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