You may have wondered how crystal structures are determined. The chief tool for inves-tigation of structure on the scale of atomic dimensions is X-ray diffraction (XRD). Certain physical laws show that the wavelength of the electromagnetic radiation must be of the order of the size of the feature to be investigated. Hence, in order to investigate atomic dimensions in the rang e of 0.5–50`, electromagnetic radiation wit h an energy character -istic of X rays is required. X rays were discovered in the 19th century, and von Laue and Bragg developed the theory of XRD in the early 20th century. It is a relatively simple technique, yet it can provide a wealth of accurate information on interplanar spacings and atomic positions.
Let us review the wave interaction mechanism known as interference illustrated in Figure 3.12–1. It is based on the observation that amplitudes of interferingwaves add to yield the amplitude of a composite wave. When waves of equal amplitude are in phase, their maxima and minima are aligned, and the amplitude of the composite wave is twice that of its components (Figure 3.12–1b). This is known as constructive interference. If, however, as shown in Figure 3.12–1c, one of the two waves is shifted by a distance of half a wavelength, the amplitude of the composite wave is equal to zero at all points. This phenomenon is known as destructive interference and is the operative principle of an-tireflective coatings (used, for example, on a camera lens).
The inteference phenomenon is central to understandingthe interaction of X rays with crystals. Consider a beam of parallel X rays impinging upon the surface of a crystal. To the X-ray beam, the crystal lattice appears to be a collection of atoms located in planes,
Amplitude of the composite wave is the sum of the amplitudes of its components
Constructive interference
Destructive interference (a)
(b) (c)
λ λ λ/ 2
θ λ
3a 2a
1a 2b
3c 1
2 3
θ B
C A
d Phase
shift Pathlength beam 2a – Pathlength beam 1a = 0
Pathlength beam 2b – Pathlength beam 1a = AB + BC Pathlength beam 3c – Pathlength beam 1a = 2 (AB + BC )
AB + BC = 2d sinθ Constructive interference if and only if
2d sinθ= nλ
Source Detector
as shown in Figure 3.12–2. The wavelength of the radiation is, the angle of incidence is, and the perpendicular distance between parallel planes, the interplanar spacing, is d.
Consider two parallel waves of X rays. The first wave is reflected from the top plane of atoms, and the second is reflected from the next lower parallel plane of atoms. These are waves 1a and 2b in Figure 3.12–2. The difference in the distance traveled, from source
FIGURE 3.12–1 The phenomenon of wave interference: (a) the general case, (b) constructive interference, and (c) destructive interference.
FIGURE 3.12–2 The geometry of X rays reflected from parallel atomic planes. The angle of incidence is, the wavelength of the X rays is, and the spacing between parallel planes of atoms is d. The important point is that there is a path-length difference between X rays reflected from parallel planes.
20.0 40.0 60.0 80.0 100.0 120.0 Diffraction angle, 2θ(degrees)
0.40 0.80 1.20 1.60 2.00 2.40 2.80 3.20 3.60 4.00
Intensity
to detector, between X-ray beams 1a and 2b is given by the expression:
AB⫹ BC ⫽ 2d sin (3.12–1)
where AB and BC represent the length of the line segments between the indicated points.
Because of this path-length difference, there will be a phase shift between the two X rays when they reach the detector. The two waves interfere constructively only if the path-length difference is an integral multiple of the X-ray wavepath-length. When constructive interference occurs in an X-ray experiment, we call itdiffraction. The diffraction condi-tion is:
2d sin ⫽ n (3.12–2)
where n is an integer. This equation, calledBragg’s law, is the fundamental equation of diffraction. It states that diffraction occurs at specific values of, whereis determined by the wavelength of the radiation and the interplanar spacing. Hence, ifis known and the diffracted beam intensity is measured over a range of, then d can be determined. The value of n in Equation 3.12–2 determines the order of the reflection; when n⫽ 1, it is a first-order reflection; n⫽ 2 is a second-order reflection; and n ⫽ i is the ith order reflection. Using XRD, lattice parameters can be determined with high accuracy—in angstroms to four or five digits to the right of the decimal place!
