If the lattice parameter for a crystalline structure is $3.6 ext{ \AA}$, then the atomic radius in fcc crysta

Question

If the lattice parameter for a crystalline structure is $3.6	ext{ \AA}$, then the atomic radius in fcc crystal is:

Accepted Answer

1. **Identify the Crystal Structure:** The problem specifies a face-centered cubic (fcc) crystal.
2. **Recall the Formula:** For an fcc lattice, the atoms touch each other along the face diagonal. The length of the face diagonal is $\sqrt{2}a$, where $a$ is the lattice parameter (edge length). Since there are three atoms along the face diagonal (one at each corner and one at the face center), the face diagonal equals $4r$, where $r$ is the atomic radius. Thus, the relationship is $4r = \sqrt{2}a$.
3. **Calculate Atomic Radius ($r$):**
   Rearranging the formula to solve for $r$ gives $r = \frac{\sqrt{2}a}{4} = \frac{a}{2\sqrt{2}}$.
   Substitute the given value $a = 3.6	ext{ \AA}$:
   $$r = \frac{3.6}{2 	imes 1.414} = \frac{3.6}{2.828} \approx 1.27	ext{ \AA}$$
4. **Conclusion:** The atomic radius in the fcc crystal is approximately $1.27	ext{ \AA}$.

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