材料科學導論 Introduction of Materials Science 許正興國立聯合大學電機工程學系
1. Introduction of Materials Science and Engineering 2. Atomic Structure and Bonding 3. Crystal Structures and Crystal Geometry 4. Solidification, Crystalline Imperfections and Diffusion in Solids 5. Phase Diagrams 6. Electronic Materials
Crystal Structures and Crystal Geometry Solid may be categorized broadly into crystalline and amorphous solids. Crystalline solids, due to orderly structure of their atoms, molecules, or ions possess well-defined shapes. => Metals are crystalline and are composed of welldefined crystals or grains. the grains are small and not clearly observable due to the opaque pq nature of metals. The Space Lattice and Unit Cells If the atoms or ions form a solid are arrange in a pattern that repeats itself in three dimensions, they form a solid that has long-range order (LRO) and is referred to as a crystalline solid or crystalline material. => metals, alloys, and some ceramic materials. Some materials whose atoms and ions are not arranged in a long-range, periodic, and repeatable manner and possess only short-range range order (SRO) => this means that order exists only in the immediate neighborhood of an atom or a molecule. ; such as liquid water => amorphous or noncrystalline
Atomic arrangements in crystalline solids can be described by referring the atoms to the points of intersection of a network of lines in three dimensions. Such a network is called a space lattice. => described as an infinite three-dimensional array of points. In an ideal crystal the grouping of lattice points about any given point are identical with the grouping about any other lattice point in the crystal lattice. Each space lattice can thus be described by specifying the atom positions in a repeating unit cell. A group of atoms organized in a certain arrangement relative to each other and associated with lattice points constitues the motif or basis. The size and shape of the unit cell can be described by three lattice vectors a, b and c (axial length a, b and c ; interaxial angle α,, β and γ => lattice constants of unit cell )
Crystal Systems and Bravais Lattice By assigning specific values for axial lengths and interaxial angles, unit cells of different types can be constructed. => seven different types of unit cells are necessary to create all space lattices. A. J. Bravais show 14 standard unit cells could describe all possible lattice networks. => there are four basic types if unit cells: (i) simple (ii) body-centered (iii) face-centered (iv) basecentered 立方 正方 斜方 菱形 六方 單斜 三斜
Principle Metallic Crystal Structure Most element metal (90 percent) crystallize upon solidification into three densely packed crystal structures: body-centered cubic (BCC), face-centered (FCC) and hexagonal close-packed (HCP).
The atomic radii of selected metals are listed in table.
Body-Centered Cubic (BCC) Crystal Structure In this unit cell we see that the central atom is surrounded by eight nearest neighbors and is said to have a coordination number (CM) of 8. => Each of these cells has the equivalent of two atoms per unit cell. One complete atom is located at the center of the unit cell, and an eighth of a sphere is located at each corner of the cell, making another atom => 1 (at the center) + 8 (1/8) (at the corners) = 2 The APF for the BCC unit cell is calculated to be 68 percent. => Such as Fe, Cr, W, Mo,V
Ex.: Calculate the atomic packing factor (APF) for the BCC unit cell, assuming the atoms to be hard spheres. <Sol> APF = (volume of atoms in BCC unit cell)/(volume l of BCC unit cell) => V atoms = (2)(4πR 3 /3) = 8.373R 3 => The volume of the BCC unit cell is :V unit cell =a 3 where a = 4R/(3) 1/2 => V unit cell =a 3 = 12.32R 3 => APF = (8.373R 3 )/(12.32R 3 ) = 0.68
Face-Centered Cubic (FCC) Crystal Structure In this unit cell there is one lattice point at each other corner of the cube and one at the center of each cube face. => the APF for this close-packed structure is 0.74.=> CM=12 The equivalent of four atoms per unit cell which has the eight corner octants account for one atom (8 1/8 = 1), and the six half-atom on the cube faces contribute another three atoms. => such as Al, Cu, Ni, Pb..
Hexagonal Close-Packed (HCP) Crystal Structure The APF of the HCP crystal structure t is 0.74, the sameasthat t for the FCC crystal structure since in both structure the atoms are packed as tightly as possible. In both the HCP and FCC crystal structures each atom is surrounded by 12 other atoms => CM = 12 the marked 1 contribute 1/6 of an atom to the unit cell ; the marked 2 contribute 1/2 of an atom to the unit cell; the marked 3 is centered inside. => the total number of atoms inside an HCP unit cell is 1 (1/6) 12+1 (1/2) 2+3=6 The rti ratio of the hihtc height c of the hexagonal prism of the HCP crystal structure to its basal side a is called the c/a ratio => the ideal HCP crystal structure consisting of uniform spheres packed as tightly together as possible is c/a = 1.633 Ca and Zn : higher the ideality=>slightly elongated along the c axis Mg, Co, Zr, Ti : less the ideality => slightly compressed in the direction along c axis. 1 1 3 2 2 1 1 1 1
Atom Positions in Cubic Unit Cells To locate atom positions in cubic unit cells, we use rectangular x, y, and z axes.
