Question

Consider the following statements and arrange their values in decreasing order:
A. The interplanar spacing in fcc for \((111)\) is \(d_{111}\),
B. The interplanar spacing in fcc for \((110)\) is \(d_{110}\),
C. The interplanar spacing in fcc for \((100)\) is \(d_{100}\).

• A. A \(>\) B \(>\) C
• B. A \(>\) C \(>\) B
• C. C \(>\) A \(>\) B
• D. C \(>\) B \(>\) A

Hint:

For cubic crystals, larger value of \(\sqrt{h^2+k^2+l^2}\) gives smaller interplanar spacing.

Answer 1

The Correct Option is D

Concept:
For a cubic crystal, the interplanar spacing is given by: \[ d_{hkl}=\frac{a}{\sqrt{h^2+k^2+l^2}} \] where \(a\) is the lattice parameter and \((hkl)\) are the Miller indices.

Step 1: Find \(d_{100}\). \[ d_{100}=\frac{a}{\sqrt{1^2+0^2+0^2}} \] \[ d_{100}=a \]

Step 2: Find \(d_{110}\). \[ d_{110}=\frac{a}{\sqrt{1^2+1^2+0^2}} \] \[ d_{110}=\frac{a}{\sqrt{2}} \]

Step 3: Find \(d_{111}\). \[ d_{111}=\frac{a}{\sqrt{1^2+1^2+1^2}} \] \[ d_{111}=\frac{a}{\sqrt{3}} \]

Step 4: Compare the values. \[ a>\frac{a}{\sqrt{2}}>\frac{a}{\sqrt{3}} \] Therefore, \[ d_{100}>d_{110}>d_{111} \] Using the given symbols: \[ C>B>A \] \[ \therefore \text{Correct Answer is (D)} \]