Question

The number of carbon atoms per unit cell of diamond unit cell is:
A. $4$
B. $8$
C. $6$
D. $1$

Answer 1

Hint: A Diamond completely consists of carbon atoms and is an example of face-centered cubic unit cell that means the carbon atoms are present at the corners as well as face centers, and the carbon atoms are also present in the alternating tetrahedral voids of the unit cell.

Complete step by step solution:
Firstly, A unit cell is referred to as the smallest repeating unit in a crystal lattice. Unit cells occur in many different varieties. And the cubic crystal system is one of them which has three types of unit cells and one of them is face-centered unit cell and Diamond is an example of fcc lattice.
As we know, that diamond is a face-centered cubic unit cell; the atoms in the lattice here are present at the face centers ,corners and at the alternating tetrahedral voids of the lattice. The tetrahedral voids present in a unit cell are “ $2n$ ”, where ” $n$ ” is the number of atoms present in a unit cell and for a fcc lattice is $n = 4$.
Before calculating the number of carbon atoms in the unit cell one must know about the contribution of the atoms towards a unit cell in the lattice. An atom present at the corner contributes $\dfrac{1}{8}$to a unit cell as it is shared to 8 different unit cells attached together, whereas the one present at the face-center contributes $\dfrac{1}{2}$to a unit cell.
Now, let’s calculate the number of carbon atoms present in diamond:
As there are 8 atoms at corners in a unit cell with the contribution of $\dfrac{1}{8}$therefore,
$8 \times \dfrac{1}{8} = 1$
As there are 6 atoms at the face centers in a unit cell with the contribution of $\dfrac{1}{2}$therefore,
$6 \times \dfrac{1}{2} = 3$
And the carbon atoms are present at alternating tetrahedral voids or half of tetrahedral voids therefore,
$8(2n) \times \dfrac{1}{2} = 4$,where $n = 4$ for fcc lattice.
So, the total number of carbon atoms are $4 + 4 = 8$ (Option B).




Note: A student can be confused about the number of octahedral (“ $n$ ”) and tetrahedral voids (“ $2n$ ”) present in a lattice so a student should keep the values in mind. And when it comes to the contribution of the octahedral and the tetrahedral voids they have a contribution of 1, as they are totally a part of the lattice.