Question

The lattice constant of a BCC unit cell with an atomic radius of 1.24 Å is

• A. 1.432
• B. 2.864
• C. 1.754
• D. 1.432

Hint:

For a BCC unit cell, the lattice constant is related to the atomic radius by \( a = \frac{4r}{\sqrt{3}} \). For an FCC unit cell, it is \( a = \frac{2\sqrt{2}r}{1} \).

Answer 1

The Correct Option is B

In a Body-Centered Cubic (BCC) structure, the relationship between the lattice constant (\( a \)) and the atomic radius (\( r \)) is given by: \[ a = \frac{4r}{\sqrt{3}} \] where: - \( a \) is the lattice constant, - \( r \) is the atomic radius, - The factor \( \frac{4}{\sqrt{3}} \) comes from the geometry of a BCC unit cell.
Step 1: Substituting the given atomic radius Given \( r = 1.24 \) Å, \[ a = \frac{4 \times 1.24}{\sqrt{3}} \]


Step 2: Calculating the lattice constant \[ a = \frac{4.96}{1.732} \] \[ a \approx 2.864 \, \text{Å} \]


Step 3: Evaluating the Options - Option (A) - Incorrect: 1.432 Å is not the correct calculation. - Option (B) - Correct: 2.864 Å matches our calculation. - Option (C) - Incorrect: 1.754 Å is not derived from the formula. - Option (D) - Incorrect: 1.432 Å is repeated and incorrect.


Step 4: Conclusion Since the calculated lattice constant is 2.864 Å, the correct answer is option (B).