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e-1

x

z

y

Atoms – 2014 -- start with single electron: H-atom (Atkins Ch.9) 3-D problem - free move in x, y, z - easier if change coord. systems:

Cartesian Spherical Coordinate

(x, y, z) (r, , )

Reason: V(r) = -Ze2/r depend on separation not orientation

electrostatic potential – basis of chemistry Note: (if proton is fixed at 0,0,0 then |r| = [x

2+y

2+z

2]1/2

) r = (xe – xp)i + (ye – yp)j + (ze – zp)k --vector, p e

|r = [(xe – xp)2 + (ye – yp)

2 + (ze – zp)

2]1/2

--length (scalar)

Goal – separate variables in V(r) x,y,z mixed, since r mix by √‾‾‾

– no problem for K.E. – T already separated

First step – just formal: reduce from 6-coord: xe,ye,ze & xp,yp,zp

to 3-internal coordinates. Eliminate center of mass

-- see notes (Center of Mass handout) on the Web site whole atom: R = X+Y+Z X = (mexe+mpxp)/(me+mp) . . etc. this normalizes the position – correct for mass.(weighted) for H-atom these are almost equal to: xp, yp, zp but process is general – balance masses – i.e. no torque

other 3 coord: relative x = xe – xp etc. ideal for V(r)

Problem separates, V = V(r) only depend on internal coord.

H = [-2/2M R

2 + -2

/2 r2 + V(r)] (R,r) = E

H - sum independent coord. M = me + mp = me•mp/(me+mp)

(R,r) = (R) (r) product wave function separates

summed like before (eg.3-D p.i.b.)

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i.e. Operate H on (R,r) and operators pass through

R and r dependent terms, (R) and (r), to give:

First term: + second two terms: (= E [(R) (r)])

(-2/2M)(r)R

2(R) and (R)[(-2

/2r2+V(r)](r)

so divide through by (R,r) = (R) (r) and results are independent (R and r) sum equal constant, E

R-dependent equation: (-2/2M)(R))R

2(R) = ET

Motion (T) of whole atom – free particle -not quantized - ignore

r-dependent equation: (r))[(-2/2r

2+V(r)](r) = Eint

relative (internal) coord. - formal: we let E = ET + Eint

internal equation simplify, convert to spherical coord.: x,y,z r,,

internal Hamiltonian: H(r)(r) = [(-2/2r

2–Ze

2/r](r) = Eint(r)

(idea – potential only depends on r,

so other two coordinates, –only contribute K.E.) new form 2:

r,,2 = 1/r

2{/r(r

2/r)+[1/sin]/(sin /) + [1/sin

2]

2/

2}

Separation, one coord. at time. Easy to separate dependent terms

to get only in 1 term, multiply by r2sin

2 through H = E

AAssiiddee,, just for information, not in class, Step by step separate, start:

[(-2/2r,,

2–Ze

2/r](r,,) =

(-2/2r

2){/r(r

2/r)+ [1/sin]/(sin /) +

[1/sin2]

2/

2}(r,,) – Ze

2/r (r,,) = Eint (r,,)

multiply by r2sin

2 :

1. (-2/2 sin

2/r(r

2/r)+ [1/sin]/(sin /) +

2/

2 – [r

2sin

2e

2/r]}(r,,) = r

2sin

2Eint (r,,)

isolate the dependent terms:

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2. (-2/2 {sin

2 /r(r

2/r)+ sin /(sin /) –

(2/2) r

2sin

2(Eint + Ze

2/r)}(r,,) = (2

/22/

2(r,,)

see all operators one side

Use (r, , ) = R(r) and cancel out 2/2

3. {sin2 /r(r

2/r)+ sin /(sin /)} R(r) –

r2sin

2(2/2

)(Eint + Ze2/r) R(r) = -

2/

2 R(r)

Recall 2/

2 only operate on part, rest pass through

Same the other side, pass through/r and /

{sin2 /r(r

2/r)+ sin /(sin /)} R(r)

– r2sin

2(2/2

)(Eint + Ze2/r)R(r) = - R(r)

2/

2

divide through as before by r

5. [R(r) {sin2 /r(r

2/r)+sin /(sin /)} R(r)

–r2sin

2(2/2

)(Eint+Ze2/r)R(r) = - [

2/

2 const.

