Planar and linear density of BCC iron

Planar density = atoms centred on the plane per unit area of the plane; linear density = atoms centred on the line per unit length.

(a) Planar density

(100) plane: a square of side $a$. Four corner atoms each contribute $\tfrac14$, giving 1 atom.

$$\text{PD}_{(100)} = \frac{1}{a^2} = \frac{1}{(0.2866)^2} = 12.17\ \text{atoms/nm}^2$$

(110) plane: a rectangle $a \times \sqrt{2}\,a$. Four corner atoms ($4\times\tfrac14 = 1$) plus the body-centre atom, which lies on this plane, give 2 atoms.

$$\text{PD}_{(110)} = \frac{2}{\sqrt{2}\,a^2} = \frac{2}{1.4142 \times 0.08214} = 17.22\ \text{atoms/nm}^2$$

(b) Linear density

For [111], the body diagonal passes through two corner atoms and the centre atom, so 2 atoms are counted in the repeat length $\sqrt{3}\,a$.

Close-packed plane and direction

The densest plane is (110) (17.22 atoms/nm$^2$) and the densest direction is [111] (4.03 atoms/nm). Slip needs the least shear stress on close-packed planes and along close-packed directions, because the atomic spacing along such a direction is smallest (small Burgers vector) and such planes are the most widely spaced. Hence in BCC metals slip occurs on {110}<111> systems (and also on {112} and {123} planes).

Answer: PD(100) = 12.17 atoms/nm², PD(110) = 17.22 atoms/nm²; LD[100] = 3.49, LD[110] = 2.47, LD[111] = 4.03 atoms/nm; close-packed plane (110), direction [111].