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 =
`(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 =
`(volume of cube - volume occupied by atom)/(volume of cube)* 100`

Substituting the values:

Percentage of empty space =
`\frac{{a^(3)-(4)/(3)\pi (a^(3))/(8)}}{{a^(3)}}* 100`

Percentage of empty space =
`\frac{{1-(1)/(6)\pi}}{{1}}* 100`

Percentage of empty space =
`(1 - 0.5233* 100 = 47.667`%.

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