2 Quantization: Planck, Einstein, Energy, and Photons

6.2 Quantization: Planck, Einstein, Energy, and Photons  







269



Check Your Understanding

(a)



Which color in the visible spectrum has the highest frequency? Which has the lowest frequency?



(b) Is the frequency of the radiation used in a microwave oven (2.45 GHz) higher or lower than

that from your favorite FM radio station (for example, 91.7 MHz).

(c)



Are the wavelengths of x-rays longer or shorter than those of ultraviolet light?



Energy Increases

1024



1022



1020



x-rays



␥-rays

10−16



10−14



1018



10−12



10−10



1016



1014



UV

10−8



1012

IR



10−6



10−4



1010



108



Microwave

10−2



106



FM

AM

Radio waves

100



102



500

Energy increases



102



100



(Hz)



Long radio waves

104



106



108



(m)



FiguRE6.2 The electromagnetic spectrum. Visible light



Visible spectrum



400



104



600

Wavelength increases



700



␭ (nm)



(enlarged portion) is a very small part

of the entire spectrum. The radiation’s energy increases from the

radio-wave end of the spectrum (low

frequency, ν, and long wavelength, λ)

to the γ-ray end (high frequency and

short wavelength).



rEvIEW & cHEcK FOr SEctIOn 6.1

Which of the following types of electromagnetic radiation has the longest wavelength?

(a)



ultraviolet light



(b) x-rays



(c)



red light from a laser pointer



(d) microwaves from a microwave oven



© Cengage Learning/Charles D. Winters



6.2 Quantization:Planck,Einstein,Energy,

andPhotons

Planck’s Equation

If a piece of metal is heated to a high temperature, electromagnetic radiation is

emitted with wavelengths that depend on temperature. At lower temperatures, the

color is a dull red (Figure 6.3). As the temperature increases, the red color brightens, and at even higher temperatures a brilliant white light is emitted.

Your eyes detect the emitted radiation in the visible region of the electromagnetic spectrum. Although not seen, both ultraviolet and infrared radiation are also

given off by the hot metal (Figure 6.3). In addition, it is observed that the wavelength of the most intense radiation is related to temperature: As the temperature

of the metal is raised, the maximum intensity shifts toward shorter wavelengths, that

is, toward the ultraviolet. This corresponds to the change in color observed as the

temperature is raised.

At the end of the 19th century, scientists were not able to explain the relationship

between the intensity and the wavelength for radiation given off by a heated object (often called blackbody radiation). Theories available at the time predicted that the intensity

should increase continuously with decreasing wavelength, instead of reaching a maximum and then declining as is actually observed. This perplexing situation became

known as the ultraviolet catastrophe because predictions failed in the ultraviolet region.



kotz_48288_06_0266-0299.indd 269



Blackbody radiation. In physics a

blackbody is a theoretical concept in

which a body absorbs all radiation

that falls on it. However, it will

emit energy with a temperaturedependent wavelength. The light

emanating from the spaces between

the burning charcoal briquets in this

photo is a close approximation to

blackbody radiation. The color of the

emitted light depends on the temperature of the briquets.



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270



c h a p t er 6   The Structure of Atoms



0



80



00



© Cengage Learning/Charles D. Winters



covering a spectrum of wavelengths. For a given temperature, some of the radiation is emitted at long wavelengths and some at short wavelengths. Most, however,

is emitted at some intermediate wavelength, the

maximum in the curve. As the temperature of the object

increases, the maximum moves from the red end of the

spectrum to the violet end. At still higher temperatures,

intense light is emitted at all wavelengths in the visible

region, and the maximum in the curve is in the ultraviolet region. The object is described as “white hot.”

(Stars are often referred to as “red giants” or “white

dwarfs,” a reference to their temperatures and relative

sizes.)



