A beam of x rays with wavelengths ranging from$\mathbf{0}\mathbf{.}\mathbf{120}\mathbf{}\mathit{n}\mathit{m}$ to $\mathbf{0}\mathbf{.}\mathbf{07}\mathbf{}\mathit{n}\mathit{m}$ scatters from a family of reflecting planes in a crystal. The plane separation is$\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{}\mathit{n}\mathit{m}$. It is observed that scattered beams are produced for$\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{}\mathit{n}\mathit{m}$andlocalid="1664277381313" $\mathbf{0}\mathbf{.}\mathbf{075}\mathbf{}\mathit{n}\mathit{m}$. What is the angle between the incident and scattered beams?

The scattered wavelengths are

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

The plane separation is

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

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

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

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

Dividing the above equations,

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

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" $$\begin{gathered} 2x0.25nm\times \sin \theta =3x0.1nm \\ sin\theta =0.6 \\ \theta ={sin}^{-1}0.6 \\ \approx {37}^{°} \end{gathered}$$

Separation between incident and scattered beams is

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

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