Question

How much volume of the simple cubic unit cell is empty space (i.e. not filled by the atoms)?

Answer 1

For simple cubic unit cell:

Let the edge length = a

Such that a = 2r  (where r is the radius of the atom)

Then radius of the atom = $\frac{a}{2}$

Volume of the unit cell = $a^{3}$

Number of atoms in simple cubic unit cell = 1.

Therefore, empty space = volume of cube - volume occupied by atom

Percentage of empty space = $\frac{volume of cube - volume occupied by atom}{volume of cube}\times 100$

Substituting the values:

Percentage of empty space = $\frac{{a^{3}-\frac{4}{3}\pi \frac{a^{3}}{8}}}{{a^{3}}}\times 100$

Percentage of empty space = $\frac{{1-\frac{1}{6}\pi}}{{1}}\times 100$

Percentage of empty space = $(1 - 0.5233\times 100 = 47.667$%.

Hence, $47.667$% is the empty space of the simple cubic unit cell.