Intensity of Emitted Light



Figure 6.3   The radiation given off by a heated

body.  When an object is heated, it emits radiation



K



6000 K

4000 K



200



300



400



500



600



700



800



900



1000



Wavelength (nm)



In 1900, a German physicist, Max Planck (1858–1947), offered an explanation

for the ultraviolet catastrophe: The electromagnetic radiation emitted originated in

vibrating atoms (called oscillators) in the heated object. He proposed that each oscillator had a fundamental frequency (ν) of oscillation and that the atoms could only

oscillate at either this frequency or whole-number multiples of it (nν). Because of

this, the emitted radiation could have only certain energies, given by the equation

E ​= ​nhν



where n must be a positive integer. That is, Planck proposed that the energy is quantized. Quantization means that only certain energies are allowed. The proportionality constant h in the equation is now called Planck’s constant, and its experimental

value is 6.6260693 ​× ​10−34 J ∙ s. The unit of frequency is 1/s, so the energy calculated using this equation is in joules ( J). If an oscillator changes from a higher energy to a lower one, energy is emitted as electromagnetic radiation, where the difference in energy between the higher and lower energy states is

∆E ​= ​Ehigher n ​− ​Elower n ​= ​∆nhν



If the value of ∆n is 1, which corresponds to changing from one energy level to the

next lower one for that oscillator, then the energy change for the oscillator and the

electromagnetic radiation emitted would have an energy equal to





•  The relationship of energy,

wavelength, and frequency

As frequency (␯) increases, energy (E) increases



E = h␯ =



hc

␭



As wavelength (␭) decreases, energy (E) increases



kotz_48288_06_0266-0299.indd 270



E ​= ​hν







(6.2)



This equation is called Planck’s equation.

Now, assume as Planck did that there must be a distribution of vibrations of atoms in an object—some atoms are vibrating at a high frequency, some are vibrating

at a low frequency, but most have some intermediate frequency. The few atoms with

high-frequency vibrations are responsible for some of the light, say in the ultraviolet

region, and those few with low-frequency vibrations for energies in the infrared region. However, most of the light must come from the majority of the atoms that have

intermediate vibrational frequencies. That is, a spectrum of light is emitted with a

maximum intensity at some intermediate wavelength, in accord with experiment.

The intensity should not become greater and greater on approaching the ultraviolet region. With this realization, the ultraviolet catastrophe was solved.

The key aspect of Planck’s work for general chemistry is that Planck introduced

the idea of quantized energies based on the equation E = hν, an equation that was



11/18/10 2:35 PM



6.2  Quantization: Planck, Einstein, Energy, and Photons







271



to have a profound impact on the work of Albert Einstein in explaining another

puzzling phenomenon.



Einstein and the Photoelectric Effect

A few years after Planck’s work, Albert Einstein (1879–1955) incorporated Planck’s

ideas into an explanation of the photoelectric effect and in doing so changed the description of electromagnetic radiation.

In the photoelectric effect, electrons are ejected when light strikes the surface

of a metal (Figure 6.4), but only if the frequency of the light is high enough. If light

with a lower frequency is used, no electrons are ejected, regardless of the light’s intensity (its brightness). If the frequency is at or above a minimum, critical frequency,

increasing the light intensity causes more electrons to be ejected.

Einstein decided the experimental observations could be explained by combining Planck’s equation (E ​= ​hν) with a new idea, that light has particle-like properties. Einstein characterized these massless particles, now called photons, as packets

of energy, and stated that the energy of each photon is proportional to the frequency of the radiation as defined by Planck’s equation. In the photoelectric effect,

photons striking atoms on a metal surface will cause electrons to be ejected only if

the photons have high enough energy. The greater the number of photons that

strike the surface at or above the threshold energy, the greater the number of electrons dislodged. The metal atoms will not lose electrons, however, if no individual

photon has enough energy to dislodge an electron from an atom.



•  Einstein and Digital Cameras 

When you take a photograph with a

digital camera the detector in the camera (a CCD or “charged-coupled detector”) works on a principle much like

the photoelectric effect.



Energy and Chemistry: Using Planck’s Equation

Compact disc players use lasers that emit red light with a wavelength of 685 nm.

What is the energy of one photon of this light? What is the energy of 1 mol of photons of red light? To answer these questions, first convert the wavelength to the frequency of the radiation,

For λ ​= ​685 nm, ν ​= ​4.38 × 1014 1/s (calculated using Equation 6.1)

Light (photons)

Cathode (−)

Highfrequency light



Meter to

measure current



e−



Highintensity light



e−



e−

e−



e−



e−



e−



e−



Electron (e−)



Frequency of light incident on photocell



(a) A photocell operates by the photoelectric

effect. The main part of the cell is a lightsensitive cathode. This is a material, usually

a metal, that ejects electrons if struck by

photons of light of sufficient energy.

