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Chapter 3: Problem 24

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

Short Answer

Expert verified
Answer: In a face-centered orthorhombic crystal structure unit cell, there are four atoms, and the coordination number is 12.

Step by step solution

01

Draw the outlines of the orthorhombic unit cell

To start, draw a rectangular box representing the orthorhombic unit cell. Be sure to show the different lengths a, b, and c for the sides of the unit cell. All angles should be 90掳.
02

Place atoms at the corners

Add an atom at each of the eight corners of the unit cell (cube). Label these atoms as A, B, C, D, E, F, G, and H. Point out that these atoms will be shared with neighboring unit cells.
03

Place atoms at the center of each face

Add an atom at the center of each of the six faces of the unit cell (rectangular box). Label these atoms as I, J, K, L, M, and N. Point out that these atoms will also be shared with neighboring unit cells.
04

Determine coordination number

Identify the coordination number for this face-centered orthorhombic crystal structure, which is the number of neighboring atoms each atom has. In this case, the coordination number is 12.
05

Calculate the number of atoms per unit cell

Calculate the number of atoms per unit cell by considering how much of each atom is within the unit cell. The atoms at the corners are each inside \(\frac{1}{8}\) of the unit cell, and the atoms at the center of each face are each inside \(\frac{1}{2}\) of the unit cell. So the total atoms per unit cell would be: \(8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4\). There are four atoms per unit cell in the face-centered orthorhombic crystal structure. With these steps, you will have successfully sketched a unit cell for the face-centered orthorhombic crystal structure.

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Most popular questions from this chapter

Convert the \([110]\) and \([00 \overline{1}]\) directions into the four-index Miller-Bravais scheme for hexagonal unit cells.

Show that the atomic packing factor for HCP is \(0.74\).

Sketch a tetragonal unit cell, and within that cell indicate locations of the \(1 \frac{1}{2} \frac{1}{2}\) and \(\frac{1}{2} \frac{1}{4} \frac{1}{2}\) point coordinates.

Using the Molecule Definition Utility found in the "Metallic Crystal Structures and Crystallography" and "Ceramic Crystal Structures" modules of \(V M S E\) located on the book's web site [www.wiley.com/ college/callister (Student Companion Site)], generate (and print out) a three-dimensional unit cell for \(\beta\) tin (Sn), given the following: (1) the unit cell is tetragonal with \(a=0.583 \mathrm{~nm}\) and \(c=0.318 \mathrm{~nm}\), and (2) \(\mathrm{Sn}\) atoms are located at the following point coordinates: \(\begin{array}{lllll}0 & 0 & 0 & & 011 \\ 1 & 0 & 0 & & \frac{1}{2} 0 \frac{3}{4} \\ 1 & 1 & 0 & & \frac{1}{2} 1 \frac{3}{4} \\ 0 & 1 & 0 & & 1 \frac{1}{2} \frac{1}{4} \\ 0 & 0 & 1 & 0 & 1 \frac{1}{2} \frac{1}{4} \\ 1 & 0 & 1 & & \frac{1}{2} \frac{1}{2} \frac{1}{2} \\ 1 & 1 & 1 & & \end{array}\)

(a) What are the direction indices for a vector that passes from point \(\frac{1}{3} \frac{1}{2} 0\) to point \(\frac{2}{3} \frac{3}{4} \frac{1}{2}\) in a tetragonal unit cell? (b) Repeat part (a) for a rhombohedral unit cell.
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