Lecture 3 – August 25, 2010
Review
• Crystal = lattice + basis
• there are 5 non-equivalent 2D lattices: 5 Bravais lattices
• divided into 4 systems (symmetries; “point groups”)
– oblique, square, rectangular (2), and hexagonal.
Actual formation (self-assembly) of these lattices depends
on details –
boundary conditions (marble demo)
1
Key to understanding relationships:
start with cube, break symmetries
3D systems(symmetries)
System
(point group)
Number of
Lattices
Lattice Symbol
Restriction on
crystal cell angle
Cubic
48=3!x23 elements
3
P or sc, I or bcc,F
or fcc
a=b=c
α =β =γ=90°
Tetragonal
2
P, I
a=b≠c
α=β =γ=90°
Orthorhombic
4
P, C, I, F
a≠b≠ c
α=β =γ=90°
Monoclinic
2
F, C
a≠b≠ c
α=β=90 °≠β
Triclinic
1
P
a≠b≠ c
α≠β≠γ
Trigonal
1
R
a=b=c
α=β =γ <120°
,≠90°
Hexagonal
1
P
a=b≠c
α =β =90°
γ=120°
Table 1. Seven crystal systems make up fourteen Bravais lattice types in three dimensions.
P - Primitive: simple unit cell
F - Face-centred: additional point in the centre of each face
I - Body-centred: additional point in the centre of the cell
from http://britneyspears.ac;
2
C - Centred: additional point in the centre of each end
similar to Christman handout
R - Rhombohedral: Hexagonal class only
Lecture 4 - Aug 27, 2010
Review
• Crystal = lattice + basis
• there are 5 non-equivalent (Bravais) lattices in 2D, 14 in 3D
• divided into 4 systems (symmetries; “point groups”), 7 in 3D
– Cubic, tetragonal, orthorhombic, monoclinic, triclinic; hex.,trigonal
Close-packed crystals
Stacking of these planes:
ABABAB... = hexagonal close packed (hcp)
C
C
B
B
A
ABCABC... = face centered cubic (fcc)
A
3
We will skip Miller indices here (they are easier to
understand after we discuss the reciprocal lattice), and go
on to:
Real crystals
Simple cubic w/o basis: none. Unstable.
Uncharged atoms prefer close-packed structures, many near
neighbors (12 n.n. in hcp or fcc).
Ions: NaCl structure (fcc with 2-ion basis, 6 n.n.)
CsCl structure (sc with 2-ion basis) (“2-ion unit cell”, 8 n.n.)
Covalently bonded structures: Diamond.
fcc, 2 atoms/primitive unit cell = 8 atoms/conventional unit
cell. Basis at (000) and (1/4,1/4,/1/4)
(fcc translations (1/2,1/2,0), (1,0,0).
4
PH 481/581 Lecture 6, Sept. 1, 2010
Bragg Scattering of x-rays
Bragg’s parallel plane picture
is mnemonic, not derivation.
Extra path length must
be integer number of
wavelengths:
2 d sin q = n l
q
Figure Courtesy Wikipedia
5
Better treatment: x-rays are
scattered by electron density n(r)
x-ray
r
k
k'
In 1D, n(x) = n(x+a) (periodic)
“lattice” means:
Expand in Fourier series
will be real if n = n
closed under
n(x) = Σ nG eiGx n(x)
(complex conjugates)
addition
where G=integer * 2p/a
These G’s form a “reciprocal lattice” – is also
{G: eiGx is periodic in the direct lattice}
where “direct lattice” is ..., -2a, -a, 0, a, 2a, ...
We want to generalize this to 3D.
-G
*
G
6
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 + u3a3.
Any periodic function n(r) can be expressed as a Fourier
series n(r) = Σ nG eiG·r where each eiG·r is periodic in the
lattice:
eiG·(r+R) = eiG·r ◄▬► eiG·R = 1 ◄▬► G·R = 2p*integer.
The set of such G’s is closed under addition – a lattice,
called the “reciprocal lattice” of the “direct lattice”
determined by a1, a2, and a3.
You can show that the vectors b1, b2, and b3 determined by
ai · bj = 2p dij are primitive translation vectors of this
reciprocal lattice (RL). [Start by showing b1 is in RL: b1·R = 2p*integer]
This means b1 must be perpendicular to a2 and a3.
7
Explicit formulas are: b  2pa 2  a3 , b  2pa3  a1 , b  2pa1  a 2
1
a1  a 2  a 3 
2
a 2  a 3  a1 
3
a 3  a1  a 2 
Examples of reciprocal lattices
Orthorhombic direct lattice:
a1  a1xˆ , a 2  a2 yˆ , a3  a3zˆ .
Easy to check that
Reciprocal lattice is
ai · bj = 2p dij
2p
2p
2p
b1 
xˆ , b 2 
yˆ , b 3 
zˆ
a1
a2
a3
a2
Hexagonal direct lattice:
a1
a1  axˆ , a2  a( 12 xˆ  23 yˆ ), a3  czˆ.
Reciprocal lattice is
2pa 2  a 3
2pa 3  a1
b2
from b1 
a1  a 2  a 3 
, b2 
a 2  a 3  a1 
b1
8
Calculating FT of n(r)
n(r) = Σ nG eiG·r
nG = Fourier transform of scattering density n(r).
Only nonzero at points G of reciprocal lattice.
1
Explicitly,
iGr
3
nG 
e
n
(
r
)
d
r