No current is observed until the critical

frequency is reached.



Current (number of e−

ejected by cathode)



Critical frequency



Current (number of e−

ejected by cathode)



Current (number of e−

ejected by cathode)



Anode (+)



Frequency of light incident on photocell



(b) When light of higher frequency than

the minimum is used, the excess energy of the

photon allows the electron to escape the

atom with greater velocity. The ejected

electrons move to the anode, and a current

flows in the cell. Such a device can be used

as a switch in electric circuits.



High-intensity light

Low-intensity light



Frequency of light incident on photocell



(c) If higher intensity light is used, the only

effect is to cause more electrons to be

released from the surface. The onset of

current is observed at the same frequency

as with lower intensity light, but more

current flows with more intense light.



Figure 6.4   A photoelectric cell.



kotz_48288_06_0266-0299.indd 271



11/18/10 2:35 PM



272 



c h a p t er 6 The Structure of Atoms



Strategy Map: Planck’s Equation



and then use the frequency to calculate the energy per photon.



PROBLEM



Calculate energy of red light

photons.



KNOWN DATA/INFORMATION



• Wavelength of red light (685 nm)



E per photon  =  hν  =  (6.626 × 10−34 J ∙ s/photon)  ×  (4.38  ×  1014 1/s)

 



= 2.90 × 10−19 J/photon



Finally, calculate the energy of a mole of photons by multiplying the energy per

photon by Avogadro’s number:

E per mole  =  (2.90 × 10−19 J/photon) × (6.022 × 1023 photons/mol)



STEP 1. Convert wavelength to

frequency. (Equation 6.1)



Frequency in cycles per second

STEP 2. Calculate energy using

Planck’s equation. (Equation 6.2)



Obtain energy in J/photon.

STEP 3. Use Avogadro’s number

to convert J/atom to J/mol.



 



= 1.75 × 105 J/mol (or 175 kJ/mol)



The energy of red light photons with a wavelength of 685 nm is 175 kJ/mol,

whereas the energy of blue light photons (λ  =  400 nm) is about 300 kJ/mol. The

energy of the blue light photons is in the range of the energies necessary to break

some chemical bonds in proteins. Higher energy ultraviolet photons are even more

likely to cause chemical bond breaking. This is what happens if you spend too much

time unprotected in the sun. In contrast, light at the red end of the spectrum and

infrared radiation has a lower energy, and, although it is generally not energetic

enough to break chemical bonds, it can affect the vibrations of molecules. We sense

infrared radiation as heat, such as the heat given off by a glowing burner on an electric stove or a burning log in a fire.



Obtain energy in J/mol.



rEvIEW & cHEcK FOr SEctIOn 6.2

Various manufacturers have developed mixtures of compounds that protect skin from UVA

(400-320 nm) and UVB (320-280 nm) radiation. These sunscreens are given “sun protection

factor” (SPF) labels that indicate how long the user can stay in the sun without burning.

1.



Calculate the energy of a mole of photons of UVB light with a wavelength of 310 nm.

(a)



2.



3.9 × 105 J/mol



(b) 1.2 × 105 J/mol



(c)



0.00039 J/mol



Which has the greater energy per photon, UVB light at 310 nm or microwave radiation

having a frequency of 2.45 GHz (1 GHz  = 109 s−1)?

(a)



UVB



(b) microwave



JGI/Jamie Grill/Blend Images/Jupiter Images



6.3 AtomicLineSpectraandNielsBohr



Sunscreens and damage from

radiation. Higher-energy sunlight

falling on the Earth is often classified

as UVA and UVB radiation. UVB

radiation has a higher energy than

UVA radiation and is largely responsible for sunburns and tanning.

Sunscreens containing organic

compounds such as 2-ethylhexyl-pmethoxycinnamate and oxybenzone

absorb UV radiation, preventing it

from reaching your skin.



kotz_48288_06_0266-0299.indd 272



If a high voltage is applied to atoms of an element in the gas phase at low pressure,

the atoms absorb energy and are said to be “excited.” The excited atoms can then

emit light, and a familiar example is the colored light from neon advertising signs.

