In the $\mathrm{x}$-ray reflection $(\mathrm{n}=1)$, the distance between two parallel planes of…

In the $\mathrm{x}$-ray reflection $(\mathrm{n}=1)$, the distance between two parallel planes of $\mathrm{NaCl}$ is $280 \mathrm{pm}$ and diffraction angle is $5.2^{\circ}$. What is the wavelength of it's light radiation (Sin $\left.5.2^{\circ}=0.09\right)$
  1. $0.504 \mathrm{~A}^{\circ}$
  2. $5.04 \mathrm{~A}^{\circ}$
  3. $50.4 \mathrm{~A}^{\circ}$
  4. $504 \mathrm{~A}^{\circ}$

Solution

From Bragg's equation :- $\begin{aligned} \mathrm{n} \lambda & =2 \mathrm{~d} \sin \theta \\ \Rightarrow \quad & \lambda=\frac{2 \mathrm{~d} \sin \theta}{\mathrm{n}}=\frac{2\left(280 \times 10^{-12}\right)(0.09)}{1} \\ & =5.04 \times 10^{-11} \mathrm{~m} \\ & =0.504 \times 10^{-10} \mathrm{~m} \\ & =0.504 Å\end{aligned}$

Asked in: AP EAMCET 2023 (16 May Shift 2)

Practice more Solid State questions on Aicharya