Lecture 3 ndash August 25 2010

Reviewbull Crystal = lattice + basisbull there are 5 non-equivalent 2D lattices 5 Bravais latticesbull divided into 4 systems (symmetries ldquopoint groupsrdquo)

ndash oblique square rectangular (2) and hexagonal

Actual formation (self-assembly) of these lattices depends on details ndash

boundary conditions (marble demo)

1

3D systems(symmetries)System Number of

Lattices Lattice Symbol Restriction on

crystal cell angle

Cubic 3 P or sc I or bccF or fcc

a=b=cα =β =γ=90deg

Tetragonal 2 P I a=bnecα=β =γ=90deg

Orthorhombic 4 P C I F anebne cα=β =γ=90deg

Monoclinic 2 F C anebne cα=β=90 degneβ

Triclinic 1 P anebne cαneβneγ

Trigonal 1 R a=b=cα=β =γ

lt120deg ne90deg

Hexagonal 1 P a=bnecα =β =90degγ=120deg Table 1 Seven crystal systems make up fourteen Bravais lattice types in three dimensions

P - Primitive simple unit cellF - Face-centred additional point in the centre of each faceI - Body-centred additional point in the centre of the cellC - Centred additional point in the centre of each endR - Rhombohedral Hexagonal class only

from httpbritneyspearsac similar to Christman handout

Key

to

unde

rsta

ndin

g re

latio

nshi

ps

star

t w

ith c

ube

bre

ak s

ymm

etrie

s

(point group)

48=3x23 elements

2

3D systems(symmetries)System Number of

Lattices Lattice Symbol Restriction on

crystal cell angle

Cubic 3 P or sc I or bccF or fcc

a=b=cα =β =γ=90deg

Tetragonal 2 P I a=bnecα=β =γ=90deg

Orthorhombic 4 P C I F anebne cα=β =γ=90deg

Monoclinic 2 F C anebne cα=β=90 degneβ

Triclinic 1 P anebne cαneβneγ

Trigonal 1 R a=b=cα=β =γ

lt120deg ne90deg

Hexagonal 1 P a=bnecα =β =90degγ=120deg Table 1 Seven crystal systems make up fourteen Bravais lattice types in three dimensions

P - Primitive simple unit cellF - Face-centred additional point in the centre of each faceI - Body-centred additional point in the centre of the cellC - Centred additional point in the centre of each endR - Rhombohedral Hexagonal class only

from httpbritneyspearsac similar to Christman handout

Key

to

unde

rsta

ndin

g re

latio

nshi

ps

star

t w

ith c

ube

bre

ak s

ymm

etrie

s

(point group)

48=3x23 elements

2

Lecture 4 - Aug 27 2010

Reviewbull Crystal = lattice + basisbull there are 5 non-equivalent (Bravais) lattices in 2D 14 in 3Dbull divided into 4 systems (symmetries ldquopoint groupsrdquo) 7 in 3D

ndash Cubic tetragonal orthorhombic monoclinic triclinic hextrigonal

Close-packed crystals

CB

A

CB

A

Stacking of these planesABABAB = hexagonal close packed (hcp)

ABCABC = face centered cubic (fcc)

3

Real crystals Simple cubic wo basis none Unstable

Uncharged atoms prefer close-packed structures many near neighbors (12 nn in hcp or fcc)

Ions NaCl structure (fcc with 2-ion basis 6 nn)

CsCl structure (sc with 2-ion basis) (ldquo2-ion unit cellrdquo 8 nn)

We will skip Miller indices here (they are easier to understand after we discuss the reciprocal lattice) and go on to

Covalently bonded structures Diamond

fcc 2 atomsprimitive unit cell = 8 atomsconventional unit cell Basis at (000) and (141414)

(fcc translations (12120) (100)

4

Bragg Scattering of x-rays

Extra path length must be integer number of wavelengths

2 d sin = n

Braggrsquos parallel plane picture is mnemonic not derivation

PH 481581 Lecture 6 Sept 1 2010

Figure Courtesy Wikipedia

5

Better treatment x-rays are scattered by electron density n(r)

In 1D n(x) = n(x+a) (periodic)Expand in Fourier seriesn(x) = Σ nG eiGx

where G=integer 2aThese Grsquos form a ldquoreciprocal latticerdquo ndash is also G eiGx is periodic in the direct latticewhere ldquodirect latticerdquo is -2a -a 0 a 2a We want to generalize this to 3D

ldquolatticerdquo means closed under addition

6

k

x-raykr

n(x) will be real if n-G = nG

(complex conjugates)

Periodic functions in a lattice To calculate scattering from a periodic electron density n(r) we need to describe what it means to be periodic n(r+R) = n(r) for R in lattice R = u1a1 + u2a2 + u3a3Any periodic function n(r) can be expressed as a Fourier series n(r) = Σ nG eiGr where each eiGr is periodic in the latticeeiG(r+R) = eiGr eiGR = 1 GR = 2integer The set of such Grsquos is closed under addition ndash a lattice called the ldquoreciprocal latticerdquo of the ldquodirect latticerdquo determined by a1 a2 and a3

You can show that the vectors b1 b2 and b3 determined by

ai bj = 2 ij

are primitive translation vectors of this reciprocal lattice (RL)

This means b1 must be perpendicular to a2 and a3 Explicit formulas are 213

213

132

132

31

31

2

2

2

aaa

aab

aaa

aab

aaa

aab

2

2

7

[Start by showing b1 is in RL b1R = 2integer]

Examples of reciprocal lattices

Orthorhombic direct latticeˆˆˆ 332211 zayaxa aaa

Reciprocal lattice is

zbybxb ˆ2

ˆ2

ˆ2

33

22

11 aaa

Easy to check that

ai bj = 2 ij

Hexagonal direct latticeˆ)ˆˆ(ˆ 32

321

21 zayxaxa caa Reciprocal lattice is

a1

a2

b1

b2

132

132

31

31

2

2 from

aaa

aab

aaa

aab

2

2

8

Calculating FT of n(r)

nG = Fourier transform of scattering density n(r)Only nonzero at points G of reciprocal latticeExplicitly

9

n(r) = Σ nG eiGr

rrrGG

3

cellunit

)(1

dneV

n i

where V is the volume of the unit cell

Calculating x-ray scattering intensity

10

Incoming amplitude (electric field)

Lecture 7 Sept 3 2010 PH 481581

kx-ray

r

n(r) = Σ nG eiGr

rrsquo

detector

Amplitude reaching detector =rkie Re

rposition Source

n(r)d3 r rrike

rkr

k

number of scatterers

Phase change

3 ( )

Source position

3 ( )

Source position

( )

( ) ( )

i i

i i i

e n d e

e n d e e F

k r k r r

r

k r k k r k r

r

r r

r r k k

where F is called the scattering amplitude (in Kittelrsquos book)

22 2 2 2

2 2

2( ) 2 1

rr r r

r r

r rr r r r

122

2 2 2

21 1 2nd order

rr r

r r r

r r r rr r

( )k kr kr

r rr r k r k r k r r

rrsquo - r

G

Conclusion x-ray scattering probes reciprocal lattice (RL)

If incoming wavevector is k amplitude for outgoing wavevector krsquo ~ Fourier component nkrsquo-k In a perfect crystal this is nonzero only if krsquo ndash k is a reciprocal lattice vector call it G

krsquo

k

You have to rotate the crystal powder it

Also this is elastic scattering |k|=|krsquo| so krsquo must lie on a sphere The probability that one of the vectors k+G lies exactly on the sphere is zero (the sphere has zero thickness)

Powder pattern crystallites are in all possible orientations which rotates each G over the surface of a sphereSphere of possible Grsquos intersects sphere of possible krsquo s in a circle (ring) seen on screen

Gk

krsquo

scre

en

12

Powder pattern

Pattern for the mineral olivine httpndbserverrutgersedummcifcbf

httpwwwccp14acukccpweb-mirrorsarmelegypteconf2theoryhtml

Ball-milled

Hand-crushed

d

Relating krsquo ndash k = G to the Bragg equation

13

G

krsquo

k

Bragg picture is krsquok

leading to

2 d sin = n

2 2( ) k G k

In reciprocal lattice picture krsquo = k + G so elastic scattering condition is

2 2 22 k k G G k22 0 k G k

Geometrically we have 2 2

sin sin 2 sin2

Gk

G

Consistent only if (so smallest G goes with n=1 )Gn

d 2 G

d2

Predicting angles of powder diffraction rings

14

G

krsquo

k

Recall picture of krsquo = k + G

amp from top triangle sin2

kG

so )(2

and 2

but lkha

k

G

2

sin 12222 lkh

a

5012

4002

3111

2011

1001

21222 lkhlkh

Lecture 8 PH 481581

So the Bragg planes are related to reciprocal lattice vectors and the plane spacing d is related to the RLV magnitude G = |G| by d = 2G

15

Schematically RLV G lattice planes

or more precisely a plane and a vector G each determine a direction in space There are many Gs along this direction and many planes normal to it so the relation is really between families of Grsquos and families of planesfamily 0plusmn G plusmn 2G of parallel RLVs family of parallel planesOnly some of these planes go through lattice points and these are spaced d = 2G apart(G must be the smallest length in the family)

16

0

Direct lattice

Reciprocal Lattice

Drawing low-index planes (far apart)high-index planes (close together)

Calculating scattering amplitude from atomic form factors

17

Consider a crystal with a basisr1

r2

Plot density along dashed linen(r) = n1(r-r1) + n2(r-r2)= n1() + n2(2)

rx

x 2x

r2xr1x

where = r - r1 and = r ndash r2

n1 n2

r

The scattering amplitude FG is defined by

Lecture 10 PH 481581 Sept 13 2010

crystal entire

3)( rdenF iG

rGr

But the integrand is periodic so the result is proportional to the number N of unit cells The amplitude per unit cell is called the structure factor SG = FG N

Calculating scattering amplitude (structure factor)

18

Structure factor

jj

iii

iiii

ii

ii

iG

fefefe

denedene

denden

rdenrden

rdenS

j )()()(

)()(

)()(

)()(

)(

21

32

31

3)(2

3)(1

322

311

cell

3

21

21

21

GGG

ρρ

ρρ

rrrr

r

rGrGrG

ρGrGρGrG

ρrGρrG

rGrG

rG

r1r2

r

rdenf iii

3)()( ρGρGwhere

Example CsCl simple cubic r1= = (000) r2 = (a2a2a2)Some algebra gives

odd if

)points fcc(even if)()1()(

21

2121 l khff

l khffffS lkh

G

GG

In limit f1 f2(ie CsCl FeFe bcc) this becomes fcc RL

Another example (NaCl structure)

19

KCl isoelectronic f1 = f2 lose reflections due to extra symmetry

NaCl is fcc basis is r1= = (000) r2 = (a200)

centers) (cell odd all if

points)(corner even all if)()1()(

21

2121 lkhff

lkhffffS h

G

GG

(if some even and some odd eg (100) these points are not even ON the reciprocal lattice so we donrsquot calculate SG)

Example in Fig 2-17 p 42 KBr has 111 200 220 as predicted

Chapter 3 BindingTypes of bonding

bull van der Waalsbull Ionic bull Covalentbull metallic

20

Start with van der Waals because all pairs of atoms have vdW attraction For neutral atoms (so there is no electrostatic force) it always dominates at long distances

Ionic as a Na and a Cl atom approach each other the extra electron on Na tunnels to Cl

Covalent as 2 atoms approach each with extra electron bonding and anti-bonding orbitals are formed

Metallic long-wavelength plane waves play the role of the bonding orbitals shorter-wavelength (higher energy) waves are like anti-bonding orbitals

Lecture 12Sept 17 2010PH 481581

21

Comparison of different core potentials V=r-n

for n = 1236 and12n=1 is said to be ldquosoftrdquon=12 is ldquohardrdquo

n=infinity

22

END

• Lecture 3 ndash August 25 2010
• 3D systems(symmetries)
• Lecture 4 - Aug 27 2010
• Real crystals
• Bragg Scattering of x-rays
• Better treatment x-rays are scattered by electron density n(r)
• Periodic functions in a lattice
• Examples of reciprocal lattices
• Calculating FT of n(r)
• Calculating x-ray scattering intensity
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22

Real crystals Simple cubic wo basis none Unstable

Uncharged atoms prefer close-packed structures many near neighbors (12 nn in hcp or fcc)