There are literally dozens of techniques used in XRD. Perhaps the most easily under-stood method is the powder diffraction method. A sample is ground into a powder and mixed with a noncrystalline binder. Powder and binder are subjected to a monochromatic (single-wavelength) source of coherent (in-phase) X radiation. Diffracted intensity is measured as a function ofwith an instrument called a diffractometer. The results of a diffractometer experiment on sodium chloride are shown in Figure 3.12–3. The intensity maxima represent conditions where Bragg’s law is satisfied for some plane in one of the
FIGURE 3.12–3 A plot of X-ray diffraction data, in the form of beam intensity versus incidence angle, for
polycrys-many ground crystal fragments. There are polycrys-many families of planes in a crystal—兵1 0 0其, 兵1 1 0其, 兵1 1 1其, 兵2 0 0其, 兵2 1 0其, 兵2 1 1其, 兵2 2 0其, 兵2 2 1其, 兵2 2 2其, and so on. Each peak must be identified or indexed according to the plane and d spacing that produce the diffraction. Once all the d spacings are determined, the crystal structure can be assigned.
The d spacing, or interplanar spacing of crystals, can be determined geometrically for a known crystal. For example, the interplanar spacing for (1 1 1) in an FCC crystal can be determined by inspection of Figure 3.5–3a. The normal to the plane intersects three parallel members of the 兵1 1 1其 family of planes per unit cell. Hence, the interplanar spacing is:
d共1 1 1兲 ⫽ a0
兹3
3 (3.12–3)
where in this notation a0⫽ d共1 0 0兲. In cubic systems, the geometrical relations can be expressed relatively simply:
d共h k l兲 ⫽ a0
共h2⫹ k2⫹ l2兲0.5 (3.12–4)
where d共h k l兲 is the d spacing for plane 共h k l兲. More complex equations have been developed for other crystal systems.
...
EXAMPLE 3.12–1
a. Calculate the interplanar spacing of the兵1 1 1其 family of planes in an FCC material with atomic radius r.
b. Use the result of part a to verify that the c兾a ratio in the HCP structure is equal to 1.633.
Solution
a. Using Equation 3.12–4 for a cubic crystal, d共1 1 1兲 ⫽ a0
共12⫹ 12⫹ 12兲0.5⫽ a0
兹3
In the FCC structure, the a0-r relationship is a0⫽ 4r兾兹2. Combining these two equa-tions yields:
d共1 1 1兲 ⫽ 4r 兹6
b. The quantity d共1 1 1兲 represents the distance between two close-packed planes. In the HCP structure the height of the unit cell is equal to twice the distance between close-packed planes of atoms. That is,
c⫽ 2d共1 1 1兲 ⫽ 2冉兹64r冊⫽ 兹68r
In the HCP structure, the a0-r relationship is simply a0⫽ 2r. Therefore, we find:
c⫽ 8r 兹6⫽ 4a0
兹6⇒ c a0⫽ 4
兹6⫽ 1.633
...
Crystallographers have had a great deal of experience indexing diffraction patterns. One way to expedite the process is through the use of “missing” diffraction peaks. Consider the 兵0 0 1其 diffraction from a simple cubic structure as shown in Figure 3.12–4a. An intensity
Pathlength 2 – Pathlength 1 = nλ (a)
1
2
2 1 λ
θ
(b) 1
2
2 1 λ
θ
1
3
2 1
θ
(c)
Pathlength 3 – Pathlength 1 = nλ Pathlength 2 – Pathlength 1 = Pathlength 3 – Pathlength 2 = 2n – 1 λ
2
2 3
FIGURE 3.12–4 The geometry of diffraction for several families of crys-tal planes: (a) the兵0 0 1其 family in simple cubic; (b) the兵0 0 1其 family in BCC, neglecting the contributions from the parallel兵0 0 2其 family; and (c) the com-plete兵0 0 2其 family in BCC.
