CHAPTER 7
Quantum Theory of the Atom
■ SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.
7.1. Rearrange the equation c = , which relates wavelength to frequency and the speed of light (3.00
108 m/s):
c 3.00 108 m/s
= = = 7.672 10−7 = 7.67 10−7 m, or 767 nm
3.91 1014 /s
7.2. Rearrange the equation c = , which relates frequency to wavelength and the speed of light
(3.00 108 m/s). Recognize that 456 nm = 4.56 10−7 m.
c 3.00 108 m/s
= = = 6.578 1014 = 6.58 1014 /s
4.56 107 m
7.3. First, use the wavelengths to calculate the frequencies from c = . Then calculate the energies
using E = h.
c 3.00 108 m/s
= = = 3.00 1014 /s
1.0 106 m
c 3.00 108 m/s
= = = 3.00 1016 /s
1.0 108 m
c 3.00 x 108 m/s
= = = 3.00 1018 /s
1.0 x 10-10 m
E = h = 6.63 10−34 Js 3.00 1014 /s = 1.989 10−19 = 2.0 10−19 J (IR)
E = h = 6.63 10−34 Js 3.00 1016 /s = 1.989 10−17 = 2.0 10−17 J (UV)
E = h = 6.63 10−34 Js 3.00 1018 /s = 1.989 10−15 = 2.0 10−15 J (x ray)
The x-ray photon (shortest wavelength) has the greatest amount of energy; the infrared photon
(longest wavelength) has the least amount of energy.
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 7: Quantum Theory of the Atom 243
7.4. From the formula for the energy levels, E = −RH/n2, obtain the expressions for both Ei and Ef.
Then calculate the energy change for the transition from n = 3 to n = 1 by subtracting the lower
value from the upper value. Set this equal to h. The result is
RH RH 8RH
9 − 1 = 9 = h
The frequency of the emitted radiation is
8RH 8 2.179 1018 J
= = = 2.921 1015 = 2.92 1015 /s
9 9 6.63 1034 J s
Since = c/,
3.00 108 m/s
= = 1.027 10−7 = 1.03 10−7 m, or 103 nm
2.92 1015 /s
7.5. Calculate the frequency from c = , recognizing that 589 nm is 5.89 10−7 m.
c 3.00 108 m/s
= = = 5.093 1014 = 5.09 1014 /s
5.89 107 m
Finally, calculate the energy difference.
E = h = 6.63 10−34 Js 5.093 1014 /s = 3.3766 10−19 = 3.38 10−19 J
7.6. To calculate wavelength, use the mass of an electron (m = 9.11 10−31 kg), and Planck's constant
(h = 6.63 10−34 Js, or 6.63 10−34 kgm2/s).
h 6.63 1034 kg m 2 /s
= = = 3.323 10−10 m (332 pm)
mv 9.11 1031 kg 2.19 106 m/s
7.7. a. The value of n must be a positive whole number greater than zero. Here, it is zero. Also, if
n is zero, there are no allowed values for l and ml.
b. The values for l can range only from zero to (n − 1). Here, l has a value greater than n.
c. The values for ml range from −l to +l. Here, ml has a value greater than l.
d. The value for ms is either + ½ or −½. Here, it is zero.
■ ANSWERS TO CONCEPT CHECKS
7.1. The frequency and wavelength are inversely related. Therefore, if the frequency is doubled, the
wavelength is halved. Red light has a wavelength around 700 nm, so doubling its frequency
halves its wavelength to about 350 nm, which is in the ultraviolet range just beyond the visible
spectrum.
7.2. Since the transitions are between adjacent levels, the energy-level diagram must look something
like the following diagram, with the red transition between two close levels and the blue
transition between two more widely spaced levels. (The three levels could be spaced so the red
and blue transitions are interchanged, with the blue transition above the red one.)
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 244 Chapter 7: Quantum Theory of the Atom
Level 3
750 nm (red line)
Level 2
400 nm (blue line)
Level 1
The transition from the top level to the lowest level would correspond to a transition that is
greater in energy change than either of the other two transitions. Thus, the three transitions, from
lowest to highest energy change, are in this order: red, blue, and the transition from the highest to
lowest level. The last transition would have the highest frequency and therefore the shortest
wavelength. It would lie just beyond the blue portion of the visible spectrum in the ultraviolet
region.
