Question
Iron has a density of 7.86 g/cm3 and crystallines in a body-centered cubic lattice. Show that only 68% of a body-centered lattice is actually occupied by atoms, and determine the atomic radius of iron.
Answers
bcc means 2 atoms per unit cell.
mass of unit cell is
2*55.847/6.022 x 10^23 = ?? g/unit cell.
mass = volume x density or
volume = mass/density = ??g/7.86 g/cc = xx cc
a (the edge length) = volume<sup>1/3</sup>
for a bcc crystal, radius = (a/4)*3<sup>1/2</sup>. Check my thinking. I obtained 2.87 x 10^-8 cm for the radius. Check my arithmetic and algebra.
mass of unit cell is
2*55.847/6.022 x 10^23 = ?? g/unit cell.
mass = volume x density or
volume = mass/density = ??g/7.86 g/cc = xx cc
a (the edge length) = volume<sup>1/3</sup>
for a bcc crystal, radius = (a/4)*3<sup>1/2</sup>. Check my thinking. I obtained 2.87 x 10^-8 cm for the radius. Check my arithmetic and algebra.
For the volume occupied, find the volume of the atoms (4/3)*pi*r^3 (you know r from the first calculation to be 1.24 x 10^-8 cm). Multiply that volume by 2 (because there are 2 atoms/unit cell), then divide by the volume of the unit cell (which I found to be 2.36 x 10^-23 cc) which gives 67.7% which rounds to 68% of the space actually occupied by the atoms. The other 32% is space.
I quoted radius above that obtained 2.86 x 10^-8 but that isn't what I found. I found radius of the iron atom to be 1.24 x 10^-8 cm.