crystal structure & atomic packing - nptelnptel.ac.in/courses/113105023/lecture3.pdf · crystal...

NPTEL Phase II : IIT Kharagpur : Prof. R. N. Ghosh, Dept of Metallurgical and Materials Engineering ||

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Module 3

Crystal structure & atomic packing

Lecture 3

Crystal structure & atomic packing


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Keywords: Point Lattice, unit cell, lattice parameter, crystal planes & direction, Miller indices, common crystal structure of metals, atomic packing, close packed planes, packing

sequence, interstitial sites,

Introduction

Metals are crystalline although they do not exhibit well developed crystal faces like inorganic compounds. Most metals we use are made of several crystals having different orientations. These are separated by irregular boundaries. Properties of metals are strong function of its crystal structure and orientation. This depends on the transformation processes the metal undergoes right from solidification and subsequent shaping & treating steps it passes through before being put to use. Therefore in order to understand & predict performance of metallic material it is necessary to have some idea about its crystal structure & orientation. This module gives a broad overview of relevant portions of crystal structure & orientation that would help you understand the basic concepts of physical metallurgy. Point lattice & unit cell

Atoms in crystals are arranged in a periodic fashion. Their locations are represented with the help of a regular array of points in space called point lattice. This consists of a number of points arranged in such a fashion in 3D that every point has identical surrounding. Figure 1 represents such an array. The smallest building block of this array consisting of eight points is called unit cell. The cell is completely defined by the length of its three edges and their subtended angles. This is illustrated in figure 2. These are: a, b, c (denoting the edges) & �� (denoting angles between the edges called crystal axes) are commonly known as lattice parameters.

Figure 1: the top left hand corner shows the smallest building block which if arranged in three directions as shown in top right hand corner would generate an a regular array of points as shown in bottom left hand corner. This is known as point lattice.


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Figure 2: Top left hand corner shows the smallest building block of a point lattice. The three arrows indicate three axes named a, b, c. Cells having points only at its corner is called primitive lattice. A corner of the cell is equally shared by 8 neighboring cells. Since there are eight corners in a unit cell number of lattice point per unit cell Np = 1. The sketches at the bottom shows possible additional locations of lattice points in the case of simple cubic lattice where a = b = c and �� = 90°.

In addition to the points at the corner of the cells it is possible to place additional points without violating the definition of point lattice. These are body center, base centre and face centre. The bottom row of figure 2 shows two more unit cells with additional points marked by black solid spheres. Consider the body centered cubic cell. It has an additional point right at its centre. In an array this would belong exclusively to this cell. This is why number of points / unit cell in a body centered lattice Np= (1/8) x 8+1 = 2 Problem: Show that the number of points / unit cell in a face centered cubic lattice is 4. Answer: Point at the center of a face is shared equally by two adjacent cells. There are six faces in a unit cell. Therefore Np= (1/8) x 8+(1/2) x 6 = 4

Figure 3: Six parameters namely lengths of the edges (a, b, c) and the angles between axes (��) as shown in the figure are needed to describe a unit cell of the lattice.


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Depending on the relations between the edges or the angles unit cells can have 7 different shapes each representing a specific crystalline shape. This is illustrated in the following table.

Table 1: Unit cells representing different types of crystals The unit cells listed in table 1 represents primitive cells where there are lattice points only at the corners of each cell. In addition there can be a few more lattice type where there are additional points at body centre, face center or base center without violating the definition of point lattice. For example cubic unit cell can have two additional unit cells one with a point at the center and the other at face centers. It is possible to represent all known crystal structure using such lattices. The total number of point lattice (also known as Bravais lattice) is 14. Fortunately the crystal structure of metal is much simpler in comparison to those of the other crystalline solids. Crystal structure of common of metals

Three most common crystal structures of metals are body centered cubic (bcc), face centered cubic (fcc) and hexagonal close packed (hcp). In terms of point lattice these are represented by body centered cubic lattice, face centered cubic lattice and primitive hexagonal lattice. In the case of the first two each point represents one metal atom, whereas in the case of hcp structure each point represents a pair of points. The way the atoms are arranged in these three are illustrated in figure 4. It only shows the location of atoms but does not reveal the way these are packed.

