Question

An element has a face-centered cubic (fcc) structure with a cell edge of a. The distance between the centers of two nearest tetrahedral voids in the lattice is:
a.) $\dfrac { a }{ 2 } $
b.) $a$
c.) $\dfrac { 3 }{ 2 } { a }$
d.) $\sqrt { 2 } { a }$

Answer 1

Hint: A face-centered cubic (fcc) unit cell:- It is defined as the unit cell that contains atoms at all the corners and at the center of all the faces of the cube.

Complete Solution :
As we know that for face-centered cubic (z = 4).
The relation between the radius of the constituent particle, r, and edge length, a for the face-centered unit cell.

In triangle ABC,
${ AB }^{ 2 }{ =AC }^{ 2 }{ +BC }^{ 2 }$
${ b }^{ 2 }{ =a }^{ 2 }{ +a }^{ 2 }$
b = ${ 4 }$
a = ${ 2 }\sqrt { 2 } { a }$

- We know that, in the fcc unit cell, tetrahedral voids are located at body diagonal at a distance $\dfrac { 3 }{ 4 } { a }$ from the corners.

- Now, in triangles, ACE and BCD are similar triangles,
Therefore, $\dfrac { AE }{ AC } { = }\dfrac { BD }{ BC } $
$\dfrac { { a\times 2 } }{ \sqrt { 3 } { a } } { = }\dfrac { BD\times 4 }{ \sqrt { 3 } { a } } $
So, BD = $\dfrac { a }{ 2 } $
Hence, the distance between the two nearest tetrahedral voids in the lattice is $\dfrac { a }{ 2 } $.
So, the correct answer is “Option A”.

Additional Information:
The unit cell is the smallest 3D part of a crystal lattice which when repeated in different directions, generates the entire lattice.
- An FCC unit cell has a total of six faces and each face has one atom. But the facial atom is always divided between two unit cells. This means that only half part of the facial atom belongs to one unit cell.
Hence, it is clear that the number of atoms in FCC cell
=\[\dfrac{1}{8}\times \text{ }8\text{ }+\text{ 3}\]
= 4 atoms
So, the correct answer is “Option A”.

Note: The possibility to make a mistake is that in face-centered unit cells the atoms are closely packed in sequence and the number of atoms in fcc is ${ 4 }$.