V unit cell
where V is the volume of the unit cell.
9
Lecture 7 Sept. 3, 2010 PH 481/581
Calculating x-ray scattering intensity
x-ray
detector
k
r’
Amplitude reaching detector =

Re e
Source
positionr
ikr
n(r)d3 r
e


Phase change
number of scatterers
eik r n(r )d 3reik '(r ' r ) 
Source
position r
eik r

r
ik r '  r
Incoming amplitude
(electric field?)
k
k '  r'
r'
n(r) = Σ nG eiG·r
 2r  r r 2 
(r  r)  r   2r  r  r  r  1  2  2 
r
r 

1/2
 2r  r r 2 
 r  r

r  r  r  1  2  2   r  1  2  2nd order 
r
r 
r



r  r
k r  r  kr   k
 k   r  k   r  k   (r  r )
r
2
2
2
2
n(r )d 3rei (k k )r  eik r F (k  k )
Source
position r
where F is called the scattering amplitude (in Kittel’s book)
10
Conclusion: x-ray scattering probes
reciprocal lattice (RL)
If incoming wavevector is k, amplitude for outgoing
wavevector k’ ~ Fourier component nk’-k.
In a perfect crystal, this is nonzero only if k’ – k
is a reciprocal lattice vector, call it G:
Also, this is elastic scattering: |k|=|k’|, so k’ 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)
You have to rotate the crystal, powder it, ...
Powder pattern: crystallites are in all possible
k’
orientations, which rotates each G over the
k
surface of a sphere:
Sphere of possible G’s intersects sphere of
possible k’ s in a circle (ring), seen on screen.
k’
k
....
....
. . . .G
....
....
....
G
Powder pattern
Pattern for the mineral olivine,
http://ndbserver.rutgers.edu/mmcif/cbf/
Ball-milled
Hand-crushed
http://www.ccp14.ac.uk/ccp/webmirrors/armel/egypte/conf2/theory.html
12
Relating k’ – k = G to the Bragg equation
Bragg picture is
k
leading to
2 d sin q = n l
k’
q
d
In reciprocal lattice picture, k’ = k + G, so elastic scattering
condition is
(k  G )2  k 2
k’
k 2  2k  G  G 2  k 2
2k  G  k 2  0
q
q
G
Geometrically, we have
k
G
2p
2p
 k sin q 
sin q  l  2
sin q
2
l
G
2p
d
2
p
d

