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Chapter 3: Problem 4

For the HCP crystal structure, show that the ideal \(c / a\) ratio is \(1.633\).

Short Answer

Expert verified
Answer: The ideal c/a ratio for a Hexagonal Close Packed (HCP) crystal structure is 1.633. It is determined using the geometry of the unit cell and trigonometry, primarily focusing on the relationship between the edge lengths of the unit cell and the height of the tetrahedron that forms the HCP structure.

Step by step solution

01

Understand the Hexagonal Close Packed Structure

The HCP crystal structure is formed by stacking layers of atoms in a close-packed arrangement. The atoms in one layer are situated on top of the triangular voids in the next layer, resulting in a repeating ABABAB... pattern. The unit cell of the HCP structure has edges a, b, and c, with angles alpha=90°, beta=90°, and gamma=120°. The c-axis is perpendicular to the hexagonal layers, thus the c/a ratio will represent the relative height of the HCP unit cell.
02

Tetrahedron in HCP Structure

Considering the HCP unit cell, we can focus on one tetrahedron formed by the layers A and B, where one vertex is at the center of the unit cell and the other three vertices are at three adjacent atoms in one of the layers. We will denote the side length of this tetrahedron as 'd' and the height of the tetrahedron from the vertex at the center of the unit cell as 'h'.
03

Calculate 'd' using edge length 'a'

The side length 'd' is equal to the distance between two adjacent atoms in the same hexagonal layer. This is equal to the edge length 'a' of the unit cell. Therefore, we have \(d = a\).
04

Calculate the height 'h' of the tetrahedron

From the tetrahedron, we can form a right triangle with base 'd/2' and height 'h', where the hypotenuse is the distance between the vertex at the center of the unit cell and the midpoint of the opposite edge. As we found in Step 3, 'd' is equal to 'a', so we can calculate 'h' using the Pythagorean theorem and the fact that we have \(120^{\circ}\) angles in the hexagonal layer: \(h^2 + (\dfrac{d}{2})^2 = d^2\) \(h^2 + (\dfrac{a}{2})^2 = a^2\) \(h^2 = a^2 - (\dfrac{a^2}{4})\) \(h^2 = \dfrac{3}{4}a^2\) \(h = \dfrac{\sqrt{3}}{2}a\)
05

Relate 'h' to 'c'

Now, we need to relate the height of the tetrahedron to the height of the unit cell (c). In HCP structure, 3 layers A, B and A align on top of each other, hence the height 'c' of the HCP structure will be 2 times the height of the tetrahedron 'h'. \(c = 2h\) \(c = 2 * \dfrac{\sqrt{3}}{2}a\) \(c = \sqrt{3}a\)
06

Calculate the ideal c/a ratio

Finally, to get the ideal c/a ratio, we will divide the value of 'c' derived in Step 5 by 'a': \(c/a = \dfrac{\sqrt{3}a}{a}\) \(c/a = \sqrt{3} \approx 1.633\) Thus, the ideal c/a ratio for a Hexagonal Close Packed (HCP) crystal structure is 1.633.

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Most popular questions from this chapter

(a) Derive linear density expressions for BCC [110] and [111] directions in terms of the atomic radius \(R\). (b) Compute and compare linear density values for these same two directions for iron \((\mathrm{Fe})\).

Sketch a unit cell for the face-centered orthorhombic crystal structure.

(a) Derive planar density expressions for \(\mathrm{BCC}\) (100) and (110) planes in terms of the atomic radius \(R\). (b) Compute and compare planar density values for these same two planes for molybdenum (Mo).

Sketch the atomic packing of the following: (a) The (100) plane for the FCC crystal structure (b) The (111) plane for the BCC crystal structure (similar to Figures \(3.12 b\) and \(3.13 b\) ).

Cite the indices of the direction that results from the intersection of each of the following pairs of planes within a cubic crystal: (a) The (110) and (111) planes (b) The (110) and (1\overline{10) planes } (c) The \((11 \overline{1})\) and \((001)\) planes
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