pertemuan ii struktur kristal.pptx
TRANSCRIPT
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Istilah Material kristal adalah material padat dimana atom-
atomnya tersusun dalam susunan yang berulang . Contoh: Semua logam, sebagian besar keramik, beberapa polymer.
Non kristalin atau amorphous adalah material yang tidak memiliki keteraturan atom dalam jarak panjang
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Istilah Lattice/kisi: susunan berulang dari titik-titik yang
menunjukkan jarak
Unit cell/sel satuan: sub bagian kisi yang masih memiliki karakteristik kisi atau memiliki kesamaan sifat dengan seluruh kisi kristal.
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7 Crystal Systems:
aa
a
Cubic: a = b = c; = = = 90°
ba
c Tetragonal: a =b c = = = 90°
ab
c
Orthorhombic: a b c = = = 90°
Rhombohedral: a = b = c 90°
a
cHexagonal: a=bc = = 90°; = 120°
Monoclinic: abc = = 90°
Triclinic: a b c 90°
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• Sangat jarang karena rendahnya densitas penumpukan hanya Polonium yang memiliki struktur ini
• Arah penumpukan pada sisi kubus.
• Bilangan koordinasi= 6 (# jumlah atom terdekat) untuk tiap atom
(Courtesy P.M. Anderson)
Struktur Kristal LogamStruktur kubus sederhana/Simple Cubic Structure (SC)
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• APF untuk struktur simple cubic = 0.52
APF = a3
4
3 p(0.5a)31
atom
unit selatom
volume
unit selvolume
Atomic Packing Factor (APF)
APF = Volume atom pada unit sel*
Volume unit cell
*assumsi bola padat
Adapted from Fig. 3.23, Callister 7e.
close-packed directions
a
R=0.5a
Isi (8 x 1/8) =
1 atom/unit selDengan : a = Rat*2
Where Rat is the ‘handbook’ atomic radius
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• Bilangan koordinasi # = 8
(Courtesy P.M. Anderson)
• Atom-atom saling bersentuhan disepanjang diagonal.--Note: Atom-atom tersebut identik
Struktur Body Centered Cubic (BCC)
contoh: Cr, W, Fe (), Tantalum, Molybdenum
2 atom/unit sel: (1 pusat) + (8 sudut x 1/8)
Struktur Kristal Logam
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Atomic Packing Factor: BCC
a
APF =
4
3p ( 3a/4)32
atom
Unit sel atomvolume
a3unit sel
volume
panjang = 4R =Arah penumpukan:
3 a
• APF untuk struktur body-centered cubic = 0.68
aR
Adapted from Fig. 3.2(a), Callister 7e.
a 2
a 3
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• Bilangan koordinasi# = 12
(Courtesy P.M. Anderson)
• Atom saling bersentuhan disepanjang diagonal bidang.--Note: atom-atom identik
Struktur Face Centered Cubic (FCC)
contoh: Al, Cu, Au, Pb, Ni, Pt, Ag
4 atom/unit sel: (6 face x ½) + (8 corners x 1/8)
Struktur Kristal Logam
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TABEL KONSTANTA KISI (a) DAN RADIUS ATOM (R) STRUKTUR BCC
Metal a (nm) R (nm)
Chromium 0,289 0,125
Iron 0,287 0,124
Molybdenum 0,315 0,136
Potassium 0,533 0,231
Sodium 0,429 0,186
Tantalum 0,33 0,143
Tungsten 0,316 0,137
Vanadium 0,304 0,312
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• APF struktur face-centered cubic = 0.74
Atomic Packing Factor: FCC
Maximum APF dari beberapa struktur logam
APF =
4
3 p ( 2a/4)34
atom
unit sel atomvolume
a3unit sel
volume
Arah penumpukan: panjang = 4R =2 a
Isi unit sel: 6 x 1/2 + 8 x 1/8 = 4 atom/unit sela
2 a
Adapted fromFig. 3.1(a),Callister 7e.
