written 8.3 years ago by | • modified 8.3 years ago |
Mumbai university > FE > SEM 1 > Applied Physics 1
Marks: 5M
Year: Dec 2012
written 8.3 years ago by | • modified 8.3 years ago |
Mumbai university > FE > SEM 1 > Applied Physics 1
Marks: 5M
Year: Dec 2012
written 8.3 years ago by |
(Note: Use different line patterns to show different ions)
BaTiO3 (Barium Titanate) structure is a Tetragonal type in which a = b ≠ c and α, β, and γ are all right angles.
Here, a = b = 3.98 $A^0$ and c = 4.03 $A^0$.
In BaTiO3 unit cell, $Ba^{2+}$ ions occupy the corner, each ion shared by 8 cells; thus contributing 1/8th of the mass each and total ion in a unit cell being 8/8 = 1.
$O^{2-}$ ions are placed at the centre of each face, each shared by two unit cells; thus contributing 1/2 the mass each and total ion in a unit cell being 6/2 = 3.
At the centre of each cell is one $Ti^{4+}$ ion contained completely within a unit cell.
Thus, number of ions per unit cell will be,
n = (1 ion of Barium) + (3 ions of Oxygen) +(1 ion of Titanium)
So, n = 5