A certain element crystallises in a body-centred cubic (bcc) lattice. The mass of 5 10²³ atoms of this element is 27 g . If the density of the crystal is 4.0 g cm ⁻³ , the edge length of the unit cell is ______ pm.

Step-by-step solution

For a body-centred cubic (bcc) lattice, the number of atoms per unit cell, Z = 2 . The mass of 5 10²³ atoms is 27 g . Mass of one atom, m = 27 5 10²³ = 5.4 10⁻²³ g . The density of the crystal is given by: = Z m a^3 Substituting the given values: 4.0 = 2 5.4 10⁻²³ a^3 a^3 = 10.8 10⁻²³ 4.0 = 2.7 10⁻²³ cm ^3 a^3 = 27 10⁻²⁴ cm ^3 Taking the cube root on both sides: a = 3 10⁻⁸ cm Converting to picometers ( 1 cm = 10¹⁰ pm ): a = 3 10⁻⁸ 10¹⁰ pm = 300 pm . Answer: 300