Ions NaCl structure (fcc with 2-ion basis 6 nn)

CsCl structure (sc with 2-ion basis) (ldquo2-ion unit cellrdquo 8 nn)

We will skip Miller indices here (they are easier to understand after we discuss the reciprocal lattice) and go on to

Covalently bonded structures Diamond

fcc 2 atomsprimitive unit cell = 8 atomsconventional unit cell Basis at (000) and (141414)

(fcc translations (12120) (100)

4

Bragg Scattering of x-rays

Extra path length must be integer number of wavelengths

2 d sin = n

Braggrsquos parallel plane picture is mnemonic not derivation

PH 481581 Lecture 6 Sept 1 2010

Figure Courtesy Wikipedia

5

Better treatment x-rays are scattered by electron density n(r)

In 1D n(x) = n(x+a) (periodic)Expand in Fourier seriesn(x) = Σ nG eiGx

where G=integer 2aThese Grsquos form a ldquoreciprocal latticerdquo ndash is also G eiGx is periodic in the direct latticewhere ldquodirect latticerdquo is -2a -a 0 a 2a We want to generalize this to 3D

ldquolatticerdquo means closed under addition

6

k

x-raykr

n(x) will be real if n-G = nG

(complex conjugates)

Periodic functions in a lattice To calculate scattering from a periodic electron density n(r) we need to describe what it means to be periodic n(r+R) = n(r) for R in lattice R = u1a1 + u2a2 + u3a3Any periodic function n(r) can be expressed as a Fourier series n(r) = Σ nG eiGr where each eiGr is periodic in the latticeeiG(r+R) = eiGr eiGR = 1 GR = 2integer The set of such Grsquos is closed under addition ndash a lattice called the ldquoreciprocal latticerdquo of the ldquodirect latticerdquo determined by a1 a2 and a3

You can show that the vectors b1 b2 and b3 determined by

ai bj = 2 ij

are primitive translation vectors of this reciprocal lattice (RL)

This means b1 must be perpendicular to a2 and a3 Explicit formulas are 213

213

132

132

31

31

2

2

2

aaa

aab

aaa

aab

aaa

aab

2

2

7

[Start by showing b1 is in RL b1R = 2integer]

Examples of reciprocal lattices

Orthorhombic direct latticeˆˆˆ 332211 zayaxa aaa

Reciprocal lattice is

zbybxb ˆ2

ˆ2

ˆ2

33

22

11 aaa

Easy to check that

ai bj = 2 ij

Hexagonal direct latticeˆ)ˆˆ(ˆ 32

321

21 zayxaxa caa Reciprocal lattice is

a1

a2

b1

b2

132

132

31

31

2

2 from

aaa

aab

aaa

aab

2

2

8

Calculating FT of n(r)

nG = Fourier transform of scattering density n(r)Only nonzero at points G of reciprocal latticeExplicitly

9

n(r) = Σ nG eiGr

rrrGG

3

cellunit

)(1

dneV

n i

where V is the volume of the unit cell

Calculating x-ray scattering intensity

10

Incoming amplitude (electric field)

Lecture 7 Sept 3 2010 PH 481581

kx-ray

r

n(r) = Σ nG eiGr

rrsquo

detector

Amplitude reaching detector =rkie Re

rposition Source

n(r)d3 r rrike

rkr

k

number of scatterers

Phase change

3 ( )

Source position

3 ( )

Source position

( )

( ) ( )

i i

i i i

e n d e

e n d e e F

k r k r r

r

k r k k r k r

r

r r

r r k k

where F is called the scattering amplitude (in Kittelrsquos book)

22 2 2 2

2 2

2( ) 2 1

rr r r

r r

r rr r r r

122

2 2 2

21 1 2nd order

rr r

r r r

r r r rr r

( )k kr kr

r rr r k r k r k r r

rrsquo - r

G

Conclusion x-ray scattering probes reciprocal lattice (RL)

If incoming wavevector is k amplitude for outgoing wavevector krsquo ~ Fourier component nkrsquo-k In a perfect crystal this is nonzero only if krsquo ndash k is a reciprocal lattice vector call it G

krsquo

k

You have to rotate the crystal powder it

Also this is elastic scattering |k|=|krsquo| so krsquo must lie on a sphere The probability that one of the vectors k+G lies exactly on the sphere is zero (the sphere has zero thickness)

Powder pattern crystallites are in all possible orientations which rotates each G over the surface of a sphereSphere of possible Grsquos intersects sphere of possible krsquo s in a circle (ring) seen on screen

Gk

krsquo

scre

en

12

Powder pattern

Pattern for the mineral olivine httpndbserverrutgersedummcifcbf

httpwwwccp14acukccpweb-mirrorsarmelegypteconf2theoryhtml

Ball-milled

Hand-crushed

d

Relating krsquo ndash k = G to the Bragg equation

13

G

krsquo

k

Bragg picture is krsquok

leading to

2 d sin = n

2 2( ) k G k

In reciprocal lattice picture krsquo = k + G so elastic scattering condition is

2 2 22 k k G G k22 0 k G k

Geometrically we have 2 2

sin sin 2 sin2

Gk

G

Consistent only if (so smallest G goes with n=1 )Gn

d 2 G

d2

Predicting angles of powder diffraction rings

14

G

krsquo

k

Recall picture of krsquo = k + G

amp from top triangle sin2

kG

so )(2

and 2

but lkha

k

G

2

sin 12222 lkh

a

5012

4002

3111

2011

1001

21222 lkhlkh

Lecture 8 PH 481581

So the Bragg planes are related to reciprocal lattice vectors and the plane spacing d is related to the RLV magnitude G = |G| by d = 2G

15

Schematically RLV G lattice planes

or more precisely a plane and a vector G each determine a direction in space There are many Gs along this direction and many planes normal to it so the relation is really between families of Grsquos and families of planesfamily 0plusmn G plusmn 2G of parallel RLVs family of parallel planesOnly some of these planes go through lattice points and these are spaced d = 2G apart(G must be the smallest length in the family)

16

0

Direct lattice

Reciprocal Lattice

Drawing low-index planes (far apart)high-index planes (close together)

Calculating scattering amplitude from atomic form factors

17

Consider a crystal with a basisr1

r2

Plot density along dashed linen(r) = n1(r-r1) + n2(r-r2)= n1() + n2(2)

rx

x 2x

r2xr1x

where = r - r1 and = r ndash r2

n1 n2

r

The scattering amplitude FG is defined by

Lecture 10 PH 481581 Sept 13 2010

crystal entire

3)( rdenF iG

rGr

But the integrand is periodic so the result is proportional to the number N of unit cells The amplitude per unit cell is called the structure factor SG = FG N

Calculating scattering amplitude (structure factor)

18

Structure factor

jj

iii

iiii

ii

ii

iG

fefefe

denedene

denden

rdenrden

rdenS

j )()()(

)()(

)()(

)()(

)(

21

32

31

3)(2

3)(1

322

311

cell

3

21

21

21

GGG

ρρ

ρρ

rrrr

r

rGrGrG

ρGrGρGrG

ρrGρrG

rGrG

rG

r1r2

r

rdenf iii

3)()( ρGρGwhere

Example CsCl simple cubic r1= = (000) r2 = (a2a2a2)Some algebra gives

odd if

)points fcc(even if)()1()(

21

2121 l khff

l khffffS lkh

G

GG

In limit f1 f2(ie CsCl FeFe bcc) this becomes fcc RL

Another example (NaCl structure)

19

KCl isoelectronic f1 = f2 lose reflections due to extra symmetry

NaCl is fcc basis is r1= = (000) r2 = (a200)

centers) (cell odd all if

points)(corner even all if)()1()(

21

2121 lkhff

lkhffffS h

G

GG

(if some even and some odd eg (100) these points are not even ON the reciprocal lattice so we donrsquot calculate SG)

Example in Fig 2-17 p 42 KBr has 111 200 220 as predicted

Chapter 3 BindingTypes of bonding

bull van der Waalsbull Ionic bull Covalentbull metallic

20

Start with van der Waals because all pairs of atoms have vdW attraction For neutral atoms (so there is no electrostatic force) it always dominates at long distances

Ionic as a Na and a Cl atom approach each other the extra electron on Na tunnels to Cl

Covalent as 2 atoms approach each with extra electron bonding and anti-bonding orbitals are formed

Metallic long-wavelength plane waves play the role of the bonding orbitals shorter-wavelength (higher energy) waves are like anti-bonding orbitals

Lecture 12Sept 17 2010PH 481581

21

Comparison of different core potentials V=r-n

for n = 1236 and12n=1 is said to be ldquosoftrdquon=12 is ldquohardrdquo

n=infinity

22

END

• Lecture 3 ndash August 25 2010
• 3D systems(symmetries)
• Lecture 4 - Aug 27 2010
• Real crystals
• Bragg Scattering of x-rays
• Better treatment x-rays are scattered by electron density n(r)
• Periodic functions in a lattice
• Examples of reciprocal lattices
• Calculating FT of n(r)
• Calculating x-ray scattering intensity
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22

Bragg Scattering of x-rays

Extra path length must be integer number of wavelengths

2 d sin = n

Braggrsquos parallel plane picture is mnemonic not derivation

PH 481581 Lecture 6 Sept 1 2010

Figure Courtesy Wikipedia

5

Better treatment x-rays are scattered by electron density n(r)

In 1D n(x) = n(x+a) (periodic)Expand in Fourier seriesn(x) = Σ nG eiGx

where G=integer 2aThese Grsquos form a ldquoreciprocal latticerdquo ndash is also G eiGx is periodic in the direct latticewhere ldquodirect latticerdquo is -2a -a 0 a 2a We want to generalize this to 3D

ldquolatticerdquo means closed under addition

6

k

x-raykr

n(x) will be real if n-G = nG

(complex conjugates)

Periodic functions in a lattice To calculate scattering from a periodic electron density n(r) we need to describe what it means to be periodic n(r+R) = n(r) for R in lattice R = u1a1 + u2a2 + u3a3Any periodic function n(r) can be expressed as a Fourier series n(r) = Σ nG eiGr where each eiGr is periodic in the latticeeiG(r+R) = eiGr eiGR = 1 GR = 2integer The set of such Grsquos is closed under addition ndash a lattice called the ldquoreciprocal latticerdquo of the ldquodirect latticerdquo determined by a1 a2 and a3

You can show that the vectors b1 b2 and b3 determined by

ai bj = 2 ij

are primitive translation vectors of this reciprocal lattice (RL)

This means b1 must be perpendicular to a2 and a3 Explicit formulas are 213

213

132

132

31

31

2

2

2

aaa

aab

aaa

aab

aaa

aab

2

2

7

[Start by showing b1 is in RL b1R = 2integer]

Examples of reciprocal lattices

Orthorhombic direct latticeˆˆˆ 332211 zayaxa aaa

Reciprocal lattice is

zbybxb ˆ2

ˆ2

ˆ2

33

22

11 aaa

Easy to check that

ai bj = 2 ij

Hexagonal direct latticeˆ)ˆˆ(ˆ 32

321

21 zayxaxa caa Reciprocal lattice is

a1

a2

b1

b2

132

132

31

31

2

2 from

aaa

aab

aaa

aab

2

2

8

Calculating FT of n(r)

nG = Fourier transform of scattering density n(r)Only nonzero at points G of reciprocal latticeExplicitly

9

n(r) = Σ nG eiGr

rrrGG

3

cellunit

)(1

dneV

n i

where V is the volume of the unit cell

Calculating x-ray scattering intensity

10

Incoming amplitude (electric field)

Lecture 7 Sept 3 2010 PH 481581

kx-ray

r

n(r) = Σ nG eiGr

rrsquo

detector

Amplitude reaching detector =rkie Re

rposition Source

n(r)d3 r rrike

rkr

k

number of scatterers

Phase change

3 ( )

Source position

3 ( )

Source position

( )

( ) ( )

i i

i i i

e n d e

e n d e e F

k r k r r

r

k r k k r k r

r

r r

r r k k

where F is called the scattering amplitude (in Kittelrsquos book)

22 2 2 2

2 2

2( ) 2 1

rr r r

r r

r rr r r r

122

2 2 2

21 1 2nd order

rr r

r r r

r r r rr r

( )k kr kr

r rr r k r k r k r r

rrsquo - r

G

Conclusion x-ray scattering probes reciprocal lattice (RL)

