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

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Fundamentals of Condensed Matterand Crystalline PhysicsSolution Manual(rev. Dec. 2011)

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CHAPTER 11-1.Show that the volume of the primitive cell of a BCC crystal lattice isa3/2 whereaisthe lattice constant of the conventional cell.1-1.Solution:From Fig. 1.9, the primitive axes are!!a1=a(1/2) ˆx+(1/2) ˆy"(1/2)ˆz()!a2=a(1/2) ˆx"(1/2) ˆy+(1/2)ˆz()!a3=a"(1/2) ˆx+(1/2) ˆy+(1/2) ˆz()!V=!a1"!a2()#!a3=12ˆxˆyˆz11$11$11%&'''()***#!a3=a3/21-2.Show that the volume of the primitive cell of a FCC crystal lattice isa3/4 whereaisthe lattice constant of the conventional cell.1-2.Solution:From Fig. 1.9, the primitive axes are!!a1=a(1/2) ˆx+(1/2) ˆy()!a2=a(1/2) ˆx+(1/2) ˆz()!a3=a(1/2) ˆy+(1/2) ˆz()!V=!a1"!a2()#!a3=12ˆxˆyˆz110101$%&&&'()))#!a3=a3/41-3.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 aballoon is!R=3a/4. There are a total of 2 balloons in each BCC unit cell for totaloccupied volume!8"3R3and packing fraction!8"R3/3a3=8"(3 /4)33=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 aballoon is!R=a/22. There are a total of 4 balloons in each BCC unit cell for totaloccupied volume!16"3R3and packing fraction!16"R3/3a3=16"3122#$%&'(3="32=2"61-5.The 2D crystal shown in Fig. 1-14 contains three atoms with a chemical formulaABC2. Illustrated in the figure are several possible tiles. (a) Identify which of the tilesare primitive cells. (b) Identify which of the tiles are conventional cells. (c) Identify anytiles that are unable to correctly fill the space.(d) For each primitive cell, provideexpressions 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 spaced) 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 cellj) primitive cell1-6.Consider again the 2D crystal shown in Fig. 1-14. Describe all the basic symmetryoperations (translational, rotational and mirror) satisfied by this lattice.1-6.Solution:Translational symmetry and 4-fold rotational symmetry. Mirror symmetry is found alongany dashed line and any parallel line midway in between. Mirror symmetry is also foundalong any diagonal.1-7.For the HCP crystal structure, show that the idealc/aratio is 1.633.1-7.Solution:Imagine balloons inflated about each site of the primitive cell shown in Fig. 1-12. Thelimitation arises from contact made in the half of the primitive cell containing a midpointsite. This half cell assumes the form of two tetrahedra (equilateral pyramids), one ofwhich is inverted. Thus c/2 is the height of a pyramid of base a,!c2=a23or!ca=223=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 3.12g/cc, how many bromine atoms reside in a single unit cell? (b) Which type oforthorhombic 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=3.12g/cc!N"8

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(b) Since bromine has a chemical formula of Br2, there are 4 molecules of bromine in aunit cell, suggesting a FC orthorhombic structure.(c)PF=8(4!1.5()3/ 3)(4.65x6.73x8.70)=0.3331-9.Shown in Fig. 1-15 is the unit cell of a monatomic crystal. (a) How would youdescribe this particular crystal structure? (b) What is the maximum packing fraction youshould expect for this specific structure?Fig. 1-151-9.Solution:(a) This unit cell has all 90° angles and two sides equal in length. From Fig. 1-7 we canidentify it as tetragonal and, since there is a centered site inside, it would be referred to asbody-centered tetragonal. (b) Imagine inflating balloons at each lattice site until contactis first made. One finds that the body diagonal is 5.831 Å and contact along this linewould occur at an atom radius of r = 5.831/4 Å = 1.46 Å (just slightly shorter than halfthe distance in the base plane (1.50Å). The packing fraction is thenPF=2(4!1.46()3/ 3)(3x3x4)=0.724

