Transcript Quantum Mechanics

Alright class we are going back
to quantum numbers.
I decided this would be a better lead
into dipole then just throwing you
into the mess.
Do you remember Wave and
Frequency?
• Do you remember plank. Well he is back it
is time to work on his constant and what it
means.
• Also the electromagnetic spectrum.
Properties of Waves
Wavelength (l) is the distance between identical points on
successive waves.
Amplitude is the vertical distance from the midline of a
wave to the peak or trough.
Frequency (n) is the number of waves that pass through a
particular point in 1 second (Hz = 1 cycle/s).
The speed (u) of the wave = l x n
3
Maxwell (1873), proposed that visible light consists of
electromagnetic waves.
Electromagnetic
radiation is the emission
and transmission of energy
in the form of
electromagnetic waves.
Speed of light (c) in vacuum = 3.00 x 108 m/s
All electromagnetic radiation
lxn=c
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5
A photon has a frequency of 6.0 x 104 Hz. Convert this
frequency into wavelength (nm). Does this frequency fall in
the visible region?
6
The wavelength of the green light from a traffic signal is
centered at 522 nm. What is the frequency of this
radiation?
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Problem 1
What is the wavelength (in meters) of an
electromagnetic wave whose frequency is
3.64 x 107 Hz?
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Mystery #1, “Heated Solids Problem”
Solved by Planck in 1900
When solids are heated, they emit electromagnetic radiation
over a wide range of wavelengths.
Radiant energy emitted by an object at a certain temperature
depends on its wavelength.
Energy (light) is emitted or
absorbed in discrete units
(quantum).
E=hxn
Planck’s constant (h)
h = 6.63 x 10-34 J•s
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Calculate the energy (in joules) of a photons with a wavelength
of 5.00 x 104 nm (infrared region).
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Calculate the energy (in joules) of a photons with a wavelength
of 5.00 x 10-2 nm (x-ray region).
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Problem 2
The energy of a photon is 5.87 x 10-20 J. What
is its wavelength in nanometers?
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Mystery #2, “Photoelectric Effect”
Solved by Einstein in 1905
Light has both:
1. wave nature
2. particle nature
hn
KE e-
Photon is a “particle” of light
hn = KE + W
KE = hn - W
where W is the work function and
depends how strongly electrons
are held in the metal
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When copper is bombarded with high-energy electrons, X rays
are emitted. Calculate the energy (in joules) associated with
the photons if the wavelength of the X rays is 0.154 nm.
14
The work function of cesium metal is 3.42 x 10-19 J. Calculate
the minimum frequency of light necessary to eject electrons
from the metal.
15
The work function of cesium metal is 3.42 x 10-19 J. Calculate
the kinetic energy of the ejected electron if light of frequency
1.00 x 1015 s-1 is used for irradiating the metal.
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Problem 3
The work function of titanium metal is 6.93 x
10-19 J. Calculate the energy of the ejected
electrons if light of frequency 2.50 x 1015 s-1 is
used to irradiate the metal.
KE = _____x 10-19 J
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Mystery #3, “Emission Spectra”
Line Emission Spectrum of Hydrogen Atoms
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Bohr’s Model of
the Atom (1913)
1. e- can only have specific
(quantized) energy
values
2. light is emitted as emoves from one energy
level to a lower energy
level
En = -RH (
1
n2
)
n (principal quantum number) = 1,2,3,…
RH (Rydberg constant) = 2.18 x 10-18J
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E = hn
E = hn
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Ephoton = DE = Ef - Ei
ni = 3
ni = 3
ni = 2
nf = 2
1
Ef = -RH ( 2
nf
1
Ei = -RH ( 2
ni
1
DE = RH( 2
ni
)
)
1
n2f
nnf f==11
22
)
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Calculate the wavelength (in nm) of a photon emitted
by a hydrogen atom when its electron drops from the
n = 5 state to the n = 3 state.
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Problem 4
What is the wavelength (in nm) of a photon
emitted during a transition from ni=6 to nf=4
in the H atom?
ν = _____x 103 nm
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Why is e- energy quantized?
De Broglie (1924) reasoned
that e- is both particle and
wave.
2pr = nl
h
l = mu
u = velocity of em = mass of e26
What is the de Broglie wavelength (in nm) associated
with a 2.5 g Ping-Pong ball traveling at 15.6 m/s?
27
Calculate the wavelength of the ‘particle’ in each of
the following two cases: (a) The fastest serve ever
recorded in tennis was about 150 miles per hour, or
68 m/s. Calculate the wavelength associated with a
6.0 x 10-2 kg tennis ball travelling at this speed.
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Calculate the wavelength of the ‘particle’ in each of
the following two cases: (b) Calculate the wavelength
associated with an electron (9.1094 x 10-31kg)
travelling at this speed.
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Problem 5
Calculate the wavelength (in nm) of a H
atom (mass = 1.674 x 10-27 kg) moving at
7.00 x 102 cm/s.
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Chemistry in Action: Laser – The Splendid Light
Light Amplification by Stimulated Emission of Radiation
Laser light is (1) intense, (2) monoenergetic, and (3) coherent
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Chemistry in Action: Electron Microscopy
le = 0.004 nm
Electron micrograph of a normal
red blood cell and a sickled red
blood cell from the same person
STM image of iron atoms
on copper surface
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Shortcomings of Bohr’s model
•Did not account for the emission spectra of atoms
containing more than on electron.
•Did not explain extra lines in the emission spectra for
hydrogen when magnetic field is applied.
•Conflict with discovery of ‘wavelike’ properties – how can
you define the location of a wave?
Heisenberg Uncertainty Principle
It is impossible to know simultaneously both the
momentum p (defined as mass times velocity) and the
position of a particle with certainty.
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Schrodinger Wave Equation
In 1926 Schrodinger wrote an equation that described both
the particle and wave nature of the e-
 h

