Chapter 7 Atomic Structure and the Periodic Table
I.
Electromagnetic Radiation = Light
A.
What is Light?
1. Visible light is a particular kind of electromagnetic radiation
2. X-rays, UV, Infrared, Microwaves, and Radio waves are all light forms
3. Light is a form of Energy
4. All light travels in waves at “the speed of light” = c = 3 x 108 m/s
B. Characteristics of Light
1. Wavelength = distance between two peaks in a wave
a. l (lambda) is the symbol
b. Meters = m is the unit
2. Frequency = number of complete waves passing a given point per second
a. n (nu) is the symbol
b. Hertz = Hz = s-1 is the unit
3. Amplitude = A = measure of the intensity of the wave, “brightness”
4. The speed of light is constant: c = l x n
a. l and n are inversely proportional
b. If one increases, the other decreases
c. Their product is always the speed of light = c = l x n = 3 x 108 m/s
5.
Example: What is the frequency of red light with l = 650 nm?
3.0 x 108 m / s
14 1
c  l    

4
.
61
x
10
s
-9
l 650 x 10 m
c
II.
The Nature of Matter
A.
Classical Physics ~1900
1. Matter is composed of particles that have mass and a known position
2. Light is a form of energy without mass and uncertain position
3. These two phenomena were thought to be distinct
B.
Planck’s Revelation
1. Studied the light given off by heated objects
2. Found that classical physics couldn’t explain his observations
3. Showed that light could be thought of as particles for certain applications
a. Energy can only be gained or lost as light in whole-number multiples
b. Particles of light have fixed energies = Basis of quantum theory
c. Energy is quantized = can only occur in discrete units = photons
d. The energy of the photon is directly proportional to the frequency of
light (Higher frequency = More energy in photons)
E  hn
E
hc
l
h  Planck' s constant  6.626 x 10-34 Js
4. Example: How much E is being emitted by light of 450 nm?
E  hn 
hc
l

6.626 x 10 Js 3.0 x 10 m / s 

 4.41 x 10
450 x 10 m
-34
8
-9
-19
J
C. Einstein’s Contribution
1. We can view any kind of light as a stream of particles = photons as well as
a wave = Dual Nature of Light
2. Special Theory of Relativity: E = mc2
Energy has mass that we can calculate from wavelength or frequency
E hc / l
h
m 2  2 
c
c
lc
3. de Broglie’s Equation showed that particles have wavelength too
h
h
h
m

l 
lc lv
mv
4. Example: l = ? for e- (v = 1 x 107 m/s) and 0.10 kg ball (v = 35 m/s)
h
6.626 x1034 (kgm2 / s 2 ) s
11
lelectron 


7
.
27
x
10
m
31
7
mv 9.11x10 kg 1.0 x10 m / s
h 6.626 x1034 (kgm2 / s 2 )s
lball 

 1.9 x1034 m
0.10kg35m / s 
mv



5. Electron wavelength is on the order of the spacing of atoms in a crystal; the
ball’s wavelength is very tiny (it doesn’t behave much like a wave)
D. X-Ray Diffraction
1. Diffraction = scattering light by an array of particles (rainbow on CD)
2. Diffraction proves electrons (particles) can behave like waves
a. Davisson and Germer aimed electrons at a Nickel crystal in 1927
b. They observed a diffraction pattern, showing electrons are wavelike
E. Conclusions
1. Energy is a form of matter
2. Both matter and energy can behave as particles and as waves
3. Massive objects (baseball) behave mostly like particles
4. Tiny objects (electron, photons) behave mostly like waves
III. Atomic Spectra and the Bohr Model
A. Atoms can absorb or give off energy in the form of light
1. Light given off = emission spectrum
2. light energy gained = absorption spectrum
3. The light atoms give off or gain is of very specific wavelengths called a
line spectrum
4. Each element has its own line spectrum which can be used to identify it
B. Energy and Spectra
1. The line spectrum must be related to energy transitions in the atom.
a. Absorption = atom gaining energy
b. Emission = atom releasing energy
2. Since all samples of an element give the exact same pattern of lines, every
atom of that element must have only certain, identical energy states
3. The atom is quantized
4. All possible energies, would result in continuous spectrum instead of lines
Continuous Spectrum of white light
Line spectrum of Hydrogen
C. Bohr’s model of the atom explained spectra of hydrogen
1. The Hydrogen electron moves around the nucleus in specific allowed
circular orbits (Neils Bohr, 1885-1962)
2. Energy of an atom is related to the distance electron is from the nucleus
3. Energy of the atom is quantized
a. atom have specific energy states called energy levels
b. when atom gains energy, electrons “moves” to a higher quantum level
c. when atom loses energy, electrons “moves” to a lower energy level
d. lines in spectrum correspond to the difference in energy between levels
4. Atoms have a minimum energy called the ground state
a. Therefore, electrons do not crash into the nucleus
b. The ground state of H has its one electron closest to the nucleus
c. Energy levels higher than the ground state are called excited states
d. The farther the energy level is from the nucleus, the higher its energy
5. We can calculate the energies of the Energy Levels of Hydrogen
2


