An element has bcc structure with atomic radius 200 pm. Find the volume of unit cell.

Step-by-step solution

For a bcc unit cell, the relation between edge length a and atomic radius r is given by: 3 a = 4r a = 4r 3 Substituting r = 200 pm = 2 10⁻⁸ cm : a = 4 2 10⁻⁸ 1.732 4.618 10⁻⁸ cm Volume of the unit cell, V = a^3 V = (4.618 10⁻⁸)^3 9.848 10⁻²³ cm ^3 Answer: 9.848 10⁻²³ cm ^3