CHAPTER 7
QUANTUM THEORY AND ATOMIC STRUCTURE
The Nature of Light
Electromagnetic radiation (also called electromagnetic energy
or radiant energy) – consists of energy propagated by electric
and magnetic fields that increase and decrease in intensity as
they move through space
 This classical wave model explains why rainbows form,
how magnifying glasses work, and many other familiar
observations.
The Wave Nature of Light
The wave properties of electromagnetic radiation are described
by three variables and one constant…
Frequency – the number of cycles the
wave undergoes per second [s-1, or hertz
(Hz)]
Wavelength – the distance between any
point on a wave and the corresponding
point on the next crest or trough of a
wave (the distance the wave travels in
one cycle) [m, nm, pm, angstroms (A)]
Speed – the distance it moves per unit
time (meters/second). Electromagnetic radiation moves at
2.99792458x108 m/s (3.00x108 m/s) in a vacuum. This is a
physical constant and is referred to as the speed of light.
Amplitude – the height of
the crest (or depth of the
trough) For an
electromagnetic wave, the
amplitude is related to the
intensity of the radiation.
Light of a particular color has
a specific frequency (and
thus, wavelength) but it can
be dimmer (lower amplitude,
less intense) or brighter (higher amplitude, more intense)
The Electromagnetic Spectrum
Visible light is a small region of the electromagnetic spectrum
 All waves in the spectrum travel at the same speed through
a vacuum but differ in frequency and wavelength
Long-wavelength, low frequency radiation…
 Used in microwave ovens, radios, and cell phones
Electromagnetic emissions are everywhere…
 Human artifacts such as light bulbs, x-ray equipment, and
car motors
 Natural sources such as the Sun, lightning, radioactivity,
and even the glow of fireflies!
Sample Problem 7.1 p. 263
Interconverting Wavelength and Frequency
1.00 A 1.00 m
= 1.00x10-10 m
1x1010 A
(x-ray)
Speed(c) = frequency(v) x wavelength(λ)
3.00x108 m/s = frequency(v) x 1.00x10-10 m
frequency(v) = 3.00x1018 s-1
(radio signal)
325 cm 1 m
100 cm
= 3.25 m
Speed(c) = frequency(v) x wavelength(λ)
3.00x108 m/s = frequency(v) x 3.25 m
frequency(v) = 9.23x1018 s-1
(Blue sky)
473 nm
1.00 m
= 4.73x10-7 m
1x109 nm
Speed(c) = frequency(v) x wavelength(λ)
3.00x108 m/s = frequency(v) x 4.73x10-7 m
frequency(v) = 6.34x1014 s-1
Follow-Up Problem 7.1 p. 263
Speed(c) = frequency(v) x wavelength(λ)
3.00x108 m/s = 7.23x1014 Hz x wavelength(λ)
wavelength(λ) = 4.15x10-7 m
4.15x10-7 m 1x1010 A = 4150 A
1m
4.15x10-7 m 1x109 nm = 415 nm
1m
The Classical Distinction Between Energy and Matter
Some distinctions between the behavior of waves and matter…
 Refraction and dispersion
o Light of a given wavelength travels at different speeds
through various transparent media – vacuum, air (even
different temperatures of air), water, quartz, etc.
 When a light wave passes from one medium into
another, the speed of the wave changes
 Refraction – change in speed causes a
change in direction (figure below)
 Dispersion – white light separates
(disperses) into its component colors when it
passes through a prism (or other refracting
object) – rainbow is an example
In contrast to a wave of light, a particle of matter, like a pebble
(above figure) does not undergo refraction…
 Diffraction and interference
o Diffraction – the bending of a wave when it strikes
the edge of an object
 If a wave passes through a slit about as wide as
its wavelength, it bends around both edges of the
slit and forms a semicircle on the other side of the
opening
In contrast… if you throw a collection of particles (a hand full
of sand for example) at a small opening, some hit the edge of the
opening, while others go through the opening and continue in a
narrower group.
