# 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. Consider the tetrahedron labeled as JKLM, which is reconstructed as 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, where R is the atomic radius. Furthermore, from triangle JHM, (JM = OH + (MH or

Now, we can determine the JH length by consideration of triangle JKL, which is an equilateral triangle, 3 cos 300 = 2, and Substituting this value for JH in the above expression yields 2 c2 o a 02 30 and, solving for c/a 8 2 1. 633 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 04TIR30 = 60 3 = 8TIR3 base area can be calculated as follows.

The following fgure shows an HCP unit cell and the basal plane. The base area is equal to six times the area of the equilateral triangle, OAB. c o The area of equilateral triangle, OAB = h x AB x AO 32 4 = 0. 5 x op Thus, the area of the basal plane 32332 Further, as can be seen from the figure of the basal plane, a = 2R. Therefore, the base area = 6R2 3 Now, using the result of Problem 1, given above, the relation between the unit cell height, c, and the basal plane edge length, a, is given as: c = 1. 63 Thus, c = 1. 63a = 3. 26R The unit cell volume can now be calculated as:

VC=C x base area = 3. 26R x 10. 392R 2 = 33. 878R 3 Thus, APE = = 0. 74 VC 33. 878R 3 3. Rhenium has an HCP crystal structure, an atomic radius of 0. 137 nm, and a c/a ratio of 1. 615. Compute the volume of the unit cell for Re. This problem asks that we calculate the unit cell volume, VC, for Re which has an HCP crystal structure. Now, VC = c x base area, and the base area has been calculated in Problem 2 above as 6R2 3 . Thus, VC=C x 6R23 In this problem, c = 1. 61 5a, and a = 2R. Therefore VC=1. 615X2X63XR3 = 1. 615 x 2 x 63 x (1. 37 x 10-8 cm)3 = 8. 63 x 10-23 crn3 = 8. 63 x 10-2 nrn3