Question

A solid forms fcc unit cell with \( 2\text{A} \) side length. If another solid with exactly same atomic radius (\( r \)) as that of the above solid forms a simple cubic lattice, the side length of the unit cell will be:

• A. \( \sqrt{2} \, \text{A} \)
• B. \( 2 \, \text{A} \)
• C. \( \frac{2}{\sqrt{2}} \, \text{A} \)
• D. \( \frac{1}{\sqrt{2}} \, \text{A} \)

Hint:

For different lattice types, use the known relations between atomic radius and side length. For fcc: \( a = 2\sqrt{2}r \), and for simple cubic: \( a = 2r \).

Answer 1

The Correct Option is A

For a solid with face-centered cubic (fcc) structure, the relation between the side length \( a \) and atomic radius \( r \) is: \[ a = 2\sqrt{2}r. \] For the solid with simple cubic structure, the relation between the side length \( a' \) and atomic radius \( r \) is: \[ a' = 2r. \] Now, let the side length of the simple cubic unit cell be \( a' \). Given the relation for the fcc unit cell side length \( a = 2\text{A} \), we can solve for the side length \( a' \) of the simple cubic unit cell: \[ 2\sqrt{2}r = 4r \quad \Rightarrow \quad a' = \sqrt{2} \, \text{A}. \] Thus, the side length of the unit cell for the simple cubic lattice will be \( \sqrt{2} \, \text{A} \).