Pertemuan II Struktur Kristal.pptx

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Struktur Kristal

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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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Kisi Kristal

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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


• 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


• 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)


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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


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


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TABEL KONSTANTA KISI (a) DAN RADIUS ATOM (R) STRUKTUR HCP

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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 =>


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 Planes -- families


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Crystallographic Planesz

x

ya b

c

4. Miller Indices (110)

example a b cz

x

ya b

c


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


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


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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.


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


(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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Diffraction in Cubic Crystals:

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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.

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SUMMARY

• Material tersusun dalam single crystals atau polycrystalline. • Difraksi sinar X digunakan untuk menentukan jenis

bidang struktur kristal.

Pertemuan II Struktur Kristal.pptx

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