DEPARTMENT OF CHEMISTRY
CHEM-GA 2400-001
The Science of Materials
Fall 2023
Problem Set 3
120 points total
Due Date: Tuesday, October 10
On or before 12:30 pm
Submit completed Problem Set to Mail Pouch
outside Professor Ward’s Office Brown 554
Name: ___________________________________
Problem 3.1. The unit cell of CaTiO3 structure is given below.
Ti4+
O2Ca2+
(a) What is the Bravais lattice?
(b) What is the basis?
(c) How many atoms are there per unit cell ?
Problem 3.2. Show that the atomic packing factor for an ideal HCP crystal is 0.74.
Problem 3.3. Zirconium has an HCP crystal structure and a density of 6.51 g/cm3.
(a) What is the volume of its unit cell in cubic meters?
(b) If the c/a ratio is 1.593, compute the values of c and a.
Problem 3.4. The unit cell for tin (Sn) has tetragonal symmetry, with a and b lattice parameters of 0.583
and 0.318 nm, respectively. If its density, atomic weight, and atomic radius are 7.30 g/cm3, 118.69 g/mol,
and 0.151 nm, respectively, compute the atomic packing factor.
Problem 3.5. List the point indices of all atoms that are associated with the diamond cubic unit cell
(Callister, Figure 12.16. Note: The figure number may depend on the edition).
Problem 3.6. Derive the structure factor and state the reflection conditions for diamond.
Problem 3.7. Compute the percent ionic character of the interatomic bonds for each of the following
compounds: TiO2, ZnTe, CsCl, InSb, and MgCl2.
Problem 3.8. Calculate the number of vacancies per cubic meter in iron at 850°C. The energy for vacancy
formation is 1.08 eV/atom. The density and atomic weight for Fe are 7.65 g/cm3 and 55.85 g/mol,
respectively.
Problem 3.9. Calculate the activation energy for vacancy formation in aluminum, given that the equilibrium
number of vacancies at 500°C (773 K) is 7.57 ´ 1023 m-3. The atomic weight and density (at 500°C) for
aluminum are, respectively, 26.98 g/mol and 2.62 g/cm3.
Problem 10 (30 points). For both FCC and BCC crystal structures containing atoms of a single kind, there
are two different types of interstitial sites, the void spaces between the spherical atoms. These void spaces
are critical to the incorporation of “guest” atoms that are required to form alloys (we will see this later). In
each case, one site is larger that the other, this site being the one normally occupied by impurity atoms. For
FCC, this larger one is located at the center of each edge of the unit cell; it is termed an “octahedral”
interstitial site. On the other hand, the larger site in the BCC lattice is found at 0, 1/2, 1/4 positions – that
is, lying on {100} faces and situated midway between two unit cell edges on this face and one-quarter of
the distance between the other two unit cell edges; it is termed a “tetrahedral” interstitial site. For both FCC
and BCC crystal structures, compute the radius r of an impurity atom that just fit into one of these sites.
Give the radius in terms of the atomic radius R of the host atom.