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Solid State Physics

Homework Set 2

SolutionsGreg Balchin

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Ibach and Luth Problem 2.3

In order to answer this question we must distinguish between an ordered system and a disorderedsystem. An alloy, consisting of atoms A and B, is disordered if the A and B atoms are arrangedrandomly on lattice sites. A disorderd alloy will occur for a general value of the concentration, x, inthe compound AxB1x. However, for certain values of x, for example x =

1

4, 12

, or 34

, the alloy will

be ordered [1]. It is important to note that a disordered alloy possesses a lattice structure. This is incontrast to an amorphous system, such as a glass, which has no underlying crystal structure.

A method for determining the effect of alloying is to measure the residual resistivity of the com-pound as a function of the concentration, x. Residual resistivity is defined as the resistivity in the limitas temperature goes to zero [1]. The resistivity increases as the alloy becomes more disordered anddecreases as the alloy becomes more ordered. Thus, a minimum in the resistivity versus concentrationindicates an ordered alloy. The data by Johansson and Linde [2] on the binary system CuxAu1xshows two minima in the resistivity as shown in the figure below.

Figure 1: Resistivity versus atomic percent of gold. The original data and diagram are from Johanssonand Linde. The plot above is from Kittel.

The first minimum occurs at 25% Au and the second minimum occurs at 50% Au. This impliesthere are two ordered systems of the form Cu3Au and CuAu. However, this tells us nothing aboutwheather these ordered phases are fcc based structures or some other type of structure. To get this

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information we must look at the Au-Cu phase diagram shown below. This diagram is from Massalski[3].

Figure 2: Phase diagram for the Au-Cu system.

The phase diagram shows the Cu3Au phase around 25% Au and the CuAu phase around 50%Au. There are actually two CuAu phases referred to as CuAu I and CuAu II. There is also a thirdphase CuAu3, which did not show up in the resistivity measurement of Johansson and Linde. This isprobably because the diffusion time for this phase is longer than the other two. Hanson [4] indicatesthat the Cu3Au phase is an fcc based structure. This is referred to as the L12 structure for the Au-Cualloy. The CuAu I phase has a tetragonal based structure, referred to as the L1 0 type structure. CuAuII has an orthorhombic based structure which is closely related to the CuAu I structure [4]. Massalski[3] indicates that the CuAu3 based structure is L12. For the problem we are considering here we willignore the CuAu3 phase because there are many unanswered questions concerning this phase.

Thus, we see that the Au-Cu system will form an ordred alloy with an fcc based lattice of the formCu3Au (75% Cu and 25% Au) below 400

C. The Au atoms will occupy the 000 position and the Cuatoms will occupy the 1

2

1

20, 1

2012

, and 0 12

1

2positions [1].

In order to experimentally determine wheather the alloy is ordered or not we see that we canget some information from a measure of the residual resistivity. However, a better way is to use themethod of x-ray diffraction. Remember that a disordered alloy possesses a lattice structure, thus, adisordered alloy will produce sharp x-ray diffraction lines as indicated in the x-ray powder photographshown below [1].

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Figure 3: X-ray powder photograph for the disordered Cu3Au alloy. This photograph is from Kittel.

The above x-ray powder photograph for the disordered Cu3Au alloy is what you would expect ifeach lattice site were occupied by only one type of atom. Thus the effective scattering power of eachplane is equal to the average of the Au and Cu scattering powers [1].

An x-ray powder photograph of the ordered Cu3Au phase shows extra diffraction lines which arenot present in the disorderd phase as shown below [1].

Figure 4: X-ray powder photograph for the ordered Cu3Au alloy. This photograph is from Kittel.

The extra diffraction lines are referred to as superstructure lines.

As an aside, refer back to the Au-Cu phase diagram.

Beginning at high temperatures and going down the diagram we see that we are, at first, in aliquid phase. This is indicated by L in the diagram. As we continue down the diagram we cometo two lines which begin just below 1100 C. These lines are phase boundaries. Between the twolines we have a phase which consists of both a liquid and a solid disordered alloy. This is just likehaving a stable ice/water mixture. Continuing down, below 899 C at 56.5% Au, we have only a soliddisordered phase. As we continue down the diagram at around 400 C we encounter the solid orderedalloy phases of Au-Cu. Here, we see, several phases exist depending on the concentrations of Cu andAu.

Not indicated in the phase diagram is the behavior of the phase boundary lines for the solidordered phases as we continue down in temperature. According to the third law of thermodynamicsthe entropy of a system approaches a constant value in the limit as temperature goes to 0 K. Anotherway to say this, according to Nernst, is that at 0 K the entropy difference vanishes between all of

the configurations of a system which are in internal thermal equilibrium. See Kittel and Kroemer[5]. Thus, we would expect the solid phase boundary curves to taper inward as the temperatureapproaches 0 K. This is indicated below.

