Question

A metal of density 7.5 X 10^3 kg m^-3 has an fcc crystal structure with lattice parameter a = 400
pm. Calculate the number of unit cells present in 0.015 kg of the metal
(7.5x10 kg
m odi aried and foc psas gare are goilu Rarel a = 400 pm . 0.015kg​

Answer 1

Given:

Density, d = 7.5 × 10³ kg/m³ = 7.5 gm/cm³

Edge length, a = 400 pm = 4 × 10⁻⁸ cm

Mass, M = 0.015 kg = 15 gm

To Find:

The number of unit cells present in 0.015 kg of the metal.

Calculation:

- The volume of 1 unit cell, V = a³

⇒ V = (4 × 10⁻⁸)³

⇒ V = 6.4 × 10⁻²³ cm³

- Now, calculate the volume in the given mass:

d = M/V'

⇒ V' = M/d

⇒ V' = 15/7.5 = 2 cm³

- No of unit cells =  V'/V

⇒ N = 2/6.4 × 10⁻²³

⇒ N = 0.3125 × 10²³

⇒ N = 3.125 × 10²²

- So, the number of unit cells present in 0.015 kg of the metal is 3.125 × 10²².