A certain metal can crystallise in simple cubic (sc), body-centred...

The relationship between the edge length $a and the atomic radius$r$for the three cubic lattices is as follows:<br>For simple cubic (sc):$a_1 = 2r$<br>For body-centred cubic (bcc):$\sqrt{3}a_2 = 4r \Rightarrow a_2 = \dfrac{4r}{\sqrt{3}}$<br>For face-centred cubic (fcc):$\sqrt{2}a_3 = 4r \Rightarrow a_3 = 2\sqrt{2}r$<br><br>Substitute these into the given expression$\dfrac{a_2 \cdot a_3}{a_1^2}$:<br>$\dfrac{a_2 \cdot a_3}{a_1^2} = \dfrac{\left(\dfrac{4r}{\sqrt{3}}\right) \cdot (2\sqrt{2}r)}{(2r)^2}$<br><br>$\dfrac{a_2 \cdot a_3}{a_1^2} = \dfrac{\dfrac{8\sqrt{2}r^2}{\sqrt{3}}}{4r^2} = \dfrac{2\sqrt{2}}{\sqrt{3}}$<br><br>Now, square this term:<br>\left( \dfrac{a_2 \cdot a_3}{a_1^2} \right)^2 = \left( \dfrac{2\sqrt{2}}{\sqrt{3}} \right)^2 = \dfrac{8}{3}$

Finally, multiply by 3:
$3 \times \dfrac{8}{3} = 8$

Answer: $8$