Question

The lattice constant of a simple cubic lattice having interplanar spacing 3\AA{ for (002) plane is:}

• A. \( 4.2 \, \text{\AA} \)
• B. \( 6.0 \, \text{\AA} \)
• C. \( 6.2 \, \text{\AA} \)
• D. \( 4.0 \, \text{\AA} \)

Hint:

For any cubic crystal system (Simple, BCC, or FCC), the formula for interplanar spacing \( d = a / \sqrt{h^2 + k^2 + l^2} \) remains the same. Memorizing this single formula is sufficient for all cubic lattice problems involving Miller indices and lattice parameters.

Answer 1

The Correct Option is B

Step 1: Recall the formula for interplanar spacing in a cubic lattice. The interplanar spacing \(d_{hkl}\) for a cubic lattice with lattice constant \(a\) is given by: \[ d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}} \] where \((h, k, l)\) are the Miller indices of the plane.
Step 2: Identify the given values. We are given: - Interplanar spacing, \(d = 3 \, \text{\AA}\) - Miller indices for the plane, \((hkl) = (002)\)
Step 3: Substitute the values into the formula and solve for \(a\). \[ 3 \, \text{\AA} = \frac{a}{\sqrt{0^2 + 0^2 + 2^2}} \] \[ 3 \, \text{\AA} = \frac{a}{\sqrt{4}} \] \[ 3 \, \text{\AA} = \frac{a}{2} \] \[ a = 2 \times 3 \, \text{\AA} = 6 \, \text{\AA} \] Thus, the lattice constant is \( 6.0 \, \text{\AA} \).