Question

Aluminium (a.w =27) crystallized in a cubic unit cell with edge length (a) = 100 pm, with density ‘d’ = 180 $g/c{{m}^{3}}$, then type of unit cell is:
A. SCC
B. BCC
C. FCC
D. HCP

Answer 1

Based on the number of unit cells in the crystal we can find the type of unit cell is present in the crystal.
There is a formula to calculate the number of unit cells present and it is as follows.
$\rho =\dfrac{Z\times M}{{{N}_{A}}\times {{a}^{3}}}$
Where $\rho$ = Density of the crystal
Z = number of unit cells in an atom
M = Molecular weight
${{N}_{A}}$ = Avogadro's number
a = Edge length of unit cell

Complete step by step answer:
- In the question it is given that Molecular weight of Aluminium = 27 g/mole
Edge length of unit cell = 100 pm = $100\times {{10}^{-10}}cm={{10}^{-8}}cm$ (1 pm = ${{10}^{-10}}cm$)
Density = 180 $g/c{{m}^{3}}$
- Substitute all the above given values in the following equation to get the number of unit cells present in the crystal.

& \rho =\dfrac{Z\times M}{{{N}_{A}}\times {{a}^{3}}} \\\ & Z=\dfrac{\rho \times {{N}_{A}}\times {{a}^{3}}}{M} \\\ & \Rightarrow Z=\dfrac{180(6.022\times {{10}^{23}})({{10}^{-8}})}{27} \\\ & \Rightarrow Z=4.0166\approx 4 \\\ \end{aligned}$$ \- The number of atoms per unit cell from the above calculations is 4, Means Z is 4. \- Therefore the number of atoms per unit cell is 4 and the type of unit cell is Face centered cubic (FCC). **So, the correct answer is “Option C”.** **Note:** In FCC (Face centered cubic) the atoms per unit cell is 4, in BCC (Body centered cubic) the atoms per unit cell are 2, in SCC (Simple closest packing) the atoms per unit cell is 1 and in HCP (Hexagonal closest packing) the atoms in per unit cell is 6.