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Chapter 3 -

In the Name of God

1

Materials Science

DrDrDrDr Mohammad Mohammad Mohammad Mohammad HeidariHeidariHeidariHeidari----RaraniRaraniRaraniRarani

Chapter 3 - 2

ISSUES TO ADDRESS...

• How do atoms assemble into solid structures?

• How does the density of a material depend onits structure?

• When do material properties vary with thesample (i.e., part) orientation?

Chapter 3: Structures of Metals & Ceramics

• How do the crystal structures of ceramic materials differ from those for metals?

Chapter 3 - 3

• Non dense, random packing

• Dense, ordered packing

Dense, ordered packed structures tend to havelower energies.

Energy and PackingEnergy

r

typical neighborbond length

typical neighborbond energy

Energy

r

typical neighborbond length

typical neighborbond energy

Chapter 3 - 4

• atoms pack in periodic, 3D arraysCrystalline materials...

-metals-many ceramics-some polymers

• atoms have no periodic packingNoncrystalline materials...

-complex structures-rapid cooling

crystalline SiO2

noncrystalline SiO2"Amorphous" = NoncrystallineAdapted from Fig. 3.40(b),Callister & Rethwisch 3e.

Adapted from Fig. 3.40(a),Callister & Rethwisch 3e.

Materials and Packing

Si Oxygen

• typical of:

• occurs for:

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Chapter 3 - 5

Metallic Crystal Structures

• How can we stack metal atoms to minimize empty space?

2-dimensions

vs.

Now stack these 2-D layers to make 3-D structures

Chapter 3 - 6

• Tend to be densely packed.

• Reasons for dense packing:- Typically, only one element is present, so all atomicradii are the same.

- Metallic bonding is not directional.- Nearest neighbor distances tend to be small inorder to lower bond energy.

- Electron cloud shields cores from each other

• Have the simplest crystal structures.

We will examine three such structures...

Metallic Crystal Structures

Chapter 3 - 7

• Rare due to low packing density (only Po has this structure)• Close-packed directions are cube edges.

• Coordination # = 6(# nearest neighbors)

(Courtesy P.M. Anderson)

Simple Cubic Structure (SC)

Chapter 3 - 8

• APF for a simple cubic structure = 0.52

APF = a3

4

3π (0.5a) 31

atomsunit cell

atomvolume

unit cellvolume

Atomic Packing Factor (APF)

APF = Volume of atoms in unit cell*

Volume of unit cell

*assume hard spheres

Adapted from Fig. 3.42,Callister & Rethwisch 3e.

close-packed directions

a

R=0.5a

contains 8 x 1/8 = 1 atom/unit cell

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Chapter 3 - 9

• Coordination # = 8

• Atoms touch each other along cube diagonals.--Note: All atoms are identical; the center atom is shaded

differently only for ease of viewing.

Body Centered Cubic Structure (BCC)

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

2 atoms/unit cell: 1 center + 8 corners x 1/8

Chapter 3 - 10

Atomic Packing Factor: BCC

a

APF =

4

3π ( 3a/4)32

atoms

unit cell atomvolume

a3unit cell

volume

length = 4R =Close-packed directions:

3 a

• APF for a body-centered cubic structure = 0.68

aR

a2

a3

Chapter 3 - 11

• Coordination # = 12

• Atoms touch each other along face diagonals.--Note: All atoms are identical; the face-centered atoms are shaded

differently only for ease of viewing.

Face Centered Cubic Structure (FCC)

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

4 atoms/unit cell: 6 face x 1/2 + 8 corners x 1/8

Chapter 3 - 12

• APF for a face-centered cubic structure = 0.74Atomic Packing Factor: FCC

maximum achievable APF

APF =

4

3π ( 2a/4)34

atoms

unit cell atomvolume

a3unit cell

volume

Close-packed directions: length = 4R = 2 a

Unit cell contains:6 x 1/2 + 8 x1/8

= 4 atoms/unit cella

2 a

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Chapter 3 - 13

A sites

B B

B

BB

B B

C sites

C C

CA

B

B sites

• ABCABC... Stacking Sequence• 2D Projection

• FCC Unit Cell

FCC Stacking Sequence

B B

B

BB

B B

B sitesC C

CA

C C

CA

AB

C

Chapter 3 - 14

• Coordination # = 12

• ABAB... Stacking Sequence

• APF = 0.74

• 3D Projection • 2D Projection

Hexagonal Close-Packed Structure (HCP)

