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)= (a)/(√(h^2+k^2+l^2))`
For Chromium where
`R=0.1249 nm` the lattice constant is

`a=4R/ √(3)=4*0.1249 / √(3)=0.288 nm`
Distance between (310) planes is

`d_((310)) = (0.288)/( √(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]