Fundamentals of Condensed Matter and Crystalline Physics

Instructor's Solution Manual

(rev. Dec. 2011)

CHAPTER 1

1-1. Show that the volume of the primitive cell of a BCC crystal lattice is a3/2 where a is the lattice constant of the conventional cell. 1-1. Solution: From Fig. 1.9, the primitive axes are

!

! a 1 = a (1/2) ˆ x + (1/2) ˆ y " (1/2)ˆ z ( )! a 2 = a (1/2) ˆ x " (1/2) ˆ y + (1/2)ˆ z ( )! a 3 = a "(1/2) ˆ x + (1/2) ˆ y + (1/2)ˆ z ( )

!

V =! a 1 "! a 2( ) #! a 3 =

12

ˆ x ˆ y ˆ z 1 1 $11 $1 1

%

&

' ' '

(

)

* * * #! a 3 = a3 /2

1-2. Show that the volume of the primitive cell of a FCC crystal lattice is a3/4 where a is the lattice constant of the conventional cell. 1-2. Solution: From Fig. 1.9, the primitive axes are

!

! a 1 = a (1/2) ˆ x + (1/2) ˆ y ( )! a 2 = a (1/2) ˆ x + (1/2)ˆ z ( )! a 3 = a (1/2) ˆ y + (1/2)ˆ z ( )

!

V =! a 1 "! a 2( ) #! a 3 =

12

ˆ x ˆ y ˆ z 1 1 01 0 1

$

%

& & &

'

(

) ) ) #! a 3 = a3 /4

1-3. Show that the packing fraction of a BCC crystal lattice is

!

3" /8 = 0.680 . 1-3. Solution: From table 1-1, the nearest neighbor distance is

!

3a /2 , and so the maximum radius of a balloon is

!

R = 3a /4 . There are a total of 2 balloons in each BCC unit cell for total

occupied volume

!

8"3R3 and packing fraction

!

8"R3 /3a3

=8" ( 3 /4)3

3=

3"8

1-4. Show that the packing fraction of a FCC crystal lattice is

!

2" /6 = 0.740 . 1-4. Solution: From table 1-1, the nearest neighbor distance is

!

a / 2 , and so the maximum radius of a balloon is

!

R = a /2 2 . There are a total of 4 balloons in each BCC unit cell for total

occupied volume

!

16"3R3 and packing fraction

!

16"R3 /3a3

=16"3

12 2#

$ %

&

' ( 3

="3 2

=2"6

1-5. The 2D crystal shown in Fig. 1-14 contains three atoms with a chemical formula ABC2. Illustrated in the figure are several possible tiles. (a) Identify which of the tiles are primitive cells. (b) Identify which of the tiles are conventional cells. (c) Identify any tiles that are unable to correctly fill the space. (d) For each primitive cell, provide expressions for the appropriate basis vectors describing the basis set of atoms. Fig. 1-14

1-5. Solution: a) primitive cell with Black (0,0), Gray ((1/2),0), White (0, (1/2)), ((1/2),(1/2)) b) not primitive and will not tile space (note bad chemical formula) c) conventional cell, will tile space d) not primitive and will not tile space (note bad chemical formula) e) primitive cell with Black ((1/2),0), Gray ((1/2),(1/2)), White ((1/4), (1/4)), ((3/4),(1/4)) f) not primitive and will not tile space (note bad chemical formula) g) not primitive and will not tile space (note bad chemical formula) h) primitive cell with Black (0,(1/2)), Gray (0,0), White ((1/4), (1/4)), ((1/4),(3/4)) i) primitive cell j) primitive cell 1-6. Consider again the 2D crystal shown in Fig. 1-14. Describe all the basic symmetry operations (translational, rotational and mirror) satisfied by this lattice. 1-6. Solution: Translational symmetry and 4-fold rotational symmetry. Mirror symmetry is found along any dashed line and any parallel line midway in between. Mirror symmetry is also found along any diagonal. 1-7. For the HCP crystal structure, show that the ideal c/a ratio is 1.633. 1-7. Solution: Imagine balloons inflated about each site of the primitive cell shown in Fig. 1-12. The limitation arises from contact made in the half of the primitive cell containing a midpoint site. This half cell assumes the form of two tetrahedra (equilateral pyramids), one of which is inverted. Thus c/2 is the height of a pyramid of base a,

!

c2

= a 23

or

!

ca

= 2 23

=1.633 .

1-8. Bromine has an orthorhombic lattice structure with

!a1 = 4.65 Å, !a1 = 6.73Å,

!a1 = 8.70Å. (a) The atomic weight of bromine is 79.9 g/mol. If it has a density is 3.12 g/cc, how many bromine atoms reside in a single unit cell? (b) Which type of orthorhombic lattice (i.e, BC, FC etc.) is suggested your finding in part (a)? Explain. (c) If the atomic radius of bromine is 1.51Å, determine the packing fraction. 1-8. Solution:

(a) ! = MV=N(79.9 / 6.02x1023)(4.65x6.73x8.70)

g / Å3 = 3.12 g / cc! N " 8

(b) Since bromine has a chemical formula of Br2, there are 4 molecules of bromine in a unit cell, suggesting a FC orthorhombic structure.

