4 Particle–Wave Duality: Prelude to Quantum Mechanics

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



kotz_48288_06_0266-0299.indd 278



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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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