Directions in Cubic Unit Cells Often it is necessary to refer to specific directions in crystal lattices. This is especially important for metals and alloys with properties that vary with crystallographic orientation. => direction index are the vector components of the direction resolved along each of the coordinate axes and reduced to the smallest integers. The position coordinates of the unit cell where the direction vector emerges from the cube surface after bi being converted to integers are the directioni indices. The direction indices are enclosed by square brackets with no separating commas. => All parallel direction vectors have the same direction indices. Equivalent directions are called indices of a family : [100], [010], [001], [010], [001], [100] = <100>
Miller Indices for Crystallographic Planes in Cubic Unit Cells Sometimes it is necessary to refer to specific lattice plan of within a crystal structure, or it may be of interest to know the crystallographic orientation of a plane or group of planes in a crystal lattice. => to identify crystal planes in cubic crystal structures, the Miller notation system is used. Miller indices of a crystal plane are defined as the reciprocals of the fractional intercepts that the plane makes with crystallographic x, y, and z axes of the three nonparallel edges of the cubic unit cell. To procedure for determining the Miller indices for a cubic crystal plane is as follows: 1. Choose a plane that does not pass through the origin at (0, 0, 0) 2. Determine the intercepts of the plane in termsofthecrystallographic x, y, and z axes for a unit cube. These intercepts may be fractions. 3. Form the reciprocals of these intercepts 4. Clear fractions and determine the smallest set of whole numbers that are in the same ratio as the intercepts. The notation (hkl) is used to indicate Miller indices.
Intercepts : 1/3, 2/3, 1 reciprocals of these intercepts : 3, 3/2, 1 multiplied by 2 : 6, 3, 2 Miller indices : (632) If sets of equivalent lattice planes are related by the symmetry of the crystal system, they are called planes of a family or form, and the indices of one plane of the family are enclosed in braces as {hkl} to represent the indices of a family of symmetrical planes. (100), (010), and (001) => {100}
In cubic crystal structures the interplanar spacing between two closest parallel plans with the same Miller indices is designatedd hkl, where h, k, and l are the Miller indices of the planes => this spacing represents the distance from a selected origin containing one plane and another parallel plane with the same indices that is closest to it. 立方晶系之面的間距 d hkl (interplanar spacing) 表示兩個米勒指數完全相同, 在空間中最接近且平行之晶面間距 The distance between (110) planes 1 and 2 d 110 is AB ; the distance between (110) planes 2 and 3 is d 110 and is length BC.
Ex. Copper has an FCC crystal structure and a unit cell with a lattice constant of 0.361 nm. What is its interplanar space d 220? <Sol> d =a/(h hkl 2 +k 2 +l 2 ) 1/2 = 0.361 nm/(2 2 +2 2 +0 2 ) 1/2 = 0.128 nm
Volume, Planar, and Linear Density Unit-Cell Calculations Volume Density Using the hard-sphere atomic model for the crystal structure unit cell of a metal and a value for the atomic radius of the metal obtained from X-ray diffraction analysis, a value for the volume density of ametal can be obtained by using the equation
Ex. Copper has an FCC crystal structure and an atomic radius of 0.1278 nm. Assuming the atoms to be hard spheres that touch each other along the face diagonal of the FCC unit cell, calculate a theoretical value for the density of copper in megagrams per cubic meter. The atomic mass of copper is 63.54 g/mol. <Sol> The slightly lower density of For the FCC unit cell, 2a = 4R the experimental value (8.96 4R (4)(0.1278nm) => a = = = 0.361 nm Mg/m 3 or 8.96 g/cm 3 ) could 2 2 be attributed to the absence mass / unit cell volume density of copper = ρ v = of atoms at some atomic sites volume / unit cell In the FCC unit cell there are four atoms/unit cell. (vacancies), line defects, and 23 Each copper atom has a mass of (63.54 g/mol)/(6.02 10 atoms/mol) mismatch where grains meet (grain boundaries). (4 atoms)(63.54 g / mol) => m = 23 6.02 10 atoms / mol 10 ( 6 Mg ) = 4.22 10 g -9 3 10 m 29 The volume V of the Cu unit cell is V = a = 0.361nm = 4.7 10 m nm 28 m 4.22 10 Mg 3 3 => the density of copper is : ρv = = = 8.98 Mg/m (8.98 g/cm ) 29 3 V 4.7 10 m 28 Mg 3 3
Planar Atomic Density Sometimes it is important to determine the atomic densities on various crystal planes. To do this a quantity called the planar atomic density For the lattice constant of α iron is 0.287 nm ρ p 1 1+ 1 4 = 4 = 2a a 2 atoms 2(0.287 nm) 2 = 17.2 10 13 atoms/mm 2
Linear Atomic Density Sometimes it is important to determine the atomic densities on various directions in crystal structures. To do this a quantity called the linear atomic density
Crystal Structure Analysis Our present knowledge of crystal structures has been obtained mainly by X-ray diffraction techniques that use x-rays whose wavelength are the same as the distance between crystal lattice planes. X-ray Source X-ray used for diffraction are electromagnetic waves with wavelengths in the range 0.05 to 0.25 nm. <visible light is of the order of 600 nm> ; a voltage of about 35 kv When the electrons strike the target metal (e.g., Mo), x-rays are given off.
