Download Chapter 7: Electrons in Atoms Electromagnetic Radiation

Chapter 7: Electrons in Atoms
Dr. Chris Kozak
Memorial University of Newfoundland, Canada
1
Electromagnetic Radiation
•  Electric and magnetic
fields propagate as waves
through empty space or
through a medium.
•  A wave transmits energy.
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1
EM Radiation
Low ν
High ν
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Frequency, Wavelength and Velocity
•  Frequency (ν) in Hertz—Hz or s-1.
•  Wavelength (λ) in meters—m.
•  cm
µm
nm
Å
(10-2
m)
(10-6
m)
(10-9
m)
(10-10
pm
m) (10-12 m)
•  Velocity (c)—2.997925 x 108 m s-1.
c = λν
λ = c/ν
ν= c/λ
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2
What is the frequency of light corresponding to blue
light? λ = 473 nm. c = 2.998x108 m s-1.
A. 
B. 
C. 
D. 
E. 
F. 
G. 
H. 
1.58x10-15 s-1
1.42x10-15 s-1
6.33x10-14 s-1
1.42x10-2 s-1
1.42x102 s-1
6.33x1014 s-1
1.42x1015 s-1
1.58x1015 s-1
Electromagnetic Spectrum
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3
Constructive and Destructive Interference
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Refraction of Light
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4
Atomic Spectra
Hydrogen
Helium
Lithium
Sodium
Potassium
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Atomic Spectra
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5
Blackbody Radiation
3 phenomena confounded physicists at the turn of the 20th century:
1)  Atomic Spectra
2)  Photoelectric effect
3)  Blackbody radiation
When a solid object is heated to 1000K, visible light is emitted. As T increases, intensity and
wavelength change. Why?
This is characteristic of blackbody radiation (a hypothetical body that absorbs and emits all
frequencies of radiation). We see the “colour” and intensity of light increase in each of
these increase.
1000 K
Embers in a fire
1500 K
Stove heating element
2000 K
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Light bulb filament
Quantum Theory
Blackbody Radiation:
•  So, heated bodies emit light.
•  In Blackbody Radiation, Intensity (I) is inversely
proportional to wavelength (λ), or, it increases with
increasing frequency (v).
•  I proportional to 1/λ.
•  Classical theory predicts continuous increase of
intensity with frequency.
•  However, in reality the intensity drops off after
specific wavelengths.
Max Planck, 1900:
ΔE = Δnhν
Energy, like matter, is discontinuous.
E is energy, n is a positive integer, h = 6.62607 x 10-34 J.s (Planck’s
constant) and ν is frequency in hertz (s-1)
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6
What is the energy corresponding to blue light?
λ = 473 nm h= 6.626 x 10-34 J s, c=2.998x108 m s-1.
ΔE = hν , ν = c / λ
Enter your answer between 0 and 1000 kJ mol-1
Rank
Responses
1
2
3
4
5
6
Other
The Photoelectric Effect
•  Light striking the surface of certain metals causes ejection of
electrons
•  Wave properties of light is unable to explain some observations
•  ν > νo threshold frequency (Light must have a minimum frequency or
no current flows)
•  ne- a I # of e- depends on intensity (Absence of Lag time. Current flow
is immediate. The metal does not “accumulate” energy to eject the
electron as predicted by wave theory)
•  Ek α ν kinetic energy depends on frequency (Wave theory says that
amplitude, not frequency is responsible… so any colour of light
should be able to do this so long as it is bright enough. This can’t be
right!)
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7
The Photoelectric Effect
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The Bohr Atom
• 
• 
• 
• 
Electrons move in circular orbits about the nucleus (like in classical physics)
Fixed set of stationary states (allowed orbits)
Governed by angular momentum: nh/2π, n=1, 2, 3….
Energy packets (quanta) are absorbed or emitted when electrons change
stationary states
•  The integral values that are allowed are called quantum numbers
•  The energy of an electron becomes increasingly negative the closer it gets to
the nucleus according to E = -RH/n2. RH is a constant
•  Why is electron E a negative value?
E=
-RH
n2
RH = 2.179 x 10-18 J
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8
Energy-Level Diagram
ΔE = Ef – Ei =
= RH (
-RH
-RH
–
nf2
n i2
1
1
–
) = hν = hc/λ
ni2 nf2
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Ionization Energy of Hydrogen
ΔE = RH (
1
1
–
) = hν
2
ni
nf2
As nf goes to infinity for hydrogen starting in the ground state:
hν = RH (
1
) = RH
n i2
This also works for hydrogen-like (1-electron) species such as
He+ and Li2+
hν = -Z2 RH
Z is the nuclear charge (number of protons)
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9
Great, but what do we really need to
know?
