IV-1 1

PART FOUR: Introduction to Quantum Concepts in Chemistry The Wave Particle Duality (DeK. & G. p. 20)

• 1924: De Broglie hypothesized that all matter possesses wave-like properties.

• In quantum mechanics (& this course) this is

most important when dealing with electrons.

ph mν

h λ ==

mν ≡ momentum (measure of inertia or tendency of particle to remain in motion)

• Prince Louis-Victor Pierre Raymond de Broglie : Nobel Prize in Physics 1929

• http://www.nobel.se/physics/laureates/1929/broglie-lecture.html

IV-2 2

Can be rationalized as follows:

• E = hν = mc2 ; c = λν

•

λh mc ; mc

λhc 2 ==

by analog, de Broglie proposed that a material particle with mass m and speed ν would have a wavelength p

h mνh λ == .

• de Broglie wavelength of a electron moving at 1.0×106 m/s is

7 m/s) 10kg)(110(9.1

Js106.6 λ 631-34- =××

×= Å

• This is the right order of magnitude for

molecules. http://www.colorado.edu/physics/2000/quantumzone/debroglie.html

IV-3 3

• de Broglie λ for a 1.0 g particle moving at

1.0 m/s is 7×10-27 cm.

• Quantum effects are negligible for macroscopic particles.

1927: Clinton, Davisson & Germer demonstrated (experimentally) that electrons are diffracted by crystals just like X-rays

IV-4 4

It is interesting to recall that G.P. Thompson, who shared the 1937 Nobel Prize with Davisson for these experiments which proved that electrons are waves, is the son of J.J. Thompson who received the Nobel Prize in 1906 for proving that cathode rays were actually particles - electrons! And the amazing thing is that they were both right.

IV-5 5

Light:

When light from a point source passes through a small circular aperture, it does not produce a bright dot as an image, but rather a diffuse circular disc known as Airy's disc surrounded by much fainter concentric circular rings.

IV-6 6

This diagram shows the phenomenon of light interference. A screen with a double slit, equally separated, is illuminated with a bright light, the two slits cause diffraction of the light waves, The two slits, produce two sets of wave fronts which overlap. There will be either Constructive Interference ( bright area ) or Destructive Interference ( dark area ).

IV-7 7

The Heisenberg Uncertainty Principle • Impossible to know simultaneously both the

momentum and position of a particle. Electrons behave strangely as we will see.

π4h x)(∆ p) (∆ x ≥

uncertainty in position

uncertainty in momentum (wave-like)

• h is Planck’s constant 6.6×10-34 Js

IV-8 8

This “thought” experiment introduced by Heisenberg, helps clarify this idea, is introduced with the following illustration. For the electron to be visible, at least one photon of light must bounce off of it, and then pass through the microscope into your eye. A problem occurs here, as the photon transfers some unknown amount of its momentum to the electron. Thus, in the process of finding the position of the electron the same light that allows you to see it changes the electron's momentum

IV-9 9

Heisenberg Uncertainty principle is one of the major results of Quantum Mechanics.

• Impossible to specify position & momentum of a particle simultaneously.

• DENIES possibility of ascribing electrons to particular orbits like planets around the sun.

This uncertainty leads to many strange things. For example, in a Quantum Mechanical world, I (or anyone else) cannot predict where a particle will be with 100 % certainty. I can only speak in terms of probabilities. For example, I can say that an atom will be at some location with a 99 % probability, but there will be a 1 % probability it will be somewhere else (in fact, there will be a small but finite probability that it will be found across the Universe). This is strange. “God does not play dice” We do not know if this indeterminism is actually the way the Universe works because the theory of Quantum Mechanics is probably incomplete. That is, we do not know if the Universe actually behaves in a probabilistic manner (there are many possible paths a particle can follow and the observed path is chosen probabilistically) or if the Universe is deterministic in the sense that I can predict the path a particle will follow with 100 % certainty.

IV-10 10

“God does not play dice”. Albert Einstein

"God not only plays dice, he sometimes throws the dice where they cannot be seen" Stephen Hawking This argument is not our concern. We wish to learn quantum mechanics to understand bonding and reactivity.

