3 The structure of crystalline solids ()

Why important : Properties of materials are directly related to their crystal structure Ex. C graphite Speciality - Diamond The unique of MSE

3.2 Fundamental Concepts

Crystalline material is one in which the atoms are situated in a repeating or periodic array over large atomic distance (periodic structure in 3-D) Non-crystalline / Amorphous : glass, polymer() crystalline structure, we use Atomic hard sphere model Fig 3.2 Well-defined bonding , but in real world, there is no clear bonding!

Lattice Three-dimensional array of points coinciding with atom position ( or sphere centers) show fig. 3.1

3.3 Unit cell

The smallest repetitive volume that comprises the complete lattice pattern of a crystal.

3.4 Metallic crystal structures BCC (Body-centered cubic)

Or FCC (Face-centered cubic)

. Coordinated Number (CN) The number of nearest neighbor or touching atoms. Ex CN

3.7 Crystal systems lattice ?

1. cubic a=b=c ===90 (a) Simple cubic (S.C)

1 atom/unit cell 8*1/8 CN : 6 P (primitive ) One lattice point(atom)/unit cell

(b) Face-centered cubic (FCC)

4 atoms/unit cell 1/8*8 + 1/2*6 = 4 CN : 12 Ex. Al, Cu, Ni, Pb

Question How does E in figure as shown relate to & melting point.

Answer

2

2

rU

rFE

=

E Melting , , E

7 crystal systems 14 braveries lattice

Review structure Unit cell CN 7 crystal system 14 braveries lattices

1st system

Cubic crystals (a) Simple cubic (P)

1 atom/cell CN = 6

(b) Face-centered Cubic (FCC) 4 atoms/cell

CN = 12 (c) Body-centered cubic (BCC)

2 atoms/unit cell 1/8*8 + 1 = 2 I (interior) Ex. Cr, Fe(), K

2nd system

Tetragonal () a = b c ===90 P simple Body-centered Ex. In, B, Sn(white )

3rd system

Orthorhombic a b c === 90 P, F, I, C Base-centered Ex. I, P

4th system

Monoclinic a b c == 90 P, C

mp E Hg Pb Al Cu Fe W

SC

FCC BCC

5th system Triclinic a b c 90

6th system

Hexagonal a = b c == 90, = 120 P Ex. C(graphite) c/a = 1.63

7th system

Rhombohedral a = b = c == 90 P Ex. Hg

Conclusion

7 crystal system 14 bravais lattice

Atomic Packing Factor (AFP) ()

volumecellunit totalcellunit ain atoms of volumeAPF =

(a) BCC 3 a = 4r , a = 4r/3 Two atoms/ u.c , CN = 8

%68)3/4(

3/42a

r4/32APF3

3

3

3

=

=

=r

1. Cubic a = b = c === 90 P. F. I 2. Tetragonal a =b c === 90 P. I 3. Orthorhombic a b c === 90 P. F. I. C 4. Rhombohedral a = b = c == 90 P 5. Hexagonal a =b c == 90, = 120 P 6. Monoclinic a b c == 90 P 7. Triclinic a b c 90 P

Total 14

BCC

(b) FCC Four atoms/ u.c , CN = 12

ra 42 = 2

4ra =

%74a

r4/34APF3

3

=

=

FCC , ()

Metallic Crystal Structure : FCC, BCC, HCP 3.5 Density computations

Know crystal structure find true density

Ac NVnA

=

n# of atom/ u.c Aatomic weight Vcvolume of unit cell NAAvogadros number

Ex. 3.3 Cu, which R = 0.128nm, FCC A = 63.5 g/mole Ans

222 4Raa =+ 222 Ra =

( ) 21622 333 RRaVc === ( ) ( )

( )[ ] ( )[ ] ( )3

2338Ac

/89.8/10023.6/1028.1216

/5.63../4NV

cmgmolatomsuccm

molgcuatomsnA=

==

Literature value 8.94 g/cm3, very low. unit cell

( )[ ]381028.116 cm Volume 1 cm3 ?

( )( )[ ]

2238

3

1051028.116

1=

cmcm per unit cell doping , doping core

HW #1Primitive cell of HCP

3.9 Crystallographic directions 3.10 Crystallographic planes

1. Location 2. Direction 3. Plane

1. Location ( coordination ) : unit translation , should not separate by Columns 0 0 0 , 1/2 1/2 1/2 atoms

0 1 0 , 1/2 3/2 1/2 atoms translation

( I ) Translation in simple cubic 0 0 0 ( atomic position) ( II ) Translation in BCC 0 0 0 1/2 1/2 1/2 ( III ) FCC 0 0 0 1/2 1/2 0 1/2 0 1/2 0 1/2 1/2 1/2 1/2 1/2 Direction [ -1 -1 0 ] Plane ( 1 1 0 )

2. Crystal Directions

[ u v w ] pass through the origin projection on each of the three axis reduce them to the smallest integers

Ex. 3.4

x y z Step1 projection a/2 b 0c Projection (in terms of a, b, &c) 1/2 1 0 Reduction 1 2 0 Enclosure [1 2 0]

FCC

BCC

b

a c

3. Crystal planes (Miller indices) (h k l) For cubic crystals

normal to the plane {h k l} is [h k l] [1 0 0] {1 0 0} [1 1 0] {1 1 0} [1 1 1] {1 1 1}