Cubic a = b = c � = β = γ = 90º

Tetragonal a = b ≠ c α=β=γ=90º

Hexagonal a=b≠c α=β=120º,γ=90º

Orthorhombic a≠b≠c α=β=γ=90º

Rhombohedral a=b=c α=β=γ≠90º

Monoclinic a≠b≠c α=γ=90º≠β

Triclinic a≠b≠c α≠β≠γ


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Figure 4: Shows the way metal atoms are arranged in bcc, fcc & hcp crystal. It also indicates names of common metals having such crystal structure.

The atoms are placed in the lattice as closely spaced as possible. Assume these to be hard spheres. In bcc structures the atoms touch each other along its diagonal. Whereas in fcc the atoms touch each other along its face diagonal. This is shown in figure 5.

Figure 5: Shows the way atoms touch each other in fcc, bcc & hcp structure. In the case of hcp structure only one atom has been shown in the top layer for reasons explained later.

Problem: Assuming atoms to be hard spheres estimate packing density (PD) of atoms in bcc structure. Answer: Let the radius of atom be R. There are two atoms in a unit cell having lattice parameter = a. In a bcc structure atoms occupy the 8 corners and the center of a cube. The number of atoms / unit cell = 1+ (1/8) x 8 = 2. Atoms touch each other along the cube diagonal. Here as well the lattice parameter represents the distance between two consecutive atoms along the edge of a cube. Therefore the diameter of an atom = a√3/2. The packing density = Net volume of all atoms in a unit cell / volume of unit cell = {(2π/6) (a√3/2)^3 /a^3} = 2πx3√3/6 = 0.68. The figure below would help you understand the basis of this calculation.


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Figure 6: Packing density of fcc structure: There are 8 corners and 6 faces in a cube. The number of atoms / unit cell in fcc crystal is 8 x (1/8)+6 x (1/2) = 4. The distance between two consecutive atoms along the edge of the cube is called its lattice parameter. Often it is denoted as ‘a’. Since atoms touch each other along the face diagonal of the cube, the diameter of an atom (assuming it to be a hard sphere) is therefore equal to (a/√2). The packing density = Net volume of all atoms in a unit cell / volume of unit cell = {(4π/6) (a/√2) ^3 /a^3} = 4πx2√2/6 = 0.74. Hexagonal close packed structure is also as closely packed as fcc. This clearly shows that the packing density of bcc metal is less than that of hcp & fcc. It also gets reflected on its coordination number. The atom at the center of bcc structure has 8 nearest atoms located at the corner of the cube. By simple construction involving joining of atoms it is possible to show that all atoms in this crystal have identical surrounding having 8 nearest neighbors. It is also evident from figure 4. Arrangement of atoms in a close packed plane: If you try to place identical solid spheres in a plane as closely packed as possible you would notice that each sphere is surrounded by six neighbors. Let this layer be denoted as ‘A’. Try to place a second layer of similarly arranged atoms on top of the layer ‘A’. If you want them to be as closely spaced as possible all atoms must occupy the valleys. You would notice that an atom in the next layer would block one of the valleys completely and a part the three nearest valleys. Let us call this layer as ‘B’ and the sites which are partially blocked by the atoms as ‘B’. This would now provide two distinct types of valleys for the next layer of atoms; they are ‘C’ and ‘A’. If the third layer of atoms happens to occupy site ‘A’ you would get a hexagonal close packed structure. Whereas if it occupies site ‘C’ you would get a face centered cubic structure. Since each layer is packed as closely as possible both the arrangements should have identical packing density (volume fraction occupied by hard spherical atoms). Look at the arrangement of

