Match the following columns.
 

	Column I (Radius of unit cell)		Column II (Unit cell)
A	$\frac{\mathrm{a}}{2\sqrt{2}}$	1
B	$\frac{\sqrt{3}}{4}\mathrm{a}$	2
C	$\frac{\mathrm{a}}{2}$	3

$\begin{array}{l}\mathrm{A}\to 3;\mathrm{B}\to 1;\mathrm{C}\to 2\\ [\because 2 \mathrm{is} \mathrm{simple} \mathrm{cubic},3 \mathrm{is} \mathrm{fcc} \mathrm{and} 1 \mathrm{is} \mathrm{bcc}]\mathrm{. }\end{array}$

(1) For bcc:

In body centered cubic unit cell, one atom is located at body center apart from corners of the cube. Let us take a unit cell of edge length “a”. Length of body diagonal, c can be calculated with help of Pythagoras theorem,

In $∆\mathrm{AFD}$, $c^2=a^2+b^2$

Where b is the length of face diagonal, thus b = $\sqrt{2}a$

   $c^2=3a^2$ ; c = $\sqrt{3}$ a

From the figure, radius of the sphere, r = 1/4 × length of body diagonal, c

 $\Rightarrow \mathrm{r}=\frac{\mathrm{c}}{4}$ = $\frac{\sqrt{3}}{4}a$ ; a = $\frac{4}{\sqrt{3}}$ r

(2) For simple cubic lattice:

In simple cubic unit cell, atoms are located at the corners of cube. Let us take a unit cell of edge length “a”. Radius of atom can be given as,

r = $\frac{a}{2}$ ; a = 2r

(3)  For ccp :

Hexagonal close packing (hcp) and cubic close packing (ccp) have same packing efficiency. Let us take a unit cell of edge length “a”. Length of face diagonal, b can be calculated with the help of Pythagoras theorem,

In $∆\mathrm{ABC}$,  ${\mathrm{b}}^2={\mathrm{a}}^2+{\mathrm{a}}^2$

The radius of the sphere is r
The face diagonal (b) = r + 2r + r = 4r

$\therefore (4r{)}^2=a^2+a^2$

$\Rightarrow (4r{)}^2=2a^2$$\Rightarrow a=\sqrt{\frac{16r^2}{2}}$ = $2\sqrt{2}\mathrm{r}$