Q: The empty space in the body centred cubic lattice is

Explanation:
To understand this, let’s break it down step by step:

1. In a BCC lattice, there are atoms at each corner of the cube and one atom at the center of the cube.

2. Total number of atoms in one unit cell:
– Corner atoms: (1/8 × 8) = 1 atom (since each corner atom is shared by 8 unit cells)
– Center atom: 1 atom (fully within the unit cell)
– Total = 2 atoms per unit cell

3. Volume calculations:
– If ‘a’ is the edge length of the cube, the total volume of the unit cell = a³
– Each atom has a radius r, where r = (√3a)/4 (this comes from the fact that atoms touch along the body diagonal)
– Volume of each atom = (4/3)πr³

4. The actual space occupied by atoms:
– Volume occupied by 2 atoms = 2 × (4/3)πr³
– Substituting r = (√3a)/4
– Volume occupied = 2 × (4/3)π × ((√3a)/4)³
– This works out to about 0.68 × a³

5. Therefore:
– Fraction of space occupied by atoms = 0.68 or 68%
– Empty space (void fraction) = 1 – 0.68 = 0.32 or 32%

This is why option D (0.32) is the correct answer. The BCC structure has a packing efficiency of 68%, leaving 32% as empty space between the atoms.