Introduction to Solid State Physics (9th Edition) by Charles Kittel – Complete Solutions Manual for Solid State Physics Problems

ALL 22 CHAPTERS COVERED




SOLUTIONS MANUAL

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

,CHAPTER 1

1. The vectors x̂ + yˆ + zˆ and −xˆ − yˆ + zˆ are in the directions of two body diagonals of a
cube. If  is the angle between them, their scalar product gives cos  = –1/3, whence
 = cos−11/ 3 = 90 +19 28 ' = 109 28 ' .

2. The plane (100) is normal to the x axis. It intercepts the a' axis at 2a' and the c' axis
at 2c' ; therefore the indices referred to the primitive axes are (101). Similarly, the plane
(001) will have indices (011) when referred to primitive axes.

3. The central dot of the four is at distance


cos 60 a
a = a ctn 60 =
cos 30 3

from each of the other three dots, as projected onto the basal plane. If
the (unprojected) dots are at the center of spheres in contact, then
2
 a  c
2

a2 =   +  ,
 3  2

or

2 1 c 8
a 2 = c2; = 1.633.
3 4 a 3




1-1

, CHAPTER 2

1. The crystal plane with Miller indices hkA is a plane defined by the points a1/h, a2/k, and a3 / A . (a)
Two vectors that lie in the plane may be taken as a1/h – a2/k and a1 /h − a3 / A . But each of these vectors
gives zero as its scalar product with G = ha1 + ka2 + Aa3 , so that G must be perpendicular to the plane
hkA . (b) If n̂ is the unit normal to the plane, the interplanar spacing is n̂  a1/h . But n̂ = G / | G | ,
whence d(hkA) = G  a1 / h|G| = 2 / | G| . (c) For a simple cubic lattice G = (2 / a)(hx̂ + kyˆ + Aẑ ) ,
whence

G2 h + k + A
2 2 2
1
= = .
d2 42 a2
1 1
3a a 0
2 2
1 1
2. (a) Cell volume a  a  a = − 3a a 0
1 2 3
2 2
0 0 c


1
= 3 a2c.
2

x̂ yˆ zˆ
a2  a3 1 1
(b) b = 2 = 4 − 3a a 0
1
| a a a | 2
3a c 2 2
1 2 3
0 0 c
2 1
= ( x̂ + ŷ ), and similarly for b 2 , b3.
a 3

(c) Six vectors in the reciprocal lattice are shown as solid lines. The broken
lines are the perpendicular bisectors at the midpoints. The inscribed hexagon
forms the first Brillouin Zone.

3. By definition of the primitive reciprocal lattice vectors
(a2 a 3 )  (a3 a 1 ) (a1  a 2 )
V = (2)3 = (2)3 / | (a  a  a ) |
BZ
| (a  a a ) | 3 1 2 3
1 2 3

= (2)3 / VC .

For the vector identity, see G. A. Korn and T. M. Korn, Mathematical handbook for scientists and
engineers, McGraw-Hill, 1961, p. 147.

4. (a) This follows by forming



2-1