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Structural Chemistry-2014 -- Atkins Ch. 7, alternate: Hanna, House, Blinder

To understand structure must come to terms with

atoms ↔ make molecules and then with electrons/nuclei ↔ make up atoms!

Study of interaction of small particles Quantum Mechanics

Here “Quantum Chemistry” Molecular scale theory explains: Spectroscopy, Dynamics, Binding, Conformation…

big issue particles ↔ waves: frequency = c/wavelength

Why a special mechanics for small particles? 19

th century “everything worked out”

Newton mechanics – continuous (particle - all energies) Maxwell relations (E & M – waves describe light)

Some problems with this: a. Excite elements in discharge

emission spectra, only at specific wavelength/frequency

1/ = /c = ˜

H-atom: 1/ = R (1/n12 – 1/n2

2)

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Balmer-vis n1=2; Lyman – uv n1=1; Paschen near-IR n1=3 R = 109,677 cm

-1 – why discrete?

frequency

Engel Fig. 12.6 picture of line spectra

b. Black-body radiation

If all equally probable, - probability density would blow

up at short since more short waves in a cavity –Rayleigh-Jeans UV catastrophe

Actual Shape (energy distribution with or )

observed rises with dec. (inc. ) to maximum,

then steep fall off at short high

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Observations:

Stefan-Boltzmann – Energy density ~ T4 (integral of )

M = T4 ~ 5.67 x 10

-8 Wm

-2 K

-4

Wein – distribution maximum depend inversely on T

maxT = constant = 2.9 x 10-3

m K Several people tried to derive it classically – fail

Planck assume oscillators could only have energies

E ~ 0, h, 2h, 3h, … (hc/) – restriction, non-classical h – constant – function to distributions

= c/ frequency Then used Statistical-Thermo ideas to get distribution

energy density from “hole”

()= as function (and T) like Boltzman probability

Test: long – h/kT 0

Mult. top/bot. by e-hc/kT, expand e

-x = 1 – x + ½ x

2 etc.

() ~

... 1 1

... 1hc8

kThc

kThc

5

() ~

hc

kThc8

5 ~ 8kT/4

classical (Wein displacement) result fits long behavior,

i.e. hc/kt is very small (short 8kT/4 ∞,UV catastrophe)

Short (rewrite): () (8hc/5)(e

-hc/kT)/(1- e-hc/kT)

–small exp

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but as increase exponential term builds from zero to 1, denom. collapse – blow up

In between long - short , () must have a maximum

max: (d/d) = 0 = [-5/6 – (hc/kT

5 )(-1/

2)]

maxT = constant = hc/5k -Wein: max fit vs.1/T, slope ~ h Result: a) h = 6.626 x 10

-34 J

.s - note: h is an angular momentum

b) form of () fits observation “perfectly”

Justify assumption E = 0, h, 2h, etc. by fit observation (not “proven”)

Photoelectric Effect – Einstein explain--light impinge on metal,

electrons emitted, accel. in V, grid can stop e- if eVg = KE

K.E. of electrons depend on – long no elect.

th – threshold for appearance

K.E. linear dep. on Intensity not affect E,

just number of electrons (current) light acting like a particle

K.E. = ½ m e2 = h( - Th)

= h –

– work function

no e- escape e

- with KE=h-

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So far: Energy limited to discrete values – Planck h Light (wave) behave particle – Einstein photelectric

key – threshold – frequency energy – intensity (classically E-field)

number of electrons, not energy

Complement: Davisson-Germer e-beam – fixed energy onto (metal)

ordered surface

diffract at specific angles due to periodic structure

analogous to diffract x-rays or light w/ grating

electrons are particles behave as a wave like light (wave behavior) diffract from grating

Combine Louis de Broglie (PhD thesis): = h/p or p = h/

momentum of particle relate to 1/ = /c (or frequency of light relate to its momentum)

↔ p Wave-Particle Duality – Central to Schrödinger QM

– property of small particles / high energies

Catch: a wave has no position so specify x problem

Also: if particle is localized can’t oscillate problem Heisenberg propose this to be fundamental limit

pxx ħ/2 position and momentum complementary variables

Uncertainty principle – cannot know both x,px precisely (works for complementary variables, also E,t )

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So now: have energy restricted (quantized) have light behave like particles particle like light (quantized)

Big Deal – not continuous like classical

So what about those line spectra? Somehow light / frequency limited -- discrete

Atoms (Rutherford) became clear dense nucleus (+) electrons must be outside – postulate in “orbits”?

