Chapter 1: Matter and Measurement

General Chemistry
Principles and Modern Applications
Petrucci • Harwood • Herring
8th Edition
Chapter 9: Electrons in Atoms
Philip Dutton
University of Windsor, Canada
N9B 3P4
Prentice-Hall © 2002
(modified 2003 by Dr. Paul Root and 2005 by Dr. David Tramontozzi)
Prentice-Hall © 2002
General Chemistry: Chapter 9
Unofficial Midterm Marks
# of students vs. mark out of 30
50
45
number of students
40
35
30
25
20
15
10
5
0
0-3.00
3.01-6.00 6.01-9.00 9.01-12.0 12.01-15.0 15.01-18.0 18.01-21.0 21.01-24.0 24.01-27.0 27.01-30
mark out of 30
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General Chemistry: Chapter 9
Contents
9-1
9-2
9-3
9-4
9-5
9-6
9-7
Electromagnetic Radiation
Atomic Spectra
Quantum Theory
The Bohr Atom
Two Ideas Leading to a New Quantum Mechanics
Wave Mechanics
Quantum Numbers and Electron Orbitals
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General Chemistry: Chapter 9
Contents
9-8
9-9
Quantum Numbers
Interpreting and Representing Orbitals of the
Hydrogen Atom
9-9 Electron Spin
9-10 Multi-electron Atoms
9-11 Electron Configurations
9-12 Electron Configurations and the Periodic Table
Focus on Helium-Neon Lasers
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General Chemistry: Chapter 9
9-1 Electromagnetic Radiation
• Electric and magnetic fields
propagate as waves through
empty space or through a
medium.
• A wave transmits energy.
wavelength
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General Chemistry: Chapter 9
EM Radiation
Low 
Low frequency, high wavelength
High 
High frequency, low wavelength
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General Chemistry: Chapter 9
Frequency, Wavelength and Velocity
• Frequency () in Hertz—Hz or s-1.
• Wavelength (λ) in meters—m.
• cm
m
nm
Å
(10-2 m)
(10-6 m) (10-9 m)
pm
(10-10 m) (10-12 m)
• Velocity (c) = 2.997925 × 108 m s-1.
• constant velocity
c = λ
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λ = c/
General Chemistry: Chapter 9
= c/λ
Example 9-1
Most of the light from a sodium vapour lamp
has a wavelength of 589nm. What is the
frequency of this radiation?
1 x 10-9m
λ = 589 nm x
1 nm
c = 2.998 x 108m/s
v=
c
l
= 5.89 x 10-7m
2.998 x 108 m/s
14 s-1
14 Hz
5.09
x
10
=
=
=
5.09
x
10
5.89 x 10 -7 m
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General Chemistry: Chapter 9
Electromagnetic Spectrum
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General Chemistry: Chapter 9
ROYGBIV
Red
Orange
Yellow
700 nm
450 nm
Green
Blue
Indigo
Violet
© 2002
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General
GeneralChemistry:
Chemistry: Chapter
Chapter99
Slide 8
Constructive and Destructive Interference
Constructive
Interference
Destructive
Interference
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General Chemistry: Chapter 9
Examples of Interference
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General Chemistry: Chapter 9
Refraction of Light
Speed of light slows in any media other than a vacuum
therefore, light gets refracted.
Waves of differingGeneral
λ differ
inChapter
speeds
in other media.
Chemistry:
9
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9-2 Atomic Spectra
H2(g)
He(g)
Li
Na
Each element has its own distinctive line
spectrum-akin to a fingerprint!!
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General Chemistry: Chapter 9
K
Atomic Spectra
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General Chemistry: Chapter 9
9-3 Quantum Theory
Blackbody Radiation:
Max Planck, 1900:
Energy, like matter, is discontinuous.
E = h
h = 6.62607 x 10-34 J s
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General Chemistry: Chapter 9
The Photoelectric Effect
• Light striking the surface of certain metals
causes ejection of electrons.
•  > o
• e- α I
• ek a 
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- threshold frequency
- # e- emitted depends on intensity
- kinetic energy of emitted edepends on frequency of light
General Chemistry: Chapter 9
The Photoelectric Effect
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General Chemistry: Chapter 9
The Photoelectric Effect
• At the stopping voltage the kinetic energy of the
ejected electron has been converted to potential.
1
mu2 = eVs
2
• At frequencies greater than o:
Vs = k ( - o)
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General Chemistry: Chapter 9
The Photoelectric Effect
Ek = eVs
Eo = ho
eVo
o =
h
eVo, and therefore o, are characteristic of the metal.
Conservation of energy requires that:
Ephoton = Ek + Ebinding
Ek = Ephoton - Ebinding
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1
mu2 + eVo
h =
2
1
eVs =
mu2 = h - eVo
2
General Chemistry: Chapter 9
9-4 The Bohr Atom
-RH
En = 2
n
RH = 2.179  10-18 J
rn = n2a0
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= 0.53Å (53 pm)
General Chemistry: Chapter 9
Example 9-3
Is it likely that there is an energy level for the
hydrogen atom En = -1.00 x 10-20 J ?
En = -RH / n2
n2 = -RH / En
n2 = -2.179 x 10-18 J / -1.00 x 10-20 J = 217.9
n = 14.79
….. because n is not an integer, this is not an allowed
energy level for hydrogen …..
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General Chemistry: Chapter 9
Energy-Level Diagram
-RH -RH
– 2
ΔE = Ef – Ei =
2
nf
ni
1
1
–
= RH ( 2
) = h = hc/λ
2
ni
nf
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General Chemistry: Chapter 9
Example 9-4
Determine the wavelength of the Balmer series of hydrogen
corresponding to the transition from n = 5 to n = 2
DE = E5 – E2 = RH ( 1 – 1 ) = RH ( 1 – 1 )
ni2 nf2
52 22
= -4.576 x 10-19 J
E = hv
λ=
and
the negative sign indicates that
energy is emitted
v=c/l
then E = hc / l
(6.626 x 10-34 J s photon-1)(2.998 x 108 m s-1)
4.576 x 10-19 J
= 434.1 nm
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General Chemistry: Chapter 9
Ionization Energy of Hydrogen
1
1
–
ΔE = RH ( 2
) = h
2
ni
nf
As nf goes to infinity for hydrogen starting in the ground state:
1
h = RH ( 2 ) = RH
ni
This also works for hydrogen-like species such as He+ and Li2+.
En = -Z2 RH
n2
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Where Z = atomic
number
General Chemistry: Chapter 9
Example 9-5
• Determine the kinetic energy of the electron ionized from
a Li2+ ion from its ground state, using a photon frequency
of 5.000 x 1016 s-1.
En = -Z2RH / n2
En = [-(32)(2.179 x 10-18J) / 12 ] = -1.961 x 10-17 J
(energy required to remove electrons creating Li2+)
E = hv = (6.626 x 10-34 J s photon-1)(5.000 x 1016 s-1)
= 3.313 x 10-17 J photon-1
(energy carried by the photon)
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General Chemistry: Chapter 9
Example 9-5
• The ionization energy (the energy required to remove the
electron) is Ei = -E1 = 1.961 x 10-17J.
• The extra energy from the photon is converted into kinetic
energy
Kinetic energy = 3.313 x 10-17 J – 1.961 x 10-17 J
= 1.352
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x 10-17 J
General Chemistry: Chapter 9
HAVE A GOOD
HALLOWE’EN
WEEKEND
(DON’T FORGET TO TURN YOUR CLOCKS
BACK AND ENJOY THAT EXTRA HOUR
OF SLEEP ON SUNDAY!!)
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General Chemistry: Chapter 9

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