Homework #2

1. Suppose identical solid spheres are distributed through space in such a way that their
centers are lie on the points of a lattice, and spheres on neighboring points just touch
without overlapping. (Such an arrangement of spheres is called a close-packing
arrangement.) Assuming that the spheres have unit density, show that the density of a
set of close-packed spheres on each of the four structures (the "packing fraction") is:
fcc: 2π / 6 = 0.74
bcc: 3π / 8 = 0.68
sc: π / 6 = 0.52
diamond: 3π / 16 = 0.34

Solution:
For fcc structure, the nearest neighbor distance is a / 2 , thus R = a /(2 2 ) .
Here a is the lattice constant of the fcc lattice and R is the radius of the sphere. Since
there are four lattice sites per fcc cubic cell, the density should be
⎛ 4πR 3 ⎞ 3 16π ⎛ 1 ⎞ π 2π
3

⎜⎜ 4 × ⎟⎟ / a = ⎜ ⎟ = = = 0.74 .
⎝ 3 ⎠ 3 ⎝2 2⎠ 3 2 6

For bcc structure, the nearest neighbor distance is 3a / 2 , thus R = 3a / 4 .

Here a is the lattice constant of the bcc lattice and R is the radius of the sphere. Since
there are two lattice sites per bcc cubic cell, the density should be
3
⎛ 4πR 3 ⎞ 3 8π ⎛ 3 ⎞ 3π
⎜⎜ 2 × ⎟⎟ / a = ⎜ ⎟ = = 0.68 .
3 ⎠ ⎜ ⎟
⎝ 3 ⎝ 4 ⎠ 8

For sc structure, the nearest neighbor distance is a , thus R = a / 2 . Here a is the

lattice constant of the sc lattice and R is the radius of the sphere. Since there are one
lattice sites per sc cubic cell, the density should be
⎛ 4πR 3 ⎞ 3 4π ⎛ 1 ⎞ π
3

⎜⎜1× ⎟/a = ⎜ ⎟ = = 0.52 .

⎝ 3 ⎟⎠ 3 ⎝2⎠ 6

The diamond structure is two sets of fcc lattices shifted along the diagonal
direction by a quarter of the diagonal distance. The nearest neighbor distance is
3a / 4 , thus R = 3a / 8 . Here a is the lattice constant of the fcc lattice and R is the
radius of the sphere. Since there are 4+4=8 lattice sites within the fcc cubic cell for
diamond structure, the density should be
3
⎛ 4πR 3 ⎞ 3 32π ⎛ 3 ⎞ 3π
⎜⎜ 8 × ⎟⎟ / a = ⎜ ⎟ = = 0.34 .
⎜ ⎟
⎝ 3 ⎠ 3 ⎝ 8 ⎠ 16
2. (a) Prove that the ideal c/a ratio for the hexagonal close-packed structure is
8 / 3 = 1.633 .
(b) Sodium transforms from bcc to hcp at about 23K (the "martensitic"
transformation). Assuming that the density remains fixed through this transition, find
the lattice constant a of the hexagonal phase, given that a=4.23Å in the cubic phase
and that c/a ratio is ideal in the hcp phase.

Solution:

ahcp
c/2

2 3a hcp
⋅
ahcp 3 2 ahcp

(a) Let ahcp be the hcp lattice constant, i.e., the edge length of the hexagon. Then
2
c ⎛ 2 3ahcp ⎞
= ahcp
2
−⎜ ⋅ ⎟ = 2 a . Thus c = 2 2 = 8 = 1.633
2 ⎜3 2 ⎟ 3
hcp
ahcp 3 3
⎝ ⎠
3 3 2
(b) The volume of a hcp cell is ahcp c = 3 2ahcp
3
. There are 6 atoms with each
2
6w 2w
hcp cell. Thus the density is ρ = 3
= 3 . Here w is the weight one
3 2ahcp ahcp
sodium atom. For bcc sodium, there are 2 atoms within each bcc cubic cell. The
2w
density should be ρ = 3 . If the density doesn’t change during the hcp-bcc
abcc
2 w 2w a 4.233 A
transition, there is 3
= 3 . Thus a hcp = 1bcc
/6
= = 3.77 A .
ahcp abcc 2 1.122

3. Under what conditions, will the body centered tetragonal lattice become
(a) a bcc structure?
(b) an fcc structure?

Solution:
 a a
c

a a a/ 2

bct bcc fcc

(a) bcc is a special case of bct with c / a = 1 .

(b) fcc is also a special case of bct structure (see red color) with c / a = 2 .

4. Figure out the indexes of the following lattice planes. (The arrows are the basic
vectors of the lattice.)

r r r
a3 a3 r a3
r r r
r a r
a1 2 a1 a2 a1
a2

(a) (b) (c)

Solution:
⎛ 1 1 1⎞
(a). ⎜ , , ⎟ = (6,3,4 )
⎝ 2 4 3⎠
⎛1 1 1⎞
(b). ⎜ , , ⎟ = (2,4,1)
⎝2 1 4⎠
⎛1 1 1 ⎞
(c). ⎜ , , ⎟ = (1,1,0)
⎝3 3 ∞⎠

5. Make a drawing of the (110) plane of a bcc lattice. What's the distance between
adjacent (110) planes if the lattice constant is a?

Solution:
 a

bcc
The shaded is a (110) plane. The next (110) plane contains the site with red color.
Thus the distance between the adjacent (110) plane (red line) is a / 2 .

6. Make a drawing of the (111) plane of an fcc lattice. What's the distance between
adjacent (111) planes if the lattice constant is a?

Solution:

fcc
The shaded are two adjacent (111) planes. Their separation distance equals 1/3 of the
diagonal distance, i.e., equals to 3a / 3 .