Question

Copper has an FCC crystal structure and a unit cell with a lattice constant of \(0.361 \, \text{nm}\). The interplanar spacing \(d_{111}\) is:

• A. \(d_{111} = 3.128 \, \text{nm}\)
• B. \(d_{111} = 1.128 \, \text{nm}\)
• C. \(d_{111} = 0.628 \, \text{nm}\)
• D. \(d_{111} = 0.128 \, \text{nm}\)

Answer 1

The Correct Option is D

The interplanar spacing for cubic materials is calculated as:
\[d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}}\]
Using these parameters for an FCC lattice (\(h = 1\), \(k = 1\), \(l = 1\), \(a = 0.361 \, \text{nm}\)), the result is:
\[d_{111} = \frac{0.361}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{0.361}{\sqrt{3}} \approx 0.128 \, \text{nm}\]