Quantum Theory and the Electronic
Structure of Atoms
Chapter 7
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Properties of Waves
Wavelength () is the distance between identical points on
successive waves.
Amplitude is the vertical distance from the midline of a
wave to the peak or trough.
Frequency () is the number of waves that pass through a
particular point in 1 second (Hz = 1 cycle/s).
The speed (u) of the wave = x 2
Maxwell (1873), proposed that visible light consists of
electromagnetic waves.
Electromagnetic
radiation is the emission
and transmission of energy
in the form of
electromagnetic waves.
Speed of light (c) in vacuum = 3.00 x 108 m/s
All electromagnetic radiation
x c 3
4
A photon has a frequency of 6.0 x 104 Hz. Convert this
frequency into wavelength (nm). Does this frequency fall in
the visible region?
x=c
= c/
= 3.00 x 108 m/s / 6.0 x 104 Hz
= 5.0 x 103 m
= 5.0 x 1012 nm
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Mystery #1, “Heated Solids Problem”
Solved by Planck in 1900
When solids are heated, they emit electromagnetic radiation
over a wide range of wavelengths.
Radiant energy emitted by an object at a certain temperature
depends on its wavelength.
Energy (light) is emitted or
absorbed in discrete units
(quantum).
E=hx
Planck’s constant (h)
h = 6.63 x 10-34 J•s 6
Mystery #2, “Photoelectric Effect”
Solved by Einstein in 1905 h
Light has both:
KE e-
1. wave nature
2. particle nature
Photon is a “particle” of light
h = KE + W
KE = h - W
where W is the work function and
depends how strongly electrons
are held in the metal 7
When copper is bombarded with high-energy electrons, X rays
are emitted. Calculate the energy (in joules) associated with
the photons if the wavelength of the X rays is 0.154 nm.
E=hx
E = h x c /
E = 6.63 x 10-34 (J•s) x 3.00 x 10 8 (m/s) / 0.154 x 10-9 (m)
E = 1.29 x 10 -15 J
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Line Emission Spectrum of Hydrogen Atoms
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Bohr’s Model of
the Atom (1913)
1. e- can only have specific
(quantized) energy
values
2. light is emitted as e-
moves from one energy
level to a lower energy
level
1
En = -RH ( )
n2
n (principal quantum number) = 1,2,3,…
RH (Rydberg constant) = 2.18 x 10-18J
11
E = h
E = h
12
Ephoton = E = Ef - Ei
ni = 3 ni = 3 1
Ef = -RH ( 2 )
nf
ni = 2 1
Ei = -RH ( 2 )
nf = 2 ni
1 1
E = RH( 2 )
ni n2f
nnf f==11
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Calculate the wavelength (in nm) of a photon emitted
by a hydrogen atom when its electron drops from the
n = 5 state to the n = 3 state.
1 1
Ephoton = E = RH( )
n2i n2f
Ephoton = 2.18 x 10-18 J x (1/25 - 1/9)
Ephoton = E = -1.55 x 10-19 J
Ephoton = h x c /
= h x c / Ephoton
= 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J
= 1280 nm
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Why is e- energy quantized?
De Broglie (1924) reasoned
that e- is both particle and
wave.
h
2r = n = mu
u = velocity of e-
m = mass of e-
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What is the de Broglie wavelength (in nm) associated
with a 2.5 g Ping-Pong ball traveling at 15.6 m/s?
= h/mu h in J•s m in kg u in (m/s)
= 6.63 x 10-34 / (2.5 x 10-3 x 15.6)
= 1.7 x 10-32 m = 1.7 x 10-23 nm
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Chemistry in Action: Laser – The Splendid Light
Laser light is (1) intense, (2) monoenergetic, and (3) coherent
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Chemistry in Action: Electron Microscopy
e = 0.004 nm STM image of iron atoms
on copper surface
Electron micrograph of a normal
red blood cell and a sickled red
blood cell from the same person
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Schrodinger Wave Equation
In 1926 Schrodinger wrote an equation that
described both the particle and wave nature of the e-
Wave function () describes:
1. energy of e- with a given
2. probability of finding e- in a volume of space
Schrodinger’s equation can only be solved exactly
for the hydrogen atom. Must approximate its
solution for multi-electron systems.
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Schrodinger Wave Equation
is a function of four numbers called
quantum numbers (n, l, ml, ms)
principal quantum number n
n = 1, 2, 3, 4, ….
distance of e- from the nucleus
n=1 n=2 n=3
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Where 90% of the
e- density is found
for the 1s orbital
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Schrodinger Wave Equation
quantum numbers: (n, l, ml, ms)
angular momentum quantum number l
for a given value of n, l = 0, 1, 2, 3, … n-1
l=0 s orbital
n = 1, l = 0
l=1 p orbital
n = 2, l = 0 or 1
l=2 d orbital
n = 3, l = 0, 1, or 2
l=3 f orbital
Shape of the “volume” of space that the e- occupies
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l = 0 (s orbitals)
l = 1 (p orbitals)
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l = 2 (d orbitals)
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Schrodinger Wave Equation
quantum numbers: (n, l, ml, ms)
magnetic quantum number ml
for a given value of l
ml = -l, …., 0, …. +l
if l = 1 (p orbital), ml = -1, 0, or 1
if l = 2 (d orbital), ml = -2, -1, 0, 1, or 2
orientation of the orbital in space
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ml = -1, 0, or 1 3 orientations is space
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ml = -2, -1, 0, 1, or 2 5 orientations is space
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Schrodinger Wave Equation
(n, l, ml, ms)
spin quantum number ms
ms = +½ or -½
ms = +½ ms = -½
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Schrodinger Wave Equation
quantum numbers: (n, l, ml, ms)
Existence (and energy) of electron in atom is described
by its unique wave function .
