hw1 soln

hw1 soln - 3.7" This problem calls for a computation...

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Unformatted text preview: 3.7" This problem calls for a computation of the density of molybdemun. According to Equation 3.5 2le _ Mo F3 — ,—, (“WA For BCC, n = 2 atoms-:‘imit cell. and V = "43:"? C ‘13.; I Thus. 9 = HAMU - .3 ‘ 4R I V ‘I A 4.! _. {2 atoms-"unit cell)(95.94 gr’inol) [(4)001363 x 10‘? cm)3.-"-v"§]3.-’(uuit cell}{6.023 >< 1033 atoms.-"1nol) = 10.21 gfcn'ij The value given inside the front cover is 10.33 g.-"c1n3. 3.14 For each of these three alloys we need. trial and error. to calculate the density using Equation 3.5, and compare it to the value cited in the problem. For SC, BCC. and FCC crystal structures. the respective values of 412 a? are l. 2. and 4, whereas- the expressions for .1: (since VC = (13) are ER. 2191'? . and {.3 . 1' For alloy A, let us calculate p assluning a BCC crystal structure. MAA ! .r ICAA p: HAA 42"3 . [E] “A (2 atoms-"mm cell)l[43.1 g-"moD -3 (430.22 =< 10-3 cm) I'_ . .-’(1mitcell) {6.023 =< 1f}23 atoms-find) N3 _| = 6.40 g-"ctn3 Therefore, its crystal structure is BCC. For alloy 13, let us calculate [3 assuming a simple cubic crystal structure. _ 13AB 9 ‘ 3 .- (2a) AA (1 atom-"wait cell)(184.4 g-"inoD [(2)046 X 10‘8 c111)]7).-"(un.itcell) {15.023 >< 1033 atoms-incl) = 12 .3 g-"cm3 Therefore, its crystal structure is simple cubic. For alloy C, let us calculate [3 assuming a BCC crystal structure. (2 atoms.-’unit cell){91.6 g-"moD ‘3 . -3 ‘ ..-"(Lmit cell)(6_023 :< 10—3 atoms-"11101) "J3 | = 9.60 g-"cm3 Therefore, its crystal stmcmrc is BCC. 3.3]. Direction A is a [T10] direction, which determination is smmnarized as follows. We first of all position the origin of the coordinate system at the tail of the direction yector; then in terms of this new coordinate system l l' 2 Projections — a b 06 Projections in terms ofa. b, and c —l 1 0 Reduction to integers not necessary Enclosure [110] Direction B is a [121] direction, which determination is summarized as follows. The yector passes through the origin of the coordinate system and thus no translation is necessary. Therefore. 1 .‘L' Q Projections E b E 2 2 . . . 1 1 PrOjections n1 terms ofa. b, and c E 1 : Reduction to integers 1 2 1 Enclosure [1'21] Direction C is a [01:] direction. which determination is summarized as follows. we first of all. position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system i l' E . . E! PrOjections 00 —E — c . . . l PrOjections n1 terms ofa. b, and c 0 —: —1 Reduction to integers O —1 —2 Enclosure [GTE] Direction D is a [1:1] direction, which determination is suimnarized as follows. we first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system i l' E Projections E —b E 2 3 . . . 1 1 PrOjections n1 terms ofa. b, and c E —1 : Reduction to integers 1 —'2 1 Enclosure [131] 3.32 Direction A is a [BET] direction, which determination is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system A l' £ Projections a b _% Projections in terms of a. b, and C l l —% Reduction to integers 3 3 —l Enclosure [33T] Direction B is a [10:] direction. which determination is summarized as follows. we first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system 1 I Z . . 2a Pm] ections — — 05 —£ 3 3 . . . 2 1 PIOJections n1 terms ofa. in, and c —E 0 —: Reduction to integers —4 O —3 Enclosure [10R] Direction C is a [Rol] direction, which determination is summarized as follows. we first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system l l' S Projections — E b 3 2 6 . . . 1 l Pro_]ections n1 terms of a. in, and c —E l E Reduction to integers —3 6 l Enclosure [361] Direction D is a [lil] direction. which determination is summarized as follows. we first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system l J.‘ i . . 45 Pro; ections — E — —£ 2 2 3 . . . l l l 3rojections in terms ofa. 3:, and c —E 7 —: Reduction to integers —l l —l Enclosure [l l l] 3.40 For plane A we will leave the origin at the unit cell as shown. If we extend this plane back into the plane of the page, then it is a (1 II) plane. as summarized below. i l' ; Intercepts a b — c Intercepts in terms of a. b. and c l l — l Reciprocals of intercepts l l — 1 Reduction not necessary Enclosure [1 1T) [Nata If we move the origin one unit cell distance parallel to the x axis and then one unit cell distance parallel to the _1-' axis. the direction becomes (II For plane B we will leave the origin of the unit cell as shown; this is a (230) plane. as summarized below. 1 l' i I) Intercepts E _ 3x 2 3 . 1 l Intercepts 1n terms of a. b. and c 3 E on: Reciprocals of intercepts- 2 3 0 Enclosure {230) 3.42 For plane A since the plane passes through the origin of the coordinate system as shown, we will move the origin of the coordinate system one unit cell distance vertically along the : axis; thus, this is a (Ell) plane, as surmnarizecl below. i l E a Intercepts T E) — c . l Intercepts in terms of a. 3:. and c E l — l Reciprocals of intercepts '2 l — 1 Reduction not necessary Enclosure (2 1 I) For plane 13. since the plane passes through the origin of the coordinate system as shown. we will move the origin one unit cell distance vertically along the : axis: this is a (CDT) plane. as summarized below. 1 I 2 b Intercepts a: a j — c Intercepts in terms of a. b. and c :2: % — l Reciprocals of intercepts O 2 — 1 Reduction not necessary Enclosure (02 l) ...
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This note was uploaded on 04/09/2008 for the course MSE 104 taught by Professor Lin during the Spring '08 term at UCLA.

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hw1 soln - 3.7" This problem calls for a computation...

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