1. Fisika Zat Padat
Dr. Lutfi Rohman

2. Symmetry & Crystal Structures
Solid State Physics :Topic 1
Symmetry & Crystal Structures

3. Diffraction covered in detail in the next chapter.
Why are planes in a lattice important?
(A) Determining crystal structure
Diffraction methods directly measure the distance between parallel planes of lattice points. This information is used to determine the lattice parameters in a crystal & measure the angles between lattice planes.
(B) Plastic deformation
Plastic (permanent) deformation in metals occurs by the slip of atoms past each other in the crystal. This slip tends to occur preferentially along specific lattice planes in the crystal. Which planes slip depends on the crystal structure of the material.
     
(C) Transport Properties
In certain materials, the atomic structure in certain planes causes the transport of electrons and/or heat to be particularly rapid in that plane, and relatively slow away from the plane.
Example: Graphite
Conduction of heat is more rapid in the sp2 covalently bonded lattice planes than in the direction perpendicular to those planes.
Example: YBa2Cu3O7 superconductors
Some lattice planes contain only Cu and O. These planes conduct pairs of electrons (called Cooper pairs) that are responsible for superconductivity. These superconductors are electrically insulating in directions perpendicular to the Cu-O lattice planes.
Diffraction covered in detail in the next chapter.
Thermal conductivity is 400 times greater in a-b plane than in the c direction of graphite.
Planes and chains structure was determined by my co-workers at Argonne in We will make some of this later in the semester when we study superconductors.

4. Example. Determining Miller Indices of Planes
Determine the Miller indices of planes A, B, and C in Figure 3.22.
(c) 2003 Brooks/Cole Publishing / Thomson Learning™
Figure Crystallographic planes and intercepts (for Example)

5. Example, SOLUTION
Plane A
1. x = 1, y = 1, z = 1
2.1/x = 1, 1/y = 1,1 /z = 1
3. No fractions to clear
4. (111)
Plane B
1. The plane never intercepts the z axis, so x = 1, y = 2, and z =
2.1/x = 1, 1/y =1/2, 1/z = 0
3. Clear fractions:
1/x = 2, 1/y = 1, 1/z = 0
4. (210)
Plane C
1. We must move the origin, since the plane passes through , 0, 0. Let’s move the origin one lattice parameter in the y-direction. Then, x = , y = -1, and z =
2.1/x = 0, 1/y = -1, 1/z = 0
3. No fractions to clear.

7. Example. Drawing Direction and Plane
Draw (a) the direction and (b) the plane in a cubic unit cell.
(c) 2003 Brooks/Cole Publishing / Thomson Learning™
Figure Construction of a (a) direction and (b) plane within a unit cell (for Example )

8. Example, SOLUTION
a. Because we know that we will need to move in the negative y-direction, let’s locate the origin at 0, 0, 0. The ‘‘tail’’ of the direction will be located at this new origin. A second point on the direction can be determined by moving +1 in the x-direction, -2 in the y-direction, and +1 in the z direction [Figure 3.24(a)].
b. To draw in the plane, first take reciprocals of the indices to obtain the intercepts, that is:
x = 1/-2 = -1/2 y = 1/1 = 1 z = 1/0 =
Since the x-intercept is in a negative direction, and we wish to draw the plane within the unit cell, let’s move the origin +1 in the x-direction to 1, 0, 0. Then we can locate the x-intercept at 1/2 and the y-intercept at +1. The plane will be parallel to the z-axis [Figure 3.24(b)].

9. (GPa)

10. The interplanar spacing (aka d-spacing)
perpendicular distance between planes in a given family (hkl)
symbolically designated as dhkl      
The d-spacing decreases as the Miller indices increase
The density of lattice points in a plane---i.e. the number of lattice points per unit area in the plane--- decreases as the Miller indices increase

11. d-spacing formula For orthogonal crystal systems (i.e. ===90) :-
For cubic crystals (special case of orthogonal) a=b=c :-
e.g. for (1 0 0) d = a
(2 0 0) d = a/2
(1 1 0) d = a/2 etc.