(c) PF =8(4! 1.5( )3 / 3)(4.65x6.73x8.70)

= 0.333

1-9. Shown in Fig. 1-15 is the unit cell of a monatomic crystal. (a) How would you describe this particular crystal structure? (b) What is the maximum packing fraction you should expect for this specific structure? Fig. 1-15

1-9. Solution: (a) This unit cell has all 90° angles and two sides equal in length. From Fig. 1-7 we can identify it as tetragonal and, since there is a centered site inside, it would be referred to as body-centered tetragonal. (b) Imagine inflating balloons at each lattice site until contact is first made. One finds that the body diagonal is 5.831 Å and contact along this line would occur at an atom radius of r = 5.831/4 Å = 1.46 Å (just slightly shorter than half the distance in the base plane (1.50Å). The packing fraction is then

PF =2(4! 1.46( )3 / 3)

(3x3x4)= 0.724

CHAPTER 2

2-1. An ideal gas consists of non-interacting, point particles that move about in rapid and incessant fashion. Sketch the form of g(r) for this ideal gas and discuss its features. 2-1. Solution: Ideal gas consists of point particles that have no attractive interaction. Probability of finding a second particle is just that of the density, n.

2-2. Make a xerox reproduction of Fig. 2-11 below that represents the atoms in a amorphous solid and, using a compass, manually calculate g(r) for a single ensemble using the dark particle as the central particle. Do this with a dr no larger than the particle radius, b. Plot your result and identify the first and second coordination spheres. 2-2. Solution: Draw rings about the central particle like so:

Count the number of particles centers found in each ring of size dr (here equal to b) and make a table:

For 2D case here, g(r) = dN / (2!rdr) n = dN / (2!mb2 ) n . Including the central atom,

there are 98.5 centers inside the m = 15th ring of area ! (15b)2 . A histogram of the g(r) (for only this one ensemble) looks like this:

Monday, October 17 1:12 PM 2011Page #1 - “HW2_1”

JIHGFEDg(r)dNring #0.00000.00001.000000.00000.00002.000011.69685.00003.00002

0.254531.00004.000031.01815.00005.00004

0.933265.50006.000050.509053.50007.000061.272610.0008.00007

0.509054.50009.000081.221712.00010.0009

0.647887.000011.000100.721158.500012.000110.8614711.00013.000120.9090212.50014.000130.8144812.00015.00014

2-3. Figure 2-12 shows the nematic phases of two liquid crystals: (a) a discotic liquid crystal and (b) a lipid liquid crystal. For each of these partially amorphous systems, discuss the symmetry properties including both translational and rotational symmetries.

2-3. Solution: (a) The discotic phase illustrated possesses a two-fold rotational symmetry about either of the two axes that are perpendicular to the common normal of the face of the discs. (b) The lipid phase illustrated possesses a two-fold rotational symmetry about either of the two axes that are perpendicular to the common direction of alignment. 2-4. The pair distribution function for a bag of marbles is shown in Fig. 2-13. From the figure, (a) determine the nearest and next-nearest separation distances corresponding to the first and second coordination shells, and (b) estimate the coordination number for the first and second coordination shells.

2-4. Solution: Each small square in the figure is (5 marbles/cm) X (0.25 cm) = 1.25 marbles per square. The first coordination shell is approximated by the shaded region in the above figure that peaks near 1.5 cm and contains a total of roughly 4 boxes or 5 marbles. The second coordination shell peaking at 3 cm contains a total of roughly 18 boxes or 22.5 marbles. 2-5. A common chalcogenide glass is As2Ge3. Determine the average coordination number for this system. 2-5. Solution: The formation of covalent bonds forces the coordination near an As to be 3 and that near Ge to be 4. Since there are two As for every 3 Ge, 40% of the atoms are As and 60% are Ge. The average coordination is the weighted value: (0.4 x 3) + (0.6 x 4) = 3.6. 2-6. Typical window glass is formed by a mixture of approximately 70% SiO2, 20% Na2O and 10% CaO, known as soda-lime-silicate. How does the addition of Na2O and CaO affect the CRN of SiO2 if the O donated by either is to end up bonded with a Si atom? 2-6. Solution:

Addition of Na2O and CaO both lead to the formation of non-bridging oxygen bonds that serve to weaken the network structure. The O contributed by either Na2O or CaO replaces the missing oxygen on one of two Si units when the bridging bond is broken. The pair of terminal oxygens are charge compensated by the two Na cations or the ça ion and produce a weaker ionic crosslink.