Answer
$d_{211} = 0.1347 nm$
Work Step by Step
Required:
The metal niobium (Nb) has a BCC crystal structure. If the angle of diffraction for the (211) set of planes occurs at 75.99° (first-order reflection) when monochromatic x-radiation having a wavelength of 0.1659 nm is used, compute the interplanar spacing for this set of planes
Solution:
Using Equation 3.21 and considering that 2θ = 75.99°,
$d_{211} = \frac{nλ}{2sinθ} = \frac{(1)(0.1659 nm)}{(2) (sin \frac{75.99°}{2})} = 0.1347 nm$
