Answer
$R = 0.1345 nm$
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 atomic radius of Rh. Value of $d_{311} = 0.1147 nm$
Solution:
Using Equation 3.22 to calculate the value of $a$,
$a = d_{311} \sqrt (3^{2} + 1^{2} + 1^{2}) = (0.1147)\sqrt 11 = 0.3804 nm$
Using Equation 3.1 to calculate value of $R$,
$R = \frac{a}{2\sqrt 2} = \frac{0.3804 nm}{2\sqrt 2} = 0.1345 nm$