12. d spacing
Example
The lattice constant for aluminum is angstroms. What is d220?
Answer
Aluminum has an fcc structure, so a = b = Å

13. A cubic crystal has a=5. 2 Å (=0. 52nm)
A cubic crystal has a=5.2 Å (=0.52nm). Calculate the d-spacing of the (1 1 0) plane
Answer
A cubic crystal structure, so a = b = 5.2 Å
A tetragonal crystal has a=4.7 Å, c=3.4 Å. Calculate the separation of the:
(1 0 0)
(0 0 1)
(1 1 1) planes

14. Question (exercise):
If a = b = c = 8 Å, find d-spacings for planes with Miller indices (1 2 3)
Calculate the d-spacings for the same planes in a crystal with unit cell a = b = 7 Å, c = 9 Å.
Calculate the d-spacings for the same planes in a crystal with unit cell a = 7 Å, b = 8 Å, c = 9 Å.

15. Interstitials
FCC
BCC
Within any crystal structure, there are void spaces between the atoms
These are called interstices, or interstitial voids, or interstitial sites.
Each crystal structure has specific types of voids
geometry determined by the nearest neighbors
size determined by atomic/ionic radius and void geometry
Interstitial sites are important because they are possible sites for other types of atoms
- formation of compounds (NaCl, CsCl)
- alloying agents (hydrogen occupies interstitial sites to make metal hydrides)
- impurity atoms

16. What types/sizes of atoms or ions can fit in a given interstitial site?
Answer:
Calculate the effective radius ρ of the void space using trigonometry AND the hard-sphere model with the radius r for the host atom (for metals) or ions (for ionic crystals).
Any element with atomic/ionic radius less than or equal to ρ can occupy that interstitial site.
If the element has radius larger than ρ, then it will cause some distortion to the crystal structure
The increased energy due to distortion will limit the number of interstitial sites that can be occupied (example: carbon in iron).

17. Often described as 2 interpenetrating FCC lattices
Examples of common structures: (1) The Sodium Chloride (NaCl) Structure (LiH, MgO, MnO, AgBr, PbS, KCl, KBr)
The NaCl structure is FCC
The basis consists of one Na atom and one Cl atom, separated by one-half of the body diagonal of a unit cube
There are four units of NaCl in each unit cube
Atom positions:
Cl : 000 ; ½½0; ½0½; 0½½
Na: ½½½; 00½; 0½0; ½00
Each atom has 6 nearest neighbours of the opposite kind
Often described as 2 interpenetrating FCC lattices

18. a NaCl structure Crystal a LiH 4.08 Å MgO 4.20 MnO 4.43 NaCl 5.63 AgBr
5.77
PbS
5.92
KCl
6.29
KBr
6.59 a

19. (2) The Cesium Chloride (CsCl) structure
(CsBr, CsI, RbCl, AlCo, AgZn, BeCu, MgCe, RuAl, SrTl)
The CsCl structure is BCC
The basis consists of one Cs atom and one Cl atom, with each atom at the center of a cube of atoms of the opposite kind
There is on unit of CsCl in each unit cube
Atom positions:
Cs : 000
Cl : ½½½ (or vice-versa)
Each atom has 8 nearest neighbours of the opposite kind

20. CsCl structure
Crystal a
BeCu
2.70 Å
AlNi
2.88
CuZn
2.94
CuPd
2.99
AgMg
3.28
LiHg
3.29
NH4Cl
3.87
TlBr
3.97
CsCl
4.11
TlI
4.20 a
Why are the a values smaller for the CsCl structures than for the NaCl (in general)?

21. Closed-packed structures
(or, what does stacking fruit have to do with solid state physics?)

22. Closed-packed structures
There are an infinite number of ways to organize spheres to maximize the packing fraction.
The centres of spheres at A, B, and C positions (from Kittel)
There are different ways you can pack spheres toge-ther. This shows two ways, one by putting the spheres in an ABAB… arrangement, the other with ACAC…. (or any combination of the two works)

