An element has a face-centred cubic (fcc ) structure with a cell edge of a. The distance between the centres of two nearest tetrahedral voids in the lattice is :

Question

An element has a face-centred cubic (fcc ) structure with a cell edge of a. The distance between the centres of two nearest tetrahedral voids in the lattice is : (A) (a/2) (B) a (C) (3/2) a  (D) √2 a

Tardigrade Admin · Accepted Answer

Distance between two nearest tetrahedral void = ( (a/2) )

Q. An element has a face-centred cubic (fcc ) structure with a cell edge of a. The distance between the centres of two nearest tetrahedral voids in the lattice is :

$\frac{a}{2}$

a

$\frac{3}{2} a$

$2 a$

Distance between two nearest tetrahedral void $= (\frac{a}{2})$

Q. An element has a face-centred cubic (fcc ) structure with a cell edge of a. The distance between the centres of two nearest tetrahedral voids in the lattice is :

2a​

a

23​a

2​a

Distance between two nearest tetrahedral void =(2a​)

$\frac{a}{2}$

$\frac{3}{2} a$

$2 a$

Distance between two nearest tetrahedral void $= (\frac{a}{2})$