LN 4 IDLE MIND SOLUTIONS. N A n. vol unit cell. xa 3 (m 3 mole) AW (g mole) (g cm 3 )

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1 LN 4 IDLE MIND SOLUTIONS 1. To do this, the equalit of the molar volume and v # unit cells mole vol unit cell N A n a 3 (m 3 mole) v atomic wt. densit AW (gmole) (gcm 3 ) is used, remembering to convert the volume obtained to m 3 /mole using the atomic weight: AW (gmole) (gcm 3 ) 10 6m3 cm 3 N A (atomsmole) n (atomsunit cell) a3 (m 3 unit cell) Ni: AW = g/mole ρ = 8.90 g/cm 3 FCC, so n = 4 atoms/unit cell a 3 (58.70 gmole) (4 atomsunit cell) (8.90 gcm 3 ) ( atomsmole) 106m3 cm 3 = m 3 a = ( ) 1/3 = m Cr: AW = 52.0 g/mole ρ = 7.19 g/cm 3 BCC, so n = 2 atoms/unit cell a 3 (52.0 gmole) (2 atomsunit cell) (7.19 gcm 3 ) ( atomsmole) 106m3 cm 3 = m 3 a = ( ) 1/3 = m ρ =

2 2. Mo: AW = g/mole ρ = 10.2 g/cm 3 BCC, so n = 2 atoms/unit cell a 3 (95.94 gmole) (2 atomsunit cell) (10.2 gcm 3 ) ( atomsmole) 106m3 cm 3 = m 3 a = m For BCC, a 3 4r, so r m 3. BCC structure, so n = 2 a = 3.31 Å = m ρ = 16.6 g/cm 3 AW 10 6 N A n a 3 AW ( atomsmole) ( m) 3 (2 atomsunit cell) (10 6 m 3 cm 3 ) 16.6 gcm 3 = g/mole 4. Cu is FCC, so: n = 4 a = 3.655Å = m AW = g/mole AW 10 6 N A n a 3 (63.55 gmole) (4 atomsunit cell) ( atomsmole) ( m 3 ) 8.64 gcm3 ρ =

3 5. Ni: n = 4 AW = g/mole ρ = 8.83 g/cm 3 For a face centered cubic structure, the second nearest neighbor distance equals a (see LN4 11). AW 10 6 N A n a 3 a 3 (58.70 gmole) (106 m 3 cm 3 ) (4 atomsunit cell) ( atomsmole) (8.83 gcm 3 ) m 3 a m 1012 pm m pm 1st nearest neighbor 2nd nearest neighbor 6. Vanadium: AW = g/mole ρ = 5.8 g/cm 3 BCC, so n = 2 The highest densit would be found in the [111] direction. To find a : AW a 3 N A n a a cm m The length in the [111] direction is a 3, so there are: 2 atomsa 3 2atoms( m) = atoms/m ρ =

4 7. This problem can easil be solved b relating it to the unit cell dimensions. (We must watch out for dimensions!) BCC, n = 2 a = m ρ = 5.81 g/cm 3 At.Wt. Mol.Vol. N A n a 3 At.Wt. N A n a ( ) at mole 1 atu.c. m3 U.C. g cm 3 At.Wt a at mole 1 atu.c. m3 U.C. g cm 3 = 49.1 g/mole appears to be vanadium?? 8. P.D. P.D. P.D. = 68% Vol. atomsu.c. Vol.U.C. (8r3 ) r 3 (64r 3 ) r for BCC P.D. (8r3 )3 a a 3 4r a (4r) 3 a 3 (64r 3 ) (a) To answer this question we can look at the lecture notes (#4, p. 11) and find for the FCC sstem (Ni) the second nearest neighbor distance given as a! Alternatel we can look at two adjacent unit cells and confirm a to be the distance to the net nearest neighbor. The problem is reduced to the determination of a, the lattice constant: a ρ =

