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```Electronic Structure of Atoms
Chapter 6
Light is a Wave
•
•
•
•
Electromagnetic wave
Wavelength (l), m
Frequency (n), Hz or s-1
Travels at c (3.00 X 108 m/s) in a
vacuum
• Wavelength and frequency are inversely
proportional
• c = ln
•How many complete waves are shown
above?
•What is the wavelength of light shown
above?
Electromagnetic Spectrum
Blu-Ray = 405 nanometers (blue light)
DVD = 650 nanometers (red light)
• Visible light = 4 X 10-7 m to 7X 10-7 m (400 to 700
nm)
Waves: Ex 1
Calculate the wavelength of a 60 Hz EM wave
Waves: Ex 2
Calculate the wavelength of a 93.3 MHz FM radio
station
Waves: Ex 3
Calculate the frequency of 500 nm blue light.
Planck’s Quantum Hypothesis:
Light is a Particle
• Photon –light particle
• Photons emitted in “packets” (whole numbers)
• Light is both a wave and a particle
E = hn (for one photon)
h = 6.63 X 10-34 J s (Planck’s constant)
Photons: Ex 1
Calculate the energy of a photon of wavelength 600
nm.
600 nm 1 X 10-9m = 6 X10-7 m
1 nm
Photons: Ex 2
Calculate the energy of a photon of wavelength 450
nm (blue light). Also, calculate the energy per
mole.
Ans: (4.42 X 10-19 J , 266 kJ/mol)
Photons: Ex 3
Calculate the energy of laser light with a frequency
of 4.69 X 1014 s-1 . Also, calculate the energy per
mole.
Ans: (3.11 X 10-19 J, 187 kJ/mol)
Photons: Ex 3a
If the laser emits 1.3 X 10-2 J per pulse, how many
photons (quanta) are released during this pulse?
Ans: 4.18 X 1016 photons
Photoelectric Effect (Einstein)
• When light shines on a metal, electrons are
emitted
• Can detect a current from the electrons
• Used in light meter, scanners, digital cameras
Three Key Points
1. Below a certain frequency, no electrons are
emitted
2. Greater intensity light produces more electrons
3. Greater Frequency light produces no more
electrons, but they come off with greater speed
Low Frequency
Not enough energy to
eject electron
High Frequency
Can eject electron
Energy of photon is
greater than W (Work
function)
2. More intensity
– More photons
– More electrons ejected with same KE
3. Greater
Frequency
– No more
electrons
ejected
– Electrons ejected
with greater
speed (KE)
Photon/Matter Interactions
1. Electron excitation (photon disappears)
2. Ionization/photoelectric effect (photon
disappears)
3. Scattering by nucleus or electron
4. Pair production (photon disappears)
Electron Excitation
Ionization/Photoelectric Effect
•Photon is absorbed (disappears)
•Photon is absorbed (disappears)
•Electron jumps to an excited state
•Electron is propelled out of the atom
Scattering
•Photon collides with a nucleus or
electron
•Photon loses some energy
•Speed does not change, but the
wavelength increases
Pair Production
•Photon closely approaches a nucleus
•Photon disappears
•An electron and positron are created.
Principle of Complimentarity
• Any experiment can only observe light’s wave or
particle properties, not both
• Different “faces” that light shows
Line Spectra
• Discharge tube
– Low density gas (acts like isolated atoms)
– high voltage
• Light emitted only at certain (discrete)
wavelengths
Hydrogen
Helium
Solar absorption spectrum
• Electrons orbit in ground
energy)
• Jumps to excited state by
absorbing a photon
• Returns to ground state by
emitted a photon
hn = Ee - Eg
4. Ways to make something glow
Bohr
Model
Photon Absorption
-Glow in the dark
Collision
-Heat
-Electricity
-Chemical Reaction
Photon Absorption
Collision
Bohr’s Equation
• Works only for H
and other 1electron atoms
(He+, Li2+, Be3+,
etc…)
En = -RH
n2
En
RH
n
= Energy of an orbital
= Rydberg constant (2.18 X 10-18 J)
= Principal quantum number of orbital
Bohr’s Equation: Ex 1
Calculate the energy of the first three orbitals of
hydrogen
En = - 2.18 X 10-18 J
n2
E1 = - 2.18 X 10-18 J
12
E1 = - 2.18 X 10-18 J
E2 = - 2.18 X 10-18 J
22
E2 = - 5.45 X 10-19 J
E3 = - 2.18 X 10-18 J
32
E3 = - 2.42 X 10-19 J
Bohr’s Equation: Ex 2
What wavelength of light is emitted if a hydrogen
electron drops from the n=2 to the n=1 orbit?
