ALL 22 CHAPTERS COVERED
d d d
SOLUTIONS MANUAL
d
,TABLEOFCONTENTS
d d
Chapter 1 - 1-1
d d d
Chapter 2 - 2-1
d d d
Chapter 3 - 3-1
d d d
Chapter 4 - 4-1
d d d
Chapter 5 - 5-1
d d d
Chapter 6 - 6-1
d d d
Chapter 7 - 7-1
d d d
Chapter 8 - 8-1
d d d
Chapter 9 - 9-1
d d d
Chapter 10 - 10-1
d d d
Chapter 11 - 11-1
d d d
Chapter 12 - 12-1
d d d
Chapter 13 - 13-1
d d d
Chapter 14 - 14-1
d d d
Chapter 15 - 15-1
d d d
Chapter 16 - 16-1
d d d
Chapter 17 - 17-1
d d d
Chapter 18 - 18-1
d d d
Chapter 20 - 20-1
d d d
Chapter 21 - 21-1
d d d
Chapter 22 - 22-1
d d d
,CHAPTER 1 d
1. The vectors x̂ +yˆ+zˆ and −xˆ−yˆ+zˆ are in the directions of two body diagonals of a
d d d d d d d d d d d d d d d d d d d d d d
cube. If is the angle between them, their scalar product gives cos = –1/3, whence
d d d d d d d d d d d d d d d d
= cos−11/3 = 90+1928' =10928'.
d d d d d d d d d d d d d
2. The plane (100) is normal to the x axis. It intercepts the a' axis at 2a' and the c' axis
d d d d d d d d d d d d d d d d d d d
at 2c'; therefore the indices referred to the primitive axes are (101). Similarly, the plane
d d d d d d d d d d d d d d d
(001) will have indices (011) when referred to primitive axes.
d d d d d d d d d
3. The central dot of the four is at distance
d d d d d d d d
cos60 a
a = a ctn 60= cos
d d d
d d d d d d
30 d 3
from each of the other three dots, as projected onto the basal plane. If the
d d d d d d d d d d d d d d
(unprojected) dots are at the center of spheres in contact, then
d d d d d d d d d d d
2 2
a c d d d d
a 2 = + ,
d d
d
d
d d d
2
3 d d
d
or
1 2 c 8
a2 = c2;
d d
d d
=1.633. d
3 4 a 3
1-1
, CHAPTER 2 d
1. The crystal plane with Miller indices
d d d d d d hkℓ is a plane defined by the points a1/h, a2/k, and a3 /ℓ. (a)
d d d d d d d d d d d d d d d
Two vectors that lie in the plane may be taken as a1/h – a2/k and
d d d d d d d d d d d d d d d a1 /h −a3 /ℓ. But each of these vectors gives
d d d d d d d d d d d d
d zero as its scalar product with
d d d d d d G = ha1 +ka2 +ℓa3, so that G must be perpendicular to the plane
d d d d d d d d d d d d d d d d
hkℓ. (b) If n̂ is the unit normal to the plane, the interplanar spacing is n̂a1/h. But n̂ =G/ | G |,
d d d d d d d d d d d d d d d d d d d d d d d d d d d
whence d(hkℓ) =Ga1 /h|G|= 2/ |G|. (c) For a simple cubic lattice G = (2/a )(hx̂ +kyˆ +ℓẑ ) , whence
d d d d d
d
d d d d d d d d d d d d d d d d d d d d d d d d
G2 h +k +ℓ
2 2 2
1
= =
d d d d
d
d2 42 . d
a2
1 1
3a a 0
d
2 2
1 1
2. (a) Cell volume a a a = − 3a a 0
d
d d d d d d d d d d
1 2 3
2 2
0 0 c
1
= d 3a2c. d
2
x̂ yˆ zˆ
a2 a3 1 1
(b) b = 2 = 4 − 3a a 0
d d
d
d d d
1
|a a a |
d d d d d d
2
3a c 2 2
1 2 3 d
0 0 c
2 1
= ( x̂ +ŷ ), and similarly for b 2, b3.
d
d d d
a
d d d d d d d
d
3
(c) Six vectors in the reciprocal lattice are shown as solid lines. The broken lines
d d d d d d d d d d d d d
are the perpendicular bisectors at the midpoints. The inscribed hexagon forms the
d d d d d d d d d d d d
first Brillouin Zone.
d d d
3. By definition of the primitive reciprocal lattice vectors
d d d d d d d
(a2 a 3 )(a3 a 1 )(a1 a 2 )
V = (2)3 = (2)3 / |(a a a )|
d d d d d
d d d d d d
d d d d d d d d d d d d
BZ
|(a a a ) |
d d d d
3
d
1 2 3
1 2 3
= (2)3 /V .C
d d d d
For the vector identity, see G. A. Korn and T. M. Korn, Mathematical handbook for scientists
d d d d d d d d d d d d d d d
dand engineers, McGraw-Hill, 1961, p. 147.
