so far: we solved some of the problems with classical physics
TRANSCRIPT
SO FAR: We Solved some of the
Problems with Classical Physics
•Blackbody Radiation? Plank Quantized
Energy of Atoms:
•Photoelectric Effect? Einstein Quantized energy
states of light photons:
•Compton Effect? Light is a particle with
momentum!
•Matter Waves? deBroglie waves!
•Discrete Spectra?
•Atoms?
nE nhf
E hf
e
h
p
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Continuous vs Discrete
This is a continuous spectrum of colors: all colors are present.
This is a discrete spectrum of colors: only a few are present.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Review: Dispersion: Diffraction Gratings
How does dispersion with a grating compare with a prism?
Longer wavelength light is bent more with a grating.
Shorter wavelength light is bent more with a prism.
sind m
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Incandescent Light Bulb
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Hydrogen Spectra
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Helium Spectra
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Kirkoff’s Rules for Spectra: 1859
Bunsen
German physicist who developed the spectroscope and the science of
emission spectroscopy with Bunsen.
Kirkoff
* Rule 1 : A hot and opaque solid, liquid or highly compressed gas emits a continuous spectrum.
* Rule 2 : A hot, transparent gas produces an emission spectrum with bright lines.
* Rule 3 : If a continuous spectrum passes through a gas at a lower temperature, the transparent
cooler gas generates dark absorption lines.
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Compare absorption lines in a source with emission lines found in the laboratory!
Kirchhoff deduced that elements were present in the atmosphere of the Sun
and were absorbing their characteristic wavelengths, producing the absorption
lines in the solar spectrum. He published in 1861 the first atlas of the solar
spectrum, obtained with a prism ; however, these wavelengths were not very
precise : the dispersion of the prism was not linear at all.
Anders Jonas Ångström 1869Ångström measured the wavelengths on the
four visible lines of the hydrogen spectrum,
obtained with a diffraction grating, whose
dispersion is linear, and replaced
Kirchhoff's arbitrary scale by the
wavelengths, expressed in the metric
system, using a small unit (10-10 m) with
which his name was to be associated.
Line color Wavelength
red 6562.852 Å
blue-green 4861.33 Å
violet 4340.47 Å
violet 4101.74 Å
Balmer Series: 1885Johann Balmer found an empirical equation that correctly
predicted the four visible emission lines of hydrogen
H 2 2
1 1 1
2R
λ n
RH is the Rydberg constant
RH = 1.097 373 2 x 107 m-1
n is an integer, n = 3, 4, 5,…
The spectral lines correspond to different
values of n
Johannes Robert Rydberg generalized
it in 1888 for all transitions:
Hα is red, λ = 656.3 nm
Hβ is green, λ = 486.1 nm
Hγ is blue, λ = 434.1 nm
Hδ is violet, λ = 410.2 nm
Let no one unversed in geometry enter here.
The Universe is made of pure mathematical ideas – the Platonic
Solids. Plato believed that the stars, planets, Sun and Moon move
round the Earth in crystalline spheres.
Earth and the universe were seen as constructed out of five basic
elements: earth, water, air, fire, and ether. The natural place of the
motionless Earth was at the centre of that universe. The stars in the
heavens were made up of an indestructible substance called ether
(aether) and were considered as eternal and unchanging.
Joseph John Thomson
“Plum Pudding” Model 1904
• Received Nobel Prize in
1906
• Usually considered the
discoverer of the electron
• Worked with the
deflection of cathode rays
in an electric field
• His model of the atom
– A volume of positive
charge
– Electrons embedded
throughout the volume
1911: Rutherford’s
Planetary Model of the
Atom
(Couldn’t explain the stability or spectra of atoms.)
•A beam of positively charged alpha
particles hit and are scattered from a
thin foil target.
•Large deflections could not be
explained by Thomson’s pudding
model.
1911: Rutherford’s Planetary
Model of the Atom
(Couldn’t explain the stability or spectra of atoms.)
