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Unit cells calculations 2 - 41 nl=2dsi119,d H?”...

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Unformatted text preview: 41. nl=2dsi119,d: H?” zwi2313132313><10’10m2313pm 2sme 2><sm1422° 42. In a face-centered cubic unit cell (ccp structure), the atoms touch along the face diagonal: "u T (4r)2 :12 + 12 m. I 12rJ§ B'- L chbe:l3:(r\/8_)3:22.63r3 There are four atoms in a face-centered cubic cell. Each atom has a volume of 4/3 7113. Vatoms : 4 X £7513 : 16.7612) 3 V 16.76 3 . . . So mm = r3 = 0.7406, or 74.06% of the volume of each un1t cell 1s occupled by chbe 22.63 r atoms. A body-centered cubic unit cell contains two net atoms (8 comer atoms x 1/ 8 atom per comer + 1 center atom : 2 atoms/unit cell). In a body-centered unit cell (see Figure 16.18), the atoms touch along the body diagonal of the unit cell. Therefore, body diagonal : 4r : J3 l. The length of a cube edge (1?) is related to the radius of the atom (r) by the equation 1? = 4r/1/3 . Volume of unit cell : P : (4r/ J3 )3 : 12.32 r3 Volume of atoms in unit cell : 2 X 2719 : 8.378 r3 3 So Vamms = 8378 r3 : 0.6800 : 68.00% occupied. V 12.32 r cube In a simple cubic unit cell, the atoms touch along the cube edge (I): T ‘ T 2(radius) : 2r : l 2 r I l vwbe= [3 = (203 = 8r3 There is one atom per simple cubic cell (8 corner atoms X 1/8 atom per corner : 1 atom/unit cell). Each atom has an assumed volume of 4/3 m3 = volume of a sphere. 43. um: gas : 4.1891‘3 4. 189 1‘3 8r3 V atom V So : 0.5236, or 52.36% occupied. cube A cubic closest packed structure (face-centered cubic unit cell) packs the atoms more efficiently than a body-centered cubic unit cell, which is more efficient than a simple cubic unit cell. A cubic closest packed structure has a face-centered cubic unit cell. In a face-centered cubic unit, there are: 1/8 t 1/2 t 8 comers >< fl + 6 faces >< fl : 4 atoms comer face The atoms in a face-centered cubic unit cell touch along the face diagonal of the cubic unit cell. Using the Pythagorean formula, Where I : length of the face diagonal and r : radius of the atom: £2 + :2 = (4r)2 '1“ aa- l:r1/8_ I l: rf: 197 Md12 m >< J§: 5.57 x 10-10111 : 5.57 x108 cm Volume ofa unit cell : 13 : (5.57 X 10'8 cm)3 : 1.73 X 10'22 cm3 Mass ofa unit cell : 4 Ca atoms X % >< W : 2662 X 10*22 g Ca 6.022 x 10 atoms mol Ca . 2.662 10'22 Den51ty : mass — x—ga : 1.54 g/cm3 volume — 1.73 >< 10'22 cm 46. A face-centered cubic unit cell contains four atoms. For a unit cell: mass ofX : volume >< density : (4.09 X 10'8 cm)3 X 10.5 g/em3 : 7.18 X 10'22 g mol X = 4 atoms X >< % = 6.642 >< 10*?14 mol X 6.022 x 1023 atoms 7.18 x 10*22 gX Molar mass : — 6.642 x 10’24 molX : 108 g/mol; the metal is silver (Ag). 50. If face-centered cubic: :2 “ls :(137pm)f :387pm23.87>< 10’8cm 4 atomsW >< —6 022 11:3: x —183.9g1 W Density : ' X Satongs mo : 21.1 g/cm3 (3.87 >< 10— cm) If body-centered cubic: z: i = M =316pm=3.16 X10’8cm «5 J? 2 atoms W x —6 022 11:10:: x —183.9g1 W Density : ' >< atoms m0 : 19.4 gz’cm3 (3.16 >< 10'8 cm)3 The measured density of tungsten is consistent with a body-centered unit cell. 52. To produce an n—type semiconductor, dope Ge with a substance that has more than four valence electrons, e.g., a Group SA element. Phosphorus or arsenic are two substances that will produce n—type semiconductors when they are doped into germanium. To produce a p- type semiconductor, dope Ge with a substance that has fewer than four valence electrons, e.g., a Group 3A element. Gallium or indium are two substances that will produce p-type