Question

The mass of diamond unit cell is:
A. $96\text{ amu}$
B. $\text{96 gm}$
C. $\text{144 amu}$
D. $\text{144 gm}$

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. Calculate the number of carbon atoms present per unit cell of the diamond unit cell. Then multiply it with the mass of the carbon atom.

Complete step by step answer:
First of all; we will find out the number of carbon atoms present per unit cell in a diamond unit cell.
Firstly, a unit cell is referred to as the smallest repeating unit in a crystal lattice and occurs in many different varieties. And the cubic crystal system is one of them is face centered unit cell. Diamond is an example of FCC lattice.
As diamond is a face-centered cubic unit cell, the atoms 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 $n=4$.
In diamond, an atom present at the corner contributes $\dfrac{1}{8}$ to a unit cell as it is shared to different unit cells attached together, whereas the one present at the face-center contributes $\dfrac{1}{2}$ to a unit cell. As there are eight atoms at corners in a unit cell with the contribution of $\dfrac{1}{8}$ therefore, $8\times \dfrac{1}{8}=1$.
As there are six atoms at the face centers in a unit cell with the contribution of $\dfrac{1}{2}$ therefore, $6\times \dfrac{1}{2}=3$.
And there are carbon atoms present alternating tetrahedral voids or half of tetrahedral voids therefore, $8\times \dfrac{1}{2}=4$.
So, there are $4+4=8$ carbon atoms per unit cell in a diamond unit cell.
We know that, mass of a carbon atom is $12\text{ amu}$.
Therefore, the mass of the diamond unit cell is $(12\times 8)\text{ amu}=96\text{ amu}$.

Hence, the correct option is A.

Note: Possibly, you may get confused with the number of octahedral ($n$) and tetrahedral ($2n$) voids present in a lattice. When it comes to the contribution of octahedral and tetrahedral voids, they have a common contribution of $1$.