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Chapter 3: Problem 13

Niobium has an atomic radius of 0.1430 nm and a density of \(8.57 \mathrm{g} / \mathrm{cm}^{3} .\) Determine whether it has an FCC or BCC crystal structure.

Short Answer

Expert verified
Answer: To determine whether niobium has an FCC or BCC crystal structure, follow the steps outlined in the solution to calculate the packing fraction (or packing efficiency) for both FCC and BCC crystal structures. Compare these calculations to the given atomic radius and density of niobium. The crystal structure that matches closely to the given density represents the structure of niobium.

Step by step solution

01

Calculate the volume of the FCC and BCC unit cells

For an FCC unit cell, the edge length 'a' is related to the atomic radius 'r' as: a = 2 * √2 * r For a BCC unit cell, the edge length 'a' is related to the atomic radius 'r' as: a = 4 * r / √3 Now, we can calculate the volume of the FCC unit cell (V_FCC) and the BCC unit cell (V_BCC): V_FCC = a^3 = (2 * √2 * r)^3 V_BCC = a^3 = (4 * r / √3)^3
02

Calculate the number of atoms in each unit cell

For an FCC structure, there are 4 atoms per unit cell, and for BCC, there are 2 atoms per unit cell.
03

Calculate the mass of the unit cells

To calculate the mass of a unit cell, we need to multiply the number of atoms in the unit cell by the atomic weight of niobium (92.90638 u) and divide by Avogadro's number (6.02214076 × 10^23/mol). Mass of the FCC unit cell = (4 * 92.90638) / (6.02214076 × 10^23) Mass of the BCC unit cell = (2 * 92.90638) / (6.02214076 × 10^23)
04

Calculate the density of the FCC and BCC unit cells

Next, we need to calculate the density of the FCC and BCC unit cells. Density is the mass divided by the volume. Density_FCC = Mass_FCC / V_FCC Density_BCC = Mass_BCC / V_BCC
05

Compare the calculated densities to the given density

We are given niobium's density as 8.57 g/cm³. Compare this to the densities we calculated for the FCC and BCC unit cells. The one that matches closely to the given density is the crystal structure of niobium. If Density_FCC is close to 8.57 g/cm³, then niobium has an FCC crystal structure. If Density_BCC is close to 8.57 g/cm³, then niobium has a BCC crystal structure.

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Most popular questions from this chapter

Using the Molecule Definition Utility found in both "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 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: $$\begin{array}{ll} 000 & 011 \\ 100 & \frac{1}{2} 0 \frac{3}{4} \\ 110 & \frac{1}{2} 1 \frac{3}{4} \\ 010 & 1 \frac{1}{2} \frac{1}{4} \\ 001 & 0 \frac{1}{2} \frac{1}{4} \\ 101 & \frac{1}{2} \frac{1}{2} \frac{1}{2} \\ 111 \end{array}$$

The metal niobium has a BCC crystal structure. If the angle of diffraction for the (211) set of planes occurs at \(75.99^{\circ}\) (first-order reflection) when monochromatic x-radiation having a wavelength of \(0.1659 \mathrm{nm}\) is used, compute (a) the interplanar spacing for this set of planes and (b) the atomic radius for the niobium atom.

Figure 3.21 shows an x-ray diffraction pattern for lead taken using a diffractometer and monochromatic x-radiation having a wavelength of \(0.1542 \mathrm{nm}\); each diffraction peak on the pattern has been indexed. Compute the interplanar spacing for each set of planes indexed; also determine the lattice parameter of Pb for each of the peaks.

(a) Derive planar density expressions for FCC (100) and (111) planes in terms of the atomic radius \(R.\) (b) Compute and compare planar density values for these same two planes for aluminum.

(a) Derive linear density expressions for FCC [100] and [111] directions in terms of the atomic radius \(R\) (b) Compute and compare linear density val use for these same two planes for copper.
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