Chapter types of materials amorphous, crystalline, and polycrystalline. 5. Same as #3 for the ceramic and diamond crystal structures.


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1 Chapter Highlights: Notes: 1. types of materials amorphous, crystalline, and polycrystalline.. Understand the meaning of crystallinity, which refers to a regular lattice based on a repeating unit cell.. For fcc, bcc, and sc lattice, be able to determine the number of atoms in each cell (n) and the atomic packing factor (APF), also be able to determine the relationship between the atomic radius (R), the lattice constant (a) and the density ρ ρ (n, a, R). 4. Know the common ceramic crystal structures and the properties (ionic charge and ionic size) that determine crystal structure. Be able to use this information to predict crystal structure. 5. Same as # for the ceramic and diamond crystal structures. 6. In a cubic unit cell, be able to identify crystallographic directions and draw them. 7. In a cubic unit cell, be able to identify crystallographic planes and draw them. 8. In a cubic unit cell, be able to determine linear and planar atomic densities. 9. Be able to match an xray crystallography pattern to the crystal structure and to use the diffraction pattern to obtain the lattice constant or atomic radius. Metallic Crystal Structures Metals, ceramics, polymers, etc. may be either: 1. Crystalline Si for semiconductor and photovoltaic devices.. Amorphous corrosionresistant Ni plating, no grain boundaries.. Polycrystalline structural steels, other industrial metals (see figure).
2 c0f5 Figure.5 (above) from text, good example of polycrystalline structure. You might want a different morphology depending on the application. Example 1: Corrosion resistant Ni coatings are amorphous, since grain boundaries provide an easy path for corrosion to weaken a material. Example : Si semiconductor devices are constructed from single crystal (crystalline over long length scales) Si wafers. Show IBM wafer. The required electrical performance of cpu chips, DRAM chips, etc. depends on using single crystal Si. Lattice: Unit cell: Infinite array of points in a solid. Smallest repeating configuration of atoms from which lattice can be constructed.
3 Table below illustrates the different crystal systems, with different combinations of a,b,c,α,β,γ. In ES 60, we will focus on only one crystal system, the cubic system, which is the most important. However, many common materials fall in the hexagonal system. c0tf06
4 FCC unit cell (Figure.1) c0f01 A) # of atoms, n 8(1/8) + 6(1/) 4 B) Coordination number 1, count nearest neighbors. C) Show drawing of one face, relate lattice constant (a) to atomic radius (R), show that a R. This relies on recognizing that the atoms are touching along the diagonal of each face, as can be seen in the upper left graphic. D) Calculate density ρ from equation (.5): na ρ V c N A ρ ( 4A R ) N A 4 A R N A
5 Given A and a or R, be able to calculate ρ and vice versa. Also, be able to determine the crystal structure given A, a and ρ. E) Atomic packing fraction (APF) Volume of atoms within unit cell APF Total unit cell volume 4 n π R APF 4 n R π a a 4π (4) R APF ( R ) π APF 0.74 Example: Calculate the radius of a iridium atom, given that it has a FCC structure, with a density of.4 g/cm and an atomic weight of 19. g/mol. Start with the equation that we derived above for the density of a FCC material:.4 ρ ( g / cm R 4A R ) N 4 A 4 A R N A ( 19. g / mol ) ( R) ( 6.0x10 / mol ) 4 x cm R 1.6x10 cm 0. 16nm Next two figures illustrate bodycentered cubic (BCC) and hexagonal closepacked (HCP) crystal structures. You should be able to do the above calculations (AE) for the BCC unit cell, and for all other cubic unit cells covered in this course.
