QB_PH_093: Schrodinger equation

First-order ($n = 1$) Bragg reflection of an X-ray beam occurs from crystal lattice planes of interplanar spacing $d = 0.282\text{ nm}$ at glancing angle $\theta = 30^\circ$. What is the wavelength $\lambda$ of the X-rays in nm?

Correct Answer

### 5-Stage Derivation & Analytical Solution: 1. **Stage 1: Governing Concept & Domain Context:** * This question evaluates **Fundamentals of Fourier and Laplace transforms** under *Fourier and Laplace transforms* in GATE PH. 2. **Stage 2: Mathematical Derivation:** * By Bragg's diffraction condition: $2 d \sin\theta = n \lambda$. Substituting $n = 1$, $d = 0.282\text{ nm}$, and $\theta = 30^\circ$: $\lambda = 2(0.282)(\sin 30^\circ) = 2(0.282)(0.5) = 0.282\text{ nm}$. 3. **Stage 3: Distractor Elimination Forensics:** * The evaluated correct answer is derived using standard physical equations. * Distractor values correspond to typical unit conversion omissions or inverted formulas. 4. **Stage 4: Common Student Trap:** * *Trap:* Using $\cos\theta$ instead of $\sin\theta$ in Bragg's law. *Reference: George B. Arfken, Hans J. Weber, Mathematical Methods for Physicists (Academic Press).*