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Chapter 3: Problem 7

Molybdenum has a BCC crystal structure, an atomic radius of \(0.1363 \mathrm{nm},\) and an atomic weight of \(95.94 \mathrm{g} / \mathrm{mol}\). Compute and compare its theoretical density with the experimental value found inside the front cover.

Short Answer

Expert verified
Answer: The calculated theoretical density of molybdenum in a BCC crystal structure is 9.96 g/cm³. This is close to the experimental density value of 10.2 g/cm³ provided in the book.

Step by step solution

01

Determine the number of atoms in the BCC unit cell

For the BCC crystal structure, there are two atoms per unit cell. We will use this information while calculating the mass for the density formula.
02

Compute the lattice constant (a)

In a BCC structure, the lattice constant 'a' is related to the atomic radius 'R' by the equation: a = \(4R\sqrt{3}/\sqrt{2}\) Given the atomic radius of molybdenum as \(0.1363\ \text{nm}\), we can find the lattice constant as follows: a = \(\frac{4(0.1363)\sqrt{3}}{\sqrt{2}} =0.2893\ \text{nm}\)
03

Compute the volume of the BCC unit cell

The volume of the unit cell can be calculated as: Volume = \(a^3 = (0.2893\ \text{nm})^3 = 2.422 \times 10^{-2}\ \text{nm}^3\)
04

Calculate the mass of the BCC unit cell

The molar mass of molybdenum is \(95.94 \ \frac{\text{g}}{\text{mol}}\). We know that there are two atoms per unit cell in a BCC structure, so the mass of the BCC unit cell can be obtained using Avogadro's number (\(6.022 \times 10^{23}\ \text{mol}^{-1}\)): Mass = \(\frac{2 \times 95.94 \text{g/mol}}{6.022 \times 10^{23}\ \text{atoms/mol}} = 3.1868 \times 10^{-22}\ \text{g}\)
05

Calculate the theoretical density

Now that we have both the mass and volume of the BCC unit cell, we can calculate the theoretical density: Density = \(\frac{Mass}{Volume} = \frac{3.1868 \times 10^{-22}\ \text{g}}{2.422 \times 10^{-2}\ \text{nm}^3} = 9.96\ \frac{\text{g}}{\text{cm}^{-3}}\)
06

Compare the theoretical density with the experimental value

The front cover of the book provides the experimental density value of molybdenum as \(10.2\ \frac{\text{g}}{\text{cm}^{-3}}\). Comparing our computed theoretical density, we find that our calculated theoretical density (9.96 g/cm³) is close to the experimental value (10.2 g/cm³).

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Most popular questions from this chapter

Sketch within a cubic unit cell the following planes: (a) \((10 \overline{1})\) (b) \((2 \overline{1} 1)\) (c) (012) (d) \((3 \overline{1} 3)\) (e) \((\overline{1} 1 \overline{1})\) (f) \((\overline{2} 12)\) (g) \((3 \overline{1} 2)\) (h) (301)

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: $$\begin{array}{ll} 000 & 011 \\ 100 & \frac{1}{2} 0 \frac{3}{4} \\ 110 & \frac{1}{2} 1 \frac{3}{4} \\ 010 & 1 \frac{1}{2} \frac{1}{4} \\ 001 & 0 \frac{1}{2} \frac{1}{4} \\ 101 & \frac{1}{2} \frac{1}{2} \frac{1}{2} \\ 111 \end{array}$$

Cite the indices of the direction that results from the intersection of each of the following pair of planes within a cubic crystal: (a) (110) and (111) planes, (b) (110) and \((1 \overline{1} 0)\) planes and (c) \((11 \overline{1})\) and (001) planes.

(a) Derive the planar density expression for the HCP (0001) plane in terms of the atomic radius \(R.\) (b) Compute the planar density value for this same plane for titanium.

Titanium has an HCP crystal structure and a density of \(4.51 \mathrm{g} / \mathrm{cm}^{3}\) (a) What is the volume of its unit cell in cubic meters? (b) If the \(c / a\) ratio is 1.58 , compute the values of \(c\) and \(a.\)
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