In fcc lattice ,A, B, C,D atoms are arranged at corner , face centre , ocatahedral void and tetrahedral void respectively , then the body diagonal contains :

In a fcc lattice A, B, C and D atoms are arranged at corners, face centres octahedral voids and half of tetrahedral voids respectively and if the crystal deposited is defected such that all the particles at one body diagonal of each unit cell are missing, correspondingly then what is the resulted formula of the compound ?

In a fcc lattice A, B, C and D atoms are arranged at corners, face centres octahedral voids and half of tetrahedral voids respectively and if the crystal deposited is defected such that all the particles at one body diagonal of each unit cell are missing, correspondingly then what is the resulted formula of the compound ?

The number of tetrahedral voids present on each body diagonal ccp unit cell is

The number of tetrahedral voids present on each body diagonal ccp unit cell is

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

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

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

In a solid between A and B atoms of A are arranged in ccp array and atoms of B occupy all the octahedral voids and half of the tetrahedral voids. The formula of the compound is

In a solid between A and B atoms of A are arranged in ccp array and atoms of B occupy all the octahedral voids and half of the tetrahedral voids. The formula of the compound is