Download λ - Chemistry 7

Chapter 7:
Quantum Theory and the
Electronic Structure of
Atoms
EM radiation
is
units called
are
absorbed
a wave and a
particle
having
Quanta
amplitude
emitted
energy
frequency
related by
involve energy changes in
atoms
atoms
electrons
molecules
described by
Wave functions
e- configuration
determined by
Aufbau Rules
comprising
wavelength
related by
λ
In 1873, Maxwell found that light was an electromagnetic wave that
travels at the at the speed of light, c.
speed of light = c = λ x ν = 2.99 x 108 m/s
The wavelength, λ,
is the crest-to-crestdistance in space.
The frequency ν, is
the number of times
per second that a
crest passes a given
point on the x-axis.
Electric field is perpendicular to an
oscillating magnetic field both perpendicular
direction to both direction of propagation
λ = wavelength
λ
The wavelength, λ, is the crest-tocrest-distance in space of the
waveform taken at an instant in time.
The amplitude, A of the electric field
is the maximum disturbance of the
waveform.
The frequency ν, is the number of
times per second that a crest passes
any given point on the x-axis. It is
related to the period (sec) of the wave
by f = 1/T
The frequency ν, and wavelength are
related by a constant, the speed of
light, c
=λxν
λ
The speed of light is a constant that connects the wavelength to light
frequency.
speed of light = c = λ x ν = 2.99 x 108 m/s
10-3
0.10
10
103
105
107
1020
1018
1016
1014
1012
1010
λ (nm)
ν (Hz)
Gamma X-rays UV
IR
109
108
Microwave
1011
1013
106
104
Radio Waves
Radiation
400nm
500nm
600nm
700nm
The Visible Spectrum of Light (wavelength)
Red
Orange
Yellow
Green
Blue
Indigo
Violet
700 nm
525 nm
430 nm
370 nm
Must learn frequency-wavelength calculations
7.1
Your new cordless phone has a frequency of 2.4 GHz.
What is the wavelength of the radiation used by the
phone? In what region of the EM spectrum is this?
7.2
Yellow light has a wavelength of 598 nm. What is the
frequency of this light?
7.3
A certain type of electro-magnetic radiation has a
frequency of 3.5 × 1015 Hz. What is the wavelength of
this radiation? In what region of the spectrum is this?
Many experiments over 300 years have shown that
electromagnetic radiation (light) can exhibit properties of a
particle and at other times properties of a wave.
Light Intensity
Slit
Screen
When light is passed through a
tiny single slit one bright line
appears.
Light Intensity
Two Slits
Screen
When two slits spaced closely
together a pattern of bright and
dark appears showing that light
is an wave that interfers.
It has been known since the 1600’s that light acts like a
wave; it can be diffracted by passing light through
small slits producing an interference pattern.
Known in 1600’s
by Huygens
Wall of Screen
Projection of Wall
Bright spot
Dark spot
Light
Waves
Film
With Slit
Interference Pattern
Despite knowing about the existence of waves and what
appeared to be a particle-based nuclear atom there were
mysteries involving light and matter that the physicists
could not explain in the early 1900’s.
I. Black-body Radiation Problem - the
frequencies and intensities of light emitted by
heated solids at a given temperature.
II. The Photo-Electric Effect - the ejection of
electrons from certain metals upon exposure
of light depended on the frequency of light,
not its intensity (how bright).
III. Line Emission and Absorption Spectra
- compounds, gases emitted only emitted or
absorb very specific frequencies (colors) of
light.
Weins Displacement Law says that at some temperature T all matter
emits a distribution of radiation (blackbody) with one particular
wavelength dominating.
λmax
2898 microns
=
T(Kelvin)
The Sun
The Earth
MYSTERY #1: As solids are heated to higher
temperature they emit electromagnetic radiation of
higher frequencies (lower wavelength) and higher
intensities of light. Classical physics could not fit
experiment observations.
smoldering coal
Classical physics
predicts
electric heating element
light bulb filament
Plank
Resolves
MaxMax
Plank’s
Solution
toBlackbody
BlackbodyProblem
The Concept of the Quantum Is Born (Year 1900)
• Planck postulated that energy and light can exist in
fixed quantities or “quanta” vs continuous range
of energies that our macroscopic world seems to
be.
• Plank’s finds that the energy of electromagnetic
radiation (light) is proportional to it’s frequency (not
its intensity) and that the energies are in whole
number quantities.
ν= frequency of light
(sec-1)
E=nhν
energy
(joules)
n=
whole
number
h= Plank’s
constant =
6.63 x 10-34 J s
MYSTERY #2: When a photoelectric tube is bombarded
with electromagnetic radiation (light), only specific
frequencies eject electrons from the surface of the light
sensitive plate to generate current. Einstein finds that
light acts like a particle. He calls the light particles
“photons” and each photon has an energy = hν.
Light
Freed
e-
+
pole
battery
evacuated tube
kinetic
energy of e-
light sensitive
metal plate
Current
meter
Ephoton = hν = KE + BE
energy of
1 photon
Plank’s
constant
light
frequency
binding
energy of e-
Albert Einstein explained the
photoelectric effect in 1905.
incoming red light
collector
• Light acts like a particle and are
“quantized energy packets”
called photons.
• 1 photon has energy = E = hν.
no e- ejected
• 1 mole of photons = E x Avog. #
• Photon has to have a certain
minimum energy to eject an
electron from the metal. If the
photon does not have enough
energy the electron will remain
with the metal (no matter how
intense the light is (the number
of photons per unit time)
• Greater intensity of the wrong
frequency (more photons) does
not overcome the electron
binding energy. It is the energy
(frequency) and not the
intensity that is key.
