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Unformatted text preview: CHAPTER 1 1_.l. (a)1=rom Table 1.1...
(i) Si or Ge
(ii) AIP, AlAs, Ale, GaN, GaP, GaAs, GaSb, InP, InAs, or InSb (b) crystalline material has the same atomic pattern or order throughout the material, while
polycrystalline material has crystalline subsections that are misaligned with respect to each
other. (e) A unit cell is a small portion of a crystal that could be used to reproduce the crystal.
(The preceding is from the first sentence of Subsection 1.2.1.) (d) Unit cell Atom unit cel
simple cubic ........ 1
bcc ..... '. ............. 2
fcc ................... 4
diamond ............ 8 (e) 1.5.:108cm (f) a (one lattice constant) (g) 4 (h) As summarized in Table 1.3: ()—crystal plane, { }—equivalent planes, ~Crysta1
direction, and < >equivalent directions. (i) See Subsection 1.3.2. 1.2 In the AlojGaqus unit cell, pictured below, fcc sublattice sites containing the Column III
elements are equally occupied by Al and Ga atoms. ll L3 (3) Ge crystallizes in the diamond lattice where there are 8 atoms per unit cell (see
Subsection 1.2.3). Thus DENSITY —— —8—3 = ——8——— = 4.44 x 1022 atoms/cm3
0 (5.65 x 10'3)3 L4 (a) From Fig. 1.3(c), we conclude nearestneighbors in the bee lattice lie along the unit cell body diagonal. Since the body diagonal of a cube is equal to ~13 times a cube side length
(the lattice constant a), NearestNeighbor __ j:
( Distance ) _ 2 a (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 xii times a cube side length. Thus NearestNeighbor _ 1,2.—
( Distance ) " 2 a Li
(a) Looking at Fig. 1.4(a) one concludes 0 C O 0 ° 9  ?""T 0 ago: a . . é  +i
Ha (b) For Si at room temperature a = 5.43 x 10'8 cm. From the above ﬁgure one concludes
that there are (1/4 x 4 corner atoms) + 1 body atom = 2 atoms per an area of a2 on the
(100) surface. Thus one has 2‘ = ———2— = 6.78 x 10“ Si atomslcm2 02 (5.43 x 108)2 12 (c) For a (110) plane one has the atom placement pictured below . . .""'."""."T T a O oi..o.¢.. _L (d) On the (110) plane in the area a x {20 one has (1/4 x 4 corner atoms) + (1/2 x 2 edg
atoms) + 2 body atoms = 4 atoms. Thus one has it. = ——2£— = 9.59 x 1014 Si atoms/cm2
\i 2:22 (5.43 x 108)2 (e) MATLAB program script (paralleling Exercise 1.3)... %Solution to Problem 1.5(e) N=input('input number of atoms on (100) face of unit cell, N = ').' a=input('lattice constant in angstrom, a = '):
Surfaceden=N* (1 . Delﬁ) / (a‘Z) %number of atoms/cm? (Note: This and all other problem solutions are available on disk.) _l_._6
(a) (i) Following the procedure outlined in the text
1, 3, 1 ...intercepts (normalized)
1, 1/3, 1 ...[1f1ntercept]s
3, l, 3 "reduction to lowest whole—number set
(3 l 3) ...Miller index notation for plane (ii) As noted in the text near the end of Subsection 1.2.4, the normal to a plane in the '
cubic crystal system has the same Miller indices as the plane. [3 l 3] ...Miller index notation for normal to plane
(b) (i) Again following the Miller indexing procedure, 1, 1, 1/2 ...norma1ized intercepts 1, 1, 2 ...[1ﬁntercept]s 1, 1, 2 ...lowest wholenumber set ( l 1 2) ..Miller indices of plane 1—3 (ii) Assume the vector has a length 6!. Its projections along the x, y, and z axes are then 0, 0, and (1, respectively. Reducing to the lowest possible wholenumber set and
enclosing in square brackets, then yields
[001] ...Miller indices of direction vector 11 For each of the given planes, the Miller indexing procedure must be reversed to determine the intercepts of the given plane on the coordinate axes. Using part (c) as an example, one
proceeds as follows: (123) ...Miller indices
1, 2, 3 ...[1/intercept]s
1,112,113 ...intercepts
The plane in question intercepts the x, y, z coordinate axes at a, 0/2, and £113, respectively. Note that any multiple of the cited intercept set — such as 30, 2a, a — would also be correct.
