Question

Determine the expected diffraction angle for the first-order reflection from the (310) set of planes for bcc chromium when monochromatic radiation of wavelength 0.0777 nm is used. the atomic radius for cr is 0.1249 nm.

Answer 1

In a BCC crystal the distance between two adjacent planes is given by 
$d_{hkl}= \frac{a}{\sqrt{h^2+k^2+l^2}}$
For Chromium where $R=0.1249 nm$ the lattice constant is 
$a=4R/ \sqrt{3}=4*0.1249 / \sqrt{3}=0.288 nm$
Distance between (310) planes is
$d_{(310)} = \frac{0.288}{ \sqrt{3^2+1^2+0^2} }=0.091 nm$
The diffraction angle for the n=1 order is
$2d*\sin (\theta) =\lambda$
$\sin(\theta) =\lambda /(2d) =0.0777/(2*0.091)=0.427$
[tex] \theta =25.72 \degree[\tex]