Oct.7, 2012

Google doodle

Niels Bohr birthday

Bohr (Niels) postulate (we moved back a decade – 1913) a) Energy (light) emit only when change orbits b) Frequency of light jump orbits

E/h = (relate to energy, E – like Einstein) c) “Orbit” → relate electrostatic attraction e

- and p

+

to centripetal force

* d) Angular momentum restricted (nħ): nh/2 n = 0,1,2,. .

Look up derivation – fairly simple

E = T + V = ½ m2 – Ze

2/r Z - atomic number

c) equate coulomb-centripetal forces: -Ze2/r

2 = m(-

2/r)

Newton: force ┴ to motion for change in direction

solve: m2 = Ze

2/r then plug in: E = -Ze

2/2r

d) subst: angular momentum assume: r·p = r·mv = nħ

r = n2ħ

2/Ze

2m where: ħ = h/2m

2= Ze

2/r = n

2ħ

2/mr

2

E = - Z2e

4m/2n

2ħ

2 EH = h = R(1/n1

2 – 1/n2

2) (H-atom)

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For Z=1 Rydberg: R = e4m/2ħ

2 in J (norm. as 109,677 cm

-1)

Fits the Balmer series and Rydberg formula Bohr result explain H-atom discrete line spectra

E~1/n2 E ~ R(1/n1

2 – 1/n2

2)

Also worked for He+, Li

2+ etc. 1-e atoms (change e

4 Z

2e

4)

Depended on assuming angular momentum quantized This is same as assuming – fixed orbits

Recall de Broglie = h/p = h/m

if particle on ring 2r = ncircumference have integer number

wave lengths

plug in: 2r = n(h/m)

mr = nħ matches Bohr’s assumption (10 yrs later)

Again waves particles key With this set-up time was right to generalize

Bohr model great for H but fail for He0

Problem assumption pseudo classical

1926 Schrödinger developed a wave – mechanics

treat particles with wave properties (relate to wave equation of Maxwell but different)

Heisenberg same time did a matrix – mechanics fully consistent but judged more difficult concept

[Dirac later added relativity – source of spin]

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Postulates – simplest approach - wave–quantum mech. idea just like Thermo – postulate set of rules – derive properties – test against observables Catch – initially must just accept, no rationale, then test

little physical picture Atkins or House QM book Postulate 1: State of a system fully described by a

wavefunction: (r1, r2, … t)

> variables (r1, r2 - xyz coord. particle 1, 2, t = time) > Anything not in wave fct. cannot be known in QM

Shorthand – represent state by quant. num. n,ℓ,m,. . .

– or observables a,b,c - helpful for us, not w/f

Alternative: m n vector bracket from Heisenberg

Postulate 2: Observables correspond to operators

summary: constant: c → c concept: mult.

variable: x → x e.g. position

function: (x) → (x) e.g. PE: V(x)

momentum: px → -(ih/2)d/dx = -iħ d/dx

Operators act on w/f– typical: multiplication, derivation

[, ] = [ – ] – commutator, normally expect 0

Constraint is that [x, px] [xpx – pxx] = i ħ

recall: d/dx(y=(yd/dx)+(dy/dx)

here x, px do not commute, since [x, px]

Properties: if (x) = w (x) eigenvalue equation

if operate on (x) and result is w, a constant *(x)

is eigenfunction of with eigenvalue w

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Many possibilities examples: a) let A = d/dx

