Figure 36-46 is a graph of intensity versus angular positionfor the diffraction of an x-ray beam by a crystal. The horizontal scale is set by${\mathit{\theta }}_{\mathbf{s}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{°}$.The beam consists of two wavelengths, and the spacing between the reflecting planes is$\mathbf{0}\mathbf{.}\mathbf{94}\mathit{n}\mathit{m}$. What are the (a) shorter and (b) longer wavelengths in the beam?

Diffraction is a phenomena in which the interference or the bending of waves occurs around the corners of the opening or obstacle through or on which it strikes.

The diffraction formula is following,

$$\begin{gathered} \mathrm{n\lambda }=2\mathrm{dsin\theta } \\ \mathrm{\lambda }=\text{wavelength} \\ \mathrm{d}=\text{spacing} \\ \mathrm{\theta }=\text{bragg angle} \\ \mathrm{n}=\text{order of diffraction} \end{gathered}$$

The spacing between reflecting plans is, $d=\text{0.94} \text{nm}$.

The bragg angle is, $\theta =2.00°$.

As the graph is running from 0 degree to 2 degrees the shortest wavelength wave is present at around the 0.75 degrees.

So putting all the values we will get shortest wavelength

${\lambda }_1=2d\sin {\theta }_1$

Substitute all the value in the above equation.

$$\begin{gathered} {\lambda }_1=2\left(\text{0.94} \text{nm}\right)\sin 0.75° \\ {\lambda }_1=0.025 \text{nm} \\ {\lambda }_1=25 \text{pm} \end{gathered}$$

As the graph is running from 0 degree to 2 degrees the longest wavelength wave is present at around the 1.15 degrees.

So putting all the values we will get longest wavelength

${\lambda }_2=2d\sin {\theta }_2$

Substitute all the value in the above equation.

$$\begin{gathered} {\lambda }_2=2\left(\text{0.94} \text{nm}\right)\sin 1.15° \\ {\lambda }_2=0.038 \text{nm} \\ {\lambda }_2=38 \text{pm} \end{gathered}$$

Hence the longest wavelength is,$38 \text{pm}$ .