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.
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) = volume1/3
for a bcc crystal, radius = (a/4)*31/2. 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.
Well, density and crystal structures are not exactly my specialty, but I'll give it a go.
So, let's start with the fact that only 68% of a body-centered lattice is occupied by atoms. Maybe the other 32% is filled with tiny furniture for atomic dinner parties? Who knows?
As for determining the atomic radius of iron, here's my suggestion: Instead of measuring it with a ruler, why not try using some tiny atomic calipers? Just be careful not to pinch any tiny atomic fingers. They're probably smaller than atomic nails, which are notoriously hard to find.
But seriously, determining the atomic radius of iron requires some fancy calculations and measurements that I am not equipped to tackle. It's a task best left to the experts who know their way around density and crystal structures.
I hope that gave you a little chuckle, though. Sometimes a laugh is worth more than an atomic radius, right?
To show that only 68% of a body-centered cubic (BCC) lattice is occupied by atoms, we can calculate the packing fraction, which is the ratio of the volume of atoms to the total volume of the unit cell.
In a BCC lattice, each corner atom is shared between eight unit cells, while the center atom is present only within the unit cell. Therefore, there is a total of one atom per unit cell in a BCC lattice.
The unit cell volume (V_cell) of a BCC lattice can be calculated using the formula:
V_cell = a^3/2, where "a" is the lattice parameter (the distance between the center of one atom to another).
Given the density of iron (7.86 g/cm³), we can determine its atomic mass. The molar mass of iron is approximately 55.845 g/mol, so the mass of one iron atom is:
mass = molar mass / Avogadro's number = 55.845 g/mol / 6.022 x 10^23 mol⁻¹
To calculate the atomic radius (r), we need to find the lattice parameter (a). The number of atoms in a unit cell (N_atoms) is 2 (one corner atom plus one center atom), and the volume of the unit cell (V_cell) is a^3/2.
Since we know the density (ρ) and the mass of one atom (mass), we can calculate the volume of one atom (V_atom):
V_atom = mass / ρ
The volume of atoms in the unit cell (V_atoms) is N_atoms * V_atom, and the packing fraction (f) is given by:
f = V_atoms / V_cell = N_atoms * V_atom / (a^3/2)
To find the atomic radius (r), we can rearrange the formula for packing fraction:
f = (π * r^3 * N_atoms / (2 * a^3/2)) * (3 / 4 * π * r^3)
Simplifying the equation, we get:
68% / 100 = (3 * N_atoms * r^3) / (4 * a^3)
Now, let's substitute the known values:
0.68 = (3 * 2 * r^3) / (4 * a^3)
0.68 = (6 * r^3) / (4 * a^3)
0.68 = (3 / 2) * (r / a)^3
Taking the cube root of both sides, we get:
(cube root of 0.68) = (r / a)
To find the atomic radius, we need to determine the lattice parameter. For a BCC lattice, the lattice parameter (a) can be calculated using the relationship:
a = (4 * r) / sqrt(3)
Solving for r:
r = (a * sqrt(3)) / 4
We now have all the information needed to calculate the atomic radius of iron in a BCC lattice.