d d d d d
4. (a) This follows by forming
d d d d
2-1
d d d
SOLUTIONS MANUAL
d
,TABLEOFCONTENTS
d d
Chapter 1 - 1-1
d d d
Chapter 2 - 2-1
d d d
Chapter 3 - 3-1
d d d
Chapter 4 - 4-1
d d d
Chapter 5 - 5-1
d d d
Chapter 6 - 6-1
d d d
Chapter 7 - 7-1
d d d
Chapter 8 - 8-1
d d d
Chapter 9 - 9-1
d d d
Chapter 10 - 10-1
d d d
Chapter 11 - 11-1
d d d
Chapter 12 - 12-1
d d d
Chapter 13 - 13-1
d d d
Chapter 14 - 14-1
d d d
Chapter 15 - 15-1
d d d
Chapter 16 - 16-1
d d d
Chapter 17 - 17-1
d d d
Chapter 18 - 18-1
d d d
Chapter 20 - 20-1
d d d
Chapter 21 - 21-1
d d d
Chapter 22 - 22-1
d d d
,CHAPTER 1 d
1. The vectors x̂ +yˆ+zˆ and −xˆ−yˆ+zˆ are in the directions of two body diagonals of a
d d d d d d d d d d d d d d d d d d d d d d
cube. If is the angle between them, their scalar product gives cos = –1/3, whence
d d d d d d d d d d d d d d d d
= cos−11/3 = 90+1928' =10928'.
d d d d d d d d d d d d d
2. The plane (100) is normal to the x axis. It intercepts the a' axis at 2a' and the c' axis
d d d d d d d d d d d d d d d d d d d
at 2c'; therefore the indices referred to the primitive axes are (101). Similarly, the plane
d d d d d d d d d d d d d d d
(001) will have indices (011) when referred to primitive axes.
d d d d d d d d d
3. The central dot of the four is at distance
d d d d d d d d
cos60 a
a = a ctn 60= cos
d d d
d d d d d d
30 d 3
from each of the other three dots, as projected onto the basal plane. If the
d d d d d d d d d d d d d d
(unprojected) dots are at the center of spheres in contact, then
d d d d d d d d d d d
2 2
a c d d d d
a 2 = + ,
d d
d
d
d d d
2
3 d d
d
or
1 2 c 8
a2 = c2;
d d
d d
=1.633. d
3 4 a 3
1-1
, CHAPTER 2 d
1. The crystal plane with Miller indices
d d d d d d hkℓ is a plane defined by the points a1/h, a2/k, and a3 /ℓ. (a)
d d d d d d d d d d d d d d d
Two vectors that lie in the plane may be taken as a1/h – a2/k and
d d d d d d d d d d d d d d d a1 /h −a3 /ℓ. But each of these vectors gives
d d d d d d d d d d d d
d zero as its scalar product with
d d d d d d G = ha1 +ka2 +ℓa3, so that G must be perpendicular to the plane
d d d d d d d d d d d d d d d d
hkℓ. (b) If n̂ is the unit normal to the plane, the interplanar spacing is n̂a1/h. But n̂ =G/ | G |,
d d d d d d d d d d d d d d d d d d d d d d d d d d d
whence d(hkℓ) =Ga1 /h|G|= 2/ |G|. (c) For a simple cubic lattice G = (2/a )(hx̂ +kyˆ +ℓẑ ) , whence
d d d d d
d
d d d d d d d d d d d d d d d d d d d d d d d d
G2 h +k +ℓ
2 2 2
1
= =
d d d d
d
d2 42 . d
a2
1 1
3a a 0
d
2 2
1 1
2. (a) Cell volume a a a = − 3a a 0
d
d d d d d d d d d d
1 2 3
2 2
0 0 c
1
= d 3a2c. d
2
x̂ yˆ zˆ
a2 a3 1 1
(b) b = 2 = 4 − 3a a 0
d d
d
d d d
1
|a a a |
d d d d d d
2
3a c 2 2
1 2 3 d
0 0 c
2 1
= ( x̂ +ŷ ), and similarly for b 2, b3.
d
d d d
a
d d d d d d d
d
3
(c) Six vectors in the reciprocal lattice are shown as solid lines. The broken lines
d d d d d d d d d d d d d
are the perpendicular bisectors at the midpoints. The inscribed hexagon forms the
d d d d d d d d d d d d
first Brillouin Zone.
d d d
3. By definition of the primitive reciprocal lattice vectors
d d d d d d d
(a2 a 3 )(a3 a 1 )(a1 a 2 )
V = (2)3 = (2)3 / |(a a a )|
d d d d d
d d d d d d
d d d d d d d d d d d d
BZ
|(a a a ) |
d d d d
3
d
1 2 3
1 2 3
= (2)3 /V .C
d d d d
For the vector identity, see G. A. Korn and T. M. Korn, Mathematical handbook for scientists
d d d d d d d d d d d d d d d
dand engineers, McGraw-Hill, 1961, p. 147.
d d d d d
4. (a) This follows by forming
d d d d
2-1
