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Grade: 12th pass
        
This question is from chapter SOLID STATES
PLEASE SOLVE THIS
one year ago

Answers : (1)

Arun
22751 Points
							
Dear student
 
Since oxide ions are fcc so 4O-2 ∴ unit cell
                        A+2 are at 1/8th of the tetrahedral so 1A+2|unitcell
                        B+3 occupies ½ of the octahedral voids ∴ 2B+2/unit cell
                        After removing O-2 ions
                        Oxide ion / unit cell = 4/8 + 3 = 3.5
                        A+2 ions/ unit cell = 2/8 + 1 = 10/8 = 1.25
                        B+3 ions/unit cell = 2
                        ∴ P.F = 3.5 × 4/3 πr3 + 1.25 × 43πr3A+2 + 2 × 4/3 πr3B+3
                        We know that a = 4r / √2
                        rA+2 = 0.225,         ∴rA, = 0.225r–
                        rB+3 / r– =0.414       ∴ rB+3 = 0.414r-
                        Putting all the values
                        P.F = 0.676
 
 
 
Regards
Arun (askIITians forum expert)
one year ago
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