the proof of h/c = 1.633 for hcp
MATSE 259
Solutions to homework #2
1. For the HCP crystal structure, show that the ideal c/a ratio is 1.633.
We are asked to show that the ideal c/a ratio for HCP is 1.633. A sketch of one third of an HCP unit cell is shown below. a c
M
L
J
K
a
Consider the tetrahedron labeled as JKLM, which is reconstructed as
M
L
H
J
K
The atom at point M is midway between the top and bottom faces of the unit cell
__
that is MH = c/2. And, since atoms at points J, K, and M, all touch one another,
__
__
JM = JK = 2R = a where R is the atomic radius. Furthermore, from triangle JHM,
1
__
__
__
(JM )2 = (JH )2 + (MH )2, or
__
⎛c⎞ 2 a2 = (JH )2 + ⎜2⎟
⎝ ⎠
__
Now, we can determine the JH length by consideration of triangle JKL, which is an equilateral triangle,
L
J
30
H
a/2
K
a/2
3
cos 30° = JH = 2 , and
__
a
JH =
3
__
Substituting this value for JH in the above expression yields
2 c2
⎛ a ⎞ 2 ⎛c⎞ 2 a a2 = ⎜ ⎟ + ⎜2⎟ = 3 + 4
⎝ ⎠
⎝ 3⎠ and, solving for c/a c a =
8
3 = 1.633
2
2. Show that the atomic packing factor for HCP is 0.74.
This problem calls for a demonstration that the APF for HCP is 0.74. Again, the
APF is the ratio of the total sphere volume, VS, to the unit cell volume, VC. For
HCP, there are the equivalent of six spheres per unit cell, and thus
⎛4πR3⎞
VS = 6⎜ 3 ⎟ = 8πR3
⎝
⎠
Now, the unit cell volume is the product of the base area times the cell height, c.
The base area can be calculated as follows. The following figure shows an HCP unit cell and the basal plane. The base area is equal to six times the area of the equilateral triangle, OAB.
a
A
B
60° a c
a
F
C
O
E
a
A
P
B
60°
O
The area of equilateral triangle, OAB = 0.5 × AB × OP
= ½ × AB × AO Sin60°
3 2
= ½ × a × a Sin60° = a 4
3
D
Thus, the area of the basal plane = 6 ×
3 2 3 3 2 a =
a.
4
2
Further, as can be seen from the figure of the
Solutions to homework #2
1. For the HCP crystal structure, show that the ideal c/a ratio is 1.633.
We are asked to show that the ideal c/a ratio for HCP is 1.633. A sketch of one third of an HCP unit cell is shown below. a c
M
L
J
K
a
Consider the tetrahedron labeled as JKLM, which is reconstructed as
M
L
H
J
K
The atom at point M is midway between the top and bottom faces of the unit cell
__
that is MH = c/2. And, since atoms at points J, K, and M, all touch one another,
__
__
JM = JK = 2R = a where R is the atomic radius. Furthermore, from triangle JHM,
1
__
__
__
(JM )2 = (JH )2 + (MH )2, or
__
⎛c⎞ 2 a2 = (JH )2 + ⎜2⎟
⎝ ⎠
__
Now, we can determine the JH length by consideration of triangle JKL, which is an equilateral triangle,
L
J
30
H
a/2
K
a/2
3
cos 30° = JH = 2 , and
__
a
JH =
3
__
Substituting this value for JH in the above expression yields
2 c2
⎛ a ⎞ 2 ⎛c⎞ 2 a a2 = ⎜ ⎟ + ⎜2⎟ = 3 + 4
⎝ ⎠
⎝ 3⎠ and, solving for c/a c a =
8
3 = 1.633
2
2. Show that the atomic packing factor for HCP is 0.74.
This problem calls for a demonstration that the APF for HCP is 0.74. Again, the
APF is the ratio of the total sphere volume, VS, to the unit cell volume, VC. For
HCP, there are the equivalent of six spheres per unit cell, and thus
⎛4πR3⎞
VS = 6⎜ 3 ⎟ = 8πR3
⎝
⎠
Now, the unit cell volume is the product of the base area times the cell height, c.
The base area can be calculated as follows. The following figure shows an HCP unit cell and the basal plane. The base area is equal to six times the area of the equilateral triangle, OAB.
a
A
B
60° a c
a
F
C
O
E
a
A
P
B
60°
O
The area of equilateral triangle, OAB = 0.5 × AB × OP
= ½ × AB × AO Sin60°
3 2
= ½ × a × a Sin60° = a 4
3
D
Thus, the area of the basal plane = 6 ×
3 2 3 3 2 a =
a.
4
2
Further, as can be seen from the figure of the