o Interference – when waves of light pass through two
adjacent slits (see above diagram), the nearby
emerging semicircular waves interact through the
process of interference
 If the waves’ crests coincide (in phase), they
interfere constructively (amplitudes add together
to form brighter region)
 If the waves’ troughs coincide (out of phase),
they interfere destructively (amplitudes cancel to
form a darker region)
In contrast… particles do not exhibit interference
The Particle Nature of Light
Three observations involving matter and light confused
physicists at the turn of the 20th century…
Blackbody Radiation and the Quantum Theory of Energy
Observation When an object is
heated to about 1000K, it begins to emit
visible light (as you can see in the photo
of the smoldering coal).
Observation  When an object is
heated to about 1500K, it begins to emit
light that is brighter and more orange
(as you can see in the photo of the
electric heating element)
Observation When an object is
heated to about 2000K, it emits light
that is still brighter and whiter (as you
can see in the light bulb filament)
Explanation In 1900, Max Planck (German physicist)
assumed that the hot glowing object could emit (or absorb) only
certain quantities of energy (E = nhv)
 E = energy of the radiation
 n = is a positive integer (1, 2, 3, and so on) quantum
number
 h = Planck’s constant (6.626x10-34 J∙s)
 v = frequency
Conclusions
 hot objects emit only certain quantities of energy and that
the energy must be emitted from the object’s atoms
o this means that each atom emits only certain quantities
of energy
 this would mean that each atom has only certain
quantities of energy to start with
 thus, the energy of an atom is quantized
o energy is emitted in fixed quantities
 Each change in an atom’s energy occurs when the atom
absorbs or emits one or more “packets”, or definite
amounts, of energy.
o Each energy packet is called a quantum (fixed
quantity or quanta)
 A quantum of energy is equal to hv
 An atom changes its energy state by
emitting (or absorbing) one or more quanta
 The energy of the emitted (or absorbed)
radiation is equal to the difference in the
atom’s energy states
o ∆Eatom = Eemitted or absorbed
o ∆E = hv
 h = Planck’s constant
 v = frequency
Photoelectric Effect and the Photon Theory of Light
Despite the idea of quantization, physicist still pictured energy
traveling in waves…
 the wave model could not explain the second confusing
observation (the flow of current when light strikes a metal)
Observation  the photoelectric
effect (when monochromatic light of
sufficient frequency shines on a
metal plate, a current flows)
Possible conclusion  the current
arises because the light transfers
energy that frees electrons from the
metal surface (this conclusions has
two confusing features)
1. Presence of a threshold… for
current to flow, the light shining on
the metal must have a minimum, or
threshold, frequency, and different
metals have different minimum frequencies. The wave
theory associates the energy of light with the amplitude
(intensity), not its frequency (color). This theory predicts
that an electron would break free when it absorbs enough
energy from light of any color.
2. Absence of a time lag…current flows the moment light of
the minimum frequency shines on the metal, regardless of
the light’s intensity. (The wave theory predicts that with
dim light there would be a time lag before the current
flows, because the electrons would have to absorb enough
energy to break free)
Explanation  the photon theory… Einstein proposed that
light itself is particulate, quantized into tiny “bundles” of
energy, called photons.
 Each atom changes its energy, ∆Eatom, when it absorbs or
emits one photon, one particle of light, whose energy is
related to its frequency, not its amplitude.
How does the photon theory explain the two features of the
photoelectric effect?
 Why is there a frequency threshold?
o A beam of light consists of an enormous number of
photons. The intensity (brightness) is related to the
number of photons, but not the energy of each.
Therefore, a photon of a certain minimum energy
must be absorbed to free an electron from the surface.
Since energy depends on frequency (hv), the theory
predicts a threshold frequency.
