Ciencia De Los Materiales Askelanc 5 Chapter 3

3
Atomic and Ionic Arrangements
3–25 Calculate the atomic radius in cm for the following: (a) BCC metal with ao = 0.3294 nm and one atom per lattice point; and (b) FCC metal with ao = 4.0862 Å and one atom per lattice point. Solution: (a) For BCC metals, r =

( 3 )ao = ( 3 )(0.3294 nm) = 0.1426 nm = 1.426 × 10 −8 cm
4 4

(b) For FCC metals, r =

( 2 )ao = ( 2 )(4.0862 Å) = 1.4447 Å =1.4447 × 10 −8 cm
4 4

3–26

Determine the crystal structure for the following: (a) a metal with ao = 4.9489 Å, r = 1.75 Å and one atom per lattice point; and (b) a metal with ao = 0.42906 nm, r = 0.1858 nm and one atom per lattice point. Solution: We want to determine if “x” in the calculations below equals (for FCC) or 3 (for BCC): (a) (x)(4.9489 Å) = (4)(1.75 Å) x=
2 , therefore FCC 2

(b)(x)(0.42906 nm) = (4)(0.1858 nm) x= 3–27 3 , therefore BCC

The density of potassium, which has the BCC structure and one atom per lattice point, is 0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate (a) the lattice parameter; and (b) the atomic radius of potassium.

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The Science and Engineering of Materials Solution:

Instructor’s Solution Manual

(a) UsingEquation 3–5: 0.855 g/cm3 = (2 atoms/cell)(39.09 g/mol) (ao)3(6.02 × 1023 atoms/mol)

ao3 = 1.5189 × 10−22 cm3 or ao = 5.3355 × 10−8 cm (b) From the relationship between atomic radius and lattice parameter: r =

( 3 )(5.3355 × 10
4

−8 cm

) = 2.3103 × 10

−8 cm

3–28 The density of thorium, which has the FCC structure and one atom per lattice point, is 11.72 g/cm3. The atomic weightof thorium is 232 g/mol. Calculate (a) the lattice parameter and (b) the atomic radius of thorium. Solution: (a) From Equation 3–5: 11.72 g/cm3 = (4 atoms/cell)(232 g/mol) (ao)3(6.02 × 1023 atoms/mol)

ao3 = 1.315297 × 10−22 cm3 or ao = 5.0856 × 10−8 cm (b) From the relationship between atomic radius and lattice parameter: r =

( 2 )(5.0856 × 10
4

−8 cm

) = 1.7980 × 10

−8 cm

3–29A metal having a cubic structure has a density of 2.6 g/cm3, an atomic weight of 87.62 g/mol, and a lattice parameter of 6.0849 Å. One atom is associated with each lattice point. Determine the crystal structure of the metal. Solution: 2.6 g/cm3 = (x atoms/cell)(87.62 g/mol) (6.0849 × 10−8 cm)3(6.02 × 1023 atoms/mol)

x = 4, therefore FCC 3–30 A metal having a cubic structure has a density of1.892 g/cm3, an atomic weight of 132.91 g/mol, and a lattice parameter of 6.13 Å. One atom is associated with each lattice point. Determine the crystal structure of the metal. Solution: 1.892 g/cm3 = (x atoms/cell)(132.91 g/mol) (6.13 × 10−8 cm)3(6.02 × 1023 atoms/mol)

x = 2, therefore BCC 3–31 Indium has a tetragonal structure with ao = 0.32517 nm and co = 0.49459 nm. The density is 7.286g/cm3 and the atomic weight is 114.82 g/mol. Does indium have the simple tetragonal or body-centered tetragonal structure? Solution: 7.286 g/cm3 = (x atoms/cell)(114.82 g/mol) (3.2517 × 10−8 cm)2(4.9459 × 10−8 cm)(6.02 × 1023 atoms/mol)

x = 2, therefore BCT (body-centered tetragonal)

CHAPTER 3 3–32

Atomic and Ionic Arrangements

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Bismuth has a hexagonal structure, with ao = 0.4546 nmand co = 1.186 nm. The density is 9.808 g/cm3 and the atomic weight is 208.98 g/mol. Determine (a) the volume of the unit cell and (b) how many atoms are in each unit cell. Solution: (a) The volume of the unit cell is V = ao2cocos30. V = (0.4546 nm)2(1.186 nm)(cos30) = 0.21226 nm3 = 2.1226 × 10−22 cm3 (b) If “x” is the number of atoms per unit cell, then: 9.808 g/cm3 = x = 6 atoms/cell (xatoms/cell)(208.98 g/mol) (2.1226 × 10−22 cm3)(6.02 × 1023 atoms/mol)

3–33

Gallium has an orthorhombic structure, with ao = 0.45258 nm, bo = 0.45186 nm, and co = 0.76570 nm. The atomic radius is 0.1218 nm. The density is 5.904 g/cm3 and the atomic weight is 69.72 g/mol. Determine (a) the number of atoms in each unit cell and (b) the packing factor in the unit cell. Solution: The volume of the...