E1 = - 2.18 X 10-18 J
E2 = - 5.45 X 10-19 J
DE = (- 2.18 X 10-18 J - - 5.45 X 10-19 J)
DE = -1.64 X 10-18 J
DE = hn
n = 2.48 X 1015 s-1
c = ln
l = c/n = (3 X 108m/s)(2.48 X 1015 s-1 )
l = 122 nm
DE = hn
c = ln
DE = hc/l
l = hc
DE
l = (6.626 X 10-34 J s)(3.0 X 108 m/s)
(1.63 X 10-18J)
l = 1.22 X 10-7 m = 122 nm (UV)
Bohr’s Equation: Ex 3
Calculate the wavelength of light emitted if a
hydrogen electron drops from the n=6 to the n=2
orbit?
(ANS: 410 nm (violet))
Bohr’s Equation: Ex 4
Calculate the wavelength of light that must be
absorbed to exite a hydrogen electron from the
n=1 to the n=3 orbit?
(ANS: 103 nm (UV))
Bohr Model: Other Atoms
En = (Z2)(- 2.18 X 10-18 J)
n2
Z = Atomic number of the element (H=1, He=2, etc)
The Batcave Intruder Alert Laser uses a beam that has a frequency of 3.53 X
1014 Hz.
a) Calculate the wavelength in nanometers. (850)
b) Calculate the energy per photon. (2.34X10-19 J)
c) Calculate the energy per mole of photons.
(1.41 X 105 J/mol)
d) Calculate the number of photons in a 50.0 mJ pulse of this laser. (2.13 X
1017 photons)
e) Calculate the moles of photons in that pulse. (3.54 X 10-7 moles)
f) Identify the range of the electromagnetic spectrum of this laser.
Wave Nature of Matter
• Louis DeBroglie
• All matter has wave and particle properties
• “matter waves”
l= h
mv
momentum
• Everything has a wavelength
•Wavelike properties only
matter for small objects
•Electrons, protons, light, etc…
Diffraction pattern
of electrons
scattered off
aluminum foil
DeBroglie Wavelength: Ex 1
Calculate the wavelength of a baseball of mass 0.20
kg moving at 40.25 m/s (90 mph)
l= h
mv
l = (6.626 X 10-34 J s)
(0.20 kg)(40.25 m/s)
l = 8.2 X 10-35 m
DeBroglie Wavelength: Ex 2
Calculate the wavelength of an electron
(9.109 X 10-31 kg) moving at 2.2 X 106 m/s
l= h
mv
l = (6.626 X 10-34 J s)
(9.11 X 10-31 kg)(2.2 X 106 m/s)
l = 3.3 X 10-10 m or 0.33 nm
DeBroglie Wavelength: Ex 3
What velocity must a neutron (1.67 X 10-27 kg)
move to have a wavelength of 500 pm
(1 pm = 1X10-12 m)?
500 pm
l= h
mv
v= h
ml
1X10-12 m
1 pm
= 5.0 X 10-10 m
=
(6.626 X 10-34 J s)
= 794 m/s
(1.67 X 10-27 kg)(5.0 X 10-10 m/s)
Quantum Mechanical Model
1. Electron is both wave and particle
2. Probabilistic view – Where does the electron
“hang out” rather than where the electron “is
Quantum Mechanical Model
Electron as a particle
• Heisenberg Uncertainty Principle – can never know
both the position and velocity of an electron at the
same time
• Results
a. Electron moves randomly (not like a planet)
b. Electron cloud – 90% probability
c. Important for transistors/chips
nucleus
Random
electron cloud
Quantum Mechanical Model
Electron as Wave
• Schrodinger Wave
Equation (1926) – treats
electron solely as a
wave
Quantum Mechanical Model
Result One
Explains the forbidden zone (waves do not match)
Forbidden Zone
Quantum Mechanical Model
Result Two
Orbital are not circular
a. Orbital – region of space where there is a significant
chance of finding an electron
b. Does not move like a planet
Three Major Discoveries
Light is both a
wave and a
particle
Electron
Orbitals are
Quantized
Planck
Einstein
Bohr
The electron
(and all matter)
is both a wave
and a particle
De Broglie
(Later
Schrodinger/
Heisenberg)
Quantum Numbers
1. First (principal) QN (n)– how far the electron is
from the nucleus (larger the number, farther
away) – Level or shell
n=2
n=1
2. Second (azimuthal) QN (l) – the shape of the
orbital
Value of l
Letter used
0
s
1
p
2
d
3
f
Quantum Mechanical Model
3. Third (magnetic) QN (ml)– the suborbital
Orbital
# suborbitals
s
p
d
f
0
3 (px,py,pz)
5
7
Total e-
2
6
10
14
• In a magnetic field,
you can “see” the
three p suborbitals in a
line spectrum
4. Fourth (spin) QN (ms)– spin of the electron
Pauli Exclusion Principle – No two electrons in an atom can
have the same four quantum numbers
–
–
Two electrons in the same suborbital (ex: py) must have
opposite spins
Can have values of +1/2 or -1/2
Shows the two electrons, spin +½ and spin -½
n
Possible
Values of l
Possible Values of ml
Subshell
name
1
0
0
1s
2
0
1
0
-1, 0, +1
2s
2p
3
0
1
2
0
-1, 0, +1
-2, -1, 0, +1, +2
3s
3p
3d
4
0
1
2
3
0
-1, 0, +1
-2, -1, 0, +1, +2
-3, -2, -1, 0, +1, +2, +3
4s
4p
4d
4f
Quantum Numbers: Ex 1
What is the designation for a subshell n=5 and l =1?
