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Title: Chapter 7: Quantum Theory and


1
Chapter 7 Quantum Theory and
Atomic Structure
7.1 The Nature of Light 7.2 Atomic Spectra 7.3
The Wave - Particle Duality of Matter and
Energy 7.4 The Quantum - Mechanical Model of the
Atom
2
Electromagnetic Radiation
  • WAVELENGTH - The distance between identical
    points on successive waves. ( ? )
  • FREQUENCY - The number of waves that pass through
    a particular point per second. (?)
  • AMPLITUDE - The vertical distance from the
    midline to a peak, or trough in the wave.

c v
???
3
Fig. 7.1
4
Fig. 7.2
5
Fig. 7.3
6
The Spectrum of Electromagnetic Radiation
  • The wavelength of visible light is between 400
    and 700 nanometers
  • Radio, TV , microwave and infrared radiation have
    longer wavelengths (shorter frequencies), and
    lower energies than visible light.
  • Gamma rays and X-rays have shorter wavelengths
    (larger frequencies), and higher energies than
    visible light!

7
Calculation of Frequency from Wavelength
Problem The wavelength of an x-ray is 1.00 x10
-9 m or 1 nm, what is the frequency of this
x-ray? If the wavelength of long-wavelength
electromagnetic radiation is 7.65 x 104 m, what
is the frequency of this long-wavelength
radiation used to contact submerged nuclear
submarines at sea? Plan Use the relationship
between wavelength and frequency to obtain the
answer. wavelength x frequency speed of
light! Solution
speed of light wavelength(m)
frequency(cycles/sec)


frequency
a)

b)
frequency
8
Calculation of Frequency from Wavelength
Problem The wavelength of an x-ray is 1.00 x10
-9 m or 1 nm, what is the frequency of this
x-ray? If the wavelength of long-wavelength
electromagnetic radiation is 7.65 x 104 m, what
is the frequency of this long-wavelength
radiation used to contact submerged nuclear
submarines at sea? Plan Use the relationship
between wavelength and frequency to obtain the
answer. wavelength x frequency speed of
light! Solution
speed of light wavelength(m)
frequency(cycles/sec)

3.00 x 108 m/s 1.00 x 10 - 9 m
frequency 3.00 x
1017 cycles/sec
a)

b)
frequency
9
Calculation of Frequency from Wavelength
Problem The wavelength of an x-ray is 1.00 x10
-9 m or 1 nm, what is the frequency of this
x-ray? If the wavelength of long-wavelength
electromagnetic radiation is 7.65 x 104 m, what
is the frequency of this long-wavelength
radiation used to contact submerged nuclear
submarines at sea? Plan Use the relationship
between wavelength and frequency to obtain the
answer. wavelength x frequency speed of
light! Solution
speed of light wavelength(m)
frequency(cycles/sec)

3.00 x 108 m/s 1.00 x 10 - 9 m
frequency 3.00 x
1017 cycles/sec
a)
3.00 x 108 m/s 7.65 x 104 m
b)
frequency 3.92 x
103 cycles/s
10
The Photoelectric Effect - I
  • Below the threshold energy, nothing occurs !
  • Above the threshold, the kinetic energy of the
    ejected electrons is proportional to the
    frequency of the light.
  • Also, when above the threshold, as intensity of
    the light increases, so does the number of
    ejected electrons.
  • All metals experience this effect, but each has a
    unique threshold frequency.

