Question

Iron crystallizes in a bcc structure, with a lattice constant of 2.8685 Å. a)Find the atom density (number of atoms per cm2) for iron in the (100) plane b) Find the atom density for iron in the (011) plane  

Answer 1

In BCC atoms occupy all corners and one atom at body center of cube.

100 plane is one side of unit cell so we have 4 atoms occupied at corners ,

Number of atoms per unit cell for 100 plane = ( 4 x 1/2) = 2 ( since each corner contributes 1/2 in 100)

Given edge length = 2.8685 A = ( 2.8685 x 10^-8) cm

Area of 100 phase of one unit cell = a^2 = ( 2.8685 x 10^-8)^2 = 8.2283 x 10^ -16 cm^2

So for are of 8.2283 x 10^ -16 cm^2 we have 2 atom

For 1cm2 number of atoms = ( 2) x ( 1 /8.2283x10^-16) = 2.43 x 10^14

b)   011 plane is body dioganol plane of BCC, number of atoms per 1 unit cell in 011 plane

is   ( 4 x 1/8) + 1x 1/2 = 1 atom per 110 plane in BCC

area of 110 phase = a x ( 1.732 a)   = ( 2.8685 x 10^ -8) x ( 1.732 x 2.8685x10^-8)

        = 14.25 x 10^ -16 cm^2

1 atom per 14.25 x 10^ -16 cm^2

per 1 cm^2 number of atoms = ( 1/ 14.25x10^-16 cm^2) = 7.0175 x 10^14