How many suborbitals are in this subshell?
What are the values of ml for each of the orbitals?
Quantum Numbers: Ex 1
What is the designation for a subshell n=5 and l =1?
5p
How many orbitals are in this subshell?
3
What are the values of ml for each of the orbitals?
-1, 0, +1
Quantum Numbers: Ex 2
Which of the following sets of quantum numbers
are not allowed in the hydrogen atom?
A.
B.
C.
D.
n=2, l=0, ml =0, ms=1/2
n=1, l=0, ml=0, ms= -1/2
n=3, l=1, ml= 2, ms=1/2
n=4, l=2, ml= -2, ms=1/2
Quantum Numbers: Ex 2
Which of the following sets of quantum numbers
are not allowed in the hydrogen atom?
A.
B.
C.
D.
n=2, l=0, ml =0, ms=1/2
n=1, l=0, ml=0, ms= -1/2
n=3, l=1, ml= 2, ms=1/2
n=4, l=2, ml= -2, ms=1/2
Quantum Numbers: Ex 3
Which of the following sets of quantum numbers
are not allowed in the carbon atom?
A.
B.
C.
D.
n=2, l=0, ml =0, ms=1/2
n=1, l=1, ml=0, ms= -1/2
n=2, l=1, ml= -1, ms=1/2
n=4, l=2, ml= -3, ms=1/2
Quantum Numbers: Ex 3
Which of the following sets of quantum numbers
are not allowed in the carbon atom?
A.
B.
C.
D.
n=2, l=0, ml =0, ms=1/2
n=1, l=1, ml=0, ms= -1/2
n=2, l=1, ml= -1, ms=1/2
n=4, l=2, ml= -3, ms=1/2
Electron Configurations
1. Electron Configuration – shorthand notation to
tell you the locations of all the electrons in an
atom or ion
2. Notation
2p3
Orbit
Shape
# e-
Electron Configuration
Electron Configurations
Why is “d” one less?
• Energy is slightly
above the next s
orbital
• Complete the orbital
filling diagram for
manganese
Electron Configurations
1. Easy Examples
H
He
O
2. Write e- configuration
Li
V
N
F
Sr
Ar
P Se
Mg Kr
Fe
S
Electron Configurations
3. Which element is represented by the following
electron configurations?
1s22s22p63s23p64s23d5
1s22s22p63s23p64s23d104p65s24d7
1s22s22p63s23p64s1
1s22s22p63s23p3
1s22s22p63s1
1s22s22p63s23p2
1s22s22p63s23p64s23d104p6
Noble Gas (Condensed) Shortcut
1. Rule – Use the noble gas in the previous row
2. Examples
Ne and P
Ru
Kr
You try:
Br Ar S Ca I Xe
Electron Configuration Exceptions
• Mostly with transition metal elements
• There is a special stability to filled and half-filled
orbitals
Element
Cr
Mo
Cu
Ag
Actual configuration
[Ar]4s13d5
[Kr]5s14d5
[Ar]4s13d10
[Kr]5s14d10
[Ar]4s23d4
[Kr]5s24d4
[Ar]4s23d9
[Kr]5s24d9
Ions
p
Sr
Sr+
2+
Sr
2+
Al
Al3+
e
e- configuration
p
S
1S
S21Br
Ba
2+
Ba
B3+
e
e- configuration
Ions
Transition Metal Ions
Lose their “s” electrons first.