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CHAPTER 22-1.An ideal gas consists of non-interacting, point particles that move about in rapid andincessant 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 offinding 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 aamorphous solid and, using a compass, manually calculateg(r)for a single ensembleusing the dark particle as the central particle. Do this with adrno larger than the particleradius,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) andmake 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 = 15thring of area!(15b)2. A histogram of the g(r)(for only this one ensemble) looks like this:Monday, October 17 1:12 PPage #1 - “HW2_1”JIHGFEDg(r)dNring #0.00000.00001.000000.00000.00002.000011.69685.00003.000020.254531.00004.000031.01815.00005.000040.933265.50006.000050.509053.50007.000061.272610.0008.000070.509054.50009.000081.221712.00010.00090.647887.000011.000100.721158.500012.000110.8614711.00013.000120.9090212.50014.000130.8144812.00015.00014

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2-3.Figure 2-12 shows the nematic phases of two liquid crystals: (a) a discotic liquidcrystal 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 ofthe 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 thetwo 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 thefigure, (a) determine the nearest and next-nearest separation distances corresponding tothe first and second coordination shells, and (b) estimate the coordination number for thefirst 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 thatpeaks near 1.5 cm and contains a total of roughly 4 boxes or 5 marbles. The secondcoordination 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 coordinationnumber for this system.2-5.Solution:The formation of covalent bonds forces the coordination near an As to be 3 and that nearGe to be 4. Since there are two As for every 3 Ge, 40% of the atoms are As and 60% areGe. 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 andCaO affect the CRN of SiO2if the O donated by either is to end up bonded with a Siatom?2-6.Solution:

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Addition of Na2O and CaO both lead to the formation of non-bridging oxygen bonds thatserve to weaken the network structure. The O contributed by either Na2O or CaOreplaces 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 ionand produce a weaker ionic crosslink.

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CHAPTER 33-1.Show that the equilibrium separation of two particles interacting through theLennard-Jones potential is!req=21/ 6".3-1.Solution:U=4!"r!"#$%&12'!r!"#$%&6()*+*,-*.*,dUdr=4!6"r!"#$%&61r'12"r!"#$%&121r()*+*,-*.*=0,!req=21/ 6"3-2.Show that the equilibrium nearest neighbor separation for an ionic crystal is!req=12A"12/#kZ2e2()1/11.3-2.Solution:Utot=NpairsA!r!"#$%&12'"kZ2e2r!"#$%&()*+*,-*.*,dUtotdr=Npairs!12A!r"#$%&'121r+"kZ2e2r"#$%&'1r()*+*,-*.*=0,!req=12A"12/#kZ2e2()1/113-3.Ice floats in water because when water is frozen it expands into a lower density stateof matter. Explain this anomalous behavior on the basis of what you know about theshape of a H2O 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 andactually expands somewhat as it is cooled into the crystalline state. Below 4°C watermolecules begin to assume an increased tendency for directional bonding which leads tothe (transient) formation of so-called clathrate structures which contain more empty spacethan regular liquid.3-4.Demonstrate that, energetically, a noble gas element is slightly more stable in theFCC crystal structure than it would be in a BCC structure by calculating the ratio of thesetwo cohesive energies. The lattice sums for the BCC structure are:!A=p1j"12j=2.N#=9.11418B=p1j"6j=2.N#=12.2533.3-4.Solution:

