= Solution
For <indexing a diamond-cubic powder pattern>, $d_{hkl}=a/\sqrt{h^2+k^2+l^2}$ and <Bragg's law> gives $\sin^2\theta=\lambda^2(h^2+k^2+l^2)/(4a^2)$. Using half of each printed scattering angle, the ratios to the first peak are approximately
$$
1,\ 2.66658,\ 3.66665,\ 5.33319,\ 6.33322.
$$
Multiplying by3 gives the allowed sequence $3,8,11,16,19$. Thus
$$
\boxed{111,\quad220,\quad311,\quad400,\quad331.}
$$
The numerical calculation, expressed without assuming a historical Data Book wavelength, is
$$
\begin{array}{c|c|c|c}
2\theta\ (\mathrm{deg})&hkl&h^2+k^2+l^2&a/\lambda\\\hline
28.524&111&3&3.51534\\
47.443&220&8&3.51540\\
56.294&311&11&3.51535\\
69.351&400&16&3.51538\\
76.629&331&19&3.51537
\end{array}
$$
Therefore \b[$a\simeq3.51537\lambda$]. As a separately identified reference convention, choosing Cu K$\alpha_1$ with $\lambda\simeq0.15406\ {\rm nm}$ gives
$$
\boxed{a\simeq0.54158\ {\rm nm}.}
$$
That wavelength follows from the copper $KL_3$ transition near $8047.8227\ {\rm eV}$ in the NIST transition database, using $\lambda=hc/E$.<X-Ray Transition Energies Database by National Institute of Standards and Technology>[Copper, experimental KL3 transition energy]{ref}
The paper says Cu K$\alpha$ but does not give the line-selection convention or an explicit wavelength. A rounded unresolved-doublet wavelength, for example $0.15418\ {\rm nm}$, would instead give about $0.54200\ {\rm nm}$. These are conditional wavelength evaluations, not claims about the unavailable historical Data Book. The angles are internally consistent with one cubic parameter; they should not be replaced by literature peak positions simply to obtain a familiar <silicon> value.
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