Is there an expansion or contraction as iron transforms from FCC to BCC? The atomic radius of iron i

Question

Is there an expansion or contraction as iron transforms from FCC to BCC? The atomic radius of iron is 125 pm in FCC but 50$\sqrt{2}$ pm in BCC.

Accepted Answer

$d_{FCC} = \frac{Z.M}{N_{A}.V} = \frac{4 \times 56}{\left( 6 \times 10^{23} \right)\left( \frac{4 \times 125 \times 10^{- 10}}{\sqrt{2}} \right)^{3}} = 8.45\text{gm/c}\text{m}^{3}$

$d_{BCC} = \frac{Z.M}{N_{A}.V} = \frac{2 \times 56}{\left( 6 \times 10^{23} \right)\left( \frac{4 \times 50\sqrt{2} \times 10^{- 10}}{\sqrt{3}} \right)^{3}} = 42.87\text{gm/c}\text{m}^{3}$

As the density of iron is increased, there is contraction.

Question

Options

Solution

More Solid State Questions