If incoming wavevector is k amplitude for outgoing wavevector krsquo ~ Fourier component nkrsquo-k In a perfect crystal this is nonzero only if krsquo ndash k is a reciprocal lattice vector call it G

krsquo

k

You have to rotate the crystal powder it

Also this is elastic scattering |k|=|krsquo| so krsquo must lie on a sphere The probability that one of the vectors k+G lies exactly on the sphere is zero (the sphere has zero thickness)

Powder pattern crystallites are in all possible orientations which rotates each G over the surface of a sphereSphere of possible Grsquos intersects sphere of possible krsquo s in a circle (ring) seen on screen

Gk

krsquo

scre

en

12

Powder pattern

Pattern for the mineral olivine httpndbserverrutgersedummcifcbf

httpwwwccp14acukccpweb-mirrorsarmelegypteconf2theoryhtml

Ball-milled

Hand-crushed

d

Relating krsquo ndash k = G to the Bragg equation

13

G

krsquo

k

Bragg picture is krsquok

leading to

2 d sin = n

2 2( ) k G k

In reciprocal lattice picture krsquo = k + G so elastic scattering condition is

2 2 22 k k G G k22 0 k G k

Geometrically we have 2 2

sin sin 2 sin2

Gk

G

Consistent only if (so smallest G goes with n=1 )Gn

d 2 G

d2

Predicting angles of powder diffraction rings

14

G

krsquo

k

Recall picture of krsquo = k + G

amp from top triangle sin2

kG

so )(2

and 2

but lkha

k

G

2

sin 12222 lkh

a

5012

4002

3111

2011

1001

21222 lkhlkh

Lecture 8 PH 481581

So the Bragg planes are related to reciprocal lattice vectors and the plane spacing d is related to the RLV magnitude G = |G| by d = 2G

15

Schematically RLV G lattice planes

or more precisely a plane and a vector G each determine a direction in space There are many Gs along this direction and many planes normal to it so the relation is really between families of Grsquos and families of planesfamily 0plusmn G plusmn 2G of parallel RLVs family of parallel planesOnly some of these planes go through lattice points and these are spaced d = 2G apart(G must be the smallest length in the family)

16

0

Direct lattice

Reciprocal Lattice

Drawing low-index planes (far apart)high-index planes (close together)

Calculating scattering amplitude from atomic form factors

17

Consider a crystal with a basisr1

r2

Plot density along dashed linen(r) = n1(r-r1) + n2(r-r2)= n1() + n2(2)

rx

x 2x

r2xr1x

where = r - r1 and = r ndash r2

n1 n2

r

The scattering amplitude FG is defined by

Lecture 10 PH 481581 Sept 13 2010

crystal entire

3)( rdenF iG

rGr

But the integrand is periodic so the result is proportional to the number N of unit cells The amplitude per unit cell is called the structure factor SG = FG N

Calculating scattering amplitude (structure factor)

18

Structure factor

jj

iii

iiii

ii

ii

iG

fefefe

denedene

denden

rdenrden

rdenS

j )()()(

)()(

)()(

)()(

)(

21

32

31

3)(2

3)(1

322

311

cell

3

21

21

21

GGG

ρρ

ρρ

rrrr

r

rGrGrG

ρGrGρGrG

ρrGρrG

rGrG

rG

r1r2

r

rdenf iii

3)()( ρGρGwhere

Example CsCl simple cubic r1= = (000) r2 = (a2a2a2)Some algebra gives

odd if

)points fcc(even if)()1()(

21

2121 l khff

l khffffS lkh

G

GG

In limit f1 f2(ie CsCl FeFe bcc) this becomes fcc RL

Another example (NaCl structure)

19

KCl isoelectronic f1 = f2 lose reflections due to extra symmetry

NaCl is fcc basis is r1= = (000) r2 = (a200)

centers) (cell odd all if

points)(corner even all if)()1()(

21

2121 lkhff

lkhffffS h

G

GG

(if some even and some odd eg (100) these points are not even ON the reciprocal lattice so we donrsquot calculate SG)

Example in Fig 2-17 p 42 KBr has 111 200 220 as predicted

Chapter 3 BindingTypes of bonding

bull van der Waalsbull Ionic bull Covalentbull metallic

20

Start with van der Waals because all pairs of atoms have vdW attraction For neutral atoms (so there is no electrostatic force) it always dominates at long distances

Ionic as a Na and a Cl atom approach each other the extra electron on Na tunnels to Cl

Covalent as 2 atoms approach each with extra electron bonding and anti-bonding orbitals are formed

Metallic long-wavelength plane waves play the role of the bonding orbitals shorter-wavelength (higher energy) waves are like anti-bonding orbitals

Lecture 12Sept 17 2010PH 481581

21

Comparison of different core potentials V=r-n

for n = 1236 and12n=1 is said to be ldquosoftrdquon=12 is ldquohardrdquo

n=infinity

22

END

• Lecture 3 ndash August 25 2010
• 3D systems(symmetries)
• Lecture 4 - Aug 27 2010
• Real crystals
• Bragg Scattering of x-rays
• Better treatment x-rays are scattered by electron density n(r)
• Periodic functions in a lattice
• Examples of reciprocal lattices
• Calculating FT of n(r)
• Calculating x-ray scattering intensity
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22

Better treatment x-rays are scattered by electron density n(r)

In 1D n(x) = n(x+a) (periodic)Expand in Fourier seriesn(x) = Σ nG eiGx

where G=integer 2aThese Grsquos form a ldquoreciprocal latticerdquo ndash is also G eiGx is periodic in the direct latticewhere ldquodirect latticerdquo is -2a -a 0 a 2a We want to generalize this to 3D

ldquolatticerdquo means closed under addition

6

k

x-raykr

n(x) will be real if n-G = nG

(complex conjugates)

Periodic functions in a lattice To calculate scattering from a periodic electron density n(r) we need to describe what it means to be periodic n(r+R) = n(r) for R in lattice R = u1a1 + u2a2 + u3a3Any periodic function n(r) can be expressed as a Fourier series n(r) = Σ nG eiGr where each eiGr is periodic in the latticeeiG(r+R) = eiGr eiGR = 1 GR = 2integer The set of such Grsquos is closed under addition ndash a lattice called the ldquoreciprocal latticerdquo of the ldquodirect latticerdquo determined by a1 a2 and a3

You can show that the vectors b1 b2 and b3 determined by

ai bj = 2 ij

are primitive translation vectors of this reciprocal lattice (RL)

This means b1 must be perpendicular to a2 and a3 Explicit formulas are 213

213

132

132

31

31

2

2

2

aaa

aab

aaa

aab

aaa

aab

2

2

7

[Start by showing b1 is in RL b1R = 2integer]

Examples of reciprocal lattices

Orthorhombic direct latticeˆˆˆ 332211 zayaxa aaa

Reciprocal lattice is

zbybxb ˆ2

ˆ2

ˆ2

33

22

11 aaa

Easy to check that

ai bj = 2 ij

Hexagonal direct latticeˆ)ˆˆ(ˆ 32

321

21 zayxaxa caa Reciprocal lattice is

a1

a2

b1

b2

132

132

31

31

2

2 from

aaa

aab

aaa

aab

2

2

8

Calculating FT of n(r)

nG = Fourier transform of scattering density n(r)Only nonzero at points G of reciprocal latticeExplicitly

9

n(r) = Σ nG eiGr

rrrGG

3

cellunit

)(1

dneV

n i

where V is the volume of the unit cell

Calculating x-ray scattering intensity

10

Incoming amplitude (electric field)

Lecture 7 Sept 3 2010 PH 481581

kx-ray

r

n(r) = Σ nG eiGr

rrsquo

detector

Amplitude reaching detector =rkie Re

rposition Source

n(r)d3 r rrike

rkr

k

number of scatterers

Phase change

3 ( )

Source position

3 ( )

Source position

( )

( ) ( )

i i

i i i

e n d e

e n d e e F

k r k r r

r

k r k k r k r

r

r r

r r k k

where F is called the scattering amplitude (in Kittelrsquos book)

22 2 2 2

2 2

2( ) 2 1

rr r r

r r

r rr r r r

122

2 2 2

21 1 2nd order

rr r

r r r

r r r rr r

( )k kr kr

r rr r k r k r k r r

rrsquo - r

G

Conclusion x-ray scattering probes reciprocal lattice (RL)

If incoming wavevector is k amplitude for outgoing wavevector krsquo ~ Fourier component nkrsquo-k In a perfect crystal this is nonzero only if krsquo ndash k is a reciprocal lattice vector call it G

krsquo

k

You have to rotate the crystal powder it

Also this is elastic scattering |k|=|krsquo| so krsquo must lie on a sphere The probability that one of the vectors k+G lies exactly on the sphere is zero (the sphere has zero thickness)

Powder pattern crystallites are in all possible orientations which rotates each G over the surface of a sphereSphere of possible Grsquos intersects sphere of possible krsquo s in a circle (ring) seen on screen

Gk

krsquo

scre

en

12

Powder pattern

Pattern for the mineral olivine httpndbserverrutgersedummcifcbf

httpwwwccp14acukccpweb-mirrorsarmelegypteconf2theoryhtml

Ball-milled

Hand-crushed

d

Relating krsquo ndash k = G to the Bragg equation

13

G

krsquo

k

Bragg picture is krsquok

leading to

2 d sin = n

2 2( ) k G k

In reciprocal lattice picture krsquo = k + G so elastic scattering condition is

2 2 22 k k G G k22 0 k G k

Geometrically we have 2 2

sin sin 2 sin2

Gk

G

Consistent only if (so smallest G goes with n=1 )Gn

d 2 G

d2

Predicting angles of powder diffraction rings

14

G

krsquo

k

Recall picture of krsquo = k + G

amp from top triangle sin2

kG

so )(2

and 2

but lkha

k

G

2

sin 12222 lkh

a

5012

4002

3111

2011

1001

21222 lkhlkh

Lecture 8 PH 481581

So the Bragg planes are related to reciprocal lattice vectors and the plane spacing d is related to the RLV magnitude G = |G| by d = 2G

15

Schematically RLV G lattice planes

or more precisely a plane and a vector G each determine a direction in space There are many Gs along this direction and many planes normal to it so the relation is really between families of Grsquos and families of planesfamily 0plusmn G plusmn 2G of parallel RLVs family of parallel planesOnly some of these planes go through lattice points and these are spaced d = 2G apart(G must be the smallest length in the family)

16

0

Direct lattice

Reciprocal Lattice

Drawing low-index planes (far apart)high-index planes (close together)

Calculating scattering amplitude from atomic form factors

17

Consider a crystal with a basisr1

r2

Plot density along dashed linen(r) = n1(r-r1) + n2(r-r2)= n1() + n2(2)

rx

x 2x

r2xr1x

where = r - r1 and = r ndash r2

n1 n2

r

The scattering amplitude FG is defined by

Lecture 10 PH 481581 Sept 13 2010

crystal entire

3)( rdenF iG

rGr

But the integrand is periodic so the result is proportional to the number N of unit cells The amplitude per unit cell is called the structure factor SG = FG N

Calculating scattering amplitude (structure factor)

18

Structure factor

jj

iii

iiii

ii

ii

iG

fefefe

denedene

denden

rdenrden

rdenS

j )()()(

)()(

)()(

)()(

)(

21

32

31

3)(2

3)(1

322

311

cell

3

21

21

21

GGG

ρρ

ρρ

rrrr

r

rGrGrG

ρGrGρGrG

ρrGρrG

rGrG

rG

r1r2

r

rdenf iii

3)()( ρGρGwhere

Example CsCl simple cubic r1= = (000) r2 = (a2a2a2)Some algebra gives

odd if

)points fcc(even if)()1()(

21

2121 l khff

l khffffS lkh

G

GG

In limit f1 f2(ie CsCl FeFe bcc) this becomes fcc RL

Another example (NaCl structure)

19

KCl isoelectronic f1 = f2 lose reflections due to extra symmetry

NaCl is fcc basis is r1= = (000) r2 = (a200)

centers) (cell odd all if

points)(corner even all if)()1()(

21

2121 lkhff

lkhffffS h

G

GG

(if some even and some odd eg (100) these points are not even ON the reciprocal lattice so we donrsquot calculate SG)

