Question

An element crystallizes in bcc lattice. The atomic radius of the element is 2.598 \AA{. What is the edge length (in \AA{}) of the unit cell?}

• A. 4
• B. 8
• C. 6
• D. 5

Hint:

For BCC structures, remember the key formula: $a = \frac{4r}{\sqrt{3}}$. For FCC, use $a = \frac{2\sqrt{2}r}{}$. These are crucial shortcuts for unit cell geometry problems.

Answer 1

The Correct Option is C

Step 1: Use the BCC relation between atomic radius and edge length.
In a body-centered cubic (bcc) structure: \[ a = \frac{4r}{\sqrt{3}} \] Step 2: Substitute the given atomic radius.
Given: $r = 2.598$ \AA{} \[ a = \frac{4 \times 2.598}{\sqrt{3}} = \frac{10.392}{1.732} \approx 6.0 \text{ \AA} \] Step 3: Final Answer
Edge length = 6 \AA