Light and Atoms

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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.

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The Line Spectra of Several Elements
Hg
Fig. 7.8
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Energy Level Diagram

• Energy
• Excited States
• photons path
• Ground State

Light Emission Light Emission
Light Emission
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Figure 7.10 Energy-level diagram for the
electron in the hydrogen atom.
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Figure 7.11 Transitions of the electron in the
hydrogen atom.
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Three Series of Spectral Lines of Atomic
Hydrogen
Fig. 7.9
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A desktop analogy for the H atoms energy
Fig. 7.11
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Emission and Absorption Spectraof Sodium Atoms
Fig. 7.B
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Fig. 7.C
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Fig. 7.D
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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
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Emission Energetics - II
Problem A compact disc player uses light with a
frequency of 3.85 x 1014 per second. What is this
lights wavelength? What portion of the
electromagnetic spectrum does this wavelength
fall? What is the energy of one mole of photons
of this frequency? Plan Calculate the energy of
a photon of the light using Ehv, and wavelength
x C v . Then compare the frequency with the
electromagnetic spectrum to see what kind of
light we have. To get the energy per mole,
multiply by Avogadros number. Solution
3.00 x 108m/s 3.85 x 1014/s
wavelength c / v
7.78 x 10 -7 m 778 nm
778 nm is in the Infrared region of the
electromagnetic spectrum
Ephoton hv (6.626 x 10 -34Js) x ( 3.85 x 1014
/s) 2.55 x 10 -19 J
Emole (2.55 x 10 -19J) x (6.022 x 1023 / mole)
1.54 x 105 J/mole
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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
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Wave Motion in Restricted Systems
Fig. 7.12
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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)
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Figure 7.23 Orbital energies of the hydrogen
atom.
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Figure 7.18 Plot of y2 for the lowest energy
level of the hydrogen atom.
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Figure 7.19 Probability of finding an electron
in a spherical shell about the nucleus.
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Figure 7.21 The scanning tunneling microscope.
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Figure 7.20 Scanning tunneling microscope of
benzene molecules on a metal surface.Photo
courtesy of IBM Almaden Research Center.
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Figure 7.22 Quantum corral. Photo courtesy of
IBM Almaden Research Center research done by
Dr. Don Eigler and co-workers.
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Figure 7.24 Cross-sectional representations of
the probability distributions of S orbitals.
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Figure 7.25 Cutaway diagrams showing the
spherical shape of S orbitals.
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Figure 7.26 The 2p orbitals.
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Figure 7.27 The five 3d orbitals.
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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 !!

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de Broglie Wavelength Calc. - I
Problem Calculate the wavelength of an electron
traveling 1 of the speed of light ( 3.00 x
108m/s). Plan Use the de Broglie relationship
with the mass of the electron, and its speed.
Express the wavelength in meters and
nanometers. Solution
electron mass 9.11 x 10 -31 kg
velocity 0.01 x 3.00 x 108 m/s 3.00 x 106 m/s
h m x u
6.626 x 10 - 34Js ( 9.11 x 10 -
31kg )( 3.00 x 106 m/s )
wavelength

kg m2 s2
J
therefore
wavelength 0.24244420 x 10 - 9 m 2.42 x 10
-10 m 0.242 nm
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Fig. 7.14
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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 !

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Fig. 7.15
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A Radial Probability Distribution of Apples
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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
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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!
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Fig 7.16
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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
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Fig. 7.18
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Fig. 7.19
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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.

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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

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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

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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?
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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