Question

The third peak in the X-ray diffraction pattern of a face-centered cubic (FCC) crystal occurs at a \(2\theta = 45^\circ\). The wavelength of the monochromatic X-ray beam is \(1.54\ \text{\AA}\). Considering first-order reflection, find the lattice parameter (in \AA). (Round off to two decimal places)

Hint:

For FCC crystals, only planes with all-even or all-odd indices are allowed. Peak order corresponds to increasing values of \(h^2+k^2+l^2\).

Answer 1

Correct Answer: 5.64

For FCC crystals, allowed planes satisfy:
\[ h^2 + k^2 + l^2 = 3,\; 4,\; 8,\; 11,\; 12,\ldots \] The third peak corresponds to the third allowed value:
\[ h^2 + k^2 + l^2 = 8 \] Using Bragg’s law for first-order reflection \((n=1)\):
\[ n\lambda = 2d\sin\theta \] Given \(2\theta = 45^\circ \Rightarrow \theta = 22.5^\circ\).
Thus: \[ d = \frac{\lambda}{2\sin\theta} \] Substitute:
\[ d = \frac{1.54}{2\sin 22.5^\circ} \] \[ \sin 22.5^\circ = 0.3827 \] \[ d = \frac{1.54}{2 \times 0.3827} = \frac{1.54}{0.7654} = 2.01\ \text{\AA} \] For cubic crystals:
\[ d = \frac{a}{\sqrt{h^2+k^2+l^2}} \] Here: \[ h^2 + k^2 + l^2 = 8 \] Thus lattice parameter:
\[ a = d\sqrt{8} = 2.01 \times 2.828 \] \[ a = 5.69\ \text{\AA} \] Rounded to two decimals:
\[ a = 5.69\ \text{\AA} \]