Question

The unit cell of metallic silver is FCC, if radius of Ag atom is 144.4 Pm. Calculate
a. edge length of unit cell
b. Volume of Ag atom
c. The percent of the volume of a unit cell that is occupied by Ag atoms.
d. The percent of empty space

Answer 1

Given:

The unit cell of metallic silver is FCC

the radius of Ag atom = 144.4 pm

To Find:

a. Edge length of unit cell

b. The volume of Ag atom

c. The percent of the volume of a unit cell that is occupied by Ag atoms.

d. The percent of empty space

Solution:

What does the FCC unit cell denote?

• In a face-centered cubic(FCC) unit cell, atoms are present on all the corners and in the center of each face of the cube.
• The atoms present on the 8 corners contribute $\frac{1}{8}$ of total space as they are shared between 8 unit cells.
• The atoms present on each of the 6 faces contribute $\frac{1}{2}$ of total space as they are shared between 2 unit cells.
• Therefore,total number of atoms are calculated as:

                Z = $\frac{1}{8}$ × 8 + $\frac{1}{2}$ × 6

                Z = 1 + 3 = 4                          (1)

a. Edge Length of Unit Cell

• How to find the edge length of the unit cell

1. Establish the relationship between the edge length and the radius of the atom for the FCC lattice.

• In FCC lattice, the relation between the edge length(a) and the radius(r) is given by:

                          a =2 $\sqrt{2}$r

2. Substitute the values in the formula:

                           a = 2$\sqrt{2}$ × 144.4 × $10^{-12}$

                      ⇒ a = 407.20 pm

Therefore, the edge length of the unit cell is 407.20 pm.

b. Volume of Ag atom

• To find the volume of the Ag atom, we will assume the atom to be a perfect sphere.
• The volume of a sphere is given by:

                                    V = $\frac{4}{3}$πr³

• Substituting known values:

                   V = $\frac{4}{3}$ × 3.14 × (144.4 × $10^{-12}$)³

             ⇒ V = 12.6121808 × $10^{-30}$

            ⇒ V ≈ 12.61 × $10^-{30}$ m³

Therefore, Volume of Ag atom is 12.61 × $10^-{30}$ m³.

c. The percent of the volume of a unit cell that is occupied by Ag atoms.

• What is the packing efficiency of a unit cell?

The Packing Efficiency of a unit cell is the percentage of space occupied by the atoms in the unit cell.

• How to calculate the packing efficiency?

Packing efficiency can be calculated as:

                    Packing efficiency = $\frac{volume of all atoms}{total volume of unit cell}$ × 100

• Volume of all atoms is given by

                            v = no. of atoms(Z) × volume of 1 atom

                            v = 4 × $\frac{4}{3}$πr³ = $\frac{16}{3}$πr³

• Volume of unit cell is given by:

                                   V = a³

                              ⇒ V = (2√2 r)³

                              ⇒ V = 16√2 r³

• Dividing v by V, we get:

                Packing Efficiency = $\frac{16pi r^{3} }{16\sqrt{2}r^{3} . 3 }$  × 100

                                               = $\frac{3.14}{3\sqrt{2} }$ × 100

                                               ≈ 74%

Therefore, the percent of the volume of a unit cell occupied by Ag atoms is 74%.

d.  The percent of empty space

• The percent of empty space can be simply calculated by subtracting the percent of the volume occupied by the atoms by 100.
• Hence, empty space = Total space - Ocuupied Space

                  % of empty sppace = 100 -74 = 26%

Therefore, the percent of empty space is 26%.

Hence, answers are:

a. 407.20 pm

b. 12.61 × $10^-{30}$ m³.

c. 74%

d. 26%