6 atoms/unit cell

ex: Cd, Mg, Ti, Zn

• c/a = 1.633

c

a

A sites

B sites

A sites Bottom layer

Middle layer

Top layer

Chapter 3 - 15

Theoretical Density, ρ

where n = number of atoms/unit cellA = atomic weight VC = Volume of unit cell = a3 for cubicNA = Avogadro’s number

= 6.022 x 1023 atoms/mol

Density = ρ =

VCNA

n Aρ =

CellUnitofVolumeTotalCellUnitinAtomsofMass

Chapter 3 - 16

• Ex: Cr (BCC) A = 52.00 g/molR = 0.125 nmn = 2 atoms/unit cell

ρtheoretical

a = 4R/ 3 = 0.2887 nm

ρactual

aR

ρ = a3

52.002

atoms

unit cell molg

unit cell

volume atoms

mol

6.022x 1023

Theoretical Density, ρ

= 7.18 g/cm3

= 7.19 g/cm3

Adapted from Fig. 3.2(a), Callister & Rethwisch 3e.

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Chapter 3 - 17

• Bonding:-- Can be ionic and/or covalent in character.-- % ionic character increases with difference in

electronegativity of atoms.

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

• Degree of ionic character may be large or small:

Atomic Bonding in Ceramics

SiC: small

CaF2: large

Chapter 3 - 18

Ceramic Crystal Structures

Oxide structures– oxygen anions larger than metal cations– close packed oxygen in a lattice (usually FCC)– cations fit into interstitial sites among oxygen ions

Chapter 3 -c03tf02

Chapter 3 - 20

Factors that Determine Crystal Structure1. Relative sizes of ions – Formation of stable structures:

--maximize the # of oppositely charged ion neighbors.

Adapted from Fig. 3.4, Callister & Rethwisch 3e.

- -

- -+

unstable

- -

- -+

stable

- -

- -+

stable2. Maintenance of

Charge Neutrality :--Net charge in ceramic

should be zero.--Reflected in chemical

formula:

CaF2: Ca2+cation

F-

F-

anions+

AmXp

m, p values to achieve charge neutrality

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Chapter 3 - 21

• Coordination # increases with

Coordination # and Ionic Radii

Adapted from Table 3.3, Callister & Rethwisch 3e.

2

rcationranion

Coord #

< 0.155

0.155 - 0.225

0.225 - 0.414

0.414 - 0.732

0.732 - 1.0

3

4

6

8

linear

triangular

tetrahedral

octahedral

cubic

Adapted from Fig. 3.5, Callister & Rethwisch 3e.

Adapted from Fig. 3.6, Callister & Rethwisch 3e.

Adapted from Fig. 3.7, Callister & Rethwisch 3e.

ZnS (zinc blende)

NaCl(sodium chloride)

CsCl(cesium chloride)

rcationranion

To form a stable structure, how many anions cansurround around a cation?

Chapter 3 - 22

Computation of Minimum Cation-Anion Radius Ratio for a Coordination No. of 3

• Determine minimum rcation/ranion for an octahedral site (C.N. = 3)

155.023

231

anion

cation =−=r

r

Chapter 3 - 23

Computation of Minimum Cation-Anion Radius Ratio

• Determine minimum rcation/ranion for an octahedral site (C.N. = 6)

a = 2ranion

2ranion + 2rcation = 2 2ranion

ranion + rcation = 2ranion rcation = ( 2 −1)ranion

arr 222 cationanion =+

414.012anion

cation =−=rr

Chapter 3 - 24

Same concepts can be applied to ionic solids in general. Example: NaCl (rock salt) structure

rNa = 0.102 nm

∴ cations (Na+) prefer octahedral sites

rCl = 0.181 nm

564.0

181.0

102.0

anion

cation

=

=r

r

based on this ratio,-- coord # = 6 because

0.414 < 0.550 < 0.732

AX Crystal StructuresAX–Type Crystal Structures include NaCl, CsCl, and zinc blende

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Chapter 3 - 25

• On the basis of ionic radii, what crystal structurewould you predict for FeO?

• Answer:

5500

14000770

anion

cation

.

.

.rr

=

=

based on this ratio,-- coord # = 6 because

0.414 < 0.550 < 0.732

-- crystal structure is NaCl

Data from Table 3.4, Callister & Rethwisch 3e.

Example Problem: Predicting the Crystal Structure of FeO

Ionic radius (nm)

0.053

0.077

0.0690.100

0.140

0.181

0.133

Cation

Anion

Al3+

Fe2+

Fe3+

Ca2+

O2-

Cl-

F-Chapter 3 - 26

MgO and FeO

O2- rO = 0.140 nm

Mg2+ rMg = 0.072 nm

rMg/rO = 0.514

∴ cations prefer octahedral sites

So each Mg2+ (or Fe2+) has 6 neighbor oxygen atoms

Adapted from Fig. 3.5, Callister & Rethwisch 3e.

