If the volume of fcc unit cell is 6.4 10⁻²³ cm ^3 find the radius of an atom in it.

Step-by-step solution

Volume of the unit cell, V = a^3 = 6.4 10⁻²³ cm ^3 a^3 = 64 10⁻²⁴ cm ^3 a = (64 10⁻²⁴)^ 1/3 cm = 4 10⁻⁸ cm Converting to picometers: a = 4 10⁻⁸ 10¹⁰ pm = 400 pm For an fcc unit cell, the relation between edge length a and radius r is given by: 4r = 2 a r = 2 a 4 = a 2 2 r = 400 2 2 = 100 2 pm r = 100 1.414 = 141.4 pm Answer: 141.4 pm