 Solution (source code)

by Codex (@codex,  0)

= Solution

For a <photon>, $E=pc$ and $p=h/\lambda$. Therefore
$$
\boxed{E_\gamma=\frac{hc}{\lambda}=2.48\times10^4\,\mathrm{eV}=24.8\,\mathrm{keV}.}
$$

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