11
The Photoelectric Effect - II
  • Albert Einstein
  • Theorized Photons
  • Won Nobel prize - 1921
  • Photons have an energy equal to
  • E h?
  • h Planks Constant, and is equal to
  • 6.6260755 x 10 - 3 4Jsec

12
Demonstration of the Photoelectric Effect
Fig. 7.7
13
Photons and the Photoelectric Effect
  • Einstein also stated that the change in the
    photons energy was equal to the ejected
    electrons energy.
  • Therefore, the photons energy equaled the
    electrons kinetic energy added to the electrons
    binding energy
  • ETotal Electron binding Electron Kinetic
    energy

14
Calculation of Energy from Frequency
Problem What is the energy of a photon of
electromagnetic radiation being emitted by radio
station KBSG 97.3 FM ( 97.3 x 108
cycles/sec)? What is the energy of a gamma ray
emitted by Cs137 if it has a frequency of 1.60 x
1020/s? Plan Use the relationship between
energy and frequency to obtain the energy of the
electromagnetic radiation (E hv). Solution
EKBSG hv \
\
15
Calculation of Energy from Frequency
Problem What is the energy of a photon of
electromagnetic radiation being emitted by radio
station KBSG 97.3 FM ( 97.3 x 108
cycles/sec)? What is the energy of a gamma ray
emitted by Cs137 if it has a frequency of 1.60 x
1020/s? Plan Use the relationship between
energy and frequency to obtain the energy of the
electromagnetic radiation (E hv). Solution
EKBSG hv (6.626 x 10 -34Js)(9.73 x 109/s)
6.447098 x 10 -24J
EKBSG 6.45 x 10 - 24 J
16
Calculation of Energy from Frequency
Problem What is the energy of a photon of
electromagnetic radiation being emitted by radio
station KBSG 97.3 FM ( 97.3 x 108
cycles/sec)? What is the energy of a gamma ray
emitted by Cs137 if it has a frequency of 1.60 x
1020/s? Plan Use the relationship between
energy and frequency to obtain the energy of the
electromagnetic radiation (E hv). Solution
EKBSG hv (6.626 x 10 -34Js)(9.73 x 109/s)
6.447098 x 10 -24J
EKBSG 6.45 x 10 - 24 J
Egamma ray hv
17
Calculation of Energy from Frequency
Problem What is the energy of a photon of
electromagnetic radiation being emitted by radio
station KBSG 97.3 FM ( 97.3 x 108
cycles/sec)? What is the energy of a gamma ray
emitted by Cs137 if it has a frequency of 1.60 x
1020/s? Plan Use the relationship between
energy and frequency to obtain the energy of the
electromagnetic radiation (E hv). Solution
EKBSG hv (6.626 x 10 -34Js)(9.73 x 109/s)
6.447098 x 10 -24J
EKBSG 6.45 x 10 - 24 J
Egamma ray hv ( 6.626 x 10-34Js )( 1.60 x
1020/s ) 1.06 x 10 -13J
Egamma ray 1.06 x 10 - 13J
18
Calculation of Energy from Wavelength
Problem What is the energy of a photon of
electromagnetic radiation that is used in
microwave ovens to cook things by rotating water
molecules, the wavelength of the radiation is
122 mm. Plan Convert the wavelength into
meters, then the frequency can be calculated
using the relationshipwavelength x frequency C
(where C is the speed of light), then using Ehv
to calculate the energy. Solution
wavelength 122 mm 1.22 x 10 -1m
19
Calculation of Energy from Wavelength
Problem What is the energy of a photon of
electromagnetic radiation that is used in
microwave ovens to cook things by rotating water
molecules, the wavelength of the radiation is
122 mm. Plan Convert the wavelength into
meters, then the frequency can be calculated
using the relationshipwavelength x frequency C
(where C is the speed of light), then using Ehv
to calculate the energy. Solution
wavelength 122 mm 1.22 x 10 -1m
3.00 x108 m/s 1.22 x 10 -1m
C wavelength
frequency
2.46 x 1010 cycles/s
20
Calculation of Energy from Wavelength
Problem What is the energy of a photon of
electromagnetic radiation that is used in
microwave ovens to cook things by rotating water
molecules, the wavelength of the radiation is
122 mm. Plan Convert the wavelength into
meters, then the frequency can be calculated
using the relationshipwavelength x frequency C
(where C is the speed of light), then using Ehv
to calculate the energy. Solution
wavelength 122 mm 1.22 x 10 -1m
3.00 x108 m/s 1.22 x 10 -1m
C wavelength
frequency
2.46 x 1010 cycles/s
Energy E hv (6.626 x 10 -34Js)(2.46 x
1010/s) 1.63 x 10 - 24 J
21
Photoelectric Effect -I
  • Mininum energy to remove an electron from
    potassium metal is 3.7 x 10 -19J. Will photons of
    frequencies of 4.3 x 1014/s (red light) and 7.5 x
    1014 /s (blue light) trigger the photoelectric
    effect?
  • E red
  • E blue

