The crystalline form of iron, known as α–Fe, has a body centered cubic (bcc) unit cell with an edge — Inorganic Chemistry — Solid State Chemistry Question

Let R be the atomic radius of iron and a = 2.87 Å the length of the unit cell edge. Then, as atoms touch each other along the body diagonal and from a Pythagorean theorem in the cube:

3a = 4R ⇒ R = 1.24 Å
The Avogadro number NA can be calculated from the density (ρ) formula. The latter is obtained by finding the number of atoms per unit cell, multiplying this number by the mass of each atom and, eventually, dividing the result by the volume of the unit cell (a₃). Note that each bbc unit cell contains two whole spheres, that is 2 Fe atoms.
ρ = (2 × 55.847 / NA) / a^3 ⇒ NA = (2 × 55.847 g mol^-1) / (7.86 g cm^-3 × (2.87 × 10^-8)^3 cm^3) = 6.01 × 10^23 mol^-1

By applying the Pythagorean theorem in the cube, one finds:
a^2 + a^2 = (4R)^2 ⇒ R = 1.27 Å

(slightly different from the value found above for bcc structure, because of the different packing, having an influence on the atomic radius or at its estimation).
As for the density, recalling that each fcc unit cell contains four whole spheres, that is 4 Fe atoms, once again one has:
ρ = (4 × 55.847 g mol^-1) / (6.023 × 10^23 mol^-1 × (3.59 × 10^-8)^3 cm^3) = 8.02 g cm^-3
The higher value of γ–Fe density, as compared with α–Fe, points at the fact that the fcc structure is denser than bcc. fcc represents the, so called, cubic close packed structure which, together with the hexagonal close packed, are the most efficient ways of packing together equal sized spheres in three dimensions.

The unit cells below are illustrated by using reduced size spheres. Note that, in hard spheres packing model the represented atoms must be in contact one to each other.
According to the left figure, a perfectly fitted interstitial atom centered at (½, 0, ½) in an α–Fe cell, would have a radius of:
Rinterstitial = ½ a – RFe where a = 2.87 Å and RFe = 1.24 Å (see question 8.1)
Therefore Rinterstitial (α–Fe) = 0.20 Å

Similarly, according to the figure in right, a perfectly fitted interstitial atom centered at (½, ½, ½) in an γ–Fe cell, would have a radius of:
Rinterstitial = ½ a – RFe where a = 3.59 Å and RFe = 1.27 Å (see question 8.2)
Therefore Rinterstitial (γ–Fe) = 0.53 Å

1 nm = 10 Å. Thus:
For α-Fe: R_carbon / R_interstitial = 0.77 Å / 0.20 Å = 3.85
For γ-Fe: R_carbon / R_interstitial = 0.77 Å / 0.53 Å = 1.45
Therefore, the carbon atom is roughly four times too large to fit next to the nearest iron atoms in α–Fe without strain, while it is only 1.5 times oversize to fit in the γ–Fe structure. The above estimations explain well the low solubility of carbon in α–Fe (< 0.1 %).

The wavelength (λ) of the X–rays will be calculated from Bragg's law, assuming first order diffraction: 2d sinθ = λ, where θ is the angle of diffraction equal to 32.6° and d is the interplanar spacing of the (200) set of parallel lattice planes, that is, the perpendicular distance between any pair of adjacent planes in the set. The (200) planes are shown shaded in the figure.

Let a be a length of the cubic unit cell edge.
Then from previous data for α–Fe:
a = 2.87 Å so the distance between adjacent (200) planes is d = a/2 = 1.44 Å.
Therefore from Bragg’s law: λ = 2d sinθ = 2 × 1.44 × sin(32.6°) ⇒ λ = 1.55 Å.
This value corresponds to the Kα1 transition of iron.