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Temperature

Concentration

0 K

Figure 5: Behavior of solid ordered alloy phase boundary as temperature decreases.

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The following solution to this problem can be found in Ashcroft and Mermin [6].

Crystals can be divided into seven crystal systems with fourteen Bravais lattices. These aredescribed in the table below.

Crystal System Bravais Lattice

Cubic simple cubicbody-centered cubicface-centered cubic

Tetragonal simple tetragonalcentered tetragonal

Orthorhombic simple orthorhombicbased-centered orthorhombicbody-centered orthorhombicface-centered orthorhombic

Monoclinic simple monocliniccentered monoclinic

Triclinic triclinicTrigonal trigonal

Hexagonal hexagonal

A diagram of each crystal structure is shown in the figure below.

Figure 6: The different crystal systems: (a) cubic, (b) tetragonal, (c) orthorhombic, (d) monoclinic,(e) triclinic, (f) trigonal, (g) hexagonal. This diagram is from Ashcroft and Mermin.

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Crystallographic Point Groups

There are 32 crystallographic point groups. These arise from the following symmetry operations.

(1) Rotation through integral multiples of 2n

about some axis.(2) Rotation - Reflection which consists of a rotation followed by a reflection in a plane perpendicular to

the rotation axis.(3) Rotation - Inversion which is a rotation followed by an inversion in a point lying on the rotation axis.(4) Reflection which takes every point into its mirror image in a plane.(5) Inversion

In what follows, Schoenflies notation will be used to indicate the point groups.

O: octahedron full symmetry group of the cube.T: tetrahedron full symmetry group ot teh tetrahedron.C: cyclic

D: dihedralS: spiegel (mirror)h: horozontalv: verticald: diagonaln: n-fold axis

Cn: n-fold rotation axisCnv: n-fold rotation axis plus a mirror planeCnh: n-fold axis plus a single mirror plane perpendicular to the axis

Sn: n-fold rotation-reflection axisDn: n-fold rotation axis plus a 2-fold axis perpendicular to an n-fold axisDnh: Dn plus a mirror plane perpendicular to the n-fold axisDnd: Dn plus mirror planes containing the n-fold axis which bisect the angles between the 2-fold axesOh: full symmetry group for the cube with improper rotationsTd: full symmetry group of the tetrahedron plus improper rotationsTh: T plus inversion

The 32 point groups are:

Cubic: Oh, O, Th, Td, T

Tetragonal: C4, C4v, C4h, S4, D4, D4h, D2dOrthorhombic: C2v, D2, D2hMonoclinic: C2, C2h, C1hTriclinic: C1, S2Trigonal: C3, C3v, S6, D3, D3dHexagonal: C6, C6v, C6h, C3h, D6, D6h, D3h

These point groups can be shown diagramatically in the following figures.

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Figure 7: The cubic group. This diagram is from Ashcroft and Mermin.

Figure 8: The noncubic groups. This diagram is from Ashcroft and Mermin.

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Simple Cubic

The simple cubic structure is shown below.

a

r

Figure 9: The simple cubic structure (top view).

The radius of each sphere is

a = 2r r = a2

Each corner contains1

8 of a sphere, thus, there are 8 18

= 1 sphere in the simple cubic structure.The volume of the sphere is

Vsph = 8

1

8

4

3r3 =

4

3a

2

3=

4

3a3

8

Vsph =

6a3

The packing fraction is

f =Vsph

Vcube=

a3

6

a3=

6= 0.52

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Face-Centered Cubic

The face-centered cubic structure is shown below.

a

r

r

r

2r face diagonal

Figure 10: The face-centered cubic structure (top view).

The face diagonal has a distance of

2 a. The radius of each sphere is

2 a = 4r r =

2 a

4

Each corner contains 18

of a sphere and each face contains 12

of a sphere, thus, there are 81

8

+6

1

2

= 4

spheres in the face-centered cubic structure. The volume of the spheres is

Vsph = (1)4

3 r3 + (3)

4

3 r2 = (4)

4

3 r3

Vsph = 4

4

3

2a

4

3=

2a3

6


f =Vsph

Vcube=

2a3

6

a3=

2

6= 0.74

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Body-Centered Cubic

The body-centered cubic structure is shown below.

a

r

r

r

2r body diagonal

Figure 11: The body-centered cubic structure (top view).

