ATOMIC AND CRYSTAL

STRUCTURES

Dr. TU LE MANH
 Outline

• Electronic structure of the atom

• Atomic bonding

• Unit cells

• Points, directions and planes in the unit cell

• The recommended texts for this course are Callister and
Rethwisch : Materials Science and Engineering, and Donald R
Askeland: The Science and Engineering of Materials, 3rd
edition.
 The Atom
• atom – electrons – 9.11 x 10-31 kg
protons
neutrons
} 1.67 x 10 -27 kg
• atomic number = # of protons in nucleus of atom
= # of electrons of neutral species

• A [=] atomic mass unit = amu = 1/12 mass of 12C

Atomic wt = wt of 6.022 x 1023 molecules or atoms

1 amu/atom = 1g/mol

C 12.011
H 1.008 etc.

3
 Electronic Structure
• Electrons have wavelike and particulate properties.
– This means that electrons are in orbitals defined by a probability.
– Each orbital at discrete energy level is determined by quantum
numbers.

Quantum # Designation
n = principal (energy level-shell) K, L, M, N, O (1, 2, 3, etc.)
l = subsidiary (orbitals) s, p, d, f (0, 1, 2, 3,…, n-1)
ml = magnetic 1, 3, 5, 7 (-l to +l)
ms = spin ½, -½

4
 Electronic structure
• The electronic structure is obtained from the solutions
of the Schroedinger equation for the Coulomb
potential
• Results in three quantum numbers
– principal quantum number n
• determines energy of electron
– azimuthal quantum number l
• angular momentum
– magnetic quantum number ml
• z-component of angular momentum
• It was found that a fourth quantum number is needed
– spin ms
• Pauli’s exclusion principles : no two electrons in an
atom can have the same set of quantum numbers
 Electron Energy States
Electrons...
• have discrete energy states
• tend to occupy lowest available energy state.

4d
4p N-shell n = 4

3d
4s

Energy 3p M-shell n = 3
3s
Adapted from Fig. 2.4,
Callister & Rethwisch 8e.
2p L-shell n = 2
2s

1s K-shell n = 1
6
Shell representation of atom
 SURVEY OF ELEMENTS
• Most elements: Electron configuration not stable.
Element Atomic # Electron configuration
Hydrogen 1 1s 1
Helium 2 1s 2 (stable)
Lithium 3 1s 2 2s 1
Beryllium 4 1s 22s 2
Boron 5 1s 22s 2 2p 1 Adapted from Table 2.2,
Callister & Rethwisch 8e.
Carbon 6 1s 2 2s 2 2p 2
... ...
Neon 10 1s 2 2s 2 2p 6 (stable)
Sodium 11 1s 2 2s 2 2p 6 3s 1
Magnesium 12 1s 22s 2 2p 6 3s 2
Aluminum 13 1s 2 2s 2 2p 6 3s 2 3p 1
... ...
Argon 18 1s 2 2s 2 2p 6 3s 2 3p 6 (stable)
... ... ...
Krypton 36 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 (stable)

• Why? Valence (outer) shell usually not filled completely.

8
 Electron Configurations
• Valence electrons – those in unfilled shells
• Filled shells more stable
• Valence electrons are most available for bonding and tend to
control the chemical properties

– example: C (atomic number = 6)

1s2 2s2 2p2

valence electrons

9
 Electronic Configurations
ex: Fe - atomic # = 26 1s2 2s2 2p6 3s2 3p6 3d 6 4s2

4d
4p N-shell n = 4 valence
electrons
3d
4s

Energy 3p M-shell n = 3
3s
Adapted from Fig. 2.4,
Callister & Rethwisch 8e.
2p L-shell n = 2
2s

1s K-shell n = 1

10
 The Periodic Table
• Columns: Similar Valence Structure

inert gases
give up 1e-
give up 2e-

accept 2e-
accept 1e-
give up 3e-

H He
Li Be O F Ne
Adapted from
Na Mg S Cl Ar Fig. 2.6,
Callister &
K Ca Sc Se Br Kr
Rethwisch 8e.
Rb Sr Y Te I Xe
Cs Ba Po At Rn
Fr Ra

Electropositive elements: Electronegative elements:

Readily give up electrons Readily acquire electrons
to become + ions. to become - ions.
11
 Electronegativity
• Ranges from 0.7 to 4.0,
• Large values: tendency to acquire electrons.