iH  (r , t )
2
 2   V  (r , t ) =
2p
t
 8p m

2
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Schrodinger Wave Equation
Wave function () describes:
1. energy of e- with a given 
2. probability of finding e- in a volume of space
Schrodinger’s equation can only be solved exactly
for the hydrogen atom. Must approximate its
solution for multi-electron systems.
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Schrodinger Wave Equation
 is a function of four numbers called
quantum numbers (n, l, ml, ms)
principal quantum number n
n = 1, 2, 3, 4, ….
distance of e- from the nucleus
n=1
n=2
n=3
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Where 90% of the
e- density is found
for the 1s orbital
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Schrodinger Wave Equation
quantum numbers: (n, l, ml, ms)
angular momentum quantum number l
for a given value of n, l = 0, 1, 2, 3, … n-1
n = 1, l = 0
n = 2, l = 0 or 1
n = 3, l = 0, 1, or 2
l=0
l=1
l=2
l=3
s orbital
p orbital
d orbital
f orbital
Shape of the “volume” of space that the e- occupies
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l = 0 (s orbitals)
l = 1 (p orbitals)
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l = 2 (d orbitals)
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Schrodinger Wave Equation
quantum numbers: (n, l, ml, ms)
magnetic quantum number ml
for a given value of l
ml = -l, …., 0, …. +l
if l = 1 (p orbital), ml = -1, 0, or 1
if l = 2 (d orbital), ml = -2, -1, 0, 1, or 2
orientation of the orbital in space
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ml = -1, 0, or 1
3 orientations is space
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ml = -2, -1, 0, 1, or 2
5 orientations is space
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Schrodinger Wave Equation
(n, l, ml, ms)
spin quantum number ms
ms = +½ or -½
ms = +½
ms = -½
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Schrodinger Wave Equation
quantum numbers: (n, l, ml, ms)
Existence (and energy) of electron in atom is described
by its unique wave function .
Pauli exclusion principle - no two electrons in an atom
can have the same four quantum numbers.
Each seat is uniquely identified (E, R12, S8)
Each seat can hold only one individual at a
time
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Schrodinger Wave Equation
quantum numbers: (n, l, ml, ms)
Shell – electrons with the same value of n
Subshell – electrons with the same values of n and l
Orbital – electrons with the same values of n, l, and ml
How many electrons can an orbital hold?
47
How many 2p orbitals are there in an atom?
How many electrons can be placed in the 3d subshell?
48
List the values of n, ℓ, and mℓ for orbitals in the 4d subshell.
What is the total number of orbitals associated with the
principle quantum number n=3?
49
Problem 6
Give the values of the quantum numbers
associated with the orbitals in the 3p subshell.
n = ____
ℓ = ____
mℓ=____
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Problem 7
What is the total number of orbitals associated
with the principle quantum number n=4.
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Energy of orbitals in a single electron atom
Energy only depends on principal quantum number n
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Energy of orbitals in a multi-electron atom
Energy depends on n and l
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“Fill up” electrons in lowest energy orbitals (Aufbau principle)
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The most stable arrangement of electrons in
subshells is the one with the greatest number of
parallel spins (Hund’s rule).
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Order of orbitals (filling) in multi-electron atom
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s
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Write the four quantum numbers for an
electron in a 3p orbital.
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Problem 8
Write the four quantum numbers for an
electron in a 4d orbital.
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Electron configuration is how the electrons are
distributed among the various atomic orbitals in an
atom.
number of electrons
in the orbital or subshell
1s1
principal quantum
number n
angular momentum
quantum number l
Orbital diagram
H
1s1
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What is the electron configuration of Mg?
What are the possible quantum numbers for the last
(outermost) electron in Cl?
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What is the maximum number of electrons that can be
present in the principle level for which n=3?
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Problem 9
Calculate the total number of electrons that
can be present in the principle level for which
n=4
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An oxygen atom has a total of eight electrons. Write the four
quantum numbers for each of the electrons in the ground
state..
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Problem 10
Write a complete set of quantum numbers for
each of the electrons in boron.
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Paramagnetic
unpaired electrons
2p
Diamagnetic
all electrons paired
2p
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Outermost subshell being filled with electrons
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Write the ground state electron configuration for sulfur
Write the ground state electron configuration for palladium
which is diamagnetic
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Problem 11
Write the ground state configuration for
phosphorus.
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