Question

In a ccp type structure, if half of the face-centered atoms are removed, then percentage void in a unit cell is approximately:

• A. $54$%
• B. $46.25$%
• C. $63$%
• D. $37$%

Answer 1

Answer: (A)

$54$%

For CCP type structure, atoms are present at the corners as well as in the face centre.

So, the number of atoms present $=\frac{1}{8}\times 8+\frac{1}{2}\times 6=4$

So, when the half of the face-centred atoms are removed, the number of the atoms present $=\frac{1}{8}\times 8+\frac{1}{2}\times 3=\frac{5}{2}$

The volume of the cube $=a^3$ where a is the edge length.

The volume occupied by an atom $=\frac{4}{3}\pi r^3$ where r is the radius of atoms.

For a FCC lattice, $4r=\sqrt{2}a\Rightarrow a=2\sqrt{2}r$

Packing efficiency $=\frac{numberofatomspresent\times Volumeoccupiedbyoneatom}{Totalvolumeofcube}\times 100$

$=\frac{\frac{5}{2}+\frac{4}{3}\pi r^3}{a^3}\times 100$

$=\frac{\frac{5}{2}+\frac{4}{3}\pi r^3}{(2\sqrt{2}r{)}^3}\times 100$

$=46.4\%$

Hence the % of void $=100-46.4=53.6\approx 54\%$