Answer
56.84°
Work Step by Step
Required:
Determine the expected diffraction angle for the first-order reflection from the (111) set of planes for FCC nickel (Ni) when monochromatic radiation of wavelength 0.1937 nm is used.
Solution:
From Table 3.1, atomic radius of Nickel is 0.1246 nm. Calculating for the value of lattice parameter $a$,
$a = 2R\sqrt 2 = (2)(0.1246 nm)\sqrt 2 = 0.3524 nm$
Using Equation 3.22, $d_{111}$ is computed as,
$d_{111} = \frac{a}{\sqrt (1)^{2} + (1)^{2} + (1)^{2}} = \frac{0.3524 nm}{\sqrt 3} =0.2035 nm $
Modifying Equation 3.21 to compute the value of θ,
$sin θ = \frac{nλ}{2d_{111}} = \frac{(1)(0.1937 nm)}{(2)(0.2035 nm)} = 0.4759 $
$θ = sin^{-1}(0.4759) = 28.42°$
Thus,
$2θ = (2)(28.42°) = 56.84°$