•A beam of positively charged alpha
particles hit and are scattered from a
thin foil target.
•Large deflections could not be
explained by Thomson’s model.
Electrons exist in quantized orbitals with energies given by
multiples of Planck’s constant. Light is emitted or absorbed
when an electron makes a transition between energy levels. The
energy of the photon is equal to the difference in the energy
levels:
i fE E E hf
34
, n= 0,1,2,3,...
6.626 10
E nhf
h x Js
1. Electrons in an atom can occupy only certain discrete quantized
states or orbits.
2. Electrons are in stationary states: they don’t accelerate and they
don’t radiate.
3. Electrons radiate only when making a transition from one
orbital to another, either emitting or absorbing a photon.
Bohr’s Model 1913
, n= 0,1,2,3,...E nhf
Postulate:
The angular momentum of an electron is
always quantized and cannot be zero:
( 1,2,3,....)2
hL n n
i fE E E hf
: = ( 1,2,3,....)2
hFrom L n mvr n
Bohr’s Derivation of the Energy for Hydrogen:
E K U
F is
centripetal:
Conservation of E:
Sub back into E:
From Angular
Momentum:
(1)
Sub r back into (1):
(2)
Sub into (2):
Why is it negative?
2 2 2
0 0
2 2
4 2
e e ekq q q
vnh hn hn
Bohr Orbital Binding Energy for
The Hydrogen Atom
2
113.6nE eV
n
Ground State: n = 1
First Excited: n = 2
2nd Excited: n = 3
1. -13.6eV is the energy of the H ground state.
2. Negative because it is the Binding Energy and work must be
done on the atom (by a photon) to ionize it.
3. A 13.6eV photon must be absorbed to ionize the ground state
4. Bohr model only works for single electron atoms since it doesn’t
take into account electron-electron interaction forces.
• The frequency of the photon emitted when
the electron makes a transition from an
outer orbit to an inner orbit is
• It is convenient to look at the wavelength
instead
2
2 2
1 1
2ƒ i f e
o f i
E E k e
h a h n n
• The wavelengths are found by
2
2 2 2 2
1 1 1 1 1
2
ƒ eH
o f i f i
k eR
λ c a hc n n n n
Bohr Line Spectra of Hydrogen
2
2 2
1 1 1( )
f i
RZn n
7 11.097 10R x m
Balmer: Visible
Lyman: UV
Paschen: IR
Bohr’s Theory derived the spectra equations that Balmer,
Lyman and Paschen had previously found experimentally!
Bohr: Allowed Orbital Radii
11 2(5.9 10 )nr x m n
n = 1, 2, 3
Bohr Radius (ground state): ao = 5.9x10-11 m
2 2
21 2 3, , ,n
e e
nr n
m k e
2
0nr a n
Problem
a) Which transition, A, B, or C, emits a photon with the greatest energy?
Calculate the wavelength in m and the energy in eV.
b) What is the energy of a photon needed to ionize the atom when the
electron is initially in the third excited state? Calculate in eV.
c) What is the orbital radius of the n=4 state?
A hydrogen atom is in the third (n = 4) excited state. It can make a jump to a
different state by transition A, B, or C as shown, and a photon is either emitted
or absorbed.
2
2 2
1 1 1( )
f i
RZn n
7 1 2
2 2
1 1(1.097 10 )1 ( )
1 4x m
89.72 10x m 110284375m
2
3 213.6
3
ZE eV
2
2
113.6
3eV 1.51eV 2
0nr a n
ao = 5.9x10-11 m
Spontaneous Emission: RandomTransition Probabilities
Fermi’s Golden Rule
Transition probabilities correspond to the intensity of light emission.
IMPORTANT NOTE!!Free electrons have continuous energy states! Only bound
electrons have quantized energy states! This is because they
are not forced to fit in a confined space like a wave in a box!!!