peak will occur at an angle in accordance with Bragg’s law. Now consider 兵0 0 1其 diffraction from a BCC material. Assume conditions are appropriate so that the X rays interfere constructively, as shown in Figure 3.12–4b. The atoms in the center of this and other cells create a plane of equal density to those of the 兵0 0 1其. These are the 兵0 0 2其 planes. These planes fall midway between successive兵0 0 1其 planes. At the angle appro-priate for constructive interference of兵0 0 1其, the 兵0 0 2其 produce destructive interference (see Figure 3.12–4c). Hence, no intensity maximum is observed for 兵1 0 0其 in BCC structures.
The concept of “missing” diffraction peaks from certain crystal structures can be extended to develop some general rules for diffraction. Two of the important rules are:
1. The sum 共h ⫹ k ⫹ l兲 must be even for diffraction to occur from 共h k l兲 in BCC structures, and
2. h, k, and l must be all even or all odd for diffraction to occur from共h k l兲 in FCC structures.
X-ray diffraction is an extremely powerful tool for investigating the structure of crys-talline materials. We have presented only the fundamental relationships between X rays and crystal structure. A number of complex X-ray techniques have been developed, but all the techniques rely on the basic concepts presented above.
...
EXAMPLE 3.12–2
Calculate the spacing of planes responsible for an intensity peak at 32.2⬚, assuming first-order diffraction. The wavelength of the radiation is 1.54`.
Solution
This is a Bragg’s law problem, so the relevant equation is 3.12–2:
2d sin ⫽ n
Solving for d:
d⫽ n 2 sin
Inserting the values n⫽ 1, ⫽ 1.54`,and ⫽ 32.2⬚ yields:
d⫽ 1.54`
2 sin 32.3⬚⫽ 1.445`
...
SUMMARY
...
Most solids exhibit short- and long-range order and are called crystalline solids. The atoms that make up crystals are arranged on a crystallographic lattice. There are 14 Bravais lattices, and all crystals belong to one and only one lattice type. A material that changes lattice type with temperature is called polymorphic. Some materials change crystal structure many times prior to melting. Unit cells are arranged in a periodic fashion and repeat indefinitely in three demensions. Miller indices are a shorthand notation used to refer to directions and planes in a unit cell.
The simplest crystals are those containing one element, or pure metals. Many metal crystals are based on a cubic unit cell: FCC or BCC. However, some are based on
KEY TERMS
...
anisotropic close-packed structures interstices octahedral site
atomic packing factor crystal lattice isotropic planar density ~P!
(APF) diffraction lattice polycrystalline
basal plane face-centered cubic (FCC) lattice parameter polymorphic
basis families of directions lattice point simple cubic (SC)
body-centered cubic families of planes linear density ~L! tetrahedral site (BCC)
hexagonal close-packed liquid crystal unit cell
Bragg’s law (HCP) Miller indices volumetric density
close-packed directions highest-density plane (V!
close-packed planes
SECTION 3.1 Introduction
SECTION 3.2
Bravais Lattices and Unit Cells
hexagonal or tetragonal unit cells. The repeat distance, or lattice parameter, of most metals is small— on the order of a few angstroms.
Metals are close-packed if the atoms occupy the maximum fraction of space possible in the unit cell. The atomic packing factor of a close-packed crystal is 74% and the coordination number is 12. Both FCC and HCP structures are as dense as possible —with close-packed planes and close-packed directions. One difference between HCP and FCC is the stacking sequence of the close-packed planes.
The crystallography of two-element metals or ceramics is more complex. Typically one component is positioned in SC, BCC or FCC positions, and the other component resides in the octahedral or tetrahedral holes in the unit cell. The lattice parameter of most two-component solids, such as an MX ceramic, is typically of the order of 5 to 10`.
Polymers crystallize only partially at best. Polymer crystals are characterized by large unit cells, perhaps as large as 20` in one direction. There are par ts of four or more mers and, hence, many atoms in a unit cell. Polymeric crystals are typically of low symmetry.
