Question

An electron in a hydrogen atom is in a high n state. It drops down one state at a time. What is the rst transition to give a visible photon?

Answer 1

Energy of electron in a hydrogen atom in n state = E_n = (-13.6 eV)/n²

E₃ = (-13.6 eV)/9 = -1.51 eV

E₂ = (-13.6 eV)/4 = -3.4 eV

In dropping from the n = 3 state to the n = 2 state the electron loses 1.89 eV worth of energy. This is the energy carried away by the photon.

Converting this to joules gives E = 1.89 * 1.60 x 10^-19 J/eV = 3.024 x 10^-19 J

For a photon E = hf = hc/λ

λ = hc/E = 6.63 x 10^-34 * 3 x 10⁸ / 3.024 x 10^-19

λ = 6.58 x 10^-7 m = 658 nm

This is in the visible part of the spectrum, so it would be the nearest transition to give a visible photon.