= Solution
{c}
At each reflection, the argument of the <atomic scattering factor> is
$$
s_{hkl}=\frac{\sin\theta_{hkl}}{\lambda}
=\frac{\sqrt{h^2+k^2+l^2}}{2a},\qquad
s_{110}^2=3.55556\,\mathrm{nm^{-2}},\quad s_{111}^2=5.33333\,\mathrm{nm^{-2}}.
$$
Substitution into the given <atomic scattering factor> approximation gives
$$
\begin{aligned}
f_{\rm Au}(110)&=59.4421,&f_{\rm Cu}(110)&=23.1801,\\
f_{\rm Au}(111)&=51.5618,&f_{\rm Cu}(111)&=20.7241.
\end{aligned}
$$
Thus the <structure factors> are $S_{110}=36.2620$ and $S_{111}=113.7340$. The <powder diffraction multiplicity> is $m_{110}=12$: choose which <Miller index> is zero in three ways and the signs of the other two in four ways. For $\{111\}$, the eight sign choices give $m_{111}=8$. With the specified instrumental and angular factors omitted, the integrated <powder diffraction> <wave intensity> is proportional to $m_{hkl}|S_{hkl}|^2$, so
$$
\boxed{\frac{I_{111}}{I_{110}}=\frac8{12}
\left(\frac{113.7340}{36.2620}\right)^2\simeq6.56.}
$$
It would be incorrect to use the same <atomic scattering factors> at the two different <scattering vectors>, or to omit the <powder diffraction multiplicity>.
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