Bound-Bound Transitions: Atoms, boxes
Bound-Free Transitions: Ionization
Free-Bound Transitions: Ion captures an electron
Free-Free Transitions: Collisions, electron absorption
Continuous or Discrete?
CC
C
D
2
0 13.6E Z eV
1. Bohr model does not explain angular momentum postulate.
2. Bohr model does not explain splitting of spectral lines.
3. Bohr model does not explain multi-electron atoms.
4. Bohr model does not explain ionization energies of elements.
1. Bohr model does not explain angular momentum postulate:
The angular momentum of an electron is
always quantized and cannot be zero*:
=2
( 1,2,3,....)
hL n n
n
h
*If L=0 then the electron travels linearly and does not ‘orbit’….
But if it is orbiting then it should radiate and the atom would be unstable…eek gads!
WHAT A MESS!
E hf hp
c c
If photons can be particles, then
why can’t electrons be waves?
e
h
p
Electrons are
STANDING
WAVES in
atomic orbitals.
deBroglie Wavelength:
346.626 10h x J s
2 nr n
( 1,2,3,....)
e nL m vr n
n
h
2e n
hm vr n
1924: de Broglie Waves
Explains Bohr’s postulate of angular
momentum quantization:
h
p 2 nr n
2 n
e
h hr n n
p m v
1926:Schrodinger’s EquationRewrite the SE in spherical coordinates.
Solutions to it give the possible states of the electron.
Solution is sepearable:
Quantum Numbers and their quantization are
derived purely from Mathematical Solutions of
Schrodinger’s Equation (boundary conditions)
However, they have a phenomenological basis!!To be discussed later…
Solution:
Wave Functions given by Quantum
Numbers!!
where the first three radial wave functions of the electron in
a neutral hydrogen atom are
The probability of finding an electron within a shell of
radius r and thickness δr around a proton is
Quantum H Atom• 1s Wave Function:
• Radial Probability Density:
• Most Probable value:
• Average value:
• Probability in (0-r):
0
( )
r
P P r dr
P(r) = 4πr2 |ψ|2
0
( )aver r rP r dr
0/
13
0
1( )
r a
s r ea
0dP
solvedr
Stationary States of HydrogenSolutions to the Schrödinger equation for the hydrogen
atom potential energy exist only if three conditions are
satisfied:
1. The atom’s energy must be one of the values
where aB is the Bohr radius. The integer n is called
the principal quantum number. These energies are
the same as those in the Bohr hydrogen atom.
Stationary States of Hydrogen2. The angular momentum L of the electron’s orbit must
be one of the values
The integer l is called the orbital
quantum number.
3. The z-component of the angular
momentum must be one of the values
The integer m is called the magnetic quantum number.
Each stationary state of the hydrogen atom is identified by a
triplet of quantum numbers (n, l, m).
Interpretation:
Orbital Angular Momentum
& its Z-component are Quantized
( 1) ( 0,1,2,... 1)L l l l n h
( ,.. 1,0,1,2,... )Z l lL m m l l h
Angular momentum magnitude is quantized:
Angular momentum direction is quantized:
Note: for n = 1, l = 0. This means that the ground
state angular momentum in hydrogen is zero, not
h/2, as Bohr assumed. What does it mean for
L = 0 in this model?? Standing Wave
For each l, there are (2l+1) possible ml states.
cos zL
L
The Zeeman Effect is the splitting of spectral lines when a magnetic
field is applied. This is due to the interaction between the external
field and the B field produced by the orbital motion of the electron.
Only certain angles are allowed between the orbital angular
momentum and the external magnetic field resulting in the
quantization of space!
