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 :
A
`2A,C,2D`
B
`2A, 2B , 2C`
C
`2A, 2B,D`
D
`2A,2D`
Text Solution
Verified by Experts
The correct Answer is:
A
Topper's Solved these Questions
SOLID STATE
NARENDRA AWASTHI|Exercise Level 1 (Q.33 To Q.62)|1 Videos
SOLID STATE
NARENDRA AWASTHI|Exercise Level 3 - One Or More Answers Are Correct|1 Videos
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 distance between an ocatahral and tetrahedral void in fcc lattice would be:
Atoms A, B, C and D are present at corners, face centres, body centres and edge centres and respectively in a cubic unit cell. Find the formula of compound.
In zinc blende structure anions are arranged in ccp and cations are present in the tetrachedral voids and only half the tetrahedral voids are occupied, the coordination numbers of cation and anion are respectively
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 unit cell, atoms A, B, C and D are present at corners, face - centres, body - centre and edge - centres respectively. If atoms touching one of the plane passing through two diagonally opposite edges are removed, then formula of compound is
