Prof. Drs.H.Darsono, M.Sc

Transkripsi

1 Prof. Drs.H.Darsono, M.Sc FMIPA UNIVERSITAS AHMAD DAHLAN)

2 Crystal structure coordination # packing factor close packed directions Simple Cubic (SC) cube edges Body Centered Cubic (BCC) body diagonal Face Centered Cubic (FCC) face diagonal Hexagonal Close Pack (HCP) hexagonal side

3 Density = mass/volume mass = number of atoms per unit cell * mass of each atom mass of each atom = atomic weight/avogadro s number # atoms/unit cell Atomic weight (g/mol) Volume/unit cell (cm 3 /unit cell) n A V c N A Avogadro's number (6.023 x atoms/mol)

4 Element Aluminum Argon Barium Beryllium Boron Bromine Cadmium Calcium Carbon Cesium Chlorine Chromium Cobalt Copper Flourine Gallium Germanium Gold Helium Hydrogen Symbol Al Ar Ba Be B Br Cd Ca C Cs Cl Cr Co Cu F Ga Ge Au He H At. Weight (amu) Atomic radius (nm) Density (g/cm 3 ) Crystal Structure FCC BCC HCP Rhomb HCP FCC Hex BCC BCC HCP FCC Ortho. Dia. cubic FCC Adapted from Table, "Characteristics of Selected Elements", inside front cover, Callister 6e.

5 # atoms/unit cell Atomic weight (g/mol) Volume/unit cell (cm 3 /unit cell) n A V c N A Avogadro's number (6.023 x atoms/mol) Example: Copper Data from Table inside front cover of Callister (see previous slide): crystal structure = FCC: 4 atoms/unit cell atomic weight = g/mol (1 amu = 1 g/mol) atomic radius R = nm (1 nm -7 = 10 cm) Vc = a 3 ; For FCC, a = 4R/ 2 ; Vc = 4.75 x cm 3 Result: theoretical Cu = 8.89 g/cm 3 Compare to actual: Cu = 8.94 g/cm 3

6 metals > ceramic s > polymer s Why? Metals have... close-packing (metallic bonding) large atomic mass Ceramics have... less dense packing (covalent bonding) often lighter elements Polymers have... poor packing (often amorphous) lighter elements (C,H,O) Composites have... intermediate values (g/cm 3 ) Metals/ Alloys Platinum Gold, W Tantalum Silver, Mo Cu,Ni Steels Tin, Zinc Titanium Aluminum Magnesium Graphite/ Ceramics/ Semicond Polymers Composites/ fibers Based on data in Table B1, Callister *GFRE, CFRE, & AFRE are Glass, Carbon, & Aramid Fiber-Reinforced Epoxy composites (values based on 60% volume fraction of aligned fibers in an epoxy matrix). Zirconia Al oxide Diamond Si nitride Glass -soda Concrete Silicon Graphite PTFE Silicone PVC PET PC HDPE, PS PP, LDPE Data from Table B1, Callister 6e. Glass fibers GFRE* Carbon fibers CFRE * Aramid fibers AFRE * Wood

7 Arah Kristal kubus Arah diberi nama berdasarkan proyeksi vektor dari titik awal suatu kristal ke titik lain didalam sel.. Gunakan sistem koordinat kartesian tangan kanan PARAMETER ARAH GRAFIK KRISTAL Posisi titik Arah garis Arah bidang.

8 Pada kristal penting untuk menyatakan titik, garis dan bidang didalam sel satuan dari kisi kristal. Ada tiga indeks digunakan untuk menyatakan titik, garis dan bidang berdasarkan notasi geometri dasar. Tiga indeks ditentukan dengan menentukan titik awal pada salah satu sudut dari sel satuan dan sumbu koordinat sepanjang sisi.