Zeeman Effect Explained
( ,.. 1,0,1,2,... )Z l lL m m l l h ( 1) ( 0,1,2,... 1)L l l l n h
Stern-Gerlach 1921A beam of silver atoms sent through a non-uniform magnetic field was split into
two discrete components. Classically, it should be spread out because the
magnetic moment of the atom can have any orientation. QM says if it is due to
orbital angular momentum, there should be an odd number of components,
(2l+1), but only two components are ever seen. This required an overhaul of
QM to include Special Relativity. Then the electron magnetic ‘spin’ falls out
naturally from the mathematics. The spin magnetic moment of the electron
with two spin magnetic quantum number values:
1
2sm
https://www.youtube.com/watch?v=0_daIZjx-6E
Intrinsic Spin is Quantized
3( 1)
2 2 2
h hS s s
h 1 1 ( , )
2 2 2Z s sS m m
Fine Line Splitting: B field due to intrinsic spin interacts with B
field due to orbital motion and produces additional energy states.
Spin magnitude is quantized and fixed:
Spin direction is quantized:
spin
e
m S
Spin is the Spin
Quantum Entanglement
Quantum Computing
The Qubit
At the heart of the realm of quantum computation is the
qubit. The quantum bit, by analogy with the binary digit, the
bit, used by everyday computers, the qubit is the quantum
computer's unit of currency. Instead of being in a 1 or zero
state, a qubit can be in a superposition of both states.
https://www.youtube.com/watch?v=g_IaVepNDT4
Quantum NumbersOnce the principal quantum number is known, the values of the
other quantum numbers can be found. The possible states of a
system are given by combinations of quantum numbers!
Quantum Rules
1,2,3,4....
0,1,2.... 1
, 1,...,0,1,... 1,
1/ 2, 1/ 2
l
s
n
l n
m l l l l
m
Shells and SubshellsShells are determined by the principle quantum number, n.
Subshells are named after the type of line they produce in the
emission spectrum and are determined by the l quantum number.
s: Sharp line (l = 0)
p: Very bright principle line( l = 1)
d: Diffuse line (l = 2)
f: Fundamental (hydrogen like) ( l =3)
1s
2s
2p
Wave Functions given by Quantum
Numbers!!
where the first three radial wave functions of the electron in
a neutral hydrogen atom are
The probability of finding an electron within a shell of
radius r and thickness δr around a proton is
Wave Functions for Hydrogen
• The simplest wave function for hydrogen is the one that describes the 1s (ground) state and is designated ψ1s(r)
• As ψ1s(r) approaches zero, r approaches and is normalized as presented
• ψ1s(r) is also spherically symmetric
– This symmetry exists for all s states
1 3
1( ) or a
s
o
ψ r eπa
Radial Probability Density
• A spherical shell of radius r
and thickness dr has a volume
of 4πr2 dr
• The radial probability density
function is
P(r) = 4πr2 |ψ|2
where
2 2
1 3
1or a
s
o
ψ eπa
The radial probability density function P(r) is the probability per
unit radial length of finding the electron in a spherical shell of
radius r and thickness dr
22
1 3
4( ) or a
s
o
rP r e
a
1 3
1( ) or a
s
o
ψ r eπa
P(r) for 1s State of Hydrogen
• The radial probability density function for the hydrogen atom in its ground state is
• The peak indicates the most probable location, the Bohr radius.
• The average value of r for the ground state of hydrogen is 3/2 ao
• The graph shows asymmetry, with much more area to the right of the peak
22
1 3
4( ) or a
s
o
rP r e
a
Electron Clouds
• The charge of the electron is
extended throughout a
diffuse region of space,
commonly called an electron
cloud
• This shows the probability
density as a function of
position in the xy plane
• The darkest area, r = ao,
corresponds to the most
probable region
https://www.youtube.com/watch?v=IlkY-HtjrkA
Wave Function of the 2s state
• The next-simplest wave function for the hydrogen
atom is for the 2s state
– n = 2; ℓ = 0
• The wave function is
– ψ 2s depends only on r and is spherically symmetric
32
2
2
1 1( ) 2
4 2or a
s
o o
rψ r e
a aπ
Hydrogen 2s Radial Probability
32
2
2
1 1( ) 2
4 2or a
s
o o
rψ r e
a aπ
The next-simplest wave function for the hydrogen
atom is for the 2s state: n = 2; ℓ = 0
Comparison of 1s and 2s States
• The plot of the radial
probability density for
the 2s state has two
peaks
• The highest value of P
corresponds to the
most probable value
– In this case, r 5ao
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
1926:Schrodinger’s EquationRewrite the SE in spherical coordinates.
