6 \, \text{eV} \left( \frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}} \right)\]
where \(E\) is the energy of the photon, \(n_{i}\) is the initial energy level, \(n_{f}\) is the final energy level, and the constant 13.6 eV is the ionization energy of hydrogen.
Show more…