IV-11 11

Electrons

• Wave-Particle Duality

• If a wave-motion is confined to a finite region in space a series of patterns of standing waves result.

• Electron is confined by columbic field of the nucleus.

--- “orbitals”

• Orbitals.

Science, Vol. 187, 1975, pp. 605-612. “Of Atoms, Mountains & Stars: A Study In Qualitative Physics” Victor F. Weisskopf.

IV-12 12

Atomic Orbitals (DeK & G. p. 23 sec. 1.10)

• If electrons moved in simple orbits, p and x could be determined.

• Violates Heisenberg Uncertainty Principle.

• We can only talk about Probability of

finding electron.

• Electron Density or Probability Density. Chemical structure & Bonding, R. DeKock and H. Gray, Pub. Univ. Sci. Book.

(a) (b)

IV-13 13

The Wave Equation and the Particle in a Box

• What are we trying to do?

• We need to CONFINE an electron to an atom.

• When we do this, we find that the electron

can only have certain energies.

• Moreover, the states that the electron resides in has shape (“orbitals”).

Before we try to set this up the wave mechanics for the simplest atomic problem (i.e. H Atom) we do an even more simple problem.

IV-14 14

The Electron in a Box

• “Box” – region of space where potential energy of particle is zero.

• Potential rises to infinity at the walls.

“Square Well Potential”

∞ ∞ V

• One dimensional square well potential.

• Particle cannot get out of box.

IV-15 15

• Box can be 1, 2 or 3 Dimensional ↑ ↑

e.g. Plane e.g. Cubic Box, Sphere

• Energy of Particle inside box is all Kinetic. We can solve the particle in a box using Wave Mechanics. Waves

• You have met wave equations before.

• Look like

y = sinx A = ksinx + k’cosx A = sin(wt) A = kexp(±ikx) = cos(kx)± isin(kx)

IV-16 16

0A

x

λ

ψ

ψ(x) = Asin(2πx/λ) (As x goes form 0 → λ we go through one complete oscillation) (More General is:

ψ(x) = Asin(2πx/λ) + Bcos(2πx/λ) Classical Wave Equation

∇2ψ(x, y, z) = -(2π/λ)2ψ(x, y, z)

∇2 = (∂2/∂x + ∂2/∂y + ∂2/∂z) (more on that one a bit later)

IV-17 17

Waves in box (atom) will form as standing waves:

IV-18 18

Back to the Particle in a Box BTW there are many web pages devoted to this problem. Play around with them!! For example: http://www.chem.uci.edu/education/undergrad_pgm/applets/dwell/dwell.htm

• Particle (mass m) free to move along x-axis between x = 0 and x = L.

• No change in potential (set V=0 for 0<x<L).

• At x = 0 and x = L and beyond the particle

meets an infinite barrier.

• We now solve this problem using de Broglie’s “matter waves”.

IV-19 19

i.e. A(x) = mνhλ ;

λx2πkcos =

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

So:

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

λx2πksin ψ(x)

(we have changed A for Q)

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

λ x2πsin

λ2πk

dxψd 2

22

ψ(x) 2

λ2π ⋅−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

But p mv h/λ == (de Broglie)

IV-20 20

So:

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

λx2πksin ψ(x)

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

λ x2πsin

λ2πk -

dxd 2

22ψ

ψ(x) 2

λ2π- ⋅=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

• For a moment let us think about the number

that this has generated (-(2π/λ)2 ) • Remember the energy of the particle is All

Kinetic ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛= 2mv2

1 E . (Isaac Newton)

IV-21 21

But mv = h/8 = p (de Broglie) m2v2 = h2/82 = p2

m2v2/2m = ½ mv2 = E = p2/2m So: m2v2 = h2/82 = p2 = 2mE ˆ 1/82 = p2 / h2

We are looking for -(2π/λ)2

(2π/λ)2 = (p2 / h2 ) A (2B)2 = p2 / ħ2 (or 2mE/ ħ2)

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡ = 2πh h

IV-22 22

So the number we produced is the Energy of the electron.

