Question

A) Calculate the radius of a chromium (Cr) atom, knowing that chromium crystallizes in a body-centred cubic structure and has
a density of 7.14g/cm³?

B) Given that finding the radius of a face centered cubic structure can be found with l² + l² = (4r)², how do you find the radius of a cimple cubic structure?

Thank you if you can help!

Answer 1

Cr-metal = BCC

   a) No of atoms in BCC = 8(corners)*1/8 + 1(center) = 2

   b) mass of Cr atom in the cell = 2*56

           = 112 amu

       ( 1 amu = 1.66*10^-24 grams)

                     = 112*1.66*10^-24

                     = 1.86*10^-22 grams

   c) edge length(a) = x cm

          

       volume of unit cell = a^3

                    
e) density of solid = mass/volume

                7.14 = (1.86*10^-22)/(a^3)

            a = edge length = 2.964*10^-8 cm

   BCC edgelength(a) = sqrt(16/3)*r = 2.964*10^-8

                    r = radius = 1.283*10^-8 cm

B) for simple cubic edgelength(a) = 2r

                r = radius of the atom