Smaller electronegativity Larger electronegativity

Adapted from Fig. 2.7, Callister & Rethwisch 8e. (Fig. 2.7 is adapted from Linus Pauling, The Nature of the
Chemical Bond, 3rd edition, Copyright 1939 and 1940, 3rd edition. Copyright 1960 by Cornell University.

12
Energy level diagram
 Atomic Structure & Interatomic
Bonding

• What promotes bonding?

• What types of bonds are there?

• What properties are inferred from bonding?

14
Atomic bonding - metallic bond
Electric conduction in metals
 Ionic Bonding
Found in
Ceramic Materials
 Ionic bond – metal + nonmetal
donates accepts
electrons electrons

Dissimilar electronegativities

ex: MgO Mg 1s2 2s2 2p6 3s2 O 1s2 2s2 2p4

[Ne] 3s2

Mg2+ 1s2 2s2 2p6 O2- 1s2 2s2 2p6

[Ne] [Ne]

18
Ionic bond
Electrical conduction in ionic material
 Ionic Bonding
• Occurs between + and - ions.
• Requires electron transfer.
• Large difference in electronegativity required.
• Example: NaCl

Na (metal) Cl (nonmetal)
unstable unstable
electron

Na (cation) + - Cl (anion)
stable Coulombic stable
Attraction

21
 Ionic Bonding
• Energy – minimum energy most stable
– Energy balance of attractive and repulsive terms


A 
B
EN = EA + ER =
r rn
Repulsive energy ER

Interatomic separation r

Net energy EN
Adapted from Fig. 2.8(b),
Callister & Rethwisch 8e.

Attractive energy EA
22
 Examples: Ionic Bonding
• Predominant bonding in Ceramics
NaCl
MgO
CaF 2
CsCl

Give up electrons Acquire electrons

Adapted from Fig. 2.7, Callister & Rethwisch 8e. (Fig. 2.7 is adapted from Linus Pauling, The Nature of the
Chemical Bond, 3rd edition, Copyright 1939 and 1940, 3rd edition. Copyright 1960 by Cornell University.

23
 Covalent Bonding
• similar electronegativity  share electrons
• bonds determined by valence – s & p orbitals
dominate bonding
• Example: CH4
shared electrons
H
C: has 4 valence e-, CH 4
from carbon atom
needs 4 more
H: has 1 valence e-, H C H
needs 1 more
shared electrons
Electronegativities H from hydrogen
are comparable. atoms

Adapted from Fig. 2.10, Callister & Rethwisch 8e.

24
Covalent bond
Covalent bond in compound
 SECONDARY BONDING
Arises from interaction between dipoles
• Fluctuating dipoles
asymmetric electron ex: liquid H 2
clouds H2 H2

+ - + - H H H H
secondary secondary
bonding Adapted from Fig. 2.13,
Callister & Rethwisch 8e. bonding

• Permanent dipoles-molecule induced

secondary
-general case: + - bonding
+ -
Adapted from Fig. 2.15,
Callister & Rethwisch 8e.
secondary
-ex: liquid HCl H Cl bonding H Cl

-ex: polymer secondary bonding

27
Electronic filling scheme
Van der Waals bond
Van der Waals bond - PVC example
 Summary: Bonding
Type Bond Energy Comments
Ionic Large! Nondirectional (ceramics)

Covalent Variable Directional

large-Diamond (semiconductors, ceramics
small-Bismuth polymer chains)

Metallic Variable
large-Tungsten Nondirectional (metals)
small-Mercury
Secondary smallest Directional
inter-chain (polymer)
inter-molecular
31
Interatomic spacing
 Properties From Bonding: Tm
• Bond length, r • Melting Temperature, Tm
Energy
r

• Bond energy, Eo ro
r
Energy smaller Tm

unstretched length
ro larger Tm
r
Eo = Tm is larger if Eo is larger.
“bond energy”

33
Interatomic spacing

Modulus.mov
Moduli of elasticity
 Properties From Bonding : 
• Coefficient of thermal expansion, 
length, Lo coeff. thermal expansion
unheated, T1
L L
= (T2 -T1)
heated, T 2 Lo

•  ~ symmetric at ro
Energy
unstretched length
ro
r  is larger if Eo is smaller.

