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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 $\left.5.2^{\circ}=0.09\right)$
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The correct answer is:
$0.504 \mathrm{~A}^{\circ}$
From Bragg's equation :-
$\begin{aligned} \mathrm{n} \lambda & =2 \mathrm{~d} \sin \theta \\ \Rightarrow \quad & \lambda=\frac{2 \mathrm{~d} \sin \theta}{\mathrm{n}}=\frac{2\left(280 \times 10^{-12}\right)(0.09)}{1} \\ & =5.04 \times 10^{-11} \mathrm{~m} \\ & =0.504 \times 10^{-10} \mathrm{~m} \\ & =0.504 Å\end{aligned}$
$\begin{aligned} \mathrm{n} \lambda & =2 \mathrm{~d} \sin \theta \\ \Rightarrow \quad & \lambda=\frac{2 \mathrm{~d} \sin \theta}{\mathrm{n}}=\frac{2\left(280 \times 10^{-12}\right)(0.09)}{1} \\ & =5.04 \times 10^{-11} \mathrm{~m} \\ & =0.504 \times 10^{-10} \mathrm{~m} \\ & =0.504 Å\end{aligned}$
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