Download Atomic Physics Sections 9.1-9.7

James T. Shipman
Jerry D. Wilson
Charles A. Higgins, Jr.
Omar Torres
Chapter 9
Atomic Physics
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Atomic Physics
What evidence suggests that
matter is composed of atoms?
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Evidence for Atoms:
Law of Conservation of Mass
in a chemical reaction Mass is not gained or lost
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Evidence for Atoms:
Law of Definite Proportions
A compound always has the same relative amounts of the
elements that compose it. For example, when water is broken
down by electrolysis into oxygen and hydrogen, the mass ratio is
always 8 to 1.
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Atomic Theory
Dalton’s Atomic Theory (1803 – 1807):
1. Each element is composed of small particles called atoms.
2. All atoms of a given element are chemically identical
3. Atoms in chemical reactions combine in simple, fixed,
whole-number ratios to form compounds.
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Atomic Theory
The Discovery of Electrons By J. J. Thomson
• In 1903 J.J. Thomson discovered the negatively charged
electron. Since atoms as a whole are electrically neutral,
some other part of the atom must be positive.
• Thomson conceived the atom as a sphere of positive
charge in which negatively charged electrons were 2 embedded; his model is called “plum pudding model”
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Rutherford’s Atom: 1911
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Rutherford’s Atom: 1911
1. A piece of gold foil is bombarded with α particles
2. Most α particles went through the gold foil, some were deflected
and a few bounced backwards.
3. The deflected particles led to the hypothesis of a positive nucleus
2at the center, and electrons© occupied
the volume outside
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Rutherford’s Atom: 1911
• Rutherford envisioned the atom as having a positive charge
(the nucleus) around which the electrons orbited
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Section 9.1
Continuous Spectrum of Visible Light
Light of all colors is observed
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Section 9.3
Line Emission Spectrum for Hydrogen
• When light from a gas-discharge tube is analyzed only spectral lines
of certain frequencies are found
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Section 9.3
Line Absorption Spectrum for Hydrogen
• Results in dark lines (same as the bright lines of the line emission
spectrum) of missing colors
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Section 9.3
Spectra & the Bohr Model
• Spectroscopists did not initially understand why only
discrete, characteristic wavelengths of light were
– Emitted in a line emission spectrum, and
– Absorbed in a line absorption spectrum
• In 1913 an explanation of the observed spectral line
phenomena was advanced by the Danish physicist Niels
Bohr.
• Bohr predicted that the single hydrogen electron would only
be found in discrete orbits with particular radii
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Section 9.3
434 nm
656 nm
486 nm
410 nm
5
-e
4
434 nm
3
2
656 nm
-e
1
-e
+P
486 nm
-e
-e
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410 nm
Atomic Absorption
Atomic “line” absorption occurs when an electron absorbs a
photon and makes a transition from a lower energy to a
higher energy
e
E2
hv
E1
e-
Absorption Spectrum
656.3nm
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Atomic Emission
Atomic “line” emission occurs when an electron makes a
transition from a higher energy to a lower energy by emitting
a photon
e
E2
hv
E1
e656.3nm
Emission Spectrum
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Bohr and the Hydrogen Atom
• Bohr predicted that the single hydrogen electron would only
be found in discrete orbits with particular radii
– Bohr’s possible electron orbits were given whole-number
designations, n = 1, 2, 3, …
– “n” is called the principal quantum number
– The energy of each orbit is:
2.178 10
En 
2
n
18
(J )
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Section 9.3
2.178 10
En 
2
n
18
(J )
Energy Levels of Hydrogen Atom
Principal quantum number, n
Energy, En (J)
1
2
3
4
5
6
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Transition Energy for Hydrogen
Transition Energy
Transition
En (J)
6→2
5→2
6
5
E6
E5
4
E4
3
E3
2
E2
4 →2
3 →2
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Hydrogen emission lines
Energy Levels of Hydrogen Atom
E1 = -2.178 x 10-18/12 = -2.178 x 10-18 J
E2 = -2.178 x 10-18/22 = -5.445×10-19 J
E3 = -2.178 x 10-18/32 = -2.420×10-19 J
E4 = -2.178 x 10-18/42 = -1.361×10-19 J
E5 = -2.178 x 10-18/52 = -8.712×10-20 J
E6 = -2.178 x 10-18/62 = -6.050×10-20 J
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Photon Energy
According to Max Planck, the energy of a photon is directly
proportional to its frequency and inversely proportional to its
wavelength
Ephoton  hv 
Where
hc

Ephoton = energy of the photon (in Joules),
h = Planck’s constant (6.626 x 10-34 J.s),
ν = frequency (in Hz)
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Determining Photon Energy – Example
Find the energy in joules of the photons of blue
light of frequency 7.50×1014 Hz.
Solution:
E = hf = (6.63×10–34J s)(7.50×1014/s)
= 49.73×10–20 J
** Note the blue light has more energy than red light
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Section 9.2
Emission Wavelengths for Hydrogen
For the transition: 6  2
Energy of the Emitted photon  Ei  E f 
-6.050×10
-20
- (-5.445×10 ) 
-19
hc

hc

hc

4.840×10-19
9
10
nm
34
8
(6.626 10 J .s)(2.998 10 m / s) 
m

4.840×10-19 J
1.98 1016
 (nm) 
 410.4
-19
4.840×10
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Electron Cloud Model of an Atom
The electron cloud is actually a visual representation of the
probability distribution of finding the electron.
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Section 9.7
The Modern Periodic Table
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Elements and the Atomic Number
Sodium:
(as shown in the periodic table)
11 (Atomic number: total number of protons)
Atomic symbol
Na
22.99 (Average Atomic Mass)
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2.2 Elements and Atomic Number
Mass Number(A) = 23
(Total number of protons and neutrons)
Atomic number(Z) = 11
Na
(Total number of protons)
Protons: 11
Neutrons: 23-11=12
Electrons: 11
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Isotopes
Isotopes are atoms with identical atomic
numbers but different mass numbers.
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The Atomic Mass
Atomic Mass is a weighted average of all naturally occurring
isotopes of the atom in atomic mass units (amu). For Carbon
atom, there are 3 naturally occurring isotopes: 12, 13 &14
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What is the Atomic Mass of Mg?
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Calculate the Atomic Mass of
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+
Ni ?
Homework
Exercises:
2
3
4
10
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