E
larger 
o
E smaller 
o 36
 Summary: Primary Bonds
Ceramics Large bond energy
(Ionic & covalent bonding): large Tm
large E
small 

Metals Variable bond energy

(Metallic bonding): moderate Tm
moderate E
moderate 

Polymers Directional Properties

(Covalent & Secondary): Secondary bonding dominates
small Tm
small E
large 

37
 Crystalline

Is the glass in the window crystalline?

Is extruded aluminium crystalline?

For a material to be crystalline, it must show

1. Long range order
2. A basic repeating pattern
Different levels of order
 Energy and Packing
• Non dense, random packing Energy

typical neighbor
bond length

typical neighbor r
bond energy

• Dense, ordered packing Energy

typical neighbor
bond length

typical neighbor r
bond energy

Dense, ordered packed structures tend to have

lower energies.
40
 Materials and Packing
Crystalline materials...
• atoms pack in periodic, 3D arrays
• typical of: -metals
-many ceramics
-some polymers crystalline SiO2
Adapted from Fig. 3.23(a),
Callister & Rethwisch 8e.

Si Oxygen
Noncrystalline materials...
• atoms have no periodic packing
• occurs for: -complex structures
-rapid cooling
"Amorphous" = Noncrystalline noncrystalline SiO2
Adapted from Fig. 3.23(b),
Callister & Rethwisch 8e.

41
 Questions and Answers 1
1. Electron configurations for
Carbon 1s2 2s2p2 Neon 1s22s22p6
Sodium 1s22s22p63s1 Chlorine 1s22s22p63s23p5
2. Why are the atomic weights of the elements not integers?
Atomic weight is the weighted average of the atomic mass, that is the
sum of masses of the protons and neutrons, of an atom’s naturally
occurring isotopes.
1. Which are the valence electrons for Na and Cl
Na 3s2
Cl 3s23p5
1. Are the following crystalline or non-crystalline?
Glass is non-crystalline
extruded Aluminium is crystalline
1. What two features define crystallinity?
1. Long range order
2. A basic repeating pattern
Unit cell and lattice
 Crystal Systems
Unit cell: smallest repetitive volume which
contains the complete lattice pattern of a crystal.

7 crystal systems

14 crystal lattices

a, b, and c are the lattice constants

Fig. 3.4, Callister & Rethwisch 8e.

44
Seven crystal systems
14 Bravais lattices
 Metallic Crystal Structures
• Tend to be densely packed.
• Reasons for dense packing:
- Typically, only one element is present, so all atomic
radii are the same.
- Metallic bonding is not directional.
- Nearest neighbor distances tend to be small in
order to lower bond energy.
- Electron cloud shields cores from each other
• Have the simplest crystal structures.

We will examine three such structures...

47
 Bravais movie

Cubic unit cell

stretched to orthorhombic,
then skewed to monoclinic,
and finally to triclinic
Cubic systems
 Simple Cubic Structure (SC)
• Rare due to low packing density (only Po has this structure)
• Close-packed directions are cube edges.

• Coordination # = 6
(# nearest neighbors)

Click once on image to start animation

(Courtesy P.M. Anderson)

50
 Atomic Packing Factor (APF)
Volume of atoms in unit cell*
APF =
Volume of unit cell
*assume hard spheres
• APF for a simple cubic structure = 0.52
volume
atoms atom
a 4
unit cell 1 (0.5a) 3
3
R=0.5a APF =
a3 volume
close-packed directions
unit cell
contains 8 x 1/8 =
1 atom/unit cell
Adapted from Fig. 3.24,
Callister & Rethwisch 8e. 51
Relation between atomic radius and lattice
parameter
Coordination number

FCC?
 Body Centered Cubic Structure (BCC)
• Atoms touch each other along cube diagonals.
--Note: All atoms are identical; the center atom is shaded
differently only for ease of viewing.

ex: Cr, W, Fe (), Tantalum, Molybdenum

• Coordination # = 8

Adapted from Fig. 3.2,

Click once on image to start animation Callister & Rethwisch 8e.
(Courtesy P.M. Anderson)
2 atoms/unit cell: 1 center + 8 corners x 1/8
54
 Atomic Packing Factor: BCC
• APF for a body-centered cubic structure = 0.68
3a

2a

Close-packed directions:
Adapted from R length = 4R = 3 a
Fig. 3.2(a), Callister &
Rethwisch 8e.
a
atoms volume
4
unit cell 2  ( 3a/4) 3
3 atom
APF =
3 volume
a
unit cell 55
 Face Centered Cubic Structure (FCC)
• Atoms touch each other along face diagonals.
--Note: All atoms are identical; the face-centered atoms are shaded
differently only for ease of viewing.

ex: Al, Cu, Au, Pb, Ni, Pt, Ag

• Coordination # = 12

Adapted from Fig. 3.1, Callister & Rethwisch 8e.

