Question

Iron can exist in several crystal structures including BCC and FCC. Given the radius of iron is 126 pm for iron for the following conditions and draw the directions and/or planes:

Calculate the planar atom density for the following:

BCC lattice structure on (101)

FCC lattice structure on (101)

BCC lattice structure on (010)

FCC lattice structure on (010)

Answer 1

For BCC :
a* sqrt(3) = 4*r
a = 4*126/sqrt(3)
   = 291 pm
   = 2.91*10^-10 m

For FCC :
a* sqrt(2) = 4*r
a = 4*126/sqrt(2)
   = 356.4 pm
   = 3.564*10^-10 m

BCC lattice structure on (010) :
Number of atom : 4*1/4 = 1 atom
Area = a^2 = (2.91*10^-10)^2 = 8.47*10^-20 m^2
Planar density = number of atom / Area
                              = 1/8.47*10^-20
                              =1.18*10^19 atom/m^2

FCC lattice structure on (010) :
Number of atom : 4*1/4 + 1/2*1= 1.5 atom
Area = a^2 = (3.564*10^-10)^2 = 1.27*10^-19 m^2
Planar density = number of atom / Area
                              = 1.5/1.27*10^-19
                              =1.18*10^19 atom/m^2

BCC lattice structure on (101) :
Number of atom : 4*1/4 + 1= 2 atom
Area = a*a*sqrt(2) =1.4142* (2.91*10^-10)^2 = 1.20*10^-19 m^2
Planar density = number of atom / Area
                              = 2/1.20*10^-19
                              =1.67*10^19 atom/m^2

FCC lattice structure on (101) :
Number of atom : 4*1/4 + 2*1/2= 2 atom
Area = a*a*sqrt(2) =1.4142* (2.91*10^-10)^2 = 1.20*10^-19 m^2
Planar density = number of atom / Area
                              = 2/1.20*10^-19
                              =1.67*10^19 atom/m^2