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: Understanding the Question:
The question asks for the mathematical ratio of the interplanar spacing (\( d_{hkl} \)) for the low-index planes (100), (110), and (111) in a standard cubic crystal system.

Step 2: Key Formula or Approach:
For any cubic crystal system with a lattice parameter \( a \), the interplanar spacing \( d_{hkl} \) of a plane with Miller indices \( (hkl) \) is given by:
\[ d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}} \]

Step 3: Detailed Explanation:

• Calculation of Spacing for (100) Plane:
Using \( h=1, k=0, l=0 \):
\[ d_{100} = \frac{a}{\sqrt{1^2 + 0^2 + 0^2}} = a \]

• Calculation of Spacing for (110) Plane:
Using \( h=1, k=1, l=0 \):
\[ d_{110} = \frac{a}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{a}{\sqrt{2}} \]

• Calculation of Spacing for (111) Plane:
Using \( h=1, k=1, l=1 \):
\[ d_{111} = \frac{a}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{a}{\sqrt{3}} \]

• Determining the Ratio:
The ratio of their interplanar spacings \( d_{100} : d_{110} : d_{111} \) can be written as:
\[ d_{100} : d_{110} : d_{111} = a : \frac{a}{\sqrt{2}} : \frac{a}{\sqrt{3}} \]
Factoring out the lattice parameter \( a \) yields:
\[ d_{100} : d_{110} : d_{111} = 1 : \frac{1}{\sqrt{2}} : \frac{1}{\sqrt{3}} \]


Step 4: Final Answer:
Thus, the correct ratio of interplanar spacing is \( 1 : 1/\sqrt{2} : 1/\sqrt{3} \), matching Option (A).