22
Photoelectric Effect -I
  • Mininum energy to remove an electron from
    potassium metal is 3.7 x 10 -19J. Will photons of
    frequencies of 4.3 x 1014/s (red light) and 7.5 x
    1014 /s (blue light) trigger the photoelectric
    effect?
  • E red hv (6.626 x10 - 34Js)(4.3 x1014 /s)
  • E red 2.8 x 10 - 19 J
  • E blue

23
Photoelectric Effect -I
  • Mininum energy to remove an electron from
    potassium metal is 3.7 x 10 -19J. Will photons of
    frequencies of 4.3 x 1014/s (red light) and 7.5 x
    1014 /s (blue light) trigger the photoelectric
    effect?
  • E red hv (6.626 x10 - 34Js)(4.3 x1014 /s)
  • E red 2.8 x 10 - 19 J
  • E blue hv (6.626 x10 - 34Js)(7.5x1014 /s)
  • E blue 5.0 x 10 - 19 J

24
Photoelectric Effect - II
  • Since the binding energy of potassium is
    3.7 x 10 - 19 J
  • The red light will not have enough energy to
    knock an electron out of the potassium, but the
    blue light will eject an electron !

25
Light and Atoms
  • When an atom gains a photon, it enters an excited
    state.
  • This state has too much energy - the atom must
    lose it and return back down to its ground state,
    the most stable state for the atom.
  • An energy level diagram is used to represent
    these changes.

26
Energy Level Diagram
  • Energy
  • Excited States
  • photons path
  • Ground State

Light Emission Light Emission
Light Emission
27
The Line Spectra of Several Elements
Fig. 7.8
28
Three Series of Spectral Lines of Atomic
Hydrogen
Fig. 7.9
29
Fig. 7.10
30
A desktop analogy for the H atoms energy
Fig. 7.11
31
Emission and Absorption Spectraof Sodium Atoms
Fig. 7.B
32
Fig. 7.C
33
Fig. 7.D
34
Emission Energetics - I
Problem A sodium vapor light street light emits
bright yellow light of wavelength 589 nm. What
is the energy change for a sodium atom involved
in this emission? How much energy is emitted per
mole of sodium atoms? Plan Calculate the energy
of the photon from the wavelength, then
calculate the energy per mole of
photons. Solution
h x c wavelength
Ephoton hv

35
Emission Energetics - I
Problem A sodium vapor light street light emits
bright yellow light of wavelength 589 nm. What
is the energy change for a sodium atom involved
in this emission? How much energy is emitted per
mole of sodium atoms? Plan Calculate the energy
of the photon from the wavelength, then
calculate the energy per mole of
photons. Solution
( 6.626 x 10 -34J s)( 3.00 x 10 8m/s)
h x c wavelength
Ephoton hv

589 x 10 -9m
Ephoton 3.37 x 10 -19J
36
Emission Energetics - I
Problem A sodium vapor light street light emits
bright yellow light of wavelength 589 nm. What
is the energy change for a sodium atom involved
in this emission? How much energy is emitted per
mole of sodium atoms? Plan Calculate the energy
of the photon from the wavelength, then
calculate the energy per mole of
photons. Solution
( 6.626 x 10 -34J s)( 3.00 x 10 8m/s)
h x c wavelength
Ephoton hv