The light from excited atoms is composed of only a few different wavelengths of

light. We can demonstrate this by passing a beam of light from excited neon or hydrogen through a prism (Figure 6.5); only a few colored lines are seen. The spectrum obtained in this manner, such as that for excited H atoms, is called a line

emission spectrum. This is in contrast to the light falling on Earth from the Sun, or

the light emitted by a very hot object, which consists of a continuous spectrum of

wavelengths (Figures 6.2 and 6.3)

Every element has a unique emission spectrum, as exemplified by the spectra

for hydrogen, mercury, and neon in Figure 6.6. Indeed, the characteristic lines in

the emission spectrum of an element can be used in chemical analysis, both to identify the element and to determine how much of it is present.

A goal of scientists in the late 19th century was to explain why excited gaseous

atoms emitted light of only certain frequencies. One approach was to look for a

mathematical relationship among the observed wavelengths because a regular pattern of information implies a logical explanation. The first steps in this direction

were taken by Johann Balmer (1825–1898) and later by Johannes Rydberg



11/18/10 2:35 PM



6.3  Atomic Line Spectra and Niels Bohr







273



Gas discharge tube

contains hydrogen



Prism



Figure 6.5   The visible line emission spectrum of hydrogen.  The emitted light is passed

through a series of slits to create a narrow beam of light, which is then separated into its component

wavelengths by a prism. A photographic plate or photocell can be used to detect the separate wavelengths as individual lines. Hence, the name line spectrum for the light emitted by a glowing gas.



(1854–1919). From these studies, an equation—now called the Balmer equation

(Equation 6.3)—was found that could be used to calculate the wavelength of the

red, green, and blue lines in the visible emission spectrum of hydrogen (Figure 6.6).

1

1

1

ϭ R 2 Ϫ 2

n 

␭

2







when n Ͼ 2



(6.3)





In this equation n is an integer, and R, now called the Rydberg constant, has the

value 1.0974 ​× ​107 m−1. If n ​= ​3, for example, the wavelength of the red line in the

hydrogen spectrum is obtained (6.563 ​× ​10−7 m, or 656.3 nm). If n ​= ​4, the wavelength for the green line is calculated. Using n ​= ​5 and n ​= ​6 in the equation gives

the wavelengths of the blue lines. The four visible lines in the spectrum of hydrogen

atoms are now known as the Balmer series.



The Bohr Model of the Hydrogen Atom

Early in the 20th century, the Danish physicist Niels Bohr (1885–1962) proposed a

model for the electronic structure of atoms and with it an explanation for the emission spectra of excited atoms. Bohr proposed a planetary structure for the hydrogen

atom in which the electron moved in a circular orbit around the nucleus, similar to

a planet revolving about the sun. In proposing this model, however, he had to



␭(nm) 400



500



600



700



H



Hg



Ne



Figure 6.6   Line emission spectra of hydrogen, mercury, and neon.  Excited gaseous elements produce characteristic spectra that can be used to identify the elements and to determine how

much of each element is present in a sample.



kotz_48288_06_0266-0299.indd 273



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274



c h a p t er 6   The Structure of Atoms



contradict the laws of classical physics. According to classical theories, a charged

electron moving in the positive electric field of the nucleus should lose energy; a

consequence of the loss of energy is that the electron would eventually crash into the

nucleus. This is clearly not the case; if it were, matter would eventually self-destruct.

To solve this contradiction, Bohr postulated that there are certain orbits corresponding to particular energy levels where this would not occur. As long as an electron is in

one of these energy levels, the system is stable. That is, Bohr introduced quantization

into the description of electronic structure. By combining this quantization postulate

with the laws of motion from classical physics, Bohr derived an equation for the total

energy possessed by the single electron in the nth orbit (energy level) of the H atom.

Planck’s constant

Rydberg constant

Speed of light



Total energy of electron in the nth level = En = −

n=∞

n=6

n=5

n=4



0

1

16

1

−

9



−



1

4



n=2

E2 = −5.45 × 10−19 J/atom



(



1

E

=

n2

Rhc

−



•



•



n=1

E1 = −2.18 × 10−18 J/atom



(6.4)







Here, En is the energy of the electron (in J/atom); and R, h, and c are constants [the

Rydberg constant (R = 1.097 × 107 m−1), Planck’s constant, and the speed of light,

respectively]. The symbol n is a positive, unitless integer called the principal quantum number. It can have integral values of 1, 2, 3, and so on.

