Question

Using the given data, find the type of cubic lattice for Fe crystal.

• A. BCC
• B. FCC
• C. SC
• D. HCP

Answer 1

Answer: (A)

BCC

We know, $\rho =\frac{n\times \text{Atomic mass}}{N_0\times a^3}$

or $n=\frac{\rho \times N_0\times a^3}{\text{Atomic mass}}$
For $Fe$: $n=\frac{7.86\times 6.023\times {10}^{23}\times {(286\times {10}^{-10})}^3}{55.85}=2$

Therefore, the cubic lattice of $Fe$ is body centred.
For $V$: $n=\frac{5.96\times 6.023\times {10}^{23}\times {(301\times {10}^{-10})}^3}{50.94}=2$

Therefore, the cubic lattice of $V$ is body centred.
For $Pd$: $n=\frac{12.16\times 6.023\times {10}^{23}\times {(388\times {10}^{-10})}^3}{106.4}=4$

Therefore, the cubic lattice of $Pd$ is face centred.
For $Al$: $n=\frac{2.70\times 6.023\times {10}^{23}\times {(405\times {10}^{-10})}^3}{26.98}=4$

Therefore, the cubic lattice of $\text{Fe}$ is body centred.

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