0
battery
ammeter
no current
incoming red light
emitter
collector
e- ejected
50
battery
current flows
Learning Check: Photon-Energy Calculations
1.
Calculate the energy, in joules per photon, of
radiation that has a frequency of 4.77 x 1014 s-1.
• E = hν = 6.63 x 10-34 J s 4.77 x 1014 s-1
2.
3.
If the energy of an infrared krypton laser is 2.484
x 10-19 joules per photon, determine the
wavelength of the radiation.
A cook uses a microwave oven to heat a meal. The
wavelength of the radiation is 1.20 cm. What is the
energy of one photon of this microwave radiation?
Note: The value of Planck’s constant is 6.626 x 10-34 J/s per
photon.
MYSTERY #3: Narrow bands of colors of light are emitted
when gases are excited by high voltage. The colors are
characteristic and reproducible for different elements.
The light generated by the
hydrogen gas is dispersed
into its component
wavelengths by a prism
and projected on a screen
High voltage
tube containing
hydrogen gas
Prism
Screen
All elements display a characteristic emission spectra that we can
use to fingerprint all elements.
Li
Na
K
Ca
Sr
Ba
Zn
Cd
Hg
H
He
Ne
Ar
alkali
metals
alkaline
earth
Metals
gases
Flame color tests cause a similar electron excitement and
are characteristic for elements. The color says something
tells us electronic structure E = hν of the elements and can
help identify elements.
Hydrogen
Helium
Lithium
Sodium
Emission spectra of elements
Potassium
The beauty and intriguing color of fireworks arises from the
characteristic color of elements in their salts.
The First Explanations Were Empirical
In 1885 Balmer developed an empirical mathematical
relationship between ν and n for the visible lines of hydrogen
emission spectrum.
Balmer
equation
λ (nm) = 364.56
!
2
n
n 2 − 22
"
In 1885 Johann Rydberg developed a generalized but
empirical mathematical relationship between ν and n for the all
lines of hydrogen emission spectrum! This is freaky!
Rydberg
equation
1
λ
= R
1
n12
-
1
n22
How could scientists account for the lines if
atomic theory and classical physics were
correct?
How does the emission spectrum relate
the atomic and electronic structure of the H
atom?
λ (nm)
for the visible series, n1 = 2 and n2 = 3, 4, 5, ...
In 1913, scientist Neils Bohr uses a planetary-quantum mathematical
model that explains the emission spectrum of the hydrogen atom.
1. e- can only have specific
(quantized) energy values
(planetary like orbits)
2. light is emitted as e- moves from
one energy level to a lower energy
level
n =6
n =5
n =4
n =3
n =2
( )
En = -RH 12
n
4. the difference in energy between
any two levels:
3. the energy of a level:
ΔE = RH
(
1
n2i
1
n2
f
)
i = initial e- state and f = final e- state.
RH (Rydberg constant) = 2.18 x 10-18J
Bohr postulates that light is
emitted when an e- goes from an
orbit with a high n value to a lower
n value
The Bohr explanation of the H spectral lines.
n
n =6
ΔE = RH (
1
n2i
1
n2f
)
RH (Rydberg constant) = 2.18 x 10-18J
Quantum staircase.
Bohr’s theory worked well for hydrogen only
but could not explain the emission of other
atoms. Classical physics also says that
electrons can not accelerate around a nucleus
without eventually collapsing into the nucleus
(i.e. there is no planetary motion for charges).
How and why is e- energy quantized
and what physics can explain it?
In 1924, while working on his PhD thesis, Louis
DeBroglie reasoned that if light wave could act like a
particle then perhaps a particle could act like a wave.
He connects Plank’s Law to Einstein’s mass
equivalence.
E = hν = hc/λ =
Plank’s Law
Louis DeBroglie
mc2
Einstein’s Law of
Mass equivalence
Solving for wavelength and watching our units
h
λ = mv
λ = wavelength (m)
v = velocity (m/s)
h = Planck’s constant (6.626 × 10-34 J-s)
The wave-like
properties of matter
only manifest when
matter has a very small
mass and high velocity:
electron that move at
high speeds.
Matter can be a particle and a wave!
The de Broglie Wavelengths of Selected Matter
Speed (m/s)
de Broglie λ (m)
Substance
Mass (g)
slow electron
9 x 10-28
1.0
7x10-4
fast electron
9 x 10-28
5.9 x 106
1x10-10
alpha particle
6.6 x 10-24
1.5 x 107
7x10-15
one-gram mass
1.0
0.01
7x10-29
baseball
120
25.0
1.2x10-34
Earth
6.0 x 1027
3.0x104
4x10-63
h
λ=
mv
What is the de Broglie wavelength (in meters) of: a) pitched
baseball with a mass of 120. g and a speed of 100 mph or
44.7 m/s and b) of an electron with a speed of 1.00 x 106 m/s
(electron mass = 9.11 x 10-31 kg; h = 6.626 x 10-34 kg m2/s).
DeBroglie
wavelength
(m)
Plank’s
Constant = h =
6.63 x 10-34 J s
h
λ=
mν
mass of object
(kg)
velocity of
object (m/s)
h
6.626 × 10−34 kg m2 /s
λ=
=
= 1.24 × 10−34 m
mν
(0.120 kg) (44.7 m/s)
λ=
6.626 x 10-34 kg m2/s
(9.11x10-31 kg)(1.00x106 m/s)
= 7.27 x 10-10 m
Soon after de Broglie’s theory was put forth it was found
that e- particles could be made to diffract and interfere just
like light waves.
x-ray diffraction of aluminum foil
electron diffraction of aluminum foil