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بلور شناسی سلول واحد (مفاهیم پایه ) انواع شبکه های دو بعدی و سه بعدی

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Presentation on theme: "بلور شناسی سلول واحد (مفاهیم پایه ) انواع شبکه های دو بعدی و سه بعدی"— Presentation transcript:

1 بلور شناسی سلول واحد (مفاهیم پایه ) انواع شبکه های دو بعدی و سه بعدی
اندیس های میلر

2 تفاوت؟

3 سلول واحد در شبکه دو بعدی

4 سلول واحد در شبکه دو بعدی NaCl
Crystal Structure

5 انتخاب دلخواه سلول واحد(حجم یکسان)
Crystal Structure

6 چینش اتمها در سلول واحد مهم نیست
Crystal Structure

7 - or if you don’t start from an atom
Crystal Structure

8 ایا سلول واحد است؟ Crystal Structure

9 ایا سلول مثلثی سلول واحد است؟
Crystal Structure

10 Crystal Structure

11 پنج شبکه براوه دو بعدی Crystal Structure

12 یک ملکول هر نوع چرخشی را می تواند داشته باشد
اما شبکه؟

13 Crystal Structure

14 1-CUBIC Crystal Structure

15 a- Simple Cubic (SC) Crystal Structure

16

17 Face Centered Cubic (FCC)
4 اتم در سلول واحدش وجود دارد (Cu,Ni,Pb..etc) ساختار fcc. دارند Crystal Structure

18 3 - Face Centered Cubıc Crystal Structure Atoms are all same.

19 2 - HEXAGONAL SYSTEM سه اتم در سلول واحدش وجود دارد. Crystal Structure

20 2 - HEXAGONAL SYSTEM Crystal Structure Atoms are all same.

21 Crystal Structure

22 3 - TRICLINIC 4 - MONOCLINIC CRYSTAL SYSTEM
تری کلینیک کمترین میزان تقارن را داراست Triclinic (Simple) a ¹ ß ¹ g ¹ 90 oa ¹ b ¹ c Monoclinic (Simple) a = g = 90o, ß ¹ 90o a ¹ b ¹c Monoclinic (Base Centered) a = g = 90o, ß ¹ 90o a ¹ b ¹ c, Crystal Structure

23 5 - ORTHORHOMBIC SYSTEM Orthorhombic (FC) a = ß = g = 90o a ¹ b ¹ c
Orthorhombic (Simple) a = ß = g = 90o a ¹ b ¹ c Orthorhombic (Base-centred) a = ß = g = 90o a ¹ b ¹ c Orthorhombic (BC) a = ß = g = 90o a ¹ b ¹ c Crystal Structure

24 6 – TETRAGONAL SYSTEM Tetragonal (BC) a = ß = g = 90o
Tetragonal (P) a = ß = g = 90o a = b ¹ c Tetragonal (BC) a = ß = g = 90o a = b ¹ c Crystal Structure

25 7 - Rhombohedral (R) or Trigonal
Rhombohedral (R) or Trigonal (S) a = b = c, a = ß = g ¹ 90o Crystal Structure

26 Miller Indices اندیس های میلر نمادهایی هستند که جهت صفحات اتمی را در کریستال مشخص می کنند این اندیس ها به گونه ای مشخص می شوند که مجوعه ای بی نهایت از صفحات بلوری را شامل مشوند و نحوه انتخابشان بگونه ای است که هماره صفحه انتخاب شده داخل سلول واحد قرار می گیرد. Crystal Structure

27 Example-1 Axis 1 ∞ 1/1 1/ ∞ Miller İndices (100) (1,0,0) X Y Z
Intercept points 1 Reciprocals 1/1 1/ ∞ Smallest Ratio Miller İndices (100) Crystal Structure

28 Example-2 Axis 1 ∞ 1/1 1/ 1 1/ ∞ Miller İndices (110) (0,1,0) (1,0,0)
Y Z Intercept points 1 Reciprocals 1/1 1/ 1 1/ ∞ Smallest Ratio Miller İndices (110) Crystal Structure

29 Example-3 Axis 1 1/1 1/ 1 Miller İndices (111) (0,0,1) (0,1,0) (1,0,0)
Y Z Intercept points 1 Reciprocals 1/1 1/ 1 Smallest Ratio Miller İndices (111) Crystal Structure

30 Example-4 Axis 1/2 1 ∞ 1/(½) 1/ 1 1/ ∞ 2 Miller İndices (210) (0,1,0)
(1/2, 0, 0) (0,1,0) Axis X Y Z Intercept points 1/2 1 Reciprocals 1/(½) 1/ 1 1/ ∞ Smallest Ratio 2 Miller İndices (210) Crystal Structure

31 Example-5 Axis 1 ∞ ½ 2 Miller İndices (102) a b c 1/1 1/ ∞ 1/(½)
Intercept points 1 Reciprocals 1/1 1/ ∞ 1/(½) Smallest Ratio 2 Miller İndices (102) Crystal Structure

32 Example-6 Axis -1 ∞ ½ 2 Miller İndices (102) a b c 1/-1 1/ ∞ 1/(½)
Intercept points -1 Reciprocals 1/-1 1/ ∞ 1/(½) Smallest Ratio 2 Miller İndices (102) Crystal Structure

33 Miller Indices [2,3,3] Plane intercepts axes at
Reciprocal numbers are: Indices of the plane (Miller): (2,3,3) Indices of the direction: [2,3,3] (200) (111) (100) (110) (100) Crystal Structure

34 اندیس های میلر و جهتهای صفحات اتمی در بلور

35 اندیس های میلر و جهتهای صفحات اتمی در بلور

36 جهتهای بلوری و صفحات اتمی عمود بر انها اندیس های میلر یکسانی دارند.

