Solution
= Solution
{c}
The <primitive cubic lattice> has $G_{hkl}=(2\pi/a)\sqrt{h^2+k^2+l^2}$. Combining this with <Bragg's law> gives the <Bragg angle>
$$
\sin\theta_{hkl}=\frac{\lambda}{2a}\sqrt{h^2+k^2+l^2}.
$$
Using $a=0.375\,\mathrm{nm}$ and $\lambda=0.154\,\mathrm{nm}$ yields
$$
\boxed{2\theta_{110}=33.762^\circ\simeq33.8^\circ,\qquad
2\theta_{111}=41.666^\circ\simeq41.7^\circ.}
$$
These are the full scattering angles; the <Bragg angles> themselves are half as large.