5 9. (a) Continued. At.Wt m3 mole N A 4 a3 a m (b) FCC is a close packed structure with 12 nearest neighbors: si in one plane and three above and below that plane. 10. Au: in the P.T. we find: Structure: FCC At. Vol.: 10.2 cm 3 /mole = m 3 /mole For the FCC sstem we know that a 2 4r. Thus: = N A /4 a 3 a m a 2 4r r m 11. We can solve this problem b considering the effective volume per atom in the two unit cells, BCC and FCC. The volume change in % is then given b (V f V i ) / V i 100. (To simplif calculations, we assume the atomic radius to be unit 1Å if ou want.) BCC: 2 at./u.c. V at.(bcc) = a 3 /2 a 3 = 4 (r = unit) = 64/6 3 a 3 = 64/3 3 FCC: 4 at./u.c. V at.(fcc) = a 3 /4 a = 2 2 (r = unit) = (16 2)/4 a 3 = 16 2 = 4 2 ρ =

6 11. Continued. The volume change for the transformation (in %) is: V V f V i V i = 8.1% [101] [112] [012] [021] [100] 13. (a) The largest holes are the octahedral voids formed b eight (8) contiguous atoms, for eample, about the (ver) center of an FCC unit cell. The location of the center: 1/2, 1/2, 1/2. (b) The total number of octahedral voids per U.C.? One in the center and 1/4 void centered on each edge of the U.C. Since there are 12 edges, we have a total of (1 + 12/4) = 4 octahedral voids. ρ =

7 14. The (111) plane in an FCC structure contains si [110] directions. [110] [011] ÇÇ ÇÇÇ ÇÇÇÇÇ ÇÇÇÇÇÇ ÇÇÇÇ [101] [101] [011] [110] 15. We look at the problem on the scale of the U.C. (FCC). a 2 a 2ata 2 2 From the P.T. we know: Mol.Vol. = 9.10 cm 3 /mole FCC = 4 at./u.c = N A /4 a 3 a m 2 atoms a atoms m at m 2 ρ =

8 16. We solve this problem b determining the molar (atomic) volume. Mol.Vol. = at.wt./ρ = N A /2a 3 We need to know a given d (110) d (110) a 2 a m m Mol.Vol ( ) cm 3 mole Mol.Vol. At.Wt At.Wt cm3 mole g cm Pt: A look at the U.C.: The face centered atoms are net nearest neighbors! The are spaced at a. a, the lattice constant, can be found to be m. a ρ =

9 18. ÇÇÇÇÇ ÇÇÇÇÇÇ ÇÇÇ Ç (111) ÇÇÇ ÇÇÇÇÇÇ ÇÇÇÇÇ Ç (012) (012) Ç ÇÇ ÇÇ Ç 19. In nature most solids appear as polcrstalline sstems, which consist of interwoven conglomerates of small crstallites (grains with diameters ranging from <10 6 to >10 4 m). Anisotrop in materials is contingent on the eistence of order over macroscopic dimension. Thus single crstals will ehibit properties that are dependent on direction (anisotrop). Polcrstalline materials, on the other hand, appear isotropic because of the random spatial orientation of the crstallites. 20. Required: BCC 2 atoms/u.c. Molar Volume Atomic Weight N A 2 a3 Atomic Weight ( ) At.Wt. = 49.1 ρ =

10 21. Required: SC 1 atom/u.c. a = 2r Packing Densit Volume of AtomsUnit Cell VolumeUnit Cell 4 r3 3 a 3 4 r3 3 8r % 22. [101] [101] [210] [101] 23. Required: FCC = 4 atoms/u.c. d (hkl) = a h 2 k 2 l 2 Atomic Weight Densit Molecular Volume a 3 N A 4 3d 3 3 N A 4 Atomic Weight 3 ( ) (a d (111) 3) ρ =

11 24. It is most convenient to resort to the Unit Cell dimensions: We recognie that there is one atom per lattice constant (a) in the [100] direction [100] From the P.T.: At.Wt. = ρ = 12.0 g/cm 3 Structure = FCC N a3 A 4 3 a N A cm (We used ρ in units of g/cm 3.) = m We have one atom/( m), or: atoms/m in the [100] direction. 25. a 2 a We recognie two atoms within 2 the area a 2 2. ÉÉ ÉÉÉÉÉÉ ÉÉÉÉÉÉ ÉÉÉÉ 1/4 1/2 1/4 1/4 1/2 1/4 a Cu: At.Wt. = ρ = 8.96 g/cm 3 Structure = FCC a 3 N A 4 a N A cm m 2 atoms 2 ( ) atomsm 2 ρ =

