Question

The ratio of interplanar spacing for (100):(110):(111) planes in a cubic lattice is

• A. \( 1 : 1/\sqrt{2} : 1/\sqrt{3} \)
• B. \( 1 : \sqrt{2} : \sqrt{3} \)
• C. \( \sqrt{3} : \sqrt{2} : 1 \)
• D. \( 1 : 2 : 3 \)

Hint:

In a cubic system, as the Miller indices \( h, k, l \) increase, the interplanar spacing \( d_{hkl} \) decreases. Therefore, the (100) plane will always have a larger spacing than the (110) and (111) planes.

Answer 1

The Correct Option is A

Step 1: Set up the spacing rule.
For a cubic crystal, plane spacing follows \( d_{hkl} = \dfrac{a}{\sqrt{h^2+k^2+l^2}} \), so once the lattice parameter \( a \) is fixed, spacing depends only on the sum of squared Miller indices.

Step 2: Work out each denominator.
For (100): \( \sqrt{1^2} = 1 \). For (110): \( \sqrt{1^2+1^2} = \sqrt{2} \). For (111): \( \sqrt{1^2+1^2+1^2} = \sqrt{3} \).

Step 3: Form the ratio.
Since \( a \) is common to all three planes it cancels out, leaving the ratio as the reciprocals of these denominators.
\[ \boxed{d_{100} : d_{110} : d_{111} = 1 : 1/\sqrt{2} : 1/\sqrt{3}} \]
This matches option (A).