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Chapter 10: Problem 62

The radius of tungsten is 137 \(\mathrm{pm}\) and the density is 19.3 \(\mathrm{g} / \mathrm{cm}^{3}\) . Does elemental tungsten have a face-centered cubic structure or a body- centered cubic structure?

Short Answer

Expert verified
Elemental tungsten has a face-centered cubic (FCC) structure because the calculated density for FCC structure (19.31 g/cm³) is very close to the given density (19.3 g/cm³), while the calculated density for BCC structure (12.69 g/cm³) is far from the given density.

Step by step solution

01

Convert the radius of tungsten to cm

First, we must convert the given radius of tungsten from picometers (pm) to centimeters (cm) using the conversion factor: 1 pm = 1 × 10^-12 cm The radius of tungsten in cm is: r = 137 pm × (1 × 10^-12 cm / 1 pm) = 1.37 × 10^-8 cm
02

Calculate the unit cell edge length (a) for both FCC and BCC structures

- For a face-centered cubic (FCC) structure: The diagonal of the half cube should be equal to 4 times the radius of the atom: \(\sqrt{2}a = 4r\) Hence, we can find the edge length (a) for FCC as follows: a_FCC = (4r) / √2 = (4 × 1.37 × 10^-8 cm) / √2 = 3.88 × 10^-8 cm - For a body-centered cubic (BCC) structure: The diagonal of the entire cube should be equal to 4 times the radius of the atom: \(\sqrt{3}a = 4r\) Hence, we can find the edge length (a) for BCC as follows: a_BCC = (4r) / √3 = (4 × 1.37 × 10^-8 cm) / √3 = 3.17 × 10^-8 cm
03

Determine the number of atoms per unit cell for both FCC and BCC structures

- For a face-centered cubic (FCC) structure: There are 8 corner atoms (1/8 contribution each) and 6 face atoms (1/2 contribution each), resulting in: n_FCC = 8 × (1/8) + 6 × (1/2) = 4 atoms/unit cell - For a body-centered cubic (BCC) structure: There are 8 corner atoms (1/8 contribution each) and 1 body atom (full contribution), resulting in: n_BCC = 8 × (1/8) + 1 = 2 atoms/unit cell
04

Calculate the density for both FCC and BCC structures

Given: Mass of one tungsten atom (m) = atomic mass of tungsten / Avogadro's number = 183.84 g/mol / 6.022 × 10^23 atoms/mol = 3.05 × 10^-22 g/atom - For a face-centered cubic (FCC) structure: Density_FCC = (n_FCC × m) / (a_FCC^3) Density_FCC = (4 × 3.05 × 10^-22 g) / ((3.88 × 10^-8 cm)^3) = 19.31 g/cm³ - For a body-centered cubic (BCC) structure: Density_BCC = (n_BCC × m) / (a_BCC^3) Density_BCC = (2 × 3.05 × 10^-22 g) / ((3.17 × 10^-8 cm)^3) = 12.69 g/cm³
05

Compare the calculated densities with the given density to determine the crystal structure

The given density of tungsten is 19.3 g/cm³. - Calculated density for FCC structure: 19.31 g/cm³ (very close to the given density) - Calculated density for BCC structure: 12.69 g/cm³ (far from the given density) Therefore, the elemental tungsten has a face-centered cubic (FCC) structure.

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