 Why is there no time lag?
o An electron breaks free when it absorbs a photon of
enough energy
 It cannot break free by “saving up” energy from
several photons, each having less than the
minimum energy.
 Example… ping-pong photons (if one pingpong ball doesn’t have enough energy to
knock a book of a shelf, neither does a series
of ping-pong balls, because the book can’t
store the energy from the individual impacts.
But, one baseball traveling at the same speed
does have enough energy)
 The current is weak in dim light because fewer
photons of enough energy can free fewer
electrons per unit time, but some current flows as
soon as light of sufficient energy (frequency)
strikes the metal.
Sample Problem 7.2 p. 267
Calculating the Energy of Radiation from its Wavelength
∆E = hv
Step 1
Convert wavelength to meters
1.20 cm 1 m
= 0.0120 m
100 cm
Step 2
Calculate frequency from wavelength
Speed(c) = frequency(v) x wavelength(λ)
3.00x108 m/s = frequency(v) x 0.0120 m
frequency(v) = 2.50x1010 s-1
Step 3
Calculate energy of one photon
∆E = hv
∆E = (6.626x10-34 J∙s) (2.50x1010 s-1)
∆E = 1.66x10-23 J
Follow-Up Problem 7.2 p. 267
(a)
Speed(c) = frequency(v) x wavelength(λ)
3.00x108 m/s = frequency(v) x 1x10-8 m
frequency(v) = 3x1016 s-1
∆E = hv
∆E = (6.626x10-34 J∙s) (3x1016 s-1)
∆E = 2x10-17 J
(b)
Speed(c) = frequency(v) x wavelength(λ)
3.00x108 m/s = frequency(v) x 5x10-7 m
frequency(v) = 6x1014 s-1
∆E = hv
∆E = (6.626x10-34 J∙s) (6x1014 s-1)
∆E = 4x10-19 J
(c)
Speed(c) = frequency(v) x wavelength(λ)
3.00x108 m/s = frequency(v) x 1x10-4 m
frequency(v) = 3x1012 s-1
∆E = hv
∆E = (6.626x10-34 J∙s) (3x1012 s-1)
∆E = 2x10-21 J
As wavelength increases (frequency decreases) energy
decreases
Atomic Spectra
The third confusing observation about matter and energy
involved the light emitted when an element is vaporized and
then excited electrically.
Line Spectra and the Rydberg Equation
When light from electrically excited gaseous atoms passes
through a slit and is refracted by a prism, it does not create a
continuous spectrum or rainbow, as sunlight does.
 Instead, it creates a line spectrum, a series of lines at
specific frequencies separated by black spaces.
o Each line spectrum is characteristic of the element
producing it. (Figure 7.8 below)
Features of the Rydberg Equation
Johannes Rydberg (Swedish physicist) developed a relationship,
called the Rydberg equation that predicted the position and
wavelength of any line in a given series.
 1/λ = R (1/n21 - 1/n22)
o λ = wavelength
o n1 and n2 are positive integers with n2 > n1
o R = Rydberg constant (1.096776x107 m-1)
Problems with Rutherford’s Nuclear Model
Almost as soon as Rutherford proposed his nuclear model, a
major problem arose…
 A positive nucleus and a negative electron attract each
other, and for them to stay apart, the kinetic energy of the
electron’s motion must counterbalance the potential energy
of the attraction.
o Laws of classical physics say that a negative particle
moving in a curved path around a positive particle
must emit radiation and thus lose energy.
 If electrons behave in this way, they would spiral
into the nucleus and atoms would collapse!
 Laws of classical physics would also suggest that
the emitted radiation would create a continuous
spectrum, not a line spectrum.
