Question

In a close-packed body-centred cubic lattice of potassium, the correct relation between the atomic radius $\left(r\right)$ of potassium and the edge-length $\left(a\right)$ of the cube is:

• A. $r=\frac{a}{\sqrt{2}}$
• B. $r=\frac{a}{\sqrt{3}}$
• C. $r=\frac{\sqrt{3}}{2}a$
• D. $r=\frac{\sqrt{3}}{4}a$

Answer 1

The correct option is D $r=\frac{\sqrt{3}}{4}a$

$r$ is the atomic radius and $a$ is the edge-length.

$AB$ is the face diagonal and has length $b$.

$AD$ is the body diagonal and has length $c$. It is equal to $4r$$

In right angled triangle $ABC$, $b^2=a^2+a^2=2a^2$

In right triangle $ABD$ $c^2=a^2+b^2=a^2+2a^2=3a^2$

Hence, $c^2=3a^2$

$c=\sqrt{3}a=4r$

$r=\frac{\sqrt{3}}{4}a$