589 x 10 -9m
Ephoton 3.37 x 10 -19J
Energy per mole requires that we multiply by
____________.
37
Emission Energetics - I
Problem A sodium vapor light street light emits
bright yellow light of wavelength 589 nm. What
is the energy change for a sodium atom involved
in this emission? How much energy is emitted per
mole of sodium atoms? Plan Calculate the energy
of the photon from the wavelength, then
calculate the energy per mole of
photons. Solution
( 6.626 x 10 -34J s)( 3.00 x 10 8m/s)
h x c wavelength
Ephoton hv

589 x 10 -9m
Ephoton 3.37 x 10 -19J
Energy per mole requires that we multiply by
Avogadros number.
Emole 3.37 x 10 -19J/atom x 6.022 x 1023
atoms/mole 2.03 x 105 J/mol
Emole 203 kJ / mol
38
Using the Rydberg Equation
Problem Find the energy change when an electron
changes from the n4 level to the n2 level in
the hydrogen atom? What is the wavelength of this
photon? Plan Use the Rydberg equation to
calculate the energy change, then calculate the
wavelength using the relationship of the speed of
light. Solution
Ephoton -2.18 x10 -18J -

Ephoton

39
Using the Rydberg Equation
Problem Find the energy change when an electron
changes from the n4 level to the n2 level in
the hydrogen atom? What is the wavelength of this
photon? Plan Use the Rydberg equation to
calculate the energy change, then calculate the
wavelength using the relationship of the speed of
light. Solution
Ephoton -2.18 x10 -18J -

Ephoton -2.18 x 10 -18J -


wavelength

40
Using the Rydberg Equation
Problem Find the energy change when an electron
changes from the n4 level to the n2 level in
the hydrogen atom? What is the wavelength of this
photon? Plan Use the Rydberg equation to
calculate the energy change, then calculate the
wavelength using the relationship of the speed of
light. Solution
Ephoton -2.18 x10 -18J -

Ephoton -2.18 x 10 -18J -
- 4.09 x 10 -19J

wavelength

41
Using the Rydberg Equation
Problem Find the energy change when an electron
changes from the n4 level to the n2 level in
the hydrogen atom? What is the wavelength of this
photon? Plan Use the Rydberg equation to
calculate the energy change, then calculate the
wavelength using the relationship of the speed of
light. Solution
Ephoton -2.18 x10 -18J -

Ephoton -2.18 x 10 -18J -
- 4.09 x 10 -19J
h x c E
wavelength

42
Using the Rydberg Equation
Problem Find the energy change when an electron
changes from the n4 level to the n2 level in
the hydrogen atom? What is the wavelength of this
photon? Plan Use the Rydberg equation to
calculate the energy change, then calculate the
wavelength using the relationship of the speed of
light. Solution
Ephoton -2.18 x10 -18J -

Ephoton -2.18 x 10 -18J -
- 4.09 x 10 -19J
h x c E
(6.626 x 10 -34Js)( 3.00 x 108 m/s)
wavelength

4.09 x 10 -19J
wavelength 4.87 x 10 -7 m 487 nm
43
Wave Motion in Restricted Systems
Fig. 7.12
44
The de Broglie Wavelengths of Several Objects
Substance Mass (g) Speed (m/s)
? (m)
Slow electron 9 x 10 - 28
1.0 7 x 10 - 4 Fast
electron 9 x 10 - 28 5.9
x 106 1 x 10 -10 Alpha particle
6.6 x 10 - 24 1.5 x 107
7 x 10 -15 One-gram mass 1.0
0.01
7 x 10 - 29 Baseball 142
25.0 2
x 10 - 34 Earth 6.0 x 1027
3.0 x 104 4 x 10
- 63
Table 7.1 (p. 274)
45
Light Has Momentum
  • momentum p mu mass x velocity
  • p Planks constant / wavelength
  • or p mu h/wavelength
  • wavelength h / mu de Broglies
    equation
  • de Broglies expression gives the wavelength
    relationship of a particle traveling a velocity
    u !!

46
Fig. 7.14
47
Heisenberg Uncertainty Principle
  • It is impossible to know simultaneously both the
    position and momentum (mass X velocity) and the
    position of a particle with certainty !

