Courses
Courses for Kids
Free study material
Offline Centres
More
Store

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

Last updated date: 13th Jun 2024
Total views: 340.7k
Views today: 5.40k
Verified
340.7k+ views
Hint: There are FOUR atoms present in the FCC unit cell. All these atoms occupy some volume in the lattice. The shape of the atoms are considered to be spheres. therefore, we must use the formula to find the volume of a sphere.

Formula used: This volume can be calculated using the formula of volume of a sphere that is: $\dfrac{{4\pi {r^3}}}{3}$ where $r$ is the radius of one atom.

Each atom is considered to be a sphere. This Crystal lattice contains atoms at the $8$ corners of the crystal lattice and at the center of the $6$ faces of the cube. The atoms at the corners are shared by eight-unit cells. This means that a one-unit cell contains only ${\dfrac{1}{8}^{th}}$ of the volume of the atom in one unit cell. Now the atoms at the center of the six faces are shared between two unit cells. This means that one unit cell contains only ${\dfrac{1}{2}^{}}$ the volume of an atom. Therefore, we can derive the number of atoms in an FCC lattice through the below derivation:
$= 8 \times \dfrac{1}{8} + 6 \times \dfrac{1}{2}$
$= 1 + 3$
$= 4$
Therefore, the total volume of atoms occupying space in the FCC unit cell is derived below:
Volume $= n \times \dfrac{{4\pi {r^3}}}{3}$
Here, $n$ denotes the number of atoms present in FCC lattice and $r$denotes the radius of the atom.
Volume $= 4 \times \dfrac{{4\pi {r^3}}}{3}$
$= \dfrac{{16\pi {r^3}}}{3}$

So, the correct answer is Option C.

Note: The coordination number of the FCC unit cell is 12.
Packing efficiency of the crystal lattice is $74\%$. This means that $74\%$ of the total volume of the entire unit cell is occupied by the atoms. Out of this, the remaining $26\%$ of the unit cell is empty space also known as voids.