Question

Problem 1. Consider a crystal with HCP structure, having an ideal c/a ratio. Complete parts (a)-(c) below.

(a) (1.0 pt max) Demonstrate that the ideal c/a ratio equals 2 6/3.

(b)(0.5 pt max) Find the volume of one unit cell for a given a value.

(c) (0.5 pt max) Demonstrate that the atomic packing factor for this crystal structure is  2/6.

Hints: (a) The c/a ratio is considered ideal when the distance from some atom to its nearest neighbors within the same hexagonally-packed layer equals the distance to its nearest neighbors belonging to a different layer.

(b) The volume of a prism is the product of the area of its base face and the height. Use the c/a ratio found in part (a).

(c) Find the relationship between the lattice constant a and the atomic radius R to calculate the APF value. Use the unit cell volume found in part (b)

Answer 1

(a).   The ideal axial ratio (c/a) for a hexagonal close-packed crystal structure can be calculated by considering non-interacting identical hard spheres packed together in the h.c.p. crystal structure.

If the sphere radius is r, then the lattice parameters a (= b) and c can be written in terms of r:

These two relationships can be solved for the ideal axial ratio c/a:

a² = a²/3 + c²/4

4 = 4/3 + c²/a²

c/a = 1.633

(b). From c/a ratio -

c = 1.633 a

Volume of a prism = a² * c

= ​ a² * 1.633 a

= 0.707 a³

volume of one unit cell = 6 * volume of 1 prism.

= 6 * 0.707 a³

= 4.242 a³

(c). In hcp structure -

Number of atoms = 6, and

a = b = 2r

APF = N * V_(atom) / V_{(unit cell)}

= 6 * 4 /3 * r³ /  4.242 a³

= 8 * * (a / 2)³ / 4.242 a³

= / 4.242

= 0.74