JEE Main 2023 — Solid State Question with Solution

In a face-centered cubic (FCC) unit cell, atoms are present at the corners as well as at the centers of the faces. Hence, the diagonal of the face of the unit cell is equal to 4 times the radius of an atom.

This gives us the equation:

$$\sqrt{2}a=4r$$

Which simplifies to:

$$a=2\sqrt{2}r$$

In a body-centered cubic (BCC) unit cell, atoms are present at the corners and at the center of the unit cell. The body diagonal of the unit cell is equal to 4 times the radius of an atom.

This gives us the equation:

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

Which simplifies to:

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

Therefore, Option D is correct: $$2\sqrt{2}r=a$$ and $$4r=\sqrt{3}a$$