23. (3) The Hexagonal Closed-packed (HCP) structure
Be, Sc, Te, Co, Zn, Y, Zr, Tc, Ru, Gd,Tb, Py, Ho, Er, Tm, Lu, Hf, Re, Os, Tl
The HCP structure is made up of stacking spheres in a ABABAB… configuration
The HCP structure has the primitive cell of the hexagonal lattice, with a basis of two identical atoms
Atom positions: 000, 2/3 1/3 ½ (remember, the unit axes are not all perpendicular)
The number of nearest-neighbours is 12.
Conventional HCP unit cell

24. (looking along [111] direction
The FCC and hexagonal closed-packed structures (HCP) are formed from packing in different ways. FCC (sometimes called the cubic closed-packed structure, or CCP) has the stacking arrangement of ABCABCABC… HCP has the arrangement ABABAB….
[1 1 1]
[0 0 1]
FCC
(CCP)
(looking along [111] direction
HCP
ABAB
sequence
ABCABC
sequence

25. Example 1 Determining the Number of Lattice Points in Cubic Crystal Systems
Determine the number of lattice points per cell in the cubic crystal systems. If there is only one atom located at each lattice point, calculate the number of atoms per unit cell.
Example 1. SOLUTION
In the SC unit cell: lattice point / unit cell = (8 corners)1/8 = 1
In BCC unit cells: lattice point / unit cell = (8 corners)1/8 + (1 center)(1) = 2
In FCC unit cells: lattice point / unit cell = (8 corners)1/8 + (6 faces)(1/2) = 4
The number of atoms per unit cell would be 1, 2, and 4, for the simple cubic, body-centered cubic, and face-centered cubic, unit cells, respectively.

26. Example 2 Determining the Relationship between Atomic Radius and Lattice Parameters
Determine the relationship between the atomic radius and the lattice parameter in SC, BCC, and FCC structures when one atom is located at each lattice point.
(c) 2003 Brooks/Cole Publishing / Thomson Learning™
Figure. The relationships between the atomic radius and the Lattice parameter in cubic systems (for Example 2).

27. Example 2. SOLUTION
Referring to Figure , we find that atoms touch along the edge of the cube in SC structures.
In BCC structures, atoms touch along the body diagonal. There are two atomic radii from the center atom and one atomic radius from each of the corner atoms on the body diagonal, so
In FCC structures, atoms touch along the face diagonal of the cube. There are four atomic radii along this length—two radii from the face-centered atom and one radius from each corner, so:

28. (c) 2003 Brooks/Cole Publishing / Thomson Learning™
Figure. Illustration of coordinations in (a) SC and (b) BCC unit cells. Six atoms touch each atom in SC, while the eight atoms touch each atom in the BCC unit cell.

29. Example 3 Calculating the Atomic Packing Factor
Calculate the packing factor for the FCC cell.
Example 3 SOLUTION
In a FCC cell, there are four lattice points per cell; if there is one atom per lattice point, there are also four atoms per cell. The volume of one atom is 4πr3/3 and the volume of the unit cell is .

30. Example 4 Determining the Density of BCC Iron
Determine the density of BCC iron, which has a lattice parameter of nm.
Example 4 SOLUTION
Atoms/cell = 2, a0 = nm =  10-8 cm
Atomic mass = g/mol
Volume of unit cell = = (2.866  10-8 cm)3 =  cm3/cell
Avogadro’s number NA = 6.02  1023 atoms/mol

31. Figure. The hexagonal close-packed (HCP) structure (left) and its unit cell.

32. Tugas 1: Buktikan bahwa APF untuk SC, BCC dan HCP adalah seperti yang ada di tabel di atas!

33. HCP and FCC structures
Crystal
c/a He
1.633 Be
1.581 Mg
1.623 Ti
1.586 Zn
1.861 Cd
1.886 Co
1.622 Y
1.570 Zr
1.594 Gd
1.592 Lu
The hexagonal-closed packed (HCP) and FCC structures both have the ideal packing fraction of 0.74 (Kepler figured this out hundreds of years ago)
The ideal ratio of c/a for this packing is (8/3)1/2 = 1.633

34. Amorphous Materials Glass
The continuous random network structure of amorphous silicon dioxide, notice that each Si atom (gold spheres) has 4 bonds, and each oxygen atom (red spheres) has 2 bonds.

41. Tugas 2 bagian B: