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Lattice Constant of Iron 
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#1
Jan2511, 07:05 PM

P: 118

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. 


#2
Jan2611, 11:09 AM

P: 118

Nevermind, I figured it out. In case anyone is interested, the density of a conventional cubic cell is going to be
[tex]\frac{112 AMU}{a^3}[/tex] since the bcc lattice has to lattice points in a conventional cell. This must be equated with 7900 Kg/m^3. After conversion through dimensional analysis, you solve for a, which is the lattice parameter. 


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