[반도체] [솔루션]반도체물성과소자 semicondutor phsics and deviceDonald A. Neamen3판
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작성일 20-05-26 11:22
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레포트 > 공학,기술계열
(a) fcc: 8 corner atoms × 1/8 = 1 atom
Then
π
[참고자료] ㅇ
= × ⇒
Ratio
100%
π
1 enclosed atom = 1 atom
[이용대상] ㅇ
r
[솔루션]semicondutor phsics and deviceDonald A. Neamen3판
반도체물성과소자 입니다
3
Density = ⇒
d = 4r = a 2 ⇒ a
Unit cell vol = a = r = r 3 3 3 c2 2 h 16 2
r
IK J
r
4
3
1 atom per cell, so atom vol. = ( )F
IK J
솔루션,반도체,소자,3판,deviceDonald
H G
= = r
6 face atoms × ½ = 3 atoms
= × ⇒
8 Ge atoms per unit cell
5 65 10 8 3
2 2
Then
Unit cell vol = a = ( r) = r 3 3 3 2 8
565 10 8 3 b . x g
(a) 4 Ga atoms per unit cell
1.3
4
설명
1.2
H G I
(b) Face-centered cubic lattice
4
3
다.
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4
4
3
4
a r 3
H G I
3
[솔루션]semicondutor phsics and deviceDonald A. Neamen3판 [참고자료] ㅇ [자료범위] ㅇ [이용대상] ㅇ
Density of Ge = − 4.44 1022 3 x cm
Total of 8 atoms per unit cell
만약 내용이 다를시 해피래포트에 환불요청하시면 환불됩니다.
IK J
4
Unit cell vol. = =F
(b)
(a) Simple cubic lattice; a = 2r
2 atoms per cell, so atom vol. = F
[자료범위] ㅇ
IK
4
FH G
8
Problem Solutions
d
Density of Ga = − 2.22 1022 3 x cm
3 πr
8
K J 1
−
r
Ratio = 52.4%
2
4 As atoms per unit cell, so that
Density of As = − 2.22 1022 3 x cm
H
3
3
FH G
d = 4r = a 3 ⇒ a = r
(c) Body-centered cubic lattice
3
−
3
4 atoms per cell, so atom vol. = F
3 4
순서
6 face atoms × ½ = 3 atoms
b . x g
Ratio
(b) bcc: 8 corner atoms × 1/8 = 1 atom
16 2
[반도체] [솔루션]반도체물성과소자 semicondutor phsics and deviceDonald A. Neamen3판
1.1
Total of 2 atoms per unit cell
(c) Diamond: 8 corner atoms × 1/8 = 1 atom
4 enclosed atoms = 4 atoms
3
Total of 4 atoms per unit cell
Density = ⇒
100%
3 πr
Ratio = 74%
[解法(해법)]semicondutor phsics and deviceDonald A. Neamen3판 입니다.