While most materials are polycrystalline, some materials are specially grown in the form of single crystals. Since they lack periodic interruptions in structure —grain boundaries—single crystals have unique and useful properties.
X-ray diffraction is a powerful technique for investigating the atomic structure of materials. Bragg’s law, n⫽ 2d sin , is the fundamental equation that governs the interaction of X rays with crystallographic planes. Using X-ray diffraction, we can deter-mine the crystal structure of a material and the precise positions of the atoms in the unit cell.
HOMEWORK PROBLEMS
...
1. Discuss the effect of temperature on short- and long-range order. Include the effect of melting.
2. Go find a printed textile fabric or wallpaper. Sketch the pattern. Does it have LRO?
Does a mural have LRO?
3. How does short-range order change when a crystal melts?
4. Sketch tetragonal and orthorhombic cells.
5. Based on a sample of printed textile fabric, show the unit cell, the lattice points, and the basis, and measure the lattice parameters.
Plane E
6. Draw a base-centered cubic structure. How is this crystal system included in Fig-ure 3.2–3?
7. Redraw the FCC structure with one of the atoms in the center of a face at the origin.
8. Do the corner atoms in the BCC touch one another? Do those of the FCC structure?
9. Optical microscopy is commonly used by mineralogists to identify composition. The macroscopic form of a crystal often, but not always, reflects the symmetry of the unit cell. Quartz crystals are pointed hexagons. Which cell might characterize this crys-talline form of SiO2?
10. Calculate the length of a face diagonal in Cr, a BCC metal.
11. Calculate the density of FCC copper using data based only on atom measurements. The measured value is 8.96 g/cm3.
12. Calculate the density of magnesium, an HCP metal, from atomic data, and compare the result with the measured value, 1.74 g/cm3. Account for differences.
13. Show why the c兾a ratio in HCP materials should be close to 1.63. Why does c兾a in some metals differ from 1.63?
14. Gadolinium changes from HCP to BCC upon heating at 1260⬚C. The lattice parameters for the HCP are a⫽ 0.36745 nm and c ⫽ 1.18525 nm. The lattice parameter of the BCC structure is 0.4060 nm. Calculate the volume change associated with the change in crystal structure.
15. List the five metals with the lowest densities. These materials are prime candidates for high ratios of strength or stiffness to weight (density). Note which of these metals are ones that you are familiar with.
16. If atoms are located at all the 1, 0, 0– and 1兾2, 1兾2, 0–type positions in a cubic system, what is the Bravais lattice?
17. Identify the plane in an FCC metal formed by the intercepts 1,⫺1, ⫺2. What is the normal to this plane?
18. Calculate the angle between关1 0 0兴 and 关1 1 1兴 in Al.
19. Construct a coordinate system at the center of a cubic unit cell with the axes parallel to the〈1 0 0〉 directions. Determine the tetrahedral angle, the angle between directions from the origin to two ends of any face diagonal.
20. What is the angle between关1 1 0兴 and 关1 1 1兴 in a cubic crystal? Between (1 1 0) and (1 1 1)? Between (1 1 0) and关1 1 1兴?
21. Give the indices of the points, directions, and planes in the cubic cells shown in Figure HP3.1.
Direction C
22. Give the indices of the points, directions, and planes in the cubic cells shown in Figure HP3.2.
23. Identity the plane formed by the intercepts 1兾2, ⫺1, 2 in a cubic cell. What is the normal to this plane?
24. Sketch all the members of the兵1 0 0其 family of planes in a tetragonal unit cell. Now sketch all the members of the兵0 0 1其 family in a similar tetragonal unit cell.
25. Calculate the planar density of atoms on the (1 0 0), (1 1 0), and (1 1 1) planes in an FCC unit cell.
26. Calculate the planar density of (1 0 0) and (1 1 0) in a BCC structure.
27. Sketch the atoms in the closest-packed direction on the closest-packed plane in Cr.
28. What is the packing factor of the unit cell shown in Figure 3.7–1?
29. Calculate the separation of close-packed planes in FCC and HCP structures.
30. Compare the linear density of关1 0 0兴 and 关1 1 1兴 in FCC crystals.
31. Fiber-reinforced composites are made from fibers and resin. Determine the maximum volume fraction of monosize-diameter fibers that can be put into a composite by calcu-lating the packing factor of close-packed rods.