Example in Fig 2-17 p 42 KBr has 111 200 220 as predicted

Chapter 3 BindingTypes of bonding

bull van der Waalsbull Ionic bull Covalentbull metallic

20

Start with van der Waals because all pairs of atoms have vdW attraction For neutral atoms (so there is no electrostatic force) it always dominates at long distances

Ionic as a Na and a Cl atom approach each other the extra electron on Na tunnels to Cl

Covalent as 2 atoms approach each with extra electron bonding and anti-bonding orbitals are formed

Metallic long-wavelength plane waves play the role of the bonding orbitals shorter-wavelength (higher energy) waves are like anti-bonding orbitals

Lecture 12Sept 17 2010PH 481581

21

Comparison of different core potentials V=r-n

for n = 1236 and12n=1 is said to be ldquosoftrdquon=12 is ldquohardrdquo

n=infinity

22

END

• Lecture 3 ndash August 25 2010
• 3D systems(symmetries)
• Lecture 4 - Aug 27 2010
• Real crystals
• Bragg Scattering of x-rays
• Better treatment x-rays are scattered by electron density n(r)
• Periodic functions in a lattice
• Examples of reciprocal lattices
• Calculating FT of n(r)
• Calculating x-ray scattering intensity
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22

Periodic functions in a lattice To calculate scattering from a periodic electron density n(r) we need to describe what it means to be periodic n(r+R) = n(r) for R in lattice R = u1a1 + u2a2 + u3a3Any periodic function n(r) can be expressed as a Fourier series n(r) = Σ nG eiGr where each eiGr is periodic in the latticeeiG(r+R) = eiGr eiGR = 1 GR = 2integer The set of such Grsquos is closed under addition ndash a lattice called the ldquoreciprocal latticerdquo of the ldquodirect latticerdquo determined by a1 a2 and a3

You can show that the vectors b1 b2 and b3 determined by

ai bj = 2 ij

are primitive translation vectors of this reciprocal lattice (RL)

This means b1 must be perpendicular to a2 and a3 Explicit formulas are 213

213

132

132

31

31

2

2

2

aaa

aab

aaa

aab

aaa

aab

2

2

7

[Start by showing b1 is in RL b1R = 2integer]

Examples of reciprocal lattices

Orthorhombic direct latticeˆˆˆ 332211 zayaxa aaa

Reciprocal lattice is

zbybxb ˆ2

ˆ2

ˆ2

33

22

11 aaa

Easy to check that

ai bj = 2 ij

Hexagonal direct latticeˆ)ˆˆ(ˆ 32

321

21 zayxaxa caa Reciprocal lattice is

a1

a2

b1

b2

132

132

31

31

2

2 from

aaa

aab

aaa

aab

2

2

8

Calculating FT of n(r)

nG = Fourier transform of scattering density n(r)Only nonzero at points G of reciprocal latticeExplicitly

9

n(r) = Σ nG eiGr

rrrGG

3

cellunit

)(1

dneV

n i

where V is the volume of the unit cell

Calculating x-ray scattering intensity

10

Incoming amplitude (electric field)

Lecture 7 Sept 3 2010 PH 481581

kx-ray

r

n(r) = Σ nG eiGr

rrsquo

detector

Amplitude reaching detector =rkie Re

rposition Source

n(r)d3 r rrike

rkr

k

number of scatterers

Phase change

3 ( )

Source position

3 ( )

Source position

( )

( ) ( )

i i

i i i

e n d e

e n d e e F

k r k r r

r

k r k k r k r

r

r r

r r k k

where F is called the scattering amplitude (in Kittelrsquos book)

22 2 2 2

2 2

2( ) 2 1

rr r r

r r

r rr r r r

122

2 2 2

21 1 2nd order

rr r

r r r

r r r rr r

( )k kr kr

r rr r k r k r k r r

rrsquo - r

G

Conclusion x-ray scattering probes reciprocal lattice (RL)

If incoming wavevector is k amplitude for outgoing wavevector krsquo ~ Fourier component nkrsquo-k In a perfect crystal this is nonzero only if krsquo ndash k is a reciprocal lattice vector call it G

krsquo

k

You have to rotate the crystal powder it

Also this is elastic scattering |k|=|krsquo| so krsquo must lie on a sphere The probability that one of the vectors k+G lies exactly on the sphere is zero (the sphere has zero thickness)

Powder pattern crystallites are in all possible orientations which rotates each G over the surface of a sphereSphere of possible Grsquos intersects sphere of possible krsquo s in a circle (ring) seen on screen

Gk

krsquo

scre

en

12

Powder pattern

Pattern for the mineral olivine httpndbserverrutgersedummcifcbf

httpwwwccp14acukccpweb-mirrorsarmelegypteconf2theoryhtml

Ball-milled

Hand-crushed

d

Relating krsquo ndash k = G to the Bragg equation

13

G

krsquo

k

Bragg picture is krsquok

leading to

2 d sin = n

2 2( ) k G k

In reciprocal lattice picture krsquo = k + G so elastic scattering condition is

2 2 22 k k G G k22 0 k G k

Geometrically we have 2 2

sin sin 2 sin2

Gk

G

Consistent only if (so smallest G goes with n=1 )Gn

d 2 G

d2

Predicting angles of powder diffraction rings

14

G

krsquo

k

Recall picture of krsquo = k + G

amp from top triangle sin2

kG

so )(2

and 2

but lkha

k

G

2

sin 12222 lkh

a

5012

4002

3111

2011

1001

21222 lkhlkh

Lecture 8 PH 481581

So the Bragg planes are related to reciprocal lattice vectors and the plane spacing d is related to the RLV magnitude G = |G| by d = 2G

15

Schematically RLV G lattice planes

or more precisely a plane and a vector G each determine a direction in space There are many Gs along this direction and many planes normal to it so the relation is really between families of Grsquos and families of planesfamily 0plusmn G plusmn 2G of parallel RLVs family of parallel planesOnly some of these planes go through lattice points and these are spaced d = 2G apart(G must be the smallest length in the family)

16

0

Direct lattice

Reciprocal Lattice

Drawing low-index planes (far apart)high-index planes (close together)

Calculating scattering amplitude from atomic form factors

17

Consider a crystal with a basisr1

r2

Plot density along dashed linen(r) = n1(r-r1) + n2(r-r2)= n1() + n2(2)

rx

x 2x

r2xr1x

where = r - r1 and = r ndash r2

n1 n2

r

The scattering amplitude FG is defined by

Lecture 10 PH 481581 Sept 13 2010

crystal entire

3)( rdenF iG

rGr

But the integrand is periodic so the result is proportional to the number N of unit cells The amplitude per unit cell is called the structure factor SG = FG N

Calculating scattering amplitude (structure factor)

18

Structure factor

jj

iii

iiii

ii

ii

iG

fefefe

denedene

denden

rdenrden

rdenS

j )()()(

)()(

)()(

)()(

)(

21

32

31

3)(2

3)(1

322

311

cell

3

21

21

21

GGG

ρρ

ρρ

rrrr

r

rGrGrG

ρGrGρGrG

ρrGρrG

rGrG

rG

r1r2

r

rdenf iii

3)()( ρGρGwhere

Example CsCl simple cubic r1= = (000) r2 = (a2a2a2)Some algebra gives

odd if

)points fcc(even if)()1()(

21

2121 l khff

l khffffS lkh

G

GG

In limit f1 f2(ie CsCl FeFe bcc) this becomes fcc RL

Another example (NaCl structure)

19

KCl isoelectronic f1 = f2 lose reflections due to extra symmetry

NaCl is fcc basis is r1= = (000) r2 = (a200)

centers) (cell odd all if

points)(corner even all if)()1()(

21

2121 lkhff

lkhffffS h

G

GG

(if some even and some odd eg (100) these points are not even ON the reciprocal lattice so we donrsquot calculate SG)

Example in Fig 2-17 p 42 KBr has 111 200 220 as predicted

Chapter 3 BindingTypes of bonding

bull van der Waalsbull Ionic bull Covalentbull metallic

20

Start with van der Waals because all pairs of atoms have vdW attraction For neutral atoms (so there is no electrostatic force) it always dominates at long distances

Ionic as a Na and a Cl atom approach each other the extra electron on Na tunnels to Cl

Covalent as 2 atoms approach each with extra electron bonding and anti-bonding orbitals are formed

Metallic long-wavelength plane waves play the role of the bonding orbitals shorter-wavelength (higher energy) waves are like anti-bonding orbitals

Lecture 12Sept 17 2010PH 481581

21

Comparison of different core potentials V=r-n

for n = 1236 and12n=1 is said to be ldquosoftrdquon=12 is ldquohardrdquo

n=infinity

22

END

• Lecture 3 ndash August 25 2010
• 3D systems(symmetries)
• Lecture 4 - Aug 27 2010
• Real crystals
• Bragg Scattering of x-rays
• Better treatment x-rays are scattered by electron density n(r)
• Periodic functions in a lattice
• Examples of reciprocal lattices
• Calculating FT of n(r)
• Calculating x-ray scattering intensity
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22

Examples of reciprocal lattices

Orthorhombic direct latticeˆˆˆ 332211 zayaxa aaa

Reciprocal lattice is

zbybxb ˆ2

ˆ2

ˆ2

33

22

11 aaa

Easy to check that

ai bj = 2 ij

Hexagonal direct latticeˆ)ˆˆ(ˆ 32

321

21 zayxaxa caa Reciprocal lattice is

a1

a2

b1

b2

132

132

31

31

2

2 from

aaa

aab

aaa

aab

2

2

8

Calculating FT of n(r)

nG = Fourier transform of scattering density n(r)Only nonzero at points G of reciprocal latticeExplicitly

9

n(r) = Σ nG eiGr

rrrGG

3

cellunit

)(1

dneV

n i

where V is the volume of the unit cell

Calculating x-ray scattering intensity

10

Incoming amplitude (electric field)

Lecture 7 Sept 3 2010 PH 481581

kx-ray

r

n(r) = Σ nG eiGr

rrsquo

detector

Amplitude reaching detector =rkie Re

rposition Source

n(r)d3 r rrike

rkr

k

number of scatterers

Phase change

3 ( )

Source position

3 ( )

Source position

( )

( ) ( )

i i

i i i

e n d e

e n d e e F

k r k r r

r

k r k k r k r

r

r r

r r k k

where F is called the scattering amplitude (in Kittelrsquos book)

22 2 2 2

2 2

2( ) 2 1

rr r r

r r

r rr r r r

122

2 2 2

21 1 2nd order

rr r

r r r

r r r rr r

( )k kr kr

r rr r k r k r k r r

rrsquo - r

G

Conclusion x-ray scattering probes reciprocal lattice (RL)

If incoming wavevector is k amplitude for outgoing wavevector krsquo ~ Fourier component nkrsquo-k In a perfect crystal this is nonzero only if krsquo ndash k is a reciprocal lattice vector call it G

krsquo

k

You have to rotate the crystal powder it

Also this is elastic scattering |k|=|krsquo| so krsquo must lie on a sphere The probability that one of the vectors k+G lies exactly on the sphere is zero (the sphere has zero thickness)

Powder pattern crystallites are in all possible orientations which rotates each G over the surface of a sphereSphere of possible Grsquos intersects sphere of possible krsquo s in a circle (ring) seen on screen

Gk

krsquo

scre

en

12

Powder pattern

Pattern for the mineral olivine httpndbserverrutgersedummcifcbf

httpwwwccp14acukccpweb-mirrorsarmelegypteconf2theoryhtml

Ball-milled

Hand-crushed

d

Relating krsquo ndash k = G to the Bragg equation

13

G

krsquo

k

Bragg picture is krsquok

leading to

2 d sin = n

2 2( ) k G k

In reciprocal lattice picture krsquo = k + G so elastic scattering condition is

2 2 22 k k G G k22 0 k G k

Geometrically we have 2 2

sin sin 2 sin2

Gk

G

Consistent only if (so smallest G goes with n=1 )Gn

d 2 G

d2

Predicting angles of powder diffraction rings

14

G

krsquo

k

Recall picture of krsquo = k + G

amp from top triangle sin2

kG

so )(2

and 2

but lkha

k

G

2

sin 12222 lkh

a

5012

4002

3111

2011

1001

21222 lkhlkh

Lecture 8 PH 481581

So the Bragg planes are related to reciprocal lattice vectors and the plane spacing d is related to the RLV magnitude G = |G| by d = 2G