The Bohr Model of the Hydrogen Atom
Niels Bohr (Danish physicist) suggested a model for the
hydrogen atom that did predict the existence of line spectra…
 Postulates of the model…
o The hydrogen atom has only certain energy levels
 Bohr referred to these as stationary states
 Each state is associated with a fixed circular
orbit around the nucleus
o The higher the energy level, the further
the orbit is from the nucleus
o The atom does not radiate energy while in one of its
stationary states
 Although this violates classical physics, the atom
does not change energy while the electron moves
within an orbit
o The atom changes to another stationary state (the
electron moves to another orbit) only by absorbing or
emitting a photon. The energy of the photon (hv)
equals the difference in the energies of the two states.
 Ephoton = ∆Eatom = Efinal – Einitial = hv
Features of the
Model
Quantum numbers
and the electron
orbit… the quantum
number n is a positive
integer (1, 2, 3, …)
associated with radius
of an electron orbit,
which is directly
related to the
electron’s energy.
(the lower the n
value, the smaller the
radius of the orbit, and the lower the energy level
Ground state… when the electron is in its first orbit (n=1), it is
closest to the nucleus, and the hydrogen atom is in its lowest
energy level, called the ground state
Excited states… if the electron is in any orbit farther from the
nucleus, the atom is in its excited state. With the electron in the
second orbit (n=2), the atom is in the first excited state; when it
is in the third orbit (n=3), the atom is in its second excited state,
and so forth.
Absorption… if a hydrogen atom absorbs a photon whose
energy equals the difference between lower and higher energy
levels, the electron moves to the outer (higher energy) orbit.
Emission… if a hydrogen atom in a higher energy level
(electron in a farther orbit) returns to a lower energy level
(electron in closer orbit), the atom emits a photon whose energy
equals the difference between the two levels.
How the Model Explains Line Spectra
A spectral line results because a photon of specific energy (and
thus frequency) is emitted. The emission occurs when the
electron moves to an orbit closer to the nucleus as the atom’s
energy changes from a higher state to a lower one.
 Key point…an atomic spectrum is not continuous because
the atom’s energy is not continuous, but rather has only
certain states.
Bohr’s model accounts for three series of spectral lines of
hydrogen… When a sample of hydrogen is excited, the atoms
absorb different quantities of energy. Because there are so many
atoms in the whole sample, all the energy levels (orbits) have
electrons.
 Infrared series… when electrons drop from an outer energy
level to the n = 3
 Visible series… when electrons drop from an outer energy
level to the n = 2
 Ultraviolet series… when electrons drop from an outer
energy level to the n = 1
Limitations of the Model (Bohr’s)
Although it fails to predict the spectrum of any other atom
(mainly because hydrogen has only one electron and we now
know that electrons do not move in fixed, defined orbits), we
still use the terms ground state and excited state and retain the
central idea that the energy of an atom occurs in discrete levels
and it changes energy by absorbing or emitting a photon of
specific energy.
The Energy Levels of the Hydrogen Atom
Applying Bohr’s Equation for the Energy Levels of an Atom
Finding the difference in energy between two levels…
∆E = Efinal – Einitial = -2.18x10-18 J (1/n2final – 1/n2initial)
 When the atom emits energy… the electron moves closer to
the nucleus (nfinal < ninitial) and the atom’s final energy is a
larger negative number and ∆E is negative.
 When the atom absorbs energy…the electron moves away
from the nucleus (nfinal > ninitial) and the atom’s final energy
is a smaller negative number and ∆E is positive.
