Question

The effective radius of the iron atom is $1.42A^{\circ }$. It has FCC structure. Calculate its density :

[atomic mass of $Fe=56$ amu]

• A. $2.87$ g cm${}^{-3}$
• B. $11.48$ g cm${}^{-3}$
• C. $1.435$ g cm${}^{-3}$
• D. $5.74$ g cm${}^{-3}$

Answer 1

The correct option is D $5.74$ g cm${}^{-3}$

For fcc unit cell the radius of the atom is 0.3535 times the edge length

$r=0.3535a1.42=0.3535aa=4.017A^0=4.017\times {10}^{-8}cm$

A fcc unit cell contains 4 atoms per unit cell. Hence,
$Z=4$

The molar mass (M) of iron is 56 amu or 56 g/mol

$d=\frac{Z\times M}{N_A\times a^3}d=\frac{4\times 56}{6.023\times {10}^{23}\times (4.017\times {10}^{-8}{)}^3}d=5.74gc{m}^{-3}$