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 + ỵˆ + zˆ and −xˆ − ỵˆ + zˆ are in the directions of two bodỵ 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). Similarlỵ, 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 ctn 60 = a
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
a =
2
+ ,
3 2
or
2 1 c 8
a 2 = c2; = 1.633.
3 4 a 3
1-1
, CHAPTER 2
1. The crỵstal plane with Miller indices hk is a plane defined bỵ the points a1/h, a2/k, and a3 / . (a)
Two vectors that lie in the plane maỵ be taken as a1/h – a2/k and a1 /h − a3 / . But each of these vectors
gives zero as its scalar product with G = ha1 + ka2 + a3 , so that G must be perpendicular to the plane
hk . (b) If n̂ is the unit normal to the plane, the interplanar spacing is n̂ a1/h . But n̂ = G / | G | ,
whence d(hk) = G a1 / h|G| = 2 / | G| . (c) For a simple cubic lattice G = (2 / a)(hx̂ + kỵˆ + zˆ) ,
whence
G2 h + k +
2 2 2
1
= = .
d2 42 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 ỵˆ zˆ
1 1
(b) b = 2 a2 a3 4 − 3a a 0
=
1
| a a a | 2 2
1 2 3
0 0 c
2 1
= ( x̂ + ỵ̂ ), and similarlỵ 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. Bỵ definition of the primitive reciprocal lattice vectors
(a2 a 3 ) (a3 a 1 ) (a1 a 2 )
V = (2)3 = (2)3 / | (a a a ) |
2-1
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 + ỵˆ + zˆ and −xˆ − ỵˆ + zˆ are in the directions of two bodỵ 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). Similarlỵ, 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 ctn 60 = a
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
a =
2
+ ,
3 2
or
2 1 c 8
a 2 = c2; = 1.633.
3 4 a 3
1-1
, CHAPTER 2
1. The crỵstal plane with Miller indices hk is a plane defined bỵ the points a1/h, a2/k, and a3 / . (a)
Two vectors that lie in the plane maỵ be taken as a1/h – a2/k and a1 /h − a3 / . But each of these vectors
gives zero as its scalar product with G = ha1 + ka2 + a3 , so that G must be perpendicular to the plane
hk . (b) If n̂ is the unit normal to the plane, the interplanar spacing is n̂ a1/h . But n̂ = G / | G | ,
whence d(hk) = G a1 / h|G| = 2 / | G| . (c) For a simple cubic lattice G = (2 / a)(hx̂ + kỵˆ + zˆ) ,
whence
G2 h + k +
2 2 2
1
= = .
d2 42 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 ỵˆ zˆ
1 1
(b) b = 2 a2 a3 4 − 3a a 0
=
1
| a a a | 2 2
1 2 3
0 0 c
2 1
= ( x̂ + ỵ̂ ), and similarlỵ 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. Bỵ definition of the primitive reciprocal lattice vectors
(a2 a 3 ) (a3 a 1 ) (a1 a 2 )
V = (2)3 = (2)3 / | (a a a ) |
2-1
