Question

The total volume of atoms present in an FCC unit cell of a metal with radius \[r\] is:
A.\[\dfrac{12}{3}\pi {{r}^{3}}\]
B.\[\dfrac{16}{3}\pi {{r}^{3}}\]
C.\[\dfrac{20}{3}\pi {{r}^{3}}\]
D.\[\dfrac{24}{3}\pi {{r}^{3}}\]

Answer 1

Hint: In a Face-centered cubic (FCC) cell arrangement of lattice, the unit cell cube has an atom at the centre of each face of the cube along with the eight atoms in all the corners of the cell. Also each atom may or may not be a part of other adjoining neighbor unit cells too, so we need to calculate its contribution to a particular unit cell.

Complete step by step answer:
Number of atoms at the corners = \[8\] per unit cell.
Also each corner atom is shared by eight adjacent unit cells. So each unit cell has \[\dfrac{1}{8}th\] portion of each corner atom.
Now, number of atoms at face-centers = number of faces of cube = \[6\]
Also each face center atom is shared by two adjacent unit cells. So each unit cell has \[\dfrac{1}{2}th\] portion of each corner atom.
Thus total number of atoms in each unit cell \[=\dfrac{1}{8}\times 8+\dfrac{1}{2}\times 6\] \[=1+3=4\]
Assuming the corner atoms and face centered atoms are of the same radius \[r\]. As we know that there are \[4\] spheres in each unit cell.
Thus volume occupied by atoms in fcc unit cell = volume of each sphere times the number of spheres
\[=\dfrac{4}{3}\pi {{r}^{3}}\times 4=\dfrac{16}{3}\pi {{r}^{3}}\]

Hence, option (B) is correct.

Additional Information:
Fcc structure arrangement is one of the most tightly packed arrangements in a crystal lattice. It has a coordination number of \[12\], that is it has \[12\] atoms neighboring each atom. This makes these structures stronger. Materials with fcc arrangement include aluminum, copper, nickel, gamma iron, gold, and silver.

Note:
Hexagonal close packing (HCP) also has a very similar arrangement to that of fcc arrangement. In fact, the coordination number, number of atoms per unit cell and packing efficiency are also equal in both cases.