Answer
45.88°
Work Step by Step
Required:
Determine the expected diffraction angle for the first-order reflection from the (310) set of planes for BCC chromium (Cr) when monochromatic radiation of wavelength 0.0711 nm is used.
Solution:
From Table 3.1, atomic radius of Chromium is 0.1249 nm. Calculating for the value of lattice parameter $a$,
$a = \frac{4R}{\sqrt 3} = \frac{(4)(0.1249)}{\sqrt 3} = 0.2884 nm$
Using Equation 3.22, $d_{310}$ is computed as,
$d_{310} = \frac{a}{\sqrt (3)^{2} + (1)^{2} + (0)^{2}} = \frac{0.2884 nm}{\sqrt 10} =0.0912 nm $
Modifying Equation 3.21 to compute the value of θ,
$sin θ = \frac{nλ}{2d_{310}} = \frac{(1)(0.0711 nm)}{(2)(0.0912 nm)} = 0.3898$
$θ = sin^{-1}(0.3898) = 22.94°$
Thus,
$2θ = (2)(22.94°) = 45.88°$