MgO and FeO also have the NaCl structure

Chapter 3 - 27

939.0181.0170.0

Cl

Cs ==−

+

r

r

Cesium Chloride structure:

∴ Since 0.732 < 0.939 < 1.0, cubic sites preferred

So each Cs+ has 8 neighbor Cl-

Chapter 3 - 28

AX2 Crystal Structures

• Calcium Fluorite (CaF2)• Cations in cubic sites

• UO2, ThO2, ZrO2, CeO2

• Antifluorite structure –positions of cations and anions reversed

Fluorite structure

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Chapter 3 - 29

ABX3 Crystal Structures

• Perovskite structure

Ex: complex oxide

BaTiO3

Chapter 3 - 30

Density Computations for Ceramics

A

AC )(NV

AAn

C

Σ+Σ′=ρ

Number of formula units/unit cell

Volume of unit cell

Avogadro’s number

= sum of atomic weights of all anions in formula unit ΣAA

ΣAC = sum of atomic weights of all cations in formula unit

Chapter 3 - 31

Exp: Density of NaclrNa+ = 0.102 nm

rCl- = 0.181 nm

ANa = 22.99 g/mol

ACl = 35.45 g/mol

Chapter 3 - 32

Densities of Material Classesρmetals > ρceramics > ρpolymers

Why?

Data from Table B.1, Callister & Rethwisch, 3e.

ρ (g/

cm

)3

Graphite/ Ceramics/ Semicond

Metals/ Alloys

Composites/ fibersPolymers

1

2

20

30Based on data in Table B1, Callister

*GFRE, CFRE, & AFRE are Glass,Carbon, & Aramid Fiber-ReinforcedEpoxy composites (values based on60% volume fraction of aligned fibers

in an epoxy matrix).10

345

0.30.40.5

Magnesium

Aluminum

Steels

Titanium

Cu,Ni

Tin, Zinc

Silver, Mo

TantalumGold, WPlatinum

GraphiteSilicon

Glass -sodaConcrete

Si nitrideDiamondAl oxide

Zirconia

HDPE, PSPP, LDPE

PC

PTFE

PETPVCSilicone

Wood

AFRE*

CFRE*

GFRE*

Glass fibers

Carbon fibers

Aramid fibers

Metals have...• close-packing

(metallic bonding)• often large atomic masses

Ceramics have...• less dense packing• often lighter elements

Polymers have...• low packing density

(often amorphous)• lighter elements (C,H,O)

Composites have...• intermediate values

In general

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Chapter 3 - 33

Silicate CeramicsMost common elements on earth are Si & O

• SiO2 (silica) polymorphic forms are quartz, crystobalite, & tridymite

• The strong Si-O bonds lead to a high melting temperature (1710ºC) for this material

Si4+

O2-

Adapted from Figs. 3.10-11, Callister & Rethwisch 3e crystobalite

Chapter 3 - 34

Bonding of adjacent SiO44- accomplished by the

sharing of common corners, edges, or faces

Silicates

Mg2SiO4 Ca2MgSi2O7

Adapted from Fig. 3.12, Callister & Rethwisch 3e.

Presence of cations such as Ca2+, Mg2+, & Al3+

1. maintain charge neutrality, and2. ionically bond SiO4

4- to one another

Chapter 3 - 35

• Quartz is crystallineSiO2:

• Basic Unit: Glass is noncrystalline (amorphous)• Fused silica is SiO2 to which no

impurities have been added• Other common glasses contain

impurity ions such as Na+, Ca2+, Al3+, and B3+

(soda glass)

Adapted from Fig. 3.41, Callister & Rethwisch 3e.

Glass Structure

Si04 tetrahedron4-

Si4+

O2-

Si4+

Na+

O2-

Chapter 3 - 36

Layered Silicates• Layered silicates (e.g., clays, mica, talc)

– SiO4 tetrahedra connected together to form 2-D plane

• A net negative charge is associated with each (Si2O5)2- unit

• Negative charge balanced by adjacent plane rich in positively charged cations

Adapted from Fig. 3.13, Callister & Rethwisch 3e.

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Chapter 3 - 37

Polymorphism

• Two or more distinct crystal structures for the same material (allotropy/polymorphism)

titaniumα, β-Ti

carbondiamond, graphite

BCC

FCC

BCC

1538ºC

1394ºC

912ºC

δδδδ-Fe

γγγγ-Fe

αααα-Fe

liquid

iron system

Chapter 3 - 38

Polymorphic Forms of CarbonDiamond– tetrahedral bonding of

carbon• hardest material known• very high thermal

conductivity

– large single crystals –gem stones

– small crystals – used to grind/cut other materials

– diamond thin films• hard surface coatings –

used for cutting tools, medical devices, etc.

