91Ó°ÊÓ

The scattered wavelengths are

$$\begin{gathered} {\mathrm{λ}}_1=0.1nm \\ {\mathrm{λ}}_2=0.075nm \end{gathered}$$$$\begin{gathered} {\mathrm{λ}}_1=0.1nm \\ {\mathrm{λ}}_2=0.075nm \end{gathered}$$

The plane separation is

$d=0.25nm$$d=0.25nm$

The maximum scattering intensities of X-rays of wavelength $\mathit{λ}$scattered from Bragg planes having plane separation at an angle $\mathit{θ}$ is

$\mathbf{2}\mathit{d}\mathit{s}\mathit{i}\mathit{n}\mathit{θ}\mathbf{=}\mathit{m}\mathit{λ}$ .....(1)

From equation (I), Bragg's law for ${\mathrm{λ}}_1$${\mathrm{λ}}_1$and localid="1663063950542">λ2becomes

Dividing the above equations,

localid="1663064037950" $$\begin{gathered} \frac{m_1{\mathrm{λ}}_1}{m_2{\mathrm{λ}}_2}=1 \\ \frac{m_1}{m_2}=\frac{{\mathrm{λ}}_1}{{\mathrm{λ}}_2} \\ \frac{m_1}{m_2}=\frac{0.075nm}{0.1nm} \\ \frac{m_1}{m_2}=\frac{3}{4} \end{gathered}$$

Thus, the lowest values are

localid="1664277402658" $$\begin{gathered} m_1=3 \\ {m}_2=4 \end{gathered}$$

Substitute value of localid="1663065023789" $m_1$in equation (1) to get

localid="1663064113753" $$\begin{gathered} 2x0.25nm×\sin \mathrm{θ}=3x0.1nm \\ sin\mathrm{θ}=0.6 \\ \mathrm{θ}={sin}^{-1}0.6 \\ ≈{37}^{°} \end{gathered}$$

Separation between incident and scattered beams is

$$\begin{gathered} {180}^{°}-2\mathrm{θ}={180}^{°}-2x{37}^{°} \\ ={106}^{°} \end{gathered}$$

The separation is between the incident and the scattered beams is ${106}^{°}$.