Post on 06-Feb-2022
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§ 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