{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Solution1

# Solution1 - ECE230A Solution for Homework#1 1 Provide a...

• Notes
• ProfessorResolvePartridge10091
• 3
• 100% (4) 4 out of 4 people found this document helpful

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ECE230A Solution for Homework #1 1. Provide a photograph of yourself and a brief self-instruction (3N4 sentences about your background, current status and studyfresearch interest) 2. Many III-V compound semiconductors such as GaAs and InP have two cleavage planes (110) and (110). You can cut the crystal along these two planes to create mirror-like facets. This is how semiconductor lasers are fabricated (i.e. using the cleaved planes as reﬂecting mirrors). For a (211) GaAs wafer, ﬁnd (a) the cleavage plane(s) that can cut through the (211) wafer with a mirror-like facet, (b) the plane that is normal to both the (211) plane and the cleavage plane(s) in (a). The planes you ﬁnd in (a) and (b) are normally the “major flat” and “minor fla ” of an otherwise circular wafer. ANS: (a) The two cleavage planes belong to the family of plane {110}. The planes of the {110} are (110), (110), (101), (101), (011). (011) The cleavage plane (abc) with (211) wafer should be perpendicular to (211) plane as well. 9 [a, b, c] ' [2,1, 1] I 0, and (abc)E{110} [a, b, c] ' [2, 1, 1] : 2a+b+c : 0 (a, b, c: integer) Among possible sets of (a, b, 0), therefore, the cleavage plane is (01T) or (0T 1) (b) Suppose the other plane (a’b'c') is normal to both (211) and (011) (or (011)), then this plane can be obtained by 1' j k :> [a'b'c']=[01T]><[211]= 0 1 —1=2i—2}'—2k(i,j,k:basis) 2 1 1 Thus the one of the normal plane is(1TT), or equivalently (T11). a a IR-G 3. Prove mathematically the relation: reciprocal: e : 1 where are the lattice vector and reciprocal lattice vector, respectively. ANS: % % % I I I % % —) Let 611,012,013 are vectors that deﬁne a pr1m1t1ve cell and b1,b2,b3 are the corresponding reciprocal lattice vectors. a Xa a X61 So, 231 7271' 2 3 , 332 72:7 3 1 , 393 7271' a1°a2><a3 a1°a2><a3 a1°a2><a3 a1 xa2 aI 01).:271'whenz':j, and J aI obj :Owheniij ECE230A Solution for Homework #1 a a a a Let the lattlce vector R : 111 511+ 172 a2+ 113 513 (111,112,113 are integer) and the reelprocal lattlce —> —> —> —> I vectorG = V] [71+ v2 [72 + 173 b3 (17], v2 , 173 are 1nteger) 1:1 j:1 1:1 j:1 3 —> —> 21132} (at o bf) =27r(111v1 + 112172 + 113173) 2 27m (n is an integer) .=1 F1 551 s . Q1 Therefore 8 = enﬂimeg") =1 5 —>—>—> 4. Deﬁne the primitive cell (by its vectors a, b, c) of an FCC lattice and show that its and reciprocal lattice has a BCC structure. ANS: a, b, c are vectors of the primitive cell of the given FCC lattice as shown below, then, a = gag? + 218 = gag + 2), E = gag + 37), Volumel/ = Zr 0 (5 x E) :14 +§)-<&a2(f+f)><(f+37)} km 3? )7 (f+§)><(f+)7)= 1 0 1 1 a Reciprocal lattuce vector 51, E2 , £73 a a _ 271' l 1 271' b 22—” be -—a-—a- —7E+ﬁ+§ 2— —§c+*+§ 1 V( ) i613 2 2 ( y ) a( y ) ~ 2 a 271' 1 1 a a a 271' a a a 152:27”(C><ar)=1 3-7a-7a-(x—y+z)=i(x—y+z) Ea 2 2 0! ~ 2 ~ 271' 1 1 a a a 271' a a a :l73=27’7(61><b)=1 3-7a-7a-(x+y—Z)=i(x+y—z) 1a 2 2 a —>—>—> Thus, these b1, b2, b3 vectors are the primitive cell of a BCC lattice. ECE230A Solution for Homework #1 ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

Solution1 - ECE230A Solution for Homework#1 1 Provide a...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online