6 c0f0 Note that atoms within the BCC unit cell are touching along the body diagonal, meaning the diagonal that crosses the unit cell in all three coordinates (x,y,z). This must be recognized in order to determine the relationship between a and R. c0f0
7 c0f9 c0f0
8 c0f1 c0tf01
9 Something to think about: Why are there different metal crystal structures?? Remember that metallic bonding is mostly nondirectional, with ion cores floating in a sea of electrons. Show figure.11 (below). You might expect metallic bonding to give the closestpacked structure, with the maximum APF and maximum coordination. Both FCC and HCP have APF0.74 and coordination # of 1, which are the highest possible, so about /4 of metals are FCC or HCP. But why do other metals have the BCC structure (APF0.68, coordination # 8)? This is complicated but involves some degree of covalent bonding through the dorbitals of transition metals. Later we will introduce the diamond crystal structure (silicon, germanium, diamond), where the APF is only 0.4. This structure is favored due to covalent bonding in these materials, which requires a coordination number of only 4. c0f11 Ceramic Crystal Structures Ceramic materials are wholly or substantially ionic. For this reason and because the unit cell must contain two different species, they form different crystal structures from metals. Ceramic crystal structure is determined by the relative charge and relative size of the anion and cation. Monitor this by r C /r A, which is always less than one, since the size is determined by the number of electrons in the electron cloud. Remember that in a ceramic, the cation gives up electrons to the anion, so the anion is normally much larger. Ceramic crystal structure: total cation, anion charges must be equal.
10 1) Take the many possible geometrical arrangements of the large anions as in VSEPR (linear, triangular, tetrahedral, octahedral, etc.) ) Choose the structure where the cation just fits into the interstice. This maximizes the Coulomb attraction and binding energy. Table. below is constructed by the above two principles. The ratio r C /r A given in table. can be derived from those principles. The most common coordination numbers are 4, 6, 8. Ionic radii are given in table.4 below. c0tf0
11 c0tf04 Example: Show that the minimum cationtoanion radius ratio for a coordination number of 6 is c0f04 Draw four anions in a square (above) so that they just touch. Now put a cation into the interstice space. Now you can draw a right triangle with two side of length r A + r C and the hypotenuse of length r A. Therefore,
12 cos ( 45 ) r A r + r A C This can be solved to yield, rc r A r r C A This is the borderline between coordination numbers of 4 and 6 in table.. AX Crystal Structures First, consider ceramic crystal structures with an equal number of anions and cations. Figures below show zinc blende (CN 4), rock salt (CN 6), CsCl (CN 8) structures. The rock salt structure is like two superimposed FCC structures. The CsCl structure looks like BCC, but is a uniquely different crystal structure because two different ions are involved. Examples of these structures are ZnS (r C /r A 0.40, zinc blende structure), NaCl (r C /r A 0.564, rock salt structure), and CsCl (r C /r A 0.99, CsCl structure). c0f05 In the figure above, the ions are actually touching along each edge of the cube, so a r A + r C. Counting ions, n A 8(1/8) + 6(1/) 4 and n C 1(1/4)
13 c0f06 For the CsCl unit cell above, ions are touching along the body diagonal, similar to the BCC unit cell. Counting ions, n A 8(1/8) 1 and n C 1. c0f07 For the zinc blende structure above, ions connected by dotted lines are touching. If you can obtain the Cartesian coordinates (x i, y i, z i ) of a cation and anion that are touching, their separation distance (r A + r C ) is equal to their separation according to (Δx i + Δy i + Δz i ) 1/
14 A m X p Structures If m and/or p are not equal to 1, the number of cations and anions are unequal. The ceramic crystal structures are the same as AX, but some positions are now left empty. For CaF, show figure.8 below. Same as CsCl, but half of the cation (Ca) sites are empty. c0f08 Ceramic density calculations Similar to metals, the density of ceramic materials can be calculated if one knows the numbers of the different species in the unit cell, the ionic radii, and the atomic weights. From equation (.6): n ρ ' V ( A A ) C + A This includes summations over atomic weights, and introduces n, the number of formula units of the ceramic material within the unit cell. This is best explained with an example. C N A Example: MgO has the rock salt crystal structure and a density of.58 g/cm. What is the lattice constant, and how does this compare to the sum of the ionic radii in table.4?
