lecture 3 – august 25, 2010 review crystal = lattice + basis there are 5 non-equivalent 2d...
TRANSCRIPT
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
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
-
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
-
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
-