Question

A solid with FCC crystal structure is probed using X-rays of wavelength 0.2 nm. For the crystallographic plane given by (2, 0, 0), a first-order diffraction peak is observed for a Bragg angle of \(21^\circ\). The unit cell size is ............ nm. (Round off to 2 decimal places.)

Hint:

For FCC crystals, use \(d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}}\) and Bragg's law to calculate lattice parameters.

Answer 1

Correct Answer: 0.55

Step 1: Apply Bragg's law.
\[ 2d\sin\theta = n\lambda \] For first order (\(n=1\)): \[ d = \frac{\lambda}{2\sin\theta} = \frac{0.2}{2\sin21^\circ} = \frac{0.2}{0.716} = 0.279\, \text{nm} \]

Step 2: Relate interplanar spacing to lattice constant.
For cubic crystals: \[ d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}} = \frac{a}{\sqrt{4}} \] \[ a = 2d = 2(0.279) = 0.558 \, \text{nm} \approx 0.56\, \text{nm} \]

Step 3: Conclusion.
Hence, unit cell size \(a = 0.56\, \text{nm}\).