Question:
Solid A forms bcc lattice, Solid B forms fcc lattice. If the unit cell lengths (a) of A and B are 4 and 5 Å respectively, what is the ratio of the radii of A and B?
Solid A forms bcc lattice, Solid B forms fcc lattice. If the unit cell lengths (a) of A and B are 4 and 5 Å respectively, what is the ratio of the radii of A and B?
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For cubic lattices: bcc: \(4R = \sqrt{3}a\), fcc: \(4R = \sqrt{2}a\). Always use these formulas to find atomic radius from unit cell length.
Updated On: Jun 19, 2026
- \(5:4\sqrt{6}\)
- \(5:2\sqrt{6}\)
- \(4\sqrt{6}:5\)
- \(2\sqrt{6}:5\)
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The Correct Option is D
Solution and Explanation
Step 1: Relation between atomic radius and unit cell length for bcc.
For a bcc lattice, \[ 4R = \sqrt{3} \, a \] where \(R\) is the atomic radius and \(a\) is the unit cell length.
For Solid A (bcc) with \(a_A = 4\) Å: \[ R_A = \frac{\sqrt{3}}{4} \cdot 4 = \sqrt{3} \, \text{Å} \]
Step 2: Relation between atomic radius and unit cell length for fcc.
For an fcc lattice, \[ 4R = \sqrt{2} \, a \] For Solid B (fcc) with \(a_B = 5\) Å: \[ R_B = \frac{\sqrt{2}}{2} \cdot 5 = \frac{5\sqrt{2}}{2} \, \text{Å} \]
Step 3: Ratio of radii \(R_A : R_B\).
\[ R_A : R_B = \sqrt{3} : \frac{5\sqrt{2}}{2} = \frac{2\sqrt{3}}{5\sqrt{2}} = \frac{2\sqrt{6}}{5} \]
Step 4: Final conclusion.
Hence, the ratio of the radii of A and B is: \[ \boxed{2\sqrt{6}:5} \]
For a bcc lattice, \[ 4R = \sqrt{3} \, a \] where \(R\) is the atomic radius and \(a\) is the unit cell length.
For Solid A (bcc) with \(a_A = 4\) Å: \[ R_A = \frac{\sqrt{3}}{4} \cdot 4 = \sqrt{3} \, \text{Å} \]
Step 2: Relation between atomic radius and unit cell length for fcc.
For an fcc lattice, \[ 4R = \sqrt{2} \, a \] For Solid B (fcc) with \(a_B = 5\) Å: \[ R_B = \frac{\sqrt{2}}{2} \cdot 5 = \frac{5\sqrt{2}}{2} \, \text{Å} \]
Step 3: Ratio of radii \(R_A : R_B\).
\[ R_A : R_B = \sqrt{3} : \frac{5\sqrt{2}}{2} = \frac{2\sqrt{3}}{5\sqrt{2}} = \frac{2\sqrt{6}}{5} \]
Step 4: Final conclusion.
Hence, the ratio of the radii of A and B is: \[ \boxed{2\sqrt{6}:5} \]
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