15 In the above equation for the density of a ceramic, we need to know the number of formula (MgO) units in the unit cell. Inspection of figure.5 (above) shows that the unit cell contains 4 anions and 4 cations. Therefore, the unit cell contains 4 units of MgO and n 4. The rest is just substitution: n ρ ' V ( A A ) C + A C N A ( g / mol ) a ( 6.0x10 / mol ) 4.58 g / cm + a 7.478x10 cm One expects a 8 a 4.1x10 cm 0. 41nm ra + rc nm. This is a discrepancy of 0.7% from the calculated value. Diamond crystal structure The diamond crystal structure (below) is the crystal structure of Si, Ge and C (diamond). This is introduced mainly because of the widespread use of Si to make semiconductor devices (transistors), from which integrated circuits are built. The semiconductor chips in a personal computer (cpu, DRAM) and a cell phone are built starting from crystalline Si. This is by far the most widespread application of a crystalline material, since metals are polycrystalline in most applications. For this reason, crystalline Si is several orders of magnitude cheaper than most other crystalline materials. Carbon can also be found as graphite, C 60, and nanotubes, as shown in figures below. c0f16
16 c0f17 c0f19
17 c0f18 Point Coordinates In order to name directions and planes, we need to learn to identify the location of points within the unit cell. Go through example problems.7 and.8 on your own. The point coordinates are always given as a fraction of the lattice constants. Crystallographic Directions Any vector drawn in a crystal system can be related to a standard crystallographic direction, this provides a common language that allows communication regarding materials properties. The book gives a standard recipe for determining the crystallographic direction, which is denoted by square brackets []. I am going to give you a slightly different standard recipe that I think is easier. 1. Redraw the coordinate system so its origin coincides with the origin of the vector of interest.. The dimensional coordinates of the vector endpoint are determined in terms of the unit cell dimensions. v uax ˆ + vby ˆ + wczˆ. Multiply or divide by a common factor to reduce [uvw] to the smallest set of integers.
18 4. Refer to this direction as the [uvw] direction. [uvw] [uvw] All parallel vectors are considered to represent the same crystallographic direction. For cubic unit cells, abc. c0prob D) Taking the origin of the coordinate system as the vector origin, the vector endpoint is at a, a, a. In terms of the lattice constant (a), the vector endpoint is at 1 1 1,,. Multiply through by, and this direction is [ 111]. 1,. In terms of the lattice constant (a), the vector 1 endpoint is at 1, 1,. Multiply through by, and this direction is 1. 1 a, 0,. In terms of the lattice constant (a), the vector 1 endpoint is at, 0,. Multiply through by 6, and this direction is [ 4 0 ]. A) The vector endpoint is at a a, a B) The vector endpoint is at a
19 1 1 1, 1, 6 1, 6 C) The vector endpoint is at a a a. In terms of the lattice constant (a), the vector endpoint is at Crystallographic Planes,. Multiply through by 6, and this direction is 61. The book again gives a standard recipe for determining the crystallographic plane, but mine is slightly different. 1. Redraw the coordinate system so that its origin is in a convenient corner of the unit cell that is shown. This plane of interest cannot intersect with this corner. This corner should be chosen so that it is easy determine the three axis intercepts of the plane of interest.. Note the three axes intercepts and express them in terms of the lattice constants a,b and c.. Replace h, k and l with their reciprocals. (hax,kby,lcz) ˆ ˆ ˆ 4. Multiply or divide (hkl) by a common factor to achieve the smallest possible set of integers. 5. Express this plane as the (hkl) plane, known as the plane's Miller indices. All parallel planes represent the same crystallographic plane. For cubic unit cells, a b c. If the plane does not intercept an axis, its intercept is taken as infinity, which has a reciprocal of zero.
