The lattice parameter of face-centered cubic iron ($\gamma$-Fe) is 0.3571 nm. The radius (in nm) of the octahedral void in $\gamma$-Fe is _______________
We need to find the radius of the octahedral void in face-centered cubic (FCC) iron ($\gamma$-Fe), given the lattice parameter ($a$).
$ a = 2\sqrt{2} r_{atom} $
$ r_{void} \approx 0.414 \times r_{atom} $
Given the lattice parameter $a = 0.3571$ nm. We can rearrange the formula to find the atomic radius:
$ r_{atom} = \frac{a}{2\sqrt{2}} $
Substituting the value of $a$:
$ r_{atom} = \frac{0.3571 \text{ nm}}{2\sqrt{2}} \approx \frac{0.3571 \text{ nm}}{2.8284} \approx 0.12625 \text{ nm} $
Now, use the calculated atomic radius to find the void radius:
$ r_{void} = 0.414 \times r_{atom} $
$ r_{void} \approx 0.414 \times 0.12625 \text{ nm} \approx 0.05227 \text{ nm} $
The calculated radius of the octahedral void is approximately 0.05227 nm. This value falls within the expected range of 0.045 nm to 0.06 nm, confirming our calculation for the octahedral void size in the FCC lattice of iron.
| Column I | Column II |
|---|---|
| (P) Tetragonal | (1) $a \neq b \neq c$, $\alpha = \beta = \gamma = 90^\circ$ |
| (Q) Rhombohedral | (2) $a = b \neq c$, $\alpha = \beta = \gamma = 90^\circ$ |
| (R) Orthorhombic | (3) $a \neq b \neq c$, $\alpha = \gamma = 90^\circ \neq \beta$ |
| (S) Monoclinic | (4) $a = b = c$, $\alpha = \beta = \gamma \neq 90^\circ$ |