Question

The distance between an octahedral and tetrahedral void in an fcc lattice is?

• A. $\mathrm{a}\surd 3$
• B. $\frac{\mathrm{a}\surd 3}{2}$
• C. $\frac{\mathrm{a}\surd 3}{4}$
• D. $\frac{\mathrm{a}\surd 3}{6}$

Answer 1

The correct option is C

$\frac{\mathrm{a}\surd 3}{4}$

The explanation for the correct option:

Option(C):$\frac{\mathrm{a}\surd 3}{4}$

• The void which is known to be surrounded by four spheres existing at the corners of a regular tetrahedron is called a tetrahedral void.
• However when two tetrahedral voids from two different layers are aligned, together they form an octahedral void where the void is surrounded by $6$ atoms.
• In an FCC lattice, the atoms are present on each corner of the cube as well as in the face centre of each face, which can be represented as:

• As the number of particles in FCC is 4. Hence, the number of tetrahedral and octahedral voids present in the lattice are $8\mathrm{and}4$respectively.
  Now, we have to find the distance between the octahedral and the tetrahedral voids.
• In fcc lattice, the tetrahedral voids are located on the body diagonal of the cube at a one-fourth distance from the corner.
• Also, we know that body-diagonal length is represented as $a\sqrt{3}$ , hence the location of tetrahedral voids will be at ${\displaystyle \frac{a\sqrt{3}}{4}}$ where $\mathrm{a}$ is denoted as the edge length.
• However, the octahedral voids are located at the body centre of the cube and the edge centres.
• Hence, the location of these octahedral voids will be half that of the body-diagonal length i.e. ${\displaystyle \frac{a\sqrt{3}}{2}}$ .
• Now, we can calculate the distance between the two voids will as: ${\displaystyle \frac{a\sqrt{3}}{2}}-{\displaystyle \frac{a\sqrt{3}}{4}}={\displaystyle \frac{a\sqrt{3}}{4}}$

Hence, the distance between an octahedral and tetrahedral void in an fcc lattice is $\frac{\mathrm{a}\surd 3}{4}$, option (C) is the correct answer.