Question

A metallic element crystallizes into a lattice containing a sequence of layers of ABABAB ............ Any packing of spheres leaves out voids in the lattice. The percentage by volume of empty space of this is

• A. 26%
• B. 21%
• C. 18%
• D. 16 %

Answer 1

Answer: (A)

26%

Answer 2

The hexagonal base consists of six equilateral triangles, each with side 2r and altitude 2r sin 60°.

Hence, area of base = $6\left\lbrack \frac{1}{2}(2r)(2r\sin 60^{o}) \right\rbrack = 6\sqrt{3}.r^{2}$

The height of the hexagonal is twice the distance between closest packed layers.

The latter can be determined to a face centred cubic lattice with unit cell length a. In such a lattice, the distance between closest packed layers is one third of the body diagonal, i.e. $\frac{\sqrt{3}a}{3}$, Hence

$\text{ Height }(h) = 2\left\lbrack \frac{\sqrt{3}a}{3} \right\rbrack = \frac{2a}{\sqrt{3}}$

Now, in the face centred lattice, atoms touch one another along the face diagonal,

Thus, $4r = \sqrt{2}.a$

With this, the height of hexagonal becomes : $\text{ Height }(h) = \frac{2}{\sqrt{3}}\left\lbrack \frac{4r}{\sqrt{2}} \right\rbrack = \left\lbrack 4\frac{\sqrt{2}}{3} \right\rbrack.r$

Volume of hexagonal unit is, $V =$ (base area) × (height) $= (6\sqrt{3}r^{2})\left\lbrack \frac{4\sqrt{2}}{\sqrt{3}}.r \right\rbrack = 24\sqrt{2}.r^{3}$

In one hexagonal unit cell, there are 6 atoms as described below :

3 atoms in the central layer which exclusively belong to the unit cell.

1 atom from the centre of the base. There are two atoms of this type and each is shared between two hexagonal unit cells.

2 atoms from the corners. There are 12 such atoms and each is shared amongst six hexagonal unit cells.

Now, the volume occupied by atoms = $6\left\lbrack \frac{4}{3}\pi r^{3} \right\rbrack$

Fraction of volume occupied by atoms

$= \frac{\text{Volume occupied by atoms}}{\text{Volume of hexagonal unit cell}}$

$= \frac{6\left( \frac{4}{3}\pi r^{3} \right)}{24\sqrt{2}.r^{3}} = \pi/3\sqrt{2} = 0.74.$

Fraction of empty space = $(1.00 - 0.74) = 0.26$

Percentage of empty space = 26%