•  Calculate the energy, frequency and wavelength for any
hydrogen atom transition
•  Identify the wavelengths of the electromagnetic spectrum
as being in the UV, visible or IR regions.
•  You should be able to do Examples 7.1– 7.3 & 7.5 in Tro
(and the practice examples)
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Summary of Bohr’s Atomic Theory
Electrons are in motion around the nucleus (orbits)
But, for circular orbits, electrons would possess angular
momentum (acceleration) and therefore radiate energy!
So, using Planck’s quantum hypothesis,
1)  Electrons move in fixed orbits around the nucleus
2)  Fixed orbits (stationary states) mean properties of
individual electrons will have unique values, for example,
the angular momentum is quantized based on the orbit in
which the electron resides.
3)  Electrons only pass between allowed orbits. This means
that fixed quanta of energy are involved.
QUANTUM NUMBERS!
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10
What is Bohr’s Theory Good For?
Approximations of the energy associated with transitions
(movement) of electrons ONLY in ions with one electron!
Can’t explain:
•  spectra of species with more than one electron
•  effect of magnetic fields on emission spectra
It is an uneasy mixture of classical and non-classical physics.
Modern quantum theory replaced Bohr theory in 1926.
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Which of the following transi1ons emits a photon with the greatest wavelength? A. 
n=4 to n=2 B. 
n=2 to n=1 C. 
n=4 to n=3 11
Determine the energy of light emiAed by a hydrogen atom for the transi1on of an electron from n=3 to n=2. Input an answer in kJ mol-­‐1. Rank Responses 1 2 hAp://snews.bnl.gov/popsci/atom1.jpg 3 4 5 6 Other What is the energy of the photon emiAed associated when the electron in O7+ undergoes a n=3 to n=2 transi1on Input an answer in kJ mol-­‐1. Rank Responses 1 2 3 hAp://www.gc.maricopa.edu/earthsci/imagearchive/
Oxygen%20Spec%20sm.jpg 4 5 6 Other 12
Two Ideas Leading to a New Quantum
Mechanics
Wave-Particle Duality
Heisenberg’s
Uncertainty Principle
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Fire Photon Torpedoes!
•  Wave-Particle Duality.
–  Einstein suggested particle-like properties of light
could explain the photoelectric effect.
–  But diffraction patterns suggest photons are wavelike.
•  de Broglie, 1924
–  Small particles of matter may at times display
wavelike properties.
–  The concept of the “photon”!
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13
de Broglie and Matter Waves
E = mc2
Einstein’s Relativity Equation
hν = mc2
Planck’s Equation for Energy
hν/c = mc = p
Rearrange to give momentum, p
p = h/λ
Momentum carried by a photon in
relation to its wavelength!
λ = h/p = h/mu
The wavelength is related to the mass of the particle (particle property)
If matter waves exist for small particles, then beams of particles should
exhibit the characteristic properties of waves: diffraction.
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X-Ray Diffraction
Structure 1
Structure 2
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14
The Uncertainty Principle
•  Werner Heisenberg
Δx Δp ≥
h
4π
Δx is uncertainty in position
Δp is uncertainty in momentum
We cannot measure the exact position or
exact momentum of a subatomic particle
simultaneously
WHY?!
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Sample Problems
1. 
Some Diamonds appear yellow because they contain nitrogen compounds that
absorb purple light with a frequency of 7.23 x 1014 Hz. Calculate the
wavelength in nm of absorbed light.
2. 
Calculate the E of one photon of UV (λ = 1 x 10-8 m), visible (λ = 5 x 10-7 m)
and IR (λ = 1 x 10-4 m) light. What do the answers indicate about the
relationship between λ and E?
3. 
Calculate the Energy required to remove an electron from a hydrogen atom in
its ground state.
4. 
Calculate the wavelength of the transition from n = 4 to n = 1 in Hydrogen (one
of the Lyman series of transitions).
5. 
Calculate the de Broglie wavelengths of a 50 kg mass travelling at ¼ the speed
of light and for a proton (m = 1.673 x 10-27 kg) travelling at this speed. What do
these wavelengths say about the wave properties of matter in relation to their
size?
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15
Wave Mechanics
•  Standing waves.
–  Nodes do not undergo displacement.
λ=
2L
, n = 1, 2, 3…
n
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Wave Functions
•  ψ, psi, the wave function.
–  Should correspond to a
standing wave within the
boundary of the system
being described.
•  Particle in a box.