Consistent only if
(so smallest G goes with n=1:
)

n
G
G
13
Predicting angles of powder diffraction rings
Recall picture of k’ = k + G
& from top triangle, G  k sin q
2
2p
2p
but k 
and G 
(h, k , l ) so
l
a
sin q 
l

h
2a
2
k l
2
2q

2 1/2
h k
l
k’
q
q
h
2
k l
2
1 0 0
1
1 1 0
2
1 1 1
3
2 0 0
4
2 1 0
5
G
k

2 1/ 2
14
Lecture 8
PH 481/581,
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 = 2p/G
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 G’s and families of planes:
family {0,± G, ± 2G,...} of parallel RLVs ◄▬►
family of parallel planes.
Only some of these planes go through lattice points, and
these are spaced d = 2p/G apart.
15
(G must be the smallest length in the family.)
Drawing low-index planes (far apart)
high-index planes (close together)
Reciprocal Lattice
Direct lattice
.. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. ..
..........
.. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. ..
..........
..........
.. .. .. .. .. .. .. .. .. ..
..........
.. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. ..
..........
.. .. .. .. .. .. .. .. .. ..
..........
0
16
Lecture 10
PH 481/581, Sept. 13, 2010
Calculating scattering amplitude from atomic form factors
r r2
Consider a crystal with a basis,
Plot density along dashed line:
n(r) = n1(r-r1) + n2(r-r2)
= n1(r1) + n2(r2)
The scattering amplitude FG is
defined by F 
n(r )e iGr d 3r
G

entire crystal
r1
r2
r1x
n1
r2x
rx
n2
r
0 r1x
0 2x
where r1 = r - r1 and r2 = r – r2
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
17
Calculating scattering amplitude (structure factor)
Structure factor
SG 
 iG r 3
n
(
r
)
e
d r

r r2
cell
  n1 (r  r1 )e
 iG r
d r   n2 (r  r2 )e
3
iG r
r1
3
d r
r2
  n1 (ρ)e iG( r1 ρ ) d 3 r   n2 (ρ)e iG ( r2 ρ ) d 3 r
 e iGr1  n1 (ρ)e iGρ d 3 r  e iGr2  n2 (ρ)e iGρ d 3 r
e
iG r1
f1 (G )  e
 iG r2
f 2 (G )   e
 iG r j
f j (G )
j
n (ρ)e d r

= (0,0,0), r = (a/2,a/2,a/2)
 iGρ 3
where f i (G ) 
Example: CsCl: simple cubic, r1=
Some algebra gives
SG  f1 (G )  (1)
h  k l
f 2 (G ) 
i
2
f1  f 2
if h  k  l even (" fcc points" )
f1  f 2
if h  k  l odd
In limit f1  f2,(i.e., CsCl  FeFe bcc) this becomes fcc RL.
18
Another example (NaCl structure)
NaCl is fcc, basis is r1= = (0,0,0), r2 = (a/2,0,0)
SG  f1 (G )  (1) f 2 (G ) 
h
f1  f 2
if h, k , l all even (corner points)
f1  f 2
if h, k , l all odd (cell centers)
(if some even and some odd, e.g., (100), these points are not even
ON the reciprocal lattice so we don’t calculate SG.)
Example in Fig. 2-17, p. 42: KBr has 111, 200, 220, ... as predicted.
KCl: isoelectronic, f1 = f2 , lose reflections due to extra symmetry
19
Chapter 3: Binding
Types of bonding:
• van der Waals
• Ionic
• Covalent
• metallic
Lecture 12
Sept. 17, 2010
PH 481/581
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.
20
3
6
3
12
2.5
2
V = r -n
Comparison of different
core potentials V=r-n
for n = 1,2,3,6, and12
n=1 is said to be “soft”,
n=12 is “hard”.
n=infinity
n=1
1.5
1
0.5
r
0
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
21
1.8
2
END
22