CHAPTER 1
1. The vectors xˆ + yˆ + zˆ and − xˆ − yˆ + zˆ are in the directions of two body diagonals of a
cube. If θ is the angle between them, their scalar product gives cos θ = –1/3, whence
θ = cos −1 1/ 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 2
⎛ a ⎞ ⎛c⎞
a =⎜
2
⎟ +⎜ ⎟ ,
⎝ 3 ⎠ ⎝2⎠
or
2 2 1 2 c 8
a = c ; = 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 + ka 2 + 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 nˆ ⋅ a1/h . But nˆ = G / | G | ,
whence d(hkA) = G ⋅ a1 / h|G| = 2π / | G| . (c) For a simple cubic lattice G = (2π / a)(hxˆ + kyˆ + Azˆ ) ,
whence
1 G 2 h 2 + k 2 + A2
= = .
d 2 4π 2 a2
1 1
3a a 0
2 2
1 1
2. (a) Cell volume a1 ⋅ a 2 × a3 = − 3a a 0
2 2
0 0 c
1
= 3 a 2 c.
2
xˆ yˆ zˆ
a 2 × a3 4π 1 1
(b) b1 = 2π = − 3a a 0
| a1 ⋅ a 2 × a3 | 2
3a c 2 2
0 0 c
2π 1
= ( xˆ + yˆ ), 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
(a 2 × a 3 ) ⋅ (a 3 × a1 ) × (a1 × a 2 )
VBZ = (2π)3 = (2π)3 / | (a1 ⋅ a 2 × a 3 ) |
| (a1 ⋅ a 2 × a 3 ) |
3
= (2π)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
, 1 − exp[−iM(a ⋅ ∆k)] 1 − exp[iM(a ⋅ ∆k)]
|F|2 = ⋅
1 − exp[−i(a ⋅ ∆k)] 1 − exp[i(a ⋅ ∆k)]
1 − cos M(a ⋅ ∆k) sin 12 M(a ⋅ ∆k)
2
= = .
1 − cos(a ⋅ ∆k) sin 2 12 (a ⋅ ∆k)
1
(b) The first zero in sin Mε occurs for ε = 2π/M. That this is the correct consideration follows from
2
1 1 1
sin M(πh + ε) = sin πMh cos Mε + cos
πMh sin Mε.
2 2 2
zero, ± 1
as Mh is
an integer
−2πi(x j v1 +y j v 2 +z j v3 )
5. S (v1 v 2 v 3 ) = f Σ e
j
1 1 1
Referred to an fcc lattice, the basis of diamond is 000; . Thus in the product
4 4 4
S(v1v 2 v3 ) = S(fcc lattice) × S (basis) ,
we take the lattice structure factor from (48), and for the basis
1
−i π (v1 + v 2 + v3 ).
S (basis) = 1 + e 2
Now S(fcc) = 0 only if all indices are even or all indices are odd. If all indices are even the structure factor
of the basis vanishes unless v1 + v2 + v3 = 4n, where n is an integer. For example, for the reflection (222)
we have S(basis) = 1 + e–i3π = 0, and this reflection is forbidden.
∞
f G = ∫ 4πr 2 (πa 0 Gr)−1 sin Gr exp ( −2r a 0 ) dr
3
6.
0
= (4 G 3a 0 ) ∫ dx x sin x exp ( −2x Ga 0 )
3
= (4 G 3a 0 ) (4 Ga 0 ) (1 + r G 2 a 0 ) 2
3 2
16 (4 + G 2 a 0 ) 2 .
2
The integral is not difficult; it is given as Dwight 860.81. Observe that f = 1 for G = 0 and f ∝ 1/G4 for
Ga 0 >> 1.