Fig 6


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atoms in one hexagonal unit of an hcp crystal. The distance between two nearest atoms in the close packed plane (say layer A) is ‘a’ and the distance between two identical ‘A’ layers of atoms is ‘c’. For an ideal hexagonal close packed structure c/a ratio is 1.633. The number of nearest neighbors is a crystal is known as its co‐ordination number. In an hcp structure each atom in the close packed plane is surrounded by 6 atoms. Over and above this it has 3 atoms above it and 3 atoms below it. Therefore its co‐ordination number is 12. The packing sequence of close packed planes in an hcp crystal is ‘ABABAB‐‐‐‐‐’.

Figure 7: Shows how atoms could be arranged in a plane. Look at the top layer. The central atom is surrounded by six identical atoms. Each layer of atom is represented by different colors. Look through the gaps between the atoms. Beneath the yellow layer only pink layer is visible through three of the six gaps (valleys) and through the remaining three you would see partly pink and partly white color. This is because the pink layer blocks three valleys completely and three partially.

In face centered cubic crystal close packed layers are arranged in sequence ‘ABCABC‐‐‐‐‘. If you look at the atomic arrangement at an angle other than perpendicular to the close packed plane you would soon realize that the atoms occupy the corner and the face centers of a cube. Atoms touch each other along the face diagonal. The close packed planes are the diagonal planes of the cube. This is illustrated with help of figure 7. Ideal hexagonal close packed structure: In hcp structure atoms are arranged in a sequence ‘ABABAB‐‐‐‘. The packing is as dense as in fcc. Along the close packed plane the distance between atom is ’a’. Therefore ‘c’ which represents the distance between two identical layers ‘a’ is no longer independent. In other words hcp structure has a fixed c/a ratio. Figure 8 illustrates how this can be estimated.


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Figure 8: The close packed layers are represented by white & pink balls arranged one over the other. The bottom layer is ‘A’ the second layer is ‘B’ and the third layer is again ‘A’. To estimate c/a ratio look at the construction joining centers of three adjacent atoms p, q, r (in layer A) with s in layer B. The distance between p and r = a; whereas the distance s and the plane containing prq is su = c/2. Note that pqr is an equilateral triangle and t is the midpoint of side qr. Using the relations between the sides c/a can be shown to be equal to 1.633. Steps to do this are given above.

If atoms in metals are packed as described above it should be possible to derive a relation between its density and lattice parameters. This is based on certain assumptions. These are as follows: i) atoms are considered to be hard sphere ii) all lattice sites are occupied by atoms. If lattice parameter of a cubic crystal is ‘a’ and A is the atomic weight and No is the Avogadro number, and n in the number of atom / unit cell, the density (�) is given by

. The following figure gives illustrative calculation for a few fcc metals.


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Al 2.70 2.70 27

Ni 8.98 8.90 59

Cu 9.03 8.96 64

Pt 21.50 21.45 195

Pb 11.33 11.68 207

Figure 9: Shows estimation of density of a few common metals having fcc structure from their lattice parameter. In spite of the assumptions estimates are very close to their reported density.

Miller indices: system of representing crystal directions & plane

Since metals are crystalline and its properties are dependent not only on its crystal structure but also the way these are arranged in solid. Therefore it is necessary to know how to specify or represent crystal directions, planes and their orientation. Miller indices are used to denote these. This uses crystal axes as the frame of reference. First step to find the indices of a direction is to draw a parallel line passing through the origin. It uses three numbers each representing the number of steps one need to move along the three axes to reach the end point from the origin. The numbers are enclosed within a pair of box brackets. Figure 10 illustrates how crystal directions are represented by a set of such lines. An arbitrary direction is often denoted as [uvw].

