PHY 140A: Solid State Physics

Solution to Homework #1 TA: Xun Jia1 October 2, 2006

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Fall 2006

c Xun Jia (October 2, 2006) °

Physics 140A

Problem #1 In each of the following cases indicate whether the structure is a Bravais lattice. If it is, give 3 primitive vectors; if it is not describe it as Bravais lattice with as small as possible a basis. z

o x y

(a)

(b)

(c)

(d)

Figure 1: (a) Coordinate system, (b)Base centered cubic, (c) Sidecentered cubic, (d) Edge-centered cubic.

(a). Base-centered cubic (simple cubic with additional points in the centers of the horizontal faces of the cubic cell). (b). Side-centered cubic (simple cubic with additional points in the centers of the vertical faces of the cubic cell). (c). Edge-centered cubic (simple cubic with additional points at the midpoints of the lines join nearest neighbors). Solution: Construct the coordinate as illustrated in Fig. 1(a). Let the lattice constant for the corresponding simple cubic be a. Denote the unit vector along x, y, and z directions by x, y, and z, respectively.

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Fall 2006

c Xun Jia (October 2, 2006) °

Physics 140A

(a). It is a Bravais lattice, since from whichever point the lattice is viewed, it appears exactly the same. Indeed, if we take three primitive vectors as a1 = a2 x − a2 y, a2 = a2 x + a2 y, a3 = az, then every lattice point will be at R = n1 a1 + n2 a2 + n3 a3 , where n1 , n2 , n3 are integers. (b). It is not a Bravais lattice, since there are lattice points at R1 = a2 x + a2 z and R2 = a2 y + a2 z, but there is no point at R1 + R2 = a2 x + a2 y + az. If we consider points at R1 , R2 , and R3 = 0 as a basis, then the lattice becomes a Bravais lattice. (c). It is not a Bravais lattice, since there are lattice points at R1 = a2 x and R2 = a y, but there is no point at R1 + R2 = a2 x + a2 y. If we consider points at R1 , 2 R2 , R3 = a2 z, and R4 = 0 as a basis, then the lattice becomes a Bravais lattice. Those three types of lattices are shown in Fig. 1.

Problem #2 What is the Bravais lattice formed by all points with Cartesian coordinates (n1 , n2 , n3 ) if: (a). The ni are all even? (b). The ni are all even or all odd? (c). The sum of the ni is required to be even? Solution: (a). If the ni are all even, the lattice will be simple cubic with lattice constants a = 2. (b). If the ni are all even or all odd, the lattice will be body-centered cubic with lattice constants a = 2. Indeed, if we shift the lattice in part (a) as (n1 , n2 , n3 ) → (n1 + 1, n2 + 1, n3 + 1), we will get a lattice with ni all odd, and each new lattice point is in the center of the cubic of old lattice; hence we obtain a body-centered cubic lattice. (c). If the sum of the ni is required to be even, the lattice will be face-centered cubic with lattice constants a = 2. Indeed, if we shift the lattice in part (a) as (n1 , n2 , n3 ) → (n1 + 1, n2 + 1, n3 ), (n1 , n2 , n3 ) → (n1 , n2 + 1, n3 + 1), and (n1 , n2 , n3 ) → (n1 + 1, n2 , n3 + 1), the sum of ni are still even, and each new lattice point is in the center of each square of the old lattice; hence we obtain a face-centered cubic lattice.

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Fall 2006

c Xun Jia (October 2, 2006) °

Physics 140A

Problem #3 The face-centered cubic is the most dense and the simple cubic is the least dense of the three cubic Bravais lattices. The diamond structure is less dense than any of these. One measure of this is that the coordination numbers are : fcc, 12; bcc, 8; sc, 6; diamond, 4. Another is the following: suppose identical solid spheres are distributed through space in such a way that their centers lie on the points of each of these four structures, and spheres on (nearest) neighboring points just touch, without overlapping. (Such an arrangement of spheres is called a close-packing arrangement.) Show that the ratio of the volume contained in the spheres to the total volume (the ”packing fraction”) for each of the four structures is: √ (a). fcc: 2π/6 = 0.74 √ (b). bcc: 3π/8 = 0.68 (c). sc: π/6 = 0.52 √ (d). diamond: 3π/16 = 0.34 Solution: As in Fig. 2, Let the radius of the spheres be r, and the length of the cube be a, then: √ consider the side diagonal, as in Fig. 2(a), we have 2a = 4r, (a). For fcc lattice, √ thus a = 4r/ 2. Moreover, there are 8 spheres at the corners, and 1/8 of each is contained in the cube; there are 6 spheres on the sides, and 1/2 of each is contained in the cube; thus the number of spheres in the cube is 8× 81 +6× 12 = 4. Therefore, the packing fraction is: 4πr3 √ 3 = 2π = 0.74 6 4r ( √ )3 2

4×

(1)

√ (b). For bcc lattice,√consider the body diagonal, as in Fig. 2(b), we have 3a = 4r, thus a = 4r/ 3. Moreover, there are 8 spheres at the corners, and 1/8 of each is contained in the cube; there are 1 spheres at the center, which is totally contained in the cube; thus the number of spheres in the cube is 8× 81 +1×1 = 2. Therefore, the packing fraction is: 4πr3 √ 2× 3 = 3π = 0.68 8 4r ( √ )3 3 3

(2)

Fall 2006

c Xun Jia (October 2, 2006) °

Physics 140A

(c). For sc lattice, consider the side, as in Fig. 2(c), we have a = 2r. Moreover, there are 8 spheres at the corners, and 1/8 of each is contained in the cube; thus the number of spheres in the cube is 8 × 81 = 1. Therefore, the packing fraction is: 4πr3 3 = π = 0.52 (2r)3 6

