Fe crystallizes in FCC lattice at 750⁰C and at higher temperature (above 1100⁰C) converts to a BCC system. Calculate the ratio of " fraction of edge not covered by atoms in case of FCC " to the " fraction of edge not covered in case of BCC ".
Assume no change in atomic radius of Fe. ( Take $2$ = 1.41; $3$ = 1.73)

Question

Fe crystallizes in FCC lattice at 750⁰C and at higher temperature (above 1100⁰C) converts to a BCC system. Calculate the ratio of " fraction of edge not covered by atoms in case of FCC " to the " fraction of edge not covered in case of BCC ".

Assume no change in atomic radius of Fe. ( Take $2$ = 1.41; $3$ = 1.73)

TestHub · Accepted Answer

Fraction of edge not covered by atoms in a unit cell of edge length "a" and radius "r" is given as

$\frac{( a - 2 r )}{a} = 1 - 2 \times \frac{r}{a}$

For FCC;

$\displaylines 2 a = 4 r \frac{r}{a} = \frac{1}{2 2}$

For BCC

$\displaylines 3 a = 4 r \frac{r}{a} = \frac{3}{4}$

Ratio of fraction of edge not covered by atoms in FCC to BCC is

$\frac{( 1 - \frac{1}{2} )}{( 1 - \frac{3}{2} )} = \frac{( 2 - 2 )}{( 2 - 3 )} = 2.18$

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