Equation 6.4 has several important features (illustrated in Figure 6.7) such as

the role of n and the significance of the negative sign.







1



Principal quantum number













)







n=3

E3 = 2.42 ì 10−19 J/atom



Rhc

n2



The quantum number n defines the energies of the allowed orbits in the H atom.

The energy of an electron in an orbit has a negative value because the electron in the atom has a lower energy than when it is free. The zero of energy

occurs when n = ∞, that is, when the electron is infinitely separated from the

nucleus.

An atom with its electrons in the lowest possible energy levels is said to be in

its ground state; for the hydrogen atom, this is the level defined by the quantum

number n ​= ​1. The energy of this state is −Rhc/12, meaning that it has an

amount of energy equal to Rhc below the energy of the infinitely separated

electron and nucleus.

States for the H atom with higher energies (and n > 1) are called excited states,

and, as the value of n increases, these states have less negative energy values.

Because the energy is dependent on 1/n2, the energy levels are progressively

closer together as n increases.



Bohr also showed that, as the value of n increases, the distance of the electron

from the nucleus increases. An electron in the n ​= ​1 orbit is closest to the nucleus

and has the lowest (most negative) energy. For higher integer values of n, the electron is further from the nucleus and has a higher (less negative) energy.



Figure 6.7   Energy levels for

the H atom in the Bohr model. 

The energy of the electron in the

hydrogen atom depends on the value

of the principal quantum number n

(En = –Rhc/n2). The larger the value

of n, the larger the Bohr radius and

the less negative the value of the

energy. Energies are given in joules

per atom (J/atom). Notice that the

difference between successive energy

levels becomes smaller as n becomes

larger.



kotz_48288_06_0266-0299.indd 274



EXAMPLE 6.2



Energies of the Ground and Excited States of the H Atom



Problem  Calculate the energies of the n ​= ​1 and n ​= ​2 states of the hydrogen atom in joules per

atom and in kilojoules per mole. What is the difference in energy of these two states in kJ/mol?

What Do You Know?  The n = 1 and n = 2 states are the first and second states (lowest and next

to lowest energy state) in the Bohr description of the hydrogen atom. Use Equation 6.4 to calculate the energy of each state. For the calculations you will need the following constants: R

(Rydberg constant) = 1.097 × 107 m−1; h (Planck’s constant) = 6.626 × 10−34 J ∙ s; c (speed of

light) = 2.998 × 108 m/s; and N (Avogadro’s number) = 6.022 × 1023 atoms/mol.



11/18/10 2:35 PM



6.3  Atomic Line Spectra and Niels Bohr







275



Strategy  For each energy level, substituting the appropriate values into Equation 6.4 and

solving gives the energy in J/atom. Multiply this value by N to find the energy in J/mol.

Subtract the energy for the n = 1 level from the energy of the n = 2 level to obtain the

energy difference.

Solution  When n ​= ​1, the energy of an electron in a single H atom is





E 1 ϭ ϪRhc

E 1 ϭ Ϫ(1.097 ϫ 107 mϪ1 )(6.626 ϫ 10Ϫ34 J и s)(2.998 ϫ 10 8 m/s) = −2.179 × 10−18 J/atom



In units of kJ/mol,







E1 ϭ



Ϫ2.179 ϫ 10Ϫ18 J

6.022 ϫ 1023 atoms

1 kJ

ϫ

ϫ

= −1312 kJ/mol

1000 J

atom

mol



When n ​= ​2, the energy is







E2 ϭ Ϫ



Rhc

E

2.179 ϫ 10Ϫ18 J/atom

ϭϪ 1 ϭϪ

= −5.448 × 10−19 J/atom

22

4

4



In units of kJ/mol,







E1 ϭ



Ϫ5.448 × 10 −19 J

6.022 × 1023 atoms

1 kJ

ϫ

ϫ

= −328.1 kJ/mol

1000 J

atom

mol



The difference in energy, ∆E, between the first two energy states of the H atom is

∆E ​= ​E2 ​− ​E1 ​= ​(−328.1 kJ/mol) ​− ​(−1312 kJ/mol) ​= ​984 kJ/mol

Think about Your Answer  The calculated energies are negative, with E1 more negative than E2.