Finding the energy needed to ionize the H atom…
H(g)  H+(g) + e-
∆E = Efinal – Einitial = -2.18x10-18 J (1/n2final – 1/n2initial)
ninitial = 1 and nfinal = 0
so…
∆E = Efinal – Einitial = -2.18x10-18 J (1/02final – 1/12initial)
∆E = -2.18x10-18 J (0 – 1)
∆E = 2.18x10-18 J
Energy must be absorbed to remove an electron, so ∆E must be
positive (which it is in the above example)
Finding the wavelength of a spectral line…
∆E = hv (remember that c = v x λ , so v = c/λ)
∆E = hc/λ
λ = hc/∆E
Sample Problem 7.3 p. 272
Determining ∆E and λ of an Electron Transition
(a)
∆E = -2.18x10-18 J (1/02final – 1/12initial)
∆E = -2.18x10-18 J (1/42final – 1/12initial)
∆E = -2.18x10-18 J (1/16– 1/1)
∆E = 2.04x10-18 J
(b)
∆E = hc/λ
λ = hc/∆E
λ = (6.626x10-34J∙s)(3.00x108 m/s) / 2.04x10-18 J
λ = 9.74x10-8 m
9.74x10-8 m 1x109 nm = 97.4 nm
1m
Follow-Up Problem 7.3 p. 272
(a)
∆E = -2.18x10-18 J (1/n2final – 1/n2initial)
∆E = -2.18x10-18 J (1/32 – 1/62)
∆E = -2.18x10-18 J (1/9 – 1/36)
∆E = -1.82x10-19 J
(b)
∆E = hc/λ
λ = hc/∆E
λ = (6.626x10-34J∙s)(3.00x108 m/s) / 1.82x10-19 J
λ = 1.09x10-6 m
1.09x10-6 m 1x1010 A = 1.09x104 A
1m
The Wave-Particle Duality of Matter and Energy
One of the results of Einstein’s work was his discovery that
matter and energy are alternate forms of the same entity…
 This is embodied in his famous E = mc2 which relates the
quantity of energy equivalent to a given mass
 Results that showed energy to be particle like had to coexist with others that showed matter to be wavelike
The Wave Nature of Electrons and the Particle Nature of Photons
Louis de Broglie – French physics student
 Proposed that if energy is particle-like, perhaps matter is
wavelike
o He reasoned that if electrons have wavelike motion in
orbits of fixed radii, they would have only certain
allowable frequency and energies
 Example… because the end of a guitar string is
fixed, only certain vibrational frequencies (and
wavelengths) are allowable to create a note.
 See figure 7.12 below
o Combining the equations for mass-energy equivalence
(E = mc2) and energy of a photon (E = hv = hc/λ)
derived an equation for the wavelength of any particle
of mass m
 λ = h / mu
 λ = wavelength
 h = 6.626x10-34 kg∙m2/s
 m = mass in kg
 u = speed in m/s
 According to this equation, matter behaves as
though it moves in a wave
 An object’s wavelength is inversely
proportional to its mass
Sample Problem 7.4 p. 276
Calculating the de Broglie Wavelength of an Electron
λ = h / mu
λ = 6.626x10-34 kg∙m2/s / (9.11x10-31kg)(1.00x106 m/s)
λ = 7.27x10-10 m
Follow-Up Problem 7.4 p. 276
100. nm 1 m
= 1x10-7m
1x109 nm
λ = h / mu
1x10-7 m = (6.626x10-34 kg∙m2/s) / (9.11x10-31 kg)(u)
(1x10-7 m) (9.11x10-31 kg)(u) = (6.626x10-34 kg∙m2/s)
u = 7.27x103 m/s
So…
If electrons travel in waves, they should exhibit diffraction and
interference. A fast-moving electron has a wavelength of about
1x10-10 m, so a beam of such electrons should be diffracted by
the spaces between atoms in a crystal (about 10-10 m)
 C. Davidson and L. Germer guided a beam of x-rays and
then a beam of electrons at a nickel crystal and obtained
two diffraction patterns (thus particles with mas and charge
create diffraction patterns, just as electromagnetic waves
do)
o The electron microscope is a major application of this
understanding
Electron micrograph of blood cells
(x1200)
The Particle Nature of Photons
The de Broglie wave wavelength equation suggests that we can
calculate the momentum (p), the product of mass and speed, for
a photon.
λ = h / mc = h / p
The inverse relationship between p and λ means that shorter
wavelength (higher energy) photons have greater momentum.