1
7. (a) The basis has one atom A at the origin and one atom B at a. The single Laue equation
2
a ⋅ ∆k = 2π × (integer) defines a set of parallel planes in Fourier space. Intersections with a sphere are
a set of circles, so that the diffracted beams lie on a set of cones. (b) S(n) = fA + fB e–iπn. For n odd, S = fA –
2-2
1. The vectors xˆ + yˆ + zˆ and − xˆ − yˆ + zˆ are in the directions of two body diagonals of a
cube. If θ is the angle between them, their scalar product gives cos θ = –1/3, whence
θ = cos −1 1/ 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 2
⎛ a ⎞ ⎛c⎞
a =⎜
2
⎟ +⎜ ⎟ ,
⎝ 3 ⎠ ⎝2⎠
or
2 2 1 2 c 8
a = c ; = 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 + ka 2 + 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 nˆ ⋅ a1/h . But nˆ = G / | G | ,
whence d(hkA) = G ⋅ a1 / h|G| = 2π / | G| . (c) For a simple cubic lattice G = (2π / a)(hxˆ + kyˆ + Azˆ ) ,
whence
1 G 2 h 2 + k 2 + A2
= = .
d 2 4π 2 a2
1 1
3a a 0
2 2
1 1
2. (a) Cell volume a1 ⋅ a 2 × a3 = − 3a a 0
2 2
0 0 c
1
= 3 a 2 c.
2
xˆ yˆ zˆ
a 2 × a3 4π 1 1
(b) b1 = 2π = − 3a a 0
| a1 ⋅ a 2 × a3 | 2
3a c 2 2
0 0 c
2π 1
= ( xˆ + yˆ ), 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
(a 2 × a 3 ) ⋅ (a 3 × a1 ) × (a1 × a 2 )
VBZ = (2π)3 = (2π)3 / | (a1 ⋅ a 2 × a 3 ) |
| (a1 ⋅ a 2 × a 3 ) |
3
= (2π)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
, 1 − exp[−iM(a ⋅ ∆k)] 1 − exp[iM(a ⋅ ∆k)]
|F|2 = ⋅
1 − exp[−i(a ⋅ ∆k)] 1 − exp[i(a ⋅ ∆k)]
1 − cos M(a ⋅ ∆k) sin 12 M(a ⋅ ∆k)
2
= = .
1 − cos(a ⋅ ∆k) sin 2 12 (a ⋅ ∆k)
1
(b) The first zero in sin Mε occurs for ε = 2π/M. That this is the correct consideration follows from
2
1 1 1
sin M(πh + ε) = sin πMh cos Mε + cos
πMh sin Mε.
2 2 2
zero, ± 1
as Mh is
an integer
−2πi(x j v1 +y j v 2 +z j v3 )
5. S (v1 v 2 v 3 ) = f Σ e
j
1 1 1
Referred to an fcc lattice, the basis of diamond is 000; . Thus in the product
4 4 4
S(v1v 2 v3 ) = S(fcc lattice) × S (basis) ,
we take the lattice structure factor from (48), and for the basis
1
−i π (v1 + v 2 + v3 ).
S (basis) = 1 + e 2
Now S(fcc) = 0 only if all indices are even or all indices are odd. If all indices are even the structure factor
of the basis vanishes unless v1 + v2 + v3 = 4n, where n is an integer. For example, for the reflection (222)
we have S(basis) = 1 + e–i3π = 0, and this reflection is forbidden.
∞
f G = ∫ 4πr 2 (πa 0 Gr)−1 sin Gr exp ( −2r a 0 ) dr
3
6.
0
= (4 G 3a 0 ) ∫ dx x sin x exp ( −2x Ga 0 )
3
= (4 G 3a 0 ) (4 Ga 0 ) (1 + r G 2 a 0 ) 2
3 2
16 (4 + G 2 a 0 ) 2 .
2
The integral is not difficult; it is given as Dwight 860.81. Observe that f = 1 for G = 0 and f ∝ 1/G4 for
Ga 0 >> 1.
1
7. (a) The basis has one atom A at the origin and one atom B at a. The single Laue equation
2
a ⋅ ∆k = 2π × (integer) defines a set of parallel planes in Fourier space. Intersections with a sphere are
a set of circles, so that the diffracted beams lie on a set of cones. (b) S(n) = fA + fB e–iπn. For n odd, S = fA –
2-2