The n ​= ​2 state is higher in energy than the n ​= ​1 state by 984 kJ/mol. Also, be sure to notice

that 1312 kJ/mol is the value of Rhc multiplied by Avogadro’s number N (i.e., NRhc). This will be

useful in future calculations.

Check Your Understanding 

Calculate the energy of the n ​= ​3 state of the H atom in (a) joules per atom and (b) kilojoules per

mole.



The Bohr Theory and the Spectra of Excited Atoms

Bohr’s theory describes electrons as having only specific orbits and energies. If an

electron moves from one energy level to another, then energy must be absorbed or

evolved. This idea allowed Bohr to relate energies of electrons and the emission

spectra of hydrogen atoms.

To move an electron from the n ​= ​1 state to an excited state, such as the

n ​= ​2 state, energy must be transferred to the atom. When Efinal has n ​= ​2 and

Einitial has n ​= ​1, then 1.63 × 10−18 J/atom (or 984 kJ/mol) of energy must be

transferred (Figure 6.8). This is the difference in energy between final and initial

states:

∆E ​= ​Efinal state ​− ​Einitial state ​= ​(−Rhc/22) ​− ​(−Rhc/12) ​= ​(0.75)Rhc ​= 1.63 × 10−18 J/atom



Moving an electron from the first to the second energy state requires input of

1.63 × 10−18 J/atom, no more and no less. If 0.7Rhc or 0.8Rhc is provided, no transition between states is possible. Requiring a specific and precise amount of energy is

a consequence of quantization.

Moving an electron from a state of low n to one of higher n always requires

that energy be transferred to the atom from the surroundings. The opposite

process, in which an electron “falls” from a level of higher n to one of lower n,

leads to emission of energy, a transfer of energy, usually as radiation, from the



kotz_48288_06_0266-0299.indd 275



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276



c h a p t er 6   The Structure of Atoms



Figure 6.8   Absorption of

energy by the atom as the electron moves to an excited

state.  Energy is absorbed when an



n=5

n=2

Energy, E



electron moves from the n = 1 state

to the n = 2 state (∆E > 0). When

the electron returns to the n = 1

state from n = 2, energy is evolved

(∆E < 0). The energy emitted

is 984 kJ/mol, as calculated in

Example 6.2.



∆E = +984 kJ



∆E = −984 kJ



Energy

absorbed



Energy

emitted



n=1

Ground state



Excited state



Ground state



atom to its surroundings. For example, for a transition from the n ​= ​2 level to

n ​= ​1 level,

∆E ​= ​Efinal state ​− ​Einitial state ​= ​−1.63 × 10−18 J/atom (= ​−984 kJ/mol)



The negative sign indicates 1.63 × 10−18 J/atom (or 984 kJ/mol) is emitted.

Now we can visualize the mechanism by which the characteristic line emission

spectrum of hydrogen originates according to the Bohr model. Energy is provided

to the atoms from an electric discharge or by heating. Depending on how much

energy is added, some atoms have their electrons excited from the n ​= ​1 state to the

n ​= ​2, 3, or even higher states. After absorbing energy, these electrons can return to

any lower level (either directly or in a series of steps), releasing energy (Figure 6.9).

We observe this released energy as photons of electromagnetic radiation, and, because only certain energy levels are possible, only photons with particular energies

and wavelengths are emitted.

The energy of any emission line (in J/atom) for excited hydrogen atoms can be

calculated using Equation 6.5.

 1

1 

⌬E ϭ E final Ϫ E initial ϭ ϪRhc  2 Ϫ 2 

ninitial 

 nfinal







(6.5)





[The value of ∆E in J/atom may be related to the wavelength or frequency of radiation using Planck’s equation (∆E = hν) or converted to units of kJ/mol if ∆E is

multiplied by (6.022 × 1023 atoms/mol)(1 kJ/1000 J).]

For hydrogen, a series of emission lines having energies in the ultraviolet region

(called the Lyman series; Figure 6.10) arises from electrons moving from states with

n > 1 to the n ​= ​1 state. The series of lines that have energies in the visible region—

the Balmer series—arises from electrons moving from states with n > 2 to the

n ​= ​2 state. There are also series of lines in the infrared spectral region, arising

from transitions from higher levels to the n ​= ​3, 4 or 5 levels.