(a = 22*R)
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TABEL KONSTANTA KISI (a) DAN RADIUS ATOM (R) STRUKTUR FCC
Metal a (nm) R (nm)
Aluminium 0,405 0,143
Copper 0,3515 0,128
Gold 0,408 0,144
Lead 0,495 0,175
Nickel 0,352 0,125
Platinum 0,393 0,139
Silver 0,409 0,144
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• Bilangan koordinasi# = 12
• ABAB... Pola penumpukan berulang
• APF = 0.74
• Tampak 3D • Tampak 2D
Adapted from Fig. 3.3(a), Callister 7e.
Struktur Hexagonal Close-Packed (HCP)
6 atom/unit sel
contoh: Cd, Mg, Ti, Zn
• c/a = 1.633 (ideal)
c
a
A sites
B sites
A sitesBottom layer
Middle layer
Top layer
Struktur Kristal Logam
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Theoretical Density, r
where n = number of atoms/unit cell A = atomic weight VC = Volume of unit cell = a3 for cubic NA = Avogadro’s number = 6.023 x 1023 atoms/mol
Density = =
VC NA
n A =
Cell Unit of VolumeTotalCell Unit in Atomsof Mass
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Theoretical Density, r
• Ex: Cr (BCC) A = 52.00 g/mol R = 0.125 nm n = 2
a = 4R/3 = 0.2887 nma
R
= a3
52.002atoms
unit cell molg
unit cell
volume atoms
mol
6.023 x
1023
theoretical
ractual
= 7.18 g/cm3
= 7.19 g/cm3
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Posisi Atom dalam Unit Sel
Identifikasi posisi berdasarkan sistem sumbu pada kubus sumbu x+ adalah sumbu yang tegak lurus bidang gambar, sumbu y+ kearah kanan kertas, dan sumbu z+ kearah atas kertas.Titik tengah unit sel adalah
a/2, b/2, c/2 ½ ½ ½
Posisi pada pojok diagonal kisi adalah 111
z
x
ya b
c
000
111
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Crystallographic Directions: garis diantara dua titik atau sebuah vektor
1. Vektor dipindahkan sehingga melalui titik origin unit sel.
2. Baca proyeksi vektor untuk tiap-tiap sumbu pada unit sel x, y, dan z dalam satuan a, b, c.
3. Ketiga bilangan dikalikan atau dibagi sehingga menghasilkan bilangan bulat terkecil diantara ketiganya
4. Isikan bilangan tersebut diantara dua kurung siku tanpa koma [uvw]
ex: 1, 0, ½
=> 2, 0, 1=> [ 201
]-1, 1, 1
z
x
Metode penentuan arah kristal
Jika bernilai negatif, tanda negatif ditulis di bagian atas angka
[ 111 ]=>
y
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What is this Direction ?????
Proyeksi dalam a, b, c:Proyeksi:Reduksi:
Diberikan kurung []
x y z
a/2 b 0c
1/2 1 0
1 2 0
[120]
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ex: linear density of Al in [110] direction
a = 0.405 nm
Linear Density – considers equivalance and is important in Slip
Linear Density of Atoms LD =
a
[110]
Unit length of direction vector
Number of atoms
# atoms
length
13.5 nma2
2LD -==
# atoms CENTERED on the direction of interest!Length is of the direction of interest within the Unit Cell
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Determining Angles Between Crystallographic Direction:
1 1 2 1 2 1 2
2 2 2 2 2 21 1 1 2 2 2
u u v v w wCos
u v w u v w
Where ui’s , vi’s & wi’s are the “Miller Indices” of the directions in question
– also (for information) If a direction has the same Miller Indicies as a plane, it is NORMAL to that plane
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HCP Crystallographic Directions
1. Pindahkan vektor sehingga melewati origin.
2. Baca proyeksinya dalam dimensi unit sel a1, a2, a3, or c
3. Sesuaikan hingga menjadi bilangan bulat terkecil
4. Tambahkan kurung siku, tanpa koma
[uvtw]
[ 1120 ]ex: ½, ½, -1, 0 =>
Adapted from Fig. 3.8(a), Callister 7e.
dashed red lines indicate projections onto a1 and a2 axes a1
a2
a3
-a3
2
a 2
2a1
-a3
a1
a2
z Algorithm
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Bidang Kristal
Indeks Miller: Kebalikan dari perpotongan bidang di tiga sumbu. Bidang paralel memiliki indeks Miller yang sama.
