Q8) An element having atomic mass 60 has FCC unit cells .the edge length of the unit cell is 400pm .find the density of the element

Unit-cell edge length = 400 pm = 400×10^-10 cm

Volume of unit cell = (400×10^-10cm)³ = 64×10^-24 cm³

Mass of one atom = Atomic mass / Avogadro number ⇒ 60 / (6.023×10²³)

The unit cell, contains 4 atoms, So, Number of atoms in the fcc unit cell = 4

∴ Mass of the unit cell = 4×60 / (6.023×10²³)

∴ Density of unit cell = Mass of unit cell / Volume of unit cell

4 x 60 1
⇒ ----------------- x ----------
6.023× 10²³ 64× 10²⁴

⇒ 6.2 gcm−3 Ans.