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To obtain a diffraction peak, for a crystalline solid with interplane distance equal to the wavelength of incident $X$-ray radiation, the angle of incidence should be
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Verified Answer
The correct answer is:
$30^{\circ}$
From Bragg's equation
$\begin{array}{l}
\mathrm{n} \lambda=2 \mathrm{~d} \sin \theta \\
\mathrm{d}=\lambda, \mathrm{n}=1 \\
1 \times \lambda=2 \times \lambda \sin \theta \\
\sin \theta=\frac{1}{-} \\
\theta=30^{\circ}
\end{array}$
$\begin{array}{l}
\mathrm{n} \lambda=2 \mathrm{~d} \sin \theta \\
\mathrm{d}=\lambda, \mathrm{n}=1 \\
1 \times \lambda=2 \times \lambda \sin \theta \\
\sin \theta=\frac{1}{-} \\
\theta=30^{\circ}
\end{array}$
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