32. What is the “coordination number” of the rods described in the previous problem?
33. Why do some metals crystallize into BCC structures when FCC is more dense and has a higher CN?
34. Suppose you discover a new material that is characterized by close-packed planes stacked in the sequence ABAC, repeating indefinitely. Is this discovery significant? Can you calculate the atomic packing factor of the new material?
35. Suppose a crystal of sodium, which has a BCC structure, has 1 in every 100 atoms missing from lattice sites. Calculate the effect on the APF.
36. What are the planes of highest density in FCC, HCP, and BCC? What are the directions of highest density within these planes?
37. The large cell shown in Figure 3.3–4b has a packing factor of 74%. Calculate the pack-ing factor of the primitive unit cell shown in Figure 3.3–4a.
38. Balls are randomly placed in a bin. The balls are jiggled until one close-packed layer forms. A second set of balls is then added to the bin, and the bin is jiggled until these balls assume a close-packed plane sitting in close-packed positions over the first set.
The procedure is continued with a third set. The stacking sequence is noted and the third layer removed. The procedure is repeated 100 times. Which stacking sequence occurs more frequently—ABC or ABA? Why?
39. Fill the octahedral hole in an FCC structure with the largest sphere that will fit. Do the same for a BCC structure. How many neighbors does the sphere touch in each case?
40. If all the octahedral and tetrahedral sites in the FCC structure were filled with atoms, what would the AP F be?
41. Why is the octahedral site in the BCC structure smaller than the tetrahedral site?
42. Compare the number and size of tetrahedral sites in the BCC structure with those of the octahedral sites in the FCC structure.
43. An impurity with a radius of 0.07 nm is introduced into crystalline Ca. How many impurities can be put into the structure? Give your answer in terms of a multiple of the number of calcium atoms. Repeat for an impurity with a radius of 0.04 nm.
44. Carbon is added to iron to form steel. The BCC structure of iron, called ferrite, is stable to 912⬚C, whereupon the structure becomes FCC, called austenite. Which form of iron do you anticipate is capable of dissolving more carbon? Explain your answer.
45. Calculate the lattice parameter of silicon.
46. Figure 3.7–4a shows the diamond cubic structure, which is characteristic of a number of covalent crystals. Calculate the theoretical density of diamond and silicon.
47. Calculate the density of the (SiO4)4⫺tetrahedron, the basic building block of silicates.
48. Calculate the lattice parameter of MgO. One ion is cubic close-packed and the octahe-dral sites are filled by the other ion. What structure is this commonly known as?
49. Redraw Figure 3.7–4b with Zn2⫹at the origin.
50. Sketch (1 1 0) and (1 1 1) in MgO, which has the NaCl structure. Indicate the locations of the tetrahedral and octahedral sites.
51. Predict the structure of NiO.
52. Predict the structure of SiC.
53. Predict the structure of CaO.
54. Calculate the density of CsBr.
55. Show why materials such as ThO2, TeO2, and UO2crystallize into the fluorite structure.
56. Calculate the density of UO2.
57. Calculate the density of CaF2. Which direction is most dense? What plane is most dense?
58. Calculate the O@Si@O bond angle in crystobalite.
59. What information do you need to facilitate calculation of the density of SiO2in the form of crystobalite?
60. Calculate the density of polyethylene, realizing from Figure 3.7–9 that there are two mers (each@CH2@CH2@) per unit cell. Explain why most commercial samples of polyethylene have a lower density, about 0.97 g/cm3.
60. Calculate the density of polyethylene, realizing from Figure 3.7–9 that there are two mers (each@CH2@CH2@) per unit cell. Explain why most commercial samples of polyethylene have a lower density, about 0.97 g/cm3.