Chapter 3 Highlights: 1. 3 types of materials- amorphous, crystalline, and polycrystalline. 2. Understand the meaning of crystallinity, which refers to a regular lattice based on a

repeating unit cell. 3. 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 #3 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 x-ray crystallography pattern to the crystal structure and to use

the diffraction pattern to obtain the lattice constant or atomic radius. Notes: Metallic Crystal Structures Metals, ceramics, polymers, etc. may be either: 1. Crystalline- Si for semiconductor and photovoltaic devices. 2. Amorphous- corrosion-resistant Ni plating, no grain boundaries. 3. Polycrystalline- structural steels, other industrial metals (see figure).

c03f35 Figure 3.35 (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 2: 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: Infinite array of points in a solid. Unit cell: Smallest repeating configuration of atoms from which lattice can be constructed.

Table below illustrates the different crystal systems, with different combinations of a,b,c,α,β,γ. In ES 260, 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.

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FCC unit cell (Figure 3.1)

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A) # of atoms, n = 8(1/8) + 6(1/2) = 4 B) Coordination number = 12, count nearest neighbors. C) Show drawing of one face, relate lattice constant (a) to atomic radius (R), show that

Ra 22= . 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 (3.5):

NVnA=

Ac

ρ

NR24A=

N)R2(24A=

A3

A3

ρ

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 cell unit Totalcell unit withinatoms ofVolume = APF

a3Rn4=

a

R34 n

= APF 3

3

3

3

ππ

)R23(2R(4)4= APF

3

3π

0.74= 23

= APF π

Example: Calculate the radius of a iridium atom, given that it has a FCC structure, with a

density of 22.4 g/cm3 and an atomic weight of 192.2 g/mol. Start with the equation that we derived above for the density of a FCC material:

NR24A=

N)R2(24A=

A3

A3ρ

( )

( ) ( )molxR22

molg4= cmg/10023.6

/2.192/4.22233

3

3243 10518.2 −−= cmxR

nmcmxR 136.01036.1 8 == − Next two figures illustrate body-centered cubic (BCC) and hexagonal close-packed (HCP) crystal structures. You should be able to do the above calculations (A-E) for the BCC unit cell, and for all other cubic unit cells covered in this course.

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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.

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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 2.11 (below). You might expect metallic bonding to give the closest-packed structure, with the maximum APF and maximum coordination. Both FCC and HCP have APF=0.74 and coordination # of 12, which are the highest possible, so about 3/4 of metals are FCC or HCP. But why do other metals have the BCC structure (APF=0.68, coordination # = 8)? This is complicated but involves some degree of covalent bonding through the d-orbitals of transition metals. Later we will introduce the diamond crystal structure (silicon, germanium, diamond), where the APF is only 0.34. This structure is favored due to covalent bonding in these materials, which requires a coordination number of only 4.

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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 rC/rA, 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.

1) Take the many possible geometrical arrangements of the large anions as in VSEPR (linear, triangular, tetrahedral, octahedral, etc.) 2) Choose the structure where the cation just fits into the interstice. This maximizes the Coulomb attraction and binding energy. Table 3.3 below is constructed by the above two principles. The ratio rC/rA given in table 3.3 can be derived from those principles. The most common coordination numbers are 4, 6, 8. Ionic radii are given in table 3.4 below.

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c03tf04 Example: Show that the minimum cation-to-anion radius ratio for a coordination number of 6

is 0.414.

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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 2rA + 2rC and the hypotenuse of length 2rA. Therefore,

CA

A

r+rr=

2= )(

222245cos °

This can be solved to yield,

414.12

1 = 2= rr

A

C+

414.0= rr

A

C

This is the borderline between coordination numbers of 4 and 6 in table 3.3. A-X 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 (rC/rA=0.402, zinc blende structure), NaCl (rC/rA=0.564, rock salt structure), and CsCl (rC/rA=0.939, CsCl structure).

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In the figure above, the ions are actually touching along each edge of the cube, so a = 2rA + 2rC. Counting ions, nA = 8(1/8) + 6(1/2) = 4 and nC = 12(1/4) + 1 = 4.

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For the CsCl unit cell above, ions are touching along the body diagonal, similar to the BCC unit cell. Counting ions, nA = 8(1/8) = 1 and nC = 1.

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For the zinc blende structure above, ions connected by dotted lines are touching. If you can obtain the Cartesian coordinates (xi, yi, zi) of a cation and anion that are touching, their separation distance (rA + rC) is equal to their separation according to (Δxi

2 + Δyi2 + Δzi

2)1/2

AmXp 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 CaF2, show figure 3.8 below. Same as CsCl, but half of the cation (Ca) sites are empty.

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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 (3.6):

( )AC

AC

NVAAn ∑ ∑+=

'ρ

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. Example: MgO has the rock salt crystal structure and a density of 3.58 g/cm3. What is the

lattice constant, and how does this compare to the sum of the ionic radii in table 3.4?

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 3.5 (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:

( )AC

AC

NVAAn ∑ ∑+=

'ρ

( )( )molxa

molgcmg/10023.6

/00.1631.244/58.3 2333 +=

3233 10478.7 cmxa −=

nmcmxa 421.01021.4 8 == −

One expects nmrra CA 424.022 =+= . 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, C60, and nanotubes, as shown in figures below.

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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 3.7 and 3.8 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. 2. The 3-dimensional coordinates of the vector endpoint are determined in terms of the unit

cell dimensions. z wc+ yvb + xua= v ˆˆˆ

3. Multiply or divide by a common factor to reduce [uvw] to the smallest set of integers.

4. Refer to this direction as the [uvw] direction. [uvw] = [-u-v-w] All parallel vectors are considered to represent the same crystallographic direction. For cubic unit cells, a=b=c.

c03prob D) Taking the origin of the coordinate system as the vector origin, the vector endpoint is at

aaa21,

21,

21

−− . In terms of the lattice constant (a), the vector endpoint is at

−−

21,

21,

21

. Multiply through by 2, and this direction is [ ]111 .