_______________________________________________________________________________

right side (red) independent of left side, so must be constant

this you need to know – just like particle on a ring

a) Let part equal constant, m2:

2/

2 () = -m

2 ()

+m -mx

y

rotation about z-axis, solution: () = eim m = 0,1,2. . .

note: equivalent to particle on a ring problem const. = -m

2 important, K.E.(+), requires complex exponential

b) Can similarly separate () but arithmetic messier

1st divide through by sin

2-- separate r and terms

___________________________________________________________________________________________

1. [R(r) { /r(r2/r)+(sin/(sin /)} R(r)

– r2(2/2

) (Eint + Ze2/r) R(r)m

2/ sin

2

Note 1st & 3

rd terms have r but middle term only separate

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2. [{ (sin/(sin /) - m2/sin

2}

[R(r) {/r(r2/r)(2/2

) r2 (Eint + Ze

2/r)} R(r) = const.

_________________________________________________________________________________________________________

Result- only has on left and r on right, let const. = -ℓ(ℓ+1):

3. {(sin/(sin /) - m2/sin

2 ℓ(ℓ +1)} =

Solution: LeGendre polynomial: ℓm() = Pℓm(cos ) ℓ = 0,1,2, …

ℓ = 0 – 2 /2

ℓ = 1 m = 0 – (3/2)1/2

cos

ℓ = 1 m = 1 – (3/4)1/2

sin

ℓ = 2 m = 0 – (5/8)1/2

(3 cos2 – 1)

ℓ = 2 m = 1 – (15/4)1/2

(sin cos )

ℓ = 2 m = 1 – (15/16)1/2

(sin2) . . . polynom: P(x) but x = cos

ℓ = 3 m = 0 – ~ (5 cos3 – 3 cos ) as m inc. cos sin

c) Radial function messier, has V(r), yet but must fit B.C.

r Rnℓ(r) 0 (must be integrable) exponential decay, penetrate potential

~ e-r

damp (works, since r is +)

Must be orthogonal this works when fct. oscillate (wave-like) power series will do

radial (r) part = -ℓ(ℓ+1): in #2 above, generate radial equation by:

{ r -2/r(r

2/r)(2/2

) (Eint + Ze2/r) - ℓ(ℓ +1)/r

2} R(r) = 0

divide by r2, multiply both sides by R(r) & move l(l+1) term to left side

Plots of

Pℓm(cos )

= 0 || x

= || z

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Associated LaGuerre Polynomial solves equation:

Rnℓ = [const] (2/n)ℓ

12nL (2/n) e

-/n = Zr/a0

Quantum #: n = 1, 2, 3, … ℓ = 0, 1, 2, … n – 1, ℓ m

n = 1, ℓ = 0 ~ e-

n = 2 ℓ = 0 ~ (2 – )e-

n = 2 ℓ = 1 ~ e-/2

n = 3 ℓ = 0 ~ (1–2/3+22/27)e

-

n = 3 ℓ = 1 ~ (e-/3

n = 3 ℓ = 2 ~ e-/3

. . . . Note: n restricts ℓ (≤ n–1),

and ℓ restricts m (≤ ℓ)

Comparison of potentials – when potential finite, levels collapse, – when sides not infinite and vertical

w/f penetrates potential wall Particle in box; Stubby box;

infinite, steep potential, get expanding Energy-

levels, (wall)=0

Finite potential get collapsing energy levels continuous at top

Anharmonic oscillator and H-atom, dissociate

large q,r (q), (r) penetrate wall

Note: Atkins text p. 328

uses = 2Z/na0 so that exp

term becomes: e-/2

in each

case. Affects polynomial

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Harmonic oscillator; V(x)∞ Anharmonic oscill.,V(-∞)∞,V(+∞)0

sloped potential wavefct penetrate wall, finite pot.levels collapse,

H-atom Solutions for ℓ = 0 higher ℓ, less nodes If rotate this around r = 0, get a symmetric well and shapes look like υ = even harmonic oscillator shapes

finite potential “bends

over,” so V = 0 at r = , get collapse of

E-levels as n

Energy - in radial equation, solution same as Bohr, since correct, in q.m. varies with nodes curvature as before

Yellow (Engel) – wave function penetrate classic forbidden region

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Better pictures taken from Atkins Fig. 9-4 – need to learn

These functions (3 independent solutions) can be combined

(r,,) = Rnℓ(r) ℓm() m()

Note: only Rnℓ depend on r as does V(r)

Energy will not depend on , for H-atom, differ for He etc.