Adapted from Fig. 3.16, Callister & Rethwisch 3e.

Chapter 3 - 39

Polymorphic Forms of Carbon (cont)

Graphite– layered structure – parallel hexagonal arrays of

carbon atoms

– weak van der Waal’s forces between layers– planes slide easily over one another -- good

lubricant

Adapted from Fig. 3.17, Callister & Rethwisch 3e.

Chapter 3 - 40

Polymorphic Forms of Carbon (cont)Fullerenes and Nanotubes

• Fullerenes – spherical cluster of 60 carbon atoms, C60

– Like a soccer ball • Carbon nanotubes – sheet of graphite rolled into a tube

– Ends capped with fullerene hemispheres

Adapted from Figs. 3.18 & 3.19, Callister & Rethwisch 3e.

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Chapter 3 - 41

Fig. 3.20, Callister & Rethwisch 3e.

Crystal Systems

7 crystal systems

14 crystal lattices

Unit cell: smallest repetitive volume which contains the complete lattice pattern of a crystal.

a, b, and c are the lattice constants

Chapter 3 - 42

Chapter 3 - 43

Point Coordinates

Chapter 3 - 44

Example:

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Chapter 3 - 45

Example: For the unit cell shown, locate the point having coordinates ¼ 1 ½.

Chapter 3 - 46

Crystallographic Directions

1. Vector repositioned (if necessary) to pass through origin.

2. Read off projections in terms of unit cell dimensions a, b, and c

3. Adjust to smallest integer values4. Enclose in square brackets, no commas

[uvw]

ex: 1, 0, ½ => 2, 0, 1 => [ 201]

-1, 1, 1

families of directions <uvw>

z

x

Algorithm

where overbar represents a negative index

[ 111]=>

y

Chapter 3 - 47

Example:

Chapter 3 - 48

Example: Draw a [1 1 0] direction within a cubic unit cell.

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Chapter 3 - 49

Example:

Chapter 3 - 50

ex: linear density of Al in [110] direction

a = 0.405 nm

Linear Density

• Linear Density of Atoms ≡ LD =

a

[110]

Unit length of direction vector

Number of atoms

# atoms

length

13.5 nma2

2LD −==

Chapter 3 - 51

HCP Crystallographic Directions

1. Vector repositioned (if necessary) to pass through origin.

2. Read off projections in terms of unitcell dimensions a1, a2, a3, or c

3. Adjust to smallest integer values4. Enclose in square brackets, no commas

[uvtw]

[ 1120]ex: ½, ½, -1, 0 =>

Adapted from Fig. 3.24(a), Callister & Rethwisch 3e.

dashed red lines indicate projections onto a1 and a2 axes a1

a2

a3

-a3

2

a2

2a1

-a3

a1

a2

zAlgorithm

Chapter 3 - 52

HCP Crystallographic Directions• Hexagonal Crystals

– 4 parameter Miller-Bravais lattice coordinates are related to the direction indices (i.e., u'v'w') as follows.

==

=

'wwt

v

u

)vu( +-

)'u'v2(3

1-

)'v'u2(3

1-=

]uvtw[]'w'v'u[ →

Fig. 3.24(a), Callister & Rethwisch 3e.

-a3

a1

a2

z

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Chapter 3 - 53

Example:

Chapter 3 - 54

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.

• Algorithm1. Read off intercepts of plane with axes in

terms of a, b, c2. Take reciprocals of intercepts3. Reduce to smallest integer values4. Enclose in parentheses, no

commas i.e., (hkl)

Chapter 3 - 55

Crystallographic Planes

Adapted from Fig. 3.25, Callister & Rethwisch 3e.

Chapter 3 - 56

Crystallographic Planesz

x

ya b

c

4. Miller Indices (110)

example a b cz

x

ya b

c

4. Miller Indices (100)

1. Intercepts 1 1 ∞2. Reciprocals 1/1 1/1 1/∞

1 1 03. Reduction 1 1 0

1. Intercepts 1/2 ∞ ∞2. Reciprocals 1/½ 1/∞ 1/∞

2 0 03. Reduction 2 0 0

example a b c

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Chapter 3 - 57

Crystallographic Planesz

x

ya b

c•

••

4. Miller Indices (634)

example1. Intercepts 1/2 1 3/4

a b c

2. Reciprocals 1/½ 1/1 1/¾2 1 4/3

3. Reduction 6 3 4

(001)(010),

Family of Planes {hkl}

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

Chapter 3 - 58

Example: Determine the Miller indices for the plane shown.

