QB_PH_SEC7_004: Crystal Structure & Lattice Dynamics

A first-order X-ray diffraction peak from the $(111)$ planes of a cubic crystal with lattice constant $a = 3.0\text{ Å}$ is observed at a glancing angle $\theta = 30^\circ$. What is the wavelength $\lambda$ of the X-rays in Å? (Round off to two decimal places)

Correct Answer

### Step-by-Step Derivation & Verification: 1. **Interplanar Spacing for Cubic Lattice**: $d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}}$ For $(hkl) = (111)$ and $a = 3.0\text{ Å}$: $d_{111} = \frac{3.0\text{ Å}}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{3.0}{\sqrt{3}} = \sqrt{3}\text{ Å} \approx 1.732\text{ Å}$ 2. **Bragg's Law Formulation**: $2 d_{111} \sin\theta = n\lambda$ For first-order diffraction ($n = 1$) at glancing angle $\theta = 30^\circ$: $\lambda = 2 d_{111} \sin(30^\circ) = 2 \times \sqrt{3}\text{ Å} \times \frac{1}{2} = \sqrt{3}\text{ Å} \approx 1.732\text{ Å}$ 3. **Rounding**: To two decimal places, $\lambda = 1.73\text{ Å}$.