Question

Iron occurs as BCC as well as FCCunit cell. If the effective radius of an atom of iron is $124pm$, compute the density of iron in both these structures.

• A. $\text{Density in BCC}=7.886g/cc$
  $\text{Density in FCC}=8.89g/cc$
• B. $\text{Density in FCC}=7.886g/cc$
  $\text{Density in BCC}=8.89g/cc$
• C. $\text{Density in FCC}=14.886g/cc$
  $\text{Density in BCC}=8.89g/cc$
• D. $\text{Density in FCC}=7.886g/cc$
  $\text{Density in BCC}=14.86g/cc$

Answer 1

The correct option is A $\text{Density in BCC}=7.886g/cc$

$\text{Density in FCC}=8.89g/cc$
In a body centred unit cell, atoms touch each other along the cross diagonal.

Hence, $4r=\sqrt{3}a$
$\therefore a=\frac{4}{\sqrt{3}}r=\frac{4}{\sqrt{3}}\left(124pm\right)=286.4pm$
For BCC lattice, $Z=2$

$d_{BCC}=\frac{ZM}{a^3{N}_A}=\frac{2\times 55.8}{(286.4\times {10}^{-12}{)}^3\times 6.023\times {10}^{23}}=7.887{10}^{6}g/m^3=7.886g/cc$

In a face centred unit cell, atoms touch each other along the face diagonal.

Hence, $4r=\sqrt{2}a$
$\therefore a=\frac{4}{\sqrt{2}}r=\frac{4}{\sqrt{2}}\left(124pm\right)=350.7p{m}^3$

For FCC lattice, $Z=4$
$d_{FCC}=\frac{ZM}{a^3{N}_A}=\frac{2\times 55.8}{(350.7\times {10}^{-12}{)}^3\times 6.023\times {10}^{23}}=8.89{10}^{6}g/m^3=8.89g/cc$