Question

The distance between an octahedral and tetrahedral void in fcc lattice would be

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

Answer 1

The correct option is D

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

The correct option is option D

The explanation for the correct option:

1. A tetrahedral void is the space that is surrounded by four spheres that are positioned at the corners of a regular tetrahedron. An octahedral void is created when two tetrahedral voids from two separate levels are lined up.
2. The tetrahedral voids in the fcc lattice are situated one-fourth of the way from the corner on the cube's body diagonal.
3. The body-diagonal length is $\mathrm{a}\sqrt{3}$, and since an is the edge length, the site of tetrahedral voids will be at $\frac{\mathrm{a}\sqrt{3}}{4}$
4. The body centre of cube and edge centres are where the octahedral voids can be found. The body centre void is located exactly in the cube's centre, therefore it will be the same as for each edge centre.
5. As a result, if the octahedral void is present at the centre of the body diagonal's centre, we can say that it is located at $\frac{\mathrm{a}\sqrt{3}}{2}$, or half to the body diagonal.
6. Now, the distance between the two voids are $\frac{\mathrm{a}\sqrt{3}}{2}-\frac{\mathrm{a}\sqrt{3}}{4}=\frac{\mathrm{a}\sqrt{3}}{4}$

Therefore, the distance between an octahedral and tetrahedral void in fcc lattice is $\frac{\mathrm{a}\sqrt{3}}{4}$