The diffraction of barium with X-radiation of wavelength 2.29 $\stackrel{\mathrm{o}}{\mathrm{A}}$ gives a first-order reflection at 30°. What is the distance between the diffracted planes?
 

 

Using Bragg's equation

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

Where d is the distance between two diffracted planes, $\theta$ the angle to have maximum intensity of diffracted X-ray beam, n the order of reflection, and ), the wavelength.

$\therefore d=\frac{1\times 2.29\text{Å}}{2\times \sin {30}^{\circ }}=2.29\text{Å}\left(\sin {30}^{\circ }=\frac{1}{2}\right)$