All such planes are parallel equivalent planes. 2 X
(a)
Z
a
X
(e) 1—4 L8 Miller indices may be viewed as specifying the projection (in arbitrary units) of the to—be
pictured vectors along the coordinate axes. For example, [010] corresponds to a vector
with a unit projection along the y—axis and no projection along the x— or zaxes. In other
words, [010] is coincident with the +y coordinate axis. The other required direction
vectors are deduced in a similar manner and are as pictured below. L2 As noted in the problem statement, two directions [Mini 1] and [1121912] will be mutually
perpendicular if
h1h2 + k1k2 +1112 = 0 (a) Here [h1k111] == [100], requiring it; = 0. All directions [Okzlzl are perpendicular t.
[100]. Two specific simple examples are [001] and [011]. (b) Given [Mini 1] = [1 11], one requires the Miller indices of the perpendicular direction t
be such that h; + k2 + I2 = 0. Two speciﬁc examples are [OIT] and [112]. 15 LLQ As shown in the following left—hand ﬁgure, when the [011] and [010] directions are
pictured simultaneously, it becomes obvious that the angle between the two directions is
45°. Alternatively, the angle between the two directions can be computed using the cos(9)
relationship in Problem 1.9. Speciﬁcally, given [hlklh] = [011] and [hzkglz] = [010],
cos(6) = 1N§ and 6 = 45°. The required positioning of the "grooves" on the wafer’s
surface is pictured in the following right—hand ﬁgure. 2 wafer
§\\ [011}
'45°
[010]
grooves .1_l..1.
(a) if the Fig. P1.11 unit cell is conceptually copied and the cells stacked like blocks in a nursery, one concludes the resulting latttice is a simple cubic lattice . (b) There is one atom inside the unit cell and the unit cell volume is 03. Thus atoms/unit volume = 1/03 . (c) For a (110) surface plane the atom positioning would be as pictured below. : """"" T """"" i T : I : O : a i t i _L
F—ﬂa—i atoms 1 atom l ll
ll unit area ((1)050) 4—2612 (d) [111] The speciﬁed vector has equal projections on the three coordinate axes. 16 Li;
Equivalent planes: (a) 6, (b) 12, (c) 8.
Equivalent directions: ((1) 6, (e) 12, (f) 8. NOTE: The answers may be deduced from geometrical considerations  or — by noting the total number of possible combinations of the given, and negatives of the given, Miller
indices. L122 (21) In the simple cubic lattice the nearestneighbor distance is a, where a is the side length  of the cube, and the atomic radius r is therefore 0.12. Moreover, there is one atom per unit
Cell. Thus Occupied volume = git} 3 = git (al2)3 = at 03/6
Total cell volume = a3 Occugied volume Ratio = Total volume =E
6 (b) In the body centered cubic lattice the atom in the center and any one of the cube comer atoms are nearest neighbors. Thus 1/2 the nearest neighbor distance' is r = v 3 a/4. Also,
there are two atoms per unit cell. diagonal = 4r = «If—5a Occupied volume = 2%,; r 3) = %,c (‘5 “4);; ___ 318—3“ a3
Total cell volume = a3 R t _0ccugied volume _\[§rc
aio _ Total volume 8 17 (c) For a face centered cubic lattice, the closest atoms lie in a cube face. Also, there are
four atoms per unit cell in the fcc lattice. face diagonal=4r =ﬁa; r = via/4 Occupied volume = 4(3 1H3) = — 3:052 a/4)3 =ﬁtta3
Total volume = a3
Ratio = 12—5 (d) As emphasized in Fig. 1.4(c), the atom in the upper front corner of the unit cell and the
atom along the cube diagonal 1/4 of the way down the diagonal are nearest neighbors. Since the diagonal of the cube is eQuaI to {5 times a cube side length, the centerto—center
distance between nearestneighbor atoms in the diamond lattice is 43 (1/4, and the atomic radius r = {3 0/8. Moreover, there are eight atoms per unit cell in the diamond lattice.
Thus Occupied volume = 8% 1: r 3): "WW? 451/8)3 = ‘1’: 1: a3
Total volume =_a3 _ . J31:
Rauo = ‘13" 1—8 ...
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 Spring '08
 Taylor

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