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From Eq. 3.14 using A and B for the BCC structure:Utot,eq=!N!B22A"#$%&'=!N!8.2368().Using the value for the FCC (Eq. 3.15), the ratio of BCC to FCC is 0.957, suggesting thatthe BCC structure is only slightly less stable than that of the FCC.3-5.Imagine that a 1D chain of ions (withq = ±Ze) like that shown in Fig. 3-9 iscompressed so that the separation distance decreases from its equilibrium value by somefractional amountδ(i.e.,ro!req(1"!)). For the ionic potential of Eq. (3.18), show thatthe work done (per unit length of the chain) in compressing the crystal is quadratic (inleading order) and given as:11kq2ln 22req!2.3-5.Solution:We can approximate the cohesive energy as a Taylor expansion about the equilibriumseparation:U(req!!req)"U(req)!dU/dR()req!req+1 / 2()d2U/dR2()req!req()2+!, wherethe second term vanishes by definition ofreq. The work done is the change in cohesiveenergy:Work=U(req!!req)!U(req)"1 / 2()d2U/dR2()reqreq2!2. For a cohesive energygiven in Eq. (3.18) (with n = 12) as:Utot=NpairsArn!!kq2r"#$%&'()*+,-,dUtotdr=Npairs!n Arn+1+!kq2r2"#$%&'()*+,-from whichreqn!1=nA/!kq2, andd2Utotdr2=Npairsn(n+1)Arn+2!2!kq2r3"#$%&'()*+,-which at equilibrium is:12d2Utotdr2Roreq2=12Npairs(n!1)!kq2req3"#$$%&''()*+*,-*.*req2, from which the answer follows.3-6.Calcium oxide has the FCC structure for which the Madelung constant isα= 1.7476.Determine the cohesive energies per molecule of the hypothetical crystals Ca+O-andCa2+O2-. Be sure to account also for energy required to return the ions to neutral atoms.Assumero= 2.40 Å is the same for both forms, and neglect the repulsive energy. Theenergy 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 neutraloxygen is 1.46 and -9.0 eV, respectively. (Note that energy is actually required foradding the second electron). Which of the two hypothetical crystals is most likely tooccur 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 ofenergy input to ionize the calcium atom and -1.46 eV that is released when an electron istransferred 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 firstelectron 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 Madelungconstant in the following way: First sum only the contributions to the Madelung constantarising from the ions (or partial ions) contained within the innermost square. Note that anion on one edge of a square is counted as 50% inside and 50% outside and one on acorner is counted as 25% inside and 75% outside. Show that this first contributionamounts 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 levelthe Madelung constant is +1.6116.3-7.Solution:The inner most square has a positive contribution due to four half-atoms on the sides anda negative contribution due to quarter-atoms at each of four corners. Together thesecontribute to the Madelung constant as:!1=4 1 / 21!"#$%&+4'1 / 42!"#$%&= +1.293…. Similarly,the contribution due to fractions of atoms in the next shell are:!2=4 1 / 21!"#$%&+4'3 / 42!"#$%&()*+,-+4'1 / 22!"#$%&+4'1 / 48!"#$%&+8 1 / 25!"#$%&= +0.314…. The quantity in

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brackets corresponds the partial atoms in contact with the first square. Finally theoutermost shell:!2=4!1 / 22"#$%&'+4!3 / 422"#$%&'+8 1 / 25"#$%&'+4 1 / 23"#$%&'+4!1 / 432"#$%&'+8!1 / 210"#$%&'+8 1 / 213"#$%&'=0.0037…

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CHAPTER 44-1.A power transformer consists of two windings around a common ferromagnetic coreand is used in ac circuits to step up (or down) an ac voltage. During each cycle, theferromagnetic core repeats the process of growing and shrinking domains. (a) Show thatthe 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 smallcoercivity? 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 righthas been rotated so that M is along the x-axis. During segment ab, H > 0 and dM > 0 sopositive work is performed equal to the area AB under the curve. In segment bc, dM = 0so no work is performed. In segment cd H < 0 and dM < 0, so again positive work isperformed 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 dMare again > 0 and positive work equal to the area FA is performed. This total work mustbe performed in each cycle of the transformer and is an unavoidable energy loss. It canonly be minimized by minimizing the area of the hysteresis loop as by narrowing thewidth (da in the figure below) by decreasing the coercivity.4-2.Consider the centripetal force on the orbiting electron of the classical Bohr modelillustrated in Fig. 2. Consider both a situation with and without a magnetic field presentand show that when the field is present and directed upward (perpendicular to the orbitalplane), the electron speeds up by!"# $%eB/2mand that the change in its angularvelocity 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 orbitalspeed.4-2.Solution:

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