7.3. The de Broglie relation says the wavelength of a particle is inversely proportional to both mass
and speed. So, to maintain the wavelength constant while the mass increases would mean the
speed would have to decrease. In going from a particle with the mass of an electron to one with
that of a proton, the speed would have to decrease by a factor of about 2000 in order to maintain
the same wavelength. The proton would have to have a speed approximately 2000 times slower
than an electron of the same wavelength.
■ ANSWERS TO SELF-ASSESSMENT AND REVIEW QUESTIONS
7.1. Light is a wave, which is a form of electromagnetic radiation. In terms of waves, light can be
described as a continuously repeating change, or oscillation, in electric and magnetic fields that
can travel through space. Two characteristics of light are wavelength (often given in nanometers,
nm) and frequency.
7.2. The relationship among the different characteristics of light waves is c = , where is the
frequency, is the wavelength, and c is the speed of light.
7.3. Starting with the shortest wavelengths, the electromagnetic spectrum consists of gamma rays,
x rays, far ultraviolet (UV), near UV, visible light, near infrared (IR), far IR, microwaves, radar,
and TV/FM radio waves (longest wavelengths).
7.4. The term quantized means the possible values of the energies of an atom are limited to only
certain values. Planck was trying to explain the intensity of light of various frequencies emitted
by a hot solid at different temperatures. The formula he arrived at was E = nh, where E is
energy, n is a whole number (n = 1, 2, 3, …), h is Planck's constant, and is frequency.
7.5. Photoelectric effect is the term applied to the ejection of electrons from the surface of a metal or
from other materials when light shines on it. Electrons are ejected only when the frequency (or
energy) of light is larger than a certain minimum, or threshold, value that is constant for each
metal. If a photon has a frequency equal to or greater than this minimum value, then it will eject
one electron from the metal surface.
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Quantum Theory of the Atom
■ SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.
7.1. Rearrange the equation c = , which relates wavelength to frequency and the speed of light (3.00
108 m/s):
c 3.00 108 m/s
= = = 7.672 10−7 = 7.67 10−7 m, or 767 nm
3.91 1014 /s
7.2. Rearrange the equation c = , which relates frequency to wavelength and the speed of light
(3.00 108 m/s). Recognize that 456 nm = 4.56 10−7 m.
c 3.00 108 m/s
= = = 6.578 1014 = 6.58 1014 /s
4.56 107 m
7.3. First, use the wavelengths to calculate the frequencies from c = . Then calculate the energies
using E = h.
c 3.00 108 m/s
= = = 3.00 1014 /s
1.0 106 m
c 3.00 108 m/s
= = = 3.00 1016 /s
1.0 108 m
c 3.00 x 108 m/s
= = = 3.00 1018 /s
1.0 x 10-10 m
E = h = 6.63 10−34 Js 3.00 1014 /s = 1.989 10−19 = 2.0 10−19 J (IR)
E = h = 6.63 10−34 Js 3.00 1016 /s = 1.989 10−17 = 2.0 10−17 J (UV)
E = h = 6.63 10−34 Js 3.00 1018 /s = 1.989 10−15 = 2.0 10−15 J (x ray)
The x-ray photon (shortest wavelength) has the greatest amount of energy; the infrared photon
(longest wavelength) has the least amount of energy.
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 7: Quantum Theory of the Atom 243
7.4. From the formula for the energy levels, E = −RH/n2, obtain the expressions for both Ei and Ef.
Then calculate the energy change for the transition from n = 3 to n = 1 by subtracting the lower
value from the upper value. Set this equal to h. The result is
RH RH 8RH
9 − 1 = 9 = h
The frequency of the emitted radiation is
8RH 8 2.179 1018 J
= = = 2.921 1015 = 2.92 1015 /s
9 9 6.63 1034 J s
Since = c/,
3.00 108 m/s
= = 1.027 10−7 = 1.03 10−7 m, or 103 nm
2.92 1015 /s
7.5. Calculate the frequency from c = , recognizing that 589 nm is 5.89 10−7 m.
c 3.00 108 m/s
= = = 5.093 1014 = 5.09 1014 /s
5.89 107 m
Finally, calculate the energy difference.
