Question

X rays of wavelength$\mathbf{0}\mathbf{.}\mathbf{12}\mathbf{ }\mathbf{nm}$ are found to undergo second order reflection at a Bragg angle of$\mathbf{28}°$from a lithium fluoride crystal. What is the interplanar spacing of the reflecting planes in the crystal?

Short answer

The inter planar spacing of the reflecting planes in the crystal is $d=0.26 \text{nm}$.

Step-by-step solution

Bragg’s law

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,

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

Identification of the given data

The wavelength is, $\lambda =0.12 \text{nm}$.

The bragg angle is, $\theta =28°$.

 Step 3: Determination of the interplanar spacing of the reflecting planes in the crystal

By using the given data, the interplanar spacing of the reflecting planes in the crystal will be

For the second order reflection the n will be 2,

$d=\frac{n\lambda }{2\sin \theta }$

Substitute all the value in the above equation.

$$\begin{gathered} d=\frac{2\left(\text{0.12}\times {\text{10}}^{-9} \text{m}\right)}{2\sin 28°} \\ d=2.56\times {10}^{-10} \text{m} \\ d=0.26 \text{nm} \end{gathered}$$

Hence the interplanar spacing of the reflecting planes in the crystal is$d=0.26 \text{nm}$ .