An x-ray beam of wavelength A undergoes first-order reflection (Bragg law diffraction) from a crystal when its angle of incidence to a crystal face is , and an x-ray beam of wavelength undergoes third-order reflection when its angle of incidence to that face is . Assuming that the two beams reflect from the same family of reflecting planes, find (a) the interplanar spacing and (b) the wavelength A.

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,

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

$$\begin{gathered} wavelengthforthirdorderreflection{\lambda }_B=97\text{pm}\backslash \\ braggangleforfirstorderreflection{\theta }_1=23° \\ braggangleforthirdorderreflection{\theta }_2=60° \end{gathered}$$

From the bragg’s formula of diffraction as given above, the interplane spacing will be by using given data,

Here for 3rd order diffraction the wavelength will be 3 times.

$$\begin{gathered} d=\frac{3{\lambda }_B}{2\sin {\theta }_2} \\ d=\frac{3\left(97\text{pm}\right)}{2\sin 60°} \\ d=1.7\times {10}^2\text{pm} \end{gathered}$$

The wavelength A from the given and achieve data,

$$\begin{gathered} {\lambda }_A=2d\sin {\theta }_1 \\ {\lambda }_A=2\left(1.7\times {10}^2\text{pm}\right)\sin 23° \\ {\lambda }_A=1.3\times {10}^2\text{pm} \end{gathered}$$