12 26. In the simple cubic sstem the void at the center of the unit cell is bounded b the eight (8) corner atoms. The diameter of an interstitial atom is therefore given b the bod diagonal minus 2r. D a 3 2r 2r 3 2r 2r( 3 1) R (interstitial) r( 3 1) r 27. (111) ÉÉ ÉÉÉÉÉÉÉ ÉÉÉÉ ÉÉ Ç Ç (012) Ç Ç 28. Diffraction eperiments show us that Cu has an FCC crstal structure. This is not a primitive unit cell. {111} planes are found (b inspection of the FCC unit cell) to have the widest spacing. {100} and {110} planes are not the widest spaced planes because the additional lattice sites in the FCC structure occup positions between these planes. 29. (a) There are five was in which points can be arranged in two dimensions so that ever point has identical surroundings. The square lattice includes squares of all sies and orientation. We can also obtain such a lattice distribution b operating with a rectangular pattern. B changing to aes at some random angle (α), we obtain two additional arrangements: a rhombus (deduced from a square) and a parallelepiped (deduced from a rectangle). The fifth pattern is a rhombus for which α = 60 o (a triangular pattern). (continued) ρ =

13 29. (b) (1) (2) (3) a α a α a α b b b a=b, α=90 o a b, α=90 o (a b) a b, α 90 o (a b) (4) a b a=b, α 90 o (5) a b a=b, α=60 ο 30. Fe at room temperature is BCC (bod centered cubic); taking the atomic radius of Fe as 1.26Å, we have a face of the unit cell (with a 3 4r) at: a 2 16r2 3 (8.47Å) m 2 The face (100) of the unit cell contains 1 atom (4 1/4 atoms). We therefore have: 1 atom m 2 1 atom mm or: atoms/mm 2 ρ =

14 31. In the BCC structure of vanadium, the largest void can readil be identified as located about the site 1/2, 1/4, 0 in the unit cell. (There obviousl are a large number of such equivalent sites in each unit cell.) Look at a unit cell face for V with radius (R); the distance of the site from the origin (0) is, according to Pthagoras: But: a 4R (BCC) and 3 or : r R (R r) 2 a 2 2 a 4 2 and (R r) a 2 2 a 4 2 a 4 5 r a 4 5 4R 5 Rr 3 4 R R 5 3 R r R for vanadium = 1.34Å r Å m ρ =

15 32. ÉÉÉÉÉ ÉÉÉÉÉÉÉÉ ÉÉÉÉÉÉÉÉ ÉÉÉÉÉÉÉÉ ÉÉÉ (110) (a) (b) (c) (d) atoms/unit cell: (8 1/8) + (6 1/2) = 4 atoms/u.c. See above. The intercepts of a plane (421) on the and coordinates are the reciprocals 1/4 and 1/2; thus the ratio of the intercepts is 1/2. vol. atomsu.c. The atomic packing factor: P.D. vol.u.c. For BCC lattices we have 2 atoms/u.c.. a 3 4r or a 4r and a 3 64r Thus: P.D. 8r r r 3 364r % 33. Determine the linear densit of atoms/cm along the [110] direction in Cu. The simplest approach involves the unit cell dimensions: Cu: FCC (4 atoms/u.c.) <110> are face diagonals a 2 4r, or 2 atomsa 2 The calculation requires knowledge of a, the lattice constant: = N A /4 a 3 a m ρ =

16 33. Continued. Accordingl: 2 atoms m 1m 100 cm atomscm along Determine the planar atomic densit per m 2 for Ni (FCC) in {111} planes. In the unit cell we find (3 1/2) + (3 1/6) atoms = 2 atoms within the area of the {111} planes. First we determine a, the lattice constant: N A 4 a3 ; a m Area of equilateral triangle: A h(a 2)2h(a 2) h 2 (a 2) 2 ( a 2 )2 2a 2 a a2 h 3 a 2 A 3 a a a m 2 Planar Atomic Densit 2 atoms m Ni atomsm 2 ρ =

