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Unformatted text preview: SHW1 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 nearestneighbors 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), NearestNd h g
(—J—m “r u (b) From Fig. 1.3(d), nearestneighbors 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 NewestNeighbor _ 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 108)2 (c) For a (110) plane one has the atom placement pictured below HQnT . . ' 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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 Fall '10
 OzdalBoyraz

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