so far: we solved some of the problems with classical physics

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.

Continuous

Discrete


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


Incandescent Light Bulb


Hydrogen Spectra

http://www.assumption.edu/users/bniece/CHE131/LineSpectra/HiResolution/H2.pdf



Helium Spectra

http://www.assumption.edu/users/bniece/CHE131/LineSpectra/HiResolution/He.pdf

http://www.assumption.edu/users/bniece/CHE131/LineSpectra/HiResolution/He.pdf


Mercury Spectra


Neon Spectrum

The Evidence….


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.


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.


Kirkoff’s Rules

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

Theories….

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.

Classical Physics at the Limit

WHY IS MATTER (ATOMS)

STABLE?

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

Light Absorption & Emissioni fE E E hf

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


Numbers!!





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



Hydrogen 1s Radial Probability

Sample Problem

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


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

Hydrogen 2p Radial Probability

Hydrogen 3p Radial Probability

Hydrogen 3d Radial Probability

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:


Numbers!!





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.

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

Bohr model does not explain ionization energies of elements but QM

does. How?

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”

Synchrotron Radiation

X-Ray and Radio

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.


by film.


by film.


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



Stimulated Emission

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,

Laser Applications

Phaser?

Fluorescence

UV in, vibrant color out.

Phosphorescence

Time Delayed Fluorescence

Glow in the Dark: Day Glow

http://www.asl.ch/images/dayglow.jpg

http://www.asl.ch/images/dayglow.jpg

BTW: Iridescence: Diffraction

http://graphics.stanford.edu/courses/cs348b-competition/cs348b-01/butterfly_big.jpg

http://graphics.stanford.edu/courses/cs348b-competition/cs348b-01/butterfly_big.jpg

http://newton.ex.ac.uk/research/emag/butterflies/images/rheten6.jpg

http://newton.ex.ac.uk/research/emag/butterflies/images/rheten6.jpg

http://newton.ex.ac.uk/research/emag/butterflies/images/ole5.jpg

http://newton.ex.ac.uk/research/emag/butterflies/images/ole5.jpg

so far: we solved some of the problems with classical physics

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