§ 3 The structure of crystalline solids (固態晶體的結構)

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(高分子) 非晶形 無定型 為了描述 crystalline structure, we use Atomic hard sphere model → Fig 3.2 Well-defined bonding , but in real world, there is no clear bonding!

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

§ 3.3 Unit cell 單位晶格

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)

只要這四個原子的排列 就可以知道其它原子的排列. Coordinated Number 配位數 (CN) The number of nearest neighbor or touching atoms. 以學生座位當 Ex 跟堆積密度有關 → CN 越大 → 越寬

§ 3.7 Crystal systems 如何描述 lattice ?

1. cubic 立方體 a=b=c α=β=γ=90° (a) Simple cubic (S.C)

1 atom/unit cell 8*1/8 CN : 6 P (primitive ) 原始的 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 α & melting point.

Answer：

2

2

rU

rFE

∂∂

∝∂∂

∝=∂∂εσ

斜率大 曲率大 E 大 深、陡曲線→Melting ↑, α ↓, E ↑

7 crystal systems 14 braveries lattice

Review structure 單位晶格 Unit cell 配位電子 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) 體心立方

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

2nd system

Tetragonal (正方體) a = b ≠ c α=β=γ=90° P simple Body-centered Ex. In, B, Sn(white β)

3rd system

Orthorhombic 斜方晶 a ≠ b ≠ c α=β=γ= 90° P, F, I, C Base-centered Ex. I, P

4th system

Monoclinic 單斜晶 a ≠ b ≠ c α =γ= 90°≠β P, C

mp α E Hg Pb Al Cu Fe W

SC

FCC BCC

5th system Triclinic 三斜晶 a ≠ b ≠ c α≠β≠γ≠ 90°

6th system

Hexagonal a = b ≠ c α =β= 90°, γ= 120° P Ex. C(graphite) c/a 的理論值 = 1.63

7th system

Rhombohedral 斜方六面體 a = b = c α=β=γ≠ 90° P Ex. Hg

Conclusion：

7 crystal system 14 bravais lattice

△ Atomic Packing Factor (AFP) 原子堆積(密度)因子

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

=×

=×

=rππ

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

Total 14

BCC

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

ra 42 = ， 2

4ra =

%74a

r4/34APF3

3

=×

=π

FCC 排列較密 , 較易加工(滑動)

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

Know crystal structure → find true density

Ac NVnA

=ρ

n：# of atom/ u.c A：atomic weight Vc：volume of unit cell NA：Avogadro’s 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=

××

×==

−ρ

→Literature value 8.94 g/cm3, very low. →同時也來看一個 unit cell 體積

( )[ ]381028.116 cm−××

→Volume 1 cm3 內有多少原子 ?

( )( )[ ]

2238

3

1051028.116

1×=

×× − cmcm per unit cell → doping 例子 , doping core

HW #1：Primitive cell of HCP

§ 3.9 Crystallographic directions 3.10 Crystallographic planes

1. Location 2. Direction 3. Plane

1. Location ( coordination ) : unit translation 座標不加掛號 , should not separate by Columns 若知 0 0 0 , 1/2 1/2 1/2 有兩個 atoms

則可知 0 1 0 , 1/2 3/2 1/2 也有兩個 atoms 可以 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 座標 1/2 1/2 1/2 Direction [ -1 -1 0 ] 方向 Plane ( 1 1 0 ) 平面

2. Crystal Directions

[ u v w ] 中括弧 無點 整數 負號寫在上面 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 0×c 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 , α=β=γ= 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 在a1 a2 a3之projection

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[ ⇒

a1上的projection 23

a2上的projection 0

a3上的projection 23

c 0

[ ]0 1 0 10 , 23 , 0 ,

23

⇒

a1

a2 a3

Plane-view 俯視圖

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：穿過 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 π=

555.028

22

2

===R

RAPACPD π

a

☆ How to find h k l ? → procedure for finding (h k l) (1) if he plane pass original → 到 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

平行的平面 (0 0 1) (0 0 -1) For hexagonal xtals (h k I l) h + k = -i (0 0 1) 與 (0 0 2) 的差別 [0 0 1] 與 (0 0 2) 的差別

§ 3.12 Close-packed crystal structure 最密堆積晶體結構

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 ☆ 中間有三個原子

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 (單晶) : 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(晶粒) Grain boundaries(晶界) View graph 3.16

§ 3.14 Polycrystalline materials (多晶材料)

Isotropic (均等向性的) : 材料性質和晶相方向無關 Anisotropy (非等向性) : 材料性質和晶相方向有關 Texture 的晶材料的晶粒在某特殊方向(优選方向, preferential orientation) 有特別的性質, 稱此材料有 texture § 3.16 X-ray Diffraction : Determination of crystal structure

(1) X-ray : 波長(wavelength) 在數個 Å 的電磁波(electromagnetic wave)

λν chhE ==

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

☆ How to generate x-ray?

i. kinetic energy νhmveVkE === 2

21

撞到 target 後 99 % kE → into transfer heat 1 % kE → into transfer x-ray

ii. λswl (Shortest wavelength limit) = λmin = eVhc

hhcc

max

==νν

(All the eV transfer to target) mV

101.24V101.602

m 103106.63 -6

19-

8-34 ×=

⋅××⋅×

=

Voltage = 20 ~ 50 kV 代入得λ為數個Å iii. Characteristic lines (Mo)

(2) Diffraction : Bragg’s law(布拉格定律) For constructive diffraction

Satisfies QTSQn +=λ

∵ beam 2 比 beam 1 多了

QTSQ +

θsinhkldQTSQ ==

θθθλ sin2sinsin hklhklhkl dddn =+= In x-ray diffraction experiment λ：known θ : measurable Ex. For a BCC crystal d110 = 1.181Å kα = 1.54 Å

o

dn 8.40

181.1254.11sin

2sin 11 =

××

== −− λθ

∴在 2θ = 81.6°有一個 peak

★For a BCC crystals, only h+k+l= 2n 有 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 可測 (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

=γcos

→∵ hkldah

=acos hkldbk

=bcos hkldcl

=γcos

∴ 1coscoscos 222 =++ γba

1)( 22

2

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