Answer
$d_{311} = 0.1147 nm$
Work Step by Step
Required:
The metal rhodium (Rh) has an FCC crystal structure. If the angle of diffraction for the (311) set of planes occurs at 36.12° (first-order reflection) when monochromatic x radiation having a wavelength of 0.0711 nm is used, compute the the interplanar spacing for this set of planes
Solution:
Using Equation 3.22 and considering that 2θ = 36.12°,
$d_{311} = \frac{nλ}{2sinθ} = \frac{(1)(0.0711 nm)}{(2) (sin \frac{36.12°}{2})} = 0.1147 nm$