Pauli exclusion principle - no two electrons in an atom
can have the same four quantum numbers.
Each seat is uniquely identified (E, R12, S8)
Each seat can hold only one individual at a
time
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Schrodinger Wave Equation
quantum numbers: (n, l, ml, ms)
Shell – electrons with the same value of n
Subshell – electrons with the same values of n and l
Orbital – electrons with the same values of n, l, and ml
How many electrons can an orbital hold?
If n, l, and ml are fixed, then ms = ½ or - ½
= (n, l, ml, ½) or= (n, l, ml, -½)
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An orbital can hold 2 electrons
How many 2p orbitals are there in an atom?
n=2
If l = 1, then ml = -1, 0, or +1
2p
3 orbitals
l=1
How many electrons can be placed in the 3d subshell?
n=3 If l = 2, then ml = -2, -1, 0, +1, or +2
3d 5 orbitals which can hold a total of 10 e-
l=2 33
Energy of orbitals in a single electron atom
Energy only depends on principal quantum number n
n=3
n=2
1
En = -RH ( )
n2
n=1
34
Energy of orbitals in a multi-electron atom
Energy depends on n and l
n=3 l = 2
n=3 l = 1
n=3 l = 0
n=2 l = 1
n=2 l = 0
n=1 l = 0
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“Fill up” electrons in lowest energy orbitals (Aufbau principle)
??
Be
Li
B5
C 3
64electrons
electrons
BBeLi1s1s
1s
2s2s
2s2p
2 22 2 12 1
H
He12electron
electrons
He
H 1s
1s12 36
The most stable arrangement of electrons in
subshells is the one with the greatest number of
parallel spins (Hund’s rule).
F
O
C 97
N
Ne 6
810
electrons
electrons
electrons
Ne
C
N
O
F 1s 1s222s
2s2p2p3246
2 222 5
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Order of orbitals (filling) in multi-electron atom
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s
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Electron configuration is how the electrons are
distributed among the various atomic orbitals in an
atom.
number of electrons
in the orbital or subshell
1s1
principal quantum angular momentum
number n quantum number l
Orbital diagram
H
1s1
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What is the electron configuration of Mg?
Mg 12 electrons
1s < 2s < 2p < 3s < 3p < 4s
1s22s22p63s2 2 + 2 + 6 + 2 = 12 electrons
Abbreviated as [Ne]3s2 [Ne] 1s22s22p6
What are the possible quantum numbers for the last
(outermost) electron in Cl?
Cl 17 electrons 1s < 2s < 2p < 3s < 3p < 4s
1s22s22p63s23p5 2 + 2 + 6 + 2 + 5 = 17 electrons
Last electron added to 3p orbital
n=3 l=1 ml = -1, 0, or +1 ms = ½ or -½ 40
Outermost subshell being filled with electrons
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42
Paramagnetic Diamagnetic
unpaired electrons all electrons paired
2p 2p 43
Practice Problems
What is the wavelength (in meters) of an
electromagnetic wave whose frequency is 3.64 x 107
Hz?
Answer: 8.24 m
44
Practice Problems
Calculate the energy (in joules) of a photon with a
wavelength of 5.00 x 104 nm (infrared region)
Answer: 3.98 x 10-21 J
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Practice Problems
The work function of cesium metal is 3.42 x 10-19 J.
(a) Calculate the minimum frequency of light required
to release electrons from the metal.
(b) Calculate the kinetic energy of the ejected electron
if light of frequency 1.00 x 1015 s-1 is used for
irradiating the metal.
Answer: a. 5.16 x 1014 s-1 b. 3.21 x 10-19 J
46
Practice Problems
What is the wavelength of a photon (in nanometers)
emitted during a transition from the ni = 5 state to the
nf = 2 state in the hydrogen atom?
Answer: 434 nm
47
Practice Problems
Calculate the wavelength of the “particle” in the
following case: The fastest serve in tennis is about
150 miles per hour, or 68 m/s. Calculate the
wavelength associated with a 6.0 x 10-2 kg tennis ball
traveling at this speed.
Answer: 1.6 x 10-34 m
48
Practice Problems
List the values of n , /, and m/ for orbitals in the 4d
subshell.
Answer: n = 4, l = 2, ml = -2, -1, 0, 1, 2
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Practice Problems
What is the total number of orbitals associated with
the principal quantum number n = 3?
Answer: 9 orbitals
50
Practice Problems
Write the four quantum numbers for an electron in a
3p orbital.
Answer: n = 3, l = 1, ml = -1, 0, 1, mp = +1/2, -1/2
51
Practice Problems
What is the maximum number of electrons that can be
present in the principal level for which n = 3?
Answer: 18 electrons
52
Practice Problems
An oxygen atom has a total of eight electrons. Write
the four quantum numbers for the last electron.
Answer: n = 2, l = 1, ml = 0
53
Practice Problems
Write the four quantum numbers for the last electron
of Einsteinium.
Answer: n = 2, l = 1, ml = 0
54
Questions
Laser is an acronym for?
Answer: Light Amplification by Stimulated
Emission of Radiation
55
Questions
He discovered that atoms and molecules emit energy
only in certain discrete quantities, called quanta.
Answer: Max Planck
56
Questions
He stated that an electromagnetic wave has an
electric field component and a magnetic field
component. Who is this scientist?
Answer: James Clerk Maxwell
57
Questions
A phenomenon in which electrons are ejected from
the surface of certain metals exposed to light of at
least a certain minimum frequency, called the
threshold frequency.
Answer: Photoelectric Effect
58
Questions
A phenomenon in which electrons are ejected from
the surface of certain metals exposed to light of at
least a certain minimum frequency, called the
threshold frequency.
Answer: Photoelectric Effect
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