ψ(x) p- dxψd 2

22

2⋅=∴

h With a little insight we can see why this is important. Recall the last page of the previous lecture……. Classical Wave Equation

∇2ψ(x, y, z) = -(2π/λ)2ψ(x, y, z)

∇2 = (∂2/∂x + ∂2/∂y + ∂2/∂z) (more on that one a bit later)

• It was already known that one of the

properties of the wave equation was that upon double differentiation you got the

IV-23 23

original wave multiplied by a number – in this case -(2π/λ)2

Eψ dxψd 2m 2

22 =⋅−hSo This is the Schrödinger (more about this guy later) Equation for a Particle in a Box It is often written as: Ĥψ = Eψ) Ĥ is the Hamiltonian Operator

IV-24 24

William Rowan Hamilton (1805 –1865) If ever there was a child prodigy, Hamilton was it. He grew up with his uncle who was a bit of an eccentric; for instance, he tied a string around young William's toe at night, ran it through a hole in the wall into his own bedroom, and then early each morning he would tug on the string to wake him and start him on his studies. By the age of 12, William was fluent in 10 languages and at four could speak Greek, Latin and Hebrew. He was appointed to a Mathematics Chair at the Royal Observatory in Dublin at a youthful age.

IV-25 25

So: Eψ

dxψd 2m 2

22 =⋅−h

In the box the Schrödinger Equation is 0 E 2m

dxψd

222

=Ψ⋅+h

We must now solve this differential equation. The wavefunction (Q) will eventually give us the “shape” of the orbital and tell us where the electrons reside in space.

ψ E 2m- dxψd

222

⋅⋅=h

We need to find functions (ψ) that fit this equation.

IV-26 26

Possible solutions are: we don’t prove it here.

x2mEAsin ψ 21

2 ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

h

x2mEBcos ψ 21

2 ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

h

(A and B are constants) Recall: i.e. A(x) = mν

hλ ; λ

x2πkcos =⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

So:

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

λx2πksin ψ(x)

(we have changed A for Q)

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

λ x2πsin

λ2πk

dxψd 2

22

IV-27 27

Important Boundary Conditions

ψ = 0 for x = 0 and x = L

(function must go to zero smoothly at edges of box) ∴ only sine solution works so they will look like this

IV-28 28

Now let us look at the exact solution:

So, x2mEAsin ψ 21

2 ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

h

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

==2

2m b x bEAsin h

ψ will be zero at x = L only when L bE ⋅ = π, 2π, 3π, …. = nπ (n = 1, 2, 3, …) So:

n π L bE = and; Ln bE π=

Q = A sin (nB)x/L and 2

22

L b πn E =

(n = 1, 2, 3, …)

IV-29 29

SOLUTIONS

Energies Wavefunctions ↓ (Since xL

nπ bE ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛= )

and xbEsin ψ = and b = 2m/ħ2

xLnπAsin ψ ⎟

⎟

⎠

⎞

⎜⎜

⎝

⎛=2

222

L 2m πn E h=

222

8mLhnE =

IV-30 30

Major Points (see also deK & G. p. 24-30 â Energies of allowed states skip from one discrete value to another in units of 2

2

mL8h .

Particle Can Only Have Certain Energies

.... ,mL8h9 ,

mL2h ,

mL 8h

22

22

22

Energy is Quantized

Allowed Energy Levels for a Particle Constrained to a one dimensional BOX.

n = 1

n = 2

n = 3

IV-31 31

An easy way: p2 = 2mE h2/82 = 2mE E = h2/2m82

But 8 must be in multiples of L starting at 2L, then L, then 2/3L, then 0.5L etc…

So E(1) = h2/2m82 (8 =2L) ; E = h2/8mL2

E(2) = h2/2m82 (8 =L); E = h2/2mL2

etc…….

IV-32 32

Why are we not aware of quantization in our own experience? • Consider two cases.