A) The vector endpoint is at aaa31,, − . In terms of the lattice constant (a), the vector

endpoint is at

−

31,1,1 . Multiply through by 3, and this direction is

−

133 .

B) The vector endpoint is at aa21,0,

32

−− . In terms of the lattice constant (a), the vector

endpoint is at

−−

21,0,

32

. Multiply through by 6, and this direction is [ ]304 .

C) The vector endpoint is at aaa61,,

21

− . In terms of the lattice constant (a), the vector

endpoint is at

−

61,1,

21

. Multiply through by 6, and this direction is

−

163 .

Crystallographic Planes 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.

2. Note the three axes intercepts and express them in terms of the lattice constants a,b and c.

)zlc,ykb,x(ha ˆˆˆ

3. Replace h, k and l with their reciprocals. 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.

c03prob A) Take the coordinate system origin at the corner usually labeled (a, a, 0). From that origin,

the axes intercepts are aaa ,, −− . 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 ( )111 .

B) Take the coordinate system origin at the corner usually labeled (0, 0, 0). From that origin,

the axes intercepts are ∞,31,

21 aa . In units of lattice constant (a), this is ∞,

31,

21

.

Taking reciprocals, with no need to multiply or divide by a common factor, the final notation for this plane is ( )032 .

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 ( )011 plane has the same atomic configuration as the ( )101 , ( )110 ,

( )101 , ( )011 , and ( )110 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 x-ray diffraction and slip. One can also define the <100> family of directions, which includes the directions <100>, <010>, <001>.

Linear and planar atomic densities The linear and planar atomic densities are defined as:

vectordirectionofLengthvectordirectiononcenteredatoms of= Atomic Linear #ρ

planeofAreaplaneaoncenteredatoms of= Atomic Planar #ρ

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, 2)4/1(41 =+=n . 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 ( ) 222 aaaA == . Now, determine the planar atomic density (PAD):

222a

PAD =

Remember, in a BCC unit cell3

4Ra = , so substitution yields:

22 1623

342

2RR

PAD =

=

Problem: What is the linear atomic density along the BCC [001] and

− 111 directions?

You should be able to draw in these directions in the figure above.

Along the

− 111 direction, this is the body diagonal, and:

RRaLAD

21

343

232

=

==

Along the [001] direction (this is one side of the unit cell), the linear atomic density (LAD) is:

RRaLAD

43

3/411

===

X-ray 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 x-rays. The utility of x-ray 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 x-ray diffraction patterns for Pb (FCC) and W (BCC):

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.

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In the last figure, the extra segment in the light path is SQT. If SQT is an integral # of wavelengths, constructive interference occurs.

QT + SQ= nλ

θθθλ sinsinsin d= 2 d + d= n hklhklhkl This is Bragg's law for diffraction, where the quantity dhkl is the distance between parallel (hkl) planes. For cubic systems,

l + k + h

a= d222hkl

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 x-ray 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 2θ.

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Example: Label the first 6 lines of the x-ray diffraction pattern of Pb, shown above. Use the

first three lines to calculate dhkl and a. Assume λ = 0.1542 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 x-ray diffraction pattern, and then calculate the interplanar spacing and the lattice constant from the first 3 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 x-ray diffraction pattern. To minimize θ in equation (3.15) we must maximize dhkl. From equation (3.16) this occurs for the minimum values of h,k and l. I will use equation (3.11) to calculate dhkl for several combinations of h,k and l.

832201

a= d ,4

a= d ,a= d 20011

911233112

1a= d ,

1a= d ,a= d 3122

18a= d ,

1a= d ,a= d 420400333

627

a) Choosing the 6 maximum values of dhkl from above, the first 6 peaks in order are

(111), (200), (220), (311), (222), and (400).

b) Using equation (3.15):

nm 0.22= 2

nm 0.1542= )( 2

= d 11 797

232sinsin1

°θ

λ

nm 0.= 2

nm 0.1542= )( 2

= d 200 2430

237sinsin

°θ

λ

nm 0.1= 2

nm 0.1542= )( 2

= d 21 728

253sinsin0

°θ

λ

c) After rearrangement, equation (3.16) yields the lattice constant:

l+k+h da 222hkl=

For the first 3 diffraction lines one obtains a = 0.4844, 0.4860, and 0.4888 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 Ra 22= , one obtains R = 0.172 nm, close to that given in the front cover of the textbook (0.175 nm).

This is similar to how x-ray 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 X-ray Diffraction X-ray 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. X-ray diffractometers are commonly found in academic and industrial research laboratories. One exotic example of the utility of x-ray crystallography is the determination of protein crystal structures. Known information about protein structures is summarized in the protein data bank: http://www.pdb.org/pdb/home/home.do 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 x-ray 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 x-ray diffraction is to monitor recrystallization of photovoltaic thin films, which absorb photons and convert them to electrons in solar cells. Here are x-ray diffraction patterns of particulate CIGS films before (A-C) and after (E-F) annealing:

The x-ray 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 x-ray diffraction peaks. The example below illustrates the use of x-ray diffraction to provide real time monitoring of Cu receyrstallization, where the Cu thin films will be employed as ULSI interconnect materials.

Immediately following electrodeposition, the Cu thin film above exhibits grain boundaries, which will increase the Cu film resistance (introduced in Chapter 12). The Cu grain size will increase slowly at room temperature, and more rapidly with low temperature annealing. This can be monitored by time resolved x-ray diffraction, as shown below: