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Solution Manual for Fundamentals of Condensed Matter and Crystalline Physics: An Introduction for Students of Physics and Materials Science, 1st Edition

Solution Manual for Fundamentals of Condensed Matter and Crystalline Physics: An Introduction for Students of Physics and Materials Science, 1st Edition offers the best solutions to textbook problems, helping you prepare for exams and assignments.

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    Fundamentals of Condensed Matter and Crystalline Physics Solution Manual (rev. Dec. 2011)

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    CHAPTER 1 1-1. Show that the volume of the primitive cell of a BCC crystal lattice is a 3 /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 = 1 2 ˆ x ˆ y ˆ z 1 1 $ 1 1 $ 1 1 % & ' ' ' ( ) * * * # ! a 3 = a 3 /2 1-2. Show that the volume of the primitive cell of a FCC crystal lattice is a 3 /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 = 1 2 ˆ x ˆ y ˆ z 1 1 0 1 0 1 $ % & & & ' ( ) ) ) # ! a 3 = a 3 /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 ! 3 a /2 , and so the maximum radius of a balloon is ! R = 3 a /4 . There are a total of 2 balloons in each BCC unit cell for total occupied volume ! 8 " 3 R 3 and packing fraction ! 8 " R 3 /3 a 3 = 8 " ( 3 /4) 3 3 = 3 " 8

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    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 " 3 R 3 and packing fraction ! 16 " R 3 /3 a 3 = 16 " 3 1 2 2 # $ % & ' ( 3 = " 3 2 = 2 " 6 1-5. The 2D crystal shown in Fig. 1-14 contains three atoms with a chemical formula ABC 2 . 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

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    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, ! c 2 = a 2 3 or ! c a = 2 2 3 = 1.633 . 1-8. Bromine has an orthorhombic lattice structure with ! a 1 = 4.65 Å, ! a 1 = 6.73Å, ! a 1 = 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) ! = M V = N (79.9 / 6.02 x 10 23 ) (4.65 x 6.73 x 8.70) g / Å 3 = 3.12 g / cc ! N " 8

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    (b) Since bromine has a chemical formula of Br 2 , there are 4 molecules of bromine in a unit cell, suggesting a FC orthorhombic structure. (c) PF = 8(4 ! 1.5 ( ) 3 / 3) (4.65 x 6.73 x 8.70) = 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) (3 x 3 x 4) = 0.724

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    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:

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    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 ! mb 2 ) n . Including the central atom, there are 98.5 centers inside the m = 15 th ring of area ! (15 b ) 2 . A histogram of the g(r) (for only this one ensemble) looks like this: Monday, October 17 1:12 P Page #1 - “HW2_1” J I H G F E D g(r) dN ring # 0.0000 0.0000 1.0000 0 0.0000 0.0000 2.0000 1 1.6968 5.0000 3.0000 2 0.25453 1.0000 4.0000 3 1.0181 5.0000 5.0000 4 0.93326 5.5000 6.0000 5 0.50905 3.5000 7.0000 6 1.2726 10.000 8.0000 7 0.50905 4.5000 9.0000 8 1.2217 12.000 10.000 9 0.64788 7.0000 11.000 10 0.72115 8.5000 12.000 11 0.86147 11.000 13.000 12 0.90902 12.500 14.000 13 0.81448 12.000 15.000 14

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    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.

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    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 As 2 Ge 3 . 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% SiO 2 , 20% Na 2 O and 10% CaO, known as soda-lime-silicate. How does the addition of Na 2 O and CaO affect the CRN of SiO 2 if the O donated by either is to end up bonded with a Si atom? 2-6. Solution:

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    Addition of Na 2 O and CaO both lead to the formation of non-bridging oxygen bonds that serve to weaken the network structure. The O contributed by either Na 2 O 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.

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    CHAPTER 3 3-1. Show that the equilibrium separation of two particles interacting through the Lennard-Jones potential is ! r eq = 2 1/ 6 " . 3-1. Solution: U = 4 ! " r ! " # $ % & 12 ' ! r ! " # $ % & 6 ( ) * + * , - * . * , dU dr = 4 ! 6 " r ! " # $ % & 6 1 r ' 12 " r ! " # $ % & 12 1 r ( ) * + * , - * . * = 0 , ! r eq = 2 1/ 6 " 3-2. Show that the equilibrium nearest neighbor separation for an ionic crystal is ! r eq = 12 A " 12 / # kZ 2 e 2 ( ) 1/11 . 3-2. Solution: U tot = N pairs A ! r ! " # $ % & 12 ' " kZ 2 e 2 r ! " # $ % & ( ) * + * , - * . * , dU tot dr = N pairs ! 12 A ! r " # $ % & ' 12 1 r + " kZ 2 e 2 r " # $ % & ' 1 r ( ) * + * , - * . * = 0 , ! r eq = 12 A " 12 / # kZ 2 e 2 ( ) 1/11 3-3. Ice floats in water because when water is frozen it expands into a lower density state of matter. Explain this anomalous behavior on the basis of what you know about the shape of a H 2 O molecule and the nature of the hydrogen bond. 3-3. Solution: Water is unique among liquids in that it attains a maximum density around +4°C and actually expands somewhat as it is cooled into the crystalline state. Below 4°C water molecules begin to assume an increased tendency for directional bonding which leads to the (transient) formation of so-called clathrate structures which contain more empty space than regular liquid. 3-4. Demonstrate that, energetically, a noble gas element is slightly more stable in the FCC crystal structure than it would be in a BCC structure by calculating the ratio of these two cohesive energies. The lattice sums for the BCC structure are: ! A = p 1 j " 12 j = 2. N # = 9.11418 B = p 1 j " 6 j = 2. N # = 12.2533 . 3-4. Solution:

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    From Eq. 3.14 using A and B for the BCC structure: U tot, eq = ! N ! B 2 2 A " # $ % & ' = ! N ! 8.2368 ( ) . Using the value for the FCC (Eq. 3.15), the ratio of BCC to FCC is 0.957, suggesting that the BCC structure is only slightly less stable than that of the FCC. 3-5. Imagine that a 1D chain of ions (with q = ±Ze ) like that shown in Fig. 3-9 is compressed so that the separation distance decreases from its equilibrium value by some fractional amount δ (i.e., r o ! r eq (1 " ! ) ). For the ionic potential of Eq. (3.18), show that the work done (per unit length of the chain) in compressing the crystal is quadratic (in leading order) and given as: 11 kq 2 ln 2 2 r eq ! 2 . 3-5. Solution: We can approximate the cohesive energy as a Taylor expansion about the equilibrium separation: U ( r eq ! ! r eq ) " U ( r eq ) ! dU / dR ( ) r eq ! r eq + 1 / 2 ( ) d 2 U / dR 2 ( ) r eq ! r eq ( ) 2 + ! , where the second term vanishes by definition of r eq . The work done is the change in cohesive energy: Work = U ( r eq ! ! r eq ) ! U ( r eq ) " 1 / 2 ( ) d 2 U / dR 2 ( ) r eq r eq 2 ! 2 . For a cohesive energy given in Eq. (3.18) (with n = 12) as: U tot = N pairs A r n ! ! kq 2 r " # $ % & ' ( ) * + , - , dU tot dr = N pairs ! n A r n + 1 + ! kq 2 r 2 " # $ % & ' ( ) * + , - from which r eq n ! 1 = nA / ! kq 2 , and d 2 U tot dr 2 = N pairs n ( n + 1) A r n + 2 ! 2 ! kq 2 r 3 " # $ % & ' ( ) * + , - which at equilibrium is: 1 2 d 2 U tot dr 2 R o r eq 2 = 1 2 N pairs ( n ! 1) ! kq 2 r eq 3 " # $ $ % & ' ' ( ) * + * , - * . * r eq 2 , from which the answer follows. 3-6. Calcium oxide has the FCC structure for which the Madelung constant is α = 1.7476. Determine the cohesive energies per molecule of the hypothetical crystals Ca + O - and Ca 2+ O 2- . Be sure to account also for energy required to return the ions to neutral atoms. Assume r o = 2.40 Å is the same for both forms, and neglect the repulsive energy. The energy needed to ionize the first and second electrons of Ca are 6.11 and 11.87 eV, respectively. The energy released when a first and second electron is added to neutral oxygen is 1.46 and -9.0 eV, respectively. (Note that energy is actually required for adding the second electron). Which of the two hypothetical crystals is most likely to occur in Nature and why? 3-6. Solution:

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    In the monovalent case, U/N from Eq. 3.20 is -10.48 eV. To this we add 6.11 eV of energy input to ionize the calcium atom and -1.46 eV that is released when an electron is transferred to the oxygen atom. The total cohesion is -4.45 eV. In the divalent case, U/N = 4 x (-10.48 eV) = -41.95 eV. To this we add again the (6.11 eV – 1.46 eV) for the first electron transferred and (11.87 eV + 9.0 eV) for the second to obtain a total cohesion of - 16.41 eV. Due to its larger cohesion, the divalent form is more favorable. 3-7. Figure 3-9 shows a hypothetical 2D square lattice of ions. Estimate the Madelung constant in the following way: First sum only the contributions to the Madelung constant arising from the ions (or partial ions) contained within the innermost square. Note that an ion on one edge of a square is counted as 50% inside and 50% outside and one on a corner is counted as 25% inside and 75% outside. Show that this first contribution amounts to +1.2939. Continue on to the next layer (indicated by the hashed region). Show that this region alone contributes to the Madelung constant by an amount +0.3140. Lastly, compute the contribution from the outermost layer and show that out to this level the Madelung constant is +1.6116. 3-7. Solution: The inner most square has a positive contribution due to four half-atoms on the sides and a negative contribution due to quarter-atoms at each of four corners. Together these contribute to the Madelung constant as: ! 1 = 4 1 / 2 1 ! " # $ % & + 4 ' 1 / 4 2 ! " # $ % & = + 1.293 … . Similarly, the contribution due to fractions of atoms in the next shell are: ! 2 = 4 1 / 2 1 ! " # $ % & + 4 ' 3 / 4 2 ! " # $ % & ( ) * + , - + 4 ' 1 / 2 2 ! " # $ % & + 4 ' 1 / 4 8 ! " # $ % & + 8 1 / 2 5 ! " # $ % & = + 0.314 … . The quantity in

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    brackets corresponds the partial atoms in contact with the first square. Finally the outermost shell: ! 2 = 4 ! 1 / 2 2 " # $ % & ' + 4 ! 3 / 4 2 2 " # $ % & ' + 8 1 / 2 5 " # $ % & ' + 4 1 / 2 3 " # $ % & ' + 4 ! 1 / 4 3 2 " # $ % & ' + 8 ! 1 / 2 10 " # $ % & ' + 8 1 / 2 13 " # $ % & ' = 0.0037 …

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    CHAPTER 4 4-1. A power transformer consists of two windings around a common ferromagnetic core and is used in ac circuits to step up (or down) an ac voltage. During each cycle, the ferromagnetic core repeats the process of growing and shrinking domains. (a) Show that the work done in one cycle of a hysteresis loop equals the area enclosed by the loop. (b) Which would be better for a power transformer: a core with a large coercivity or a small coercivity? Why? (Hint: you might make use of the following thermodynamic relation: dW = HdM ) 4-1. Solution: Consider dW = HdM around the idealized hysteresis loop shown below that on the right has been rotated so that M is along the x-axis. During segment ab, H > 0 and dM > 0 so positive work is performed equal to the area AB under the curve. In segment bc, dM = 0 so no work is performed. In segment cd H < 0 and dM < 0, so again positive work is performed equal to the area CD under the curve between cd. Similarly, in segment de, positive work is performed equal to the area DE. In the remaining segment fa, H and dM are again > 0 and positive work equal to the area FA is performed. This total work must be performed in each cycle of the transformer and is an unavoidable energy loss. It can only be minimized by minimizing the area of the hysteresis loop as by narrowing the width (da in the figure below) by decreasing the coercivity. 4-2. Consider the centripetal force on the orbiting electron of the classical Bohr model illustrated in Fig. 2. Consider both a situation with and without a magnetic field present and show that when the field is present and directed upward (perpendicular to the orbital plane), the electron speeds up by ! " # $ % eB /2 m and that the change in its angular velocity then equals the Larmor frequency given in Eq. (4.9). As a follow up to this, determine how large of magnetic field is needed to cause the 1% change in the orbital speed. 4-2. Solution:

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