Click once on image to start animation
(Courtesy P.M. Anderson) 4 atoms/unit cell: 6 face x 1/2 + 8 corners x 1/8

56
 Atomic Packing Factor: FCC
• APF for a face-centered cubic structure = 0.74
maximum achievable APF
Close-packed directions:
length = 4R = 2 a
2a
Unit cell contains:
6 x 1/2 + 8 x 1/8
= 4 atoms/unit cell
a
Adapted from
Fig. 3.1(a),
Callister & atoms volume
Rethwisch 8e. 4
unit cell 4 ( 2a/4) 3
3 atom
APF =
3 volume
a
unit cell
57
 Hexagonal Close-Packed Structure
(HCP)
• ABAB... Stacking Sequence
• 3D Projection • 2D Projection

A sites Top layer

c
B sites Middle layer

A sites Bottom layer

a Adapted from Fig. 3.3(a),
Callister & Rethwisch 8e.

• Coordination # = 12 6 atoms/unit cell

• APF = 0.74 ex: Cd, Mg, Ti, Zn
• c/a = 1.633 58
 Theoretical Density, 

Mass of Atoms in Unit Cell

Density =  =
Total Volume of Unit Cell

nA
 =
VC NA

where n = number of atoms/unit cell

A = atomic weight
VC = Volume of unit cell = a3 for cubic
NA = Avogadro’s number
= 6.022 x 1023 atoms/mol

59
 Theoretical Density, 
• Ex: Cr (BCC)
A = 52.00 g/mol
R = 0.125 nm
n = 2 atoms/unit cell

R
Adapted from a a = 4R/ 3 = 0.2887 nm
Fig. 3.2(a), Callister &
Rethwisch 8e.
atoms
g
unit cell 2 52.00 theoretical = 7.18 g/cm3
mol
= actual = 7.19 g/cm3
a3 6.022 x 1023
volume atoms
unit cell mol 60
 Densities of Material Classes
In general Graphite/
metals > ceramics > polymers
Metals/ Composites/
Ceramics/ Polymers
Alloys fibers
Semicond
30
Why? Platinum
Based on data in Table B1, Callister
*GFRE, CFRE, & AFRE are Glass,
20 Gold, W
Metals have... Tantalum Carbon, & Aramid Fiber-Reinforced
Epoxy composites (values based on
• close-packing 60% volume fraction of aligned fibers
10 Silver, Mo in an epoxy matrix).
(metallic bonding) Cu,Ni
Steels
• often large atomic masses Tin, Zinc
Zirconia

(g/cm
5
Ceramics have... 3 4
Titanium
Al oxide
• less dense packing 3
Diamond
Si nitride
Aluminum Glass -soda
• often lighter elements Concrete
Glass fibers
)

Silicon PTFE GFRE*

2 Carbon fibers
Polymers have... Magnesium Graphite
Silicone CFRE*
Aramid fibers
PVC
• low packing density PET
PC
AFRE*
1 HDPE, PS
(often amorphous) PP, LDPE
• lighter elements (C,H,O)
0.5
Composites have... 0.4
Wood

• intermediate values 0.3

Data from Table B.1, Callister & Rethwisch, 8e.
61
BCC
FCC

Fcc_clip.mov
ABCABC stacking

Hex_fcc.mov
ABABAB stacking

Hex_cp.mov
Coordination number

Clospack.mov
 Questions and Answers 2
1. What is the electron configuration for Mg2+
– 1s22s22p6
2. Why do metals not have the simple cubic structure?
- The electron cloud allows the metal ion to pack tightly and simple cubic is
the least efficiently packed structure.
3. How many electrons are in filled d and f shells?
- d 10 electrons
- f 14 electrons
4. Which is the most densely packed structure?
- FCC
5. How many atoms in the FCC structure?
-4
6. Iron expands continuously all the way to the M.Pt. - False
7. Ionic bonds are formed between a metal and a non-metal
8. Show the difference in M.Pts. high and low on the Energy vs interatomic
distance graph.
Direction in a unit cell - Miller indices
• Using a right-handed coordinate system, determine
the coordinates of two points that lie on the direction.
• Subtract the coordinates of the "tail" point from the
coordinates of the "head" point to obtain the number
of lattice parameters travelled in the direction of each
axis of the coordinate system.
• Clear fractions and/or reduce the results obtained
from the subtraction to lowest integers.
• Enclose the numbers in square brackets [ ]. If a
negative sign is produced, represent the negative
sign with a bar over the number.
 Point Coordinates
z
111
c