15

Schematically RLV G lattice planes

or more precisely a plane and a vector G each determine a direction in space There are many Gs along this direction and many planes normal to it so the relation is really between families of Grsquos and families of planesfamily 0plusmn G plusmn 2G of parallel RLVs family of parallel planesOnly some of these planes go through lattice points and these are spaced d = 2G apart(G must be the smallest length in the family)

16

0

Direct lattice

Reciprocal Lattice

Drawing low-index planes (far apart)high-index planes (close together)

Calculating scattering amplitude from atomic form factors

17

Consider a crystal with a basisr1

r2

Plot density along dashed linen(r) = n1(r-r1) + n2(r-r2)= n1() + n2(2)

rx

x 2x

r2xr1x

where = r - r1 and = r ndash r2

n1 n2

r

The scattering amplitude FG is defined by

Lecture 10 PH 481581 Sept 13 2010

crystal entire

3)( rdenF iG

rGr

But the integrand is periodic so the result is proportional to the number N of unit cells The amplitude per unit cell is called the structure factor SG = FG N

Calculating scattering amplitude (structure factor)

18

Structure factor

jj

iii

iiii

ii

ii

iG

fefefe

denedene

denden

rdenrden

rdenS

j )()()(

)()(

)()(

)()(

)(

21

32

31

3)(2

3)(1

322

311

cell

3

21

21

21

GGG

ρρ

ρρ

rrrr

r

rGrGrG

ρGrGρGrG

ρrGρrG

rGrG

rG

r1r2

r

rdenf iii

3)()( ρGρGwhere

Example CsCl simple cubic r1= = (000) r2 = (a2a2a2)Some algebra gives

odd if

)points fcc(even if)()1()(

21

2121 l khff

l khffffS lkh

G

GG

In limit f1 f2(ie CsCl FeFe bcc) this becomes fcc RL

Another example (NaCl structure)

19

KCl isoelectronic f1 = f2 lose reflections due to extra symmetry

NaCl is fcc basis is r1= = (000) r2 = (a200)

centers) (cell odd all if

points)(corner even all if)()1()(

21

2121 lkhff

lkhffffS h

G

GG

(if some even and some odd eg (100) these points are not even ON the reciprocal lattice so we donrsquot calculate SG)

Example in Fig 2-17 p 42 KBr has 111 200 220 as predicted

Chapter 3 BindingTypes of bonding

bull van der Waalsbull Ionic bull Covalentbull metallic

20

Start with van der Waals because all pairs of atoms have vdW attraction For neutral atoms (so there is no electrostatic force) it always dominates at long distances

Ionic as a Na and a Cl atom approach each other the extra electron on Na tunnels to Cl

Covalent as 2 atoms approach each with extra electron bonding and anti-bonding orbitals are formed

Metallic long-wavelength plane waves play the role of the bonding orbitals shorter-wavelength (higher energy) waves are like anti-bonding orbitals

Lecture 12Sept 17 2010PH 481581

21

Comparison of different core potentials V=r-n

for n = 1236 and12n=1 is said to be ldquosoftrdquon=12 is ldquohardrdquo

n=infinity

22

END

• Lecture 3 ndash August 25 2010
• 3D systems(symmetries)
• Lecture 4 - Aug 27 2010
• Real crystals
• Bragg Scattering of x-rays
• Better treatment x-rays are scattered by electron density n(r)
• Periodic functions in a lattice
• Examples of reciprocal lattices
• Calculating FT of n(r)
• Calculating x-ray scattering intensity
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22

Calculating FT of n(r)

nG = Fourier transform of scattering density n(r)Only nonzero at points G of reciprocal latticeExplicitly

9

n(r) = Σ nG eiGr

rrrGG

3

cellunit

)(1

dneV

n i

where V is the volume of the unit cell

Calculating x-ray scattering intensity

10

Incoming amplitude (electric field)

Lecture 7 Sept 3 2010 PH 481581

kx-ray

r

n(r) = Σ nG eiGr

rrsquo

detector

Amplitude reaching detector =rkie Re

rposition Source

n(r)d3 r rrike

rkr

k

number of scatterers

Phase change

3 ( )

Source position

3 ( )

Source position

( )

( ) ( )

i i

i i i

e n d e

e n d e e F

k r k r r

r

k r k k r k r

r

r r

r r k k

where F is called the scattering amplitude (in Kittelrsquos book)

22 2 2 2

2 2

2( ) 2 1

rr r r

r r

r rr r r r

122

2 2 2

21 1 2nd order

rr r

r r r

r r r rr r

( )k kr kr

r rr r k r k r k r r

rrsquo - r

G

Conclusion x-ray scattering probes reciprocal lattice (RL)

If incoming wavevector is k amplitude for outgoing wavevector krsquo ~ Fourier component nkrsquo-k In a perfect crystal this is nonzero only if krsquo ndash k is a reciprocal lattice vector call it G

krsquo

k

You have to rotate the crystal powder it

Also this is elastic scattering |k|=|krsquo| so krsquo must lie on a sphere The probability that one of the vectors k+G lies exactly on the sphere is zero (the sphere has zero thickness)

Powder pattern crystallites are in all possible orientations which rotates each G over the surface of a sphereSphere of possible Grsquos intersects sphere of possible krsquo s in a circle (ring) seen on screen

Gk

krsquo

scre

en

12

Powder pattern

Pattern for the mineral olivine httpndbserverrutgersedummcifcbf

httpwwwccp14acukccpweb-mirrorsarmelegypteconf2theoryhtml

Ball-milled

Hand-crushed

d

Relating krsquo ndash k = G to the Bragg equation

13

G

krsquo

k

Bragg picture is krsquok

leading to

2 d sin = n

2 2( ) k G k

In reciprocal lattice picture krsquo = k + G so elastic scattering condition is

2 2 22 k k G G k22 0 k G k

Geometrically we have 2 2

sin sin 2 sin2

Gk

G

Consistent only if (so smallest G goes with n=1 )Gn

d 2 G

d2

Predicting angles of powder diffraction rings

14

G

krsquo

k

Recall picture of krsquo = k + G

amp from top triangle sin2

kG

so )(2

and 2

but lkha

k

G

2

sin 12222 lkh

a

5012

4002

3111

2011

1001

21222 lkhlkh

Lecture 8 PH 481581

So the Bragg planes are related to reciprocal lattice vectors and the plane spacing d is related to the RLV magnitude G = |G| by d = 2G

15

Schematically RLV G lattice planes

or more precisely a plane and a vector G each determine a direction in space There are many Gs along this direction and many planes normal to it so the relation is really between families of Grsquos and families of planesfamily 0plusmn G plusmn 2G of parallel RLVs family of parallel planesOnly some of these planes go through lattice points and these are spaced d = 2G apart(G must be the smallest length in the family)

16

0

Direct lattice

Reciprocal Lattice

Drawing low-index planes (far apart)high-index planes (close together)

Calculating scattering amplitude from atomic form factors

17

Consider a crystal with a basisr1

r2

Plot density along dashed linen(r) = n1(r-r1) + n2(r-r2)= n1() + n2(2)

rx

x 2x

r2xr1x

where = r - r1 and = r ndash r2

n1 n2

r

The scattering amplitude FG is defined by

Lecture 10 PH 481581 Sept 13 2010

crystal entire

3)( rdenF iG

rGr

But the integrand is periodic so the result is proportional to the number N of unit cells The amplitude per unit cell is called the structure factor SG = FG N

Calculating scattering amplitude (structure factor)

18

Structure factor

jj

iii

iiii

ii

ii

iG

fefefe

denedene

denden

rdenrden

rdenS

j )()()(

)()(

)()(

)()(

)(

21

32

31

3)(2

3)(1

322

311

cell

3

21

21

21

GGG

ρρ

ρρ

rrrr

r

rGrGrG

ρGrGρGrG

ρrGρrG

rGrG

rG

r1r2

r

rdenf iii

3)()( ρGρGwhere

Example CsCl simple cubic r1= = (000) r2 = (a2a2a2)Some algebra gives

odd if

)points fcc(even if)()1()(

21

2121 l khff

l khffffS lkh

G

GG

In limit f1 f2(ie CsCl FeFe bcc) this becomes fcc RL

Another example (NaCl structure)

19

KCl isoelectronic f1 = f2 lose reflections due to extra symmetry

NaCl is fcc basis is r1= = (000) r2 = (a200)

centers) (cell odd all if

points)(corner even all if)()1()(

21

2121 lkhff

lkhffffS h

G

GG

(if some even and some odd eg (100) these points are not even ON the reciprocal lattice so we donrsquot calculate SG)

Example in Fig 2-17 p 42 KBr has 111 200 220 as predicted

Chapter 3 BindingTypes of bonding

bull van der Waalsbull Ionic bull Covalentbull metallic

20

Start with van der Waals because all pairs of atoms have vdW attraction For neutral atoms (so there is no electrostatic force) it always dominates at long distances

Ionic as a Na and a Cl atom approach each other the extra electron on Na tunnels to Cl

Covalent as 2 atoms approach each with extra electron bonding and anti-bonding orbitals are formed

Metallic long-wavelength plane waves play the role of the bonding orbitals shorter-wavelength (higher energy) waves are like anti-bonding orbitals

Lecture 12Sept 17 2010PH 481581

21

Comparison of different core potentials V=r-n

for n = 1236 and12n=1 is said to be ldquosoftrdquon=12 is ldquohardrdquo

n=infinity

22

END

• Lecture 3 ndash August 25 2010
• 3D systems(symmetries)
• Lecture 4 - Aug 27 2010
• Real crystals
• Bragg Scattering of x-rays
• Better treatment x-rays are scattered by electron density n(r)
• Periodic functions in a lattice
• Examples of reciprocal lattices
• Calculating FT of n(r)
• Calculating x-ray scattering intensity
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22

Calculating x-ray scattering intensity

10

Incoming amplitude (electric field)

Lecture 7 Sept 3 2010 PH 481581

kx-ray

r

n(r) = Σ nG eiGr

rrsquo

detector

Amplitude reaching detector =rkie Re

rposition Source

n(r)d3 r rrike

rkr

k

number of scatterers

Phase change

3 ( )

Source position

3 ( )

Source position

( )

( ) ( )

i i

i i i

e n d e

e n d e e F

k r k r r

r

k r k k r k r

r

r r

r r k k

where F is called the scattering amplitude (in Kittelrsquos book)

22 2 2 2

2 2

2( ) 2 1

rr r r

r r

r rr r r r

122

2 2 2

21 1 2nd order

rr r

r r r

r r r rr r

( )k kr kr

r rr r k r k r k r r

rrsquo - r

G

Conclusion x-ray scattering probes reciprocal lattice (RL)

If incoming wavevector is k amplitude for outgoing wavevector krsquo ~ Fourier component nkrsquo-k In a perfect crystal this is nonzero only if krsquo ndash k is a reciprocal lattice vector call it G

krsquo

k

You have to rotate the crystal powder it

Also this is elastic scattering |k|=|krsquo| so krsquo must lie on a sphere The probability that one of the vectors k+G lies exactly on the sphere is zero (the sphere has zero thickness)

Powder pattern crystallites are in all possible orientations which rotates each G over the surface of a sphereSphere of possible Grsquos intersects sphere of possible krsquo s in a circle (ring) seen on screen

Gk

krsquo

scre

en

12

Powder pattern

Pattern for the mineral olivine httpndbserverrutgersedummcifcbf

httpwwwccp14acukccpweb-mirrorsarmelegypteconf2theoryhtml

Ball-milled

Hand-crushed

d

Relating krsquo ndash k = G to the Bragg equation

13

G

krsquo

k

Bragg picture is krsquok

leading to

2 d sin = n

2 2( ) k G k

In reciprocal lattice picture krsquo = k + G so elastic scattering condition is

2 2 22 k k G G k22 0 k G k

Geometrically we have 2 2

sin sin 2 sin2

Gk

G

Consistent only if (so smallest G goes with n=1 )Gn

d 2 G

d2

Predicting angles of powder diffraction rings

14

G

krsquo

k

Recall picture of krsquo = k + G

amp from top triangle sin2

kG

so )(2

and 2

but lkha

k

G

2

sin 12222 lkh

a

5012

4002

3111

2011

1001

21222 lkhlkh

Lecture 8 PH 481581

So the Bragg planes are related to reciprocal lattice vectors and the plane spacing d is related to the RLV magnitude G = |G| by d = 2G