1) An electron trapped in a 1Å wide box. 2) A 10 g. marble moving in a 10 cm wide box. Case One 2

22n mL 8

hn E ⋅=

21031

234JS

2

)10(1kg109.18)10(6.6n

−−

−

×××××⋅

=

= 6×10-18 ⋅ n2 J = 38n2 eV The minimum energy that the electron can have

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

===

3(n 3422)(n 152Then

1)(n eV 38 is

Quantization would be obvious here

IV-33 33

Case Two (The Marble) m = 10 g, L = 10 cm, En = 5.5×10-64 n2 J (48 orders of magnitude smaller than the electron in case 1) A marble with this KE has a speed of 3.3×10-31 m/s (experimentally indistinguishable from a stationary marble).

• Even for a marble moving at 1/3 m/s, n = 1030.

• Permissible energy levels are so close

together that there is no way of determining whether the marble can take an energies predicted by solutions of particle-in-a-box or any energy it likes.

• In the domain of every-day effects quantum

effects are imperceptible – so we use Newtonian Mechanics.

IV-34 34

So, Electron Marble

V = m/s31

1,000,000,000, 000,000,000,

000,000,000, 000

energy levels V = 10-31 m/s

n = 3 342

n = 2 152

n = 1 38 eV

Also 2mE pbut

λhp 2 ==

For a box L2 λ = 2

2222

4Lh

λhp ==

∴ 2

22

mL 8h E 2m

p ==

IV-35 35

“I don’t like it, and I’m sorry I ever had anything to do with it.”

ERWIN SCHRODINGER 1887-1961

“Nothing is real.” JOHN LENNON

1940-1980

“Anyone who is not shocked by quantum theory has not understood it.”

NIELS BOHR 1885-1962

IV-36 36

The Scientist and Electron or A Tale of Duality (with Apologies to John Godfrey Saxe)

Two curious young scientists. Both of inquiring mind. Did study the electron. Withwhatever they could find. So that by the ways of science, Each might satisfy his mind. The first, through much experiment, Did measure m and e. With cathode rays and oildrops; “It does appear to be, The electron is a particle, Like a pallet or a pea.” The second coughed and cleared his throat. His visage duly grave, “Experiments indiffraction This answer to me gave, It seems that the electron, Is very like a wave.” The scientist sat and pondered, “Why cannot both be true? Electrons look like waves tome And particles to you. Until a better comes along, This theory has to do.” And so our two young scientists. Their theories both to save, Combinedideas and simply said, “Electrons do behave In some ways like a particleAnd in some ways like a wave.”

Janice Kelland, Memorial University of Newfoundland, St. John’s,Newfoundland, Canada A1B 3X7

Particle in a box: See http://web.umr.edu/~gbert/Awave.HTML

IV-37 37

Particle in a Box: WavefunctionsAlso see:DeKock & Gray Fig 1-13

L xnπAsin ψ =

Allowed Wavelengths: n

2L λ = mL8

hn m2p E ; L 2

nh λh P 222

====

IV-38 38

Physical Meaning of Wavefunctions

• ψ2 measures probability of finding the particle.

• In this course we will often think of the square of the wavefunction as meaning electron density.

• Q: In n=2 how does particle cross the box?

IV-39 39

Although PIB does not correspond to any real physical system, it shows how Schrödingder Equation applies to several concepts that will be used for hydrogen atom.

1) Boundary conditions ⇒ Quantization of

Energy. 2) Ideas of Probability & Nodes. 3) Concept of Normalization ∫ =⋅∞

02 1 dT ψ

4) Concept of Zero Point Energy. Normalized Wavefunctions Since the particle must be in the box. ∫ =⋅L

02 1 dx ψ

So 1 dx x L

nπsin A L0

22 =⋅∫

L xnπsin L

2 ψ ; L2 A 1, 2

L A 21

21

2⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ ===

(more of this a bit later)

IV-40 40

Particle in a 3 Dimensional Box

• In order to extend our ideas of electrons that are quantum confined, we need to move into 3 dimensions.

Schrödingder Equation We saw earlier for 1 dimension (x)

Eψ dxd

m8πh

2ψ2

22 =−

in 3- dimensions (x,y,z)

Eψ ψzyxm8π

h2222

2 =∂∂+

∂∂+

∂∂−

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

and ψ vanishes (is zero) at the box edges as before.

