Question

In a hexagonal close-packed (HCP) material, let a represent the interatomic spacing within a close-packed layer (plane), and let c represent the lattice parameter normal (perpendicular) to these layers. There are two close-packed layers stacked along c in the conventional unit cell, as described in Lecture 3. (i) Find an expression for a in terms of the atomic radius R. (ii) Now imagine the same close-packed layers, made from the same atoms as before (i.e. R does not change), but now stacked to form a cubic close packed (CCP) material. There are three close-packed layers stacked along the body diagonal of the conventional face centered cubic (FCC) unit cell of CCP material. Find an expression that relates the atomic radius R to the close-packed layer spacing in this unit cell. (Hint: both R and the close-packed layer spacing can be written in terms of the length of any edge of the FCC unit cell.) (iii) Use your answer from (ii) to obtain an expression that relates c to R in the HCP structure. (iv) Use your answers from (i) and (iii) to show that c/a ~ 1.633 in perfect HCP materials. (v) Many HCP materials do not have this ideal value of c/a. What is the practical significance / consequence of departures from the ideal c/a value?

Answer 1

SOLUTION:-

i) We can find expression by-

In the cubic system all the faces of the cube are equivalent,

that is, they have similar indices.

However, this is not the case in the hexagonal system. The six prism faces for example have indices (1 0 0), (0 1 0),(1 1 0 ), (1 0 0 ), (0 1 0), (1 1 0 ), which are not same.

In order to address this, a fourth axis (a3) which is opposite to the vector sum of a1and a2 is used and a corresponding fourth index i is used along with hkl. Therefore the indices of a plane is given by (hkil) where i = -(h+k). Sometime i is replaced with a dot and written as (h k . l).

The spacing between planes in a crystal is known as interplanar spacing and is denoted as dhkl. In the cubic system spacing between the (hkl) planes is given as

1/d²= (h²+k²+l²)/a²

ii) 1/R² = (4/a²) (h²+k²+l²)

iii) For Hexagonal System in terms of R, we get -

1/R² = (16/3a²) (h²+hk+k²) +(l²/C²)

iv) solution is given below:- For c/a =1.633 in HCP

proof:-

Let consider,

A sketch of one third of an HCP unit cell is shown below.

Consider the tetrahedron labeled as JKLM, which is reconstructed

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,

JM²= JH²+ MH², or

a²= JH² + (c/2)²

Now, we can determine the JH

length by consideration of triangle JKL, which is an equilateral triangle,

Substituting this value for JH in the above expression yields

c/a = (8/3)

so that we have

c/a = 1.633

v) Slip should depend on geometrical conditions especially because they are related to bondings. However, in case of that HCP Materials you have two different atoms so that your anisotropic properties will change the c/a ratio.