Question

Determine the expected diffraction angle for the first-order diffraction from the (111) set of planes for FCC nickel (Ni) when monochromatic radiation of wavelength 0.1937 nm is used. The lattice parameter for Ni is 0.3524 nm

Answer 1

Answer:

56°

Explanation:

First calculate $a:$

$a=2 R \sqrt{2}=2(0.1246) \sqrt{2}=0.352 \mathrm{nm}$

The interplanar spacing can be calculated from:

$d_{111}=\frac{a}{\sqrt{1^{2}+1^{2}+1^{2}}}=\frac{0.352}{\sqrt{3}}=0.203 \mathrm{nm}$

The diffraction angle is determined from:

$\sin \theta=\frac{n \lambda}{2 d_{111}}=\frac{1(0.1927)}{2(0.2035)}=0.476$

Solve for $\theta$

$\theta=\sin ^{-1}(0.476)=28^{\circ}$

The diffraction angle is:

$2 \theta=2\left(28^{\circ}\right)=56^{\circ}$