Question

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 following:
(a) the interplanar spacing for this set of planes, and (b) the atomic radius for a rhodium atom.

Answer 1

Answer:

• the interplanar spacing for this set of planes is 0.0603 nm

• the atomic radius for a rhodium atom is 0.0707 nm

Explanation:

Part (a) the interplanar spacing for this set of planes

Applying Bragg's law

$d = \frac{n \lambda}{2sin \theta}$

where;

d is the distance between set of planes

λ is the wavelength of  monochromatic x-radiation = 0.0711 nm

n is the reaction order = 1

θ is the diffraction angle = 36.12°

$d = \frac{1(0.0711)}{2*sin(36.12)} \\\\d = 0.0603 nm$

Part (b) the atomic radius for a rhodium atom.

$d = \frac{a}{\sqrt{3^2 +1^2+1^2}} \\\\0.0603 = \frac{a}{\sqrt{11}} \\\\a = 0.0603 *\sqrt{11} \\\\a = 0.19999 nm$

Also, a = 2R√2

R = a/(2√2)

R = 0.19999/(2√2)

R = 0.0707 nm