1×

(3)

(d). For diamond lattice, it requires thinking carefully. √ Consider the body diagonal, √ as in Fig. 2(d). From simple geometry, we have 3a = 8r, thus a = 8r/ 3. (In fact, if you take a plane containing the side diagonals on the top surface and bottom surface, you will find why those 5 big spheres in Fig. 2(d) are packing along the body diagonal. This actually follows from the similarity among triangles.) Moreover, there are 8 spheres at the corners, and 1/8 of each is contained in the cube; there are 6 spheres on the sides, and 1/2 of each is contained in the cube; there are 4 spheres at the center of each tetrahedron,

(a)

(b)

(c)

(d)

Figure 2: (a) Face-centered cubic. (b)Body-centered cubic. (c) Simple cubic, (d) Diamond.

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Fall 2006

c Xun Jia (October 2, 2006) °

Physics 140A

which are all contained in the cube; thus the number of spheres in the cube is 8 × 18 + 6 × 12 + 4 × 1 = 8. Therefore, the packing fraction is: 4πr3 √ 3 = 3π = 0.34 16 8r ( √ )3 3

8×

(4)

Problem #4 Draw a picture that explains the first end note in Takemura and Fujihisa’s article, ”High-pressure structural phase transition in indium,” from Phys. Rev. B, 47, 8465(1993). Label the a and b lattice parameters for the fct and bct unit cells. Starting values can be found at Webelements. Why is fct not considered a Bravais lattice? http://prola.aps.org/pdf/PRB/v47/i14/p8465_1 http://www.webelements.com/webelements/elements/text/In/xtal.html Solution: The first end note in the paper is quoted here: ”The correct unit cell for indium is body-centered tetragonal (bct) belonging to 17 the space group D4h − I 4/mmm (No. 139). However, since the deviation from fcc is small (∼ 7%), we prefer to use a face-centered tetragonal (fct) cell. The fct cell is easy to compare with the fcc as well as with the presently found fco structure.” By looking up at the webelements webpage, the starting lattice parameters are a = b = 325.23pm and c = 494.61pm.the bct lattice are as in Fig. 3(a). This is a Bravais lattice, and is the unit cell for indium. However, we can consider this lattice structure in a different way. Consider Fig. 3(c), which is just the bct lattice of indium with 4 unit cells. If we focus on those atoms and bonds in dark color, we will really get a fct lattice structure. The lattice constants a0 and c0 for this fct are: √ a0 = 2a = 459.94pm (5) 0 c = c = 494.61pm (6) and hence c0 /a0 = 1.07, that is why this fct structure differs from fcc by only 7%. For the purpose of making comparison with fcc and fco, they chose use this fct structure, as shown in Fig. 3(b), to represent indium lattice in the paper. Face-centered tetragonal is, however, not considered to be a Bravais lattice. The preferred unit cell is the smallest one that will reproduce the lattice when repeated in space. Since the fct unit cell constructed from bct lattice will always be larger than the bct unit cell, see Fig. 3(c), we consider bct, rather than fct, to be the Bravais 5

Fall 2006

c Xun Jia (October 2, 2006) °

Physics 140A

(a)

(b)

(c) Figure 3: (a)Body-centered tetragonal. (b)Face-centered tetragonal. (c) Construction of fct based on bct lattice.

lattice for such kind of lattice structure, though those two views are equivalent to each other.

Problem #5 Potassium hydroxide (KOH) etches silicon planes with rate ratios {110} > {100} > {111} ∼ 400 : 200 : 1. Approximate these large numbers as infinity. Say you have a 200 micron thick wafer oriented in the (100) direction, coated with silicon oxide (not etched by KOH). Calculate the dimensions of the initial (front side) rectangular hole in the oxide required to produce a 5 micron by 100 micron suspended oxide film on the backside after a KOH etch. Write a sentence or two commenting on the veracity (or truthiness) of the following statement: the {111} planes of crystalline silicon are 6

Fall 2006

c Xun Jia (October 2, 2006) °

Physics 140A

etched the most slowly by KOH because they are close-packed. Solution:

part to be etched

SiO

Si wafer x

h=200

m

{111} {100}

x

0

SiO

Figure 4: The sketch of the structure in etching the wafer.

If we approximate those large numbers in the etching rate ratios as infinity, the etch will only occur along direction {110} and {100}, that is, the surface in the crystal we obtain after etching will be {111} plane. The side view of Si wafer, which is covered by SiO on both sides, is illustrated in Fig. 4. Then the question is, if we want a hole with dimension x0 at the bottom side after etching, how large the size x should be for the initial hole on the top, suppose the angle θ is known. From simple geometry, the answer is: 2h (7) x = x0 + tan θ Since the angle θ is the angle between {100} and {111} plane, which is also the angle between those two vectors, we can compute as: ! Ã {100} {111} · = 54.7◦ (8) θ = arccos |{100}| |{111}| Therefore, since we want the dimension of rectangle on bottom side to be 5µm by 100µm, from equation (7), let x0 = 5µm, we get x = 287.8µm; let x0 = 100µm, we get x = 382.8µm. Namely, the required rectangular hole on the top should be 287.8µm by 382.8µm. The {111} planes are not close-packed, because silicon is diamond structure, not fcc. Nonetheless, these planes are more close packed than the others, and thus have fewer dangling bonds vulnerable to chemical attack.

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