Z
18
E  2.178 x10 J 2 
n 
a.
b.
c.
d.
Z = nuclear charge of the atom = +1 for Hydrogen
n = an integer (ground state has n = 1)
The negative sign indicates a favorable energy of interaction
If n = ∞, the interaction is zero (e- infinitely far away from nucleus)
6. We can also calculate the change in energy of a transition (Ex. n=1 to n=6)
2
2

 

18  1 
18  1 
E  E6  E1   2.178 x10 J 2    2.178 x10 J 2   2.117 x10 18 J
 6  
 1 

a. Electron has gained energy going from ground to excited state
b. The energy would have to come from the absorption of light
7. We can calculate the wavelength of the absorbed photon



hc
6.626 x 10-34 Js 3.0 x 108 m / s
-8
E 
l 


9
.
383
x
10
m
-18
l
E
2.117 x 10 J
hc


8. Example: Calculate the energy needed to remove the ground state e-.
a. A general equation for any two states can be derived
2
2

 


 1



1
1
1
18 

18

18
   2.178 x10 J

  2.178 x10 J
E   2.178 x10 J

2
2
2
2


n

n


 ninitial 
 final  
 final ninitial 
b. Solve for final n = ∞ and initial n = 1
1
 1
E  2.178 x10 18 J 2  2   2.178 x10 18 J0  1  2.178 x10 18 J
 1 
9. Problems with Bohr’s Model
a. Only explains hydrogen atom spectrum and other 1 electron systems
b. Neglects interactions between electrons
c. Assumes circular or elliptical orbits for electrons - which is not true
IV. Quantum Mechanical Atomic Model
A.
Standing Waves
1. A string is limited to specific (quantized) vibrations
2. To Schrödinger (1887-1961), electrons quantized around a nucleus seemed similar
3. Experiments showed that electrons could be treated as waves
4. The quantum mechanical model treats electrons as waves and uses wave
mathematics to calculate probability densities of finding the electron in a
particular region in the atom
5. Schrödinger Wave Equation: Hˆ y  Ey
a. can only be solved for simple systems, but approximated for others
b. y = wavefunction, E = energy, H = operator
c. Many Energies can be found, each corresponding to an orbital
d. Orbitals correspond to a standing electron wave of a specific shape
B. Results from the Quantum Mechanical Model
1. Heisenberg uncertainty principle: can’t know both position and momentum
a. Waves don’t have specific locations
b. Since electrons behave like waves, neither can electrons
x mv  h
4
2. y2 gives the probability of finding the electron in a given location
a. Probability of where the electron is around the nucleus
b. Imagine time-lapse picture of e- in n = 1 of a hydrogen atom
0.529Å
c. This plot gives the shape of the n = 1 orbital of Hydrogen = Sphere
d. Why isn’t the radial probability highest right next to the nucleus?
i. Probability is largest closest to nucleus
ii. Volume of the “shell” is very small close to the nucleus
e. The sphere enclosing 90% of the electron probability = orbital picture