Question

A fcc system contains 4 octahedral and 8 tetrahedral voids. Nearest distance possible between octahedral and
tetrahedral void is (a is edge length of unit cell)​

Answer 1

Answer:

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

ANSWER

Tetrahedral voids (THV) are present at

4

a

3

from body corner and octahedral voids are present at body center and edge centers.

d

THV⟶OHV

=

4

a

3

solution

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Answer 2

The nearest distance possible between octahedral and tetrahedral void is$\frac{a\sqrt{3} }{4}$

Explanation:

• Given -A fcc system contains 4 octahedral and 8 tetrahedral voids.

• In fcc crystal, there are one corner sphere and three face spheres which form one tetrahedral void.

• The two tetrahedral voids are obtained along with one cube diagonal.

• In all, a total of four cube diagonals are present in one unit cell. so a total of eight tetrahedral voids are in the FCC system.

• In the FCC system, octahedral voids are present at the body centres and at the centres of the 12 edge of the cube.

• The octahedral voids lie at the body centre of the cube and the tetrahedral voids lie at the fourth distance from the respective corners of the body.
• The nearest distance possible between octahedral and tetrahedral void is$\frac{a\sqrt{3} }{4}$