Question

Find the radius of a BCC molecule having an edge length of \( 2.0 \times 10^{-11} \) m.

• A. \( 1.0 \times 10^{-11} \) m
• B. \( 1.5 \times 10^{-11} \) m
• C. \( 2.0 \times 10^{-11} \) m
• D. \( 3.0 \times 10^{-11} \) m

Hint:

In a BCC unit cell, the body diagonal connects two corner atoms and the center atom. The relationship between the edge length and the radius of the atoms is given by \( r = \frac{\sqrt{3}}{4} \, a \).

Answer 1

The Correct Option is A

In a Body-Centered Cubic (BCC) unit cell, atoms are positioned at each corner and one at the cube's center. The distance between the centers of two atoms along the body diagonal is four times the atomic radius.
Step 1: Edge length-radius relationship Let the edge length of the BCC unit cell be \( a \). The body diagonal of a BCC unit cell, determined by the Pythagorean theorem, is \( \sqrt{3} \, a \). This diagonal traverses two atoms (center and corner), so it also equals \( 4r \), where \( r \) is the atomic radius. Therefore: \[ \sqrt{3} \, a = 4r \] 
Step 2: Radius calculation Rearranging to solve for \( r \): \[ r = \frac{\sqrt{3}}{4} \, a \] 
Step 3: Edge length substitution Given the edge length \( a = 2.0 \times 10^{-11} \, \text{m} \), substitute this value: \[ r = \frac{\sqrt{3}}{4} \times 2.0 \times 10^{-11} = 1.0 \times 10^{-11} \, \text{m} \] The radius of the BCC molecule is consequently \( {1.0 \times 10^{-11}} \, \text{m} \).