IV-41 41

ψ(x, y, z) = ψx(x) ψy(y) ψz(z) Results Are:

Lz πnsin L

yπnsin LxπnAsinψ zyx=

nx = 1, 2, 3, … ny = 1, 2, 3, … nz = 1, 2, 3, … 2

22z2y2x mL8h)nn(nE ++=

notice that these resemble s,p and d-orbitals

IV-42 42

Degeneracies Energies of particle in 3D box are. E1(1) = 3h2/8 mL2; E2(3) = 6h2/8 mL2

E3(3) = 9h2/8 mL2; E4(3) = 11h2/8 mL2

Number in brackets is how many ways this energy can be arrived at e.g. 9h2/8 mL2

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

===

===

===

2n1n2n

2n2n1n

1n2n2n

z

y

x

z

y

x

z

y

x

3 different wavefunctions ψ221 = Asin 2πx/L sin 2πy/L sin πz/L ψ122 = Asin πx/L sin 2πy/L sin 2πz/L ψ212 = Asin 2πx/L sin πy/L sin 2πz/L

IV-43 43

STM Image Gallery http://www.almaden.ibm.com/vis/stm/gallery.html

Title: Quantum Corral Media: Iron on Copper(111)

IV-44 44

Electron Waves

IV-45 45

How Does It Work?

Tip is brought very close to sample surface. As it scans over surface it passes through regions of high and low electron density. A voltage is placed across the tip and sample and thus a current flows. More e density –gives more current.

IV-46 46

A new kind of microscope reveals the structures of surfaces atom by atom. The instrument’s versatility may extend to investigators in the field of physics, chemistry and biology.

DNA

IV-47 47

The Hydrogen Atom

• Our Introduction to Quantum Ideas via wave mechanics will now allow us to solve the electronic structure properly.

• Recall Bohr Atom: simple idea of orbits.

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=−= r

mvre From r

ePE 2222

• Predicted energy levels of hydrogen but did

not work for multi-electron atoms and ions.

The Bohr atom.

IV-48 48

We now apply wave mechanics to H atom. • Solve Schrödinger Equation For Hydrogen

Atom.

• Potential Energy is not zero but is given by

re2− .

• For PIB all energy was Kinetic and we

equated momentum with wavelength. RECALL 3D Box Eψ ψ

zyxm8πh

22222 =

∂∂+

∂∂+

∂∂−

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

H Atom V)ψ- (E ψ

m8πh 22

2 =∇−

IV-49 49

Why?

ψλ

2πψ2

2⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+−=∇

Before λ = p

h and p2 = 2 mE

NOW ETOTAL = EKIN + EPOT

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−+= remv2

1 22

E = K + V 22v22 Pmmv2

12mV)2m(E ===−⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

But

V)2m(Eh

phλ

−==

↑”new” λ

IV-50 50

NOW

ψ h

V-2m(E 2πψ 2

2

2

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=∇

V)ψ(E

hm8π

22 −−=

giving: EψVψψ

m8πh 22

2 =+∇−

Schrödinger Equation For H Atom zE ψ Eψ ψ ψ

m8π22

22 =∇

⎜⎜⎜⎛

re h +−

⎟⎟⎟

⎠

⎞

⎝⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

Del Squared r = proton-electron Kinetic Energy Potential Operator Energy Operator

Ĥ re

m8πh 222

2+∇−=

IV-51 51

Solutions (Atomic Orbitals)

• Solutions are best represented by “polar coordinates”.

The Quantum Numbers.

Z

Electronr

z

xy

r

X

φ

θ

• The wavefunctions are specified by THREE

quantum number (n, R, m).

• Total Wave Function is Written as product of angular and radial parts.

That is:

IV-52 52

) , (θ m (n),n,m),(n, Y R ψ φl

lll ⋅⋅=

Radial Angular For Example 1s: π2

1 2eψ r1s ⋅= −

Radial Angular 2px π2

)cos (sin3 re 62

1 ψr/2-re

2sφθ⋅=

Radial Angular

IV-53 53

• As before energies are quantized. •

2nk- E =

• n is the Principal Quantum Number

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

242

he me 2 πk ← Same as Bohr Atom

k = 13.6 eV (n = 1 1st 1P of H = 13.6 eV) R Quantum Number

• Determines Angular Momentum

• Can be thought of as shape quantum number

• Runs in integral steps from 0 to n-1. n and R determine Spatial Properties of the Orbital.