37 Crystal Structure

38 Example-7 Crystal Structure

39 Indices of a Family or Form
این {hkl} نماد کلیه اندیس های میلر مربوط به صفحات (hkl) را شامل می شود که بوسیله چرخش به همدیگر مر بوط می شوند Crystal Structure

40 3D – 14 BRAVAIS LATTICES AND THE SEVEN CRYSTAL SYSTEM
تنها 14 شبکه براوه وجود دارد که فضای سه بعدی را می پوشاند. این 14 شبکه نیز در هفت سیستم بلوری معرفی شده گنجانده می شوند. Cubic Crystal System (SC, BCC,FCC) Hexagonal Crystal System (S) Triclinic Crystal System (S) Monoclinic Crystal System (S, Base-C) Orthorhombic Crystal System (S, Base-C, BC, FC) Tetragonal Crystal System (S, BC) Trigonal (Rhombohedral) Crystal System (S) Crystal Structure

41 3–Hexagonal Close-Packed Str.
Crystal Structure

42 Hexagonal Close-packed Structure
a=b a=120, c=1.633a,  basis : (0,0,0) (2/3a ,1/3a,1/2c) Crystal Structure

43 Packing A B A Close pack B C Sequence AAAA… Sequence ABABAB..
- simple cubic Sequence ABABAB.. hexagonal close pack Sequence ABAB… - body centered cubic Sequence ABCABCAB.. ?? Crystal Structure

44 First Brillouin Zone: Two Dimensional Oblique Lattice

45 First Four Brillouin Zones: Square Lattice

46 All Brillouin Zones: Square Lattice

47 Primitive Lattice Vectors: BCC Lattice

48 First Brillouin Zone: BCC

49 Primitive Lattice Vectors: FCC

50 Brillouin Zones: FCC

51 First Brillouin Zone BCC

52 First Brillouin Zone FCC

53 X-ray Diffraction Typical interatomic distances in solid are of the order of an angstrom. Thus the typical wavelength of an electromagnetic probe of such distances Must be of the order of an angstrom. Upon substituting this value for the wavelength into the energy equation, We find that E is of the order of 12 thousand eV, which is a typical X-ray Energy. Thus X-ray diffraction of crystals is a standard probe.

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59 Wavelength vs particle energy

60 Bragg Diffraction: Bragg’s Law

61 Bragg’s Law The integer n is known as the order of the corresponding
Reflection. The composition of the basis determines the relative Intensity of the various orders of diffraction.

62 Many sets of lattice planes produce Bragg diffraction

63 d Deviation = 2 Ray 1 Ray 2 dSin
BRAGG’s EQUATION Deviation = 2 Ray 1 Ray 2 d dSin The path difference between ray 1 and ray 2 = 2d Sin For constructive interference: n = 2d Sin

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67 Bragg Spectrometer

68 Bragg Peaks

69 A beam of X-rays directed at a crystal interacts with the electrons of the atoms in the crystal
The electrons oscillate under the influence of the incoming X-Rays and become secondary sources of EM radiation The secondary radiation is in all directions The waves emitted by the electrons have the same frequency as the incoming X-rays  coherent The emission will undergo constructive or destructive interference with waves scattered from other atoms Secondary emission Incoming X-rays

70 Sets Electron cloud into oscillation
Sets nucleus (with protons) into oscillation Small effect  neglected

71 Oscillating charge re-radiates  In phase with the incoming x-rays

72 von Laue Formulation of X-Ray Diffraction

73 Condition for Constructive Interference

74 Bragg Scattering =K

75 The Laue Condition

76 Ewald Construction

77 Crystal structure determination
Many s (orientations) Powder specimen POWDER METHOD Monochromatic X-rays Single  LAUE TECHNIQUE Panchromatic X-rays ROTATING CRYSTAL METHOD  Varied by rotation Monochromatic X-rays

78 THE POWDER METHOD Cone of diffracted rays

79 Different cones for different reflections
POWDER METHOD Diffraction cones and the Debye-Scherrer geometry Different cones for different reflections Film may be replaced with detector

80 Schematic X-Ray Diffractometer
Detector X-Ray Source Powdered sample

81 Sample XRD Pattern

82 strong intensity = prominent crystal plane
weak intensity = subordinate crystal plane background radiation

83 Determine D-Spacing from XRD patterns
Bragg’s Law nλ = 2dsinθ n = reflection order (1,2,3,4,etc…) λ = radiation wavelength (1.54 angstroms) d = spacing between planes of atoms (angstroms) θ = angle of incidence (degrees)

84 strong intensity = prominent crystal plane
nλ = 2dsinθ (1)(1.54) = 2dsin(15.5 degrees) 1.54 = 2d(0.267) d = 2.88 angstroms background radiation

85 d-spacing Intensity 2.88 100 2.18 46 1.81 31 1.94 25 2.10 20 1.75 15 2.33 10 2.01 1.66 5 1.71

86 The Bragg equation may be rearranged (if n=1)
from to If the value of 1/(dh,k,l)2 in the cubic system equation above is inserted into this form of the Bragg equation you have Since in any specific case a and l are constant and if l2/4a2 = A pma 2010

87 Insert the values into a table and compute sin and sin2.
Since the lowest value of sin2 is 3A and the next is 4A the first Entry in the Calc. sin2 column is ( /3)*4 etc. d/Å Sin Sin2 Calc. Sin2 (h, k, I) 2.338 (1,1,1) 2.024 (2,0,0) 1.431 (2,2,0) 1.221 (3,1,1) 1.169 (2,2,2) 1.0124 (4,0,0) 0.9289 (3,3,1) 0.9055 (4,2,0) The reflections have now been indexed. pma 2010


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