15

Schematically RLV G lattice planes

or more precisely a plane and a vector G each determine a direction in space There are many Gs along this direction and many planes normal to it so the relation is really between families of Grsquos and families of planesfamily 0plusmn G plusmn 2G of parallel RLVs family of parallel planesOnly some of these planes go through lattice points and these are spaced d = 2G apart(G must be the smallest length in the family)

16

0

Direct lattice

Reciprocal Lattice

Drawing low-index planes (far apart)high-index planes (close together)

Calculating scattering amplitude from atomic form factors

17

Consider a crystal with a basisr1

r2

Plot density along dashed linen(r) = n1(r-r1) + n2(r-r2)= n1() + n2(2)

rx

x 2x

r2xr1x

where = r - r1 and = r ndash r2

n1 n2

r

The scattering amplitude FG is defined by

Lecture 10 PH 481581 Sept 13 2010

crystal entire

3)( rdenF iG

rGr

But the integrand is periodic so the result is proportional to the number N of unit cells The amplitude per unit cell is called the structure factor SG = FG N

Calculating scattering amplitude (structure factor)

18

Structure factor

jj

iii

iiii

ii

ii

iG

fefefe

denedene

denden

rdenrden

rdenS

j )()()(

)()(

)()(

)()(

)(

21

32

31

3)(2

3)(1

322

311

cell

3

21

21

21

GGG

ρρ

ρρ

rrrr

r

rGrGrG

ρGrGρGrG

ρrGρrG

rGrG

rG

r1r2

r

rdenf iii

3)()( ρGρGwhere

Example CsCl simple cubic r1= = (000) r2 = (a2a2a2)Some algebra gives

odd if

)points fcc(even if)()1()(

21

2121 l khff

l khffffS lkh

G

GG

In limit f1 f2(ie CsCl FeFe bcc) this becomes fcc RL

Another example (NaCl structure)

19

KCl isoelectronic f1 = f2 lose reflections due to extra symmetry

NaCl is fcc basis is r1= = (000) r2 = (a200)

centers) (cell odd all if

points)(corner even all if)()1()(

21

2121 lkhff

lkhffffS h

G

GG

(if some even and some odd eg (100) these points are not even ON the reciprocal lattice so we donrsquot calculate SG)

Example in Fig 2-17 p 42 KBr has 111 200 220 as predicted

Chapter 3 BindingTypes of bonding

bull van der Waalsbull Ionic bull Covalentbull metallic

20

Start with van der Waals because all pairs of atoms have vdW attraction For neutral atoms (so there is no electrostatic force) it always dominates at long distances

Ionic as a Na and a Cl atom approach each other the extra electron on Na tunnels to Cl

Covalent as 2 atoms approach each with extra electron bonding and anti-bonding orbitals are formed

Metallic long-wavelength plane waves play the role of the bonding orbitals shorter-wavelength (higher energy) waves are like anti-bonding orbitals

Lecture 12Sept 17 2010PH 481581

21

Comparison of different core potentials V=r-n

for n = 1236 and12n=1 is said to be ldquosoftrdquon=12 is ldquohardrdquo

n=infinity

22

END

• Lecture 3 ndash August 25 2010
• 3D systems(symmetries)
• Lecture 4 - Aug 27 2010
• Real crystals
• Bragg Scattering of x-rays
• Better treatment x-rays are scattered by electron density n(r)
• Periodic functions in a lattice
• Examples of reciprocal lattices
• Calculating FT of n(r)
• Calculating x-ray scattering intensity
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22

G

Conclusion x-ray scattering probes reciprocal lattice (RL)

If incoming wavevector is k amplitude for outgoing wavevector krsquo ~ Fourier component nkrsquo-k In a perfect crystal this is nonzero only if krsquo ndash k is a reciprocal lattice vector call it G

krsquo

k

You have to rotate the crystal powder it

Also this is elastic scattering |k|=|krsquo| so krsquo must lie on a sphere The probability that one of the vectors k+G lies exactly on the sphere is zero (the sphere has zero thickness)

Powder pattern crystallites are in all possible orientations which rotates each G over the surface of a sphereSphere of possible Grsquos intersects sphere of possible krsquo s in a circle (ring) seen on screen

Gk

krsquo

scre

en

12

Powder pattern

Pattern for the mineral olivine httpndbserverrutgersedummcifcbf

httpwwwccp14acukccpweb-mirrorsarmelegypteconf2theoryhtml

Ball-milled

Hand-crushed

d

Relating krsquo ndash k = G to the Bragg equation

13

G

krsquo

k

Bragg picture is krsquok

leading to

2 d sin = n

2 2( ) k G k

In reciprocal lattice picture krsquo = k + G so elastic scattering condition is

2 2 22 k k G G k22 0 k G k

Geometrically we have 2 2

sin sin 2 sin2

Gk

G

Consistent only if (so smallest G goes with n=1 )Gn

d 2 G

d2

Predicting angles of powder diffraction rings

14

G

krsquo

k

Recall picture of krsquo = k + G

amp from top triangle sin2

kG

so )(2

and 2

but lkha

k

G

2

sin 12222 lkh

a

5012

4002

3111

2011

1001

21222 lkhlkh

Lecture 8 PH 481581

So the Bragg planes are related to reciprocal lattice vectors and the plane spacing d is related to the RLV magnitude G = |G| by d = 2G

15

Schematically RLV G lattice planes

or more precisely a plane and a vector G each determine a direction in space There are many Gs along this direction and many planes normal to it so the relation is really between families of Grsquos and families of planesfamily 0plusmn G plusmn 2G of parallel RLVs family of parallel planesOnly some of these planes go through lattice points and these are spaced d = 2G apart(G must be the smallest length in the family)

16

0

Direct lattice

Reciprocal Lattice

Drawing low-index planes (far apart)high-index planes (close together)

Calculating scattering amplitude from atomic form factors

17

Consider a crystal with a basisr1

r2

Plot density along dashed linen(r) = n1(r-r1) + n2(r-r2)= n1() + n2(2)

rx

x 2x

r2xr1x

where = r - r1 and = r ndash r2

n1 n2

r

The scattering amplitude FG is defined by

Lecture 10 PH 481581 Sept 13 2010

crystal entire

3)( rdenF iG

rGr

But the integrand is periodic so the result is proportional to the number N of unit cells The amplitude per unit cell is called the structure factor SG = FG N

Calculating scattering amplitude (structure factor)

18

Structure factor

jj

iii

iiii

ii

ii

iG

fefefe

denedene

denden

rdenrden

rdenS

j )()()(

)()(

)()(

)()(

)(

21

32

31

3)(2

3)(1

322

311

cell

3

21

21

21

GGG

ρρ

ρρ

rrrr

r

rGrGrG

ρGrGρGrG

ρrGρrG

rGrG

rG

r1r2

r

rdenf iii

3)()( ρGρGwhere

Example CsCl simple cubic r1= = (000) r2 = (a2a2a2)Some algebra gives

odd if

)points fcc(even if)()1()(

21

2121 l khff

l khffffS lkh

G

GG

In limit f1 f2(ie CsCl FeFe bcc) this becomes fcc RL

Another example (NaCl structure)

19

KCl isoelectronic f1 = f2 lose reflections due to extra symmetry

NaCl is fcc basis is r1= = (000) r2 = (a200)

centers) (cell odd all if

points)(corner even all if)()1()(

21

2121 lkhff

lkhffffS h

G

GG

(if some even and some odd eg (100) these points are not even ON the reciprocal lattice so we donrsquot calculate SG)

Example in Fig 2-17 p 42 KBr has 111 200 220 as predicted

Chapter 3 BindingTypes of bonding

bull van der Waalsbull Ionic bull Covalentbull metallic

20

Start with van der Waals because all pairs of atoms have vdW attraction For neutral atoms (so there is no electrostatic force) it always dominates at long distances

Ionic as a Na and a Cl atom approach each other the extra electron on Na tunnels to Cl

Covalent as 2 atoms approach each with extra electron bonding and anti-bonding orbitals are formed

Metallic long-wavelength plane waves play the role of the bonding orbitals shorter-wavelength (higher energy) waves are like anti-bonding orbitals

Lecture 12Sept 17 2010PH 481581

21

Comparison of different core potentials V=r-n

for n = 1236 and12n=1 is said to be ldquosoftrdquon=12 is ldquohardrdquo

n=infinity

22

END

• Lecture 3 ndash August 25 2010
• 3D systems(symmetries)
• Lecture 4 - Aug 27 2010
• Real crystals
• Bragg Scattering of x-rays
• Better treatment x-rays are scattered by electron density n(r)
• Periodic functions in a lattice
• Examples of reciprocal lattices
• Calculating FT of n(r)
• Calculating x-ray scattering intensity
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22

12

Powder pattern

Pattern for the mineral olivine httpndbserverrutgersedummcifcbf

httpwwwccp14acukccpweb-mirrorsarmelegypteconf2theoryhtml

Ball-milled

Hand-crushed

d

Relating krsquo ndash k = G to the Bragg equation

13

G

krsquo

k

Bragg picture is krsquok

leading to

2 d sin = n

2 2( ) k G k

In reciprocal lattice picture krsquo = k + G so elastic scattering condition is

2 2 22 k k G G k22 0 k G k

Geometrically we have 2 2

sin sin 2 sin2

Gk

G

Consistent only if (so smallest G goes with n=1 )Gn

d 2 G

d2

Predicting angles of powder diffraction rings

14

G

krsquo

k

Recall picture of krsquo = k + G

amp from top triangle sin2

kG

so )(2

and 2

but lkha

k

G

2

sin 12222 lkh

a

5012

4002

3111

2011

1001

21222 lkhlkh

Lecture 8 PH 481581

So the Bragg planes are related to reciprocal lattice vectors and the plane spacing d is related to the RLV magnitude G = |G| by d = 2G

15

Schematically RLV G lattice planes

or more precisely a plane and a vector G each determine a direction in space There are many Gs along this direction and many planes normal to it so the relation is really between families of Grsquos and families of planesfamily 0plusmn G plusmn 2G of parallel RLVs family of parallel planesOnly some of these planes go through lattice points and these are spaced d = 2G apart(G must be the smallest length in the family)

16

0

Direct lattice

Reciprocal Lattice

Drawing low-index planes (far apart)high-index planes (close together)

Calculating scattering amplitude from atomic form factors

17

Consider a crystal with a basisr1

r2

Plot density along dashed linen(r) = n1(r-r1) + n2(r-r2)= n1() + n2(2)

rx

x 2x

r2xr1x

where = r - r1 and = r ndash r2

n1 n2

r

The scattering amplitude FG is defined by

Lecture 10 PH 481581 Sept 13 2010

crystal entire

3)( rdenF iG

rGr

But the integrand is periodic so the result is proportional to the number N of unit cells The amplitude per unit cell is called the structure factor SG = FG N

Calculating scattering amplitude (structure factor)

18

Structure factor

jj

iii

iiii

ii

ii

iG

fefefe

denedene

denden

rdenrden

rdenS

j )()()(

)()(

)()(

)()(

)(

21

32

31

3)(2

3)(1

322

311

cell

3

21

21

21

GGG

ρρ

ρρ

rrrr

r

rGrGrG

ρGrGρGrG

ρrGρrG

rGrG

rG

r1r2

r

rdenf iii

3)()( ρGρGwhere

Example CsCl simple cubic r1= = (000) r2 = (a2a2a2)Some algebra gives

odd if

)points fcc(even if)()1()(

21

2121 l khff

l khffffS lkh

G

GG

In limit f1 f2(ie CsCl FeFe bcc) this becomes fcc RL

Another example (NaCl structure)

19

KCl isoelectronic f1 = f2 lose reflections due to extra symmetry

NaCl is fcc basis is r1= = (000) r2 = (a200)

centers) (cell odd all if

points)(corner even all if)()1()(

21

2121 lkhff

lkhffffS h

G

GG

(if some even and some odd eg (100) these points are not even ON the reciprocal lattice so we donrsquot calculate SG)

Example in Fig 2-17 p 42 KBr has 111 200 220 as predicted

Chapter 3 BindingTypes of bonding

bull van der Waalsbull Ionic bull Covalentbull metallic

20

Start with van der Waals because all pairs of atoms have vdW attraction For neutral atoms (so there is no electrostatic force) it always dominates at long distances

Ionic as a Na and a Cl atom approach each other the extra electron on Na tunnels to Cl

