Question

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

• A. Simple cubic
• B. Body-centered cubic
• C. Hexagonal closed packing
• D. Face-centered cubic

Answer 1

The correct option is D Face-centered cubic

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 $\text{Al}$ is face centred.

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