Fe
[Ar]4s23d6
Fe1+
Fe2+
Fe3+
Cu
Cu2+
Ru
Ru3+
Transition Metal Ions
Zn2+
Co1+
Co2+
V3+
Sc2+
Sc3+
4. a) Increase b) Decrease
c) line spectrum(forbidden zone)
14. a) Gamma < (d)yellow < (e)red < (b)93.1 radio <
(c)680 AM
16. a) 3.00 X 1017 Hz
b) 3.94 X 10-3 m
c) (a) is X-Ray and (b) is microwave
d) 7.64 X 10-6 m
18. UV has a higher frequency and shorter l than
IR. UV produces more energy
22.a) 5.09 X 1014Hz
b) 20.3 kJ
c) 3.37 X 10-19 J
d) Na+
24. AM: 6.69 X 10-28 J FM: 6.51 X 10-26 J
26.1.56 X 10-18 J/photon, 127 nm
28. a) microwave
b) 6.4 X 10-11 J/hr
34.a) Absorbed b) Emitted
c) Absorbed
36.a) 9.7 X10-8 m, emitted b) 434 nm, emitted
c) 1.06 X 10-6m, absorbed
38. a) n=1 larger DE b) 121 nm, 103 nm, 97.2 nm
40. a)2.626X10-6m (IR) b) 6 4 transition
42. 3.97 X 10-10 m (3.97 A)
44. 7.75 X 10-11 m (0.775 A)
50. a) n = 3 (l=2,1,0) (9 ml values)
b) n = 5 (l = 4,3,2,1,0) (25 ml values)
52. a) 2,1,1 2,1,0 2,1,-1
b) 5,2,2 5,2,1 5,2,0 5,2,-1 5,2,-2
54. 2p
not allowed
1s
4d
not allowed
5f
3d
not allowed
Solution to 41c
6.941 g X 1kg =0.006941kg/mol
1mol
1000g
0.006941 kg X 1mol
= 1.15 X 10-26kg/atom
1mol
6.022X1023 atoms
l=
h
(1.15 X 10-26kg)(2.5 X 105 m/s)
= 2.3 X 10-13 m
66
C
P
Ne
Config
[He]2s2sp2
[Ne]3s23p3
[He]2s22p6
Core
2
10
2
Valence
4
5
8
Unpaired
2
3
0
68. a) [Ar]4s23d104p1 (1 unpaired)
b) [Ar]4s2 (0 unpaired)
c) [Ar]4s23d3 (3 unpaired)
d) [Kr]5s24d105p5 (1 unpaired)
e) [Kr]5s24d1 (1 unpaired)
f) [Xe]6s14f145d9 (2 unpaired)
g) [Xe]6s24f145d1 (1 unpaired)
70. a) [Ar]3d10 b) [Xe]4f145d8
c) [Ar]3d3 d) [Ne]3s23p6
72.a) 7A
b) 4B
c) 3A
d) Sm and Pm (f-block)
74.a) [He]2s22p3 b) [Ar]4s23d104p4
c) [Kr]5s24d7
76.a) Ba Ca K Na b) Au (shortest) Na(longest)
c) 455 nm, Ba
78 a) 9.37X1014 s-1 b) 374 kJ/mol
c) UV-B Shorter l, higher energy d)UV-B
100.
O3  O2 + O
DH = 105.2 kJ/mol
DH = 1.052 X 105 J/mol
1.052 X 105 J 1 mol
=
1.75 X 10-19 J
1 mol
6.022X1023 molec.
DE = hu
l = c/ u
u = 2.63 X 1014 Hz
l = 1140 nm
Extra Problem
A certain biomolecule requires 598 kJ/mol to break
one of its bonds. What wavelength of light
(nm)would a single photon need to break this
bond?
What region of the EM spectrum is this?
A photon of wavelength 550 nm will break a bond
in a synthetic dye. Calculate the energy per mole
of that bond (kJ/mol)
(218 kJ/mol)
a) Calculate the wavelength of light emitted from a 72 transition in
a hydrogen atom. (398 nm)
b) Calculate the speed an electron would need to have the wavelength.
(1830 m/s)
c) Calculate the wavelength of a proton moving at that same speed.
(2.17 X 10-10 m or 0.217 nm)
d) Why is the proton’s wavelength so much smaller? Which has more
energy, e- or p+?
me = 9.109 X 10-31 kg
mp = 1.673 X 10-27 kg
a) Calculate the wavelength of a proton that has been accelerated to 2 X
106 m/s. (1.98X10-13 m)
b) Calculate the frequency. (1.51 X 1021 Hz)
c) Calculate the energy. (1.00 X 10-12 J/photon)
mp = 1.673 X 10-27 kg
```

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