Fundamentals Of Condensed Matter And Crystalline Physics .

Fundamentals of Condensed Matterand Crystalline PhysicsInstructor's Solution Manual(rev. Dec. 2011)Full file at tom

CHAPTER 11-1. Show that the volume of the primitive cell of a BCC crystal lattice is a3/2 where a isthe lattice constant of the conventional cell.1-1. Solution:From Fig. 1.9, the primitive axes are!a1 a((1/2) xˆ (1/2) yˆ " (1/2) zˆ )!a2 a((1/2) xˆ " (1/2) yˆ (1/2) zˆ )!a3 a("(1/2) xˆ (1/2) yˆ (1/2) zˆ )!!% xˆ yˆ zˆ ('1* !! ! !V ( a1 " a2 ) # a3 ' 1 1 1 * # a3 a 3 /2' 2 1 1 1 *&)1-2. Show that the volume of the primitive cell of a FCC crystal lattice is a3/4 where a isthe lattice constant of the conventional cell.1-2. Solution:FromFig. 1.9, the primitive axes are!a1 a((1/2) xˆ (1/2) yˆ )!a2 a((1/2) xˆ (1/2) zˆ )!a3 a((1/2) yˆ (1/2) zˆ )! xˆ&1! ! !V ( a1 " a2 ) # a3 & 1&2 1%yˆ zˆ ') !1 0 ) # a3 a 3 /40 1 )(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 aballoon is R 3a /4 . There are a total of 2 balloons in each BCC unit cell for total8" 3R and packing fractionoccupied volume3!8"R 3 /3 8" ( 3 /4) 33"! 3 a38!!Full file at tom

1-4. Show that the packing fraction of a FCC crystal lattice is2" /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 /2 2 . There are a total of 4 balloons in each BCC unit cell for total16" 3R and packing fractionoccupied volume3!33#&16"R/316"1"2"! %( a33 2 2 ' 3 26!1-5. 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. 1-14Full file at tom

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 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. 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,c2c2or 2 a 1.633 .23a3!!!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) ! M N(79.9 / 6.02x10 23 ) V(4.65x6.73x8.70)g / Å 3 3.12 g / cc ! N " 8Full file at tom

(b) Since bromine has a chemical formula of Br2, there are 4 molecules of bromine in aunit cell, suggesting a FC orthorhombic structure.38(4! (1.5) / 3) 0.333(c) PF (4.65x6.73x8.70)1-9. 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 then32(4! (1.46 ) / 3)PF 0.724(3x3x4)Full file at tom

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 calculate g(r) for a single ensembleusing the dark particle as the central particle. Do this with a dr no 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:Full file at tom

Page #1Countthe- “HW2 1”number of particles centers found in each ring of size dr (here equal to b) andMonday, October 17 1:12 Pmake a table:ring .81448DEFGHIJFor 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 15th ring of area ! (15b)2 . A histogram of the g(r)(for only this one ensemble) looks like this:Full file at tom

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.Full file at tom

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 SiO2 if the O donated by either is to end up bonded with a Siatom?2-6. Solution:Full file at tom

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.Full file at tom

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