The energy levels of an atom is shown is figure. Which one of these transitions will result in the emission of a photon of wavelength 124 . 1   nm ? Given ( h   =   6 . 62   ×   10 – 34   J   s )

Step-by-step solution

The change in energy is related to wavelength of photon emitted as: ∆ E = h c λ = 6 . 62 × 10 - 34   J   s × 3 × 10 8   m   s - 1 1 . 6 × 10 - 19   J × 124 . 1   nm = 1242   eV   nm 124 . 1   nm = 10   eV Hence, option D is correct.