Often write: Yℓ m(,) = ℓm() m() spherical harmonics

Angular parts give shapes but not size

these are eigenfunctions of Angular Momentum

recall : L = r x p, L2 ~ -2

r2 , Lz = xpy-ypx, Lz = -i/

L2 Yℓ m (,) = ℓ(ℓ + 1) 2

Yℓ m [H,L2] = 0 commute:

Lz Yℓ m () = mYℓ m [H,Lz] = 0 simul.sol’n

This is source of familiar orbitals, ℓ = 0,1,2,3. . or s,p,d,f

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Solving Rnℓ equation En = -Z2e

4/22

n2 = -Z

2R/n

2

exactly Bohr solution, R Rydberg, (must be, since works)

R dependent, En independent of l,m

Familiar: n = 1 ℓ = 0 m = 0 – 1s n = 2 ℓ = 0 m = 0 – 2s

ℓ = 1 m = 0,1 – 2p (2p0 + 2p1) n = 3 ℓ = 0 m = 0 – 3s

ℓ = 1 m = 0,1 – 3p (3p0 + 3p1)

ℓ = 2 m=0,1,2 – 3d (3d0, 3d1, 3d2)

(lt) compare 1s and 2s w/f (rt) probability ( and

radial distribution (4r2

Note: 1s decays, 2s has node(2-) term, 2p starts at 0

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Compare probability distributions, see expansion in size with n, ℓ

Note: 2s node makes dip Note: # radial nodes (in Rnℓ) in probability (e- density) decrease with ℓ: #nodes = n - ℓ - 1

radial nodes affect the distribution in angular functions, 1s - gradually decay density, darklight 2s decay sharply to zero (white ring), rise and slowly decay result - larger size

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Linear combinations give real orbitals from complex components,

Ylm(): pz = p0 = Y10 px = p1+p-1 = Y11+Y1-1 py = i(p1-p-1)=i(Y11-Y1-1)

Similarly mix Y±2 to get dxy, dx2-y2 and

Y±1 for dxz, dyz

Angular functions have no radial value --> just surface, combine with radial function to get magnitude or e

- density,

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Best represented as a contour map or probability surface

Real orbitals, take linear combinations of ±m values,

eliminate i dependent terms, get x,y,z functions

Cartesian form: eim+ e

-imcosi sincosi sin = 2 cos ~ x

i (eim- e

-im2 sin ~ y

recall: x = r cos and y = r sin

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H-atom solutions, complex orbitals, eigen values of angular momentum:

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Linear combinations of H-atom solutions—Real orbitals

Cartesian representation polar representation

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Contour plots of orbitals

Radial function effect represented by contours, each line represents lower e- density – sign change makes node between lobes or radially (e.g. see top, 3p vs. 2p)

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Linear comb. of degenerate orbitals also solutions H-atom – show effect of mixing s and p orbitals:

Hybrids – linear comb. of s and p – orient for bonding

sp 2 orbitals / opposite direction, (sp) are 180

sp2 3 orbitals / in plane / 120 apart

sp3 4 orbitals / 4 vertices of tetrahedron (109)

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Energy level diagram – H-atom

Spectral transitions match Balmer series

but also must account for , functions Allowed selection rules (see box)

n n' = 1 – Lyman must start p orbital --> end 1s

n n' = 2 – Balmer must start d or s orbital end 2p

or start in p orbital end in 2s etc.

Test with Zeeman effect – mℓ H = E' add E from field More complex than Bohr, but same energies and angular momenta, similar restriction on solution to line spectra

Allowed – any n change: n 0

– ℓ, mℓ as before: ℓ = 1 ,

mℓ = 0, 1 no energy dependence on ℓ, but spectral transitions do depend on ℓ

E n