The coloration of flames by atoms

Consider the case of the yellow flame of sodium. We started with sodium chloride. Formula?

$ Na^+ Cl^- $

Ions are transformed into atoms. Equation?

$ Na^+Cl^- \rightarrow Na + Cl $

The atom $Na$ has 11 electrons (see the periodic table ). Electronic structure? $ 1s^2 2s^2 2p^6 3s^1 $

Valence electron: $ 3s^1$. Here are its energy states:

The energy of the combustion excites this electgron (a) from the ground state (1) to the first excited level (2). Then it falls again (b), emitting a photon. Energy of this photon?

$ E = -3.03 - (-5.14) = 2.11 eV $

Formula to calculate the wavelength of the photon?

$ \lambda = h\nu = \frac{hc}{E} $

$ h $ = Planck constant;
$ \nu $ = frequency;
$ c $ = speed of light;
$ E $ = energy;

Calculation?

$ \lambda $ = $ \frac{6.63 \cdot 10^{-34} \cdot 3.00 \cdot 10^8} {2.11 \cdot 1.60 \cdot 10^{-19}} $ =
$ 5.89 \cdot 10^{-7} m $
= $ 589 nm $

It's yellow light !