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Chapter 10: Problem 54

Nickel has a face-centered cubic unit cell. The density of nickel is 6.84 \(\mathrm{g} / \mathrm{cm}^{3} .\) Calculate a value for the atomic radius of nickel.

Short Answer

Expert verified
The atomic radius of nickel is approximately \(1.24 \times 10^{-8}\) cm.

Step by step solution

01

List the given information

Density of nickel (ρ) = 6.84 g/cm³ Unit cell type: face-centered cubic (fcc)
02

Calculate the number of atoms in a face-centered cubic unit cell

In an fcc unit cell, each corner atom is shared by 8 adjacent unit cells, and each face atom is shared by 2 adjacent unit cells. Therefore, the number of atoms in an fcc unit cell can be calculated as: Number of atoms = (Number of corner atoms × 1/8) + (Number of face atoms × 1/2) Number of atoms = (8 × 1/8) + (6 × 1/2) Number of atoms = 4
03

Write down the formula for the density of the face-centered cubic unit cell

The formula for the density of an fcc unit cell is: \(ρ = \frac{4 \times M}{a^{3} \times N_{A}}\) Where: ρ = Density M = Molar mass of nickel (g/mol) a = Edge length of the unit cell (cm) Nₐ = Avogadro's number (6.022 x 10²³ atoms/mol) We need to find the edge length (a) and use it to determine the atomic radius.
04

Find the molar mass of nickel

The molar mass of nickel (Ni) is 58.69 g/mol.
05

Rearrange the formula to solve for the edge length (a) of the unit cell

We can rearrange the density formula to solve for a: \(a^{3} = \frac{4 \times M}{ρ \times N_{A}}\) Next, we plug in the values and solve for the edge length: \(a^{3} = \frac{4 \times 58.69 \text{ g/mol}}{6.84 \text{ g/cm}^{3} \times 6.022 \times 10^{23} \text{ atoms/mol}}\) \(a^{3} = \frac{233.76 \text{ g/mol}}{6.84 \text{ g/cm}^{3} \times 6.022 \times 10^{23} \text{ atoms/mol}}\) \(a^{3} \approx 5.94 \times 10^{-23} \text{ cm}^{3}\) \(a \approx \sqrt[3]{5.94 \times 10^{-23} \text{ cm}^{3}}\) \(a \approx 3.93 \times 10^{-8} \text{ cm}\)
06

Calculate the atomic radius using the edge length

In an fcc unit cell, the relationship between edge length (a) and atomic radius (r) is: \(a = 2 \sqrt{2}r\) Now, we can solve for the atomic radius: \(r = \frac{a}{2 \sqrt{2}}\) \(r = \frac{3.93 \times 10^{-8} \text{ cm}}{2 \sqrt{2}}\) \(r \approx 1.24 \times 10^{-8} \text{ cm}\) So, the atomic radius of nickel is approximately \(1.24 \times 10^{-8}\) cm.

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