Question

$\mathrm{Na}$ and $\mathrm{Mg}$ crystallize in Bcc and Fcc type of crystals respectively. What will be the number of atoms of $\mathrm{Na}$ and $\mathrm{Mg}$ present in the unit cell of their Respective crystal

Answer 1

Step 1 Number of atoms in $\mathrm{Na}$

$\mathrm{Na}$ crystallizes in BCC crystals.

The BCC cell consists of one atom at the center and eight atoms at the corners

In the BCC lattice, 8 atoms are present at 8 corners of the BCC unit cell.

So each atom contributes ${\frac{1}{8}}^{\mathrm{th}}$ to the unit cell,

The number of corner atoms in $\mathrm{Na}$ is $8\times \frac{1}{8}=1$

The number of center atoms in $\mathrm{Na}$ = 1

The total number of atoms in $\mathrm{Na}$ is $1+1=2$

Therefore, the number of atoms of $\mathrm{Na}$ present in the unit cell of their respective crystal will be 2.

Step 2 Number of atoms in $\mathrm{Mg}$

$\mathrm{Mg}$ crystallizes in FCC crystals.

The FCC cell consists of one atom at each of the six faces and eight atoms at eight corners.

In the FCC lattice, 8 atoms are present at 8 corners of the BCC unit cell.

So each atom contributes ${\frac{1}{8}}^{\mathrm{th}}$ to the unit cell,

The number of corner atoms in $\mathrm{Mg}$ is $8\times \frac{1}{8}=1$

In the FCC lattice, 6 atoms are present at 6 corners of the FCC unit cell.

So each atom contributes ${\frac{1}{2}}^{\mathrm{th}}$ to the unit cell.

Number of atoms present at the edges in $\mathrm{Mg}$ = $6\times \frac{1}{2}=3$

The total no of atoms(n) in $\mathrm{Mg}$ is $1+3=4$

Therefore, the number of atoms of $\mathrm{Mg}$ present in the unit cell of their respective crystal will be 4.