Wave-Particle Duality
Classical experiments had shown matter to be particle-like and
energy to be wavelike. Results on the atomic scale show
electrons moving in waves and photons having momentum.
Thus, every property of matter was also a property of energy.
 The truth is both matter and energy show both behaviors
o Matter is both wave-like and particle like
o Energy is both wave-like and particle like
Heisenberg’s Uncertainty Principle
In classical physics, a moving particle has a definite location at
any instant, whereas a wave is spread out in space. If an
electron has the properties of both a particle and wave, can we
determine its position in the atom?
 In 1927, Werner Heisenberg (German physicist) postulated
the uncertainty principle
o It is impossible to know simultaneously the position
and momentum of a particle.
The Quantum-Mechanical Model of the Atom
Acceptance of the dual nature of matter and energy (both are
wave-like as well as particle like) and the uncertainty principle
culminated in the field of quantum mechanics
 Examines the wave nature of objects on the atomic scale
 Erwin Schrӧdinger (1926) derived an equation that is the
basis for the quantum-mechanical model of the hydrogen
atom
o The model describes quantities of energy that result
from allowed frequencies of its electron’s wavelike
motion
 The electron’s position can only be known within
a certain probability
The Atomic Orbital and the Probable Location of the Electron
Two central aspects…
The Schrӧdinger Equation and the Atomic Orbital
The electron’s matter-wave occupies the space near the nucleus
and is continuously influenced by it.
Hψ = Eψ
ψ = (Greek psi, pronounced “sigh”) is a wave function (or
atomic orbital), a mathematical description of the electron’s
matter-wave in three dimension
H = Hamiltonian operator (represents a set of mathematical
operations that, when carried out with a particular ψ, yields one
of the allowed energy states of the atom
Each solution of the equation gives an energy state associated
with a given atomic orbital
Important point…
 An “orbital” in the quantum-mechanical model bears no
resemblance to an “orbit” in the Bohr model
o An orbit is an electron’s actual path around the
nucleus
o An orbital is a mathematical function that describes
the electron’s matter-wave but has no physical
meaning
The Probable Location of the Electron
While we cannot know exactly where the electron is at any
moment, we know where it probably is (where it spends most of
its time)
 We can determine this by squaring the wave function ψ2
o This is called the probability density
 A probability of finding the electron in some tiny
volume of the atom
We depict the electron’s probable location in several ways,
which we will look at first of the hydrogen atom’s ground state
Probability of the electron being in some tiny volume of the
atom…
 For each energy level, we can create an electron
probability density diagram, or more simply an electron
density diagram
o The value of ψ2 for a given volume is shown with
dots: the greater the density of dots, the higher
probability of finding the electron in that volume
 For the ground state of hydrogen, the electron
probability density decrease with distance from
the nucleus along a line, r
 These diagrams are also called electron cloud depictions
because, if we could take a time-exposure photograph of
the electron’s wavelike motion around the nucleus, it would
appear as a “cloud” of positions.