Solutions to it give the possible states of the electron.
Solution is sepearable:
Quantum Numbers and their quantization are
derived purely from Mathematical Solutions of
Schrodinger’s Equation (boundary conditions)
However, they have a phenomenological basis!!To be discussed later…
Solution:
Wave Functions given by Quantum
Numbers!!
where the first three radial wave functions of the electron in
a neutral hydrogen atom are
The probability of finding an electron within a shell of
radius r and thickness δr around a proton is
Stationary States of HydrogenSolutions to the Schrödinger equation for the hydrogen
atom potential energy exist only if three conditions are
satisfied:
1. The atom’s energy must be one of the values
where aB is the Bohr radius. The integer n is called
the principal quantum number. These energies are
the same as those in the Bohr hydrogen atom.
Stationary States of Hydrogen2. The angular momentum L of the electron’s orbit must
be one of the values
The integer l is called the orbital
quantum number.
3. The z-component of the angular
momentum must be one of the values
The integer m is called the magnetic quantum number.
Each stationary state of the hydrogen atom is identified by a
triplet of quantum numbers (n, l, m).
Interpretation:
Orbital Angular Momentum
& its Z-component are Quantized
( 1) ( 0,1,2,... 1)L l l l n h
( ,.. 1,0,1,2,... )Z l lL m m l l h
Angular momentum magnitude is quantized:
Angular momentum direction is quantized:
Note: for n = 1, l = 0. This means that the ground
state angular momentum in hydrogen is zero, not
h/2, as Bohr assumed. What does it mean for
L = 0 in this model?? Standing Wave
For each l, there are (2l+1) possible ml states.
cos zL
L
Intrinsic Spin is Quantized
3( 1)
2 2 2
h hS s s
h 1 1 ( , )
2 2 2Z s sS m m
Fine Line Splitting: B field due to intrinsic spin interacts with B
field due to orbital motion and produces additional energy states.
Spin magnitude is quantized and fixed:
Spin direction is quantized:
spin
e
m S
Spin is the Spin
Quantum Entanglement
Quantum Computing
The Qubit
At the heart of the realm of quantum computation is the
qubit. The quantum bit, by analogy with the binary digit, the
bit, used by everyday computers, the qubit is the quantum
computer's unit of currency. Instead of being in a 1 or zero
state, a qubit can be in a superposition of both states.
https://www.youtube.com/watch?v=g_IaVepNDT4
Quantum NumbersOnce the principal quantum number is known, the values of the
other quantum numbers can be found. The possible states of a
system are given by combinations of quantum numbers!
Quantum Rules
1,2,3,4....
0,1,2.... 1
, 1,...,0,1,... 1,
1/ 2, 1/ 2
l
s
n
l n
m l l l l
m
Shells and SubshellsShells are determined by the principle quantum number, n.
Subshells are named after the type of line they produce in the
emission spectrum and are determined by the l quantum number.
s: Sharp line (l = 0)
p: Very bright principle line( l = 1)
d: Diffuse line (l = 2)
f: Fundamental (hydrogen like) ( l =3)
1s
2s
2p
Pauli Exclusion Principle
You must obey Pauli to figure out how to fill
orbitals, shells and subshells.
(This is not true of photons or other force carrying
particles such as gravitons or gluons (bosons))
No two electrons can have the same set of
quantum numbers. That is, no two electrons
can be in the same quantum state.
Maximum number of electrons in a shell is: # = 2n2
From the exclusion principle, it can be seen that only two
electrons can be present in any orbital: One electron will have
spin up and one spin down.