000
y
a b
x

Point coordinates for unit cell center are

a/2, b/2, c/2 ½½½

Point coordinates for unit cell corner are 111

69
 Crystallographic Directions
z Algorithm
1. Vector repositioned (if necessary) to pass
through origin.
2. Read off projections in terms of
unit cell dimensions a, b, and c
y 3. Adjust to smallest integer values
4. Enclose in square brackets, no commas

x [uvw]

ex: 1, 0, ½ => 2, 0, 1 => [ 201 ]

-1, 1, 1 => [ 111 ] where overbar represents a
negative index
families of directions <uvw>
70
Unit cell directions - example
Direction A
Direction C
Direction B
 Crystallographic planes
• Identify the points at which the planes intercepts the
x, y and z axis in terms of the lattice parameters (If
the plane passes through the origin, the origin of the
coordinate system must be moved)
• Take reciprocals of these intercepts
• Clear fractions but do not reduce to the lowest
integers
• Enclose the resulting numbers in parentheses ( ).
Again, negative numbers should be written with a bar
over the number
• (Planes and their negatives are identical)
 Crystallographic Planes
• Miller Indices: Reciprocals of the (three) axial intercepts for a
plane, cleared of fractions & common multiples. All parallel
planes have same Miller indices.

• Algorithm
1. Read off intercepts of plane with axes in
terms of a, b, c
2. Take reciprocals of intercepts

3. Enclose in parentheses, no
commas i.e., (hkl)

76
 Crystallographic Planes

Adapted from Fig. 3.10,

Callister & Rethwisch 8e.
77
 Crystallographic Planes
z
example a b c
1. Intercepts 1 1  c
2. Reciprocals 1/1 1/1 1/
1 1 0
3. Reduction 1 1 0 y
a b
4. Miller Indices (110)
x
z
example a b c
1. Intercepts 1/2   c
2. Reciprocals 1/½ 1/ 1/
2 0 0
3. Reduction 2 0 0
y
4. Miller Indices (200) a b
x
78
 Crystallographic Planes
z
example a b c c
1. Intercepts 1/2 1 3/4 
2. Reciprocals 1/½ 1/1 1/¾
2 1 4/3  y

3. Reduction 6 3 4 a b

4. Miller Indices (634) x

Family of Planes {hkl}

Ex: {100} = (100), (010), (001), (100), (010), (001)

79
• Diffraction gratings must have spacings comparable to the wavelength of
diffracted radiation.
• Can’t resolve spacings  
• Spacing is the distance between parallel planes of atoms.

80
 Microscopic Examination
• Crystallites (grains) and grain boundaries. Vary considerably in
size. Can be quite large.
– ex: Large single crystal of quartz or diamond or Si
– ex: Aluminum light post or garbage can - see the individual
grains
• Crystallites (grains) can be quite small (mm or less) – necessary
to observe with a microscope.

81
 Optical Microscopy
• Useful up to 2000X magnification.
• Polishing removes surface features (e.g., scratches)
• Etching changes reflectance, depending on crystal
orientation.

crystallographic planes
Adapted from Fig. 4.13(b) and (c), Callister
& Rethwisch 8e. (Fig. 4.13(c) is courtesy
of J.E. Burke, General Electric Co.)

