Question

Aluminum foil is .003 cm thick. Assume that all of the unit cells are arranged perpendicular to the foil surface. For a 10cm x 10cm spuare of the foil, determine:

a. The total number of unit cells in the foil

b. The thickness of the foil in nmber of unit cells

c. The total number of atoms in the foil

Answer 1

area of Al foil = 0.003 * 10 * 10 = 0.3 cm^3

diameter of Al atom = 210 pm then r = 105 pm

Al has FCC crystal structure then edge length a = 4r / square root (3)

                                                                                  a = 4 * 105 * 10^-12 / squre root (3)

                                                                                  a = 2.4249 * 10^-10 m

Volume of unit cell = a^3 = (2.4249 * 10^-10)^3 = 1.4259*10^-29 m^3

                                                                                    = 1.4259 * 10^-23 cm^3

total number of unit cells= 0.3 cm^3 / (1.4259 * 10^-23 cm^3)

total number of unit cells = 2.1039 * 10^22 unit cells

The thickness of the foil in nmber of unit cells =0.003cm / 2.4249 * 10^-8 cm = 1.237 * 10^5 unit cells

each unit cell has 4 atoms of Al the in given foil contains = 4 * 2.1039 * 10^22 atoms

                                                                                                      = 8.4156 * 10^22 atoms