Aeax

= aeax

a const., = eax

eigenfunction of A

Aeax2

= 2ax eax2

2ax not const., eax2

not eigenfct. of A onstant

Asin ax = -a cos ax sin ax ≠ cos ax not eigenfct of A b) or B = d

2/dx

2

Bsin bx = -b2 sin bx g = sin bx is eigenfunction B

Beax

= a2e

ax g' = e

ax also eigenfunction B

Could also do linear combinations

{n} set of eigenfunction of

if complete set g(x) = 1n

cnn and let n = wn n

then g = cnn = n

cn n = n

cn wn n

This is not an eigenfct relation unless wn = w all wn

g = c'nn cg

= w cnn = wg if all wn equal

degenerate (product of symmetry)

Born (Max) model: Wave function relates to probability For volume element dV, probability of finding the particle

is * dV, so probability between a and b is

integral from a to b: ∫* dV key to solving problems Evaluating Observables

Postulate 3: If system is described by i and i is an eigen-

function of where operator corresponds to observ. a

Then every measurement yields ai, the eigenvalue i = aii

i – set of eigenfunction

ai – corresponding eigenvalue, for operator

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Postulate 4: If is not eigenfunction then the average observed value is “expectation value”

*- complex conjugate ,change i -i

= – integrate over all variables

ex: a) let = eax

eval. px – momentum, operate: px on - iħ (d/dx)(e

ax)= (- iħ a) e

ax

this is eigenvalue equation = - iħ a (const. mom.)

b) let = cos ax - iħ d/dx cos ax = iħ a sin ax not eigenvalue

-

2

-

2x

dxax cos

dxax sinax cos i

a

dx xacos

dx )ax cos( dx

d

i )ax (cos

p

= 0

(numerator odd function, symmetric integral zero)

[meaning: px is or so average cancels out] Schrödinger equation – many ways to get

– here use postulates create an Energy operator “Hamiltonian” – classical operator for total Energy

H = T + V = ½ mv2 + V(x) = p

2/2m + V(x) KE+PE

Quantum Mechanics energy operator H = T + V = (-ħ

2/2m) d

2/dx

2 + V(x) 1-D

H = -ħ2/2m

2 + V(x,y,z) 3-D Hamiltonian operator

2 = d

2/dx

2 + d

2/dy

2 + d

2dz

2 Laplacian =

r = xi + yj + zk = d/dx i+ d/dy j+ d/dz k - vector operator

H = E = [(-ħ2/2m)

2 + V(x,y,z)] = E "gradient”

If is eigenfct of H then eigen values (E)total energy

d *

d *

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Use these eigenvalue properties:

Expand arbitrary wavefcts in set of eigenfct {n} of operator

n = a n {n} complete set of functions

= n

cn n arbitrary function rep.as linear comb. of fn

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so in general for H without time dependence can just use time independent result

Spectroscopy light interaction with matter is time dependent Properties of operators (bit detailed – can skip)

Hermitian operators – have real egienvalues (i.e. can be measured) hence all observables correspond to Hermitian operators (not vice versa)

Hermitian definition: m* n d = { n* m d}*

equivalent to: m* n d = m)* n d

to simplify n m = m* n d

change notation bracket – Dirac notation

Orthonormality mn = nm = n* n d

Hermiticity n m = m R n *

a) Prove: eigenvalues of Hermitian operator are real

n = n n operator left by n

n = n n = n n * = * real

b) Prove: eigenfunction of Hermitian op. are orthogonal

n = n n m = m m

m n = n nm

n m = m nm

c.c. and subtract (recall, c.c --> take neg. of all i)

n m – m n * = (n – m) mn

n = m or mn = 0 – orthogonal

important result - generalize operator form

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Important c) If two operators have simultaneous, arbitrarily precise

eigenvalues in one state, then this is an eigenfunction of both and the operators commute.

i.e. A = a and B = b

AB = Ab = bA = ba = ab = a(B)= B(a) = BA

[A, B] = 0 * prove to yourself the converse –

[A, B] = 0 eigenfunction of both A, B

Relate back to Uncertainty This proves that momentum, position of the same variable can not be measured with arbitrary accuracy because [x, px] = iħ x, px said to be complementary

however since [x,y] = 0 can be measured x,y are independent variables

complementary operators do not commute and involve an uncertainly relationship, such as x,px

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