Question

The d-spacing of an FCC lattice in [110] plane if lattice parameter (a)=3.60 A is:

• A. 2.55 A
• B. 1.80 A
• C. 0.39 A
• D. 5.09 A

Hint:

Always ensure that you use the correct plane indices in the denominator. The formula remains identical for SC, BCC, and FCC systems, but the allowed reflections in XRD will depend on the extinction rules of each lattice.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks to calculate the interplanar d-spacing (\( d_{hkl} \)) for the (110) plane family in a face-centered cubic (FCC) crystal lattice with a given lattice parameter \( a = 3.60 \text{ \AA} \).
Note: The question uses the notation [110], which typically represents a direction, but the context clearly refers to the (110) crystallographic plane.

Step 2: Key Formula or Approach:
The interplanar spacing \( d_{hkl} \) for a cubic system is calculated using the standard formula:
\[ d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}} \]

Step 3: Detailed Explanation:

• Substituting the Values:
Given lattice parameter: \( a = 3.60 \text{ \AA} \)
Miller indices of the plane: \( h = 1, k = 1, l = 0 \)

• Calculation:
\[ d_{110} = \frac{3.60 \text{ \AA}}{\sqrt{1^2 + 1^2 + 0^2}} \]
\[ d_{110} = \frac{3.60 \text{ \AA}}{\sqrt{1 + 1 + 0}} = \frac{3.60 \text{ \AA}}{\sqrt{2}} \]
Knowing that \( \sqrt{2} \approx 1.4142 \):
\[ d_{110} \approx \frac{3.60}{1.4142} \approx 2.5456 \text{ \AA} \]
Rounding this value to two decimal places gives \( 2.55 \text{ \AA} \).


Step 4: Final Answer:
Therefore, the interplanar d-spacing is approximately 2.55 , matching Option (A).