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Chapter 10: Problem 50

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 \(1.246 \times 10^{-8}\) cm.

Step by step solution

01

Write down known values and formula relating density, edge length, and atomic radius

Given information: Density of Nickel = 6.84 g/cm鲁 Unit cell: Face-centered cubic (fcc) The formula for density (蟻) is: \(蟻 = \frac{mass}{volume}\) For a face-centered cubic unit cell, the number of atoms in the unit cell is 4. We need the formula relating the edge length (a) of the FCC unit cell with the atomic radius (r) which can be derived from geometry: \(a = 2\sqrt{2}r\) We also need the formula relating edge length, mass, and volume: \(\rho = \frac{4 \times mass\ of\ one\ Ni\ atom}{a^3}\)
02

Calculate the molar mass of Nickel

The molar mass of Nickel (Ni) can be found in the periodic table. Molar mass of Ni = 58.69 g/mol
03

Calculate the mass of one Nickel atom

We will use Avogadro's number (6.022 x 10虏鲁 atoms/mol) to determine the mass of one Nickel atom. \(mass\ of\ one\ Ni\ atom = \frac{molar\ mass\ of\ Ni}{Avogadro's\ number}\) \(mass\ of\ one\ Ni\ atom = \frac{58.69 \mathrm{~g/mol}}{6.022 \times 10^{23}\ \mathrm{atoms/mol}} = 9.75 \times 10^{-23}\ \mathrm{g}\)
04

Calculate the edge length of the unit cell

Now, we use the formula for density, which we already derived earlier: \(\rho = \frac{4 \times mass\ of\ one\ Ni\ atom}{a^3}\) Rearrange the formula to find the edge length (a): \(a = \sqrt[3]{\frac{4 \times mass\ of\ one\ Ni\ atom}{\rho}}\) Plug in the density (6.84 g/cm鲁) and mass of one Ni atom (9.75 x 10^{-23} g) to solve for edge length: \(a = \sqrt[3]{\frac{4 \times 9.75 \times 10^{-23}\ \mathrm{g}}{6.84\ \mathrm{g/cm^{3}}}} = 3.522 \times 10^{-8}\ \mathrm{cm}\)
05

Calculate the atomic radius of Nickel

Use the relationship between the edge length (a) and the atomic radius (r) for a face-centered cubic unit cell: \(a = 2\sqrt{2}r\) Rearrange the formula to find the atomic radius (r): \(r = \frac{a}{2\sqrt{2}}\) Plug in the calculated edge length (3.522 x 10^{-8} cm) to solve for the atomic radius of Nickel: \(r = \frac{3.522 \times 10^{-8}\ \mathrm{cm}}{2\sqrt{2}} = 1.246 \times 10^{-8}\ \mathrm{cm}\) So, the atomic radius of Nickel is approximately 1.246 x 10^{-8} cm.

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