9 Setiap titik di dalam sel satuan menyatakan sebagai perkalian fraksi dari panjang sisi sel satuan. Position P specified as q r s; convention: coordinates not separated by commas or punctuation marks

10 Locate the point (1/4 1 ½) Specify point coordinates for all atom positions for a BCC unit cell Answer: 0 0 0, 1 0 0, 1 1 0, 0 1 0, ½ ½ ½, 0 0 1, 1 0 1, 1 1 1, 0 1 1

11 Titik kisi ditulis dalam bentuk h,k,l, dimana ketiga indeks berhubungan dengan fraksi parameter kisi pada arah x,y,z.

12 Pilih titik awal kisi pada garis katakanlah titik O. Pemilihan titik awal sembarang karena titik kisi identik (sifat simetri) Kemudian hubungkan vektor kisi dari titik kesembarang katakanlah T maka vektor dapat ditulis sbb; R = ha + kb + lc [... ] menyatakan arah garis pada kristal sedangkan [h k l] menyatakan titik yang merupakan bilangan bulat terkecil.

13 Ketika menulis [h k l] tergantung titik awal perjanjian, arah negatip bisa ditulis [ R ha kb lc] h k l harus lebih kecil dari nol bilangan integer

14 X = 1, Y = 0, Z = 0 [1 0 0]

15 210 X = 1, Y = ½, Z = 0 [1 ½ 0] [2 1 0]

16 Kumpulan garis ekuivalen <100> adalah [100], [010], [001]

17 Draw a [1-1 0] direction within a cubic unit cell Determine the indices for this direction Answer: [120]

18 The calculation of the miller indices using vectors proceeds in the following manner: We are given three points in a plane for which we want to calculate the Miller indices: P 1 (022), P 2 (202) and P 3 (210) We now define the following vectors: r 1 =0i+2j+2k, r 2 =2i+0j+2k, r 3 =2i+j+0k and calculate the following differences: r - r 1 =xi + (2-y)j + (2-z)k r 2 - r 1 =2i - 2j + 0k r 3 -r 1 = 2i j - 2k We then use the fact that: (r-r 1 ). [(r 2 -r 1 ) (r 3 -r 1 )] =A. (B C)= 0

19 We now use the following matrix representation, that gives The end result of this manipulation is an equation of the form: 4x+4y+2z=12 The intercepts are located at: x=3, y=3, z=6 The Miller indices of this plane are then: (221) ) ( z y x C C C B B B A A A C B A

20 Within a crystal lattice it is possible to identify sets of equally spaced parallel planes. These are called lattice planes. In the figure density of lattice points on each plane of a set is the same and all lattice points are contained on each set of planes. b b a a

21 William HallowesMiller in 1839 was able to give each face a unique label of three small integers, the Miller Indices Definition: Miller Indices are the reciprocals of the fractional intercepts (with fractions cleared) which the plane makes with the crystallographic x,y,z axes of the three nonparallel edges of the cubic unit cell.

22 Miller Indices are a symbolic vector representation for the orientation of an atomic plane in a crystal lattice and are defined as the reciprocals of the fractional intercepts which the plane makes with the crystallographic axes. To determine Miller indices of a plane, we use the following steps 1) Determine the intercepts of the plane along each of the three crystallographic directions 2) Take the reciprocals of the intercepts 3) If fractions result, multiply each by the denominator of the smallest fraction

23 When a plane is parallel to any axis,the intercept of the plane on that axis is infinity.so,the Miller index for that axis is Zero A bar is put on the Miller index when the intercept of a plane on any axis is negative The normal drawn to a plane (h,k,l) gives the direction [h,k,l]

24 (1,0,0)

25 (1,0,0) (0,1,0)

26 (0,0,1) (0,1,0) (1,0,0)

27 (1/2, 0, 0) (0,1,0)

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29 THREE IMPORTANT CRYSTAL PLANES ( 1 0 0) (1 1 0) (1 1 1)

30 Parallel planes are equivalent

31 Tuliskan nama bidang A dan B Gambar bidang: (011),(020),(121),(111)

32 Consider (110) plane Atomic packing different in the two cases Family of planes: all planes that are crystallographically equivalent that is having the same atomic packing, indicated as {hkl} For example, {100} includes (100), (010), (001) planes {110} includes (110), (101), (011), etc.

33 a d hkl = h 2 + k 2 +l 2 Where dhkl = inter-planar spacing between planes with Miller indices h, k and l. a = lattice constant (edge of the cube) h, k, l = Miller indices of cubic planes being considered.

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