20 c0prob A) Take the coordinate system origin at the corner usually labeled (a, a, 0). From that origin, the axes intercepts are a, a, a. In units of the lattice constant, this is ( 1, 1, 1). Taking reciprocals has no effect. No need to multiple or divide by a common factor. The final notation for this plane is ( ) B) Take the coordinate system origin at the corner usually labeled (0, 0, 0). From that origin, the axes intercepts are a, a,. In units of lattice constant (a), this is,, Taking reciprocals, with no need to multiply or divide by a common factor, the final 0. notation for this plane is ( ) Families of planes and families of directions Some planes contain identical configurations of atoms, so for practical purposes they are equivalent. For example, the BCC ( 110) plane has the same atomic configuration as the ( 101), ( 011), ( 1 01), ( 110), and ( 011) planes. Since materials properties depend on the atoms from which they are composed, these 6 planes are equivalent. Therefore, these 6 planes are referred to as the {110} family of planes. The idea of a family of planes is important in xray diffraction and slip. One can also define the <100> family of directions, which includes the directions <100>, <010>, <001>..
21 Linear and planar atomic densities The linear and planar atomic densities are defined as: Linear Atomic ρ # of atoms centered on direction Length of direction vector vector Planar Atomic ρ # of atoms centered on Area of plane a plane Note that even though these two quantities are described as densities, they are really more like atomic packing fractions in one and two dimensions. We may not have time to cover this in class. If not, you will be responsible for learning this material on your own. The following two examples should help. Example: What is the planar atomic density in the BCC (110) plane? First, count the number of atoms in the rectangle ACDF, n 1 + 4(1/ 4). The atoms in the corner count as1/4 instead of 1/8 since in two dimensions, they are only cut in half by two planes. We also need to know the area of the rectangle ACDF, which is A a ( a) a. Now, determine the planar atomic density (PAD): PAD a
22 Remember, in a BCC unit cell a 4R, so substitution yields: PAD 4R 16 R Problem: What is the linear atomic density along the BCC [001] and 1 11 directions? You should be able to draw in these directions in the figure above. Along the 1 11 direction, this is the body diagonal, and: LAD a 4R 1 R Along the [001] direction (this is one side of the unit cell), the linear atomic density (LAD) is: LAD 1 a 1 4R / 4R Xray diffraction From basic physics, diffraction occurs when light encounters a series of regularly spaced obstacles with the spacing comparable in magnitude to the light s wavelength. Since atomic dimensions are extremely small, you need very short wavelength, high energy xrays. The utility of xray diffraction is that the line pattern reflects the crystal structure, whereas the line spacing allows calculation of the atomic radius. In other words, you can determine the crystal structure by simply looking at the pattern of lines, and then calculate the atomic radius by noting the specific diffraction angle at which each line occurs. Here are the xray diffraction patterns for Pb (FCC) and W (BCC):
23 Notice that for FCC metals, the first two diffraction lines are closely spaced, and then there is a significant gap. On the other hand, the diffraction lines are more equally spaced for BCC metals. An expert can simply conclude by inspection that the first pattern corresponds to FCC, and the second to BCC crystal structure. The figures below illustrate the physics of the diffraction process.
