# Energy & Refraction

For most atoms, the structures are so complicated that one can not write down the energies in closed form (i.e. using a simple algebraic equation one can insert in a spreadsheet such as Excel). Even for the simplest atom, hydrogen, there are some details that require fairly complicated mathematics from the field of quantum electrodynamics. For the level of detail relevent to elemental analysis, ordinary quantum mechanics is adequate, and hydrogen atom energy levels can be described quite accurately with just one quantum number, n. The energy difference between two levels is given by the Rydberg formula,

$\Delta E = \dfrac{hc}{\lambda} = R_e\left(\dfrac{1}{n_l^2} - \dfrac{1}{n_u^2} \right )$

The value of the Rydberg constant (Re in the equation) depends on exactly what isotope of hydrogen is being observed. For protium (just a bare proton for the hydrogen nucleus), the energy is 109677.583 cm-1. For deuterium (add one neutron to protium), the energy is 109707.419 cm-1. For tritium, it's 109717.349 cm-1, while for a mythical infinitely heavy nucleus, the value is 109737.316cm-1.

Exercise $$\PageIndex{1}$$

In a vacuum, energy (in cm-1) = 1/λ. If a photon with just enough energy to cause ionization is desired, what wavelengths should be used for each of the 4 different hydrogen nuclei (including the fictitious "infinitely heavy" nucleus)?

The atomic emission lines of hydrogen occur at wavelengths that are straight-forward to determine. One simply inserts n for each of the energy levels between which light moves the atoms, learns E (typically in cm-1), and then computes the line's wavelength as was just done in the exercise. In the laboratory, the wavelength actually observed is shifted from the simply computed value because the refractive index of air is slightly above 1. λair = λvacuum/{refractive index}. To the immense irritation of many, the symbol for refractive index is n, which has absolutely nothing to do with quantum numbers! The dispersion of air (the relationship between refractive index and wavelength) containing no water vapor, 450 parts per million CO2 and at 15°C is approximately:

$n_{air,\:standard} = 1 + \dfrac{5.792105\times 10^6}{2.380185\times 10^{10} - v^{-2}} + \dfrac{1.67912 \times 10^5}{5.6362 \times 10^9 - v^{-2}}$

Here, the energy is in cm-1 computed from the Rydberg formula. If you want details on the refractive index of air,