Question

An element has a body centred cubic (bcc) structure with a cell edge of $288pm$. The atomic radius is :
(NEET 2020)

• A. $\frac{\sqrt{2}}{4}\times 288pm$
• B. $\frac{4}{\sqrt{2}}\times 288pm$
• C. $\frac{4}{\sqrt{3}}\times 288pm$
• D. $\frac{\sqrt{3}}{4}\times 288pm$

Answer 1

The correct option is D $\frac{\sqrt{3}}{4}\times 288pm$

In the bcc crytsal lattice, the atoms are present at corners of the cube and at the centres of the cube.

If the edge length of cell is 'a', its face diagonal will be $\sqrt{2}a$ and its body diagonal will $\sqrt{3}a$

Half the distance of body diagonal is $\frac{\sqrt{3}a}{2}$

Here, the corner atoms and the centre atoms are in contact along the body diagonal.

Thus, the distance between these two atom is 2r, where r is the radius of sphere.
$\therefore 2r=\frac{\sqrt{3}a}{2}$

$\Rightarrow r=\frac{\sqrt{3}}{4}\times 288pm$

Hence, option (c) is correct.