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Chapter 9: Problem 50

Nickel has an atomic radius of \(0.162 \mathrm{~nm}\). The edge of its cubic unit cell is \(0.458 \mathrm{~nm}\). What is the geometry of the nickel unit cell?

Short Answer

Expert verified
Answer: The geometry of the Nickel unit cell is face-centered cubic (FCC).

Step by step solution

01

Understand the types of cubic unit cells

There are three types of cubic unit cells: 1. Simple cubic (SC) - atoms are present only at the corners of the cube. 2. Body-centered cubic (BCC) - atoms are present at the corners and the center of the cube. 3. Face-centered cubic (FCC) - atoms are present at the corners and the centers of each face of the cube.
02

Compute the number of atoms in each unit cell type

Let's calculate the number of atoms per unit cell for each type: 1. Simple cubic (SC): 1 atom per unit cell (\\(\frac{1}{8}\\) atom per corner × 8 corners = 1) 2. Body-centered cubic (BCC): 2 atoms per unit cell (\\(\frac{1}{8}\\) atom per corner × 8 corners + 1 atom in the center = 2) 3. Face-centered cubic (FCC): 4 atoms per unit cell (\\(\frac{1}{8}\\) atom per corner × 8 corners + \\(\frac{1}{2}\\) atom per face × 6 faces = 4)
03

Compute the atomic packing factors for each unit cell type

The atomic packing factor (APF) represents the fraction of the unit cell volume occupied by the atoms. We'll compute the APF for each type of unit cell, using the given atomic radius r and edge length a: 1. Simple cubic (SC): APF = \\(\\\frac{1 \cdot \frac{4}{3}\pi r^3}{a^3}\\\) 2. Body-centered cubic (BCC): APF = \\(\frac{2 \cdot \frac{4}{3}\pi r^3}{a^3}\\\) 3. Face-centered cubic (FCC): APF = \\(\frac{4 \cdot \frac{4}{3}\pi r^3}{a^3}\\\)
04

Compute the APF for each unit cell type using the given r and a

Atomic radius of Nickel: r = \(0.162 \mathrm{~nm}\). Edge length of the cubic unit cell: a = \(0.458 \mathrm{~nm}\). 1. Simple cubic (SC): APF = \\(\frac{1 \cdot \frac{4}{3}\pi (0.162)^3}{(0.458)^3} \approx 0.26\\\) 2. Body-centered cubic (BCC): APF = \\(\frac{2 \cdot \frac{4}{3}\pi (0.162)^3}{(0.458)^3} \approx 0.52\\\) 3. Face-centered cubic (FCC): APF = \\(\frac{4 \cdot \frac{4}{3}\pi (0.162)^3}{(0.458)^3} \approx 1.04\\\)
05

Determine the geometry of the Nickel unit cell

Given the computed APFs, we can see that the acceptable geometry for the Nickel unit cell is face-centered cubic (FCC) since its APF is closest to 1 (meaning it's the most closely packed). The other two unit cells - SC and BCC - have lower APFs, which indicates that they are not as closely packed as the FCC unit cell. Therefore, the geometry of the Nickel unit cell is face-centered cubic (FCC).

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Most popular questions from this chapter

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