Form < u v w > family [ 1 0 0 ], [ -1 0 0 ], [ 0 -1 0 ], [0 1 0], [0 0 1], [0 0 -1] For cubic crystals, they are equivalent. For Tetragonal

a = b c , === 90 [1 0 1] = [0 1 1] [1 1 0]

For hexagonal[u v w] or [u v t w] u + v = -t a1 a2 a3 a1 a2 a3 c not projection a1 a2 a3projection

projection 1, -1/2, -1/2, 0

]0 0 1[]0 1 1 2[1 =a

]0 1 2 1[2 =a

]0 2 1 1[3 =a

]1 0 0 0[c =

EX. ]0 1 2 1[]0 1 0[

a1projection 23

a2projection 0

a3projection 23

c 0

[ ]0 1 0 10 , 23 , 0 ,

23

a1

a2 a3

Plane-view

relationship between [U V W] & [u v t w] ( )

32u'-v'nu = t-uu'=

( )3

2v'-u'nv = t-vv'=

v)-(ut += ww'= nWw = n = a factor that may be required to reduce u,

v, t, w to the smallest integer.

3.11 Linear & planar atomic density

(a) Linear density (LD) Ex. 3.8 BCC in [1 0 0] Ans atomic centers

34RaLl ==

866.034

2==

RRL D

(b) Planar Density (PD) fig. 3.9 of the (1 1 0)

Cross-sectional view

RAC 4= 22RAD =

Unit cell plane area, Ap

( )( ) 28224 2RRRADACAp === Total circle area, Ac

22 RAc =

555.028

22

2

===R

RAPACPD

a

How to find h k l ? procedure for finding (h k l) (1) if he plane pass original cornor (2) determine intersection for each axis in term of a b c. (3) take reciprocals of the number (4) change to the set of smallest integers (5) (h k l) no commas

(0 0 1) (0 0 -1) For hexagonal xtals (h k I l) h + k = -i (0 0 1) (0 0 2) [0 0 1] (0 0 2)

3.12 Close-packed crystal structure

FCC & HCP : the most efficient packing of eqial-sized sphere or atoms. 74% In FCC structure, the most close-packed plane is (1 1 1) Difference between FCC & HCP

(a) For HCP A B A B A B A B. The atoms of 3rd layer are aligned directly above the original positions. Atomic alignment repeats every second plane View graph for Fig 3.12, 3.13, see fig 3.3

Fig 3.12

Fig 3.13

(b) For FCC A B C A B C A B C Atomic alignment repeats every third plane . View graph 3.12, 3.14 see 33-1

Fig 3.14

Crystalline & non-crystalline materials How to measure vacancy ?

3.13 single crystal

Single crystal () : periodic arrangement of atoms extends throughout the entire specimen without interruption. Ex. Single crystal Si Show TEM picture

Composed of a collection of many small crystals or grains() Grain boundaries() View graph 3.16

3.14 Polycrystalline materials ()

Isotropic () : Anisotropy () : Texture (, preferential orientation) , texture 3.16 X-ray Diffraction : Determination of crystal structure

(1) X-ray : (wavelength) (electromagnetic wave)

chhE ==

Why use x-ray to determine crystal structure? OM resolution (wavelength) x-ray

How to generate x-ray?

i. kinetic energy hmveVkE === 221

target 99 % kE into transfer heat 1 % kE into transfer x-ray

ii. swl (Shortest wavelength limit) = min = eVhc

hhcc

max

==

(All the eV transfer to target) mV

101.24V101.602

m 103106.63 -619-

8-34 =

=

Voltage = 20 ~ 50 kV iii. Characteristic lines (Mo)

(2) Diffraction : Braggs law() For constructive diffraction

Satisfies QTSQn +=

beam 2 beam 1

QTSQ +

sinhkldQTSQ ==

sin2sinsin hklhklhkl dddn =+= In x-ray diffraction experiment known : measurable Ex. For a BCC crystal d110 = 1.181 k = 1.54

o

dn 8.40

181.1254.11sin

2sin 11 =

==

2 = 81.6 peak

For a BCC crystals, only h+k+l= 2n peak (1 0 0), (1 1 0), (1 1 1), (2 0 0)

For a FCC crystal, h k l all odd or even (1 0 0), (1 1 0), (1 1 1), (2 0 0), (1 3 1), (2 1 0)

For a cubic crystal, 222 lkh

adhkl++

= (distance between (h k l))

Prove! See Fig. 3.19, 3.20

x-ray (1) lattice constant (2) crystal structure (bcc or fcc,.) (3) chemical composite (eg. AB or A2B)

Prove that, 222 lkh

adhkl++

=

hkldha

=acos hkldkb

=bcos hkldlc

=cos

hkldah

=acos hkldbk

=bcos hkldcl

=cos

1coscoscos 222 =++ ba

1)( 222

2

2

2

2

=++ hkldcl

bk

ah

21

2

2

2

2

2

2

)(

1

cl

bk

ah

dhkl++

=

For cubic system 222 lkh

adhkl++

=

3.16 Non-crystalline solids (Amorphous)

Non-crystalline solids lack to systemic and regular arrangement of atoms over relatively large atomic distance.

Ex. Super-cooled liquids Ex. SiO2 may exist in crystalline and amorphous states. Crystalline (quartz) Amorphous (glass) Density 2.65 g/cm3 700~1550 MP 1710

2.14 2.16 2.18 2.22 3.1 3.3 3.14 3.21 3.31 3.33 3.37 3.40