(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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).

2. Relevant equations

3. 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.

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# Lattice Constant of Iron

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