Question

1. The face-centered gold crystal has an edge length of 407 pm. Based on the unit cell, calculate the density of gold.

2. Gallium crystallizes in a primitive cubic unit cell. The length of an edge of this cube is 362 pm . What is the radius of a gallium atom?

Answer 1

(1) We know that density ,

d = (Z x atomic mass of gold ) / ( N_o x a ³ x 10 ^-30 cm ³ )    ---(1)

Where Z = No . of atoms in a unit cell = 4                    Since it is a fcc structure   

           atomic mass of gold = 197 g

           No = avagadro number = 6.023x10 ²³

           a = edge length = 407 pm

Plug these values in (1) we get

d = (4x197 ) / ( 6.023x10 ²³ x 407³x 10 ^-30 cm ³ )

   = 19.4 g / cm ³

(2) radius , r = edge length / 2

                    = 362 / 2 pm

                    = 181 pm