48
Fig. 7.15
49
A Radial Probability Distribution of Apples
50
Fig 7.16
51
Radial probability Accurate
Stylized Combined
area distribution
representation of the 2p
of the three 2p
of the 2p
distribution orbitals 2px,
2py
distribution
and 2pz orbitals


Fig. 7.17
52
Fig. 7.18
53
Fig. 7.19
54
Quantum Numbers - I
  • 1) Principal Quantum Number n
  • Also called the energy quantum number,
    indicates the approximate distance from the
    nucleus .
  • Denotes the electron energy shells around the
    atom, and is derived directly from the
    Schrodinger equation.
  • The higher the value of n , the greater the
    energy of the orbital, and hence the energy of
    electrons in that orbital.
  • Positive integer values of n 1 , 2 , 3 , etc.

55
Quantum Numbers - II
  • 2) Azimuthal
  • Denotes the different energy sublevels within the
    main level n
  • Also indicates the shape of the orbitals around
    the nucleus.
  • Positive interger values of L are 0
    ( n-1 )
  • n 1 , L 0 n 2 ,
    L 0 and 1
  • n 3 , L 0 , 1 , 2

56
Quantum Numbers - III
  • 3) Magnetic Quantum Number - mL Also called
    the orbital orientation quantum
  • denotes the direction or orientation in a
    magnetic field - or it denotes the different
    magnetic geometriesound the nucleus - three
    dimensional space
  • values can be positive and negative (-L 0
    L)
  • L 0 , mL 0 L 1 , mL
    -1,0,1
  • L 2 , mL -2,-1,0,1,2

57
Quantum Numbers
Allowed Values
n
1 2 3
4
L
0 0 1 0 1
2 0 1 2 3
mL
0 0 -1 0 1 0 -1 0 1
0 -1 0 1
-2 -1
0 1 2 -2 -1 0 1 2
-3 -2
-1 0 1 2 3
58
Determining Quantum Numbers for an Energy Level
(Like S.P. 7.5)
Problem What values of the azimuthal (L) and
magnetic (m) quantum numbers are allowed for a
principal quantum number (n) of 4? How many
orbitals are allowed for n4? Plan We determine
the allowable quantum numbers by the rules given
in the text. Solution The L values go from 0 to
(n-1), and for n3 they are L
0,1,2,3. The values for m go from -L to zero to
L For 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 There are 16 mL values, so there are
16 orbitals for n4! as a check, the total
number of orbitals for a given value of n is n2,
so for n 4 there are 42 or 16 orbitals!
59
Quantum Numbers Noble Gases
Electron Orbitals
Number of Electrons Element
1s2
2
He
1s2 2s22p6
10
Ne
1s2 2s22p6 3s23p6
18 Ar
1s2 2s22p6 3s23p6 4s23d104p6
36 Kr
1s2 2s22p6 3s23p6 4s23d104p6 5s24d105p6
54 Xe
1s2 2s22p6 3s23p6 4s23d104p6 5s24d105p6
6s24f14 5d106p6 86 Rn
1s2 2s22p6 3s23p6 4s23d104p6 5s24d105p6
6s24f145d106p6 7s25f146d10?
60
The Periodic Table of the Elements
Electronic Structure
He
H
Ne
F
O
N
C
B
Li
Be
Ar
Cl
S
P
Si
Al
Na
Mg
Kr
Zn
Cu
Ni
Co
Fe
Mn
Cr
V
Ti
Sc
Br
Se
As
Ge
Ga
K
Ca
Xe
Cd
Ag
Pd
Rh
Ru
Tc
Mo
Nb
Zr
Y
I
Te
Sb
Sn
In
Rb
Sr
Rn
Hg
Au
Pt
Ir
Os
Re
W
Ta
Hf
La
At
Po
Bi
Pb
Tl
Cs
Ba
Ac
Rf
Ha
Fr
Ra
Sg
Ce
Pr
Nd
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
Th
Pa
U
Np
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
Lr
S Orbitals
P Orbitals f Orbitals
d Orbitals
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