Question

A topaz crystal has an interplanar spacing (d) of $\mathbf{1}\mathbf{.}\mathbf{36}\mathbf{}\mathbf{\AA }\mathbf{}\mathbf{(}\mathbf{1}\mathbf{}\mathbf{\AA }\mathbf{=}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{}\mathbf{m}\mathbf{)}$. Calculate the wavelength of the X-ray that should be used if $\mathbf{\theta }\mathbf{=}\mathbf{15}\mathbf{°}$(assume n=1).

Short answer

The wavelength of the X-ray is $\mathrm{\lambda }=70.4\mathrm{pm}$.

Step-by-step solution

Given Information

Interplanar Spacing of topaz crystal is $\mathrm{d}=1.36\mathrm{\AA }\left(\frac{1\times {10}^{-10}\mathrm{m}}{1\mathrm{\AA }}\right)=1.36\times {10}^{-10}\mathrm{m}$

The angle made by the X-ray is$\mathrm{\theta }=15°$

Concept Introduction

According to Bragg’s equation,

$\mathbf{n\lambda }\mathbf{=}\mathbf{2}\mathbf{dsin\theta }$

Where n is aninteger, d is the distance between the atomic layers and $\mathbf{\lambda }$is thewavelength of the incident X-ray.

Step 3: Determine the wavelength λ

Substituting the given values in Bragg's equation, we get,

$$\begin{gathered} \mathrm{n\lambda }=2\mathrm{dsin\theta } \\ 1\times \mathrm{\lambda }=2\times 1.36\times {10}^{-10}\mathrm{m}\times \sin \left(15°\right) \\ \mathrm{\lambda }=0.704\times {10}^{-10}\mathrm{m}\left(\frac{1\mathrm{pm}}{{10}^{-12}\mathrm{m}}\right) \\ \mathrm{\lambda }=70.4\mathrm{pm} \end{gathered}$$