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Chapter 3: Problem 19

Rhenium has an HCP crystal structure, an atomic radius of \(0.137 \mathrm{~nm}\), and a \(c / a\) ratio of 1.615. Compute the volume of the unit cell for Re.

Short Answer

Expert verified
Answer: The volume of the unit cell for Rhenium in the HCP structure is approximately 0.08143 nm^3.

Step by step solution

01

Write the formula to calculate the volume of an HCP unit cell

The HCP unit cell has a hexagonal prism shape, so we can use the formula for the volume of a hexagonal prism to find the volume of the unit cell. The volume is given by the following formula: Volume = Area of the base × Height In the case of HCP, the base is a regular hexagon. The formula for the area of a regular hexagon of side length "a" is: Area = (3√3 × a^2) / 2
02

Determine the side length (a) from the given atomic radius

For an HCP structure, the side length "a" equals to 2 times the atomic radius. So, we can calculate the side length using the given atomic radius of Rhenium: a = 2 * r a = 2 * 0.137 nm a = 0.274 nm
03

Determine the height (c) using the given c/a ratio

The c/a ratio is given as 1.615. We can find the height "c" by multiplying the side length "a" by the c/a ratio: c = 1.615 * a c = 1.615 * 0.274 nm c = 0.4431 nm
04

Calculate the area of the base (hexagon)

Now, we can use the side length "a" to calculate the area of the hexagonal base using the formula mentioned in Step 1: Area = (3√3 × a^2) / 2 Area = (3√3 × (0.274 nm)^2) / 2 Area = 0.18385 nm^2
05

Calculate the volume of the HCP unit cell

Finally, we can now calculate the volume of the unit cell by multiplying the area of the hexagonal base and the height "c": Volume = Area × c Volume = 0.18385 nm^2 × 0.4431 nm Volume = 0.08143 nm^3 Therefore, the volume of the unit cell for Rhenium in the HCP structure is approximately 0.08143 nm^3.

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Most popular questions from this chapter

For which set of crystallographic planes will a first-order diffraction peak occur at a diffraction angle of \(46.21^{\circ}\) for BCC iron when monochromatic radiation having a wavelength of \(0.0711 \mathrm{~nm}\) is used?

Sketch within a cubic unit cell the following planes: (a) \((0 \overline{1} \overline{1})\) (c) \((10 \overline{2})\) (e) \((\overline{1} 1 \overline{1})\) (g) \((\overline{1} 2 \overline{3})\) (b) \((11 \overline{2})\) (d) \((1 \overline{3} 1)\) (h) \((0 \overline{1} \overline{3})\)

Would you expect a material in which the atomic bonding is predominantly ionic in nature to be more or less likely to form a noncrystalline solid upon solidification than a covalent material? Why? (See Section 2.6.)

Figure \(3.22\) shows an \(\mathrm{x}\)-ray diffraction pattern for \(\alpha\)-iron taken using a diffractometer and monochromatic \(\mathrm{x}\)-radiation having a wavelength of \(0.1542 \mathrm{~nm}\); each diffraction peak on the pattern has been indexed. Compute the interplanar spacing for each set of planes indexed; also determine the lattice parameter of Fe for each of the peaks.

The unit cell for tin has tetragonal symmetry, with \(a\) and \(b\) lattice parameters of \(0.583\) and \(0.318 \mathrm{~nm}\), respectively. If its density, atomic weight, and atomic radius are \(7.27 \mathrm{~g} / \mathrm{cm}^{3}\), \(118.71 \mathrm{~g} / \mathrm{mol}\), and \(0.151 \mathrm{~nm}\), respectively, compute the atomic packing factor.
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