Figure 10: The lines; a, b, c are the three crystal axes. The distance between the grid lines on the plane containing axes a & b are unit distances along these axes. A crystal direction is denoted within box bracket as shown. For example [110] denotes a line passing from origin. The notation gives the displacement along the three crystal axes. Note that the third digit which denotes displacement along c axis is zero. The direction lies on the plane containing the axes a & b. The first digit is the displacement along the axis a and the second digit is the displacement along the axis b. This is illustrated by red arrows. Likewise [210] denotes two steps along a and one step along b.

Problem: What are the Miller indices of the directions represented by the line A & B shown in the figure 10?

A

B


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Answer: Note that to represent direction A you need to move two steps against axis a and one step along axis b. Movement in negative direction is represent by a bar drawn on the top of the respective digit. Thus A is denoted by 210 & B is 110

Figure 11: Problem: Find the Miller indices of the directions A, B, C, & D. The dotted lines represent lines joining lattice points. Answer: Note three directions A, B, & C are lines passing through origin. In case of D you need to imagine a line parallel to D but passing

through O. A: 120 ; B: 211 ; C: 211 ; D: 110

Note that the indices as described above only give an idea about how a direction is aligned with respect to the three crystal axes. We shall be dealing mostly with cubic lattice where the axes are orthogonal. In such a case these indices denote a vector in Cartesian coordinate. Consider a general crystal vector denoted as: ̅ . Its magnitude is

given by| | . There are similar expressions for every crystal other than the cubic crystals. Interested reader can refer to books on crystallography for such expressions.

Figure 12: Illustrates how to find the Miller indices of crystal planes. Three crystal axes are denoted by a, b & c. The plane enclosed by a set of red lines intersects the respective axes at ma, nb & pc. It suggests that m, n & p are fractions. However it need not always be so. The steps to convert these into Miller indices are as follows:

convert the intercepts as reciprocals:

next try reduce these to integers by multiplying with a suitable number say M so that

: The plane is represented as

The following table gives a few examples of how Miller indices are determined for a set of common planes in crystals.

O

A

B

C

D

c

ba


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Table 2: A few examples of how to find Miller indices of planes

Intercepts: m ∞ ∞; reciprocal 1/m 0 0; indices (100)

Intercepts: 1 1 1; reciprocal 111; indices (111)

Intercepts: 1 1 ∞; reciprocals 1 1 0; indices (110)

Problem: Find out the Miller indices of the planes whose sides are represented by red & green line as shown in figure 10. It also illustrates that all parallel planes have identical indices.

Figure 13: Indices of the plane with red border: note that the intercepts on the axes a, b, c are 1,

1 & ‐1. Therefore its indices are 111 . Indices of the plane with green border: note that the plane as marked is too small to find its intercepts. This either needs to be extended or another plane which is parallel can be constructed. Triangle with green dotted line as shown is one of the parallel planes intercept with the axes is readily visible. Its indices is (111)

Three most common crystal structures found in metallic materials are fcc, bcc and hcp. Let us try to find the indices of their close packed planes and directions. It would also be interesting to find out the number of such planes and directions.

Figure 14: This shows the location of atoms in a fcc structure. (111) is the indices of one close packed plane. Two of these having identical indices are shown as triangles. There are other similar planes as well. These are 111 , 111 ,

& 111 . Note that the plane with indices

111 is parallel to the plane 111 . There are four close packed planes. One of the close packed directions is shown as a line with an arrow head. Its indices are [110]. You could try to arrange the indices as shown for the planes & show that there are six possible close packed directions.


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Figure 15: This shows the locations of atoms in a bcc structure. (110) is the indices of one the close packed planes. There are several other similar planes. These are (011),

(101), 110 , 101 , & 011 . Note that plane

with indices 110 is parallel to the

plane 110 . Thus there are six close packed planes. One of the close packed directions is shown as a line with an arrow head. Its indices are [111]. You could try to arrange these as shown for the planes & show that there are four possible indices.

