Gold crystallizes as fcc unit cell, the edge length of unit cell is 408 pm. What is the radius of gold atom?

Step-by-step solution

For a face-centered cubic (fcc) unit cell, the relationship between the edge length a and the radius of the atom r is given by: 4r = 2 a r = a 2 2 Given a = 408 pm: r = 408 2 2 = 204 2 = 102 2 pm Substituting 2 = 1.414 : r = 102 1.414 = 144.228 pm 144.2 pm Answer: 144.2 pm