electrons.
A beam of electrons was passed
through a thin film of MgO. The
atoms in the MgO lattice diffracted
the electron beam, producing this
pattern. Diffraction is best explained
by assuming electrons have wave
properties.
De Broglie called the wave corresponding to the wavelength calculated from this
equation a “matter wave.”
This revolutionary idea linked the particle properties of the electron (mass and
velocity) with a wave property (wavelength). Experimental proof was soon produced. In 1927, C. J. Davisson (1881–1958) and L. H. Germer (1896–1971) found
that diffraction, a property of waves, was observed when a beam of electrons was
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279
6.4 Particle–Wave Duality: Prelude to Quantum Mechanics
CASE STUDY
What Makes the Colors in Fireworks?
Typical fireworks have several important chemical components. For example, there must be an oxidizer. Today, this is usually potassium perchlorate (KClO4), potassium chlorate (KClO3),
or potassium nitrate (KNO3). Potassium salts
are used instead of sodium salts because the
latter have two important drawbacks. They
are hygroscopic—they absorb water from the
air—and so do not remain dry on storage.
Also, when heated, sodium salts give off an
intense, yellow light that is so bright it can
mask other colors.
The parts of any fireworks display we
remember best are the vivid colors and brilliant flashes. White light can be produced by
oxidizing magnesium or aluminum metal at
high temperatures. The flashes you see at
rock concerts or similar events, for example,
are typically Mg/KClO4 mixtures.
Yellow light is easiest to produce because
sodium salts give an intense light with a
wavelength of 589 nm. Fireworks mixtures
Quick-burning fuse
Colored paper
fuse end
Twine
Delay fuses
(slow burning)
Cross fuse (fast fuse)
Paper wrapper
Red star composition
(KClO3/SrCO3)
Heavy cardboard
barriers
Blue star composition
(KClO4/CuCO3)
Side fuse (fast fuse)
“Flash and sound”
mixture (KClO4/S/Al)
Black powder propellant
Steel mortar buried
in ground
© Cengage Learning/Charles D. Winters
FiguREB The design of an aerial rocket for a fi reworks display. When the fuse is ignited, it burns quickly
to the delay fuses at the top of the red star mixture as well as to the black powder propellant at the bottom.
The propellant ignites, sending the shell into the air. Meanwhile, the delay fuses burn. If the timing is correct,
the shell bursts high in the sky into a red star. This is followed by a blue burst and then a flash and sound.
FiguREA Emission of light by excited
atoms. Flame tests are often used to identify
elements in a chemical sample. Shown here are the
colors produced in a flame (burning methanol) by
NaCl (yellow), SrCl2 (red), and boric acid (green).
usually contain sodium in the form of nonhygroscopic compounds such as cryolite,
Na3AlF6. Strontium salts are most often
used to produce a red light, and green is
produced by barium salts such as Ba(NO3)2.
The next time you see a fireworks display, watch for the ones that are blue. Blue
has always been the most difficult color to
produce. Recently, however, fireworks
designers have learned that the best way to
get a really good “blue” is to decompose
copper(I) chloride at low temperatures. To
achieve this effect, CuCl is mixed with
KClO4, copper powder, and the organic chlorine-containing compound hexachloroethane, C2Cl6.
Questions:
1. The main lines in the emission spectrum
of sodium are at wavelengths (nm) of
313.5, 589, 590, 818, and 819. Which one
or ones are most responsible for the
characteristic yellow color of excited
sodium atoms?
2. Does the main emission line for SrCl2
(in Figure A) have a longer or shorter
wavelength than that of the yellow line
from NaCl?
3. Mg is oxidized by KClO4 to make white
flashes. One product of the reaction is KCl.
Write a balanced equation for the reaction.
Answers to these questions are available in
Appendix N.
directed at a thin sheet of metal foil. Furthermore, assuming the electron beam to
be a matter wave, de Broglie’s relation was followed quantitatively. The experiment
was taken as evidence that electrons can be described as having wave properties in
certain situations.
De Broglie’s equation suggests that, for the wavelength of a matter wave to be
measurable, the product of m and v must be very small because h is so small. A 114-g
baseball, traveling at 110 mph, for example, has a large mv product (5.6 kg ∙ m/s)
and therefore an incredibly small wavelength, 1.2 × 10−34 m! Such a small value
cannot be measured with any instrument now available, nor is such a value meaningful. As a consequence, wave properties are never assigned to a baseball or any
other massive object. It is possible to observe wave-like properties only for particles
of extremely small mass, such as protons, electrons, and neutrons.
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280
c h a p t er 6 The Structure of Atoms
Cathode ray tubes, such as were found in television sets before the advent of
LCD and plasma TVs, generate a beam of electrons. When the electrons impact
the screen, the beam gives rise to tiny flashes of colored light. In contrast to this
effect, best explained by assuming electrons are particles, diffraction experiments suggest that electrons are waves. But, how can an electron be both a particle and a wave? In part, we are facing limitations in language; the words particle
and wave accurately describe things encountered on a macroscopic scale. However, they apply less well on the submicroscopic scale associated with subatomic
particles. In some experiments, electrons behave like particles. In other experiments, we find that they behave like waves. No single experiment can be done to
show the electron behaving simultaneously as a wave and a particle. Scientists now
accept this wave–particle duality—that is, the idea that the electron has the properties of both a wave and a particle. This concept is central to an understanding
of the modern model of the atom that we take up in the remainder of the
chapter.
