This solution explains how to calculate the interplanar spacing for Aluminium's (111) planes using its atomic radius.
The interplanar spacing ($d_{hkl}$) for cubic crystals is:
$d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}}$
Where '$a$' is the lattice parameter and '$h, k, l$' are Miller indices.
For Face-Centered Cubic (FCC) structures like Aluminium, the relationship between atomic radius ($r$) and lattice parameter ($a$) is:
$a = 2\sqrt{2} r$
Given Data:
Step 1: Determine Lattice Parameter ($a$)
Use the FCC formula:
$a = 2\sqrt{2} \times 1.431 \text{ Å} \approx 4.048 \text{ Å}$
Step 2: Calculate Interplanar Spacing ($d_{111}$)
Apply the interplanar spacing formula:
$d_{111} = \frac{a}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{a}{\sqrt{3}}$
$d_{111} \approx \frac{4.048 \text{ Å}}{\sqrt{3}} \approx 2.337 \text{ Å}$
The calculated interplanar spacing for Aluminium's (111) planes is approximately 2.337 Å.
| 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) $a = b = c$, $\alpha = \beta = \gamma \neq 90^\circ$ |
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 _______________