Maximum number of electrons in a subshell is: # = 2(2l+1)
Hund’s Rule
• Hund’s Rule states that when an atom has
orbitals of equal energy, the order in which
they are filled by electrons is such that a
maximum number of electrons have
unpaired spins
– Some exceptions to the rule occur in elements
having subshells that are close to being filled or
half-filled
Hund’s Rule
The filling of electronic states must obey both the Pauli
Exclusion Principle and Hund’s Rule.
Hund’s Rule states that when an
atom has orbitals of equal energy,
the order in which they are filled
by electrons is such that a
maximum number of electrons
have unpaired spins
Some exceptions to the rule
occur in elements having
subshells that are close to
being filled or half-filled
https://www.youtube.com/watch?v=Aoi4j8es4gQ
How many electrons in n=3 Shell?
2e 6e 10e =18e
= 2n2
3
0,1,2
2, 1,0,1,2
1/ 2, 1/ 2
l
s
n
l
m
m
Quantum Rules
1,2,3,4....
0,1,2.... 1
, 1,...,0,1,... 1,
1/ 2, 1/ 2
l
s
n
l n
m l l l l
m
2(2 1)l
Spontaneous Emission: RandomTransition Probabilities
Fermi’s Golden Rule
Transition probabilities correspond to the intensity of light emission.
Excited States and SpectraAn atom can jump from one stationary state, of energy E1,
to a higher-energy state E2 by absorbing a photon of
frequency
In terms of the wavelength:
Note that a transition from a state in which the valence
electron has orbital quantum number l1 to another with
orbital quantum number l2 is allowed only if
Hydrogen Energy Level Diagram
Selection Rules• The selection rules for allowed
transitions are
– Δℓ = ±1
– Δmℓ = 0,±1
• Transitions in which ℓ does not change are very unlikely to occur and are called forbidden transitions
– Such transitions actually can occur, but their probability is very low compared to allowed transitions
X-Ray ProductionThe discrete lines are called characteristic x-rays. These are created when
• A bombarding electron collides with a target atom
• The electron removes an inner-shell electron from orbit
• An electron from a higher orbit drops down to fill the vacancy
• Typically, the energy is greater than 1000 eV
• The continuous spectrum is called bremsstrahlung, the German word
for “braking radiation”
X Ray Imaging
when X-ray light shines on us, it goes through our skin, but allows shadows of our bones to be projected onto and captured
by film.
When X-ray light shines on us, it goes through our skin, but allows
shadows of our bones to be projected onto and captured by film.
when X-ray light shines on us, it goes through our skin, but allows shadows of our bones to be projected onto and captured
by film.
when X-ray light shines on us, it goes through our skin, but allows shadows of our bones to be projected onto and captured
by film.
when X-ray light shines on us, it goes through our skin, but allows shadows of our bones to be projected onto and captured
by film.
Lasers
Light Amplification by
Stimulated Emission of Radiation
"A splendid light has dawned on me about the absorption and
emission of radiation..."
Albert Einstein, 1916
Stimulated Emission
1 photon in, 2 out
"When you come right down
to it, there is really no such
thing as truly spontaneous
emission; its all stimulated
emission. The only distinction
to be made is whether the field
that does the stimulating is one
that you put there or one that
God put there..."
David Griffths, Introduction to
Quantum Mechanics
https://www.youtube.com/watch?v=y3SBSbsdiYg
https://www.youtube.com/watch?v=lW4Uq_2VPhE
Sample Problem
hc
E
124 10590 10 590 nm
67.
. eV m
2.10 eV m
What average wavelength of visible
light can pump neodymium into
levels above its metastable state?
From the figure it takes 2.1eV to
pump neodymium into levels
above its metastable state. Thus,
Phosphorescence
Time Delayed Fluorescence
Glow in the Dark: Day Glow
BTW: Iridescence: Diffraction