Covalent as 2 atoms approach each with extra electron bonding and anti-bonding orbitals are formed

Metallic long-wavelength plane waves play the role of the bonding orbitals shorter-wavelength (higher energy) waves are like anti-bonding orbitals

Lecture 12Sept 17 2010PH 481581

21

Comparison of different core potentials V=r-n

for n = 1236 and12n=1 is said to be ldquosoftrdquon=12 is ldquohardrdquo

n=infinity

22

END

• Lecture 3 ndash August 25 2010
• 3D systems(symmetries)
• Lecture 4 - Aug 27 2010
• Real crystals
• Bragg Scattering of x-rays
• Better treatment x-rays are scattered by electron density n(r)
• Periodic functions in a lattice
• Examples of reciprocal lattices
• Calculating FT of n(r)
• Calculating x-ray scattering intensity
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22

d

Relating krsquo ndash k = G to the Bragg equation

13

G

krsquo

k

Bragg picture is krsquok

leading to

2 d sin = n

2 2( ) k G k

In reciprocal lattice picture krsquo = k + G so elastic scattering condition is

2 2 22 k k G G k22 0 k G k

Geometrically we have 2 2

sin sin 2 sin2

Gk

G

Consistent only if (so smallest G goes with n=1 )Gn

d 2 G

d2

Predicting angles of powder diffraction rings

14

G

krsquo

k

Recall picture of krsquo = k + G

amp from top triangle sin2

kG

so )(2

and 2

but lkha

k

G

2

sin 12222 lkh

a

5012

4002

3111

2011

1001

21222 lkhlkh

Lecture 8 PH 481581

So the Bragg planes are related to reciprocal lattice vectors and the plane spacing d is related to the RLV magnitude G = |G| by d = 2G

15

Schematically RLV G lattice planes

or more precisely a plane and a vector G each determine a direction in space There are many Gs along this direction and many planes normal to it so the relation is really between families of Grsquos and families of planesfamily 0plusmn G plusmn 2G of parallel RLVs family of parallel planesOnly some of these planes go through lattice points and these are spaced d = 2G apart(G must be the smallest length in the family)

16

0

Direct lattice

Reciprocal Lattice

Drawing low-index planes (far apart)high-index planes (close together)

Calculating scattering amplitude from atomic form factors

17

Consider a crystal with a basisr1

r2

Plot density along dashed linen(r) = n1(r-r1) + n2(r-r2)= n1() + n2(2)

rx

x 2x

r2xr1x

where = r - r1 and = r ndash r2

n1 n2

r

The scattering amplitude FG is defined by

Lecture 10 PH 481581 Sept 13 2010

crystal entire

3)( rdenF iG

rGr

But the integrand is periodic so the result is proportional to the number N of unit cells The amplitude per unit cell is called the structure factor SG = FG N

Calculating scattering amplitude (structure factor)

18

Structure factor

jj

iii

iiii

ii

ii

iG

fefefe

denedene

denden

rdenrden

rdenS

j )()()(

)()(

)()(

)()(

)(

21

32

31

3)(2

3)(1

322

311

cell

3

21

21

21

GGG

ρρ

ρρ

rrrr

r

rGrGrG

ρGrGρGrG

ρrGρrG

rGrG

rG

r1r2

r

rdenf iii

3)()( ρGρGwhere

Example CsCl simple cubic r1= = (000) r2 = (a2a2a2)Some algebra gives

odd if

)points fcc(even if)()1()(

21

2121 l khff

l khffffS lkh

G

GG

In limit f1 f2(ie CsCl FeFe bcc) this becomes fcc RL

Another example (NaCl structure)

19

KCl isoelectronic f1 = f2 lose reflections due to extra symmetry

NaCl is fcc basis is r1= = (000) r2 = (a200)

centers) (cell odd all if

points)(corner even all if)()1()(

21

2121 lkhff

lkhffffS h

G

GG

(if some even and some odd eg (100) these points are not even ON the reciprocal lattice so we donrsquot calculate SG)

Example in Fig 2-17 p 42 KBr has 111 200 220 as predicted

Chapter 3 BindingTypes of bonding

bull van der Waalsbull Ionic bull Covalentbull metallic

20

Start with van der Waals because all pairs of atoms have vdW attraction For neutral atoms (so there is no electrostatic force) it always dominates at long distances

Ionic as a Na and a Cl atom approach each other the extra electron on Na tunnels to Cl

Covalent as 2 atoms approach each with extra electron bonding and anti-bonding orbitals are formed

Metallic long-wavelength plane waves play the role of the bonding orbitals shorter-wavelength (higher energy) waves are like anti-bonding orbitals

Lecture 12Sept 17 2010PH 481581

21

Comparison of different core potentials V=r-n

for n = 1236 and12n=1 is said to be ldquosoftrdquon=12 is ldquohardrdquo

n=infinity

22

END

• Lecture 3 ndash August 25 2010
• 3D systems(symmetries)
• Lecture 4 - Aug 27 2010
• Real crystals
• Bragg Scattering of x-rays
• Better treatment x-rays are scattered by electron density n(r)
• Periodic functions in a lattice
• Examples of reciprocal lattices
• Calculating FT of n(r)
• Calculating x-ray scattering intensity
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22

Predicting angles of powder diffraction rings

14

G

krsquo

k

Recall picture of krsquo = k + G

amp from top triangle sin2

kG

so )(2

and 2

but lkha

k

G

2

sin 12222 lkh

a

5012

4002

3111

2011

1001

21222 lkhlkh

Lecture 8 PH 481581

So the Bragg planes are related to reciprocal lattice vectors and the plane spacing d is related to the RLV magnitude G = |G| by d = 2G

15

Schematically RLV G lattice planes

or more precisely a plane and a vector G each determine a direction in space There are many Gs along this direction and many planes normal to it so the relation is really between families of Grsquos and families of planesfamily 0plusmn G plusmn 2G of parallel RLVs family of parallel planesOnly some of these planes go through lattice points and these are spaced d = 2G apart(G must be the smallest length in the family)

16

0

Direct lattice

Reciprocal Lattice

Drawing low-index planes (far apart)high-index planes (close together)

Calculating scattering amplitude from atomic form factors

17

Consider a crystal with a basisr1

r2

Plot density along dashed linen(r) = n1(r-r1) + n2(r-r2)= n1() + n2(2)

rx

x 2x

r2xr1x

where = r - r1 and = r ndash r2

n1 n2

r

The scattering amplitude FG is defined by

Lecture 10 PH 481581 Sept 13 2010

crystal entire

3)( rdenF iG

rGr

But the integrand is periodic so the result is proportional to the number N of unit cells The amplitude per unit cell is called the structure factor SG = FG N

Calculating scattering amplitude (structure factor)

18

Structure factor

jj

iii

iiii

ii

ii

iG

fefefe

denedene

denden

rdenrden

rdenS

j )()()(

)()(

)()(

)()(

)(

21

32

31

3)(2

3)(1

322

311

cell

3

21

21

21

GGG

ρρ

ρρ

rrrr

r

rGrGrG

ρGrGρGrG

ρrGρrG

rGrG

rG

r1r2

r

rdenf iii

3)()( ρGρGwhere

Example CsCl simple cubic r1= = (000) r2 = (a2a2a2)Some algebra gives

odd if

)points fcc(even if)()1()(

21

2121 l khff

l khffffS lkh

G

GG

In limit f1 f2(ie CsCl FeFe bcc) this becomes fcc RL

Another example (NaCl structure)

19

KCl isoelectronic f1 = f2 lose reflections due to extra symmetry

NaCl is fcc basis is r1= = (000) r2 = (a200)

centers) (cell odd all if

points)(corner even all if)()1()(

21

2121 lkhff

lkhffffS h

G

GG

(if some even and some odd eg (100) these points are not even ON the reciprocal lattice so we donrsquot calculate SG)

Example in Fig 2-17 p 42 KBr has 111 200 220 as predicted

Chapter 3 BindingTypes of bonding

bull van der Waalsbull Ionic bull Covalentbull metallic

20

Start with van der Waals because all pairs of atoms have vdW attraction For neutral atoms (so there is no electrostatic force) it always dominates at long distances

Ionic as a Na and a Cl atom approach each other the extra electron on Na tunnels to Cl

Covalent as 2 atoms approach each with extra electron bonding and anti-bonding orbitals are formed

Metallic long-wavelength plane waves play the role of the bonding orbitals shorter-wavelength (higher energy) waves are like anti-bonding orbitals

Lecture 12Sept 17 2010PH 481581

21

Comparison of different core potentials V=r-n

for n = 1236 and12n=1 is said to be ldquosoftrdquon=12 is ldquohardrdquo

n=infinity

22

END

• Lecture 3 ndash August 25 2010
• 3D systems(symmetries)
• Lecture 4 - Aug 27 2010
• Real crystals
• Bragg Scattering of x-rays
• Better treatment x-rays are scattered by electron density n(r)
• Periodic functions in a lattice
• Examples of reciprocal lattices
• Calculating FT of n(r)
• Calculating x-ray scattering intensity
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22

Lecture 8 PH 481581

So the Bragg planes are related to reciprocal lattice vectors and the plane spacing d is related to the RLV magnitude G = |G| by d = 2G

15

Schematically RLV G lattice planes

or more precisely a plane and a vector G each determine a direction in space There are many Gs along this direction and many planes normal to it so the relation is really between families of Grsquos and families of planesfamily 0plusmn G plusmn 2G of parallel RLVs family of parallel planesOnly some of these planes go through lattice points and these are spaced d = 2G apart(G must be the smallest length in the family)

16

0

Direct lattice

Reciprocal Lattice

Drawing low-index planes (far apart)high-index planes (close together)

Calculating scattering amplitude from atomic form factors

17

Consider a crystal with a basisr1

r2

Plot density along dashed linen(r) = n1(r-r1) + n2(r-r2)= n1() + n2(2)

rx

x 2x

r2xr1x

where = r - r1 and = r ndash r2

n1 n2

r

The scattering amplitude FG is defined by

Lecture 10 PH 481581 Sept 13 2010

crystal entire

3)( rdenF iG

rGr

But the integrand is periodic so the result is proportional to the number N of unit cells The amplitude per unit cell is called the structure factor SG = FG N

Calculating scattering amplitude (structure factor)

18

Structure factor

jj

iii

iiii

ii

ii

iG

fefefe

denedene

denden

rdenrden

rdenS

j )()()(

)()(

)()(

)()(

)(

21

32

31

3)(2

3)(1

322

311

cell

3

21

21

21

GGG

ρρ

ρρ

rrrr

r

rGrGrG

ρGrGρGrG

ρrGρrG

rGrG

rG

r1r2

r

rdenf iii

3)()( ρGρGwhere

Example CsCl simple cubic r1= = (000) r2 = (a2a2a2)Some algebra gives

odd if

)points fcc(even if)()1()(

21

2121 l khff

l khffffS lkh

G

GG

In limit f1 f2(ie CsCl FeFe bcc) this becomes fcc RL

Another example (NaCl structure)

19

KCl isoelectronic f1 = f2 lose reflections due to extra symmetry

NaCl is fcc basis is r1= = (000) r2 = (a200)

centers) (cell odd all if

points)(corner even all if)()1()(

21

2121 lkhff

lkhffffS h

G

GG

(if some even and some odd eg (100) these points are not even ON the reciprocal lattice so we donrsquot calculate SG)

Example in Fig 2-17 p 42 KBr has 111 200 220 as predicted

Chapter 3 BindingTypes of bonding

bull van der Waalsbull Ionic bull Covalentbull metallic

20

Start with van der Waals because all pairs of atoms have vdW attraction For neutral atoms (so there is no electrostatic force) it always dominates at long distances

Ionic as a Na and a Cl atom approach each other the extra electron on Na tunnels to Cl

Covalent as 2 atoms approach each with extra electron bonding and anti-bonding orbitals are formed

Metallic long-wavelength plane waves play the role of the bonding orbitals shorter-wavelength (higher energy) waves are like anti-bonding orbitals

Lecture 12Sept 17 2010PH 481581

21

Comparison of different core potentials V=r-n

for n = 1236 and12n=1 is said to be ldquosoftrdquon=12 is ldquohardrdquo

n=infinity

22

END

• Lecture 3 ndash August 25 2010
• 3D systems(symmetries)
• Lecture 4 - Aug 27 2010
• Real crystals
• Bragg Scattering of x-rays
• Better treatment x-rays are scattered by electron density n(r)
• Periodic functions in a lattice
• Examples of reciprocal lattices
• Calculating FT of n(r)
• Calculating x-ray scattering intensity
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22

