Fcc Lattice Interstitial Site Analysis

This solution determines the maximum radius ratio ($r/R$) for an atom that can fit into a specific interstitial site in an fcc lattice without causing distortion.

Lattice Geometry and Nearest Neighbours

• The crystal structure is face-centered cubic (fcc).
• Atoms are hard spheres with radius $R$.
• The lattice constant is $a$.
• In an fcc lattice, the nearest neighbour distance is $a/\sqrt{2}$.
• Since nearest neighbours touch, the distance between their centres is $2R$.
• Therefore, $2R = a/\sqrt{2}$, which implies $a = 2\sqrt{2}R$.

Interstitial Site and Constraints

• An interstitial site is located at coordinates $(0, a/2, 0)$.
• An atom with radius $r$ needs to fit into this site.
• The position $(0, a/2, 0)$ lies exactly halfway between the atom at the origin $(0,0,0)$ and the atom at $(0, a, 0)$.
• The distance from the origin $(0,0,0)$ to the interstitial site $(0, a/2, 0)$ is $a/2$.
• For the lattice not to be distorted, the interstitial atom must touch the surrounding lattice atoms without overlapping.
• The primary constraint comes from the atoms at $(0,0,0)$ and $(0, a, 0)$.
• The distance condition is: distance from origin centre to interstitial centre $\ge R + r$.
• So, $a/2 \ge R + r$.

Calculating Maximum Radius Ratio ($r/R$)

1. Substitute the relationship $a = 2\sqrt{2}R$ into the inequality: $ \frac{2\sqrt{2}R}{2} \ge R + r $
2. Simplify the expression: $ \sqrt{2}R \ge R + r $
3. Rearrange to find the condition for $r$: $ r \le \sqrt{2}R - R $ $ r \le (\sqrt{2} - 1)R $
4. Calculate the maximum ratio $r/R$: $ \frac{r}{R} \le \sqrt{2} - 1 $
5. Using the approximate value $\sqrt{2} \approx 1.414$: $ \frac{r}{R} \le 1.414 - 1 $ $ \frac{r}{R} \le 0.414 $

The maximum value of $r/R$ is approximately $0.414$. This value lies within the given range of 0.4 to 0.42.