Question

The interplanar distance for (100) planes in a rocksalt crystal with \( a = 2.814 \)Å is

• A. 0.612 Å
• B. 1.224 Å
• C. 2.814 Å
• D. 1.926 Å

Hint:

For cubic crystals, the interplanar spacing is determined using the formula \( d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}} \), where \( a \) is the lattice parameter and \( (hkl) \) are the Miller indices of the plane.

Answer 1

The Correct Option is C

The interplanar spacing \( d \) for cubic crystals can be calculated using Bragg’s law. The formula for interplanar spacing is: \[ d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}} \] For the (100) planes in a rock salt crystal, where \( a = 2.814 \)Å (lattice constant) and the Miller indices \( (hkl) = (100) \), we calculate: \[ d_{100} = \frac{2.814}{\sqrt{1^2 + 0^2 + 0^2}} = \frac{2.814}{1} = 2.814 \text{ Å}. \] Thus, the interplanar distance for the (100) planes is 2.814 Å.