o The electron cloud is an imaginary picture of the
electron changing its position rapidly over time; it
does not mean that an electron is a diffuse cloud of
charge
Total probability density at some distance from the nucleus…
 To find the radical probability distribution (the total
probability of finding the electron at some distance r from
the nucleus, we first mentally divide the volume around the
nucleus into thin, concentric, spherical layers, like the
layers of an onion (Figure 7.16 above)
o The falloff in probability density with distance has
important effect…
 Near the nucleus, the volume of each layer
increases faster than its density of dots decreases
 The result…
o The total probability peaks in a layer
near, but not at, the nucleus
Probability contour and the size of the atom…
 We can visualize an atom with a 90% probability contour
o The electron is somewhere within that volume 90% of
the time (Figure 7.16 E above)
Quantum Numbers of an Atomic Orbital
An atomic orbital is specified by three quantum numbers…
 Are part of the solution of the Schrӧdinger equation
 Indicate the size, shape, and orientation in space
Principle quantum number (n)
 Positive integer (1, 2, 3, and so forth)
 It indicates the relative size of the orbital (the relative
distance from the nucleus )
 Specifies the energy level of the hydrogen atom
 The higher the n value, the higher the energy level
 When the electron occupies an orbital n = 1, the hydrogen
atom is in its ground state
o When the electron occupies an orbital with n = 2 (first
excited state), the atom has more energy
Angular momentum quantum number (l)
 Is an integer from 0 to n-1
 Is related to the shape of the orbital
 The principle quantum number sets a limit on the angular
momentum number
o n = 1, l can only equal 0
o n = 2, l can equal 0 or 1
o n = 3, l can equal 0, 1, or 2
Magnetic Quantum Number (ml)
 Is an integer from -1 through 0 to +1
 Prescribes the three dimensional orientation of the orbital in
the space around the nucleus
 l limits ml
o l = 0 can only have ml = 0
o l = 1 can have three ml values, -1, 0, +1
 The number of ml values equals 2l + 1
 The total number or orbitals, for a given n value
is n2
Sample Problem 7.6 p. 282
Determining Quantum Numbers for an Energy Level
n = 3, l = 0, 1, 2
l = 0, ml = 0
l = 1, ml = -1, 0, +1
l = 2, m2 = -2, -1, 0, +1, +2
There are nine ml values, so there are nine (9) orbitals with n = 3
Follow-Up Problem 7.6 p. 282
n = 4, l = 0, 1, 2, 3
l = 0, ml = 0
l = 1, ml = -1, 0, +1
l = 2, ml = -2, -1, 0, +1, +2
l = 3, ml = -3, -2, -1, 0, +1, +2, +3
Quantum Numbers and Energy Levels
Level
 The atom’s energy levels, or shells, are given by the n
value
o The smaller the n value, the lower the energy level
and the greater the probability that the electron is
closer to the nucleus
Sublevel
 The atom’s levels are divided into sublevels, or subshells,
that are given by the l value. Each designates the orbital
shape with a letter
o l = 0 is the s sublevel
o l = 1 is the p sublevel
o l = 2 is the d sublevel
o l = 3 is the f sublevel
 sublevels with l values greater than 3 are
designated by consecutive letters after f: g
sublevel, h sublevel, and so on…
 A sublevel is named with its n value and letter designation
o n = 2 and l = 0 is called 2s
Orbital
Each combination of n, l, and ml specifies the size (energy),
shape, and spatial orientation of one of the atom’s orbitals.
Sample Problem 7.7
Determining Sublevel Names and Orbital Quantum Numbers
(a)
n = 3, l = 2
3d
-2, -1, 0, +1, +2
5 orbitals
2s
0
1 orbital
5p
-1, 0, +1
3 orbitals
(b)
n = 2, l = 0
(c)
n = 5, l = 1
(d)
n = 4, l = 3
4f
-3,-2, -1, 0, +1, +2, +3
7 orbitals
Follow-Up Problem 7.7 p. 283
2p n = 2, l = 1, ml = -1, 0, +1
5f
n = 5, l = 3, ml = -3, -2, -1, 0, +1, +2, +3
Sample Problem 7.8 p. 283
Identifying Incorrect Quantum Numbers
(a)
(b)
(c)
n
1
4
3
l
1
3
1
ml
0
+1
-2
name
1p
4d
3p
(a) A sublevel with n = 1 can have only l = 0, not 1 = 1 The
only possible sublevel name would be 1s
(b) A sublevel with l = 3 is an f sublevel. The name should
be 4f
(c) A sublevel with 1 = 1 can only have -1, 0, or +1 for ml
Follow-Up Problem 7.8 p. 283
(a)
(b)
(c)
(d)
n
4
2
3
2
l
1
1
2
0
ml
0
0
-2
0
name
4p
2p
3d
2s