The examples in figures 14‐15 show that in a crystal there are several planes and directions where atoms are arranged in an identical fashion. Crystals having maximum symmetry the number of such planes and directions is significantly large. For example in a cubic crystal there are 4 identical planes of type (111). A set of such planes is represented as {111}. It includes all possible combinations having both positive and negative indices. Similarly a set of directions having identical atomic arrangements is represented as <111>.

Figure 16: Illustrates common planes & directions in a hexagonal crystal. If the normal Miller indices are used similar planes do not have similar indices. To overcome this problem an additional axis is used. Thus instead of using (hkl) it is preferable to use (hkil) for planes. Likewise direction [uvw] is replaced by [uvtw]. The fourth index is only a dummy. It can be calculated using the expression as shown in the adjacent figure. Note that in a hcp structure basal plane (0001) is the close packed plane and there are three close packed directions lying on it. Close packed direction: [100] or [110] using conventional Miller indices whereas using four indices these are

2110 1120 The reason for using four indices is evident from the illustration given in figure 16. In terms of conventional Miller indices (100) and 110 denote prism planes. However from the indices it does not appear that these are similar planes. If you use four indices these become 1010 & 1100 . They display their identical character. Likewise the pyramidal planes in terms of conventional Miller indices are (101)& 111 . Although identical it does not appear to be so; whereas in terms of four indices these become 1011 & 1011 ; both look to be similar.


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Summary

In this chapter we learnt about common crystal structures of metallic materials; fcc, bcc & hcp. How are the atoms arranged in the lattice? What are the relations between lattice parameter and atomic radius? Estimation of packing density & coordination number revealed that both fcc & hcp are equally packed whereas bcc has relatively lower packing density. The concept of Miller indices for the representation of crystal planes & directions has been introduced. Indices of close packed directions in the three common crystal structures found in metals were also determined. The need for using four digit Miller indices for a hexagonal crystal has been explained. Exercise:

1. Sketch an unit cell and show the following planes (a) (112) (b) (101) (c) 111 (d) (123)

2. Find out the indices of the direction joining following points in a cubic lattice: (a) 1,1,1 with 1,1,2 (b) ‐1,1,1 with ‐3, 2, 1 (c) 1,1,2 with 3,2,‐1

3. Show the atomic arrangements in (111) plane of face centre cubic structure and

show the following directions 110 , 101 , 011 , 211 , 121 , 112

4. Estimate the density of platinum and lead from their lattice parameters at room temperature. Both are FCC. Compare the theoretical density with experimental values. Which is closer? Why?

Answer:

1.

2.

If the coordinate of first point is u1, v1, w1 & the second point is u2, v2, w2 the indices of the line joining the two points can easily be shown with the help of the diagram on the left is [u2-u1, v2-v1, w2-w1]. Line joining point 111 with 112 is shown try others.

111

112

[001]

b

c

a

(112)

(101)

(123)

c

b a

111


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3.

4. The relation between density (�) & lattice parameter (a) is given by where n=

number of atom /unit cell, A = atomic weight & N=Avogrado number. For platinum

A=192.09, n=4 fcc, N=6.02x1023 & a=3.9239 Angstrom. On substitution�� = 21.45 gm/cc.

Experimental density of Pt = 21.47. For lead A=207.2, n=4 fcc, a=4.9502Angstrom. On

substitution in the expression for density �= 11.35 Experimental density =11.34. The

estimation of x‐ray density in based on assumptions that all sites are occupied and atoms are

hard. If there are vacancies in the lattice real density should be less than x‐ray density. If

atoms are soft the density should be higher.

In a cubic crystal a direction [uvw] lies on a plane (hkl) then hu+kv+lw=0. Using this three close packed direction lying on (111) are [-110], [-101] & {0-11]. These are shown by firm line. Three [112] directions are [11-2], [-211] & [1-21]. These are shown as dotted lines.

110

101

011

211

121

211

crystal structure & atomic packing - nptelnptel.ac.in/courses/113105023/lecture3.pdf · crystal...

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