Micrograph of
brass (a Cu-Zn alloy)

0.75mm
82
 Optical Microscopy
Grain boundaries...
• are imperfections,
• are more susceptible
to etching,
• may be revealed as polished surface
dark lines,
• change in crystal surface groove
orientation across grain boundary
(a)
boundary. Adapted from Fig. 4.14(a)
and (b), Callister &
ASTM grain Rethwisch 8e.
(Fig. 4.14(b) is courtesy
size number of L.C. Smith and C. Brady,
the National Bureau of

N = 2n-1 Standards, Washington, DC

[now the National Institute of
Standards and Technology,
Gaithersburg, MD].)
number of grains/in2 Fe-Cr alloy
at 100x (b)
magnification 83
 Microscopy
Optical resolution ca. 10-7 m = 0.1 m = 100 nm
For higher resolution need higher frequency
– X-Rays? Difficult to focus.
– Electrons
• wavelengths ca. 3 pm (0.003 nm)
– (Magnification - 1,000,000X)
• Atomic resolution possible
• Electron beam focused by magnetic lenses.

84
 X-Ray Diffraction

• Structure determination using X-rays

• It is essential to understand the BRAGG Law
• You will see the importance of crystal planes
 X-Rays to Determine Crystal Structure
• Incoming X-rays diffract from crystal planes.

reflections must
be in phase for
a detectable signal
extra
 Adapted from Fig. 3.20,
 
distance
travelled Callister & Rethwisch 8e.
by wave “2” spacing
d between
planes

Measurement of X-ray
n
critical angle, c, intensity d
(from 2 sin c
allows computation of
detector)
planar spacing, d.

c
86
 X-ray diffraction



Bragg’s law sin
2d
Diffractometer
Diffractometer results
 X-Ray Diffraction Pattern
z z z
c c c

y (110) y y
a b a b a b
Intensity (relative)

x x x (211)

(200)

Diffraction angle 2

Diffraction pattern for polycrystalline -iron (BCC)

Adapted from Fig. 3.22, Callister 8e.

90
 Interpretation of diffraction results
• It can be shown (notes) that the interplanar spacing
dhkl of a set of hkl planes for a cubic crystal of lattice
parameter a0 is given by

• Therefore in Bragg’s law

a
d 
hkl
0
2 2 2
hkl


sin
2d


2
sin

 
2
h l
2 22

k 2
4
a0
 Interpretation of diffraction results II
• In a simple cubic crystal,
– all planes will diffract and h,k and l can take on all
possible values
– therefore h2+k2+l2 range from 1,2,3,4,5,6,8…
• In a BCC crystal,
– some planes will have destructive interference and
h,k and l cannot take on all possible values
– h2+k2+l2 have values 2,4,6,8 …
• In a FCC crystal,
– h2+k2+l2 have values 3,4,8,11,12,16 …
 X-ray diffraction example

Results


0
.0710
nm
Analysis

2
sin

 
2
h l
2 22

k 2
4
a0

Find common denominator

BCC
Calculation of the lattice parameter


At 2 59
.
42we observe reflection (400)

From 
2
sin

 
2
h l
2 22

k2
4
a0

a 
0 4


sin

h2 2 2
k l 
0
.071072 2 2
 4 0 0
4 sin(
59.42/2
)
0.286nm
 SUMMARY
• Atoms may assemble into crystalline or
amorphous structures.
• Common metallic crystal structures are FCC, BCC, and
HCP. Coordination number and atomic packing factor
are the same for both FCC and HCP crystal structures.
• We can predict the density of a material, provided we
know the atomic weight, atomic radius, and crystal
geometry (e.g., FCC, BCC, HCP).
• Crystallographic points, directions and planes are
specified in terms of indexing schemes.
Crystallographic directions and planes are related
to atomic linear densities and planar densities.

96
 SUMMARY

• Optical Microscopy is used to look at the microstructure

of materials, in particular, the grain structure in metals.
Resolution is limited by the wavelength of the light.

X-ray diffraction is used for determination of crystal

structure and interplanar spacing determinations.

97
 Q and A 3
• 1. Draw the following planes and directions in separate unit
cells
(100) (110) (210)

• 2. Which plane is the most densely packed (111) or (634)?

(111)
• 3. What kind of microscope would you use to resolve atomic
structure?
Electron Microscope
• 4. In the Bragg Law λ = 2dsinO, what does d stand for?
Inter planar spacing
• 5. What is X-ray diffraction used for?
Determination of Structure, lattice parameter
Direction in a unit cell - Miller indices 2
Crystallographic planes II

Bcc_pln.mov

Miller.mov

111_plns.mov Fcc_plns.mov
Coordination number movies

Fcc_12.mov

Tetrah_4.mov Octah_6.mov Cubc_8.mov

Hcp_12.mov