Question

ASSIGNMENT PROBLEMS
1. The atomic packing efficiency of a crystal is 68%, and the closest distance
of approach between its neighbouring atom is 2.62 A. Calculate density of
the crystal if the mass of its one atom is 50.04 amu. [Solution: 5995 kg/ml​

Answer 1

Given info : The atomic packing efficiency of a crystal is 68%, and the closest distance

of approach between its neighbouring atom is 2.62 A.

To find : The density of the crystal if the mass of its one atom is 50.04 amu.

solution : since the packing efficiency of a crystal is 68 %, the crystal must be body centered cubic lattice (BCC).

so, no of atoms per unit cell of BCC , Z = 2

closest distance of approach, r = 2.62 A°

in BCC atom,

2r = a√3

⇒2 × 2.62 A° = a√3

⇒a = 3.025A°

atomic weight , M = 50.04 amu = 50.04 × 10¯³ kg/mol

now density = ZM/Na³

= (2 × 50.04 × 10¯³)/{6.023 × 10²³ × (3.0253 × 10^-8 cm)³}

= (100.08 × 10¯³}/(6.023 × 10²³× 27.7× 10¯²⁴)

= 0.5998 × 10¯² Kg/cm³

= 5998 Kg/m³ ≈ 5995 kg/m³ [ as given in answer. you can assume because very little difference ]

Therefore the density of the crystal is 5995 kg/m³ [ approx]