EXAMPLE 6.4
Using de Broglie’s Equation
Problem Calculate the wavelength associated with an electron of mass m = 9.109 × 10−28 g
that travels at 40.0% of the speed of light.
What Do You Know? The equation proposed by de Broglie, λ = h/mv, relates wavelength to the
mass and velocity of a moving particle. Here you are given the mass of the electron and its velocity;
you will need Planck’s constant, h = 6.626 × 10−34 J ∙ s. Note also that 1 J = 1 kg ∙ m2/s2 (◀ page 29).
Finally, you will need to manipulate units of m and h to be consistent in the solution.
Strategy
•
Express the electron mass in kg and calculate the electron velocity in m ∙ s−1.
•
Substitute values of m (in kg), v (in m/s), and h in the de Broglie equation and solve for λ.
Solution
Electron mass = 9.109 × 10−31 kg
Electron speed (40.0% of light speed) = (0.400)(2.998 × 108 m ∙ s−1) = 1.20 × 108 m ∙ s−1
Substituting these values into de Broglie’s equation, we have
λ ϭ
h
6.626 ϫ 10Ϫ34 (kg и m2 /s2 )(s)
ϭ
ϭ 6.07 × 10−12 m
mv
(9.109 ϫ 10Ϫ31 kg)(1.20 ϫ 10 8 m/s)
In nanometers, the wavelength is
λ = (6.07 × 10−12 m)(1.00 × 109 nm/m) = 6.07 × 10−3 nm
Think about Your Answer The calculated wavelength is about 1/12 of the diameter of the H
atom. Also notice the care taken to monitor units in this problem.
Check Your Understanding
Calculate the wavelength associated with a neutron having a mass of 1.675 × 10−24 g and a
kinetic energy of 6.21 × 10−21 J. (Recall that the kinetic energy of a moving particle is E = 1⁄2mv2.)
rEvIEW & cHEcK FOr SEctIOn 6.4
The wavelength associated with an electron moving at 40% of the speed of light is 6.07 × 10−3
nm (see Example 6.4). How will the wavelength be changed if the velocity increased to 80% of
the speed of light?
kotz_48288_06_0266-0299.indd 280
(a)
Wavelength will be longer.
(b)
Wavelength will be shorter.
(c)
Wavelength will not change.
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6.5 The Modern View of Electronic Structure: Wave or Quantum Mechanics
281
6.5 T
he Modern View of Electronic Structure:
Wave or Quantum Mechanics
How does wave–particle duality affect our model of the arrangement of electrons
in atoms? Following World War I, German scientists Erwin Schrödinger (1887–
1961), Max Born (1882–1970), and Werner Heisenberg (1901–1976) provided
the answer.
Erwin Schrödinger received the Nobel Prize in physics in 1933 for a comprehensive theory of the behavior of electrons in atoms. Starting with de Broglie’s hypothesis that an electron could be described as a matter wave, Schrödinger developed a model for electrons in atoms that has come to be called quantum mechanics
or wave mechanics. Unlike Bohr’s model, Schrödinger’s model can be difficult to
visualize, and the mathematical approach is complex. Nonetheless, the consequences of the model are important, and understanding its implications is essential
to understanding the modern view of the atom.
Let us start by thinking of the behavior of an electron in the atom as a standing
wave. If you tie down a string at both ends, as you would the string of a guitar, and
then pluck it, the string vibrates as a standing wave (Figure 6.11). There are only
certain vibrations allowed for the standing wave formed by a plucked guitar string.
That is, the vibrations are quantized. Similarly, as Schrödinger showed, only certain
matter waves are possible for an electron in an atom.
Following the de Broglie concept, we can adopt the quantum-mechanical view
that an electron in an atom behaves as a wave. To describe this wave, physicists introduced the concept of a wavefunction, which is designated by the Greek letter ψ
(psi). Schrödinger built on the idea that the electron in an atom has the characteristics of a standing wave and wrote an equation defining the energy of an electron
in terms of wavefunctions. When this equation was solved for energy—a monumental task in itself—he found the following important outcomes:
•
•
• Wave Functions and Energy
In Bohr’s theory, the electron energy for
the H atom is given by En = −Rhc/n2.
Schrödinger’s electron wave model
gives the same result.
Only certain wavefunctions are found to be acceptable, and each is associated
with an allowed energy value. That is, the energy of the electron in the atom is
quantized.
The solutions to Schrödinger’s equation for an electron in three-dimensional space
depend on three integers, n, ℓ, and mℓ, which are called quantum numbers. Only
certain combinations of their values are possible, as we shall outline below.
The next step in understanding the quantum mechanical view is to explore the
physical significance of the wavefunction, ψ (psi). Here we owe much to Max Born’s
interpretation. He said that
(a) the value of the wavefunction ψ at a given point in space is the amplitude
(height) of the electron matter wave. This value has both a magnitude and
a sign that can be either positive or negative. (Visualize a vibrating string
Figure 6.11 Standing
waves. A two-dimensional standing
1/
2
1
Node
3/
2
Node
kotz_48288_06_0266-0299.indd 281
Node
wave such as a vibrating string must
have two or more points of zero
amplitude (called nodes), and only
certain vibrations are possible. These
allowed vibrations have wavelengths
of n(λ/2), where n is an integer
(n = 1, 2, 3, . . .). In the first vibration
illustrated here, the distance between
the ends of the string is half a wavelength, or λ/2. In the second, the
string’s length equals one complete
wavelength, or 2(λ/2). In the third
vibration, the string’s length is 3(λ/2).
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