E = h = 6.63 10−34 Js 5.093 1014 /s = 3.3766 10−19 = 3.38 10−19 J
7.6. To calculate wavelength, use the mass of an electron (m = 9.11 10−31 kg), and Planck's constant
(h = 6.63 10−34 Js, or 6.63 10−34 kgm2/s).
h 6.63 1034 kg m 2 /s
= = = 3.323 10−10 m (332 pm)
mv 9.11 1031 kg 2.19 106 m/s
7.7. a. The value of n must be a positive whole number greater than zero. Here, it is zero. Also, if
n is zero, there are no allowed values for l and ml.
b. The values for l can range only from zero to (n − 1). Here, l has a value greater than n.
c. The values for ml range from −l to +l. Here, ml has a value greater than l.
d. The value for ms is either + ½ or −½. Here, it is zero.
■ ANSWERS TO CONCEPT CHECKS
7.1. The frequency and wavelength are inversely related. Therefore, if the frequency is doubled, the
wavelength is halved. Red light has a wavelength around 700 nm, so doubling its frequency
halves its wavelength to about 350 nm, which is in the ultraviolet range just beyond the visible
spectrum.
7.2. Since the transitions are between adjacent levels, the energy-level diagram must look something
like the following diagram, with the red transition between two close levels and the blue
transition between two more widely spaced levels. (The three levels could be spaced so the red
and blue transitions are interchanged, with the blue transition above the red one.)
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 244 Chapter 7: Quantum Theory of the Atom
Level 3
750 nm (red line)
Level 2
400 nm (blue line)
Level 1
The transition from the top level to the lowest level would correspond to a transition that is
greater in energy change than either of the other two transitions. Thus, the three transitions, from
lowest to highest energy change, are in this order: red, blue, and the transition from the highest to
lowest level. The last transition would have the highest frequency and therefore the shortest
wavelength. It would lie just beyond the blue portion of the visible spectrum in the ultraviolet
region.
7.3. The de Broglie relation says the wavelength of a particle is inversely proportional to both mass
and speed. So, to maintain the wavelength constant while the mass increases would mean the
speed would have to decrease. In going from a particle with the mass of an electron to one with
that of a proton, the speed would have to decrease by a factor of about 2000 in order to maintain
the same wavelength. The proton would have to have a speed approximately 2000 times slower
than an electron of the same wavelength.
■ ANSWERS TO SELF-ASSESSMENT AND REVIEW QUESTIONS
7.1. Light is a wave, which is a form of electromagnetic radiation. In terms of waves, light can be
described as a continuously repeating change, or oscillation, in electric and magnetic fields that
can travel through space. Two characteristics of light are wavelength (often given in nanometers,
nm) and frequency.
7.2. The relationship among the different characteristics of light waves is c = , where is the
frequency, is the wavelength, and c is the speed of light.
7.3. Starting with the shortest wavelengths, the electromagnetic spectrum consists of gamma rays,
x rays, far ultraviolet (UV), near UV, visible light, near infrared (IR), far IR, microwaves, radar,
and TV/FM radio waves (longest wavelengths).
7.4. The term quantized means the possible values of the energies of an atom are limited to only
certain values. Planck was trying to explain the intensity of light of various frequencies emitted
by a hot solid at different temperatures. The formula he arrived at was E = nh, where E is
energy, n is a whole number (n = 1, 2, 3, …), h is Planck's constant, and is frequency.
7.5. Photoelectric effect is the term applied to the ejection of electrons from the surface of a metal or
from other materials when light shines on it. Electrons are ejected only when the frequency (or
energy) of light is larger than a certain minimum, or threshold, value that is constant for each
metal. If a photon has a frequency equal to or greater than this minimum value, then it will eject
one electron from the metal surface.
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
