Question

Iron at 20◦C is BCC with atoms of atomic radius 0.124 nm. The lattice constant a for the cube edge of the unit cell is

• A. a = 0.2864 nm
• B. a = 1.2864 nm
• C. a = 2.2864 nm
• D. a = 3.2864 nm

Hint:

In BCC, the relationship a = √4r/3 comes from geometry, as atoms touch along the cube diagonal. Memorize this for quick lattice constant calculations.

Answer 1

The Correct Option is A

For a BCC structure, the lattice constant \(a\) and atomic radius \(r\) are related as:
\[a = \frac{4r}{\sqrt{3}}\]
With \(r = 0.124 \, \text{nm}\), \(a\) becomes:
\[a = \frac{4 \times 0.124}{\sqrt{3}} \approx 0.2864 \, \text{nm}\]