X-ray diffraction Since the wavelengths of some x-rays are about equal to the distance between planes of atoms in crystalline solids, reinforces diffraction peaks of radiation of varying intensities can be produced when a beam of x-ray strikes a crystalline solid. Let us now consider incident x-rays 1 and 2 -> these rays to be in phase, the extra distance of travel of ray 2 is equal to MP+PN, which must be an integral number of wavelength. (n = 1, 2, 3, and is called the order of the diffraction. => both MP and PN equal d hkl sinθ where d hkl is the interplanar spacing of the crystal planes of indices (hkl) -> constructive interference. The horizontal lines represent a set of parallel crystal planes with Miller indices (hkl) Bragg s law
Ex. A simple BCC iron was placed in an x-ray diffractometer using incoming x-rays with a wavelength λ=0.1541 nm. Diffraction from the {110} planes was obtained at 2θ=44.704 o. Calculate a value for the lattice constant a of BCC iron. (Assume firstorder diffraction with n = 1.) <Sol> o 2 θ = 44.704 θ = 22.35 => λ = 2d => a = d hkl hkl h sinθ => 2 + k 2 + l 2 d o 110 λ = = 2sinθ => a( Fe ) = d ( 110 0.1541 nm 2 sin22.35 o ( ) 1 2 + 1 2 + 0 2 0.1541 = = 0.2026 nm 2 0.3803 = 0.2026 1.414 = 0.287 nm
X-ray diffraction analysis of crystal structure The powder pwd method of x-ray ydiffraction analysis The most commonly used x-ray diffraction technique is the powder method. in this technique a powdered specimen is utilized so that there will be random orientation of many crystals to ensure that some of the particles will be oriented in the x-ray beam to satisfy the diffraction conditions of Bragg s law. => Modern x-ray crystal analysis uses an x-ray diffractometer t r thatt has a rditin radiation counter to dt detectthe angle and intensity of the diffracted beam. then plots the intensity
Diffraction Conditions for Cubic Unit Cells X-ray diffraction techniques enable the structure of crystalline solids to be determined. The interpretation of x-ray diffraction data for most crystalline substances is complex and beyond the scope, and so only the simple case of diffraction in pure cubic metals will be considered. The analysis of x-ray diffraction data for cubic unit cell can be simplified by combing This equation can be used along with x-ray diffraction data to determine if a cubic crystal structure is body-centered or face-centered cubic.
For simple cubic lattice, reflections from all (hkl) planes are possible. For BCC structure diffraction occurs only on planes whose Miller indices when added together (h+k+l) total to even number ex. {110}, {200}, {211} For FCC crystal structure, the principal p diffracting planes are those Miller indices are either all even or all odd (zero is considered even). ex. {111}, {200}, {220}
From We can eliminate these quantities by forming the ratio of two sin 2 θ values as For the BCC crystal structure the first two sets of principal diffracting planes are the {110} and {200} planes. => => If the crystal tlstructure t of the unknown cubic metal tlis BCC, the ratio of the sin 2 θ values that correspond to the first two principal diffracting planes will be 0.5 For the FCC crystal structure the first two sets of principal diffracting planes are the {111} and {200} planes. => => If the crystal structure of the unknown cubic metal is FCC, the ratio of the sin 2 θ values that correspond to the first two principal diffracting planes will be 0.75
Amorphous materials Some materials are called amorphous or noncrystalline because they lack longrange order in their atomic structure. It should be noted that in general materials have a tendency to achieve a crystalline state because that is the most stable state and corresponds to the lowest energy level. => atoms in amorphous materials are bonded in a disordered manner because of factors that inhibit the formation of a periodic arrangements. => Most polymers, glasses, and some metals are members of the amorphous class of materials.