Question

X-rays from a copper x-ray tube $\mathbf{\lambda }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{54}\mathbf{}\mathbf{\AA }$were diffracted at an angle of role="math" localid="1663780778365" $\mathbf{14}\mathbf{.}\mathbf{22}\mathbf{°}$by a crystal of silicon. Assuming first-order diffraction ( n=1 in the Bragg equation), what is the interplanar spacing in silicon?

Short answer

The interplanar spacing in silicon is$\mathrm{d}=3.13\times {10}^{-10}\mathrm{m}$

Step-by-step solution

Given Information

The wavelength of the x-ray from the copper x-ray tube is $\mathrm{\lambda }=1.54\mathrm{\AA }\left(\frac{{10}^{-10}\mathrm{m}}{1\mathrm{\AA }}\right)=1.54\times {10}^{-10}\mathrm{m}$

The angle for the first order (n=1) diffraction is${\mathrm{\theta }}_1=14.22°$

Concept Introduction

According to Bragg’s equation,

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

Where n is the order of diffraction,d is the interplanar spacing and$\mathbf{\lambda }$ is the wavelength of the incident X-ray.

Determine the spacing between the crystal planes d

By using Bragg’s equation, we can write the expression for d, such that,

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

Substituting the given values in the above expression, we get,

role="math" localid="1663781134898" $$\begin{gathered} \mathrm{d}=\frac{1\times 1.54\times {10}^{-10}\mathrm{m}}{2\times \sin \left(14.22°\right)} \\ \mathrm{d}=3.13\times {10}^{-10}\mathrm{m} \end{gathered}$$