16

0

Direct lattice

Reciprocal Lattice

Drawing low-index planes (far apart)high-index planes (close together)

Calculating scattering amplitude from atomic form factors

17

Consider a crystal with a basisr1

r2

Plot density along dashed linen(r) = n1(r-r1) + n2(r-r2)= n1() + n2(2)

rx

x 2x

r2xr1x

where = r - r1 and = r ndash r2

n1 n2

r

The scattering amplitude FG is defined by

Lecture 10 PH 481581 Sept 13 2010

crystal entire

3)( rdenF iG

rGr

But the integrand is periodic so the result is proportional to the number N of unit cells The amplitude per unit cell is called the structure factor SG = FG N

Calculating scattering amplitude (structure factor)

18

Structure factor

jj

iii

iiii

ii

ii

iG

fefefe

denedene

denden

rdenrden

rdenS

j )()()(

)()(

)()(

)()(

)(

21

32

31

3)(2

3)(1

322

311

cell

3

21

21

21

GGG

ρρ

ρρ

rrrr

r

rGrGrG

ρGrGρGrG

ρrGρrG

rGrG

rG

r1r2

r

rdenf iii

3)()( ρGρGwhere

Example CsCl simple cubic r1= = (000) r2 = (a2a2a2)Some algebra gives

odd if

)points fcc(even if)()1()(

21

2121 l khff

l khffffS lkh

G

GG

In limit f1 f2(ie CsCl FeFe bcc) this becomes fcc RL

Another example (NaCl structure)

19

KCl isoelectronic f1 = f2 lose reflections due to extra symmetry

NaCl is fcc basis is r1= = (000) r2 = (a200)

centers) (cell odd all if

points)(corner even all if)()1()(

21

2121 lkhff

lkhffffS h

G

GG

(if some even and some odd eg (100) these points are not even ON the reciprocal lattice so we donrsquot calculate SG)

Example in Fig 2-17 p 42 KBr has 111 200 220 as predicted

Chapter 3 BindingTypes of bonding

bull van der Waalsbull Ionic bull Covalentbull metallic

20

Start with van der Waals because all pairs of atoms have vdW attraction For neutral atoms (so there is no electrostatic force) it always dominates at long distances

Ionic as a Na and a Cl atom approach each other the extra electron on Na tunnels to Cl

Covalent as 2 atoms approach each with extra electron bonding and anti-bonding orbitals are formed

Metallic long-wavelength plane waves play the role of the bonding orbitals shorter-wavelength (higher energy) waves are like anti-bonding orbitals

Lecture 12Sept 17 2010PH 481581

21

Comparison of different core potentials V=r-n

for n = 1236 and12n=1 is said to be ldquosoftrdquon=12 is ldquohardrdquo

n=infinity

22

END

• Lecture 3 ndash August 25 2010
• 3D systems(symmetries)
• Lecture 4 - Aug 27 2010
• Real crystals
• Bragg Scattering of x-rays
• Better treatment x-rays are scattered by electron density n(r)
• Periodic functions in a lattice
• Examples of reciprocal lattices
• Calculating FT of n(r)
• Calculating x-ray scattering intensity
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22

Calculating scattering amplitude from atomic form factors

17

Consider a crystal with a basisr1

r2

Plot density along dashed linen(r) = n1(r-r1) + n2(r-r2)= n1() + n2(2)

rx

x 2x

r2xr1x

where = r - r1 and = r ndash r2

n1 n2

r

The scattering amplitude FG is defined by

Lecture 10 PH 481581 Sept 13 2010

crystal entire

3)( rdenF iG

rGr

But the integrand is periodic so the result is proportional to the number N of unit cells The amplitude per unit cell is called the structure factor SG = FG N

Calculating scattering amplitude (structure factor)

18

Structure factor

jj

iii

iiii

ii

ii

iG

fefefe

denedene

denden

rdenrden

rdenS

j )()()(

)()(

)()(

)()(

)(

21

32

31

3)(2

3)(1

322

311

cell

3

21

21

21

GGG

ρρ

ρρ

rrrr

r

rGrGrG

ρGrGρGrG

ρrGρrG

rGrG

rG

r1r2

r

rdenf iii

3)()( ρGρGwhere

Example CsCl simple cubic r1= = (000) r2 = (a2a2a2)Some algebra gives

odd if

)points fcc(even if)()1()(

21

2121 l khff

l khffffS lkh

G

GG

In limit f1 f2(ie CsCl FeFe bcc) this becomes fcc RL

Another example (NaCl structure)

19

KCl isoelectronic f1 = f2 lose reflections due to extra symmetry

NaCl is fcc basis is r1= = (000) r2 = (a200)

centers) (cell odd all if

points)(corner even all if)()1()(

21

2121 lkhff

lkhffffS h

G

GG

(if some even and some odd eg (100) these points are not even ON the reciprocal lattice so we donrsquot calculate SG)

Example in Fig 2-17 p 42 KBr has 111 200 220 as predicted

Chapter 3 BindingTypes of bonding

bull van der Waalsbull Ionic bull Covalentbull metallic

20

Start with van der Waals because all pairs of atoms have vdW attraction For neutral atoms (so there is no electrostatic force) it always dominates at long distances

Ionic as a Na and a Cl atom approach each other the extra electron on Na tunnels to Cl

Covalent as 2 atoms approach each with extra electron bonding and anti-bonding orbitals are formed

Metallic long-wavelength plane waves play the role of the bonding orbitals shorter-wavelength (higher energy) waves are like anti-bonding orbitals

Lecture 12Sept 17 2010PH 481581

21

Comparison of different core potentials V=r-n

for n = 1236 and12n=1 is said to be ldquosoftrdquon=12 is ldquohardrdquo

n=infinity

22

END

• Lecture 3 ndash August 25 2010
• 3D systems(symmetries)
• Lecture 4 - Aug 27 2010
• Real crystals
• Bragg Scattering of x-rays
• Better treatment x-rays are scattered by electron density n(r)
• Periodic functions in a lattice
• Examples of reciprocal lattices
• Calculating FT of n(r)
• Calculating x-ray scattering intensity
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22

Calculating scattering amplitude (structure factor)

18

Structure factor

jj

iii

iiii

ii

ii

iG

fefefe

denedene

denden

rdenrden

rdenS

j )()()(

)()(

)()(

)()(

)(

21

32

31

3)(2

3)(1

322

311

cell

3

21

21

21

GGG

ρρ

ρρ

rrrr

r

rGrGrG

ρGrGρGrG

ρrGρrG

rGrG

rG

r1r2

r

rdenf iii

3)()( ρGρGwhere

Example CsCl simple cubic r1= = (000) r2 = (a2a2a2)Some algebra gives

odd if

)points fcc(even if)()1()(

21

2121 l khff

l khffffS lkh

G

GG

In limit f1 f2(ie CsCl FeFe bcc) this becomes fcc RL

Another example (NaCl structure)

19

KCl isoelectronic f1 = f2 lose reflections due to extra symmetry

NaCl is fcc basis is r1= = (000) r2 = (a200)

centers) (cell odd all if

points)(corner even all if)()1()(

21

2121 lkhff

lkhffffS h

G

GG

(if some even and some odd eg (100) these points are not even ON the reciprocal lattice so we donrsquot calculate SG)

Example in Fig 2-17 p 42 KBr has 111 200 220 as predicted

Chapter 3 BindingTypes of bonding

bull van der Waalsbull Ionic bull Covalentbull metallic

20

Start with van der Waals because all pairs of atoms have vdW attraction For neutral atoms (so there is no electrostatic force) it always dominates at long distances

Ionic as a Na and a Cl atom approach each other the extra electron on Na tunnels to Cl

Covalent as 2 atoms approach each with extra electron bonding and anti-bonding orbitals are formed

Metallic long-wavelength plane waves play the role of the bonding orbitals shorter-wavelength (higher energy) waves are like anti-bonding orbitals

Lecture 12Sept 17 2010PH 481581

21

Comparison of different core potentials V=r-n

for n = 1236 and12n=1 is said to be ldquosoftrdquon=12 is ldquohardrdquo

n=infinity

22

END

• Lecture 3 ndash August 25 2010
• 3D systems(symmetries)
• Lecture 4 - Aug 27 2010
• Real crystals
• Bragg Scattering of x-rays
• Better treatment x-rays are scattered by electron density n(r)
• Periodic functions in a lattice
• Examples of reciprocal lattices
• Calculating FT of n(r)
• Calculating x-ray scattering intensity
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22

Another example (NaCl structure)

19

KCl isoelectronic f1 = f2 lose reflections due to extra symmetry

NaCl is fcc basis is r1= = (000) r2 = (a200)

centers) (cell odd all if

points)(corner even all if)()1()(

21

2121 lkhff

lkhffffS h

G

GG

(if some even and some odd eg (100) these points are not even ON the reciprocal lattice so we donrsquot calculate SG)

Example in Fig 2-17 p 42 KBr has 111 200 220 as predicted

Chapter 3 BindingTypes of bonding

bull van der Waalsbull Ionic bull Covalentbull metallic

20

Start with van der Waals because all pairs of atoms have vdW attraction For neutral atoms (so there is no electrostatic force) it always dominates at long distances

Ionic as a Na and a Cl atom approach each other the extra electron on Na tunnels to Cl

Covalent as 2 atoms approach each with extra electron bonding and anti-bonding orbitals are formed

Metallic long-wavelength plane waves play the role of the bonding orbitals shorter-wavelength (higher energy) waves are like anti-bonding orbitals

Lecture 12Sept 17 2010PH 481581

21

Comparison of different core potentials V=r-n

for n = 1236 and12n=1 is said to be ldquosoftrdquon=12 is ldquohardrdquo

n=infinity

22

END

• Lecture 3 ndash August 25 2010
• 3D systems(symmetries)
• Lecture 4 - Aug 27 2010
• Real crystals
• Bragg Scattering of x-rays
• Better treatment x-rays are scattered by electron density n(r)
• Periodic functions in a lattice
• Examples of reciprocal lattices
• Calculating FT of n(r)
• Calculating x-ray scattering intensity
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22

Chapter 3 BindingTypes of bonding

bull van der Waalsbull Ionic bull Covalentbull metallic

20

Start with van der Waals because all pairs of atoms have vdW attraction For neutral atoms (so there is no electrostatic force) it always dominates at long distances

Ionic as a Na and a Cl atom approach each other the extra electron on Na tunnels to Cl

Covalent as 2 atoms approach each with extra electron bonding and anti-bonding orbitals are formed

Metallic long-wavelength plane waves play the role of the bonding orbitals shorter-wavelength (higher energy) waves are like anti-bonding orbitals

Lecture 12Sept 17 2010PH 481581

21

Comparison of different core potentials V=r-n

for n = 1236 and12n=1 is said to be ldquosoftrdquon=12 is ldquohardrdquo

n=infinity

22

END

• Lecture 3 ndash August 25 2010
• 3D systems(symmetries)
• Lecture 4 - Aug 27 2010
• Real crystals
• Bragg Scattering of x-rays
• Better treatment x-rays are scattered by electron density n(r)
• Periodic functions in a lattice
• Examples of reciprocal lattices
• Calculating FT of n(r)
• Calculating x-ray scattering intensity
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22

21

Comparison of different core potentials V=r-n

for n = 1236 and12n=1 is said to be ldquosoftrdquon=12 is ldquohardrdquo

n=infinity

22

END

• Lecture 3 ndash August 25 2010
• 3D systems(symmetries)
• Lecture 4 - Aug 27 2010
• Real crystals
• Bragg Scattering of x-rays
• Better treatment x-rays are scattered by electron density n(r)
• Periodic functions in a lattice
• Examples of reciprocal lattices
• Calculating FT of n(r)
• Calculating x-ray scattering intensity
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22

22

END

• Lecture 3 ndash August 25 2010
• 3D systems(symmetries)
• Lecture 4 - Aug 27 2010
• Real crystals
• Bragg Scattering of x-rays
• Better treatment x-rays are scattered by electron density n(r)
• Periodic functions in a lattice
• Examples of reciprocal lattices
• Calculating FT of n(r)
• Calculating x-ray scattering intensity
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22