Question

In modeling solid-state structures, atoms and ions are most often modeled as spheres. A structure built using spheres will have some empty, or void, space in it. A measure of void space in a particular structure is the packing efficiency, defined as the volume occupied by the spheres divided by the total volume of the structure. Given that a solid crystallizes in a body centered cubic structure that is 3.12 Å on each side, answer questions (a)-(e). (1 Å = 1 10-10 m.) (a) How many total atoms are there in each unit cell? atom(s) (b) What is the volume of one unit cell in Å3? WebAssign will check your answer for the correct number of significant figures. Å3 (c) Assuming that the atoms are spheres and the radius of each sphere is 1.35 Å, what is the volume of one atom in Å3? (Vsphere = 4/3πr3.) WebAssign will check your answer for the correct number of significant figures. Å3 (d) Therefore, what volume of atoms are in one unit cell? WebAssign will check your answer for the correct number of significant figures. Å3 (e) Based on your results from parts (b) and (d), what is the packing efficiency of the solid expressed as a percentage?

Answer 1

There are 2 atoms in each unit cell. The volume of 1 unit cell is 30.4 ų. The volume of each atom is 10.3 ų. The volume of atoms in the cell is 20.6 ų. The packing efficiency of the solid is 67.8 %.

(a) How many total atoms are there in each unit cell?

In body centered cubic cells, there are 2 atoms, 1 in the center and 1 resulting from the sum of the 1/8 atom of each of the 8 corners.

$N = 1 + 1/8 \times 8 = 2$

(b) What is the volume of one unit cell in ų?

Since the cell is a cube with a side of 3.12 Å (s), we can calculate its volume (V) using the following expression.

$V = s^{3} = 3.12^{3} = 30.4 A^{3}$

(c) Assuming that the atoms are spheres and the radius (r) of each sphere is 1.35 Å, what is the volume of one atom in ų?

We will use the following expression for the volume of the sphere.

$V = \frac{4}{3} \times \pi \times r^{3} = \frac{4}{3} \times \pi \times (1.35A)^{3}=10.3 A^{3}$

(d) Therefore, what volume of atoms are in one unit cell?

There are 2 atoms in the unit cell, each one with a volume of 10.3 ų.

$2 \pi 10.3 A^{3} = 20.6 A^{3}$

(e) Based on your results from parts (b) and (d), what is the packing efficiency of the solid expressed as a percentage?

The volume of 1 unit cell is 30.4 ų and the volume occupied by atoms in the cell is 20.6 ų. The packing efficiency of the solid is:

$E = \frac{20.6A^{3} }{30.4A^{3}} \times 100\% = 67.8\%$

There are 2 atoms in each unit cell. The volume of 1 unit cell is 30.4 ų. The volume of each atom is 10.3 ų. The volume of atoms in the cell is 20.6 ų. The packing efficiency of the solid is 67.8 %.

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