Sodium has body-centered packing. The distance between the two nearest atoms is $3.7 ext{ \AA}$. The lattice

Question

Sodium has body-centered packing. The distance between the two nearest atoms is $3.7	ext{ \AA}$. The lattice parameter is:

Accepted Answer

1. **Identify the Given Data:**
   * Crystal structure: Body-centered cubic (bcc)
   * Nearest neighbor distance, $d = 3.7	ext{ \AA}$
2. **Formula for bcc Lattice:**
   * In a bcc lattice, the nearest neighbor distance $d$ is the distance between a corner atom and the atom at the body center. This distance is half of the body diagonal.
   * The relationship between the nearest neighbor distance $d$ and the lattice parameter $a$ (edge length) is given by $d = \frac{\sqrt{3}a}{2}$.
3. **Calculate Lattice Parameter ($a$):**
   * Rearranging the formula to solve for $a$: $a = \frac{2d}{\sqrt{3}}$
   * Substitute the given value: $a = \frac{2 	imes 3.7}{\sqrt{3}} = \frac{7.4}{1.732} \approx 4.27	ext{ \AA} \approx 4.3	ext{ \AA}$
4. **Conclusion:** The lattice parameter is approximately $4.3	ext{ \AA}$.

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