# Density of Crystal

**Theoretical density of the crystal**

If the length of edge of the unit cell is known we can calculate the density of the crystal as follows:

Let length of edge of the unit cell be l.

Volume of the unit cell =

Density of the unit cell =

Mass number of the unit cell = number of atoms in a unit cell × Mass of one atom

= n × mg

But mass of one atom(m) =

Mass of the unit cell Number of atoms in a unit cell Mass of one atom

**Example 1**. An element (atomic mass = 60) having face-centred cubic crystal has a density of . What is the edge length of the unit cell. (Avogadro constant,).

Solution: since element has fcc structure hence there are 4 atoms in a unit cell (or n = 4). Atomic mass is 60 (or M = 60)

Let ‘l’ be the length of the edge of the unit cell.

**Example 2**. The density of KBr is . The length of the edge of the unit cell is 654 pm. Show that KBr has a face centred cubic structure.

*Solution,*

Length of the edge of the unit cell = 654 pm =

Volume (V) of the unit cell =

Molecular mass of KBr = 39 + 80 =

Density of KBr =

[/latex]

Since number of atoms (or ions) in a unit cell is four hence the crystal must be **face centred cubic.**

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