Chapter 3 - 59

Crystallographic Planes (HCP)

• In hexagonal unit cells the same idea is used

example a1 a2 a3 c

4. Miller-Bravais Indices (1011)

1. Intercepts 1 ∞ -1 12. Reciprocals 1 1/∞

1 0 -1-1

11

3. Reduction 1 0 -1 1

a2

a3

a1

z

Adapted from Fig. 3.24(b), Callister & Rethwisch 3e.

Chapter 3 - 60

Crystallographic Planes

• We want to examine the atomic packing of crystallographic planes

• Iron foil can be used as a catalyst. The atomic packing of the exposed planes is important.

a) Draw (100) and (111) crystallographic planes for Fe.

b) Calculate the planar density for each of these planes.

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Chapter 3 - 61

Planar Density of (100) IronSolution: At T < 912°C iron has the BCC structure.

(100)

Radius of iron R = 0.1241 nm

R3

34a =

Adapted from Fig. 3.2(c), Callister & Rethwisch 3e.

2D repeat unit

= Planar Density =a2

1

atoms2D repeat unit

= nm2

atoms12.1

m2

atoms= 1.2 x 1019

12

R3

34area

2D repeat unitChapter 3 - 62

Planar Density of (111) IronSolution (cont): (111) plane 1 atom in plane/ unit surface cell

333 2

2

R3

16R3

42a3ah2area =

===

atoms in plane

atoms above plane

atoms below plane

ah2

3=

a2

1

= = nm2

atoms7.0m2

atoms0.70 x 1019

3 2R3

16Planar Density =

atoms2D repeat unit

area2D repeat unit

Chapter 3 - 63

• Some engineering applications require single crystals:

• Properties of crystalline materials often related to crystal structure.

(Courtesy P.M. Anderson)

-- Ex: Quartz fractures more easily along some crystal planes than others.

-- diamond singlecrystals for abrasives

-- turbine blades

Fig. 9.40(c), Callister & Rethwisch 3e. (Fig. 9.40(c) courtesy of Pratt and Whitney).

(Courtesy Martin Deakins,GE Superabrasives, Worthington, OH. Used with permission.)

Crystals as Building Blocks

Chapter 3 - 64

Polycrystals

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Chapter 3 - 65

• Single Crystals-Properties vary withdirection: anisotropic.

-Example: the modulusof elasticity (E) in BCC iron:

• Polycrystals

-Properties may/may notvary with direction.

-If grains are randomlyoriented: isotropic.(Epoly iron = 210 GPa)

-If grains are textured,anisotropic.

200 µm

Single vs PolycrystalsE (diagonal) = 273 GPa

E (edge) = 125 GPa

Chapter 3 - 66

X-Ray Diffraction

• Diffraction gratings must have spacings comparable to the wavelength of diffracted radiation.

• Can’t resolve spacings < λ• Spacing is the distance between parallel planes of

atoms.

Chapter 3 - 67

X-Rays to Determine Crystal Structure

X-ray intensity (from detector)

θ

θc

d = nλ2 sin θc

Measurement of critical angle, θc, allows computation of planar spacing, d.

• Incoming X-rays diffract from crystal planes.

Adapted from Fig. 3.37, Callister & Rethwisch 3e.

reflections must be in phase for a detectable signal

spacing between planes

d

θλ

θextra distance travelled by wave “2”

Chapter 3 - 68

X-Ray Diffraction Pattern

Adapted from Fig. 3.20, Callister 5e.

(110)

(200)

(211)

z

x

ya b

c

Diffraction angle 2θ

Diffraction pattern for polycrystalline α-iron (BCC)

Inte

nsity

(re

lativ

e)

z

x

ya b

cz

x

ya b

c

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Chapter 3 - 69

• Atoms may assemble into crystalline or amorphous 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).

SUMMARY

• 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.

• 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.

• Ceramic crystal structures are based on:-- maintaining charge neutrality-- cation-anion radii ratios.

• Interatomic bonding in ceramics is ionic and/or covalent.

Chapter 3 - 70

• Some materials can have more than one crystal structure. This is referred to as polymorphism (or allotropy).

SUMMARY

• Materials can be single crystals or polycrystalline. Material properties generally vary with single crystal orientation (i.e., they are anisotropic), but are generally non-directional (i.e., they are isotropic) in polycrystals with randomly oriented grains.

• X-ray diffraction is used for crystal structure and interplanar spacing determinations.