24 c0f6 c0f7
25 In the last figure, the extra segment in the light path is SQT. If SQT is an integral # of wavelengths, constructive interference occurs. nλ SQ + QT nλ d hkl sinθ + d hkl sinθ d hkl sinθ This is Bragg's law for diffraction, where the quantity d hkl is the distance between parallel (hkl) planes. For cubic systems, d a hkl h + k + l Bragg's law is a necessary but not sufficient condition for diffraction to occur. It is only rigorous for simple cubic (sc) systems, which do not exist. For FCC and BCC systems there are extra scattering centers that are not at the corners of the cube. These provide additional constraints: BCC: h+k+l must be even. FCC: h,k,l either all even or all odd. All FCC metals have the same xray diffraction pattern; it is just expanded or compressed according to the size of the lattice constant. Similarly, all BCC metals have the same pattern (different from FCC). As shown below, if one keeps λ constant and varies θ, you only see a diffracted beam at certain θ. Note that as the sample rotates by θ, the detector must rotate by θ. c0f8
26 Example: Label the first 6 lines of the xray diffraction pattern of Pb, shown above. Use the first three lines to calculate d hkl and a. Assume λ nm. We will do this problem knowing that Pb has a FCC crystal structure, but nobody has assigned the diffraction peaks to particular planes. This problem now requires us to name the first 6 peaks in the xray diffraction pattern, and then calculate the interplanar spacing and the lattice constant from the first peaks. Remember that for a FCC unit cell hkl must be all odd or all even for diffraction to occur. First, we must use trial and error to find the 6 peaks which occur at smallest θ in the xray diffraction pattern. To minimize θ in equation (.15) we must maximize d hkl. From equation (.16) this occurs for the minimum values of h,k and l. I will use equation (.11) to calculate d hkl for several combinations of h,k and l. a a a d 111, d 00, d a, 1 d d 11 a, 11 d 1 a 19 a a d, d 400, d a 18 a) Choosing the 6 maximum values of d hkl from above, the first 6 peaks in order are (111), (00), (0), (11), (), and (400). b) Using equation (.15): λ nm d nm sin( θ ) sin λ nm d nm sin( θ ) 7 sin λ nm d nm sin( θ ) 5 sin c) After rearrangement, equation (.16) yields the lattice constant: a d hkl h + k + l
27 For the first diffraction lines one obtains a , , and nm. This is pretty good agreement given the very crude readings of the diffraction angles. If we take the average of these values (0.486 nm) and determine the atomic radius (R) from a R, one obtains R 0.17 nm, close to that given in the front cover of the textbook (0.175 nm). This is similar to how xray diffraction is used in real life. From the overall look of a diffraction pattern, an expert can identify the crystal structure involved (bcc, fcc, diamond, etc.). Then the lowest few lines are assigned, and the atomic radius calculated for those lines. If all the atomic radii agree, then the original assumption for the crystal structure was correct. The atomic radius is then taken as the average from the lowest diffraction lines. Practical Utility of Xray Diffraction Xray crystallography is a standard method for characterizing samples of metal and semiconductor materials. This can be used to identify the chemical composition of an unknown mixture, the degree of crystallinity of a sample, or to determine the crystal structure of a newly fabricated material. Xray diffractometers are commonly found in academic and industrial research laboratories. One exotic example of the utility of xray crystallography is the determination of protein crystal structures. Known information about protein structures is summarized in the protein data bank: Searching through this database, by far the most common method for studying protein structure is x ray diffraction. Determination of protein structure in solution is greatly complicated by the heterogeneous structure of proteins in solution, and by their continuous rotational and vibrational motion. Therefore, protein size and shape in solution is often inferred from xray diffraction measurements on protein crystals, despite the obvious possibility of different protein conformations in the solution and solid phases. Another example of the utility of xray diffraction is to monitor recrystallization of photovoltaic thin films, which absorb photons and convert them to electrons in solar cells. Here are xray diffraction patterns of particulate CIGS films before (AC) and after (EF) annealing:
28 The xray diffraction peaks appear in proportion to the extent of recrystallization of the CIGS thin films. This is taken from: G.S. Chojecki, D.H. Rasmusxsen, K.B. Albaugh and I.I. Suni, Recrystallization of Porous Particulate CIGS Precursors into Dense Thin Films, submitted to Sol. Energy Mater. Sol. Cells. It is sometimes possible to estimate the average grain size (introduced in Chapter 5) from the width of the xray diffraction peaks. The example below illustrates the use of xray diffraction to provide real time monitoring of Cu receyrstallization, where the Cu thin films will be employed as ULSI interconnect materials.
29 Immediately following electrodeposition, the Cu thin film above exhibits grain boundaries, which will increase the Cu film resistance (introduced in Chapter 1). The Cu grain size will increase slowly at room temperature, and more rapidly with low temperature annealing. This can be monitored by time resolved xray diffraction, as shown below:
30
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