17 35. (111) (102) É ÉÉÉ ÉÉÉ ÉÉ ÉÉ É ÉÉÉÉÉ ÉÉÉÉÉÉ ÉÉÉÉÉ 36. a (2.02Å) Å m n = 2 for the BCC structure p = 7.87 g/cm 3 = g/m 3 A.W. N A n a 3 p atomsmol 2 atomsunit cell ( m) gm 3 = g/mol 37. (a) To find the Miller indices of,2,1, take the reciprocal: Multipl b 2 so that the are all integers: Put parenthesis around them: (012) (continued) ρ =

18 37. Continued. (b) (012) ÇÇ ÇÇ Ç (c) In the cubic sstem, the indices of directions normal to a plane are given b the Miller indices of the plane: [012] 38. This is the same problem as #6 of this lecture note! 39. We know: d (hkl) a h 2 k 2 l 2 h 2 k 2 l 2 a d (hkl) h 2 + k 2 + l 2 = 8 = ( ) The famil of planes is {220}. 40. FCC; determine the molar volume and relate to the # uc s; getting, a, the lattice constant permits determination of r (a 2 4r) molar volume = g mole 1cm3 9.12cm3 1m g mole 10 6 cm 3 N A 4 a a = m a 2 4r r a a 3 r = m ρ =

19 41. Determine the atomic radius to get the Vol atoms/mole and subtract that from the molar volume. Alternatel, we ma determine the packing densit and use that (determined in school as 74%) to get the void volume (26% of the molar volume). To make the solution simpler, I would use the data from the previous question. From above 9.12 cm 3 /mole (1 0.74) = 2.37 cm 3 /mole void volume or ( N a3 ) N 4r3 3 = void volume/ mole ( ) ( ) = 2.35 cm 3 /mole 42. I would go and read the lecture notes! (LN4, page 1). Accordingl, I would put some Ge into a quart glass, heat it up to above the melting point just a bit so that all is molten ( I would have to do that in an inert gas, otherwise, that Ge would be oidied in air Ge + O 2 GeO 2, GeO 2 + Ge 2 GeO the final compound is volatile and escapes all that would happen: a lot of smoke is generated and there would be no Ge left to melt). I would etract heat somehow from the bottom until all the melt is cooled to below the melting point and then I would pra that onl a single crstal was formed in the process. 43. Self eplanator: I hope ou did it! 44. The simplest wa to differentiate the 2 samples is to put them into a dissolving solution (dilute nitric acid, HF and the like, could do it ou do not have to know this). On the single crstal sphere ou would notice etch pits with distinct smmetr features which reflect internal atomic order; there would be no geometric patterns on the surface of the etched glass sphere. You could also take a hammer and smash the sample epect on the glass, a tpical conchoidal fracture on the crstal appearance of some facets flat areas reflecting eposure of low inde planes. 45. {110} = (110), (110), (101), (110) (101), (110), (011), (101) (011), (101), (011), (011) (see lecture notes) ρ =

20 <100> linear densit = 2 atoms a 1atom a [Sr: FCC; n=4 At.wt = 87.62; ρ = 2.6g/cm 3 ] mol vol = a 3 At.wt 4 At.wt N N 4 a cm 1m 100cm linear densit = 1atom m = atoms/m 47. Ta: BCC; n=2 ; At.wt. = ; ρ = 16.6g/cm 3 a 2 a planar densit = 2 atoms/a 2 2 at. vol = N 2 a3 ; a = m planar atom densit = 2atoms a 2 2 = 2atoms = /m 2 2 ( m) (a) Ag: FCC nearest neighbors are along a 2 at a 2 2 ; second nearest neighbors are spaced at a. molar volume = At.wt N 4 a3 ; a at.wt 4 N a = m = net nearest neighbor or distance (b) d {110} Ag: a a = m ρ =

21 49. planar densit in {111} = 2atoms a2 2 h 2 (a 2) 2 2 a2 2 h h 2 a 2 a3 2 a a2 a planar atom densit = = 2 atoms/ a = atoms/m 2 = m 2atoms a m2 ρ =

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