semiconductors when they are doped into germanium. Doping germanium with phosphorus (or arsenic) produces an n-type semiconductor. The phosphorus adds electrons at energies near the conduction band of germanium. Electrons do not need as much energy to move from filled to unfilled energy levels, so conduction increases. Doping germanium with gallium (or indium) produces a p-type semiconductor. Because gallium has fewer valence electrons than germanium, holes (unfilled energy levels) at energies in the previously filled molecular orbitals are created, which induces greater electron movement (greater conductivity). (6.626 x 10-34 Js)(2.298 x 103 m/s) 2 730 10_9 2.72 X 10’19 J = energy ofband gap . X m 58. E=E= 9t 64. From Figure 16.42, MgO has the NaCl structure containing 4 Mg2+ ions and 4 02— ions per face-centered unit cell. lmol MgO x 40.31 g MgO — : 2.678 >< 10‘22 g MgO 6.022 x 1023 atoms lmolMgO 4 MgO formula units X 1 cm3 3.58g Volume ofunit cell : 2.678 >< 10’22 g MgO >< : 7.48 x 10’23 cm3 Volume ofunit cell : 13, l : cube edge length; l: (7.48 X 10'23 cm3)U3 : 4.21 X 10'8 cm For a face-centered unit cell, the 02— ions touch along the face diagonal: _ J— x 4.21 x 10’“ cm 4 J21 = 4r0,_, r0,_ : 1.49 >< 10'8cm The cube edge length goes through two radii of the 02’ anions and the diameter of the Mg2+ cation. So: 1: 2r0,_ 4 2ng,., 4.21 >< 10—86111: 2(149 >< 10—86111) + 2ng,., ng,. : 6.15 >< 10‘9 cm 68. a. The NaCl unit cell has a face-centered cubic arrangement of the anions with cations in the octahedral holes. There are four NaCl formula units per unit cell, and since there is a 1:1 ratio of cations to anions in MnO, then there would be four MnO formula units per unit cell assuming an NaCl-type structure. The CsCl unit cell has a simple cubic structure of anions with the cations in the cubic holes. There is one CsCl formula unit per unit cell, so there would be one lVInO formula unit per unit cell if a CsCl structure is observed. Formula units of MnO = (4 47 X 104, cm)3 >< 5.28gMnO >< lmol MnO Unit cell ' cm3 70.94 g MnO 6.022 x 1023 formula units lVInO X mol lVInO4 ’ 4.00 formula units MnO From the calculation, MnO crystallizes in the NaCl type structure. b. From the NaCl structure and assuming the ions touch each other, then 1 = cube edge length : 2rMn2, + 2r02, . I: 4.47 ><10’8 cm : 2rMI12+ + 2(l.40 x10’8 cm), ran+ : 8.35 ><10’8 cm : 84 pm 114. TMH2+ _ 84pm :0.60 r02- — 140.pm From Table 16.6 of the text, octahedral holes should be filled when 0.414 < r+/r_ < 0.732. Because the calculated radius ratio falls within the prescribed limits, we would expect lVIn2+ to occupy the octahedral holes formed by the cubic closest packed array of 02' ions (as predicted in part a). The ionic radius is 148 pm for Rb+ and 181 pm for C1_. Using these values, let’s calculate the density of the two stmctures. Normal pressure: Rb+ and C1‘ touch along cube edge (form NaCl structure). Cube edge =1: 2(148 +181): 658 pm = 6.58 X 10'8 cm; there are four RbCl units per unit cell. 4(85.47) + 405.45) — =2.82 fcm3 6.022 x 1023(658 x 10*)3 g Density = d = High pressure: Rb+ and C1‘ touch along body diagonal (form CsCl structure). 2r. + 2r+ = 658 pm = body diagonal = [J3 , I: 658 pm/a/g= 380. pm . . . 85.47 +35.45 Each unit cell contams 1 RbCl unit: d = fl = 3.66 g/cm3 6.022 ><10 (3.80 x 10 ) The high-pressure form has the higher density. The density ratio is 366/282 : 1.30. We would expect this because the effect of pressure is to push things closer together and thus increase density. ...
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