SHW1 - SHW-1 L3(a)Gecrystallizes in thediamond lattice...

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Unformatted text preview: SHW-1 L3 (a)Gecrystallizes in thediamond lattice wherethereaie 8atomsperunit cell (see Subsection 1.2.3). Thus DENSITY = 53 = —8—— = 4.44 x 1022 “0ch a (5.65 x10‘8)3 1A (a) From Fig. 1.3(c), we conclude nearest-neighbors in the bee lattice lie along the unit cell body diagonal. Since the body diagonal ofacube is equal to ‘5 times acube side length (the lattice constant a), Nearest-Nd h g (—J—m “r u (b) From Fig. 1.3(d), nearest-neighbors in the fcc lattice are concluded to lie along a cube- face diagonal. The diagonal of a cube face is equal to J? times a cube side length. Thus Newest-Neighbor _ E 1.5 . (a) Looking at Fig. 1.4(a) one concludes 0 O O O (b) For Si at room temperature a = 5.43 x 10'8 cm. From the above figure one concludes thattherearc(ll4x4comeratoms)+ l bodyatom=2atomsperanareaofazonthe (100) surface. Thus one has .2. 2 = —-—— = 6.78 x 10“ Si atoms/cm2 42 (5.43x 10-8)2 (c) For a (110) plane one has the atom placement pictured below H-Qn-T . . ' a O O Owl—J. I T20—| (d)Onthe(110)planeinthearea axJ—Za onehas(l/4x4corneratoms)+(l/2x2edgc atoms) + 2 body atoms =4atoms. Thus one has 4 = A ‘JZaZ (5.43x10'8)2 = 9.59 x 1014 Si atoms/cm2 L2 As noted in the problem statement. two directions [hjkllll and [Iizkzlz] will be mutually perpendicular if hlhz + klkz +1112 = 0 (a) Here [hi/c111] = [100], requiring h2 = 0. All directions [016212] are perpendicular to [100]. Two specific simple examples an: [001] and [011]. (b) Given U: ml.) .. [1 1 i), one tuquircs the Miller indices of the perpendicular direction to be such that It: 4- k2 + l; = 0. Two specific examples are [011'] and [112]. ...
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