Algorithma1. Baca perpotongan bidang pada ketiga sumbu
dalam a, b, c2. Ubah menjadi kebalikan nilainya3. Kurangi menjadi bilangan bulat terkecil4. Beri kurung, tanpa koma contoh: (hkl) families
{hkl}
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Crystallographic Planesz
x
ya b
c
4. Miller Indices (110)
example a b cz
x
ya b
c
4. Miller Indices (200)
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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Crystallographic Planes
z
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),
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x y z Intercepts Intercept in terms of lattice parameters Reciprocals Reductions
Enclosure
a -b c/2 -1 1/2
0 -1 2N/A
(012)
Determine the Miller indices for the plane shown in the accompanying sketch (a)
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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. 3.8(a), Callister 7e.
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Planar Density of (100) Iron
Solution: At T < 912C iron has the BCC structure.
(100)
Radius of iron R = 0.1241 nm
R334
a=
Adapted from Fig. 3.2(c), Callister 7e.
2D repeat unit
= Planar Density = a2
1
atoms2D repeat unit
= nm2
atoms12.1
m2
atoms= 1.2 x 1019
12
R3
34area2D repeat unit
Atoms: wholly contained and centered in/on plane within U.C., area of plane in U.C.
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Planar Density of (111) IronSolution (cont): (111)
plane1/2 atom centered on plane/ unit cell
atoms in plane
atoms above plane
atoms below plane
ah2
3=
a 2
2D rep
eat un
it
3*1/6= =
nm2
atoms7.0m2
atoms0.70 x 1019Planar Density =
atoms2D repeat unit
area2D repeat unit
28
3
R
Area 2D Unit: ½ hb = ½*[(3/2)a][(2)a]=1/2(3)a2=8R2/(3)
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Section 3.16 - X-Ray Diffraction
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.
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Figure 3.32 Relationship of the Bragg angle (θ) and the experimentally measured diffraction angle
(2θ).
X-ray intensity (from detector)
qqc
d = n l2 sinqc
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X-Rays to Determine Crystal Structure
2 2 2hkl
ad
h k l
X-ray intensity (from detector)
q
qc
d =n l
2 sinqc
Measurement of critical angle, qc, allows computation of planar spacing, d.
• Incoming X-rays diffract from crystal planes.
Adapted from Fig. 3.19, Callister 7e.
reflections must be in phase for a detectable signal!
spacing between planes
d
incoming
X-rays
outg
oing
X-ra
ys
detector
ql
qextra distance traveled by wave “2”
“1”
“2”
“1”
“2”
For Cubic Crystals:
h, k, l are Miller Indices
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Figure 3.34 (a) An x-ray diffractometer. (Courtesy of Scintag, Inc.) (b) A schematic of the experiment.
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X-Ray Diffraction Pattern
Adapted from Fig. 3.20, Callister 5e.
(110)
(200)
(211)
z
x
ya b
c
Diffraction angle
2q
Diffraction pattern for polycrystalline a-iron (BCC)
Inte
nsi
ty (
rela
tive)
z
x
ya b
cz
x
ya b
c
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• Atom tersusun menjadi kristal atau non kristalin (amorphous).
• Densitas material dapat dipekirakan jika diketahui berat atom, radius atom, dan geometri kristal (e.g., FCC, BCC, HCP).
SUMMARY
• Struktur kristal logam yang umum FCC, BCC, dan HCP. Bilangan koordinasi dan APF FCC dan HCP sama.
• Titik, arah, dan bidang pada kristal ditunjukkan dengan indeks.
Arah kristal berkaitan dengan linear densities dan bidang kristal berkaitan dengan planar densities.