# Lattice Constant of Iron

## Homework Statement

The density of bcc iron is 7900 kg/m3, and its atomic wieght is 56 amu. Using this information, calculate the lattice constant of iron's cubic unit cell and the interatomic spacing (i.e. nearest neighbor distance).

## The Attempt at a Solution

I thought this was mostly unit conversion, with the only relevant knowledge being that there are 2 atoms per cubic unit cell.

7900 kg/m^3 = 4.757*10^30 amu/m^3
4.757*10^30 amu/m^3 = 8.49552*10^28 atoms of iron / m^3
8.49552*10^28 atoms of iron / m^3 = 4.396057*10^9 atoms of iron / m
4.396057*10^9 atoms of iron / m = 0.439606 atoms of iron / angstrom

Taking the reciprocal of the final answer should get us the space between atoms, which gives me 2.27 angstroms between atoms of iron. I thought this should be the lattice constant, however I'm guessing I've done something wrong since I believe the answer should be 2.86 angstroms.

Any help would be appreciated. Thanks.

$$\frac{112 AMU}{a^3}$$