ψ =
2
⎛ n π x ⎞
sin ⎜
⎟
L
⎝ L ⎠
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16
Probability of Finding an Electron
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Wave Functions for Hydrogen
•  Schrödinger, 1927
Eψ = Hψ
–  H (x,y,z) or H (r,θ,φ)
ψ(r,θ,φ) = R(r) Y(θ,φ)
R(r) is the radial wave function.
Y(θ,φ) is the angular wave
function.
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17
Principle Shells and Subshells
•  Principle electronic shell, n = 1, 2, 3…
•  Angular momentum quantum number,
l = 0, 1, 2…(n-1)
I = 0, s
l = 1, p
l = 2, d
l = 3, f
•  Magnetic quantum
number,
•  ml= - l…-2, -1, 0, 1, 2…+l
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Orbital Energies
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18
9-8 Interpreting and Representing the
Orbitals of the Hydrogen Atom.
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s orbitals
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19
p Orbitals
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p Orbitals
48
20
d Orbitals
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Electron Spin: A Fourth Quantum
Number
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21
Electronic Structure of the H atom
We have 3 quantum numbers for H
n=1
1s orbital
l=0
ml = 0
ms
Only one type of orbital orientation/
symmetry
Only one electron (can be
either +1/2 or -1/2)
Ground State Configuration:
1s1
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Which of the following best represents an orbital with principal and angular momentum quantum numbers 4 and 1, respec1vely? B. A. C. D. E. 22
Multi-electron Atoms
•  Schrödinger equation was for only one e-.
•  Electron-electron repulsion in multi-electron
atoms.
•  Assume they have Hydrogen-like orbitals
(by approximation).
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Electron Configurations
Three Main Principles
•  Aufbau process.
–  Build up and minimize energy.
•  Pauli exclusion principle.
–  No two electrons can have all four quantum
numbers alike.
•  Hund’s rule.
–  Degenerate orbitals are occupied singly
first and with parallel spins.
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23
Which of the following orbital designa1ons is impossible? A. 
n=2, l = 0, ml = 0 B. 
n=3, l = 0, ml = 0 C. 
n=3, l = 1, ml = 1 D. 
n=3, l = 2, ml = -­‐1 E. 
n=3, l = 3, ml = -­‐3 Which of the following sets of quantum numbers is NOT matched with the orbital designa1on. A. 
n=1, l = 0: a 1s orbital B. 
n=2, l = 0: a 2p orbital C. 
n=3, l = 0: a 3s orbital D. 
n=3, l = 1: a 3p orbital E. 
n=3, l = 2: a 3d orbital 24
Orbital Energies
57
How many electrons can have the following set of quantum numbers? n = 3 Rank Responses 4d
4d
4d
4d
4d
3d
3d
3d
3d
3d
5s
4p
1 4p
4p
4s
2 3p
3p
3p
2p
2p
2p
3 3s
5 6 Other Energy
4 2s
1s
Multi-Electron Atom Atom
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Orbital Filling for Atoms Only
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Aufbau Process and Hund’s Rule
spdf notation: C (carbon) 1s22s22p2
Expanded notation:
1s22s22px1py1
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26
Filling p Orbitals (Electrons in Boxes)
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Filling the d Orbitals
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27
To which element does the
following orbital filling diagram
belong?
A.  C
B.  N
C.  Si
D.  P
E.  None of these.
Which of the following is the condensed electron configura1on for chlorine? A. 
[Ne]3p7 B. 
[Ne]3s23p5 C. 
[Ne]3s23p6 D. 
[Ne]3s23d5 E. 
[Ne]3s23p33d2 Cl2(g) Cl2(l) 28
The electron configura1on for Ca is A. 
[Ar]3s2 B. 
[Ar]3s4s C. 
[Ar]4s2 D. 
[Ar]3s4 E. 
[Ar]4p2 The electron configura1on for Br is A.  [Ar] 4s23d104p5 B.  [Ar] 4s24d104p5 C.  [Ar] 4s25d104p5 D.  [Ar] 5s24d105p5 E.  [Ar] 4s25d94p6 29
The electron configura1on for Mo is A. 
[Kr] 4s23d4 B. 
[Kr] 4s24p4 C. 
[Kr] 5s25d4 D. 
[Kr] 5s24d4 E. 
[Kr] 5s15d5 The electron configura1on for Sn is A.  [Kr] 4s23d105p2 B.  [Kr] 5s24d105p2 C.  [Kr] 5s25d105p2 D.  [Kr] 5s24d105p2 E.  [Kr] 5s26d55p2 30
Which of the following represents an excited state electron configura1on? A. C A.  . B.  . B. N C.  . D.  . C. O E.  . 2s 2p 2s 2p 2s 2p 3s 3p 3s 3p D. Si E. P 8-12 Electron Configurations and the Periodic Table
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