Density change Watch

Serendreamers
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Report Thread starter 5 years ago
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Pure iron is polymorphic and changes from a BCC to FCC lattice structure at 912 degrees. Calculate the density change. The lattice parameters are 0.293nm and 0.363nm respectively.

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I know that for BCC there are 2 atoms in a unit cell and there are 4 atoms for FCC. The first method calculated density as the volume of atoms present in a unit cell.

BCC: R= \frac {\sqrt 3}{4}a
FCC: R= \frac {\sqrt 2}{4}a

\rho_{BCC} = \frac {2 \cdot \frac{4}{3} \pi (\frac{\sqrt 3}{4}a)^3}{a^3} = 0.680

\rho_{FCC} = \frac {4 \cdot \frac{4}{3} \pi (\frac{\sqrt 2}{4}a)^3}{a^3} = 0.740

So the density change is about +8.8%

But if I use a different method by calculating density as the number of atoms per unit cell...

BCC: R= \frac {2}{(0.293 \cdot 10^-9)^3}
FCC: R= \frac {4}{(0.363 \cdot 10^-9)^3}

The density change is +5.16%

Which method is correct???
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EierVonSatan
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#2
Report 5 years ago
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Wow, A-level is really branching out huh :p:

I'm not sure I follow your first calculation - what is R referring to? the radius of an atom?

The density of a unit cell is given by \rho = \frac{ZM_r}{Na^3}, where Z is the number of atoms per cell.

So the ratio should be =\frac{4}{2} \times \frac{0.293^3}{0.363^3} = 1.0517...
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