hw sol ch3

# hw sol ch3 - 3.3 For this problem, we are asked to...

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Unformatted text preview: 3.3 For this problem, we are asked to calculate the volume of a unit cell of aluminum. Aluminum has an FCC crystal structure {Table 3.1). The FCC unit cell volume may be computed from Equation [3.4) as VC 2 16R3'V'E = (’16)(0.143 x 10'9 mi3-C2 : 5.62 x 10'29 m3 3.10 This problem asks for us to calculate the radius of a vanadium atom. For BCC. n = 2 atomsfunit cell. and _ 4R 3 _ a4e3 VC — —_ — _ "J3 3V3 Since. p: niﬂiV VCNA and solving for R 7 H3 R 2 3w 3mV B4pNA — its 2 53V 31l2 atomsfunit cellilSDB gi‘moll (muses grcm3)(6.023 x 1023 atomsfmol) 8 21.32 x10‘ cm 2 0.132 nm 3.15 For each of these three alloys we need to, by trial and error, calculate the density using Equation (3.5), and compare it to the value cited in the problem. For SC, ECG, and FCC crystal structures, the respective values of n are 1, 2, and 4, whereas the expressions for a {since VC 2 a3) are 2R, 2R5, and Aim-F3. For alloy A, let us calculate p assuming a simple cubic crystal structure. I): nAA VCNA {i atomfunit cell)(??.4 glmol) 3 [(2)(‘| .25 x 10‘3 cm)] funit cell(6.023 x 1023 atomstmoll = 8.22 glcms Therefore, its crystal structure is SC. For alloy B, let us calculate p assuming an FCC crystal structure. = (4 atomsfunit celljf’l 0?.6 gtmol) p _r‘ . a 3 . 23 (2)v2(|.33x10 cm) fumtcelliBUZBx ’10 atomsfmol) = 13.42 grcm3 Therefore, its crystal structure is FCC. For alloy C, let us calculate p assuming an SC crystal structure. p _ (‘l atomfunit cell)(’127’.3 gtmol) ' 3 [(2)(1.42 x 10'3" cmj] .tunit cell(ﬁ.023 x 1023 atomsfmol) = 9.23 grcm3 Therefore, its crystal structure is SC. 3.34 We are asked to calculate the density of FeO. This density may be computed using Equation as r1(Ammo) p: VCNA Since the crystal structure is rock salt, n' : 4 formula units per unit cell, and v —a3— 2r +2r 3—[moorrnmi+2w140nmﬂ3 o" ‘ ( Fe2+ 02-) ‘ - - s s =0cm7 Um =8J7x1023 9“ unit cell unit cell Thus, 4 formula units-"unit cell 55.85 i‘mol + 16.00 .-"mol p_ _(847x'uf23 3 23 cm r'unit cell)(6.023 x ‘ID formula units-“mo” = 5.84 g.-"cm:3 3.46 The silicate materials have relatively low densities because the atomic bonds are primarily covalent in natureMand, therefore, directional. This limits the packing efficiency of the atoms, and therefore, the magnitude of the density. 3.51 3.5? For plane A we will move the origin of the coordinate system one unit cell distance to the upward along the z axis; thus, this is a (322) plane, as summarized below. i E E Interce ts g E -E p 3 2 2 Intercepts in terms of a, bi and c 1— 1— - l 3 2 2 Reciprocals of intercepts 3 2 —2 Enclosure {322) For plane B we will move the original of the coordinate system on unit cell distance along the x axis; thus, this is a (I01) plane: as summarized below. 1 E E Q 00 E Intercepts — 2 b 2 Intercepts in terms of a, bi and c - 1 co 1— 2 2 Reciprocals of intercepts -2 U 2 Reduction —‘| D 1 Enclosure {701) ...
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## This note was uploaded on 03/27/2010 for the course ES 162 taught by Professor Ang during the Spring '10 term at Yeditepe Üniversitesi.

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hw sol ch3 - 3.3 For this problem, we are asked to...

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