Question

Iron occurs as bcc as well as fcc unit cell. If the effective radius of an atom of iron is $124$ pm, compute the density (in g cm${}^{-3}$) of iron in bcc and fcc respectively :

• A. $8.4,9.8$
• B. $9.3,4.6$
• C. $8.6,7.9$
• D. $7.9,8.6$

Answer 1

The correct option is D $7.9,8.6$

The expression for density is as follows:

Density $=d=\frac{zM}{V{N}_o}$

For bcc, $z=2$

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

So, density $=\frac{2\times 56}{a^3(6\times {10}^{23})}=7.9$ g cm${}^{-3}$

For fcc, $n=4$

$a=2\sqrt{2}r$

So, density $=\frac{4\times 56}{a^3(6\times {10}^{23})}=8.6$ g cm${}^{-3}$

Hence, the density of iron in bcc and fcc are $7.9$ g cm${}^{-3}$ and $8.6$ g cm${}^{-3}$ respectively.

Hence, the correct option is $\text{D}$