IV-54 54

Summary

n = 1

n = 2

n = 3

R= 0

R = 1

R = 2

IV-55 55

mR Quantum Number

• Magnetic Quantum Number

• Orientation of Orbital

Runs from -R to +R in integer steps

(mR = -R, -R+1 … 0 …R-1, R)

It is often thought that mR values correspond to the orbitals. e.g. when R = 1

44 344 21lzzP,yP,xP NOT

11,0,m −+=

• Real Orbitals are combinations of these.

Try this web page: have some fun http://www.nsci.plu.edu/~klarson/223/hydrogenic.html

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The Spin Quantum Number (ms)

ms = "21

Schrödinger Equation can be set up for any atom, but IT can only be solved exactly for one electron systems.

• Concept of quantum number and orbitals is still useful.

• Modifications involve e-e repulsion

energies.

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7_24

px py pz

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NOTE sign of Wavefunction.

• Important when adding wavefunctions to

form bonds. e.g. H2 ++ + +

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Electron Probability Density

• Bohr Atom: e- rotates about nucleus.

• ψ2 is proportional to Probability of Observing the Electron.

• Note 1ψspace

2 all∫ =

FOR 1s. 0ar2

3

0ea

1Nx−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

a0 = Bohr radius

π1N=

• Probability of finding electron in a “Shell” is

determined by volume of shell (= 4π2)

(See Over ⇒)

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dr

←volume of shell = dV P = 4π2|Ral|2 dr See chapter 1 deKock and Gray

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Squaring the wavefunction gives us:

This is not particularly revealing to us. What we really want to see is the electron probablity (density if you like). To get that we need to multiply be 4Br2

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This will give us the radial wavefunction. For 1s this looks like.

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For the 2s

• Number of Nodes = n - R -1 (=1)

• Only s orbitals have finite ψ2 at nucleus. • See R vs r plots.

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Radial Wavefunctions:

Note the nodes and that the wavefunction does not go to zero at large r.

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Also see these nice web pages. http://webphysics.davidson.edu/physletprob/ch10_modern/radial.html and http://www.shef.ac.uk/chemistry/orbitron/AOs/1s/radial-dist.html

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Further Insights Li(1s2 2s1) → Li+(1s2) +e IE1 = 5.392 eV Li+(1s2) → Li2+(1s1) + e IE2 = 75.638 eV Li(1s2 2s1) → Li+(1s1 2s1) + e IE1s = 64.84

• In the case of IE1s the energy is less than IE2.

• 1s electron is ejected from Li atom.

• less coulombic attraction. From data above we see that 2s screens 1s.

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Contour Plots

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Many Electron Atoms

• Electronic configurations are not the best model.

• Electron-Electron Repulsion is again important.

• Several states with same configuration have different energies.

• Furthermore SPIN-ORBIT coupling splits states into components with different angular moments.

One Electron Atom Many Electron Atom

• Can solve Schrödinger

• Must use approximate methods

• R + s = j (Total ORB angmmtm)

• ∑ +=

n

1i ki sl couple all

electrons • configurations • Terms R1+R2+R3..=L

S1+S2+S3 … S L + S = J

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Schrödinger Equation & Many Electrons 2 electrons

r2 Oe ↑ z+ r12

12

2

ATT2

2

1

2op r

e rZe - r

ZeV +−=4444 34444 21

↓ r1 Oe- REP

Third term makes it impossible to solve exactly.

• Electrons try to avoid each other – correlated.

• If electron correlations is ignored Schrödinger can be split into 2 parts (1st and 2nd electrons).

• Results in orbitals with exactly the same symmetries as for H. (E is 4 times lower).

Eψ ψr2e - Zme ψr

2e - 2me22

2222

12 =∇+∇−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ hh

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1st electron 2nd electron • Analogous results for other many electron

atoms.

• So we continue to use 1s, 2s … 2p . As before the e-e repulsion term is something we take care of later.

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