Pls solve
Q. A cubic unit cell has Fee symmetry of A atoms and B atom lies at all octahedral voids. If atoms bisecting on axis in two halves with a plane are all removed formula will be
(1) AB (2) AB2 (3) A2B (4) A2B3
Dear student
A has fcc symmetry.
Thus no.of A atoms : 8 x 1/8 +6 x1/2 = 4
No.of octahedral voids occupied by B atoms = 4
If atoms bisecting on axis in two halves with a plane are all removed the
No.of A atoms : 8 x 1/8 + 4 x 1/2 x1/2 =2
Only one B atom will be effected, So 4-1 =3
Thus the formula becomes: A2B3
option (4) is correct
Regards
A has fcc symmetry.
Thus no.of A atoms : 8 x 1/8 +6 x1/2 = 4
No.of octahedral voids occupied by B atoms = 4
If atoms bisecting on axis in two halves with a plane are all removed the
No.of A atoms : 8 x 1/8 + 4 x 1/2 x1/2 =2
Only one B atom will be effected, So 4-1 =3
Thus the formula becomes: A2B3
option (4) is correct
Regards