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: Identify the plane and formula.
For a cubic lattice, \( d_{hkl} = \dfrac{a}{\sqrt{h^2+k^2+l^2}} \), and here \( h=1, k=1, l=0 \) with \( a = 3.60 \) \AA.

Step 2: Substitute the values.
\[ d_{110} = \frac{3.60}{\sqrt{1^2+1^2+0^2}} = \frac{3.60}{\sqrt{2}} \]

Step 3: Simplify the fraction.
Since \( \sqrt{2} \approx 1.414 \), dividing gives \( d_{110} \approx \frac{3.60}{1.414} \approx 2.55 \) \AA.
\[ \boxed{2.55 \text{ \AA}} \]
This matches option (A).