The energy of the emitted photon is:
\[E = 0.05 \times \text{eV} = 0.05 \times (20 \, \text{kV})\]
Using \(E = \frac{hc}{\lambda}\), where \(h = 6.626 \times 10^{-34} \, \text{Js}\) and \(c = 3 \times 10^8 \, \text{m/s}\):
\[\lambda = \frac{hc}{E} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{0.05 \times 20 \times 10^3 \times 1.602 \times 10^{-19}} \approx 6.24 \, \text{nm}\]