Packing Efficiency in BCC Structures
In BCC structure, the atom at the centre is in touch with other two atoms which are diagonally arranged as shown in figure.
In △EFD,
b2 = a2 + a2 = 2a2
or, b = √2 a
In △AFD,
c2 = a2 + b2 = a2 + 2a2 = 3a2
or, c = √3 a
The length of the body diagonal c is equal to 4r, r is the radius of the sphere (atom). As all the three spheres along the diagonal touch
each other,
so, c = 4r
or, c = 4r = √3 a
or, a = 4r/√3
or, r = √3 a/4
We know that the total number of atoms associated with BCC unit cell is 2 , so, the volume (v) is
v = 2 × (4/3) πr3 = 8/3 πr3
And the volume (V) of the unit cell = a3 = (4r/√3)3 = 64r3/3√3
Now, the packing efficiency = (100 × v)/V = [(8/3 πr3)/(64/3√3) × r3 ] × 100
= (√3 π × 100)/8 = 68%
Therefore, 68% of unit cell is occupied by atoms and the rest 32% is empty space.
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