Bohr’s model, introducing quantization into a description of the atom, tied the

unseen (the structure of the atom) to the seen (the observable lines in the hydrogen

Figure 6.9  Radiation emitted on changes

in energy levels.  The greater the separation in



kotz_48288_06_0266-0299.indd 276



n = 3, (E3)

hν = E3 − E2

n = 2, (E2)

hν = E3 − E1



n = 1, (E1)



Energy, E



n = 2, (E2)

Energy, E



the energy levels, the greater the energy, the

higher the frequency, and the shorter the wavelength of the emitted radiation. Notice that an

electron excited to n = 3 can return directly to

n = 1, or it can drop to n = 2 and then to n = 1.

These three possibilities are observed as three

different wavelengths of emitted radiation. The

energies and frequencies are in the order

E 3 − E 1 > E 2 − E 1 > E 3 − E 2.



n = 3, (E3)



hν = E2 − E1



n = 1, (E1)



11/18/10 2:35 PM



6.3  Atomic Line Spectra and Niels Bohr







n Energy J/atom



1875 nm



1282 nm



Lyman series of lines in the ultraviolet region results from transitions

to the n = 1 level. The Balmer series

(Figures 6.5 and 6.6) arises from

transitions from levels with values of

n > 2 to n = 2. Lines in the infrared

region result from transitions from

levels with n > 3 to the n = 3 level.

(Transitions from n = 8 and higher

levels to lower levels occur but are

not shown in this figure.)



Invisible

lines

(Infrared)



−2.18 × 10−18



1094 nm



1005 nm



−5.45 × 10−19



Invisible

lines

(Ultraviolet)



1



−2.42 × 10−19



656.3 nm



2



−1.36 × 10−19



410.2 nm

434.1 nm

486.1 nm

656.3 nm



3



−8.72 × 10−20



486.1 nm



4



−6.06 × 10−20



434.1 nm



5



−4.45 × 10−20



410.2 nm



6



Figure 6.10   Some of the

electronic transitions that can

occur in an excited H atom.  The



Balmer series



93.1 nm

93.8 nm

95.0 nm

97.3 nm

102.6 nm

121.6 nm



7



Lyman series



277



spectrum). Agreement between theory and experiment is taken as evidence that the

theoretical model is valid. It became apparent, however, that Bohr’s theory was inadequate. This model of the atom explained only the spectrum of hydrogen atoms and

of other systems having one electron (such as He+), but it failed for all other systems.

A better model of electronic structure was needed.



Strategy Map 6.3

PROBLEM



  Interactive EXAMPLE 6.3 Energies of Emission Lines

for Excited Atoms

Problem  Calculate the wavelength of the green line in the visible spectrum of excited H atoms.

What Do You Know?  The green line in the spectrum of hydrogen arises from the electron transition from the n = 4 state (ninitial) to the n = 2 state (nfinal) (Figure 6.10).

Strategy

•

Calculate the difference in energy between the states using Equation 6.5. You can simplify

this calculation by using the value of Rhc from Example 6.2 (= 2.179 ì 1018 J/photon).





Relate the difference in energy to the wavelength of light using the equation E = hc/λ. (This

equation is derived by combining Equations 6.1 and 6.2.)



Solution  Calculate ∆E.

 Rhc   Rhc 

⌬E ϭ E final Ϫ E initial ϭ  Ϫ 2  Ϫ  Ϫ 2 

 2   4 

1

1

⌬E ϭ ϪRhc  Ϫ  ϭ ϪRhc(0 .1875)

 4 16 

⌬E ϭ Ϫ(2.179 ϫ 10Ϫ18 J/photon)(0.1875) ϭ Ϫ4.086 ϫ 10Ϫ19 J/photon



Calculate energy of green line in

H spectrum.



KNOWN DATA/INFORMATION



• Green line involves transition



from n = 4 to n = 2

2

S T E P 1 . Use E = –Rhc/n to

calculate E for n = 4 and n = 2.



Obtain E2 (= Efinal) and E4 (= Einitial)

S T E P 2 . Calculate

∆E = Efinal – Einitial



Obtain ∆E = Efinal – Einitial

S T E P 3 . Use Planck’s equation to

convert Ephoton to wavelength.



Obtain photon wavelength



kotz_48288_06_0266-0299.indd 277



11/18/10 2:35 PM



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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