Solution Manual for Introduction to Solid State Physics, 9th Edition by Kittel - Complete, Verified Solutions (All Chapters)

ALL 22 CHAPTERS COVERED
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SOLUTIONS MANUAL
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,TABLE OF CONTENTS
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Chapter 1 - 1-1
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Chapter 2 - 2-1
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Chapter 3 - 3-1
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Chapter 4 - 4-1
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Chapter 5 - 5-1
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Chapter 6 - 6-1
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Chapter 7 - 7-1
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Chapter 8 - 8-1
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Chapter 9 - 9-1
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Chapter 10 - 10-1
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Chapter 11 - 11-1
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Chapter 12 - 12-1
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Chapter 13 - 13-1
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Chapter 14 - 14-1
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Chapter 15 - 15-1
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Chapter 16 - 16-1
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Chapter 17 - 17-1
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Chapter 18 - 18-1
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Chapter 20 - 20-1
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Chapter 21 - 21-1
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Chapter 22 - 22-1
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, CHAPTERk1

1. Thekvectorsk x̂kkyˆkkzˆk andk xˆkkyˆkkzˆk arekinkthekdirectionskofktwokbodykdiagonalskofka
cube.k Ifk k isk thek anglek betweenk them,k theirk scalark productk givesk cosk k =k –1/3,k whence
kkcos11/k3kk90k19k28k'kk109k28k'k.

2. Thek planek (100)k isk normalk tok thek xk axis.k Itk interceptsk thek a'k axisk atk 2a'k andk thek c'k axis
atk 2c'k;k thereforek thek indicesk referredk tok thek primitivek axesk arek (101).k Similarly,k thek plane
(001)kwillkhavekindicesk(011)kwhenkreferredktokprimitivekaxes.

3. Thekcentralkdotkofkthekfourkiskatkdistance


cosk60k ak
ak kakctnk60kk cos
k30 3

fromkeachkofkthekotherkthreekdots,kaskprojectedkontokthekbasalkplane.kIfkthe
k(unprojected)kdotskarekatkthekcenterkofksphereskinkcontact,kthen


2
k ak  kck
2

a 2 k k k kk kk ,
 3 k k2k


or

2k 2 1k 2 c 8
a k k c ; k1.633.
3 4 a 3




1-1

, CHAPTERk2

1. Thekcrystalk planek withk Millerk indicesk hkAk isk ak planek definedk byk thek pointsk a1/h,k a2/k,k andk a3k/kAk.k (a)
Twokvectorskthatkliekinkthekplanekmaykbektakenkaska1/hk–
k a2/kkandk a1k/h kka3k/kAk.kButkeachkofkthesekvectorskgiveskzerok ask itsk scalark productk withk Gkkha1kkka2kk
Aa3k,ksok thatkGk mustkbek perpendicularktokthekplane
hkAk.k (b)k Ifk n̂k isk thek unitk normalk tok thek plane,k thek interplanark spacingk isk n̂kka1/hk.k Butk n̂k kGk/ | kGk|k,
whencek d(hkA)k kGkka1k/kh|G|kk2k/k|kG|k.k(c)kForkaksimplekcubicklatticek Gkk(2k/ka)(hx̂ kkkyˆkkAẑ ) k,kw
hence

1k G2 h k kk k kA
2 2 2
 k  .
d2 42 a2
1 1k
3a a 0
2 2
1 1k
2. (a)k Cellk volumekak kak kak k k 3a a 0
1 2 3
2 2
0 0 c


1
k 3ka2c.
2

x̂ yˆ zˆ
a2k ka3 1 1k
(b) bk k2  4k3 k 3a a 0
1
|kak kak kak | a2c 2 2
1 2 3
0 0 c
2k 1k
k (k x̂kkŷ ), kandksimilarlykforkb 2k,kb3.
a 3

(c) Sixkvectorskinkthekreciprocalklatticekarekshownkasksolidklines.kThekbrokenklin
eskarekthekperpendicularkbisectorskatkthekmidpoints.kThekinscribedkhexagonkfor
mskthekfirstkBrillouinkZone.

3. Bykdefinitionkofkthekprimitivekreciprocalklatticekvectors
(a2k a 3 k)kk(a3k a 1 k)k(a1kka 2 k)k
V k(2)3k k(2)3k/k|k(ak kak kak )k|
BZ
|k(ak ka a k )3k| 1 2 3
1 2 3

k(2)3k/kVkC.

Fork thek vectork identity,k seek G.k A.k Kornk andk T.k M.k Korn,k Mathematicalk handbookk fork scientistsk andke
ngineers,kMcGraw-Hill,k1961,kp.k147.

4. (a)kThiskfollowskbykforming


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