The body diagonal has a distance of


3 a = 4r r =

3 a

4


of a sphere and the cube contains one whole sphere, therefore, there are81

8

+ 1 = 2 spheres in the body-centered cubic structure. The volume of the spheres is

Vsph = (1)

4

3 r3

+ (1)

4

3 r3

= (2)

4

3 r3

Vsph = 2

4

3

3a

4

3=

3a3

8


f =Vsph

Vcube=

3a3

8

a3=

3

8= 0.68

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Diamond

The diamond structure is shown below. Note that the spheres are not to scale in order to makethe diagram easier to see.

r

r

2r

2r

2r

body diagonal

a

Figure 12: The diamond structure (top view).

The body diagonal has a distance of


3 a = 8r r =

3 a

8


of a sphere, each face contains 12

of a sphere and the cube contains 4 wholespheres, therefore, there are 8

1

8+ 6

1

2+ 4 = 8 spheres in the diamond structure. The volume of

the spheres is

Vsph = 8

4

3r3

= 8

4

3

3 a8

3

=3a3

16


f =Vsph

Vcube=

3a3

16

a3=

3

16= 0.34

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Hexagonal Close-Packed

The hexagonal close-packed structure is shown below.

r

r

2r

a

a

Figure 13: The hexagonal close-packed structure (top view).

Each diagonal has a distance of 2a. The radius of sphere is

2a = 4r r = a2


of a sphere and the top and bottom faces each contain 12

of a sphere. Thehexagonal structure also contains 3 whole spheres, thus, there are 12

1

6

+ 2

1

2

+ 3 = 6 spheres in

the hexagonal close-packed structure. The volume of the spheres is

Vsph = 6

4

3

r3 = 8

a2

3= a3

The volume of the hexagonal structure is Vhex = cArea. Where c is the height of the hexagonalstructure as shown in the figure below.

Figure 14: Definition of c for the hexagonal close-packed structure. This diagram is from Ahcroft andMermin.

The Area of each face is given by

Area =6

4a2 cot

6

=

3

3

2a2

Thus, the volume of the hexagonal structure is

Vhex =3

3

2ca2

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Now, using

c

a=

8

3 c =

8

3a

we can rewrite the voulme as

Vhex =3

3

2a2

8

3a = 3

2a3


f = Vsph

Vhex= a

3

3

2a3=

3

2= 2

6= 0.74

Note that the hcp structure has the same packing fraction as the fcc structure. This is becauseboth the hcp and fcc structures yield the maximum packing factor for hard spheres [1].

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Derivation of the two-dimensional rotation matrix.

x

x

yy

Figure 15: Rotated coordinate system.

From the diagram above we can immediately see that

x = x cos + y sin

y = x sin + y cos This can be written in matrix form as

x

y

=

cos sin sin cos

x

y

Thus the rotation matrix is

R =

cos sin sin cos

An n-fold rotation implies = 2n

.

2-fold: = 22

=

R =

1 00 1

3-fold: = 23

R =

1

2

3

2

3

2 1

2

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4-fold: = 24

= 2

R =

0 11 0

6-fold: = 26

= 3

R =

1

2

3

2

3

2

1

2

Tinkham [7] defines a reducible matrix as one that can be put in block diagonal form. The 2-foldrotation matrix is already in block diagonal form and is, therefore, reduced.

The 3-fold rotation matrix can be diagonalized by finding the eigenvalues and eigenvectors. The

eigenvalues are = 1

2 i

3

2

which are complex, thus, we have not reduced the dimension of the

matrix. Therefore, the 3-fold rotation matrix is irreducible.

The 4-fold rotation matrix can also be diagonalized with eigenvalues = i. However, we havenot reduced the dimensionality. Therefore, the 4-fold rotation matrix is not reducible.

Using the same argument for the 6-fold rotation matrix as for the 3-fold we find that the 6-foldmatrix is irreducible.

Thus, only the 2-fold rotation matrix is reducible.

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References

[1] Kittel, Charles. Introduction to Solid State Physics. New York: John Wiley & Sons, 1986.

[2] Johansson, C. H. and J. O. Linde, Annalen Der Physik, 78, 1925, 439-460.

[3] Massalski, Thaddeus B. Binary Alloy Phase Diagrams. American Society for Metals, 1986.

[4] Hanson, Max. Constitution of Binary Alloys. New York: McGraw-Hill, 1958.

[5] Kittel, Charles and Herbert Kroemer. Thermal Physics. San Francisco: W. H. Freeman and Co.,1980.

[6] Ashcroft, Neil W. and N. David Mermin. Solid State Physics. Philadelphia: W. B. Saunders Co.,1976.

[7] Tinkham, Michael. Group Theory and Quantum Mechanics. New York: McGraw-Hill, 1964.

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