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Chapter 3: Problem 10

Calculate the radius of a tantalum (Ta) atom, given that Ta has a BCC crystal structure, a density of \(16.6 \mathrm{~g} / \mathrm{cm}^{3}\), and an atomic weight of \(180.9 \mathrm{~g} / \mathrm{mol}\)

Short Answer

Expert verified
The radius of a tantalum atom is approximately \(1.665 \times 10^{-9}\, \mathrm{cm}\).

Step by step solution

01

Calculate the number of atoms per unit cell

In a body-centered cubic (BCC) crystal structure, there is one atom at each corner of the unit cell and one in the center. So, there are a total of 1 (center) + 1/8 * 8 (corners) = 2 atoms per unit cell.
02

Calculate the mass of a single Ta atom

To calculate the mass of a single Ta atom, we can use the given atomic weight and the Avogadro's number (\(N_A = 6.022 \times 10^{23} \mathrm{atoms} / \mathrm{mol}\)). The mass of a single Ta atom is: \(\frac{180.9\, \mathrm{g} / \mathrm{mol}}{6.022 \times 10^{23}\, \mathrm{atoms} / \mathrm{mol}} = 3.0065 \times 10^{-22}\, \mathrm{g} / \mathrm{atom}\)
03

Calculate the mass of Ta in the unit cell

Since there are 2 atoms per unit cell, the mass of Ta in the unit cell is: \(2 \times 3.0065 \times 10^{-22}\, \mathrm{g} / \mathrm{atom} = 6.013 \times 10^{-22}\, \mathrm{g}\)
04

Calculate the volume of the unit cell

To calculate the volume of the unit cell, we will use the given density of Ta (\(\rho = 16.6\, \mathrm{g/cm}^3\)) and the mass of Ta in the unit cell. The volume of the unit cell is: \(\frac{6.013 \times 10^{-22}\, \mathrm{g}}{16.6\, \mathrm{g/cm}^3} = 3.619 \times 10^{-24}\, \mathrm{cm}^3\)
05

Calculate the length of the unit cell

The volume of a cube is \(V=a^3\), where \(a\) is the length of its side. Using the volume of the unit cell, we can find out the length of the unit cell (a) as: \(a = \sqrt[3]{3.619 \times 10^{-24}\, \mathrm{cm}^3} = 1.532 \times 10^{-8}\, \mathrm{cm}\)
06

Calculate the radius of the Ta atom

In the BCC structure, the body diagonal has a length of \(4r\), where \(r\) is the radius of a Ta atom. The body diagonal can also be expressed as \(a\sqrt{3}\), where \(a\) is the length of the unit cell. Therefore, we can calculate the radius of a Ta atom as: \(r = \frac{a\sqrt{3}}{4} = \frac{1.532 \times 10^{-8}\, \mathrm{cm} \times \sqrt{3}}{4} = 1.665 \times 10^{-9}\, \mathrm{cm}\) So, the radius of a tantalum (Ta) atom is approximately \(1.665 \times 10^{-9}\, \mathrm{cm}\).

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Most popular questions from this chapter

Magnesium (Mg) has an HCP crystal structure and a density of \(1.74 \mathrm{~g} / \mathrm{cm}^{3}\). (a) What is the volume of its unit cell in cubic centimeters? (b) If the \(c / a\) ratio is \(1.624\), compute the values of \(c\) and \(a\).

If the atomic radius of lead is \(0.175 \mathrm{~nm}\), calculate the volume of its unit cell in cubic meters.

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: \(\begin{array}{lllll}0 & 0 & 0 & & 011 \\ 1 & 0 & 0 & & \frac{1}{2} 0 \frac{3}{4} \\ 1 & 1 & 0 & & \frac{1}{2} 1 \frac{3}{4} \\ 0 & 1 & 0 & & 1 \frac{1}{2} \frac{1}{4} \\ 0 & 0 & 1 & 0 & 1 \frac{1}{2} \frac{1}{4} \\ 1 & 0 & 1 & & \frac{1}{2} \frac{1}{2} \frac{1}{2} \\ 1 & 1 & 1 & & \end{array}\)

Sketch an orthorhombic unit cell, and within that cell indicate locations of the \(0 \frac{1}{2} 1\) and \(\frac{1}{3} \frac{1}{4}\) point coordinates.

Within a cubic unit cell, sketch the following directions: (a) [101] (e) \([\overline{1} 1 \overline{1}]\) (b) [211] (f) \([\overline{2} 12]\) (c) \([10 \overline{2}]\) (g) [3\overline{12} ] (d) \([3 \overline{13}]\) (h) [301]
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